polygons, polyhedra, patterns & beyond
TRANSCRIPT
POLYGONS,POLYHEDRA,PATTERNS&BEYOND
EthanD.Bloch
Spring2015
i
ii
Contents
TotheStudent v
Part I POLYGONS &POLYHEDRA 1
1 GeometryBasics 31.1 EuclidandNon-Euclid . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31.2 LinesandAngles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 121.3 Distance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21
2 Polygons 252.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 252.2 Triangles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 282.3 GeneralPolygons . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 392.4 Area . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 492.5 ThePythagoreanTheorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . 63
3 Polyhedra 733.1 Polyhedra–TheBasics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 733.2 RegularPolyhedra . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 783.3 Semi-RegularPolyhedra . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 813.4 OtherCategoriesofPolyhedra . . . . . . . . . . . . . . . . . . . . . . . . . . 843.5 EnumerationinPolyhedra . . . . . . . . . . . . . . . . . . . . . . . . . . . . 873.6 CurvatureofPolyhedra . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 99
iv Contents
Part II SYMMETRY &PATTERNS 105
4 Isometries 1074.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1074.2 Isometries–TheBasics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1104.3 RecognizingIsometries . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1194.4 CombiningIsometries . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1254.5 GlideReflections . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1304.6 Isometries—TheWholeStory . . . . . . . . . . . . . . . . . . . . . . . . . . 136
5 SymmetryofPlanarObjectsandOrnamentalPatterns 1475.1 BasicIdeas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1475.2 SymmetryofRegularPolygonsI . . . . . . . . . . . . . . . . . . . . . . . . . 1545.3 SymmetryofRegularPolygonsII . . . . . . . . . . . . . . . . . . . . . . . . 1605.4 RosettePatterns . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1705.5 FriezePatterns . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1785.6 WallpaperPatterns . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1875.7 ThreeDimensionalSymmetry . . . . . . . . . . . . . . . . . . . . . . . . . . 204
6 Groups 2156.1 Thebasicidea . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2156.2 Clockarithmetic . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2186.3 TheIntegersMod n . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2236.4 Groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2286.5 Subgroups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2376.6 SymmetryandGroups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 241
SuggestionsforFurtherReading 245
Bibliography 251
TotheStudent
I havewrittentheselecturenotesbecauseI havenotfoundanyexistingtextforMath107thatwasadequateintermsofthechoiceofmaterial, levelofdifficulty, organizationandexposition.MuchoftheapproachinPartII ofthesenotesisinspiredbythetextI previouslyusedforMath107(Farmer, David, “GroupsandSymmetry,” AMS,1996). I hopethatthesenoteswillfittheneedsofthiscourse, andwillhelpyoulearnthismaterial.Ifthereareanyerrorsinthetext, oranythingthatisnotclearlywritten, pleaseacceptmyapolo-
gies. I wouldverymuchappreciateyourfeedbackonthistext, bothintermsoferrorsthatyoufindandsuggestionsforchangesoradditionsthatyoumighthave. Commentscanbeforwardedtomeinperson(afterclass, orinmyoffice, Albee317), orbyemailat [email protected].
Prerequisites
ItisassumedthatanyoneusingthistexthaspassedPartI oftheMathematicsDiagnosticExam.Noparticularbackgroundinmathematicsbeyondthatisrequired. Onsomeoccasionswewillmakeuseofhighschoolalgebra(forexample, thequadraticformula)andhighschoolgeometry(forexample, thePythagoreanTheorem). Forthemostpart, however, wewilltreatmaterialthat,whiletouchinguponsomeverysubstantialideas, doesnotrequiremuchinthewayofalgebraorgeometrybackground. Precalculus(includingtrigonometry, logarithms, andthelike)isnotrequired. Onafewoccasionswewillmentiontrigonometry, butthosebriefreferencescaneasilybeskipped.Whatisneededtoreadthistextisawillingnesstolearnnewideas, tothinkthroughsubtleties,
toworkhard, andtouseyourimagination. Muchofthematerialinthetextisveryvisual, andmakingdrawings, andmentallyimagininggeometricobjects, iscrucial. Someofthearguments
vi TotheStudent
inthetext, thoughnotrequiringmuchinthewayoftechnicalbackground, arenonethelessquitetricky, andrequirecarefulattentiontothedetails.
Exercises
Likemusicandart, mathematicsislearnedbydoing, notjustbyreadingtextsandlisteningtolectures. Inmathematicscoursesweassignexercisesnotbecausewewanttoputthestudentsthroughsomesortofmathematicalbootcamp, butbecausedoingexercisesisthebestwaytoworkwiththematerial, andtoseewhatisunderstoodandwhatneedsfurtherstudy. Doingtheexercisesisthereforeacrucialpartoflearningthematerialinthistext. Exercisesrangefromtheroutinetothedifficult. Whenansweringanexercise, youmayuseanyfactsinthetextuptillthen(includingpreviousexercises).Onefeatureoftheexercisesinthistextisworthmentioning. Inmanyhighschoolmathematics
courses, thetexthasavarietyofworked-outexamples, andthenthehomeworkexercisesareoftenvirtuallyidenticaltotheworked-outexamples, butwithdifferentnumbers. Thestudentsthendotheexercisesbysimplymimickingtheexamplesinthetext. Studentsareoftensatisfiedwiththesetypesofexercises, butfromthepointofviewofintellectualgrowth, suchexercisesaresorelylacking. Thepointoflearningmathematicsisnottolearntoimitatewhattheteacherorthebookdoes, butrathertounderstandnewconceptsandbeabletoapplythem. Hence, inthistext, manyoftheexercisesdonotaskthestudentstomimicwhatisdoneinthetext, butrathertothinkforthemselves. Many(thoughnotall)oftheexercisesinthistextare, purposely,notidenticaltoworkedoutexamplesinthetext, butratheraretobesolvedbythinkingaboutthematerial. Someoftheexercisesinthistextrequirecreativityandimagination, andothersareopenended.
Part IPOLYGONS &POLYHEDRA
1
1GeometryBasics
1.1 EuclidandNon-Euclid
Ancient Greekmathematics was put into its ultimate deductive form by Euclid, who livedroughlyaround300BCE.Inhiswork“TheElements,” Euclidtookanalreadydevelopedlargebodyofgeometry, andgaveitlogicalorderbyisolatingafewbasicdefinitionsandaxioms, andthendeducingeverythingelsefromthesedefinitionsandaxioms. Thestatements thatEucliddeducesfromhisaxiomsanddefinitionsarecalledpropositions(inmoderntextbookstheyareoftenreferredtoastheorems, whichmeansthesamething). WewilllookverybrieflyatsomeaspectsofEuclid’sElements. Thismassiveworkisdividedupinto13“Books;” ourconcernisprimarilywithBookI.Following[Har00](thoughothersemphasizethispointaswell), westressthatEuclid, andthe
ancientGreeksgenerally, viewedgeometryratherdifferentlythanwecurrentlydo. Theonlyquantities thatwereof interestweregeometricones, andof those, geometricquantities thatcouldbeconstructedwithstraightedgeandcompasswereofparticularinterest. Numbersfortheirownsakeseemedlessofinterest(perhapsbecausetheydidnothaveadevelopedunder-standingofnumbers, orperhapsthatiswhytheydidnotdevelopsuchanunderstanding); asolidunderstandingofnumberscamelaterinhistory. Forexample, considerthefamousPythagoreanTheorem, whichisdemonstratedinSection 2.2. Incontemporarylanguage, thistheoremstatesthatifarighttrianglehassidesoflength a and b, andhypotenuseoflength c, then a2+b2 = c2.Sostated, thistheoremtellsussomethingaboutthreenumbers a, b and c: ifthesethreenum-berssatisfyacertaincondition(namelybeingthelengthsofthesidesandhypotenuseofarighttriangle), then theymustalsosatisfy thealgebraicequation a2 + b2 = c2. ToEuclid, suchastatementaboutnumberswouldnothavemadesense. HisversionofthePythagoreanThe-oremis: “Inright-angledtriangles thesquareonthesidesubtendingtherightangleisequaltothesquaresonthesidescontainingtherightangle.” (AllquotesfromEuclidaretakenfrom
4 1. GeometryBasics
[Euc56], whichisthestandardEnglishtranslation.) Euclid’sversionofthetheoremisastatementaboutthreesquares, namelythatonesquare(theoneonthehypotenuse)“isequal”totwoothersquares(thoseonthesides)puttogether. Euclidisinterestedintherelationbetweenthesethreesquares, whicharegeometricobjects, andnotnumbers.ItmightseemfromamodernperspectivethatEuclid’sversionofthePythagoreanisaboutarea,
andthusreallydoesinvolvenumbers, becausetheareaofaplanarfigure(thatis, afigureintheplane)isanumber. Infact, EucliddoesnotmentiontheconceptofareaatallinhisversionofthePythagoreanTheorem. WhenEuclidsaysthattwoplanarfigures(suchassquares)areequal,heisnotmakingastatementaboutthenumericalvaluesoftheirareas(forexampleinsquareinches), butratherissayingthatonefigurecanbecutupintotriangles, andreassembledintotheotherfigure. Thus, Euclid’sversionofthePythagoreanTheoremisstrictlyaboutgeometricobjects. Today, wehavetheconceptofassigningtoeachplanarfigureitsarea(whichisanum-ber), andwerestatevariousgeometricpropertiesintermsofnumericalproperties, butthatisnotthewayEuclid(andtheotherancientGreeks)viewedthings. See[Har00]forathoroughdiscussionofthisissue, andmoregenerallyonwhatEuclidsaid, andhowheunderstoodge-ometry. (Wehighlyrecommend[Har00]asacompaniontoreadingEuclid; thoughmuchofthebookisaimedatamathematicallysophisticatedaudience, somepartsareveryaccessible, andextremelyinsightful.)Without question, “The Elements” is oneof themost important, and influential, works of
mathematicseverwritten—itisarguablyoneofthemostinfluentialintellectualachievementsofhumancivilizationasawhole, notjustofmathematics. Euclid’streatmentofgeometrybe-cametheuniversallyacceptedmethodofdoinggeometryforalmosttwomillenia, uptillthe19thcentury. Moreover, notonlywasEuclideangeometryacceptedasunquestionablytrue, butEuclid’smethodofdeductivereasoningwasconsideredamodeloflogicalargumentation, andanexampleofreasoningthatproducedtheoremsthatwereunquestionablytrue. Itturnsout,however, thatneitheroftheseattributesof“TheElements”istrue. WithoutdiscountingfromtheenormousintellectualandhistoricalimportanceofEuclid’swork, fromamodernvantagepointwecanidentifythreefundamentalflawsinEuclid, theresolutionsofwhichdidnottakeplaceuntilroughlytwomilleniaafterEuclid’stime.Euclidtriedtogiveprecisedefinitionsforgeometricconcepts; hetriedtogiveasetofaxioms
thatdescribeplanar(andspatial)geometry; andhetriedtoproveallotherresultsingeometryinarigorousfashionbasedonlyonhisdefinitionsandaxioms. ItisnowunderstoodthatthereareflawsinEuclid’sdefinitions; hisaxiomsareneithercompletenornecessarilytrue; andsomeofhisproofshavegaps. Fromamodernpointofview, Eucliddidnotreallyachievethelevelof rigor thathas traditionallybeenascribed tohim. Noneof this is todeny thegreatnessofEuclid’sachievement—itisindeedmagnificent—butunderstandingtheproblemsin“TheEle-ments”helpsgiveanaccurateassessmentofEuclid, anditpointsthewaytolaterdevelopmentsingeometry.LetusturntothefirstfourofthedefinitionsfromBookI of“TheElements.” Euclidhasmore
definitionsthanthefourwequote, buttheyaresufficienttoillustratewhatEuclidwasattemptingtoaccomplish.
1.1EuclidandNon-Euclid 5
Definitions
1. A pointisthatwhichhasnopart.
2. A lineisbreadthlesslength.
3. Theextremitiesofalinearepoints.
4. A straightlineisalinethatliesevenlywiththepointsonitself.
Euclidwantstodosomethingverynicewithhisdefinitions, namelydefineallthemathemati-caltermsthatheuses, suchaspointandline. Unfortunately, hedoesnotadequatelyaccomplishhistask. Whenhesaysthat“apointisthatwhichhasnopart,” heneveractuallysayswhatapoint is, onlywhat itdoesnothave, andeven that isunclear. Whenhe says that “a line isbreadthlesslength,” whereheusestheword“line”tomeanwhatwewouldcallacurve, hedoesnottelluswhat“breadth”is, andsowedonotreallyknowwhatalineis. Similarly, hesay“astraightlineisalinethatliesevenlywiththepointsonitself,” butwhatdoesitmeanforsomethingto“lieevenly”withthepointsonitself—otherthantobestraight, butnowwearegoingincircles.Thebottomline is that, bymodernstandards, Euclid’sdefinitionsaremeaningless. In fact,
wenowunderstandthatjustasitisnecessarytostartwithsomeunprovedaxiomsasabasisforalltheothertheoremstobeproved, sotoodoweneedtostartwithsomeundefinedtermsasabasisforallotherdefinitions. Euclid’sdefinitionsaredoomedtofail, becausehetriestodefineeverything. Inthemodernaxiomaticapproachtogeometry, westartwithsomeundefinedterms(forexample, “point”and“line”), thoughwehypothesizevariousaxiomaticpropertiesfortheseundefinedterms(forexample, weassumethateverytwodistinctpointsarecontainedinauniqueline). Whatcountsisnotwhatpointsandlinesare, buthowtheybehave. Iftheybehaveaspointsandlinesoughtto, thenwearesatisfied. AlthoughEuclid’sdefinitionsdonotworkasstated, therearemodernaxiomschemes(suchastheonesbyHilbert in1899andBirkhoffin1932)inwhichthedefinitionsareworkedoutproperly—andtheydojustwhatEuclidwasattemptingtodo. Inotherwords, itispossiblepatchupEuclid’sdefinitions, withthecaveatthatsometermsareleftundefined.WenowturntoEuclid’saxioms. Theaxiomsarebrokenupinto“commonnotions”and“pos-
tulates.” Thecompletelistofcommonnotionsandpostulatesisasfollows.
TheCommonNotions
1. Thingsthatareequaltothesamethingsareequaltooneanother.
2. Ifequalsbeaddedtoequals, thewholesareequal.
3. Ifequalsbesubtractedfromequals, theremaindersareequal.
4. Thingsthatcoincidewithoneanotherareequaltooneanother.
5. Thewholeisgreaterthanthepart.
6 1. GeometryBasics
ThePostulates
1. Todrawastraightlinefromanypointtoanypoint.
2. Toproduceafinitestraightlinecontinuouslyinastraightline.
3. Todescribeacirclewithanycenterandradius.
4. Thatallrightanglesareequaltooneanother.
5. That, ifastraightlinefallingontwostraightlinesmakestheinterioranglesonthesameside less than tworightangles, thestraight lines, ifproduced indefinitely,meetonthatsideonwhicharetheangleslessthanthetworightangles.
Thecommonnotions, whicharenotaboutgeometryperse(thoughEuclidmighthavethoughtofthemgeometrically), seemreasonableenoughasstated. Thepostulates, bycontrast, needsomeexplanationinmodernterminology; Euclid’swordingisdifferentfromwhatwewouldusetoday. Euclid, andingeneraltheancientGreeks, wereveryconcernedwithgeometricconstruc-tionsusingstraightedgeandcompass(wepurposelysay“straightedge”andnot“ruler,” becausetheydidnotallowtheuseofarulertomeasurethings, onlyastraightedgetodrawstraightlinesbetweengivenpoints).Thefirstthreepostulatesinvolvestraightedgeandcompassconstructions. TheFirstPostulate
statesthat, giventwodifferentpoints, wecanconstruct(usingastraightedge)alinesegmentfromonepointtotheother. Inmodernterminology, wherewefocusonpropertiesoflinesandpointsandnotonconstructions, theFirstPostulateisoftenrephrasedas“anytwodistinctpointsarecontainedinone, andonlyone, line.” TheSecondPostulatestatesthatifwearegivenalinesegment(whichisfiniteinlength), wecanextendthelinesegment. TheThirdPostulatestatesthatgivenapoint, andgivenaradius, wecandrawthecirclethathasthegivenpointasitscenter, andhasthegivenradius.TheFourthandFifthPostulatesarenotconcernedwithstraightedgeandcompassconstruc-
tions. TheFourthPostulatesaysthatanytworightangles, nomatterwheretheyarelocatedintheplane, areequaltooneanother. Thisstatementmayseemratherobvious, butthereisinfactsomethingtobehypothesizedhere; wewilldiscussthispostulateinmoredetailinSection 1.2,wherewediscussangles.TheFifthPostulate, bycontrast, requiressomeexplanation. Tounderstandwhatthepostulate
says, supposewearegivena line, say k, and two lines that intersect k, say m and n. SeeFigure 1.1.1. Let α and β betheanglesshowninthefigure. TheFifthPostulatestatesthatifα + β < 180◦, thenthelines m and n willeventuallyintersectonthesamesideof k as αand β. (Notethat 180◦ equals“tworightangles.”) Statedthisway, Euclid’sFifthPostulatedoesmakesenseintuitively. Itwillturnout, asdiscussedlaterinthissection, thattheFifthPostulatehasgreathistoricalsignificance, muchmorethantheotherfourpostulates.
Havingstatedhisdefinitions, commonnotionsandpostulates, Euclidgoesontoprovemanypropositions. Tomakeeverythingcompletelyrigorous, EuclidprovesProposition 1usingonlyhisdefinitions, commonnotionsandpostulates. Proposition 2isprovedusingProposition 1,
1.1EuclidandNon-Euclid 7
k
m
n
α
β
Figure1.1.1
togetherwiththedefinitions, commonnotionsandpostulates. Andsoon. Someofthepropo-sitionsin“TheElements”areveryfamiliartoustoday, forexamplethePythagoreanTheorem,whichisProposition 47inBookI of“theElements,” beingthepenultimatetheoreminthatbook;thelastpropositionisaconversetothePythagoreanTheorem(whichwestateanddemonstrateinourProposition 2.5.2). ThefirsttwopropositionsinBookI areasfollows.
Propositions
1. Onagivenfinitestraightline, toconstructanequilateraltriangle.
2. Toplaceatagivenpoint(asanextremity)astraightlineequaltoagivenstraightline.
Aswithsomeofthepostulates, thesefirsttwopropositionsinvolveconstructionswithstraight-edgeandcompass(thoughnotallthepropositionsinvolveconstructions). Proposition 1ofEu-clidsaysthat, givenalinesegment, wecanconstructanequilateraltrianglethathasthelinesegmentasoneofitsedges. Proposition 2saysthat, givenalinesegment, andapointsome-whereintheplane, wecanconstructanewlinesegmentthatisequalinlengthtothegivenone, andhasthegivenpointasoneofitsendpoints.Inmodernterminology, theproofofEuclid’sProposition 1hastwostages. Wearegivena
linesegmentintheplane. First, Euclidtellsushowtoconstructacertaintrianglethathasthelinesegmentasoneofitsedges; second, heprovesthatthetrianglesoconstructedisindeedequilateral. Whatconcernsusisthefirststageofthisproof. Theideaissimple. Supposewehavealinesegmentwithendpoints A and B, asshowninFigure 1.1.2 (i). First, drawanarc(whichissimplypartofacircle)usingacompasswithcenterat A, andwithradiusthelengthfrom A
to B; thendrawanarcwithcenter B, andthesameradius. SeeFigure 1.1.2 (ii). Let C denotethepointwherethetwoarcsintersect. Usingastraightedge, drawthelinesegmentsfrom A toC, andfrom B to C. SeeFigure 1.1.2 (iii). Wenowhaveatrianglewithvertices A, B and C. Inthesecondstageofhisproof, Euclidshowsthatthistriangleisindeedequilateral.
8 1. GeometryBasics
(i) (ii)
(iii)
A B A B
C
A B
Figure1.1.2
BEFORE YOU READ FURTHER:
Euclid’sproofofProposition 1isintuitivelycompletelycorrect, butfromarigorouspointofview, thereisaflaw. Trytoseeifyoucanfigureoutwhattheproblemwiththisproofis.
TheproblemwithEuclid’sproofofProposition 1ultimatelygoesbacktothefactthathewantshisprooftorelyonlyonhisdefinitions, postulatesandcommonnotions. ThatwecandrawthetwoarcsshowninFigure 1.1.2 (ii) indeedfollowsfromtheThirdPostulate, andthatwecandrawthelinesegmentsfrom A to C andfrom B to C followsfromtheFirstPostulate. WhatisnotexplicitlyguaranteedbyanyofEuclid’spostulatesorcommonnotionsisthatthetwoarcsweconstructedactuallyintersect. Wesimplyassumedthatthetwoarcintersect, andlabeledthepointofintersection C. Howdoweknowthatthetwoarcsreallydointersect? ItcertainlylooksasiftheydoinFigure 1.1.2 (ii), butthatisnotaconvincingargument, becausewemight
1.1EuclidandNon-Euclid 9
havedrawnthefigureincorrectly. Moreover, ifaproofisgenuinelyrigorous, itoughtnottorelyonapicture—thepictureissimplymeanttohelpourintuition.Itiscertainlynottruethatanytwocirclesintheplaneintersect, forexampletwocircleswith
radius 1 incheach, andwithcenters 10 inchesapart. Hence, toguaranteethattwoarcs(whicharepartsofcircles)intersect, wewouldneedtoknowsomethingspecificaboutthemthatinsuresintersection. Forexample, ifweknewsomethingabouttherelationbetweenthecentersofthecirclescontainingthearcsandtheradiiofthecircles, thenwemightbeabletodemonstratethatthearcsintersect; ifthedistancebetweenthecentersofthecirclesistoolargeinrelationtotheradii, thenthearcsmightnotintersect. Itis, infact, possibletogivecriteriaontheradiiandcentersoftwocirclesthatinsurethattwocirclesintersect. Thereis, however, amoresubtleproblemwiththisaspectofEuclid’sproof. Notonlydoweneedtoinsurethattheradiianddistancebetweenthecentersoftwocirclesareappropriateinordertoinsurethatthecirclesintersect, butwealsoneed toknowthatcirclesare“continuous,” that is, that theyhaveno“gaps”inthemrightwheretheintersectionissupposedtotakeplace. Thisissueofgapsisverysubtle, andwedonothavethespacetodiscussithere. Inhispostulatesandcommonnotions,Euclidaddressesneither theissueofappropriateradiiandcenters, nor theissueofnogaps,andthereforehehasnotrigorouslyprovedhisProposition 1. ModernaxiomatictreatmentsofEuclideangeometrysuccessfullyavoidEuclid’sinsufficientaxiomsbygivingmoreaxiomsthanEuclidgave. ItisnotthatwhatEuclidsaidwaswrong; itissimplyinsufficient.The twoproblemswithEuclidmentionedso far, namely thedefinitions thatdonotdefine
anything, andtheinsufficientaxioms, canbothberemedied. TwothousandyearsafterEuclid,mathematicianshaveshowedthatwithregardtothesetwoissues, Euclidwascorrect, justmiss-ingsomedetails. Thereis, however, another, moretricky, problemwithEuclid. TheproblemconcernsEuclid’sFifthPostulate. A lookatthefivepostulatesquicklyrevealsthatthefifthissomehowdifferentfromthefirstfour. Thefirstfouraresimpletostate, andimmediatelybeliev-able. Thefifth, bycontrast, ismuchlongertostate, and, whilecertainlybelievable, doesnotseemasimmediatelyobviousasthefirstfour. MathematiciansthroughoutthecenturiesafterEuclidnoticedthisproblem. Thereisnothinginherentlywrongwithapostulatethatiscompli-cated, asisEuclid’sFifthPostulate, butitisbothersome. OneofthereasonspeoplelikedEuclid’sgeometryisthat(ignoringtheflawsmentionedabove, whichseemtohavebeennoticedonlylateron), itseemedtobeamodelforprovingthatcertainfactsareindisputablytrue. Euclidwastheultimateexampleofhowhumanbeingscouldobtaincertainknowledge. Ifonestartswithindisputablytrueaxioms, andproceedsinanairtightlogicalfashiontodeducethingsfromtheaxioms, thenwhateveronededucesmustalsobeindisputablytrue. However, ifEuclidistobeviewedinthisway, thenitiscrucialthathisaxiomsareindisputablytrue. Thecommonnotionsandthefirstfourpostulatesseemquiteconvincing. TheFifthPostulate, bycontrast, doesnotseemquiteasindisputable, givenitsmorecomplicatednature.WhatcanbedoneabouttheFifthPostulate? Itcannotsimplybedropped—itismostdefinitely
usedinsomeofEuclid’sproofs. WhatanumberofpeopleafterEuclidtriedtodowastodeducetheFifthPostulatefromtheotherfour. IftheFifthPostulatecouldbededucedfromtheotherfour, thenanythingprovableusingallfivepostulatescouldbeprovedusingthefirstfour, which
10 1. GeometryBasics
wouldaddtotheindisputabilityofwhateverwasprovedbyEuclid. Overtheyears, anumberofpeopleclaimedtohavededucedEuclid’sFifthPostulatefromtheotherfour. Wenowknow,however, that theywereallmistaken. Asdiscovered independentlybyKarl FriedrichGauss(1777-1855), JanosBolyai (1802-1860)andNikolai IvanovitchLobatchevsky (1793-1856) intheearly19thcentury, itispossibletoconceiveofperfectlygoodgeometriesthatinvolvethefirst fourpostulates, but somethingother than theFifthPostulate (wewillmentionhow thiscouldhappenverybrieflybelow). Thisdiscoverywasextremelyrevolutionary. Infact, Gauss,arguablythegreatestmathematicianofalltimes, butsomeonewhoseemstohavebeenquiteconcernedabouthisreputation, didnotinitiallypublishhisdiscoveryofwhatisnowcallednon-Euclideangeometry, for fearof thepublic reaction. For the twothousandyearsprior toGauss, “TheElements”hadbeentakenassomethingapproachingasacred, non-challengeable,text. TochallengeEuclid, asGaussdidprivately, andsubsequentlyBolyaiandLobatchevskydidpublicly(andindependently)in1831and1829respectively, wasseenasalmostashereticalasDarwinwaslaterinthe19thcentury.Wecannotgiveherethedetailsofthenon-Euclideanrevolution, excepttosaythatitusheredin
acompletelyneweraofgeometry. Newapproachestogeometry, suchastheuseofisometries(acrucialtoolinourinourtreatmentofsymmetryinChapters 4 and5), andRiemanniangeometry(thatwaslaterusedbyEinsteininhisgeneraltheoryofrelativity)flourishedinthe19thcentury,oncethestrangle-holdofEuclid’sapproachwaslifted. (Ofcourse, haditnotbeenforEuclid,geometrymightnothave reached the19thcentury inasdevelopeda formas itdid, soweshouldnotbelittleEuclid’sfundamentalimportancetogeometry, andallofmathematics; therearesimplyotherapproachestogeometryaswell.) Thediscoveryofnon-Euclideangeometryhadphilosophicalimportancebeyondjustgeometry, orevenmathematics. IfEuclidwasonceheldupasamodelforabsolutetruth, andifwenowknowthatothertypesofgeometryarepossible, thenweneedtorethinkwhat, ifanything, wecanknowwithabsolutecertainty. See[Tru87]or[Gre93]formoredetailsaboutthenon-Euclideanrevolution.TogetabitmoreofafeelforwhatmakesEuclideangeometrydistinctfromnon-Euclidean
geometry, wementionanimportantresultinEuclideangeometryknownasPlayfair’sAxiom. AdemonstrationofPlayfair’sAxiomwillbegiveninSection 1.2.
Proposition 1.1.1 (Playfair’sAxiom). Supposem isaline, andA isapointnotonm. Thenthereisoneandonlyonelinethrough A thatisparallelto m.
Fromourpointofview“Playfair’sAxiom”innotanaxiomatall, butatheoreminEuclideangeometry; however, thenameistraditional, andweuseitwhetherornotitmakesliteralsense.Actually, Playfair’sAxiomisnotonlyatheoreminEuclideangeometry, but, morestrongly, itisequivalenttoEuclid’sFifthPostulate. BythatwemeanthatEuclid’sfivepostulatesimplyPlay-fair’sAxiom, andEuclid’sfirstfourpostulatestogetherwithPlayfair’sAxiomimplyEuclid’sFifthPostulate. (Infact, insomemoderngeometrytexts, thestatementofPlayfair’sAxiomisincor-rectlyreferredtoasEuclid’sFifthPostulate. AlthoughlogicallyitiscorrecttosubstitutePlayfair’sAxiomforEuclid’sFifthPostulate, becausethetwostatementsimplyeachother, historicallyitiscompletelyinaccuratetoreplacetheonestatementwiththeother.)
1.1EuclidandNon-Euclid 11
Non-Euclideangeometryresults fromtakingEuclid’sfirst fourpostulates, butreplacingtheFifthPostulateby somethingelse. It is easier tounderstandwhathappens innon-Euclideangeometry ifweconsideralternatives toPlayfair’sAxiom. Supposethat m isa line, and A isapointnoton m. IfPlayfair’sAxiomwerenottrue, thentherearetwopossiblecases: eitherthereismorethanonelinethrough A thatisparallelto m, orthereisnolinethrough A thatisparallelto m. IftheformerpossibilityistakenasanaxiominsteadofPlayfair’sAxiom, theresultinggeometryiscalledhyperbolicgeometry; ifthelatterpossibilityistakenasanaxiominsteadofPlayfair’sAxiom, theresultinggeometryiscalledsphericalgeometry. InSection 2.2,wewillseethatPlayfair’sAxiomimpliesthatthesumoftheanglesinatriangleis 180◦. Hence,inEuclideangeometrythesumoftheanglesinatriangleis 180◦. Bycontrast, itcanbeprovedthatinhyperbolicgeometry, thesumoftheanglesinatriangleisalwayslessthan 180◦ (theprecisesumcanvaryfromtriangletotriangle); insphericalgeometry, thesumoftheanglesinatriangleisalwaysgreaterthan 180◦ (again, theprecisesumcanvaryfromtriangletotriangle).Itishardtoimaginehowhyperbolicorsphericalgeometrywouldworkifweusethefamiliar
sortofstraightlinesfoundintheplane, butthereisnoneedtorestrictourselvesonlytothemostfamiliarsituation. Forexample, thinkofthesurfaceofasphereasauniverse, andthinkofgreatcirclesas“straightlines”inthisuniverse(greatcirclesarestraightfromthepointofviewofabuglivingonthesphere). Thegeometrythatusesgreatcirclesonthesphereturnsouttobewhatwenowcallsphericalgeometry. Modelsforhyperbolicgeometrycanalsobefound. A detaileddiscussionofnon-Euclideangeometrywould takeus too far afield; see [Mar75]or [Gre93]forfurtherdetails. Onepointworthmentioningisthatitcanbeproved, thoughthisisfarfromobvious, thatEuclideangeometryisnomoreorlessvalidthanhyperbolicorsphericalgeometry.Thatis, ifweacceptEuclideangeometry, weneedtoacceptnon-Euclideangeometryaswell; ifwedonotacceptnon-Euclideangeometry, thenwecannotacceptEuclideangeometryeither.Mathematicallyspeaking, thereismorethanonepossiblevalidgeometry. Whichgeometryourphysicaluniversesatisfiesisanothermatter, onewhichphysicists, notmathematicians, needtodecidebyuseofexperiments. Onadailybasis, ouruniverseiseitherEuclideanorcloseenoughtoitthatitisEuclideanforallpracticalpurposes.InthistextwewillbeworkingwithinwithintheframeworkofEuclideangeometry, though
wewillnotbeapproachingthingsaxiomatically(akasynthetically). However, wewantedtohaveabriefoverviewofwhat theaxiomaticpropertiesof Euclideangeometryare, becausethesefeaturesultimatelyunderlieeverythingthatwedo, evenwhentheyarenotmentionedexplicitly. Forexample, wewillseewhereEuclid’sFifthPostulatecomesintoplayinthestudyofparallellinesinSection 1.2. Ourstudyofsymmetry, inPartII ofthisbook, usesanapproachtogeometrythatonlycameintobeingafterthediscoveryofnon-Euclideangeometry. However,eventhoughourstudyofsymmetrymayseemdifferentfromtheapproachfoundinEuclid, ittooisultimatelyEuclidean.Finally, weend this sectionwith thehope thatour remarksabout theflaws inEuclidwill
notdiscouragethereaderfromtakinganinterestin“TheElements.” Euclidcontainsawealthofsubstantialmathematicalideas, thoughthedryandsometimestediousstyledonotalwaysmaketheseideasapparentuponfirstencounter. Anyonewishingtoread“TheElements”would
12 1. GeometryBasics
dowelltoreadboththeactualtextfoundin[Euc56], togetherwithacompaniontoEuclid, suchastheexcellent[Har00].
1.2 LinesandAngles
The twomost fundamentalnotions in thegeometryof theplaneare point and line. By theword“line”wealwaysmeanastraightline(asisstandardusagetoday, thoughdifferentfromEuclid’susage). Asdiscussed inSection 1.1, Euclidattempted todefine theseconcepts, butdidnotsucceedindoingsoinameaningfulway. Wetakethemodernapproach, andsimplyassumethattherearesuchthingsaslinesandpointsintheplane, andthatthereisarelationbetweenpointsand lines, namely thatgivenapointandgivena line, either thepoint isonthelineoritisnot. Further, werequiretherelationbetweenpointsandlinestosatisfycertainfamiliarproperties, suchasthefactthatanytwodistinctpointsarecontainedinoneandonlyoneline(thisisessentiallyEuclid’sFirstPostulate). Wedonotcaresomuchwhatpointsandlinesare, buthowtheybehave. Formostofthistext(except, forexample, inChapter 3), wewillberestrictingourattentiontopointsandlinesintheplane. Itisalsopossibletodiscusspointsandlines(andothergeometricobjects)inthreedimensionalspace, andhigherdimensionstoo.Whennototherwisenoted, thereadershouldassumewearediscussingtheplane.Westartourdiscussionoflineswithsomenotation. Giventwodistinctpoints A and B inthe
plane, weknowthatthereisauniquelinecontaining A and B. Wedenotethislineby←→AB.
WhenitisnotnecessarytospecifythepointsA and B, wewillalsousesingleletterssuchasmtodenotelines. Intuitively, aline“goesonforever”intwodirections. Giventwodistinctpoints
A and B, wecanalsohavethatpartoftheline←→AB thatstartsat A, and“goesonforever”in
thedirectionof B. Suchanobjectiscalledthe ray from A through B, andisdenotedby−→AB.
Wecall A the startingpoint oftheray−→AB. Wecanalsolookatthatpartoftheline
←→AB that
startsat A andendsat B (orvice-versa). Suchanobjectiscalledthe linesegment from A to B,andisdenotedby AB. SeeFigure 1.2.1 forallthreetypesofobjects. Wecallthepoints A andB the endpoints ofthelinesegment AB.
Oneofthemostbasicquestionaboutlinesintheplaneiswhetherornottwolinesintersect,whichmeansthatthetwolineshaveapointincommon. Ourfirstresultisthefollowingratherobviousfact; evenobviousresultsneedtobeproved, however, becauseourintuitionaboutwhatis“obvious”issometimeswrong(forexample, peopleusedtothinkthattheearthwasflat).
Proposition 1.2.1. Twodistinctlinesintersectinatmostonepoint.
Demonstration. Supposem and n aredistinctlines. Supposefurtherthattheyintersectinmorethanonepoint. Hence, thereareatleasttwodifferentpoints, sayA and B, thatarecontainedinboth m and n. Itfollowsthateachof m and n isalinecontainingthepoints A and B, whichcontradictsthefactstatedabovethatanytwodistinctpointsarecontainedinoneandonlyoneline. Henceitmustbethecasethat m and n donotintersectinmorethanonepoint.
1.2LinesandAngles 13
A B
AB
A B
AB
A B
AB
Figure1.2.1
Becauseoftheaboveproposition, ifwearegiventwodistinctlines, eithertheydonotintersect,ortheyintersectinpreciselyonepoint. Wesaythattwolinesare parallellines iftheydonotintersect; iftwolinesarenotparallel, wesaythattheyare intersectinglines. SeeFigure 1.2.2.Noticethatthedefinitionofparallellinesdoesnotmentionanythingaboutparallellines“goinginthesamedirection,” or“keepingconstantdistancefromeachother.” Bothoftheseideasaretrueaboutparallellinesintheplane, buttheyarenotpartofthedefinition, andneedtobeproved. WewillessentiallyprovethefirstoftheseideasinProposition 1.2.3, andthesecondinProposition 2.2.6. Noticealsothattwoequallinesarenotconsideredparallel, becausetheycertainlydointersect.
parallel lines intersecting lines
Figure1.2.2
Strictlyspeaking, theterm“parallel”appliesonlytolines, andnottolinesegments. However,itmakesintuitivesensetodiscusslinesegmentsbeingparallelornot, andwewillsaythattwolinesegmentsare parallel ifthelinescontainingthelinesegmentsareparallel.Havingbrieflydiscussedlines, wenowturntoanothertypeoffundamentalgeometricobject,
namelyangles. An angle isaregionoftheplanethatisbetweentworaysthatintersectinacommonstartingpoint. Forexample, theshadedregioninFigure 1.2.3 isanangle. Thepointofintersectionofthetworaysiscalledthe vertex oftheangle. Giventworaysthatintersectinacommonstartingpoint, thereareactuallytwoanglesthattheraysdetermine, forexampletheshadedregioninFigure 1.2.3, andtheunshadedregion. Itisthereforenecessarytospecify
14 1. GeometryBasics
whichoftheregionsisbeingreferredto. Thewaytoavoidthissortofambiguityistospecifyananglenotonlybytworaysthatintersectinacommonstartingpoint, butalsotospecifyadirection, eitherclockwiseorcounterclockwise. InFigure 1.2.4 (i)weseeananglespecifiedbytworaysandtheclockwisedirection(indicatedbythecurvedarrow); inFigure 1.2.4 (ii)weseeadifferentanglespecifiedbythesametworaysasinPart (i)ofthefigure, butwiththeoppositedirection.
Figure1.2.3
(i) (ii)
Figure1.2.4
Anangleisageometricobject. Justaswecanmeasurethelengthsoflinesegments, wecanalsomeasureangles, notinunitsoflength(forexample, feetormeters), butinunitsofangularmeasure. Measuringageometricobjectmeansassigningtotheobjectanumber, whichinsomesensetellsusthe“size”oftheobject. Therearetwostandardunitsofangularmeasure, degreesand radians. Degreesare simpler toexplain, andare thereforeused regularly inelementaryandsecondaryschools; radiansareprefered inadvancedmathematics, forvarious technicalreasonswecannotdiscusshere. Weassumethatthereaderisfamiliarwithdegreemeasureofangles, andwewillusedegreesinthistext. (Weshouldmentionthatthenumber 360 usedinthecontextofdegreemeasureiscompletelyarbitrary, andariseshistorically, ratherthanoutofanyrealreason. Itwouldbeequallyvalidtobreakupthetotalanglegoingaroundapointintoanyothernumberof“degrees,” but 360 isfamiliartoeveryone, andwewillstickwithit.)Theonepointaboutmeasuringangles(bydegreesoranyothermethod)thatweneedtostress
involvesourpreviousobservation that thereare two“directions” inwhichananglecanbespecified, namelyclockwiseandcounterclockwise. Bothdirectionsareequallyvalid, butitis
1.2LinesandAngles 15
convenienttopickonedirectionashavingpositivemeasure, andonedirectionhavingnegativemeasure. Wewill take theclockwisedirectionaspositive, becausethat is themorefamiliardirectionindailylife. (It iscompletelyarbitrarywhichdirectionis takenaspositive, aslongasweallagreeonthechoice. Inmoreadvancedmathematics texts, it iscustomarytohavecounterclockwisebethepositivedirection.) InFigure 1.2.5 (i)weseeananglethathasmeasure45◦, andinFigure 1.2.5 (ii)weseeananglethathasmeasure −45◦.
-45°45°
(i) (ii)
Figure1.2.5
Degreesareunitsformeasuringangles. Itisimportanttodistinguishbetweenanangleanditsmeasure. Anangleisageometricobject; itsmeasureindegreesisanumberthatweassociatewith theangle. Analogously, theheightofaperson isanumber thatweassociatewith thatperson. Justastwopeoplecanhavethesameheight, butstillbedifferentpeople, similarlytwoanglescanhavethesamemeasure, butstillbedifferentangles. Forexample, inFigure 1.2.6 weseetwodifferentangles, eachofwhichhasmeasure 60◦. Ofcourse, iftwoangleshavedifferentmeasure, theymustbedifferentangles. Iftwoangleshavethesamemeasure, theyneednotbethesameangle, butitisalwaysthecasethattheyarecongruent, whichmeansintuitivelythatoneanglecouldbe“pickedupandplacedpreciselyontopoftheother.” Wewillnotneedamoreformaldefinitionofcongruencehere, andwillsticktotheintuitivenotion. (Theconceptofisometry, whichisdiscussedindetailinChapter 4, providesonewayofdefiningcongruence.)Thebottomlineisthatwhenwesaythat“twoangleareequal”wemeanthattheyarecongruent,andinparticularthattheyhavethesamemeasureindegrees. Forthesakeofbrevity, wewillsometimes“abuse”terminologyandsay, forexample, ofananglethat“is” 90◦, whenweshouldmoreproperlysaythattheangle“hasmeasure” 90◦. Suchabuseofterminologyisverycommon,andshouldcausenoconfusion.
LetusnowreturntoEuclid’sFourthPostulate, whichsays“Thatallrightanglesareequaltooneanother.” Tounderstandthispostulate, weneedtoknowwhatarightangleis. Todaywetendtothinkofarightangleasbeingdefinedtobeanangleof 90◦, butthatisnottheproperwaytounderstandrightangles, becauseitonlytellsusthemeasureofrightangles, notwhattheyaregeometrically. Thegeometricideaofarightangleisbasedonwhathappenswhentwolinesintersectinapoint. AsweseeinFigure 1.2.7 (i), twolinesintersectingdividetheplaneintofourangles. Thefouranglesarenotnecessarilyallequaltoeachother. However, ifitdoeshappenthatallfouranglesareequaltoeachother, thenwecalleachofthefouranglesa rightangle.
16 1. GeometryBasics
Figure1.2.6
SeeFigure 1.2.7 (ii). WhatEuclid’sFourthPostulatesaysisthatanysuchrightangle, formedbyanytwointersectinglines, isequaltoanyotherrightangleformedbytwootherintersectinglines. Asforthedegreemeasureofarightangle, becausethetotalanglegoingaroundapointis 360◦, andbecausetherearefourrightanglesatapoint, itfollowsthatthemeasureofanyrightangleis 360◦/4 = 90◦. Usingthestandardabuseofterminology, wewillfollowcommoncustomandsimplysaythatanyrightangle“is” 90◦. Similarly, theanglealonganystraightlineis 180◦.
(i) (ii)
Figure1.2.7
Wenowmaketwomoregeometricdefinitionsconcerningangles. First, supposewehavetwoanglesthatare“alongaline;” thatis, twoanglesthatareformedwhenaraystartsatapointonaline. Twosuchanglesarecalled supplementaryangles. Theanglesα andβ inFigure 1.2.8 (i)aresupplementaryangles. Next, supposewehavetwointersectinglines. Ofthefouranglesformed,therearetwopairsof“oppositeangles;” thatis, anglesthatintersectonlyinacommonpoint.Suchanglesarecalled verticalangles. Theangles α and β inFigure 1.2.8 (ii)areverticalangles.
Wenowstateaverysimpleresultaboutangles, usingtheconceptsjustdefined.
Proposition 1.2.2.
1. Supplementaryanglesaddupto 180◦.
2. Verticalanglesareequal.
1.2LinesandAngles 17
(i) (ii)
αα β β
Figure1.2.8
Demonstration.
(1). Thisisevident, becausetheanglealongastraightlineis 180◦.
(2). InFigure 1.2.9 weseeangles α and β, whichareverticalangles. Noticethattheangleγ isasupplementaryangletoeachof α and β. Hence, byPart (1)ofthisproposition, weknowthat α + γ = 180◦ and γ + β = 180◦. Wededucethat α = 180◦ − γ and β = 180◦ − γ,andtherefore α = β.
α βγ
Figure1.2.9
TheproofofProposition 1.2.2 isverysimple. Moreover, itdoesnotmakeuseofthefullstrengthofEuclid’spostulates, inthattheFifthPostulateisnotused. Ournextresultaboutanglesismoresubstantial, andtheFifthPostulateiscrucialtoitsproof. Weareinterestedinthesituationwheretwolines, saym andn intersectathirdline, say k; seeFigure 1.2.10. Thelinesm and k formfourangles, andthelines n and k formfourmoreangles, allofwhicharelabeledinFigure 1.2.10.Wecallangles x and y interioralternatingangles, andwealsocallangles z and w interioralternatingangles. Similarly, wecallangles α and β exterioralternatingangles, andwealsocallangles δ and ϵ exterioralternatingangles. Finally, wecallangles x and β correspondingangles, andwealsocallangles δ and w, angles z and ϵ, andangles α and y, correspondingangles. Thefollowingpropositionsaysthattheabovesortsofanglesareparticularlynicewhenwestartwithtwoparallellinesthatintersectathirdline.
Proposition 1.2.3. Supposetwoparallellinesintersectathirdline. Thentheinterioralternatinganglesareequal; theexterioralternatinganglesareequal; thecorrespondinganglesareequal.
18 1. GeometryBasics
α
ε β
δ
y w
z x
m
n
k
Figure1.2.10
Demonstration. InFigure 1.2.10 weseeangles x and y, whichareinterioralternatingangles,angles α and β, whichareexterioralternatingangles, andangles x and β, whicharecorre-spondingangles. Wewilldemonstratethepropositionwithregardtothesethreepairsofangles;theotherappropriatepairsofanglesaresimilar, andwewillskipthedetailsforthem.Westartwiththeobservationthatoneofthefollowingthreecasesmustcertainlyhold: either
x+w < 180◦, or x+w = 180◦, or x+w > 180◦. Supposefirstthat x+w < 180◦. RecallEuclid’sFifthPostulate, whichsays“That, ifastraightlinefailingontwostraightlinesmakestheinterioranglesonthesamesidelessthantworightangles, thestraightlines, ifproducedindefinitely, meeton that sideonwhichare theangles less than the two rightangles.” Thispostulateispreciselysuitedtoourcurrentsituation, anditimpliesthat m and n intersectonthesideof x and w. Wehavethereforereachedalogicalimpossibility, because m and n areassumedtobeparallel, andhencecannotintersect. Weconcludethatitcannotbethecasethatx+w < 180◦.Nowsupposethat x+w > 180◦. Clearly y = 180◦ −w and z = 180◦ − x. Therefore
y+ z = (180◦ −w) + (180◦ − x) = 360◦ − (x+w) < 180◦.
ThenbyEuclid’sFifthPostulate, itwouldfollowthat m and n intersectonthesideof y and z,whichagaincannotbe, becausethetwolinesareparallel. Theonlyremainingoptionisthatx + w = 180◦. Therefore x = 180◦ − w. Becauseweknowthat y = 180◦ − w, itfollowsthat x = y. Hence, interioralternatinganglesareequal. Because β and y areverticalangles,weknowthat β = y. Itissimilarlyseenthat α = x. Wededucethat β = x, whichsaysthatcorrespondinganglesareequal, andthat α = β, whichsaysthatexterioralternatinganglesareequal.
TheproofofProposition 1.2.3 verymuchdependsuponEuclid’sFifthPostulate, andthereforeanyotherpropositionthatisdemonstratedusingProposition 1.2.3 mustalsodependuponEu-clid’sFifthPostulate. Forexample, Proposition 2.2.1, whichdiscussesthesumoftheanglesin
1.2LinesandAngles 19
atriangle, usesProposition 1.2.3 initsproof. Thus, weseethattheFifthPostulateiscrucialinthestudyofplanargeometry.Amongotherthings, Proposition 1.2.3 showsthatourdefinitionofparallellines, whichwas
simplyintermsofnon-intersection, correspondstoourintuitionthatparallellines“gointhesamedirection.”
Exercise 1.2.1. Findangles x and y asshowninFigure 1.2.11. Thelines m and n areparallel.
y
x
m
n45°
60°
Figure1.2.11
Exercise 1.2.2. Findangles α, β and γ asshowninFigure 1.2.12. Thelines p and q areparallel.
ItturnsoutthattheconversetoProposition 1.2.3 isalsotrue. Thatis, iftwolinesmakeap-propriateangleswithagivenline, thentheyareparallel. Moreprecisely, wehavethefollowingproposition.
Proposition 1.2.4. Supposetwodifferentlinesintersectathirdline. Ifthetwolineshaveequalinterioralternatingangles, orequalexterioralternatingangles, orequalcorrespondingangles,thenthetwolinesareparallel.
Interestingly, eventhoughProposition 1.2.3 reliescruciallyonEuclid’sFifthPostulate, itturnsoutthatProposition 1.2.4 doesnotrelyupontheFifthPostulate(andisthereforetrueinotherge-ometriesforwhichthefirstfourpostulateshold, butthefifthdoesnot). However, aparticularly
20 1. GeometryBasics
γ
β
α
p
q
80° 115°
Figure1.2.12
easydemonstrationofProposition 1.2.4 canbe foundusing theFifthPostulate; thisdemon-stration is left to the readerasExercise 2.2.1, where ituses the fact, proved inSection 2.2,thatthesumoftheanglesinatriangleaddupto 180◦. (SeeTheorem 3.4.1of[WW98]forademonstrationofProposition 1.2.4 thatdoesnotmakeuseofEuclid’sFifthPostulate.)AnimmediateconsequenceofProposition 1.2.4 is thefollowingresult. First, weneedthe
followingstandardbitofterminology. Giventwolinesintheplane, wesaythattheyare per-pendicular iftheymakerightangleswitheachother.
Proposition 1.2.5. Supposetwodifferentlinesarebothperpendiculartoathirdline. Thenthetwolinesareparallel.
WecannowgiveademonstrationofPlayfair’sAxiom (Proposition 1.1.1). Inthisdemonstra-tion, andinothersubsequentplaces, wewillmakeuseofthefactthatifwearegivenalinemandapoint A (eitheronoroff m), wecanconstructalinethrough A thatisperpendiculartom. Wetakethispropertytobeaxiomatic. Also, wenotethatthisconstructionisveryeasytodousingstraightedgeandcompass, thoughwewillnotneedthedetailshere.
DemonstrationofProposition 1.1.1. Supposethatm isaline, andthatA isapointnotonm.Weneedtoshowtwothings: (1)thereisalinethrough A thatisparallelto m; and(2)thereisonlyonesuchline.ToshowPart (1), westartbyconstructingalinethrough A thatisperpendicularto m. Call
thisnewline n. Next, constructalinethrough A thatisperpendicularto n (thefactthat A ison n causesnoproblem). Callthisnewline p. SeeFigure 1.2.13 (i). Byconstruction, weseethatboth m and p areperpendicularto n. ItfollowsfromProposition 1.2.5 that m and p areparallel. Wehavethereforeconstructedalinethrough A, namely p, thatisparallelto m.
ToshowthePart (2), weneedtoshowthattheline p weconstructedaboveistheonlylinethatcontains A andisparallelto m. Let t beanylinethatpassesthrough A otherthan p. SeeFigure 1.2.13 (ii). Giventhat t isnotthesameas p, then, asseeninthefigure, itcannotbethat tisperpendicularto n, because p isperpendicularton. Itnowfollowsthat t andm donotmake
1.3Distance 21
m
p
n
A
m
p
tn
A
(i) (ii)
Figure1.2.13
equalcorrespondingangleswith n, because m makesall 90◦ angleswith n, and t doesnotmake 90◦ angleswithn. Wededucethatm and t arenotparallel, becauseiftheywereparallel,thenbyProposition 1.2.3 itwouldfollowthatm and t wouldhaveequalcorrespondingangleswith n, whichtheydonot. Itfollowsthat p istheonlylinethrough A thatisparalleltom.
Exercise 1.2.3. Westatedabove that ifwearegivena line m andapoint A, wecanconstructalinethrough A thatisperpendicularto m. Showthatthereisonlyonesuchline. Thedemonstrationhastwosubcases, dependinguponwhether A ison m ornot.
1.3 Distance
PlanargeometryasformulatedbyEuclidrests, fundamentally, ontheideathattherearethingscalledpointsandthingscalledlines, andthatthesethingshavesomerelationtoeachother(forexample, anytwodistinctpointsarecontainedinauniqueline). Insomeofthemoremodernapproachestogeometryotherfundamentalnotionshavealsocomeintoplay. Oneoftheideasthathasprovedtobeparticularuseful, forexampleinthestudyofsymmetry(aswewillseeinChapters 4 and5), isthenotionofdistancebetweenpoints. Certainly, theideaoflengthsoflinesegmentsisinEuclid, andhencethedistancebetweentheendpointsofalinesegmentisimplicitinEuclid, butmodernmathematicsdealswiththeconceptofdistancemoreexplicitly. Wewillrestrictourattentiontothedistancebetweenpointsintheplane, thoughitisalsopossibletolookatdistanceinothersituations.Thereare, infact, twodifferentapproachestothenotionofdistancebetweenpoints. First,
wecoulduseCartesiancoordinates, accordingtowhichweassigneverypointintheplaneapairofnumbers (x, y). (Weassumethereaderis familiarwithsuchcoordinates.) Wecouldthendefine thedistancebetween twopoints (x1, y1) and (x2, y2) by thestandarddistanceformula
√(x2 − x1)2 + (y2 − y1)2. (ThisformulafollowsdirectlyfromthePythagoreanThe-
orem.) However, giventhatwearenotgoingtobeusingcoordinatesinothersituations, this
22 1. GeometryBasics
approachtodistancewillnotbetheoneweuse. Alternatively, justasweassumedsomeax-iomaticpropertiesforpointsandlines, wecansimplyhypothesizethatitispossibletoassignauniquedistancebetweenanytwopointsintheplane, andthatthismeasureofdistancesatis-fiescertainproperties. Wewilltakethelatterapproach, thoughwewillnotgiveanaxiomatictreatmentofdistance(whichissurprisinglytricky).SupposeA andB arepointsintheplane.Wethenassigntothesepointsarealnumberdenoted
d(A,B), calledthe distance betweenA and B. Amongthepropertiesthatdistancesatisfiesarethefollowing, where A, B and C arepointsintheplane.
1. d(A,B) ≥ 0;
2. d(A,A) = 0, and d(A,B) = 0 whenever A = B;
3. d(A,B) = d(B,A);
4. d(A,B) ≤ d(A,C) + d(C,B).
Thefirstthreeofthesepropertiesaresimple; thefourth(calledtheTriangleInequality)takesabitmoreofanexplanation. Consideratrianglewithvertices A, B and C, asinFigure 1.3.1.ThenProperty(4)saysthatthelengthoftheside AB ofthetriangleislessthanorequaltothesumofthelengthsoftheothertwosidesofthetriangle(wehaveequalityonlyifthetriangleis“degenerate”). Byrearrangingtheletters, weseethatthesameinequalityholdsfortheothertwosidesofthetriangleaswell. (HadwetakentheapproachofusingCartesiancoordinatestodefinedistancebetweenpoints, itwouldhavebeenpossibletoprovetheabovefourpropertiesusingthedistanceformula.)
A
B
C
Figure1.3.1
Oncewehaveanotionofdistancebetweenpoints, it ispossibletoformulatemanyotherbasicgeometricconceptssuchaslines, raysandlinesegmentsintermsofdistance. Forexample,supposewearegiventwopoints A and B intheplane.Thenthelinesegmentfrom A to B isthecollectionofallpoints X intheplanesuchthatthe
equation d(A,X)+d(X,B) = d(A,B) holds. TherayfromA through B isthecollectionofallpointsX intheplanesuchthatpreciselyoneofthetwoequationsd(A,X)+d(X,B) = d(A,B)or d(A,B)+d(B,X) = d(A,X) holds. ThelinethroughA and B isthecollectionofallpoints
1.3Distance 23
X intheplanesuchthatpreciselyoneofthethreeequations d(A,X) + d(X,B) = d(A,B),or d(A,B)+d(B,X) = d(A,X), or d(B,A)+d(A,X) = d(B,X) holds. (Ifonewantstobecompletelydetailedaxiomatically—asisimportantinmoreadvancedtreatmentsofgeometry—thereareanumberofpossibleapproacheswhen itcomes to the relationof theconceptofdistancetotheconceptsofpointsandlines: onecandefinepointsandlinesaxiomatically, thenconstructadistancefunctionfromtheaxioms, andthenshowthatourformulationoflinesintermsofdistanceisconsistentwiththelinesasgivenbytheaxioms; alternatively, onecantakedistanceas thebasicaxiomaticallydefinedconcept, thenuse theaboveapproach todefinelines, andthenshowthatlinesdefinedinthiswaybehaveaslinesoughtto; or, onecandefinebothlinesanddistanceintermsofcoordinates, andthenshowthattheaboveformulationoflinesintermsofdistanceisvalid. Wewillnotgointosuchdetailsinthistext.)Circlescanalsobedefinedusingthenotionofdistance. Givenapoint A intheplane, and
anon-negativerealnumber r, thenthecircle withcenter A andradius r isthecollectionofallpoints X in theplane such that theequation d(A,X) = r holds. It is evenpossible tocomputeanglesstrictlyintermsofdistance. Considerthetrianglewithvertices A, B and C,showninFigure 1.3.1. Itisthenpossibletocomputetheanglesateachof A, B and C usingonlythelengths d(A,B), d(A,C) and d(B,C). Themethodforsuchcalculationsuses theLawofCosines, studiedintrigonometry. ThislawisstatedinProposition 2.5.3, anditmayalsobefoundinanytextbookontrigonometry. (Ifyouareunfamiliarwiththislaw, simplyignoretheformulaweareabouttostate; wewillnotbeusingthisformulaagain.) UsingtheLawofCosines, theformulafortheangleat A isseentobe
arccos
([d(B,C)]2 − [d(A,B)]2 − [d(A,C)]2
2d(A,B)d(A,C)
).
Wenowturntoanissuethatisphrasedintermsofdistancebetweenpoints, andthatwillbeofusetousinourstudyofisometriesinChapter 4, whichinturnisusedinourstudyofsymmetryinChapter 5. Supposewearegiventwopoints A and B intheplane. Wewouldliketofindallpointsintheplanethatareequidistantto A and B; thatis, wewanttofindallpoints X intheplanesuchthat d(X,A) = d(X,B). Onesuchpointisveryeasytofind, namelythemidpointofthelinesegment AB. Thereare, however, otherpointsintheplanethatareequidistanttoA and B aswell. Thefollowingpropositiontellsusaneasywaytofindallsuchpoints. Inthispropositionweuse the followingterminology. Givena linesegment AB, the perpendicularbisector ofthelinesegmentisthelinethatcontainsthemidpointofthelinesegment, andisperpendiculartothelinesegment. SeeFigure 1.3.2. Asisstandard, weuseasmallsquaretodenotearightangleinthefigure. Wecannowstateourresult.
Proposition 1.3.1. Supposethat A and B aredistinctpointsintheplane. If X isapoint, thend(X,A) = d(X,B) ifandonlyif X isontheperpendicularbisectorof AB.
ThedemonstrationofthispropositionislefttothereaderinExercise 2.2.9 (whichisputofftillSection 2.2, becausesomefactsaboutcongruenttrianglesareneeded).
24 1. GeometryBasics
A
B
Figure1.3.2
2Polygons
2.1 Introduction
A polygon isaregionoftheplanethatisboundedbyafinitenumberoflinesegmentsthataregluedtogether. Wehavethreerequirementsaboutthewayinwhichwegluethelinesegmentstogether.
(1) Linesegmentsaregluedendpoint-to-endpoint.
(2) Everyendpointofalinesegmentisgluedtopreciselyoneotherendpointofalineseg-ment.
(3) Notwolinesegmentsintersectexceptpossiblyattheirendpointswheretheyareglued.
SomepolygonsareshowninFigure 2.1.1. Somenon-polygonsareshowninFigure 2.1.2;theobjectinPart (i)ofthisfigurehasedgesthatarenotgluedendpoint-to-endpoint, theobjectinPart (ii)hasthreeendpointsoflinesegmentsgluedtogether, andtheobjectinPart (iii)haslinesegmentsintersectingnotattheirendpoints. (Itispossibletolookatpolygonswithself-intersections, thatis, inwhichrequirement(3)isdropped, butwewillnotbelookingatsuchpolygonsinthistext.)
Foreachpolygon, the edges ofthepolygonarethelinesegmentsthatboundit; theedgesaresometimescalled“sides,” thoughwewillmostlyavoidthatterm. The vertices (thesingularofwhichis“vertex”) ofapolygonarethepointswhereedgesmeet. Forexample, thepentagonshowninFigure 2.1.1 (i)hasfiveverticesandfiveedges.Itisnothardtofigureoutwhatallpolygonsare. NoticethatallthepolygonsshowninFig-
ure 2.1.1 haveedgesthatforma“circuit.” Thatis, ifyoustartatoneedge, youcanthengoto
26 2. Polygons
(i) (ii) (iii)
Figure2.1.1
(i) (ii) (iii)
Figure2.1.2
oneoftheedgesitmeets, andfromtheretothenextedge, andthenext, andsoonuntilyoucomeallthewayaroundbacktotheedgethatyoustartedwith. Infact, allpolygonsworkthisway, whichwesummarizeinthefollowingproposition, whichwegivewithoutdemonstration.
Proposition 2.1.1. Supposewearegivenapolygon.
1. Theverticesofthepolygoncanbelabeledas A1, A2, A3, . . . , An, wheretheedgesofthepolygonare A1A2, A2A3, A3A4, . . ., An−1An, An A1.
2. Thepolygonhasthesamenumberofedgesasvertices.
SeeFigure 2.1.3 foranillustrationoftheabovepropositioninthecasewhere n = 5. Thereadermightreasonablyask, inlightofProposition 2.1.1 (1), whywedidnotjustdefinepolygonstobethesortoffiguregivenbytheproposition. Theansweristhatwecouldhavedoneso, butwegavethedefinitionofpolygonsthatwedidinordertogiveadefinitionthatismoresimilartothedefinitionofpolyhedrainSection 3.1. Inanycase, wecannowproceedwithanunderstandingofpolygonsasgivenintheaboveproposition.
Sometypesofpolygonsareveryfamiliar, suchastriangles, whicharediscussedinmorede-tailinSection 2.2. Polygonswithfoursidesarereferredtoas quadrilaterals, ofwhichsomeofthemorefamiliartypesaresquares, rhombuses, rectangles, parallelograms, andtrapezoids. Asquare isaquadrilateralinwhichallfouredgesareequalandallfouranglesareequal; a rhom-bus isaquadrilateralinwhichallfouredgesareequal, butthefouranglesarenotnecessarilyequal; a rectangle isaquadrilateralinwhichallfouranglesareequal, butthefouredgesarenotnecessarilyequal; a parallelogram isaquadrilateralinwhichbothpairsofoppositeedgesareparallel; a trapezoid isaquadrilateralinwhichonepairofoppositeedgesisparallel. AnexampleofeachofthesetypesofquadrilateralisshowninFigure 2.1.4 (i)–(v); anexampleofa
2.1Introduction 27
A1
A5
A4
A3A2
Figure2.1.3
quadrilateralthatisnoneofthesetypesisshowninPart(vi)ofthefigure. Ingeneral, polygonsarenamedbythenumbersofedgesthattheyhave. A polygonwithfiveedgesiscalleda pen-tagon, apolygonwithsixedgesiscalleda hexagon, apolygonwitheightedgesiscalledanoctagon, etc. ThenamesofpolygonswithfiveormoreedgesarebasedonGreek, ratherthanLatin, roots. A polygonwith n edges, where n isanintegersuchthat n ≥ 3, iscalledan n-gon.
(i) (ii) (iii)
(iv) (v) (vi)
Figure2.1.4
Exercise 2.1.1. [UsedinSections 2.3, 2.4 and4.5] Supposethataparallelogramhasatleastone 90◦ angle. Showthatallfouranglesmustbe 90◦, andhencetheparallelogramisarectangle. (Useonlyresultsfromthissectionandprevioussections.)
28 2. Polygons
Exercise 2.1.2. Supposethatahexagonhasthepropertythateachpairofoppositeedgesisparallel.
(1) Showthatoppositeanglesinthishexagonareequal.
(2) Mustitbethecasethatoppositeedgeshaveequallengthsinthishexagon? Ifyes,showwhy. Ifnot, giveanexample.
2.2 Triangles
Thestudyoftrianglesgoesbacktotheancientworld. Trianglesplayed, andcontinuetoplay, acentralroleingeometry. Therearemanyimportantresultsabouttriangles, ofwhichwehavethespacetomentiononlyafew. Seemoststandardgeometrytextsformorefactsabouttriangles.A triangle isapolygonwiththreeedges(andhencethreevertices). Ifatrianglehasvertices
A, B and C, wedenotethetriangleby △ABC. Atanyvertexofatriangle, thetwoedgesofthetrianglethatcontainthevertexformananglethatisinsidethetriangle, calledthe interiorangle atthevertex. SeeFigure 2.2.1. Becauseinterioranglesarethemostcommonlyusedtypesofanglesintriangles, ifwesimplyreferto“theangle”atavertex, wewillalwaysmeantheinteriorangle(andsimilarlyforotherpolygons). Ifwemeansomeotherkindofangle, wewillalwayssaysoexplicitly. Observethateachedgeofatriangleislocatedoppositepreciselyoneofthevertices(andhenceoneoftheinteriorangles)ofthetriangle. Weusethenotation |AB|
todenotethelengthoftheedgeAB, andweusethenotation ∡A todenotethemeasureoftheangleat A.
A
B
C
Figure2.2.1
Thereareanumberofspecialtypesoftriangles. An equilateral trianglehasallthreeedgesequal. An isosceles trianglehastwoequaledges. A right trianglehasoneanglearightangle(thatis, a 90◦ angle). A righttrianglecanhaveonlyonerightangle. Itisalsocommontodistinguishbetween acute triangles (thathaveallangleslessthan 90◦), and obtuse triangles (thathaveoneanglegreaterthan 90◦).
2.2Triangles 29
Inadditiontotheinteriorangleateachvertexofatriangle, wecanalsoformanotherangleateachvertex, asfollows. Ateachvertex, extendbeyondthevertexoneof theedgesof thetrianglecontainingthevertex, andformthesupplementaryangle totheinteriorangle; thisangleiscalledthe exteriorangle atthevertex. InFigure 2.2.2 (i)weseeatriangle △ABC, withtheinteriorangleat A denoted α, andtheexteriorangleat A denoted δ. Notethattheinteriorandexterioranglesatavertexaddupto 180◦. Thecarefulreadermightwellaskwhatwouldhappenifwehadextendedtheotherpossibleedgeateachvertex, resultingindifferentexteriorangles. Thereareindeedtwopossibleexterioranglesateachvertex, buttheyareverticalangles.See Figure 2.2.2 (ii), where the twopossible exterior angles at A aredenoted δ and ϵ. ByProposition 1.2.2 (2)itfollowsthatthetwochoicesofexteriorangleareequal, andhence, itdoesnotmatterwhichexterioranglewechoose.
A
B
C
α
δA
B
C
δ
ε
(i) (ii)
Figure2.2.2
Wearenowreadytostateaveryimportantresultabouttheanglesinatriangle.
Proposition 2.2.1.
1. Thesumoftheinterioranglesofatriangleis 180◦.
2. Thesumoftheexterioranglesofatriangleis 360◦.
Demonstration. Supposewehaveatriangle△ABC. AsshowninFigure 2.2.3 (i), let α, β andγ betheinterioranglesofthetriangle, andlet x, y and z betheexterioranglesofthetriangle.
(1). AsshowninFigure 2.2.3 (ii), letn bethelinecontainingtheverticesB andC. ByPlayfair’sAxiom (Proposition 1.1.1), thereisalinethroughvertexA thatisparallelton. Callthisnewlinem. Labelangles δ and ϵ asshowninFigure 2.2.3 (ii). Weseethatβ and δ areinterioralternatingangles, andthat γ and ϵ areinterioralternatingangles. ByProposition 1.2.3 wededucethatβ = δ and γ = ϵ. Because δ, α and ϵ makeupastraightline, wehave δ+ α+ ϵ = 180◦. Itfollowsthat β+ α+ γ = 180◦, whichiswhatwewantedtoprove.
30 2. Polygons
A
BC
α
β γ
x
yz
A
B C
α
β γn
mδ ε
(i) (ii)
Figure2.2.3
(2). Weknowthat α + x = 180◦, that β + y = 180◦, andthat γ + z = 180◦. Hencex = 180◦ − α, and y = 180◦ − β, and z = 180◦ − γ. UsingPart (1)ofthisproposition, wethenseethat x+ y+ z = (180◦ −α) + (180◦ −β) + (180◦ − γ) = 540◦ − (α+β+ γ) =540◦ − 180◦ = 360◦.
TheproofoftheabovepropositionusesProposition 1.2.3 andProposition 1.1.1, bothofwhichrelyuponEuclid’sFifthPostulate. Indeed, Proposition 2.2.1 doesnotholdinsphericalorhyper-bolicgeometry, wheretheFifthPostulateisreplacedwithotheraxioms(seeSection 1.1 forabriefmentionofthesealternativegeometries).
Exercise 2.2.1. [UsedinSection 1.2] UseProposition 2.2.1 (1)todemonstrateProposi-tion 1.2.4.
Exercise 2.2.2. WeknowfromProposition 2.2.1 thattheanglesinatrianglearerelatedtoeachother; inparticular, wecouldnottakethreearbitraryangles, andexpecttheretobeatrianglewiththosethreeangles. Thisexerciseconcernsrelationshipsbetweenthelengthsoftheedgesofatriangle.
(1) Isthereatrianglewithedgesoflengths 2, 3 and 4? Explainyouranswer.
(2) Isthereatrianglewithedgesoflengths 2, 3 and 6? Explainyouranswer.
(3) Whatcanyousayabouttherelationshipbetweenthelengthsoftheedgesofatri-angle. Inparticular, trytocomeupwithcriteriaonthreenumbers a, b and c thatwouldguaranteethatthereexistsatrianglewithedgesoflength a, b and c. Explainyouranswer.
2.2Triangles 31
Whatdoesitmeanfortwotrianglestobe“thesame”? Clearly, if twotriangleshaveedgesofdifferentlengths, oranglesofdifferentmeasures, thenthetwotrianglesarenotthesame.Whatabouttwotriangleswithedgesthathavethesamelengths, andanglesthathavethesamemeasures? Forexample, inFigure 2.2.4 weseetwotrianglesthathavethesamelengthsofedgesandthesamemeasuresofangles(inthiscase 30◦, 60◦ and 90◦). Thesetwotrianglesarenottheexactsame, becausetheyarelocatedindifferentplaces, buttheyare“essentiallythesame.”SimilarlytowhatwesaidaboutanglesinSection 1.2, wesaythattwotrianglesare congruent,if, intuitively, onetrianglecouldbe“pickedupandplacedpreciselyontopoftheother.” Asbefore, wewillnotneedamore formaldefinitionofcongruencehere, andwill stick to theintuitivenotion; theconceptofisometry, whichisdiscussedindetailinChapter 4, providesonerigorouswayofdefiningcongruence, thoughwewillnothavethespacetoprovidethedetails.
A
A’B
CC’
B’
Figure2.2.4
Supposethattriangles△ABC and△A ′B ′C ′ arecongruent, wherevertex A correspondstovertex A ′, wherevertex B correspondstovertex B ′, andwherevertex C correspondstovertexC ′. SeeFigure 2.2.4. Then |AB| = |A ′ B ′|, and |AC| = |A ′ C ′|, and |BC| = |B ′ C ′|, and∡A = ∡A ′, and ∡B = ∡B ′, and ∡C = ∡C ′. Theconverseisalsotrue. Thatis, supposewearegiventwotriangles △ABC and △A ′B ′C ′, andweknowthat |AB| = |A ′ B ′|, that|AC| = |A ′ C ′|, that |BC| = |B ′ C ′|, that ∡A = ∡A ′, that ∡B = ∡B ′, andthat ∡C = ∡C ′.Itwillthenbethecasethatthetwotrianglesarecongruent.Actually, wecandobetter thantheabovestatement. Wejustassertedthat ifweknowsix
thingstobetrueabouttwotriangles(namelytheequalityofthelengthsofthethreeedges, andtheequalityofthemeasuresofthethreeangles), thenthetriangleswillbecongruent. Itturnsout,thoughthisisbynomeansobvious, thatcertainpartialknowledgeaboutthesesixequalitiessufficestoguaranteethattwotrianglesarecongruent. Thefollowingpropositionisonesuchresult.
Proposition 2.2.2 (Side-Side-SideTheorem). Supposethat△ABC and△A ′B ′C ′ aretriangles.Supposethat |AB| = |A ′ B ′|, that |AC| = |A ′ C ′|, andthat |BC| = |B ′C ′|. Then ∡A =∡A ′, and ∡B = ∡B ′, and ∡C = ∡C ′.
Wewillnotdemonstratetheaboveproposition, becauseitwouldtakeustoofarafield. TheSide-Side-SideTheoremsaysthatiftwotriangleshaveedgesthathavethesamelengths, then
32 2. Polygons
thetwotrianglesarecongruent. Anotherwayofthinkingaboutthistheoremisthatitsaysthattrianglesarerigid, inthefollowingsense. Supposeyoutakethreesticks, andjointhemtogetherintoatriangle. Evenifyouweretojointhestickswithhinges, thetrianglecouldnotbede-formed. Iftriangleswerenotrigid, thenknowingthelengthsoftheedgesofatrianglewouldnotuniquelydeterminetheangles, anditwouldthenbepossibletohavetwotriangleswhoseedgeshavethesamelengths, butwithdifferentangles, andthatwouldcontradicttheSide-Side-SideTheorem. Bycontrast, ifyouweretotakefoursticks, andjointhemtogetherwithhingesintoaquadrilateral, thenthefigurecouldbedeformed. SeeFigure 2.2.5. Thefactthattrianglesarerigid, butotherpolygonsarenot, issomethingthatiswellknowninreallife. Asaresult, weoftenseetriangularformsusedinconstructionoftressesandthelike.
Figure2.2.5
TheSide-Side-SideTheoremisnottheonlyresultthatguaranteesthattrianglesarecongruent.Twootherequallyusefulcongruencetheoremsarethefollowing.
Proposition 2.2.3 (Side-Angle-SideTheorem). Supposethat △ABC and △A ′B ′C ′ aretrian-gles. Suppose that |AB| = |A ′ B ′|, that ∡A = ∡A ′, and that |AC| = |A ′ C ′|. Then∡B = ∡B ′, and |BC| = |B ′ C ′|, and ∡C = ∡C ′.
Proposition 2.2.4 (Angle-Side-AngleTheorem). Suppose that △ABC and △A ′B ′C ′ are tri-angles. Suppose that ∡A = ∡A ′, that |AB| = |A ′ B ′|, and that ∡B = ∡B ′. Then|BC| = |B ′ C ′|, and ∡C = ∡C ′, and |AC| = |A ′ C ′|.
Wewillnotdemonstratetheabovetwocongruencetheorems.
Exercise 2.2.3. Findexamplestoshowthatthereisno“Angle-Side-SideTheorem.” Thatis,findtwotriangles△ABC and△A ′B ′C ′ suchthat ∡A = ∡A ′, that |AB| = |A ′ B ′|, andthat |BC| = |B ′ C ′|, andyetthetwotrianglesarenotcongruent. Exactmeasurementsofsuchtrianglesarenotneeded; asketchofthetriangles, togetherwithadescriptionofwhatyoumean, wouldsuffice.
Exercise 2.2.4. Istherean“Angle-Angle-SideTheorem”? Eitherdemonstratewhysuchatheoremistrue, orgiveanexampletoshowthatitisnottrue.
2.2Triangles 33
Congruenceoftriangles, andtheabovethreecongruencetheoremsinparticular, areextremelyusefulinprovingmanyresultsingeometry. Westartwiththefollowingverysimpleresults, butwewill usecongruence inproofsofother, more substantial, results lateron. The followingresults, whichmightseemsoobviousthattheydonotneedproof, infactneedproofjustasanyotherfactingeometrythatisnottakenasanaxiom.Ourfirstresultinvolvesparallelograms. Recall, statedinSection 2.1, thataparallelogramisa
quadrilateralinwhichbothpairsofoppositeedgesareparallel.
Proposition 2.2.5.
1. Oppositeedgesinaparallelogramhaveequallengths.
2. Oppositeanglesinaparallelogramareequal.
Demonstration. Supposethatwehaveparallelogram ABCD, asseeninFigure 2.2.6 (i). By
definition, weknowthatthelines←→AB and
←→DC areparallel, andthatthelines
←→AD and
←→BC are
parallel. Drawthelinesegment AC. SeeFigure 2.2.6 (ii). Wethenhavethetriangles △ABC
and △CDA. Letangles α, β, γ and δ beasshowninFigure 2.2.6 (ii). Observethat α and
γ are interioralternatingangles (because lines←→AB and
←→DC areparallel), and thereforeby
Proposition 1.2.3 weknowthat α = γ. Similarly, wededucethat β = δ, usingthefactthat←→AD and
←→BC areparallel.
(i) (ii)
A
D C
B A
D C
B
α
βγ
δ
Figure2.2.6
Wenowclaimthat △ABC and △CDA arecongruent, wherevertex A in △ABC corre-spondstovertex C in△CDA, wherevertex B in△ABC correspondstovertexD in△CDA,andwherevertex C in △ABC correspondstovertex A in △CDA. ThatthesetwotrianglesarecongruentfollowsfromtheAngle-Side-AngleTheorem(Proposition 2.2.4), andthefactthatα = γ, that β = δ, andthat AC isthesameinbothtriangles. Becausethetwotrianglesarecongruent, wededucethatcorrespondingedgeshavethesamelengths. Thatis, weconcludethat |AB| = |DC|, andthat |AD| = |BC|m, whichisPart (1)oftheproposition. Part (2)alsofollowsfromthecongruenceofthetwotriangles; detailsarelefttothereader.
34 2. Polygons
Exercise 2.2.5. Supposethatinaquadrilateral, bothpairsofoppositeedgeshaveequallengths. Showthatthequadrilateralisaparallelogram. (Becarefulwithnotconfusingathe-oremanditsconverse—thisfactcannotbeprovedbysimplyquotingProposition 2.2.5 (1).)
Exercise 2.2.6. Supposethatinaquadrilateral, apairofoppositeedgesareparallelandhaveequallengths. Showthatthequadrilateralisaparallelogram.
Exercise 2.2.7. Showthatthetwodiagonalsinaparallelogrambisecteachother.
AsaconsequenceofProposition 2.2.5 (1), wecandeducethefollowingresultaboutparallellines. RecallfromSection 1.2 thattwolinesaredefinedtobeparalleliftheydonotintersect.Ourintuitivepictureofparallellines, however, involvesmorethanjustthattwolinesdonotintersect, butalsothatthey“keepconstantdistancefromeachother.” Wecannowshowinthefollowingpropositionthatthisintuitivenotionisinfactcorrectforparallellinesintheplane.WewerenotabletogivethispropositioninSection 1.2, wherewefirstdiscussedparallellines,becauseweneedcongruenttrianglestoproveit. Whenyoureadthefollowingproposition, ithelpstoconsiderFigure 2.2.7.
m
n
A
C
B
D
Figure2.2.7
Proposition 2.2.6. Supposem and n areparallellines. LetA and B bepointsonm. Drawlinesthrough A and B respectivelythatareperpendicularto n; let C and D bethepointswheretheseperpendicularlinesintersect n. Then |AC| = |BD|.
Demonstration. First, observethatbecausethelines←→AC and
←→BD arebothperpendicularto
theline n, thentheyareparalleltoeachother(thisisProposition 1.2.5). Giventhat m and n
2.2Triangles 35
areparallel, itfollowsthatthequadrilateral ABCD isaparallelogram. WenowapplyProposi-tion 2.2.5 (1)tothisparallelogram, todeducethat |AC| = |BD|.
Ournextsimpleresultinvolvesisoscelestriangles.
Proposition 2.2.7 (PonsAsinorum). Supposethat △ABC isatriangle. If |AB| = |AC|, then∡B = ∡C.
Demonstration. Wearegiventhetriangle △ABC. Wenowclaimthat △ABC iscongruentwithitself, wherevertex B in △ABC correspondstovertex C in △ABC, wherevertex C in△ABC correspondstovertex B in △ABC, andwherevertex A in △ABC correspondstovertex A in △ABC. That these two trianglesarecongruent follows fromtheSide-Side-SideTheorem(Proposition 2.2.2), andthefactthat |AB| = |AC|, andthat |BC| = |BC|. Itfollowsthat △ABC hasequalangleswithitself, wheretheangleat B correspondstotheangleat C,andvice-versa(theangleatA correspondstoitself, thoughthatisnotofanyuse). Wethereforededucethattheanglesopposite AB and AC areequal, whichiswhatwearesupposedtoshow. (ItmayseemstrangethatweareapplyingtheSide-Side-SideTheoremtoatriangleanditself, ratherthantwodistincttriangles, butnothinginthestatementofthistheoremsaysthatthetwotrianglesunderconsiderationhavetobedistinct, thoughtheymostoftenare. However,eventhoughwearecomparingatrianglewithitself, wereallyareprovingsomething, becausewearehavingdifferentverticescorrespondwitheachotherinthecongruence.)
Thename“PonsAsinorum”means“Ass’Bridge” inLatin. Therearevariousexplanationsforthisname, somereferringtotheappearanceofthetriangleunderdiscussion, otherstothestateofmindofthosetryingtounderstandtheresult.AnimmediateconsequenceofPonsAsinorumisthefirstpartofthefollowingproposition,
whichagainisveryfamiliar; thesecondpartofthepropositionfollowsfromthefirstpart, to-getherwithProposition 2.2.1 (1).
Proposition 2.2.8.
1. Allthreeanglesinanequilateraltriangleareequal.
2. Eachangleinanequilateraltriangleis 60◦.
Exercise 2.2.8. Supposewearegivenatriangle △ABC. Showthatif ∡B = ∡C, then|AB| = |AC| (sothatthetriangleisisosceles).
Exercise 2.2.9. [Used inSection 1.3] Usecongruent triangles todemonstrateProposi-tion 1.3.1.
36 2. Polygons
Exercise 2.2.10. Supposewehaveacircle, andsupposethatA, B and C arepointsonthecirclesuchthatthelinesegmentAB isadiameterofthecircle. Formthetriangle△ABC.SeeFigure 2.2.8. Showthattheangleat C is 90◦.
A B
C
Figure2.2.8
Itisimportantnottogetoverlyconfidentaboutcongruencetheorems. Justknowingthattwotriangleshavethreethings(edgesorangles)equaldoesnotalwaysguaranteethatthetrianglesarecongruent. Forexample, simplyknowingthattwotriangleshavethesameanglesdefinitelydoesnotguaranteethatthetrianglesarecongruent. InFigure 2.2.9 weseetwotriangles, labeled△ABC and △A ′B ′C ′, thathavethesameangles, butdifferentlengthsoftheiredges.
A A’
B
B’
C
C’
Figure2.2.9
Wejustobserved thatknowingonly theangles ina triangledoesnotdeterminewhat thelengthsofitsedgesare. However, eventhoughtwotriangleswiththesameanglesmighthavedifferentsizes, asinFigure 2.2.9, theyhavethesame“shape.” Thefollowingpropositionmakesthisnotionofthesameshapemoreprecise, bylookingatratiosoflengthsofedges. First, wemakethefollowingdefinition. Twotriangles△ABC and△A ′B ′C ′ arecalled similar if ∡A =∡A ′, , and ∡B = ∡B ′, and ∡C = ∡C ′.Thefollowingproposition, whichmakesprecisethenotionthatsimilar triangles“havethe
sameshape,” iswhatmakestheconceptofsimilarityoftrianglessouseful.
2.2Triangles 37
Proposition 2.2.9. Supposethattriangles △ABC and △A ′B ′C ′ aresimilar. Then
|AB|
|A ′ B ′|=
|AC|
|A ′ C ′|=
|BC|
|B ′ C ′|;
equivalently, wehave
|AB|
|AC|=
|A ′ B ′|
|A ′ C ′|,
|AB|
|BC|=
|A ′ B ′|
|B ′ C ′|,
|AC|
|BC|=
|A ′C ′|
|B ′ C ′|.
ThedemonstrationofProposition 2.2.9 willbedelayeduntilSection 2.4, bywhichpointwewillhavediscussedtheareaoftriangles, whichweuseinthedemonstration.Asanexampleof theuseofProposition 2.2.9, supposethatwehavetwosimilar triangles
△ABC and △A ′B ′C ′. Suppose further thatwe are given that the edges of △ABC havelengths |AB| = 5, and |AC| = 6, and |BC| = 8; in △A ′B ′C ′ wearegivenonlythelength|A ′ C ′| = 20. Canwefindthelengthsoftheothertwoedgesof △A ′B ′C ′? Wecan, usingProposition 2.2.9. Bythatproposition, weknowthat
|AB|
|AC|=
|A ′ B ′|
|A ′ C ′|.
Hence, wededucethat5
6=
|A ′ B ′|
20.
Solvingthisequation, weseethat |A ′ B ′| =50
3. Thereadercanusethesametypeofcalculation
toseethat |B ′ C ′| = 15.
Exercise 2.2.11. A treecastsashadowthatis 20 ft.long. Atthesametimeofday, a 3 ft.stickcastsa 5 ft.shadow. Howtallisthetree?
Havingmentionedsimilartriangles, wecannotavoidmentioningoneofthemostimportantusesofsimilartriangles: trigonometry. Noteveryonewhousestrigonometryrecognizesthefun-damental roleplayedbysimilar triangles in trigonometry, andit is that role thatwewant todiscuss. Thestudyoftrigonometrytreatsthesixstandardtrigonometricfunctions, namelysine,cosine, tangent, secant, cosecant andcotangent.Letuslookatthesinefunction(theotherfivefunctionswouldworkcompletelysimilarly).
Todefinethesinefunction, weneedthefollowingterminology. Recallthatarighttriangle isatriangleinwhichoneoftheanglesisarightangle. Inarighttriangle, thetwoedgesthatformtherightanglearecalledthe sides ofthetriangle, andtheedgethatisoppositetherightangleiscalledthe hypotenuse ofthetriangle. (Thedistinctionbetweensidesandhypotenuseholds
38 2. Polygons
onlyinrighttriangles, notinothertriangles.) Inintroductorytreatmentsoftrigonometry, itistypicaltodefinethesinefunctionasfollows. Let α beananglebetween 0◦ and 90◦. Wewanttocomputethesineof α, whichisgoingtobeanumberdenoted sinα. Wecompute sinα byplacing α inarighttriangle, asinFigure 2.2.10, andthenletting sinα betheratioofthelengthofthesideopposite α tothelengthofthehypotenuse. Thatis, welet
sinα =|opposite|
|hypotenuse|.
α
opposite
hypotenuse
Figure2.2.10
Forexample, supposewewantedtocompute sin 45◦. Wecanplace 45◦ inanisoscelesrighttriangle. Onesuchrighttrianglehassidesoflength 1. UsingthePythagoreanTheorem(assumingthereaderisfamiliarwiththisresult, anditsstatementcanbefoundinSection 2.5 forthosewhoarenot), wecomputethatthehypotenuseofthetrianglehaslength
√2. SeeFigure 2.2.11.
Itfollowsthat sin 45◦ = 1√2.
1
12
45°
Figure2.2.11
Sofarsogood. Thereis, however, onepotentiallytroublingaspecttotheaboveapproachtodefiningsine. Givenanangle, wedefinedsineoftheanglebyplacingtheangleinarighttriangle,andtakingtheratioofthelengthofthesideoppositetheangletothelengthofthehypotenuse.Thereare, ofcourse, differentpossibletrianglesthatcouldbeused. Whatwouldhappeniftheratioofthelengthofthesideoppositetheangletothelengthofthehypotenuseinonerighttrianglecontainingtheanglewerenotequaltothecorrespondingratioinanotherrighttrianglecontainingtheangle? Ifthatweretohappen, thedefinitionofsineoftheanglewouldbeinvalid,becauseitoughtonlytodependupontheangleitself, andnothingelse(suchasachoiceof
2.3GeneralPolygons 39
righttriangle). Well, itturnsoutthatthereisnoprobleminthechoiceoftriangle, andhereisthereasonwhy. Supposewehadtworighttrianglescontaininganangleα. Bothtrianglesalsohavea90◦ angle. Giventhatthesumoftheanglesinanytriangleis 180◦ (byProposition 2.2.1 (1)), thenweknowthatthethirdangleineachofthetwotrianglesis 90◦−α. Therefore, thetwotriangleshavethesameangles; hencetheyaresimilartriangles. ItnowfollowsfromProposition 2.2.9thattheratioofthelengthofthesideopposite α tothelengthofthehypotenuseisthesameinbothtriangles, andthatmeansthatthereisnoprobleminourdefinitionofsine. Thesametypeofargumentholdsfortheothertrigonometricfunctions.Wenotethatalthoughthetrigonometricfunctionsaredefinedintermsoftriangles, theyhave
importantusesinmanyareabeyondthestudyoftriangles. Forexample, thetrigonometricfunc-tionsareusedtodescribeoscillatorymotion(forexample, springsandpendulums), andwavephenomena(forexample, soundandlight).
2.3 GeneralPolygons
HavingdiscussedtrianglesinSection 2.2, wenowturntopolygonswithmorethanthreeedges.Somepropertiesoftriangleshaveanalogsforallpolygons, andothersdonot. AswementionedinSection 2.2, onepropertyoftrianglesthatdoesnotholdforallpolygonsisrigidity. Anotherway to state the same fact is to say that theanalogof theSide-Side-SideTheorem (Proposi-tion 2.2.2)doesnotholdforpolygonsotherthantriangles. Forexample, asquarewithedgesoflength 1 andarhombuswithedgesoflength 1 havealledgesofequallength, andyettheydonothaveequalangles. SeeFigure 2.2.5.Ontheotherhand, therearefeaturesofpolygonswithmorethanthreesidesthatdonotappear
inthecaseoftriangles. Wenowturntoonesuchissue. ConsiderthepolygonsinFigure 2.3.1.There isa fundamentaldifferencebetween them: thepolygon inPart (i)has, intuitively, “noindentations,” whereasthepolygoninPart (ii)doeshaveanindentation. Itisnotveryconvenienttechnicallytotrytodefinethenotionof indentationsdirectly, sowecapturetheideaofnothaving indentationsas follows. A polygoniscalled convex ifany twopoints in thepolygonarejoinedbyalinesegmententirelycontainedinthepolygon. WeseeinFigure 2.3.2 howtwopointsinthepolygoninFigure 2.3.1 (ii)arejoinedbyalinesegmentthatisnotentirelycontainedinthepolygon, andthereforethepolygonisnotconvex; thefactthatsomeotherpairsofpointsinthepolygonarejoinedbylinesegmentsentirelycontainedinthepolygondoesnotmakethepolygonsatisfythedefinitionofconvexity. Bycontrast, thepolygoninFigure 2.3.1 (i)isconvex. Wenotethatalltrianglesareconvex, sothedistinctionbetweenconvexvs.non-convexdidnotariseinourdiscussionoftriangles, butitwillbeimportantinourdiscussionofpolygonsandpolyhedra.
Aswasthecasefortriangles, atanyvertexofapolygon, thetwoedgesofthepolygonthatcontainthevertexformananglethatisinsidethepolygon, calledthe interiorangle atthevertex.InFigure 2.3.3 weseeapolygon, withitsinterioranglesdenoted α1, α2, α3, α4 and α5. Wewillshortlydefineexterioranglesforpolygons.
40 2. Polygons
(i) (ii)
convex not convex
Figure2.3.1
Figure2.3.2
α1
α2α3
α4α5
Figure2.3.3
Oneofthethingsinterioranglesareusefulforisthattheygiveaneasywayofdeterminingwhetherornotapolygonisconvex. IfyoulookattheconvexpolygonpicturedinFigure 2.3.1 (i),youwillnoticethateachinterioranglesislessthan 180◦; bycontrast, inthenon-convexpolygonpicturedinFigure 2.3.1 (ii), oneoftheinterioranglesisgreaterthan 180◦. Moregenerally, wehavethefollowingresult, thedemonstrationofwhichweomit.
Proposition 2.3.1. A polygonisconvexifandonlyifallitsinterioranglesarelessthanorequalto 180◦.
Wenow turn to thequestionof angle sums in polygons. It ismuch simpler to dealwiththisquestion forconvexpolygons, sowestartwith thatcase. Weknowthat thesumof theinterioranglesofatriangleis 180◦, andthesumoftheexterioranglesofatriangleis 360◦ (seeProposition 2.2.1). Isthereananalogoftheseresultsforgeneralconvexpolygons? Toanswerthis
2.3GeneralPolygons 41
question, wefirstneedtodefineexterioranglesforconvexpolygons. Itturnsoutthatthesamedefinitionthatworkedfortrianglesworksforconvexpolygonsingeneral. Morespecifically, ateachvertexofaconvexpolygon, wecanextendbeyondthevertexoneof theedgesof thepolygoncontainingthevertex, andformthesupplementaryangle totheinteriorangle, whichwewillcallthe exteriorangle atthevertex. InFigure 2.3.4 weseeaconvexpolygon, withoneofitsverticeslabeled A, withtheinteriorangleat A denoted α, andtheexteriorangleat Adenoted β. Aswasthecasefortriangles, theinteriorandexterioranglesatavertexaddupto180◦. (Alsojustasfortriangles, itwouldmakenodifferencehadweextendedtheotherpossibleedgeateachvertex.)
αβ
A
Figure2.3.4
BEFORE YOU READ FURTHER:
TrytofigureoutforyourselfhowtogeneralizeProposition 2.2.1 foranarbitraryconvexn-gon. Thatis, givenaconvex n-gon, whatisthesumofitsinteriorangles, andwhatisthesumofitsexteriorangles? (Theanswersmightinvolvethenumber n.)
ToseehowwemightgeneralizeProposition 2.2.1 tootherconvexpolygons, letuslook, forexample, attheanglesumsfortheoctagonshowninFigure 2.3.5. Itiseasytoseethateachinteriorangleis 135◦, andeachexteriorangleis 45◦. Thesumoftheinterioranglesistherefore8 · 135◦ = 1080◦, andthesumoftheexterioranglesis 8 · 45◦ = 360◦. Interestingly, thesumoftheexterioranglesfortheoctagonisthesameasforatriangle, butthesumoftheinterioranglesfortheoctagonismorethanforatriangle. Asseeninthefollowingproposition, itturnsoutthatthesumoftheinterioranglesofaconvexpolygondependsuponthenumberofedgesofthepolygon, whereasthesumoftheexterioranglesdoesnotdependuponthenumberofedges. Thedemonstrationofthepropositionclarifieswhythisresultholds.
Proposition 2.3.2.
1. Thesumoftheinterioranglesofaconvex n-gonis (n− 2)180◦.
2. Thesumoftheexterioranglesofaconvex n-gonis 360◦.
Demonstration. Suppose that the convex n-gon P has interior angles α1, α2, α3, . . ., αn,andexteriorangles β1, β2, β3, . . ., βn. Forexample, forthepolygonoriginallyshowninFig-ure 2.3.3 (i), weseetheexteriorangleslabeledinFigure 2.3.6 (i).
42 2. Polygons
Figure2.3.5
α1
β1
α2
β2α3 β3
α4
β4α5β5
(i) (ii)
Figure2.3.6
(1). Itispossibletobreakupthepolygon P into n − 2 triangles, wheretheverticesofthetrianglesareallverticesoftheoriginalpolygon. SeeFigure 2.3.6 (ii)foronewayofdoingthis.(Thereismorethanonewaytobreakupthepolygonintotriangles, butitwillnotmatterwhichwayischoosen.) UsingProposition 2.2.1 (1)weknowthatthesumoftheanglesineachofthesetrianglesis 180◦. Becausethereare n− 2 triangles, thesumofalltheanglesinallthetrianglesis (n − 2)180◦. However, asseeninFigure 2.3.6 (ii), puttingalltheanglesinallthetrianglestogetheristhesameasputtingalltheanglesintheoriginalpolygontogether. Hencethesumofalltheanglesintheoriginalpolygonis (n− 2)180◦.
(2). Weknowthat αi + βi = 180◦ foreach i = 1, 2, 3, . . . , n. Therefore βi = 180◦ − αi
foreach i. WethenusePart (1)ofthispropositiontoseethat
β1 + β2 + β3 + · · ·+ βn = (180◦ − α1) + (180◦ − α2) + (180◦ − α3) + · · ·+ (180◦ − αn)
= (180◦ + 180◦ + 180◦ + · · ·+ 180◦︸ ︷︷ ︸n times
)− (α1 + α2 + α3 + · · ·+ αn)
= n180◦ − (n− 2)180◦ = 2 · 180◦ = 360◦.
2.3GeneralPolygons 43
Wenowturn toanglesumsinnon-convexpolygons. Consider thepolygonshowninFig-ure 2.3.7 (i). Theinterioranglesatvertices A, B, C, D, E and F are 135◦ each; theinterioranglesatvertices G and I are 90◦ each; andtheinteriorangleatvertex H is 270◦. Thesumoftheinterioranglesistherefore 6 · 135◦ + 2 · 90◦ + 270◦ = 1260◦. Noticethatthepoly-goninthefigurehas 9 edges. Ifweuse n = 9 intheformulagiveninProposition 2.3.2 (1),wewouldobtain (9 − 2)180◦ = 1260◦, whichispreciselythesumoftheinterioranglesinthepolygoninFigure 2.3.7 (i). Now, theformulainProposition 2.3.2 (1)wasonlyprovedforconvexpolygons, butperhapsitholdsfornon-convexpolygonsaswell. Theproofweusedforconvexpolygonsdoesnotworkfornon-convexpolygons. Forexample, thepolygonshowninFigure 2.3.7 (ii)cannotbedividedupintotrianglesinthesamewaythatwedidintheproofofProposition 2.3.2 (1). Itturnsoutthatanotherproofcanbeused. Thekeyisexteriorangles.
(i) (ii)
A
H I
BC
D
E
F G
Figure2.3.7
Howdowedealwithexterioranglesofnon-convexpolygons? Simplydefiningthemaswedidforconvexpolygonsdoesnotquitework. Recallthatwedefinedtheexteriorangleatthevertexofaconvexpolygonbyextendingbeyondthevertexoneoftheedgesofthepolygoncontainingthevertex, and forming the supplementaryangle to the interiorangle, whichwecalled theexteriorangleatthevertex. SeeFigure 2.3.4. However, ifwelookatthevertexlabeled H inFigure 2.3.7 (i), weseethatextendingeitheroftheedgescontainingthevertexwillgointotheinteriorofthepolygon, asituationthatdoesnotseemquiteright(atleastuponfirstglance).Onewayaroundthisproblemistorecallthattheinteriorandexterioranglesatavertexofaconvexpolygoncanberelatednotonlygeometrically, asinthedefinitionofexteriorangles,butintermsoftheirmeasure. Moreprecisely, themeasuresoftheinteriorandexterioranglesatavertexofaconvexpolygonaddupto 180◦. Therefore, themeasureofanexteriorangleisjust180◦ minusthemeasureoftheinteriorangle. (Westresstheword“measure”heretoemphasizethatitisdistinctfromtheactualangle, but, havingemphasizedithere, wewillreverttostandardterminologyandnotmentionitfurther.) Inthenon-convexcase, wecansimplytakethisrelation
44 2. Polygons
betweeninteriorandexterioranglesasadefinition. Thatis, we define the exteriorangle atthevertexofanypolygon(convexornot)as 180◦ minustheinteriorangleatthevertex.LetuslookatthepolygoninFigure 2.3.7 (i). TheexterioranglesatverticesA, B, C,D, E and
F are 180◦ − 135◦ = 45◦ each; theexterioranglesatvertices G and I are 180◦ − 90◦ = 90◦
each; andtheexteriorangleatvertex H is 180◦ − 270◦ = −90◦. Itmayseemstrangetoobtainanegativeexteriorangleatvertex H, butlookhownicelyitworksout. ThesumoftheexterioranglesofthepolygoninFigure 2.3.7 (i)is 6 · 45◦ +2 · 90◦ − 90◦ = 360◦. Thissumispreciselythesumofexterioranglesforconvexpolygons, asgiveninProposition 2.3.2 (2). Itturnsout, aswewillseeinProposition 2.3.3 below, thatthesameformulasforsumsofinteriorandexterioranglesthatworkforconvexpolygonsalsoworkfornon-convexpolygons—aslongaswedefineexterioranglesfornon-convexpolygonsaswedid, andacceptthefactthatexterioranglescouldbenegativeaswellaspositive.BeforeweturntoProposition 2.3.3, letuslookmorecloselyattheissueofnegativevs.positive
exteriorangles. Thekeyistorecallthat, asdiscussedinSection 1.2, wedistinguishedbetweenclockwisevs.counterclockwiseangles, withtheformerhavingpositivedegreemeasure, andthelatterhavingnegativedegreemeasure. LookatthepolygonshowninFigure 2.3.8 (i). Thinkoftheedgesasgoingaroundthepolygonintheclockwisedirection, asindicatedbythearrows.Wewillonceagainobtainexterioranglesbyextendingedges, aswedidforconvexpolygons,but this timewewillalwaysextend theedge thatcomesbeforeavertex (where“before” iswithrespecttogoingaroundthepolygonintheclockwisedirection). WestartbylookingatthevertexlabeledA. Theinteriorangleatthisvertexislessthan 180◦. InFigure 2.3.8 (ii)weextendtheedgecontaining A thatcomesbefore A. Wecanthenlookattheanglefromtheextendededgetotheedgethatcomesafter A. Thisangleisclockwise, andisthereforeapositiveangle.Thisangleispreciselyequalto 180◦ minustheinteriorangleat A. Next, welookatthevertexlabeled B. Theinteriorangleatthisvertexisgreaterthan 180◦. InFigure 2.3.8 (ii)weextendtheedgecontaining B thatcomesbefore B. Wecanthenlookattheanglefromthisextendededgetotheedgethatcomesafter B. Thisangleiscounterclockwise, andisthereforeanegativeangle. Thisangleisagainpreciselyequalto 180◦ minustheinteriorangleat B. Inbothcases,weseethatitispossibletogiveageometricmeaningtotheexteriorangle, aslongaswetakeintoaccountthedifferencebetweenclockwisevs.counterclockwiseangles.
Wearenowreadyforthefollowingproposition, whichgeneralizesProposition 2.3.2.
Proposition 2.3.3.
1. Thesumoftheinterioranglesofan n-gonis (n− 2)180◦.
2. Thesumoftheexterioranglesofan n-gonis 360◦.
Demonstration. Supposewehavean n-gon P asshowninFigure 2.3.9 (i), withvertices A1,A2, A3, . . ., An, andwithinteriorangles α1, α2, α3, . . ., αn. Welet β1, β2, β3, . . ., βn betheexteriorangles. WewillfirstshowthatPart (2)holds, andthenshowthatPart (1)holdsbyusingPart (2).
2.3GeneralPolygons 45
(i) (ii)
BB
AA
Figure2.3.8
(i) (ii)
A7
A4
A5
A6
A3A2
A1
A7
A4
A5
A6
A3
A2
A1
α1
α2
α3
α4
α5
α6
α7
β1
β2
β3
β4β5
β6
β7
Figure2.3.9
(2). AsshowninFigure 2.3.9 (ii), weextendeveryedgeof P intheclockwisedirection, andlabeltheexteriorangles. Observethattheanglefromtheextendededgeat A1 totheextendededgeat A2 isequaltotheexteriorangleat A1, whichisdenoted β1. Similarly, theanglefromtheextendededgeatA2 totheextendededgeatA3 isequaltotheexteriorangleatA2, whichisdenoted β2. Hence, theanglefromtheextendededgeatA1 totheextendededgeatA3 isequalto β1 + β2. Bythesameargument, theanglefromtheextendededgeat A1 totheextendededgeat A4 isequalto β1 +β2 +β3. Ifwekeepgoingallthewayaroundthepolygon, weseethattheanglefromtheextendededgeatA1 toitself, goingallthewayaround 360◦, isequaltoβ1 +β2 +β3 + · · ·+βn. Hence β1 +β2 +β3 + · · ·+βn = 360◦. (Notethatthisargumentworksonlybecausetheedgesofthepolygondonotintersecteachother; iftheywereallowedtointersecteachother, theanglefrom A1 toitselfaftergoingallthewayaroundthepolygonmightbetwice 360◦, orthreetimes 360◦, etc.)
(1). Thispartof thedemonstration isabackwardsversionof thedemonstrationofPropo-sition 2.3.2 (2). Weknow, bydefinition, that βi = 180◦ − αi foreach i = 1, 2, 3, . . . , n.
46 2. Polygons
Therefore αi = 180◦ − βi forall i. WethenusePart (2)ofthispropositiontoseethat
α1 + α2 + α3 + · · ·+ αn = (180◦ − β1) + (180◦ − β2) + (180◦ − β3) + · · ·+ (180◦ − βn)
= (180◦ + 180◦ + 180◦ + · · ·+ 180◦︸ ︷︷ ︸n times
)− (β1 + β2 + β3 + · · ·+ βn)
= n180◦ − 360◦ = n180◦ − 2 · 180◦ = (n− 2)180◦.
Exercise 2.3.1. Showthataquadrilateralhasatmostoneinterioranglethatisgreaterthan180◦.
Exercise 2.3.2. Recall, statedinSection 2.1, thatarectangle isdefinedtobeaquadrilateralinwhichallfouranglesareequal.
(1) Showthatallfouranglesinarectangleare 90◦.
(2) Showthateveryrectangleisaparallelogram.
(3) Showthatoppositeedgesinarectanglehaveequallengths.
Exercise 2.3.3. Supposethataquadrilateralhastwopairsofequaladjacentangles. Showthatthequadrilateralisatrapezoid.
Exercise 2.3.4. Showthatanyparallelogramisconvex.
2.3GeneralPolygons 47
Exercise 2.3.5. [UsedinSection 4.5] Supposethataquadrilateralhastwooppositeedgesthathaveequal lengths, and that these twoedgesarebothperpendicular tooneof theedgesthatisinbetweenthem. Thegoalofthisexerciseistoshowthatthequadrilateralisarectangle. Weoutlinetwodemonstrations, oneusingcongruenttriangles(insteps(1)–(3)below), andtheotherusingsimilartriangles(insteps(4)–(8)below); thereaderisaskedtofillinthedetailsofeachstep.Supposethat ABCD isaquadrilateral. Supposethat |AD| = |BC|, andthatboth AD
and BC areperpendicularto AB. SeeinFigure 2.3.10. Wenowproceedasfollows.
(1) Showthattriangles△ABC and△ABD arecongruent. Deducethat |AC| = |BD|.
(2) Showthattriangles △ACD and △BCD arecongruent. Deducethat ∡D = ∡C.(3) ByProposition 2.3.3 (1)we know that the sumof the angle in ABCD is 360◦.
Deducethat ∡C = 90◦ and ∡D = 90◦. Itfollowsthatallfouranglesinthequadri-lateralareequal, andhencethequadrilateralisarectangle.
(4) Wenowgiveanotherdemonstrationofthefactthatthequadrilateralisarectangle.Asafirststep, wewanttoshowthatthequadrilateralisaparallelogram. ItfollowsfromProposition 1.2.5 that AD and BC areparallel. Nowsupposethat AB andCD arenotparallel. Thenextendthemuntiltheymeet, sayinpoint P. Wewillarriveatalogicalcontradiction.
(5) Showthatthetriangles △ADP and △BCP aresimilar.
(6) Deducethat|AD|
|BC|=
|AP|
|BP|.
(7) Use the fact that |AP| > |BP| todeduce that |AD| > |BC|. Explainwhy thisisalogicalimpossibility, giventhehypothesesofthisexercise. Deducethat AB isparallelto CD, andhencethequadrilateralisaparallelogram.
(8) NowuseExercise 2.1.1 toshowthatthequadrilateralisarectangle.
Polygonscanbeveryirregularlooking, forexample, thepolygonshowninFigure 2.3.7 (ii).Wenowwanttoturnourattentiontopolygonsinwhich, asmuchaspossible, onepointlookslikeanyotherpoint. A polygonisa regularpolygon ifthefollowingtwoconditionshold: (1)alledgeshavethesamelength; (2)allinterioranglesareequal. Forexample, asquareandanequilateraltrianglearebothregularpolygons. Thatthefirstpartoftheabovedefinitiondoesnotalonesufficecanbeseenbyconsideringarhombus, inwhichalledgeshavethesamelength,butnotallinterioranglesareequal; arhombusisnotsomethingwewishtocallregular.A squareisaregularpolygonthathasfouredges. Istherearegular n-gonforeachpossible
n? Theanswerisyes. InFigure 2.3.11 weseearegulartriangle(alsoknownasanequilateral
48 2. Polygons
A B
D C
Figure2.3.10
triangle), aregularquadrilateral(alsoknownasasquare), aregularpentagon, aregularhexagon,aregularheptagon(whichhassevenedges)andaregularoctagon. Ingeneral, foragivenpositiveintegern (whichisgreaterthanorequaltothree), wecanmakearegularn-gonasfollows. First,drawacircle. (Theradiusofthecircledoesnotmatter; differentchoicesofradiuswillyielddifferentlysizedpolygons, butthesizeofpolygonsdoesnotmattertoushere.) Next, calculate360◦/n. Thendraw n raysfromthecenterofthecircle, wheretheraysformanglesof 360◦/nbetweenthem. SeeFigure 2.3.12 (i). The n raysintersectthecirclein n points. Wetakethesepointstobetheverticesofapolygon, whichwethenconstructbyjoiningtheseverticeswithedges. Thepolygonthusconstructedisaregular n-gon. SeeFigure 2.3.12 (ii). Wementionthatthisconstructionofaregular n-gonisnotaclassicalstraightedgeandcompassconstructionofthesortusedinancientGreekmathematics, butthereisnoneedforustorestrictourselvestosuchconstructions. (It turnsout that it isnotpossible toconstructall regularpolygons intheancientGreekmanner.) Observe that for the regularpolygonsconstructedby theabovemethod, theverticesalllieonacircle. Itturnsoutthatanyregularpolygon, nomatterhowitwasconstructed, hasallitsverticesonacircle.
Asweseeinthefollowingproposition, theanglesinaregularpolygonaredeterminedbythenumberofedgesinthepolygon. ThisresultwillbeusefultousinSection 3.2.
Proposition 2.3.4.
1. Eachinteriorangleofaregular n-gonis(n− 2)180◦
n.
2. Eachexteriorangleofaregular n-gonis360◦
n.
Demonstration. BothpartsofthispropositionfolloweasilyfromProposition 2.3.3. Allinterioranglesinaregularpolygonareequal, andsoeachinteriorangleequalsthesumoftheinteriorangles, whichis (n−2)180◦, dividedbyn. A similarconsiderationholdsforexteriorangles.
Thefollowingpropositionnowfollowseasilyfromwhatwehavejustseen.
2.4Area 49
Figure2.3.11
(i) (ii)
Figure2.3.12
Proposition 2.3.5. Everyregularpolygonisconvex.
Demonstration. It followsfromProposition 2.3.4 (1) that theinterioranglesineveryregularpolygonarelessthan 180◦. WenowuseProposition 2.3.1 todeducethateveryregularpolygonisconvex.
InTable 2.3.1 wegivethevaluesoftheinterioranglesofsomeoftheregularpolygons, ob-tainedbypluggingintheappropriatevaluesof n intotheformulagiveninProposition 2.3.4 (1).
2.4 Area
Westartthissectionwithadiscussionoftheareasofpolygons. Ourdiscussionofareaultimatelyreliesuponthreebasicideasconcerningarea, whichweassumewithoutproof: (1)arectanglethathaswidth x andheight y hasarea xy; (2)congruentshapeshaveequalareas; and(3)ifa
50 2. Polygons
RegularPolygon NumberofEdges(n) InteriorAngleequilateraltriangle 3 60◦
square 4 90◦
regularpentagon 5 108◦
regularhexagon 6 120◦
regularheptagon 7 128.57◦
regularoctagon 8 135◦
Table2.3.1
shapeisbrokenupintoafinitenumberofpiecesthattogetherexactlyfilluptheoriginalshape,thentheareaoftheoriginalshapeisthesumoftheareasofthepieces.Inthefollowingproposition, wegiveformulasfortheareasoftriangles, trapezoidsandparal-
lelograms(thelasttwoofwhich, ifthereaderneedsreminding, aredefinedinSection 2.1). Inordertostatetheseformulas, weneedtodefinethenotionofanaltitudeineachofthesetypesoffigures. Inallthesecase, thecommonideaisasfollows. Supposewehavealine m andapoint P thatisnoton m. The altitude from P to m isthelinesegmentthatisperpendiculartom, andhasoneendpointin m, andtheotherendpointat P. SeeFigure 2.4.1.
m
P
altitude
Figure2.4.1
Supposewehaveatriangle, andwechooseanedgeofthetriangle. The altitude perpendiculartothatedgeisthealtitudefromthevertexoppositetheedgetothelinecontainingtheedge.Inthetriangle △ABC showninFigure 2.4.2 (i), weseethealtitude, labeled h, perpendiculartoedge AC. InFigure 2.4.2 (ii)weseethatthealtitudeperpendiculartotheedgeofatriangleneednotbeinsidethetriangle. A trianglehasthreedistinctaltitudes, oneperpendiculartoeachedge, andthesethreealtitudeswillingeneralnothavethesamelengths. Aninterestingfactisthatinanytriangle(evenanobtuseone), thelinescontainingthethreealtitudesmeetinasinglepoint, calledthe orthocenter ofthetriangle; see[WW98, Section4.6]formoredetails.
Next, supposewehavea trapezoid. The altitude of the trapezoid isconstructedby takinganypointononeofthetwoparalleledges, andconstructingthealtitudefromthatpointtotheotherparalleledge. SeeFigure 2.4.3 (i). A trapezoidhasmanyaltitudes, but it follows fromProposition 2.2.6 thattheyallhavethesamelength. Inaparallelogram, wecanconstructthe
2.4Area 51
A
B
C A
B
h h
C
(i) (ii)
Figure2.4.2
altitudesperpendiculartoeachpairofedges, justasweconstructedthealtitudeofatrapezoid(whichhasonlyonepairofparallel edges). A parallelogramhas twodistinct altitudes, oneperpendiculartoeachpairofparalleledges, andthesetwoaltitudeswillingeneralnothavethesamelengths. SeeFigure 2.4.3 (ii)foroneofthealtitudesofaparallelogram.
h h
(i) (ii)
Figure2.4.3
Wecannowstateourformulasconcerningareas.
Proposition 2.4.1.
1. Supposethataparallelogramhasanedgeoflength b, andthealtitudeperpendiculartothatedgeoflength h. Thentheareaoftheparallelogramis bh.
2. Supposethatatrapezoidhasparalleledgesoflengthm1 andm2, andaltitudeoflength
h. Thentheareaofthetrapezoidism1 +m2
2h.
3. Supposethatatrianglehasanedgeoflength b, andaltitudeperpendiculartothatedge
oflength h. Thentheareaofthetriangleis1
2bh.
Demonstration.
(1). Thebasicideaistocutupourparallelogram, andrearrangeitintoarectangle. Morepre-cisely, supposethatwehaveaparallelogramwithverticesA, B,C andD, suchthat |CD| = b,andthatthealtitudeperpendicularto CD haslength h. SeeFigure 2.4.4 (i). Drawthelines
52 2. Polygons
through A and B respectively that areperpendicular to the linecontaining CD; let E andF respectivelybe thepointsof intersectionof these lineswith the linecontaining CD. Fig-ure 2.4.4 (ii).
h
AB
C Db
h h
AB
C F D E
(i) (ii)
Figure2.4.4
Observethat BF and AE areparallel, becauseofProposition 1.2.5. Also, weknowthat AB
and F E areparallelbyhypothesis. Hencethat ABFE isaparallelogram. Moreover, becauseBF and AE arebothperpendicularto F E, weseebyExercise 2.1.1 that ABFE isarectangle.Next, byusingProposition 2.2.5 (1)weknowthat |AB| = |CD|, andthat |AD| = |BC|. It
followsfromProposition 2.2.6 that |BF| = |AE|. Moreover, because BC andAD areparallel,and BF and AE areparallelasnotedabove, itcanbeseenthat ∡CBF = ∡DAE (adetaileddemonstrationofthisequalityusesProposition 1.2.3; weleavethesedetailstothereader). ItthenfollowsfromtheSide-Angle-SideTheorem(Proposition 2.2.3)thattriangles△CBF and△DAE
arecongruent. Hence, thesetwotriangleshavethesamearea.WededucethattheparallelogramABCD hasthesameareaastherectangleABFE. WecanthinkoftheedgeAB asthebaseoftherectangleABFE, andAE asthealtitudeofthisrectangle. Usingtheobservationsatthestartofthisparagraph, weseethat |AB| = b, anditfollowsfromProposition 1.2.5 that |AE| = h.Hencetheareaoftherectangle ABFE is bh. Itfollowsthattheparallelogram ABCD hasareabh.
(2). ThereaderislefttoprovidethisdemonstrationinExercise 2.4.1.
(3). Supposethatwehaveatriangle△ABC, that |AC| = b, andthealtitudeperpendiculartoAC haslength h. SeeFigure 2.4.5 (i). WenowformaparallelogrambydrawingthelinethroughB thatisparalleltoAC, anddrawingthelinethroughC thatisparalleltoAB (Playfair’sAxiom(Proposition 1.1.1)guaranteesthatwecandrawtheselines). Let D bethepointofintersectionofthetwonewlines, sothat ABCD isaparallelogram. Figure 2.4.5 (ii).
ItisevidentthattheparallelogramABCD hasedgeAC asitsbase, andhasaltitudeoflengthh. ByPart (1)ofthisproposition, weknowthattheareaoftheparallelogramis bh. ByusingProposition 2.2.5 (1)weknowthat |AB| = |CD|, andthat |AC| = |BD|. It then followsfrom the Side-Side-SideTheorem (Proposition 2.2.2) that triangles △ABC and △BCD are
2.4Area 53
h
A
B
Cb
h
A
B D
Cb
(i) (ii)
Figure2.4.5
congruent. Hence, thesetwotriangleshavethesamearea, andsoeachhashalftheareaofthe
parallelogram ABCD. Wededucethattheareaoftriangle △ABC is1
2bh.
Observethatifwethinkofatriangleasadegeneratetrapezoidinwhichoneoftheparalleledgeshaslengthzero, thentheformulafortheareaofatriangle(Part (3)oftheabovepropo-sition)followsimmediatelyfromtheformulafortheareaofatrapezoid(Part (2)oftheaboveproposition).AnimmediateconsequenceofProposition 2.4.1 isthefollowingresult. SeeFigure 2.4.6 for
anillustrationofthisfact.
Proposition 2.4.2.
1. Supposethattwoparallelogramshaveanedgeofthesamelength, andhavethealtitudesperpendiculartothoseedgeshavingthesamelength. Thentheparallelogramshavethesamearea.
2. Suppose that two triangleshaveanedgeof the same length, andhave thealtitudesperpendiculartothoseedgesofthesamelength. Thenthetriangleshavethesamearea.
h
b b
h
Figure2.4.6
Exercise 2.4.1. [UsedinThisSection] DemonstrateProposition 2.4.2 (2).
54 2. Polygons
Exercise 2.4.2. Supposethattriangles △ABC and △A ′B ′C ′ aresimilar. Let Q and Q ′
respectivelydenotetheareasof △ABC and △A ′B ′C ′. Showthat
Q
Q ′ =|AB|
2
|A ′ B ′|2 .
Wenowhaveformulasfortheareasofrectangles, trianglesparallelogramsandtrapezoids.Whataboutmorecomplicatedpolygons? Itisnotpossibletohaveasimpleformulatocovereachpossibletypeofpolygon. However, itisalwayspossibletofindtheareaofanypolygonbychoppingitupintosimpleshapes(forexample, rectanglesandtriangles), figuringouttheareaofeachofthepieces, andthenaddingtheareasofthepiecesup. SeeFigure 2.4.7 foranexampleofacomplicatedpolygonchoppedupintorectanglesandtriangles.
Figure2.4.7
Exercise 2.4.3. FindtheareasofthepolygonsshowninFigure 2.4.8.
Weremindthereaderthattheperimeter ofapolygonisthesumofthelengthsofitsedges.
Exercise 2.4.4. Showthatiftworectangleshavethesameareaandthesameperimeter,thentheyhavethesamedimensions. (Thisoneusessomealgebra.)
2.4Area 55
(i) (ii) (iii)
23
2
2
1
1
2 2
2
2 2
Figure2.4.8
Exercise 2.4.5.
(1) Findtwopolygonsallofwhoseedgeshaveintegerlengths, whichhavethesameareasandsameperimeters, butthatarenotcongruent.
(2) IfthetwoconvexpolygonsyoufoundinPart (1)werenotconvex, findtwoconvexpolygonsallofwhoseedgeshaveintegerlengths, whichhavethesameareasandsameperimeters, butthatarenotcongruent.
Exercise 2.4.6. A kite isaquadrilateralthathastwopairsofadjacentedgeswithequallengths. Wecallthediagonalthathasonepairofadjacentedgeswithequallengthsononesideandtheotherpairontheothersidethe crossdiagonal; theotherdiagonaliscalledthe maindiagonal. (Thenomenclatureinthisexerciseisnotstandardized.) Forthesakeofthisexercise, assumethatallkitesunderdiscussionareconvex(thatobviatestheneedconsideringdifferentsubcases), thoughtherearealsonon-convexkites.
(1) Showthatthemaindiagonalinakitebreaksthekiteintotwocongruenttriangles.
(2) Showthatthemaindiagonalinakitebisectseachoftheanglesatitsendpoints.
(3) Showthatthemaindiagonalinakitebisectsthecrossdiagonal.
(4) Showthatthecrossdiagonalinakiteisperpendiculartothemaindiagonal.
(5) Showthattheareaofakiteistheonehalftheproductofthelengthsofthediagonals.
Wecanapplytheconceptofareatoregularpolygons, startingwiththefollowingexercise.
56 2. Polygons
Exercise 2.4.7. Findtheareaforeachofthefollowingregularpolygons.
(1) Anequilateraltrianglewithedgesoflength 1.
(2) A regularhexagonwithedgesoflength 1.
(3) A regularoctagonwithedgesoflength 1. (Hint: Donottrytosolvethisproblembydividingtheoctagonintoeightcongruenttriangleswithacommonvertexinthecenteroftheoctagon—thatmethodisquitetricky.)
Oneverynicepropertyofregularpolygonsisthattheyminimizeperimeteramongallpolygonswithagivennumberofedgesandagivenarea. Moreprecisely, supposeweareinterestedinpolygonswith n edges, where n issomepositiveintegergreaterthanorequaltothree. Supposefurtherthatwearegivenanarea A, where A issomepositivenumber. Amongallthe n-gonsthathavearea A, whichonehasthesmallestperimeter? Theansweristheregular n-gonwitharea A. Thoughthisresultseemsreasonableintuitively, thedemonstrationisbeyondthescopeofthistext. Alternatively, ifwearegivenaperimeter P, wecanaskwhich n-gonwithperimeterP hasthelargestarea. Again, theansweristheregular n-gonwithperimeter P.
Exercise 2.4.8.
(1) Supposewehaveacircle, andsupposethatA and B arepointsonthecirclethatarenotdiametricallyoppositeeachother. SupposefurtherthatC isanotherpointonthecirclethatisbetween A and B. Considertheareaofthetriangle △ABC. ExplainwhyofallpossiblechoicesofpointsC, theonewhere△ABC hasthemaximalareaiswhere C ismidwaybetween A and B.
(2) Giveaninformalexplanationthat, amongallpolygonswith n vertices, andwithitsverticesallonagivencircle, theregular n-gonwillhavethemaximalarea.
Wecanalsouseregularpolygonstohelpgiveusanintuitive(thoughnotrigorous)explanationregardingsomeaspectsofthenumber π. Ofcourse, thenumber π relatestocircles. However,wecanuseregularpolygonstohelpusunderstandcirclesbecause, aswementionedabove, ifwestartwithacircle, wecanformaregular n-gonwithverticesonthecircle, foranypositiveinteger n greaterthanorequaltothree. Ifwechoosethenumber n tobeverylarge, thenthen-gonapproximatesthecircleveryclosely. Thelargerthe n, thebettertheapproximation. Nopolygoneverequalsthecircle, thoughforverylarge n itmightbeimpossibletodistinguisharegular n-gonfromacirclewiththenakedeye.Todiscussthenumber π, recallthatitisusuallydefinedastheratioofthecircumferenceto
thediameterofacircle. Thatiscorrect, butthereisanaspecttothisdefinitionthatisusually
2.4Area 57
glossedover—howdoweknowthatinallcircles, thereisthesameratioofthecircumferencetothediameter. Ifthisratioweredifferentindifferentcircles, thenthedefinitionof π wouldnotmakeanysense. Infact, thisratioisthesameinallcircles, ascanbeproved. A rigorousproofcanbefoundusingcalculus, butwecanusetrianglestogiveusanintuitiveideawhythisratioisthesameinallcircles. Beforewestart, wenotethatinsteadoftheratioofthecircumferencetothediameter, wecanjustaswelllookattheratioofthecircumferencetotheradiusofthecircle,whichshouldyield 2π ratherthan π. Ifwecanshowthatthetheratioofthecircumferencetotheradiusisthesameforanytwodifferentcircles, thatwillsuffice.Supposewehavearegiven twocirclesofdifferent sizes. Ineachcircle, constructa regu-
lar n-gonwithitsverticesonthecircle. Onesuchcircleandregular n-gonisshowninFig-ure 2.4.9 (i); inthefigurewehavea 10-gon, thoughitwouldbepossibletohaveanynumberofedges. Ineachofthepolygons, drawlinesegmentsfromthecenterofthecircletotheverticesofthepolygon, thusbreakingthepolygonupinto n isoscelestriangles. InFigure 2.4.9 (ii)weseethesmallerofourtwocircles; thelengthofeachedgeofthepolygonis a, andtheradiusofthecircleis r. InFigure 2.4.9 (iii)weseethelargerofourtwocircles; thelengthofeachedgeofthepolygonis c, andtheradiusofthecircleis s.
(i) (ii) (iii)
ac
r s
Figure2.4.9
Now, comparethetwopolygonsinthedifferentsizedcircles. Theyarebothregular n-gons,eventhoughtheyareofdifferentsizes. Hence, byProposition 2.3.4 (1), bothofthese n-gonshavethesameinteriorangles. Thebaseanglesintheisoscelestrianglesintowhichthepolygonsarebrokenuparethereforealsothesameinbothpolygons. Itfollowsthattheisoscelestrianglesinthesmallerpolygonareallsimilartotheisoscelestrianglesinthelargerpolygon. Inparticular,byusingProposition 2.2.9, weseethat
a
r=
c
s.
Multiplyingeachsideby n yieldsna
r=
nc
s.
58 2. Polygons
Weobservethattheperimeterofthesmallerpolygonis na, andtheperimeterofthelargerpolygonis nc. Hence
Perimeterofthesmallerpolygon
Radiusofthesmallercircle=
Perimeterofthelargerpolygon
Radiusofthelargercircle.
If n isverylarge, thentheperimeterofthepolygonsisveryclosetothecircumferenceofthecircle.Byletting n gotoinfinity, wededucethat
Circumferenceofthesmallercircle
Radiusofthesmallercircle=
Circumferenceofthelargercircle
Radiusofthelargercircle,
whichiswhatweweretryingtoshow. Thisargumentisnotcompletelyrigorousasstated, butitdoesgiveaplausibleargument.Wecantaketheabovesortofreasoningonestepfurther. Supposethatacirclehasradius r.
Aswejustdiscussed, wehave C = 2πr, where C isthecircumferenceofthecircle. Thereis, ofcourse, anotherequallyusefulformulainvolving π, namelyA = πr2, whereA istheareaofthecircle. Wecannottakethisareaformulaasthedefinitionof π, becausewehavealreadydefinedπ in termsof thecircumference. Rather, wecangiveanintuitiveargument for this formula,giventhatwehavealreadyseenwhy C = 2πr oughttobetrue. (Again, ourargumentwillnotbecompletelyrigorous, thougharigorousargumentcanbefoundusingcalculus.)Supposewehaveacircleofradius r. Again, formaregular n-gonwithverticesonthecircle.
Thistime, makesurethat n isanevennumber. Weformisoscelestrianglesasbefore; thistime,wecoloreveryothertriangle. SeeFigure 2.4.10 (i)forthecasewhere n isten. Observethattheareaofthe n-gonisveryclosetotheareaofthecircle; if n isverylarge, thentheapproximationisquiteclose.
(i) (ii)
Figure2.4.10
Now, wetakethe n-gon, andwerearrangeitstrianglesasshowninFigure 2.4.10 (ii). Theshapethattheserearrangedtrianglesformisaparallelogram. However, if n isverylarge, thetriangleswillbeverythin, andtheparallelogramwillbealmostarectangle. Whatisthelength
2.4Area 59
ofthealtitudeoftheparallelogram? If n isverylarge, thenthetriangleswillbeextremelythin,andthealtitudeofeachtrianglewillbeapproximatelythesameasthelengthofitssides, whichisjusttheradiusofthecircle, namely r. Whatisthelengthofthebaseoftheparallelogram? Itishalfthecircumferenceofthecircle, namely πr. Hence, usingProposition 2.4.1 (1), weseethattheareaoftheparallelogramisapproximately πr · r = πr2. If n getslargerandlarger, theapproximationgetsbetterandbetter. Note, however, thattheareaoftheparallelogramisveryclosetotheareaofthecircle, becauseitisthesameastheareaofthepolygoninsidethecircle.Onceagain, as n goestoinfinity, wededucethatareaofthecircleisactuallyequaltotheareaoftheparallelogram, andhencetheareaofthecircleis πr2. Onceagain, thisargumentisnotcompletelyrigorousasstated, butgivesanintuitivepictureofwhatishappening.Weend thissectionbyusing theconceptofarea todemonstratea result thaton the face
ofithaslittletodowitharea, namelyProposition 2.2.9, whichconcernssimilartriangles. Todemonstratethisproposition, weneedtwopreliminaryresults, towhichwenowturn; itisinthedemonstrationofthefirstoftheseresultsthatweencountertheuseofarea. Ourapproachhere follows [Mey99, Section2.4]. To read the statementofourfirstpreliminary result, seeFigure 2.4.11.
A
B
D E
C
Figure2.4.11
Proposition 2.4.3. Supposethatatriangle △ABC hasapoint D onedge AB, andapoint Eonedge AC, placedsothat DE isparallelto BC. Then
|AB|
|AD|=
|AC|
|AE|.
Demonstration. Asafirststep, weaddthelinesegmentBE, asshowninFigure 2.4.11 (i). Next,drawanaltitudefrom E to AB; supposethisaltitudehaslength h, asindicatedinthefigure.
60 2. Polygons
A
B
D E
C
A
B
D E
C
h
(i) (ii)
Figure2.4.12
Considerthetriangle △ABE. UsingProposition 2.4.1 (3), weknowthattheareaofthistri-
angleis1
2|AB|h. Similarly, theareaofthetriangle △ADE is
1
2|AD|h. Itthenfollowsthat
areaof △ABE
areaof △ADE=
12|AB|h
12|AD|h
=|AB|
|AD|.
Next, returntotheoriginalsituationinFigure 2.4.11, andaddthelinesegmentDC, asshowninFigure 2.4.11 (ii). Thesamesortofreasoningasbeforeshowsthat
areaof △ACD
areaof △ADE=
|AC|
|AE|;
weleavethedetailstothereader.Asournextstep, wecomparethetwotriangles △DEB and △DEC showninthetwoparts
ofFigure 2.4.11. Wecanthinkof DE asthebaseofbothtriangles. Moreover, because DE isparallelto BC (byhypothesis), itfollowsthatthealtitudeof △DEB perpendicularto DE hasthesamelengthasthealtitudeof△DEC perpendiculartoDE (thisfactusesProposition 2.2.6).ItthenfollowsbyProposition 2.4.2 (2)that △DEB and △DEC havethesamearea.Finally, giventhattriangles △DEB and △DEC havethesameareas, weseethattriangles
△ABE and △ACD havethesameareas. Pluggingthisobservationsintothetwoformulasfortheratiosofareasthatwepreviouslysaw, wesee
|AB|
|AD|=
areaof △ABE
areaof △ADE=
areaof △ACD
areaof △ADE=
|AC|
|AE|.
Thuswehaveshownthedesiredequation.
2.4Area 61
Our secondpreliminary result is theconverse to thepreviousproposition; again, seeFig-ure 2.4.11.
Proposition 2.4.4. Supposethatatriangle △ABC hasapoint D onedge AB, andapoint Eonedge AC, placedsothat
|AB|
|AD|=
|AC|
|AE|.
Then DE isparallelto BC.
Demonstration. WeseethegivensituationinFigure 2.4.13 (i); wedonotknowyetwhetherDE isparallelto BC ornot(becausethatiswhatwearetryingtoshow), andwehavedrawnthecasewherethetwolinesegmentsarenotparallel.
A
B
DE
C
A
B
D FE
C
(i) (ii)
Figure2.4.13
ByusingPlayfair’sAxiom (Proposition 1.1.1), wecandrawalinecontainingD thatisparalleltothelinecontaining BC. ThislinethroughD intersects AC inapoint, whichwewillcall F.Then DF isparallelto BC. SeeFigure 2.4.13 (ii).WecannowapplyProposition 2.4.3 tothetriangle △ABC withthepoints D and F. We
deducethat|AB|
|AD|=
|AC|
|AF|.
Ontheotherhand, weknowbyourhypothesesthat
|AB|
|AD|=
|AC|
|AE|.
62 2. Polygons
Wecombinethesetwoequationstoderive
|AC|
|AF|=
|AC|
|AE|,
andbycancellingandrearrangingitfollowsthat |AF| = |AE|. Giventhat E and F arebothpointsin AC, weseethat E = F. ByconstructionweknowthatDF isparallelto BC, andwededucethat DE isparallelto BC.
WenowhavealltheingredientsneededforthepromiseddemonstrationofProposition 2.2.9.
DemonstrationofProposition 2.2.9. Supposethattriangles△ABC and△A ′B ′C ′ aresimilar.Wewillfirstshowthat
|AB|
|AB ′|=
|AC|
|AC ′|.
First, wenote that either |AB| = |AB ′| or |AB| = |AB ′|. If ithappens tobe thecasethat |AB| = |AB ′|, thenwecandeducethat △ABC and △A ′B ′C ′ arecongruent, usingtheAngle-Side-AngleTheorem(Proposition 2.2.4). Inthatcase, itwouldfollowthat |AC| = |AC ′|,andthenwewouldseethat
|AB|
|AB ′|= 1 =
|AC|
|AC ′|,
whichiswhatwearetryingtoshow.Nowassumethat |AB| = |AB ′|, because that is theremainingcase. Therearenowtwo
possibilities, namely |AB| > |AB ′| or |AB| < |AB ′|; wewilldiscussonlythefirstofthesetwocases, theothercasebeingvirtuallyidentical. So, assumethat |AB| > |AB ′|.Because |AB| > |AB ′|, wecanfindapoint D on AB sothat |AD| = |AB ′|. SeeFig-
ure 2.4.14. Nowfindapoint E on AC sothat
|AB|
|AD|=
|AC|
|AE|;
Suchapointcanalwaysbefound. Again, seeFigure 2.4.14.
WecannowapplyProposition 2.4.4 tothetriangle△ABC withpointsD and E. Thepropo-sitionimpliesthat DE isparallel to BC. WecannowuseProposition 1.2.3 todeducethattheangle α equalstheangleat B. Byhypothesisthetriangles △ABC and △A ′B ′C ′ aresim-ilar, andhencetheangleat B equals theangleat B ′. It followsthat theangle α equals theangleat B ′. Wealsoknowthattheangleat A equalstheangleat A ′. Giventhatwealsohave|AD| = |AB ′| (whichistruebyvirtueofourchoiceof D), weseethattriangles △ADE and△A ′B ′C ′ arecongruentbytheAngle-Side-AngleTheorem(Proposition 2.2.4). Wederivethat|AE| = |AC ′|. Finally, weknowbyconstructionthat
|AB|
|AD|=
|AC|
|AE|.
2.5ThePythagoreanTheorem 63
A
B
D E
C
A'
B' C'α
Figure2.4.14
Itfollowsthat|AB|
|AB ′|=
|AC|
|AC ′|.
Thislastequationistheoneweweresupposedtodemonstrate.Wenotethatacompletelysimilarargumentcouldbeusedtoshowthat
|AC|
|AC ′|=
|BC|
|B ′C ′|;
wewillskipthedetails. Wehavethereforeshownthat
|AB|
|AB ′|=
|AC|
|AC ′|=
|BC|
|B ′ C ′|,
whichisthefirstdisplayedequationinthestatementofProposition 2.2.9. TheseconddisplayedequationinthestatementofProposition 2.2.9 followsstraightforwardlyfromthefirstdisplayedequation, andwewillleavethattothereader. Henceourdemonstrationiscomplete.
2.5 ThePythagoreanTheorem
This section treats what is probably themost famous theorem about triangles, namely thePythagoreanTheorem. (This theorem seems tohavebeenknownempirically inbothBaby-lonandChinabeforethetimeofPythagoras, thoughthereisnoevidencethatthetheoremwasprovedpriortoPythagoras.) TherearemanyotherequallyimportanttheoremsingeometryotherthanthePythagoreanTheorem, butwefocusonitnowbecauseitissofamiliar, andbecauseitbringstogetheranumberofideaswehaveencounteredsofarabouttriangleandpolygons.
64 2. Polygons
ItisimportanttostatethePythagoreanTheoremcorrectly. WheneverI askstudentsinaclasstostatethePythagoreanTheorem, theresponseI invariablyreceiveis: “a2+b2 = c2.” However,justtostatethisequationisabsolutelynotcorrect. Itisnottruethatthisequationholdsfor anynumbers a, b and c. Theequationholdsonlyforparticularvaluesof a, b and c, namelythosethatcorrespondtothelengthsofthesidesandhypotenuseofarighttriangle. (Recallthatinarighttriangle, thetwoedgesthatformtherightanglearecalledthe sides ofthetriangle, andtheedgethatisoppositetherightangleiscalledthe hypotenuse ofthetriangle.)WearenowreadytostateanddemonstratethePythagoreanTheorem. Therearemanyproofs
ofthistheorem, goingbacktotheancientworld. ForEuclid’sproofseeProposition 47ofBookI of[Euc56], thoughyouhavetolookatsomeofEuclid’spreviouspropositionstofigureoutall thedetailsofhisproofof thePythagoreanTheorem. Wegive twodifferentproofsof thetheorem(bothquitedifferentfromEuclid’sproof), toshowhowthesameresultcanbeprovedbydifferentapproaches. Ourfirstproof(averywidelyusedone)isbasedonareaandcongruenceof triangles; thesecondisbasedonsimilarityof triangles. NeitherofourproofswouldhavemadesensetotheancientGreeks(whodidnothaveouralgebra). See[Loo40]foravarietyofproofsofthePythagoreanTheorem. A curiousfactoidisthatthereisaproofofthePythagoreanTheoremattributedtoPresidentJamesGarfield—perhapstheonlyknownmathematicalproofattributedtoapresidentoftheUnitedStates.
Proposition 2.5.1 (PythagoreanTheorem). Supposethatarighttrianglehassidesoflength a
and b, andhypotenuseoflength c. Then a2 + b2 = c2.
Demonstration. FirstProof: InFigure 2.5.1 (i)weseethetriangleunderconsideration. Giventhatthesumofthethreeanglesinthetriangleis 180◦ (aswesawinProposition 2.2.1 (1)), weknowthat α+ β = 90◦.Wenowconstructa squarewithsidesof length a + b, andbreak itupasshown inFig-
ure 2.5.1 (ii). Weseefourcopiesoftheoriginalrighttriangleinsidethelargersquare. Thatthesefourtrianglesarereallycongruenttotheoriginaltriangleisintuitivelyclear, andformallyfollowsfromtheSide-Angle-SideTheorem(Proposition 2.2.3).
Nowconsidertheangle γ, asindicatedinthefigure. Giventhat α, β and γ togethermakeupastraightline, weknowthat α + β + γ = 180◦. Because α + β = 90◦, aspreviouslymentioned, wededucethat γ = 90◦. A similarargumentshowsthat theother threeanglesbetweenthesidesoflength c arealso 90◦. Itfollowsthatthefigurethathasfouredgesoflengthc isinfactasquare.Weknowthattheentiresquarewithsidesoflength a + b hasarea (a + b)2. Ontheother
hand, wecanalsocompute theareaof theentiresquarebyaddinguptheareasof thefivepieces(onesquareandfourtriangles)intowhichwehavebrokenitup. Thesquarewithsidesoflength c hasarea c2. Eachofthefourcopiesoftheoriginaltrianglehasarea 1
2ab. Thetotal
areaistherefore c2 + 4 · 12ab = c2 + 2ab. Equatingthetwowaysofcomputingthetotalarea,
obtain
(a+ b)2 = c2 + 2ab.
2.5ThePythagoreanTheorem 65
αα
α
α
α
ββ
β
β
β
γ
a
aa
a
a
bb
b
b
b
c
cc
c
c
(i) (ii)
Figure2.5.1
Recallfromalgebratheformula (a+ b)2 = a2 + 2ab+ b2. Wethenobtain
a2 + 2ab+ b2 = c2 + 2ab.
Canceling 2ab fromeachsideofthislastequationyields a2 + b2 = c2, whichiswhatwewantedtoprove.
SecondProof: InFigure 2.5.2 (i)weseethetriangleunderconsideration, wheretheanglebe-tween the sidesof lengths a and b is a right angle. In Figure 2.5.2 (ii)wehavedrawn thealtitudeperpendiculartoedge AB. Let D bethepointwherethealtitudeintersects AB. Intriangle△ABC, let a, b and c respectivelydenotethelengthsoftheedgesoppositeangles A,B and C. Let x denotethelengthof AD, andthusthelengthof BD is c− x.
Considerthetwotriangles△ABC and△DBC. Eachtrianglehasarightangle, andeachhastheangleat B, sotheyhavetwoequalangles. Becausethesumoftheanglesineachtriangleis 180◦, the two triangles in facthaveall threeangles the same. Thus the two trianglesaresimilar, wherevertices A, B and C respectivelyin △ABC correspondtovertices C, B and D
in △DBC. ItnowfollowsfromProposition 2.2.9 that
c
a=
a
c− x.
Itcanalsobeseenthatthetwotriangles △ABC and △ADC aresimilar, wherevertices A, Band C respectivelyin △ABC correspondtovertices A, C and D in △ADC. Itfollowsfrom
66 2. Polygons
(i) (ii)
x c-xDB
b
C
A
a
cB
b
C
A
a
Figure2.5.2
Proposition 2.2.9 thatc
b=
b
x.
Ifwecrossmultiplytheabovetwoequations, weobtain
c(c− x) = a2 and cx = b2.
Multiplyingoutthefirstequation, weobtain
c2 − cx = a2 and cx = b2.
Substitutingthesecondequationintothefirst, weobtain
c2 − b2 = a2.
Moving b2 totheothersideyields a2 + b2 = c2, whichiswhatwewantedtoprove.
Exercise 2.5.1. Thetwosidesofarighttriangleare 6 and 11 inchesrespectively. Howlongisthehypotenuse?
Exercise 2.5.2. A 40 ft.wireisstretchedfromthetopofapoletotheground. Thewirereachestheground 25 ft.fromthebaseofthepole. Howhighisthepole?
Exercise 2.5.3. ProvethePythagoreanTheoremusingFigure 2.5.3 insteadofFigure 2.5.1.
2.5ThePythagoreanTheorem 67
a
ab
aa
bb
b
c
cc
c
Figure2.5.3
ThePythagoreanTheoremstatesthatifarighttrianglehassidesoflength a and b, andhy-potenuseoflength c, then a2+b2 = c2. Couldithappenthatinanon-righttrianglewithedgesoflength a, b and c, theformula a2 + b2 = c2 alsoholds? Thefollowingpropositionsaysthatitcouldnothappen; inotherwords, theformula a2+b2 = c2 istheexclusiveprovinceofrighttriangles. ItisinterestingtonotethatwewillusethePythagoreanTheoremtodemonstratethefactthetheoremdoesnotholdinnon-righttriangles.
Proposition 2.5.2 (Converse to the PythagoreanTheorem). Supposewearegivena triangle△ABC. Let a, b and c respectivelydenotethelengthsoftheedgesoppositeangles A, B andC. Supposethat a2 + b2 = c2. Then C isarightangle.
Demonstration. Wefollow[Bar01, p.10]. Theangle C couldbeeitheracute(lessthan 90◦),obtuse(greaterthan 90◦)orarightangle(equalto 90◦). Wewillprovethatthefirsttwocasescannothappen; itwillthenfollowthat C isarightangle, whichiswhatwearetryingtoprove.Supposefirstthat C islessthan 90◦. SeeFigure 2.5.4 (i). InFigure 2.5.4 (ii)wehavedrawnthe
altitudeperpendiculartoedge BC. LetD bethepointwherethealtitudeintersects BC. Let hdenotethelengthof AD, andlet x denotethelenthof CD. Weseethatthelengthof BD isa − x. Observethatbecause C islessthan 90◦, itfollowsthat x > 0. (If C werearightangle,then D wouldbethesameas C, and x wouldbe 0.)
Thetwotriangles△ADC and△ABD arerighttriangles. ApplyingthePythagoreanTheoremtoeachone, weobtainthetwoequations
x2 + h2 = b2 and (a− x)2 + h2 = c2.
Isolating h2 ineachoftheseequationsyields
h2 = b2 − x2 and h2 = c2 − (a− x)2.
68 2. Polygons
(i) (ii)
x a-xDB
hb
A
C
c
aB
b
A
C
c
Figure2.5.4
Equatingthesetwoexpressionsfor h2 givesus
b2 − x2 = c2 − (a− x)2.
Recallfromalgebratheformula (a− x)2 = a2 − 2ax+ x2. Wethereforeobtain
b2 − x2 = c2 − (a2 − 2ax+ x2).
Distributingthenegativesigninyields
b2 − x2 = c2 − a2 + 2ax− x2.
Cancelling x2 frombothsidesgiveus
b2 = c2 − a2 + 2ax.
Finally, bringthe a2 totheotherside, andwededucethat
a2 + b2 = c2 + 2ax.
Wenowhavealogicalimpossibility. Ontheonehand, wehaveassumedthat a2 + b2 = c2.Ontheotherhand, wejustdeducedthat a2 + b2 = c2 + 2ax. Giventhatneither a nor x is 0,thenneitheris 2ax, andsowehaveanimpossiblesituation. Theonlywayoutofthisproblemistoadmitthatourhypothesisthat C islessthan 90◦ isfalse.A similarargumentshowsthatthehypothesisthatC isgreaterthan 90◦ isalsofalse. (Weleave
ittothereadertosupplythedetails; thedifferenceisthatFigure 2.5.4 needstobemodifiedsothatC isgreaterthan 90◦, whichmakesthealtitudeperpendicularto BC beoutsidethetriangle△ABC.) Asaresult, theonlyremainingpossibilityisthat C isarightangle, whichiswhatwewantedtoshow.
ThePythagoreanTheoremdefinitelydoesnotholdfortrianglesthatarenotrighttriangles.WenowturntotwogeneralizationsofthePythagoreanTheoremthatdoholdforalltriangles.Thefirstofthesegeneralizations, themorewellknownandusefulofthetwo, iscalledtheLaw
2.5ThePythagoreanTheorem 69
ofCosines, anditisessentiallythePythagoreanTheoremwithacorrectionfactorthatmakesitworkforalltriangles. TheLawofCosinesisimportantintrigonometry, andshowsupinotherbranchesofmathematics, andapplicationsofmathematics. TostatetheLawofCosines, weneedtousethetrigonometricfunctioncosine. Forthosenotfamiliarwithcosine, youcansimplyskipthestatementoftheLawofCosinesgivenbelow; wewillnotbeusingthislawatanypointinthistext. However, wementionit, toshowoneofthewaysinwhichthePythagoreanTheoremcanbegeneralizedtonon-righttriangles.Justtoremindthosefamiliarwiththetrigonometricfunctions, cosineisafunctionthatassigns
toeveryangle x anumberdenoted cos x. Forexample, wehave cos 0◦ = 1, and cos 60◦ = 1/2,and cos 90◦ = 0. TheLawofCosinesisasfollows.
Proposition 2.5.3 (LawofCosines). Supposewearegivenatriangle △ABC. Let a, b and c
respectivelydenotethelengthsoftheedgesoppositeangles A, B and C. Then
c2 = a2 + b2 − 2ab cosC.
ForaproofoftheLawofCosines, seemostbooksontrigonometryfordetails. WechosetohighlighttheangleC intheabovestatementoftheLawofCosines, thoughinanon-righttrianglenooneangleisspecial, andtheLawofCosinesalsostatesthat a2 = b2 + c2 − 2bc cosA andb2 = a2 + c2 − 2ac cosB. Next, supposethat △ABC isinfactarighttriangle, with C therightangle. Then C = 90◦, and cosC = 0. Inthatcase, theLawofCosinesjustreducestothePythagoreanTheorem.Wenowturntooursecond, lesswellknown, generalizationof thePythagoreanTheorem.
ThisgeneralizationisknownasPappus’VariationonthePythagoreanTheorem IntheLawofCosines, wemaintainthe a2, b2 and c2 thatareinthestatementofthePythagoreanTheorem,butputinanextracorrectionterm(involvingtrigonometry)toaccountfornon-righttriangles.Recallthatgeometrically, theterms a2, b2 and c2 correspondtotheareasofcertainsquares.InPappus’VariationonthePythagoreanTheorem, statedbelow, wereplacesquaresbycertainparallelograms, andbysodoingwewillbeabletoallowfornon-right triangleswithout theuseofacorrectionterm(andthuswithoutanytrigonometry). WhenreadingthestatementofPappus’VariationonthePythagoreanTheorem, itwillhelptolookatFigure 2.5.5.
Proposition 2.5.4 (Pappus’VariationonthePythagoreanTheorem). Supposewearegivenatri-angle△ABC. FormparallelogramsACDE andBCFG ontheedgesAC andBC respectively.Extendthelinesegments DE and FG untiltheyintersectinthepoint H. Formtheparallelo-gram ABIJ withedgesAJ and B I thatareparalleltoandhaveequallengthasHC. Thentheareaoftheparallelogram ABIJ equalsthesumoftheareasoftheparallelograms ACDE andBCFG.
Demonstration. First, weextend HC untilitcutsthroughtheparallelogram ABIJ, breakingitupintotwosmallerparallelograms AJKL and BIKL. SeeFigure 2.5.6. WewillnowshowthattheareaofparallelogramACDE equalstheareaoftheparallelogramAJKL. A completelyidenticalargument (thedetailsofwhichwewill skip)canbeused to show that theareaof
70 2. Polygons
B
G
FH
J I
C
A
E
D
Figure2.5.5
parallelogram BCFG equalstheareaoftheparallelogram BIKL. Itwillthenfollowthatthesumoftheareasoftheparallelograms ACDE and BCFG equalsthesumoftheareasoftheparallelogramsAJKL and BIKL, whichinturnequalstheareaoftheparallelogramABIJ, thuscompletingtheargument.
B
G
FH
JK
I
C
A L
E
D
Figure2.5.6
Toshowthattheareaofparallelogram ACDE equalstheareaoftheparallelogram AJKL,weneedtomakeonemoreparallelogram. ExtendAJ untilitintersectsthelinecontainingDE
2.5ThePythagoreanTheorem 71
inpointM. WeseethatMA isparalleltoHC. SeeFigure 2.5.7. HenceACHM isaparallel-ogram. Comparetheparallelograms ACDE and ACHM. Theybothhavetheedge AC, andthenbothhavethesamealtitudeperpendiculartothisedge. ItfollowsfromProposition 2.4.2 (1)thatthesetwoparallelogramshavethesamearea.
B
G
FH
JK
I
C
A L
E
D
M
Figure2.5.7
Next, comparetheparallelogramsACHM andAJKL. ObservethattheedgeHC ofthepar-allelogram ACHM isequalinlengthandparalleltotheedge AJ oftheparallelogram AJKL.Moreover, thealtitudesperpendicular to these twoedgesareequal. It follows fromProposi-tion 2.4.2 (1)thatthesetwoparallelogramshavethesamearea. Hence, becausetheparallelo-grams ACDE and ACHM havethesameareas, andtheparallelograms ACHM and AJKL
havethesameareas, itfollowsthattheparallelograms ACDE and AJKL, whichiswhatweneededtoshow.
When we apply Pappus’Variation on the PythagoreanTheorem to a right triangle, andwe choose parallelograms that happen to be squares, then we simply obtain the standardPythagoreanTheorem.
72 2. Polygons
3Polyhedra
3.1 Polyhedra–TheBasics
A polyhedron isasolidregionofspacethatisboundedbyafinitenumberofpolygonsthataregluedtogether. Wehavethreerequirementsaboutthewayinwhichwegluethepolygonstogether.
(1) Polygonsaregluededge-to-edge(thatis, entireedgesaregluedtoentireedges), orvertex-to-vertex.
(2) Everyedgeofapolygonisgluedtopreciselyoneotheredge.
(3) Notwopolygonsintersectexceptpossiblyalongtheiredgeswheretheyareglued.
SomepolyhedraareshowninFigure 3.1.1. Somenon-polyhedraareshowninFigure 3.1.2;theobjectinPart (i)haspolygonsthatarenotgluededge-to-edge, andtheobjectinPart (ii)hasthreeedgesofpolygonsgluedtogether. Notethatthepluralof“polyhedron”is“polyhedra.”Wewillrestrictourattentiontopolyhedramadeupoutofconvexpolygons, thoughitisalsopossibletoconsiderpolyhedrawithnon-convexfaces. (Itisalsopossibletolookatpolyhedrawithself-intersections, thatis, inwhichrequirement(3)isdropped; wewillnotbelookingatsuchpolyhedrainthistext, withonebriefexceptionattheendofSection 3.6.)
Foreachpolyhedron, the faces ofthepolyhedronarethepolygonsthatboundit; the edgesarethelinesegmentswherethefacesmeet; the vertices arethepointswhereedgesmeet. Forexample, thecube showninFigure 3.1.1 hassixsquarefaces, twelveedgesandeightvertices.Thefollowingsimplefactsaboutfaces, edgesandverticesofpolyhedra, whichwillbeofuse
lateron, canbederivedfromourrequirementsonhowpolygonsaregluedtogethertomakepolyhedra.
74 3. Polyhedra
(i) (ii) (iii)
Figure3.1.1
(i) (ii)
Figure3.1.2
Proposition 3.1.1.
1. Everyedgeinapolyhedroniscontainedinprecisely 2 faces.
2. Everyvertexinapolyhedroniscontainedinatleast 3 edges.
3. Everyfaceinapolyhedroncontainsatleast 3 edges.
Just aswe had both convex and non-convex polygons, so toowe can have convex andnon-convexpolyhedra. Theideaofconvexityiscompletelythesameforpolyhedraasforpoly-gons. Intuitively, apolyhedronisconvexifithasno“indentations.” Moreformally, apolyhedronisconvexifanytwopointsinthepolyhedronarejoinedbyalinesegmentcontainedentirelyinthepolyhedron. ThepolyhedrainFigure 3.1.1 (i)and(ii)areconvex, whereasthepolyhedroninPart (iii)ofthefigureisnot. Wewillmostly, thoughnotexclusively, dealwithconvexpolyhedrainthistext. Oneusefulfactaboutconvexpolyhedron, whichwestatewithoutdemonstration,isthefollowing.
Proposition 3.1.2. Atanyvertexofaconvexpolyhedron, thesumoftheanglesatthevertexadduptolessthan 360◦.
3.1Polyhedra–TheBasics 75
Thoughsomepolyhedraarequiteirregular, therearesomenicecategoriesofpolyhedrathatareconvenienttoworkwith. A pyramid isobtainedbytakingapolygonintheplane, takingapoint“above” thepolygon, and joining thispoint to theverticesof thepolygon. Thenewvertexisoftencalledthe conepoint ofthepyramid(itisalsoknownasthe apex ofthepyramid);thepolygonthatwestartedwithiscalledthe base ofthepyramid, andweoftencallsuchapyramidasa“pyramidoverthepolygon.” ThefamouspyramidsinGiza, Egypt, areexamplesofpyramidswithsquarebases(oftencalled“squarepyramids,” or“pyramidsoversquares”);observethatthemathematicaluseofthetermpyramidallowsforpyramidswithanybase, notjustasquarebase. SeeFigure 3.1.3 (i)forapyramidoverapentagon. A bipyramid isobtainedbytakingapolygonintheplane, takingonepoint“above”thepolygonandonepoint“below”it, andjoiningbothpointtotheverticesofthepolygon; thenewverticesarebothcalled conepoints ofthebipyramid(oneisthe“top”conepoint, andoneisthe“bottom”conepoint). SeeFigure 3.1.3 (ii)forabipyramidoverapentagon. A prism isobtainedbytakingapolygonintheplane, placinganidenticalcopyofthepolygondirectlyabovethefirst, andjoiningpairsofcorrespondingverticesofthetwocopiesofthepolygon. SeeFigure 3.1.3 (iii)foraprismoverapentagon. An antiprism isobtainedbytakingaregularpolygonintheplane, placingacopyoftheregularpolygonabovethefirst, butrotatedsothateachvertexoftheupperpolygonisabovethemiddleofanedgeofthelowerpolygon, andthenjoiningeachuppervertextothetwolowerverticesclosesttoit. SeeFigure 3.1.3 (iv)foranantiprismoverapentagon.
(i) (ii)
(iii) (iv)
Figure3.1.3
Theabovecategoriesofpolyhedracanoverlap. Forexample, as thereadercanverify, thebipyramidoverasquare, calledanoctahedron, isinfactalsoanantiprismoveratriangle.
76 3. Polyhedra
Exercise 3.1.1. Foreachofthefollowingquestions, iftheanswerisyes, giveanexample,andiftheanswerisno, explainwhynot. (Toexplainwhyapolyhedroncannotbeintwodifferentcategories, itdoesnotsufficesimplytostatethatthetwocategoriesareconstructeddifferently, becausesometimestwodifferentconstructionscanyieldthesameresult, forexampletheoctahedron, whichisbothabipyramidandanantiprism.)
(1) Canapolyhedronbebothabipyramidandaprism?
(2) Canapolyhedronbebothaprismandanantiprism?
(3) Canapolyhedronbebothapyramidandeitheraprismoranantiprism?
(4) Canapolyhedronbebothapyramidandabipyramid? (Ifyouthinkthattheanswerisno, it isnot sufficient simply to say thatapyramidhasoneconepoint, andabipyramidhastwoconepoints. Perhapsifyoulookatacertainpolyhedroninonewaythereisoneconepoint, andviewedanotherwaytherearetwoconepoints;perhapsnot.)
Exercise 3.1.2. Findallpyramidsthathaveallregularfaces.
Justasthereareformulasfortheareasofsimpletypesofpolygons, therearealsoformulasforthevolumesofsimpletypesofpolyhedra. However, thedemonstrationsofthesevolumeformu-lasaremorecomplicatedthanforareaformulas, andsowewillstatethefollowingpropositionwithoutdemonstration. (See[Har00, Sections26-27]foratechnicaldiscussionofvolumes.) Justaswediscussedthenotionofthealtitudeofatriangle, parallelogramortrapezoid, wecansimi-larlydefinethenotionofanaltitudeforpolyhedrasuchaspyramidsandprisms; weomitfurtherdetails. A bipyramidismadeupoftwopyramidsgluedtogether, andeachofthesepyramidshasanaltitude.
Proposition 3.1.3.
1. Supposethataprismhasabaseofarea b, andaltitudeoflength h. Thenthevolumeoftheprismis bh.
2. Supposethatapyramidhasabaseofarea b, andaltitudeoflength h. Thenthevolumeofthepyramidis 1
3bh.
3. Supposethatabipyramidhasabaseofarea b, andaltitudesoflength h1 and h2 foreachofthetwopyramidsinthebipyramid. Thenthevolumeofthebipyramidis 1
3b(h1+h2).
Itisinterestingtocomparethevolumeformulaforpyramidswiththeareaformulafortriangles,andtocomparethevolumeformulaforprismswiththeareaformulaforparallelograms.
3.1Polyhedra–TheBasics 77
Givenaconvexpolyhedron, wecanformanewpolyhedron, calledits dualpolyhedron, asfollows. First, foreachfaceoftheoriginalpolyhedron, chooseapointinitsinterior(forexample,choosethecenterofgravityoftheface). Thesechosenpointswillbetheverticesofthedualpolyhedron, calleddualvertices. Next, consideranedgeintheoriginalpolyhedron. Thisedgeiscontainedinpreciselytwofacesoftheoriginalpolyhedron. Wethenputanedgeinthedualpolyhedronjoiningthetwodualverticesthatarecontainedinthesetwofacesoftheoriginalpolyhedron. Wethusobtaintheedgesofthedualpolyhedron. Finally, consideravertexintheoriginalpolyhedron. Thisvertexiscontainedinsomefacesoftheoriginalpolyhedron. Wethenputafaceinthedualpolyhedronthathasasitsverticesthedualverticesthatarecontainedinthesefacesoftheoriginalpolyhedron. Wethusobtainthefacesofthedualpolyhedron.Forexample, supposeouroriginalpolyhedronisacube, asshowninFigure 3.1.4 (i). Thedual
ofthecubeisshowninsidethecubeinFigure 3.1.4 (ii). Thisdualpolyhedronhassixvertices,twelveedgesandeightfaces(itiscalledanoctahedron, andwewillencounteritagaininthenextsection).
(i) (ii)
Figure3.1.4
Exercise 3.1.3.
(1) Whatisthedualtoabipyramidoveran n-gon?
(2) Whatisthedualtoapyramidoveran n-gon?
Exercise 3.1.4. Suppose P isaconvexpolyhedron. Whatistherelationbetween P andthedualofthedualof P?
78 3. Polyhedra
3.2 RegularPolyhedra
Justasregularpolygonswerethemost“uniform”polygonspossible, wewanttofindpolyhedrathatareas“uniform”aspossible. Forapolygontoberegular, itneedstosatisfytworequirement,namelythatalltheedgeshavethesamelengths, andthatalltheanglesareequal(requiringonlyedgesofequallengthdoesnotsuffice). Wewillneedsimilarrequirementstoinsurethatapolyhedroniscompletelyuniform. A convexpolyhedronisa regularpolyhedron ifthefollowingthreeconditionshold: (1)everyfaceisaregularpolygon; (2)allfacesareidentical; and(3)allverticesareidentical, whichmeansthatallverticesarecontainedinthesamenumberoffaces.Itisnothardtoseethatinaregularpolyhedron, alltheedgesmusthavethesamelength.Thatthefirsttwopartsoftheabovedefinitiondonotsufficecanbeseenbyconsideringa
bipyramidoveranequilateraltriangle, asshowninFigure 3.2.1. Ifalltheedgesofthispolyhe-dronhaveequallength, thenconditions(1)and(2)oftheabovedefinitionwillbesatisfied, butwestillwouldnotwanttocallthispolyhedronregular, becausetheverticesdonotall“lookthesame.” Moreprecisely, twooftheverticesarecontainedinthreetriangleseach, whereasthreeoftheverticesarecontainedinfourtriangleseach. Henceweneedcondition(3)ofthedefintionofregularpolyhedra.
Figure3.2.1
In thecaseofpolygons, wesawthat therewere infinitelymanydistinct regularpolygons,one foreachpossiblenumberof edges. That is, there is a regular 3-gon (alsoknownasanequilateraltriangle), aregular 4-gon(alsoknownasasquare), aregular 5-gon, aregular 6-gon,etc. (Ofcourse, eachoneofthesepolygonscouldbeconstructedindifferentsizes, butweareonlyinterestedinshapesthataregenuinelydifferent.) Thefollowingresultshows, somewhatsurprisingly, thatwhatholdsforregularpolygonsdoesnotholdforregularpolyhedra.
Proposition 3.2.1. Every regular polyhedron is one of the five polyhedra described inTa-ble 3.2.1.
Demonstration. Recall thatall regularpolyhedraareassumedtobeconvex. Thekeyto thisdemonstrationisProposition 3.1.2, whichsaysthatatanyvertexofaconvexpolyhedron, thesumoftheanglesatthevertexmustadduptolessthan 360◦.
3.2RegularPolyhedra 79
Name Faces FacesperVertextetrahedron 4 triangles 3 trianglespervertexcube(akahexahedron) 6 squares 3 squarespervertexoctahedron 8 triangles 4 trianglespervertexdodecahedron 12 pentagons 3 pentagonspervertexicosahedron 20 triangles 5 trianglespervertex
Table3.2.1
Allthefacesinaregularpolyhedronarethesame(andareregularpolygons), andallverticesarecontainedinthesamenumberoffaces. Letusstartbyexaminingthesituationwhenallthefacesofaregularpolyhedronareequilateraltriangles. Weknowthattheanglesinanequilateraltriangleareall 60◦. Howmanyequilateraltrianglescancontaineachvertexofaregularpoly-hedron, andstillhavethesumoftheanglesateachvertexadduptolessthan 360◦? Weobservethat 3 · 60◦ = 180◦, that 4 · 60◦ = 240◦, that 5 · 60◦ = 300◦ andthat 6 · 60◦ = 360◦. Hence,weseethatitmightbepossibletohavearegularpolyhedronwithfacesthatareequilateraltri-angles, andwitheither 3, 4 or 5 facescontainingeachvertex; itwouldnotbepossibletohavearegularpolyhedronwithfacesthatareequilateraltriangles, andwith 6 ormorefacescontainingeachvertex. Thereindeedexistregularpolyhedrawithfacesthatareequilateraltriangles, andwitheither 3, 4 or 5 facescontainingeachvertex, namelythetetrahedron, theoctahedronandtheicosahedron. Thereexistnootherpolyhedrasatisfyingthesamecriteriaforthetypesoffacesandvertices. Wehavethereforefoundalltheregularpolyhedrawithfacesthatareequilateraltriangles.Next, letusexaminethesituationwhenallthefacesofaregularpolyhedronaresquares. We
knowthattheanglesinasquareareall 90◦. Howmanysquarescancontaineachvertexofaregularpolyhedron, andstillhavethesumoftheanglesateachvertexadduptolessthan360◦? Weobservethat 3 · 90◦ = 270◦ andthat 4 · 90◦ = 360◦. Hence, weseethatitmightbepossibletohavearegularpolyhedronwithfacesthataresquares, andwith 3 facescontainingeachvertex; itwouldnotbepossibletohavearegularpolyhedronwithfacesthataresquares,andwith 4 ormorefacescontainingeachvertex.Weseethatacubeisaregularpolyhedronwithfacesthataresquares, andwith 3 facescontainingeachvertex; thecubeisuniqueinsatisfyingthisproperty. Wehavethereforefoundalltheregularpolyhedrawithfacesthataresquares.Now, let us examine the situationwhenall the facesof a regular polyhedronare regular
pentagons. WeknowfromTable 2.3.1 thattheanglesinaregularpentagonareall 108◦. Howmanyregularpentagonscancontaineachvertexofaregularpolyhedron, andstillhavethesumoftheanglesateachvertexadduptolessthan 360◦? Weobservethat 3 · 108◦ = 324◦ andthat 4 · 108◦ = 432◦. Hence, weseethatitmightbepossibletohavearegularpolyhedronwithfacesthatareregularpentagons, andwith 3 facescontainingeachvertex; itwouldnotbepossibletohavearegularpolyhedronwithfacesthatareregularpentagons, andwith 4 ormore facescontainingeachvertex. A dodecahedron isa regularpolyhedronwith faces thatareregularpentagons, andwith 3 facescontainingeachvertex; thedodecahedronisuniquein
80 3. Polyhedra
satisfyingthisproperty. Wehavethereforefoundalltheregularpolyhedrawithfacesthatareregularpentagons.Finally, letusexaminethesituationwhenallthefacesofaregularpolyhedronareregular
polygonswith 6 ormoreedges. WeknowfromTable 2.3.1 thattheanglesinaregularhexagonareall 120◦, andthattheanglesinaregularpolygonwithmorethan 6 edgesarelargerthan120◦. Howmanyregularpolygonswith 6 ormoresidescancontaineachvertexofaregularpolyhedron, andstillhavethesumoftheanglesateachvertexadduptolessthan 360◦? Theanswerisnone, giventhat 3 · 120◦ = 360◦, and 3 timesananglelargerthan 120◦ wouldbemorethan 360◦. Giventhateveryvertexinapolyhedronmustbecontainedinatleastthreefaces, weseethat itwouldnotbepossibletohavearegularpolyhedronwithfacesthatareregularpolygonswith 6 ormoreedges.Puttingalltheabovetogether, weseethattherearepreciselyfiveregularpolyhedra, aslisted
inTable 3.2.1.
Thefiveregularpolyhedra, whicharelistedinTable 3.2.1, areshowninFigure 3.2.2. Itcanbeshownthat theverticesofeachregularpolyhedronlieonasphere(seeTheorem44.4in[Har00]fordetails). Noticethattheregularpolyhedraarenamedbythenumberoffaceseachonehas. Thefiveregularpolyhedraarealsoknownasthe platonicsolids, inhonorofthean-cientGreekphilosopherPlato. (Forthosefansofchildren’sliterature, youmightknowoftheDodecahedronin“ThePhantomTollbooth;” ifyouarenotfamiliarwiththisbook([Jus61]), itishighlyrecommended.)
Webrieflymentionedthenotionofadualpolyhedron intheprevioussection. Letuslookatthedualsofeachofthefiveregularpolyhedra. Wealreadysawintheprevioussectionthatthedualof thecube is theoctahedron. The reader canverify, by sketching theappropriatepicture, thatthedualoftheoctahedronisthecube. Similarly, itcanbeseenthatthedualofthedodecahedronistheicosahedron, andthedualoftheicosahedronisthedodecahedron.Thedualof the tetrahedron is simply itself. We therefore see that the regularpolyhedraareself-containedinaverynicearrangementwhenitcomestoduality.
Exercise 3.2.1.
(1) Whichoftheregularpolyhedraarepyramids?
(2) Whichoftheregularpolyhedraarebipyramids?
(3) Whichoftheregularpolyhedraareprisms?
(4) Whichoftheregularpolyhedraareantiprisms?
Exercise 3.2.2. Findallconvexpolyhedrathatarebothbipyramidsandantiprisms.
3.3Semi-RegularPolyhedra 81
Tetrahedron Cube Octahedron
Dodecahedron Icosahedron
Figure3.2.2
Exercise 3.2.3. Supposethatyouhaveacubemadeoutofclay; supposefurtherthattheclayisred, buttheoutsideofthecubeispaintedblue. Youthenslicethecubeinastraightlinewithaknife, causingthecubetobreakintotwopieces. Eachpiecehasanexposedredpolygon, wherethecubewassliced. Dependinguponhowyouslicethecube, youmightgetdifferentexposedpolygons; alltheexposedpolygonswillbeconvex. Forexample, ifyousliceparalleltooneofthefacesofthecube, yourexposedpolygonwillbeasquare;ifyousliceoffacornerofthecuberightnexttoavertex, yourexposedpolygonwillbeatriangle. Whatareallthepossibleexposedpolygonsthatcouldbeobtainedbyslicingthecube?
3.3 Semi-RegularPolyhedra
Regularpolyhedra, asdiscussedintheprevioussection, arethemostuniformpolyhedra. Wenowturntoaslightlybroadercategoryofpolyhedra, namelythe semi-regularpolyhedra, whichareconvexpolyhedrathatsatisfythefollowingtwocondition: (1)everyfaceisaregularpoly-gon; (2)allverticesare identical, whichmeans thatallverticesarecontainedinsametypes
82 3. Polyhedra
ofpolygonsarrangedinthesameorder. Observethatthefacesarenotallrequiredtobethesametypeofpolygon. Itisthecasethatalledgesinasemi-regularpolyhedronhavethesamelength. Certainlyeveryregularpolyhedronisalsosemi-regular, buttherearesemi-regularpoly-hedrathatarenotregular. Forexample, weseeinFigure 3.3.1 asemi-regularpolyhedroncalledtherhombicuboctahedron, thefacesofwhichareallequilateraltrianglesandsquares, andtheverticesofwhichareeachcontainedinthreesquaresandonetriangle.
Figure3.3.1
Supposewearegivenavertexinapolyhedron. Wedefinethe vertexconfiguration ofthisvertex tobea listofnumbersof the form (a1, a2, . . . , an), where thenumbers a1 throughan arethenumbersofedgesofthepolygonscontainingthevertex, listedinorderaswegoaroundthevertex(itdoesnotmatterwhichpolygonwestartwith). Forexample, inthepolyhe-dronshowninFigure 3.3.1, everyvertexhasvertexconfiguration (3, 4, 4, 4). Thedefinitionofsemi-regularpolyhedraimpliesthatinanysemi-regularpolyhedron, allverticeshavethesamevertexconfiguration.Thefollowingpropositiontellsusallthepossiblesemi-regularpolyhedra.
Proposition 3.3.1. Everysemi-regularpolyhedronisoneofthefollowing:
(A) A regularpolyhedron.
(B) A prismoveraregularpolygon, withthesidesmadeupofsquares.
(C) Anantiprismoveraregularpolygon, withthesidesmadeupofequilateraltriangles.
(D) Oneofthe 14 polyhedradescribedinTable 3.3.1.
Wewillnotdemonstratetheaboveproposition. Thedemonstrationissimilarto, thoughmorecomplexthan, thedemonstrationshowingthatthereareonlyfiveregularpolyhedra(Proposi-tion 3.2.1); formoredetails, see, forexample, theproofofTheorem46.1in[Har00]. Picturesofthe 14 polyhedralistedinTable 3.3.1 aregiveninFigure 3.3.2. ObserveinTable 3.3.1 thattherhombicuboctahedronandpseudorhombicuboctahedronhavethesamevertexconfigurations,andthesamenumbersoffacesofeachtype; however, thesepolyhedraarenotidentical. TherhombicuboctahedronisshowninthelowerleftcornerofFigure 3.3.2, andthepseudorhom-bicuboctahedronisshowninthelowerrightcornerofthefigure. Thepseudorhombicubocta-hedroncanbeobtainedfromtherhombicuboctahedronbyrotatingthetop“cap”by 45◦.
3.3Semi-RegularPolyhedra 83
Name VertexConfig. TypesofFacestruncatedtetrahedron (3, 6, 6) 4 trianglesand 4 hexagonstruncatedoctahedron (4, 6, 6) 6 squaresand 6 hexagonstruncatedicosahedron (5, 6, 6) 12 pentagonsand 20 hexagonstruncatedcube (3, 8, 8) 8 trianglesand 6 octagonstruncateddodecahedron (3, 10, 10) 20 trianglesand 12 10-gonstruncatedcuboctahedron (4, 6, 8) 12 squares, 8 hexagonsand 6 octagonstruncatedicosidodecahedron (4, 6, 10) 30 squares, 20 hexagonsand 12 10-gonscuboctahedron (3, 4, 3, 4) 8 trianglesand 6 squaresicosidodecahedron (3, 5, 3, 5) 20 trianglesand 12 pentagonsrhombicosidodecahedron (3, 4, 5, 4) 20 triangles, 30 squaresand 12 pentagonssnubcube (3, 3, 3, 3, 4) 32 trianglesand 6 squaressnubdodecahedron (3, 3, 3, 3, 5) 80 trianglesand 12 pentagonsrhombicuboctahedron (3, 4, 4, 4) 8 trianglesand 18 squarespseudorhombicuboctahedron (3, 4, 4, 4) 8 trianglesand 18 squares
Table3.3.1
Figure3.3.2
The14polyhedralistedinTable 3.3.1 areoftencalledthe Archimedeansolids. DonotgetcaughtupindecipheringthenamesoftheArchimedeansolids; itisnotimportant. Moreover,notallauthorsusethesamenamesforthe 14 polyhedralistedinTable 3.3.1. Theonewordfromthenamesofthesemi-regularpolyhedrathatisworthmentioningis“truncated.” Totruncateapolyhedron, wesimplychopoffasmallpiecearoundeachvertex. Forexample, atruncatedcubewillhaveeightsmalltriangles(oneforeachoriginalvertexofthecube), andsixoctagons(oneforeachoriginalsquarefaceofthecube). ThereaderisencouragedtolocatethetruncatedcubeinFigure 3.3.2.
84 3. Polyhedra
The terms “semi-regularpolyhedron”and “Archimedean solid”areusedwith somevaria-tionintheliterature. Sometextsaddapropertycalled“vertextransitivity” tothedefinitionofsemi-regular (thoughwedonot). Wedonotyethave the tools toexplainvertex transitivity,thoughitwillbeexplainedinSection 5.7. ItturnsoutthatallthepolyhedralistedinProposi-tion 3.3.1 exceptforone, thepseudorhombicuboctahedron, satisfyvertextransitivity. Hence,thosetextsthatrequirevertextransitivitylistonly13Archimedeansolids.Observethatthereareinfinitelymanydifferentsemi-regularpolyhedra, becausetherearein-
finitelymanydifferentsemi-regularprismsandantiprisms(therebeingoneofeachforeachpos-sibleregularpolygon). However, becausetheregularpolyhedra, theprismsandtheantiprismsareallotherwiseknownpolyhedra, sometextsfocusontheArchimedeansolidswhentheydis-cusssemi-regularpolyhedra. Wenotethat, aswasthecaseforregularpolyhedra, theverticesofeachsemi-regularpolyhedronlieonasphere(seeCorollary46.2in[Har00]fordetails).
Exercise 3.3.1. WhatcanbesaidaboutthefacesofthedualofanArchimedeansolid?
Exercise 3.3.2. Weknowthatthedualofeachregularpolyhedronisitselfaregularpoly-hedron. Canithappenthatthedualofanon-regularsemi-regularpolyhedronisitselfasemi-regularpolyhedron? Iftheanswerisyes, giveanexample, andiftheanswerisno,explainwhynot.
3.4 OtherCategoriesofPolyhedra
Intheprevioustwosectionswediscussedregularpolyhedraandsemi-regularpolyhedra, whicharetypesofconvexpolyhedrathatsatisfycertainniceproperties. Inthissectionwediscusstwofurthertypesofrelativelynicepolyhedra. Westartwith deltahedra, whichareconvexpolyhedramadeupentirelyofequilateraltriangles. Wehavealreadyencounteredthreesuchpolyhedra,namelythetetrahedron, theoctahedronandtheicosahedron. Therearesomeotherdeltahedra,besidesthesethree, thoughtheyareneitherregularnorsemi-regular(thatis, notallverticesarecontainedinsametypesofpolygonsarrangedinthesameorder).
BEFORE YOU READ FURTHER:
Therearefivedeltahedrathatareneitherregularnorsemi-regular. Trytofindasmanyoftheseasyoucan.
Thefollowingpropositionlistsallthedeltahedra.
Proposition 3.4.1. Everyconvexdeltahedroniseitheraregularpolyhedron(atetrahedron, anoctahedronoranicosahedron), orisoneofthe 5 polyhedralistedinTable 3.4.1.
3.4OtherCategoriesofPolyhedra 85
Name Facestriangularbipyramid 6 trianglespentagonalbipyramid 10 trianglessnubdisphenoid 12 trianglestricappedtriangularprism 14 trianglesbicappedsquareantiprism 16 triangles
Table3.4.1
Wenotethatnotallauthorsusethesamenamesforthefivenon-regularconvexdeltahedrathatwehaveusedintheabovetable, thoughthereisnodisagreementovertheactualpolyhe-dra, regardlessoftheirnames. Picturesofthe 5 polyhedra listed inTable 3.4.1 areshowninFigure 3.4.1.
triangular bipyramid
tricapped triangular prism
bicapped square antiprismsnub disphenoid
pentagonal bipyramid
Figure3.4.1
Wenextturntoanevenmorebroadcategoryofpolyhedra, namelythe face-regularpolyhe-dra, whicharepolyhedrathathaveallregularfaces, thoughnotallfacesarenecessarilythesame, andnotallverticesnecessarilyhavethesameconfigurationsoffacescontainingthem.(Wefollow[Har00]inusingtheterm“face-regular.”) Thiscategoryincludesallregularpolyhe-dra, allsemi-regularpolyhedraandalldeltahedra, butthereareothersaswell. Forexample,placingapyramidwithsquarebaseandequilateralsidesontopofacubeyieldstheface-regularpolyhedronshowninFigure 3.4.2. Aswasthecaseforregularandsemi-regularpolyhedra, alltheedgesinaface-regularpolyhedronmusthavethesamelength.
86 3. Polyhedra
Figure3.4.2
Itturnsoutthattherearealimitednumberofconvexface-regularpolyhedra, thoughtheproofislengthy, andisbeyondthescopeofthisbook. Fortherecord, however, westatethefollowingProposition.
Proposition 3.4.2. Everyconvexface-regularpolyhedroniseitherasemi-regularpolyhedron,oroneof 91 otherpolyhedrathatarenotsemi-regular.
Ofthe 91 convexnon-semi-regularface-regularpolyhedramentionedintheabovePropo-sition, fivearethenon-regulardeltahedralistedinProposition 3.4.1. Weshouldmentionthatsometextslist 92 convexnon-semi-regularface-regularpolyhedra, becausetheydonotcon-siderthepseudorhombicuboctahedron tobesemi-regular(asmentionedinSection 3.3). These91 (or, insometexts, 92)polyhedraaresometimesreferredtoasthe Johnsonsolids, namedafterthepersonwhofirstpublishedthecompletelistofthesepolyhedra(see[Joh66], whichisverytechnical).HavingshownoneoftheJohnsonsolidsinFigure 3.4.2, weleaveittothereadertofindsome
othersinthefirsttwoofthefollowingExercises. Wenotethatallface-regularpyramidswerefoundinExercise 3.1.2.
Exercise 3.4.1. Findallconvexface-regularpolyhedrathathaveidenticalfaces, otherthanthedeltahedra.
Exercise 3.4.2. FindatleasttwoJohnsonsolidsthatarenotpyramids, bipyramidsordelta-hedra, andaredifferentfromtheoneshowninFigure 3.4.2.
3.5EnumerationinPolyhedra 87
Exercise 3.4.3. A face-regularpolyhedroniscalled elementary ifitcannotbebrokenupintotwoormoreface-regularpolyhedrathatarejoinedalongacommonface. Forexample,theoctahedron isnotelementary, because itcanbebrokenup into twopyramidswithsquarebases. Findatleastoneothernon-elementaryface-regularpolyhedron, andatleastoneelementaryface-regularpolyhedron.
Exercise 3.4.4. Supposewehavetwoface-regularpolyhedra, andoneofthefacesinthefirstpolyhedronisidenticaltooneofthefacesinthesecondpolyhedron. Wecanthengluethetwopolyhedraalongtheiridenticalfaces, yieldingonelargerpolyhedra. Forexample,startingwithacubeandapyramidwithasquarebaseandequilateralsides, andgluingthetwoalongasquarefaceineach, resultsinthepolyhedronshowninFigure 3.4.2.
(1) Is thepolyhedron that results from thegluing two face-regularpolyhedraby thisprocessnecessarilyface-regular? Explainyouranswer.
(2) Supposetheoriginaltwoface-regularpolyhedrawerebothconvex. Isthepolyhe-dronthatresultsfromthegluingnecessarilyconvex? If theanswerisyes, explainwhy, andiftheanswerisno, giveanexampletoshowwhynot.
Exercise 3.4.5. Showthatthereareinfinitelymanynon-convexface-regularpolyhedra.
3.5 EnumerationinPolyhedra
Oneofthenicefeaturesofpolyhedra(asopposedto“smooth”objects, suchasspheres)isthattheyoffersomethingstobecounted, namelythenumberofvertices, thenumberofedgesandthenumberoffaces. Givenapolyhedron, welet V , E and F, respectively, denotethenumberofvertices, edgesandfacesofthepolyhedron. Forexample, foracubewehave V = 8, andE = 12, and F = 6. Forconvenience, wewritethesethreenumbersas (V, E, F), calledthefacevector ofthepolyhedron. Hence, thefacevectorofthecubeis (8, 12, 6).
88 3. Polyhedra
Exercise 3.5.1.
(1) Whatisthefacevectorofeachoftheregularpolyhedra?
(2) Whatisthefacevectorofeachofthesemi-regularpolyhedra?
Exercise 3.5.2. Findaconvexpolyhedronsuchthatneither E nor F isdivisibleby 3.
Exercise 3.5.3.
(1) Whatisthefacevectorofapyramidoveran n-gon?
(2) Whatisthefacevectorofabipyramidoveran n-gon?
(3) Whatisthefacevectorofaprismoveran n-gon?
(Note: Youranswerforeachpartofthisexercisewillinvolve“n.”)
Exercise 3.5.4. Findabipyramidthathasthesamefacevectorastheicosahedron.
Exercise 3.5.5. Supposethataconvexpolyhedron P hasfacevector (V, E, F). Whatisthefacevectorofthedualof P?
Foreachpolyhedron, wecandetermine its facevector. It canhappen, however, that twodifferentpolyhedrahavethesamefacevector(justastwodifferentpeoplecanhavethesameheightandweight). Forexample, thetwopolyhedrashowninFigure 3.5.1 havethesamefacevectors. Ofcourse, iftwopolyhedrahavedifferentfacevectors, theycannotbethesame.
Whatcanbesaidaboutthenumbers V , E and F thatarisefrompolyhedra? Westartwiththefollowingverysimpleresult, whichfollowsfromthefactthatapolyhedronisasolidobjectinthreedimensionalspace.
3.5EnumerationinPolyhedra 89
(i) (ii)
Figure3.5.1
Proposition 3.5.1. Supposethat P isapolyhedron.
1. V ≥ 4.
2. E ≥ 6.
3. F ≥ 4.
Inordertosaymoreabout V , E and F, weneedthefollowingnotion. Ratherthansimplylookingatthenumberoffaces, namely F, wewanttolookmorefinely, andcountthenumberoffaceswith 3 edgeseach, denoted F3, thenumberoffaceswith 4 edgeseach, denoted F4, etc.Ingeneral, foreachpositiveinteger n (where n ≥ 3), let Fn denotethenumberoffaceswith nedgeseach. Similarly, foreachpositiveinteger n (where n ≥ 3), let Vn denotethenumberofverticescontainedin n edgeseach. Forexample, forthepolyhedronshowninFigure 3.4.2, wehave F3 = 4, F4 = 5, F5 = 0, F6 = 0, etc.; wealsohave V3 = 4, V4 = 5, V5 = 0, V6 = 0,etc. Thefollowingresultwillbeusefulinprovingvariouspropositionsofinterest.
Proposition 3.5.2. Supposethat P isapolyhedron.
1. F = F3 + F4 + F5 + F6 + · · · .
2. V = V3 + V4 + V5 + V6 + · · · .
3. 2E = 3F3 + 4F4 + 5F5 + 6F6 + · · · .
4. 2E = 3V3 + 4V4 + 5V5 + 6V6 + · · · .
Demonstration.
(1). Thisequationisevidentlytrue, becauseeveryfacehasatleastthreeedges(byProposi-tion 3.1.1 (3)), andisthuscountedpreciselyoneamong F3, F4, F5, etc.
(2). Thisequationisevidentlytrue, becauseeveryvertexiscontainedinatleastthreeedges(byProposition 3.1.1 (2)), andisthuscountedpreciselyoneamong V3, V4, V5, etc.
90 3. Polyhedra
(3). Thesum 3F3 + 4F4 + 5F5 + 6F6 + · · · countsalltheedgesthatarecontainedinallthefacesofthepolyhedron. However, eachedgeofthepolyhedroniscontainedinpreciselytwofaces, usingProposition 3.1.1 (1). Therefore, thesum 3F3+4F4+5F5+6F6+ · · · countseachedgetwice, andsoitequalstwicethenumberofedges. Inotherwords, wehave 3F3 + 4F4 +5F5 + 6F6 + · · · = 2E.
(4). ThisargumentisverymuchliketheoneforPart (3)ofthisproposition, usingthefactthateachedgeofthepolyhedroncontainstwovertices.
Part (3)oftheabovepropositionboilsdowntoamuchsimplerstatementinthecasewhereallthefaceshavethesamenumberofedges. Moreprecisely, if P isapolyhedronsuchthatallitsfacesare n-gons, then nF = 2E. ThereaderisaskedtodemonstratethisfactinExercise 3.5.6.Intheparticularcasewhere P hasalltriangularfaces, then 3F = 2E.
Exercise 3.5.6. [UsedinThisSection] Supposethat P isapolyhedron.
(1) Supposethatallthefacesof P are n-gons. Showthat nF = 2E.
(2) Supposethatallthefacesof P aretriangles. Showthat F isdivisibleby 2, and E isdivisibleby 3.
(3) Supposethateveryvertexof P iscontainedin q edges. Showthat qV = 2E.
BEFORE YOU READ FURTHER:
Arethereanyinterestingrelationsbetweenthethreenumbers V , E and F thatholdforallpolyhedra, or, atleast, allconvexpolyhedra? Trytolookforsuchrelationshipsyourself.Lookat thevaluesof V , E and F forvariousexamplesofpolyhedra, for instance theregularandsemi-regularpolyhedra; canyoufindanypatterns? Canyoufindanyrelationsbetween V , E and F thatholdsinallexamplesyouhaveexamined?
Thereareindeedrelationsbetweenthenumbers V , E and F thatholdforallpolyhedra. Oneexampleofsucharelationisgiveninthefollowingproposition.
Proposition 3.5.3. Supposethat P isapolyhedron.
1. E ≥ 3
2V .
2. E ≥ 3
2F.
Demonstration.
3.5EnumerationinPolyhedra 91
(1). UsingProposition 3.5.2 (4)(2), inthatorder, weseethat
2E = 3V3 + 4V4 + 5V5 + 6V6 + · · ·≥ 3V3 + 3V4 + 3V5 + 3V6 + · · · = 3(V3 + V4 + V5 + V6 + · · · ) = 3V.
(2). ThiscaseisverysimilartoPart (1)ofthisproposition; thedetailsarelefttothereader.
Itfollowsfromtheabovepropositionthatforanypolyhedron, thenumberofedgesisalwaysgreaterthanboththenumberofverticesandthenumberoffaces.A moresubstantial(andmoredifficulttodemonstrate)relationbetweenthenumbers V , E and
F thatholdsforanyconvexpolyhedronisthefollowingproposition, knownasEuler’sFormula;itisduetothegreatmathematicianLeonhardEuler (1707-1783).
Proposition 3.5.4 (Euler’sFormula). Foranyconvexpolyhedron, wehave
V − E+ F = 2.
Demonstration. Supposewearegivenaconvexpolyhedron P. Wewant tofigureoutwhatV − E+ F equals. Thefirststepinthisdemonstrationistoformtheprojectionof P, whichwenowdescribe.Consider, forexample, acube. Imaginethatthecubeismadenotofsquaresgluedtogether,
butisjustawireframe. Wecouldthenputalightrightabovethewireframecube, andapieceofpaperbelowthecube, asshowninFigure 3.5.2 (i). Thelightcastsashadowonthepaper; theshadowispicturedinFigure 3.5.2 (ii). Wecallthisshadowthe projection ofthecube. Noticethattheprojectionismadeupofedgesandvertices, andthattheseedgesandverticesdivideuptheplaneintoanumberofregions. Weobservefurtherthatthenumberofverticesintheprojectionisthesameasthenumberofverticesintheoriginalcube, namely 8, andthatthenumberofedges in theprojection is thesameas thenumberofedges in theoriginalcube,namely 12. Further, wenoticethattheprojectiondividesuptheplaneinto 6 regions(ofwhich5 are“bounded”and 1 is“unbounded”), andthatthisnumberofregionsequalsthenumberoffacesofthecube.
Wecouldsimilarlyformtheprojectionofanyconvexpolyhedron P. InFigure 3.5.2 (iii)weseetheprojectionoftheregulartetrahedron. Thesameobservationaboutnumbersofvertices,edgesandregionsthatheldfortheprojectionofthecubeholdsfortheprojectionofanyconvexpolygon. Thatis, thenumberofverticesoftheprojectionequals V , thenumberofedgesoftheprojectionequals E, andthenumberofregionsintheplaneformedbytheprojectionequals F.Therefore, tofigureoutV−E+F for P, wecanjustaswellfigureoutV−E+F fortheprojection(where V and E nowmeanthenumberofverticesandedgesrespectivelyoftheprojection, andF nowmeansthenumberofregions). Itturnsoutthattheprojectionismucheasiertoworkwiththantheoriginalpolyhedron.Wenowproceedbymodifyingtheprojectiononestepatatime. Wewillverifythateach
modificationdoesnotchangethesum V − E+ F, thoughitmightchangeindividualvaluesofV , E and F. First, weaskwhethertheedgesontheprojectionhaveanycomplete“loops.” For
92 3. Polyhedra
(i) (ii) (iii)
Figure3.5.2
example, supposetheprojectionisasshowninFigure 3.5.3 (i); thereareanumberofcompleteloopsinthisprojection, forexample, theedgeslabeled a, b, c and d formaloop. Weproceedasfollows. Supposewehavealoopinourprojection. Thenchooseoneoftheedgesthatmakeuptheloop(itdoesnotmatterwhich), andwewillremovethatedge. InFigure 3.5.3 (ii), wehaveremovedtheedgelabeled b. Theresultisanewconfigurationofvertices, edgesandregions.Whathappensto V , E and F asaresultofremovingoneedgefromaloop? Weobservethat Visunchanged, that E decreasesby 1, andthat F decreasesby 1. Hence, thenewvalueofthesumwearecalculatingis
V − (E− 1) + (F− 1) = V − E+ F.
Inotherwords, thesumV−E+F isthesameinthenewconfigurationasintheoldconfiguration.Nowconsiderthenewconfigurationofvertices, edgesandregions. Ithasfewerloopsthan
theoriginalprojection. Iftherearestillloops, thenremoveanotheredgefromoneoftheloops.Keepremovingoneedgeatatimeuntiltherearenomoreloopsleft. Afterallthenecessaryremovals, thevalueof V − E + F isstillunchanged, butwenowhaveasimplersituation, inthattherearenoloops. Forexample, weseeinFigure 3.5.3 (iii)onepossibleresultofremovingedgesfromalltheloopsinFigure 3.5.3 (i). (Therearedifferentchoicesforremovingedgesfromloopsateachstage, sotheresultingconfigurationcanvary, butitwillmakenodifference.)Wenowperformadifferent typeofmodification. Becauseour newconfigurationhas no
loops, itmusthave“free”vertices; thatis, verticesthataretheendpointsofonlyoneedge. Forexample, thevertexlabeled A inFigure 3.5.3 (iii)isafreevertex. Wethenchooseafreevertexinourconfigurationwithoutloops, andthenremoveboththefreevertexandthesingleedgeofwhichthevertexisanendpoint; weleaveinplacetheotherendpointoftheremovededge. InFigure 3.5.3 (iv), wehaveremovedthevertex A andtheedgethathas A asanendpoint. Whathappensto V , E and F asaresultofthisnewtypeofprocedure? Weobservethat V isdecreasesby 1, that E decreasesby 1, andthat F isunchanged. Hence, thenewvalueofthesumweare
3.5EnumerationinPolyhedra 93
calculatingis(V − 1)− (E− 1) + F = V − E+ F.
Inotherwords, thesum V − E + F isonceagainthesameinthenewconfigurationasintheoldconfiguration.Nowconsiderthenewconfigurationofvertices, edgesandregions. Ithasfeweredgesthanthe
originalprojection. Iftherearestilledges, thentheremuststillbefreevertices. Keepremovingoneedgeatatimeuntilthereisonlyoneedgeleft. Then, nomatterwhattheoriginalprojectionwas, thefinalconfigurationwill lookliketheoneshowninFigure 3.5.3 (v). Ateachstepofourprocedure, thevalueof V − E+ F neverchanged. Hence, ifwecancomputethevalueofV − E+ F forthefinalconfiguration, thatwillbethesamevalueasfortheoriginalprojection,andhencefortheoriginalpolyhedron. InFigure 3.5.3 (v)weseethat V = 2, that E = 1 andthat F = 1. Hence V − E + F = 2 − 1 + 1 = 2. Therefore V − E + F = 2 fortheoriginalpolyhedron, whichiswhatwewantedtoshow.
(i) (ii) (iii)
(iv) (v)
a
b
c
d
a c
A
Figure3.5.3
Euler’sFormulahasmanyuses. Forexample, wecanuseittodeducefurtherrelationsbetweenthenumbers V , E and F forconvexpolyhedra. Onesuchresultisthefollowingproposition.
Proposition 3.5.5. Supposethat P isaconvexpolyhedron.
94 3. Polyhedra
1. 4 ≤ F ≤ 2V − 4.
2. 4 ≤ V ≤ 2F− 4.
3.5EnumerationinPolyhedra 95
Demonstration.
(1). IfwemultiplyEuler’sFormulaby 2 wehave
2V − 2E+ 2F = 4.
Next, weknowfromProposition 3.5.3 (2)that 2E ≥ 3F. Ifwesubtracttwonumbersfromthesamenumber, thensubtractingthesmallernumbergivesabiggerresult. So, weseethat
2V − 3F+ 2F ≥ 2V − 2E+ 2F = 4.
Itfollowsthat2V − F ≥ 4.
Rearranging, weobtain2V − 4 ≥ F.
CombiningthislastresultwithProposition 3.5.1 (3), wederivethat
4 ≤ F ≤ 2V − 4.
(2). ThisargumentisverymuchliketheoneforPart (1)ofthisproposition, withtheroleoffacesandverticesinterchanged.
Exercise 3.5.7. ThisexerciseusesExercise 3.5.6.Supposethat P isaconvexpolyhedron.
(1) Supposethatallthefacesof P aretriangles. Findaformulaforeachof E and F intermsof V .
(2) Supposethatallthefacesof P arequadrilaterals. Findaformulaforeachof E andF intermsof V .
(3) Supposethatallthefacesof P arepentagons. Findaformulaforeachof E and F intermsof V .
Exercise 3.5.8. Suppose that P isaconvexpolyhedron, andthatall the facesof P aretriangles. Assumefurtherthat P hasatleastfivevertices. Istherealwaysabipyramidoversome n-gonthathasthesamefacevectoras P? Ifthereis, find n intermsofthenumberofverticesof P.
96 3. Polyhedra
Exercise 3.5.9. Supposethataconvexpolyhedron P isself-dual(thatis, thepolyhedronisit’sowndual). Findaformulaforeachof E and F intermsof V .
Exercise 3.5.10. Find all possible convex polyhedra P that are self-dual (as in Exer-cise 3.5.9), andallthefacesofwhicharetriangles.
Exercise 3.5.11. Supposethat P isaconvexpolyhedron. Showthat E ≤ 3F− 6.
Exercise 3.5.12. Supposethat P isaconvexpolyhedron. Showthat P mustcontainatleastone face thathaseither 3, 4 or 5 edges. (Inotherwords, thisexerciseshowsthat therecannotbeaconvexpolyhedronwithallfaceshaving 6 ormoreedges.)
Wearenow in aposition todiscuss a very interestingquestion regarding face vectors ofconvexpolyhedra. Weknowthateveryconvexpolyhedronhasafacevector (V, E, F). Canwegobackwardsinthisprocess? Thatis, supposewearegiventhreepositiveintegers (x, y, z);dothesethreenumbersnecessarilyformthefacevectorofaconvexpolyhedron? Theanswerisno. Forexample, supposeweweregiventhenumbers (5, 7, 1). Therecannotbeaconvexpolyhedronwiththesenumbersasitsfacevector, because F ≥ 4 foranyanyconvexpolygon(asinProposition 3.5.1 (3)). Howabout (5, 9, 7)? Wedonothaveaproblemwith V , E andF notbeinglargeenough, buttherestillcannotbeaconvexpolyhedronwiththesenumbersasitsfacevector, because 5 − 9 + 7 = 3, andsoEuler’sFormulaisnotsatisfied. Howabout(8, 11, 5)? Euler’sFormulaissatisfiedthistime, buttherestillcannotbeaconvexpolyhedrawiththesenumbersasitsfacevector, because 2 · 5 − 4 = 6, andyet 8 ≰ 6, sothesenumberdonotsatisfyProposition 3.5.5 (2).Theaboveexamplesshowussomeofthereasonswhythreepositiveintegers (x, y, z) might
notbethefacevectorofaconvexpolyhedron. ThefollowingPropositionsaysthatthesearetheonlypossiblethingsthatcangowrong.
Proposition 3.5.6. Supposewearegiventhreepositiveintegers (x, y, z). Thenthesenumbersarethefacevectorofaconvexpolyhedronifandonlyifthefollowingthreecriteriahold:
1. x− y+ z = 2.
2. 4 ≤ z ≤ 2x− 4.
3. 4 ≤ x ≤ 2z− 4.
3.5EnumerationinPolyhedra 97
Inotherwords, if (x, y, z) satisfytheabovethreecriteria, thentheyarethefacevectorofsomeconvexpolyhedron(possiblymorethanone); iftheydonotsatisfyallthreecriteria, thentheyarenotthefacevectorofaconvexpolyhedron. Thedemonstrationoftheabovepropositionisbeyondthescopeofthisbook.Forexample, supposewearegiventhenumbers (5, 9, 6). Itcanbeverifiedthatthesenumbers
satisfyallthreecriteriainProposition 3.5.6, andsotheymustbethefacevectorofsomepolyhe-dron. Thereaderisaskedtofindsuchapolyhedron(hint: tryconstructingpyramids, bipyramids,andthelikewith 5 vertices).
Exercise 3.5.13. Foreachofthesetsofthreenumbersgivenbelow, statewhetherornotitisthefacevectorofaconvexpolyhedron. Ifitisthefacevectorofaconvexpolyhedron,findsuchapolyhedron; ifnot, explainwhynot.
(1) (5, 10, 6).
(2) (12, 18, 8).
(3) (23, 33, 12).
(4) (10, 20, 12).
DoesEuler’sFormulaholdforallpolyhedra? Theanswerisdefinitelyno. Forexample, considerthepolyhedral“torus” showninFigure 3.5.4 (“torus”isthemathematicalnameforanythingshapedlikethesurfaceofabagel). Itisseenfromthefigurethatthefacevectorofthispolyhedronis (16, 32, 16), andhence V − E + F = 16 − 32 + 16 = 0. ThereforeEuler’sFormuladoesnotholdinthiscase. ItturnsoutthatEuler’sFormulaholdsforthosepolyhedrathatdonothave“holes”throughthem, thoughitisbeyondthescopeofthisbooktogivethedetailsofwhythisistrue.
Figure3.5.4
98 3. Polyhedra
EventhoughEuler’sformuladoesnotholdforallpolyhedra, itturnsoutthattheconceptofV − E + F isnonethelessusefulforallpolyhedra. Foranypolyhedron P, wedefinethe Eulercharacteristic of P, denoted χ(P), tobethenumber
χ(P) = V − E+ F.
Foranyconvexpolyhedra P, weknowfromEuler’sFormulathat χ(P) = 2. IfwedenotethepolyhedronshowninFigure 3.5.4 by T , thenwesawthat χ(T) = 0. Thoughwecannotgointofurtherdetailshere, weremarkthattheEulercharacteristic, anditsgeneralizations, isaveryimportantconceptinanumberofbranchesofmodernmathematics.Finally, weuseEuler’s formula togiveanotherproofofProposition 3.2.1, whichdescribes
the five platonic solids. What is interesting about this second proof is that it is entirelycombinatorial—thatis, itisbaseduponwholenumbers—andmakesnouseofgeometry.
DemonstrationofProposition 3.2.1. Suppose P isaregularpolyhedron. Thenbydefinition Pisconvex, allfacesof P areidentical; andallverticesof P arecontainedinthesamenumberoffaces(andhencethesamenumberofedges).Supposethateverypolygonof P has n edges, andthateveryvertexof P iscontainedin q
edges. ThenbyExercise 3.5.6 weknowthat nF = 2E and qV = 2E. Hence F = 2nE and
V = 2qE.
Wenowsubstitutetheaboveformulasfor F and V intoEuler’sformula, obtaining
2
qE− E+
2
nE = 2.
Dividingeverytermintheaboveequationby 2E andcancelingyields
1
q−
1
2+
1
n=
1
E.
Because E isapositivenumber, itfollowsthat
1
q−
1
2+
1
n> 0,
andtherefore1
q+
1
n>
1
2. (3.5.1)
Whatvaluesof n and q couldsatisfyEquation (3.5.1)? Thenumbers n and q arebothwholenumber. Moreover, weknowthat n ≥ 3, becauseeverypolygonhasat least 3 edges, andq ≥ 3, becauseinapolyhedroneveryvertexiscontainedinatleast 3 edges.Coulditbethat n ≥ 6? Supposethatistrue. Then 1
n≤ 1
6. ThenEquation (3.5.1)impliesthat
1
q+
1
6≥ 1
q+
1
n>
1
2,
3.6CurvatureofPolyhedra 99
whichimpliesthat1
q>
1
2−
1
6=
1
3.
Itwouldfollowthat q < 3, whichisimpossible. Hence n ≤ 5. Therefore n isoneof 3, 4 and5.A similarargumentshowsthat q isoneof 3, 4 and 5. Therearethenninepossiblecasesfor
n and q.
1. n = 3 and q = 3. Weverifythat 13+ 1
3= 2
3> 1
2. Thispolyhedronisatetrahedron.
2. n = 3 and q = 4. Weverifythat 13+ 1
4= 7
12> 1
2. Thispolyhedronisanoctahedron.
3. n = 3 and q = 5. Weverifythat 13+ 1
5= 8
15> 1
2. Thispolyhedronisatetrahedron.
4. n = 4 and q = 3. Weverifythat 14+ 1
3= 7
12> 1
2. Thispolyhedronisacube.
5. n = 4 and q = 4. Weobservethat 14+ 1
4= 1
2, andsoEquation (3.5.1)isnotsatisfied.
Hence, thereisnosuchregularpolyhedron.
6. n = 4 and q = 5. Weobservethat 14+ 1
5= 9
20, andsoEquation (3.5.1)isnotsatisfied.
Hence, thereisnosuchregularpolyhedron.
7. n = 5 and q = 3. Weverifythat 15+ 1
3= 8
15> 1
2. Thispolyhedronisaicosahedron.
8. n = 5 and q = 4. Weobservethat 15+ 1
4= 9
20, andsoEquation (3.5.1)isnotsatisfied.
Hence, thereisnosuchregularpolyhedron.
9. n = 5 and q = 5. Weobservethat 15+ 1
5= 2
5, andsoEquation (3.5.1)isnotsatisfied.
Hence, thereisnosuchregularpolyhedron.
WehavethereforeverifiedthattheregularpolyhedraarepreciselythefivepolyhedralistedinTable 3.2.1.
3.6 CurvatureofPolyhedra
Ifwethinkofthe“surfaceofapolyhedron,” wenoticethatatsomeverticesthesurfaceappearstobe“curving”morerapidly, andinotherplacesitappearstobecurvingless. (Theword“curved”mightseemstrangewhenappliedtosomethingthatismadeupofflatpolygons, butthesametermisalsoappliedtosmoothsurfaces(suchasasphere), anditisquitestandard.) Forexample,considerthesurfaceofthepolyhedronshowninFigure 3.6.1. Intuitively, thesurfaceismoresharplycurvedatthevertexlabeledA thanatthevertexlabeled B. Itwouldbenicetoquantifycurvature byassigningtoeachvertexanumberthattellsushowcurvedthesurfaceisatthevertex. A verynicemethodforsodoing, whichwenowdescribe, goesbacktoDescartes. (See[Fed82]foratranslationandexpositionofDescartes’workonpolyhedra.)
100 3. Polyhedra
A
B
Figure3.6.1
Theideaofcurvatureatavertexofapolyhedronistoseehowfartheneighborhoodofthevertexisfrombeingflat. Iftheneighborhoodisflat, thecurvatureshouldbe 0. Giventhataflatplanehasangle 360◦ aroundanypoint, weuse 360◦ asourbasisofcomparisonformeasuringcurvature. Foranypolyhedron, andforanyvertexofthepolyhedron, wedefinethe angledefectatthevertextobe 360◦ minusthesumofalltheangles(inthevariousfacesofthepolyhedron)thatcontainthevertex. Theangledefectisthecommonmeasureofcurvatureforverticesofpolyhedra. Forexample, if v isavertexinaregularoctahedron, thenthevertexiscontainedinfour 60◦ angles, andthereforetheangledefectatthevertexis
360◦ − (60◦ + 60◦ + 60◦ + 60◦) = 120◦.
Bycomparison, ifw isavertexinaregulardodecahedron, thenthevertexiscontainedinthree108◦ angles, andthereforetheangledefectatthisvertexis
360◦ − (108◦ + 108◦ + 108◦) = 36◦.
Thefactthattheangledefectatthevertexoftheregularoctahedronislargerthantheangledefectatthevertexoftheregulardodecahedroncorrespondstothefactthattheoctahedronisintuitivelymore“sharplypointed”atitsverticesthanthedodecahedron, ascanbeseenbylookingatpicturesofeach(or, evenbetter, lookingatmodelsofthem).
Exercise 3.6.1. Findtheangledefectateachoftheverticesofthefollowingpolyhedra.
(1) A cube.
(2) A regularicosahedron.
(3) ThepolyhedronshowninFigure 3.4.2 (assumingthatallthetrianglesareequilateral).
3.6CurvatureofPolyhedra 101
Itcanbeseen, usingProposition 3.1.2, thatinaconvexpolyhedron, theangledefectatanyvertexispositive. However, innon-convexpolyhedra, itispossibletohaveanegativeangledefectatavertex. Thereadershouldtrytofindanexampleofapolyhedronwithavertexthathasnegativeangledefect.Simplycalculatingangledefects isof interest, but there ismore to the story than that. In
particular, Descartesdiscoveredaverysubtlefactaboutangledefects, whichwenowstate.
BEFORE YOU READ FURTHER:
Descarteslookedatthesumofalltheangledefectsinaconvexpolyhedron. Examineafewexamplesofpolyhedra, bothregularandnon-regular, and, ineachexample, calculatethesumoftheangledefectsatallthevertices. Doyounoticeapattern?
Ifyoudidtheabovecalculationscorrectly, youshouldhavenoticedthat foreveryconvexpolyhedronthatyoutried, thesumoftheangledefectsatall theverticesis 720◦. Descartesshowedthatthisremarkableresultindeedholdsforallconvexpolyhedra. GiventhatweknowabouttheEulercharacteristic ofpolyhedra(definedinSection 3.5), wecangeneralizeDescartes’resultasfollows. (Descartes, wholivedwellbeforeEuler, wasmostlikelyunawareoftheEulercharacteristic, thoughthereissomedebateaboutthatintheliterature.)
Proposition 3.6.1 (GeneralizedDescartes’Theorem). Supposethat P isapolyhedron. Thenthesumoftheangledefectsatalltheverticesof P equals 360◦ · χ(P).
Demonstration. Suppose a1, a2, . . . , aV aretheverticesof K, andsupposethat t1, t2, . . . , tFarethefacesof K. Notethatthereare V verticesand F faces. Supposethatface tk hasCk edges.Weobservethat
C1 + C2 + · · ·+ CF = 2E;
thisequationcanbeshownverysimilarlytothedemonstrationofProposition 3.5.2 (3); weleavethedetailstothereader.Foravertex ak, welet dk denotetheangledefectat ak; thatis, wehave
dk = 360◦ − (sumoftheanglesatvertex ak).
Thesumoftheangledefects, whichiswhatwearetryingtoevaluate, istherefore d1 + d2 +· · ·+ dV . WenowuseProposition 2.3.3 andtheaboveformulafor C1 +C2 + · · ·+CF toseethat
d1+d2 + · · ·+ dV =
= [360◦ − (sumofanglesat a1)] + · · ·+ [360◦ − (sumofanglesat aV)]
= [360◦ + · · ·+ 360◦︸ ︷︷ ︸V times
]− [(sumofanglesat a1) + · · ·+ (sumofanglesat aV)]
= 360◦ · V − (sumofallanglesof P)
102 3. Polyhedra
= 360◦ · V − [(sumofanglesin t1) + · · ·+ (sumofanglesin tF)]
= 360◦ · V − [(C1 − 2)180◦ + · · ·+ (CF − 2)180◦]
= 360◦ · V − [(C1 + · · ·+ CF)180◦ − (2 · 180◦ + · · ·+ 2 · 180◦︸ ︷︷ ︸
F times
)]
= 360◦ · V − [2E · 180◦ − 360◦ · F]= 360◦ · V − 360◦ · E+ 360◦ · F= 360◦ · χ(P).
WenotethattheGeneralizedDescartes’TheoremultimatelyreliesuponEuclid’sFifthPostu-late, becauseintheproofofthetheoremweusetheformulainProposition 2.3.3 forthesumoftheinterioranglesinapolygon(whichinturnmakesuseofthefactthatthesumoftheinterioranglesinatriangleis 180◦, whichisprovedusingEuclid’sFifthPostulate).Finally, wecanuse theGeneralizedDescartes’Theorem togiveanotherdemonstrationof
Euler’sFormula(Proposition 3.5.4). Forthisnewdemonstration, weneedthefollowingobser-vationconcerningtheGeneralizedDescartes’Theorem. Recallthatourdefinitionofpolyhedra,asstatedinSection 3.1, involvesthreecriteriaonthefacesofeachpolyhedra, namelythat(1)facesaregluededge-to-edge; (2)everyedgeofafaceisgluedtotheedgeofpreciselyoneotherface; and(3)notwofaces intersectexceptpossiblyalongtheiredgeswhere theyareglued.Thethirdconditionimpliesthatapolyhedrondoesnothaveanyself-intersections. Actually, ifwelookcarefullyatthedemonstrationoftheGeneralizedDescartes’Theorem, itisseenthatwhereascriteria(1)and(2)arecrucial(inshowingthat C1 + C2 + · · · + CF = 2E), weneveractuallyusecriterion(3). Hence, theconclusionofGeneralizedDescartes’Theoremholdsevenforpolyhedrathatdonotsatisfycriterion(3); thatis, forpolyhedrainwhichfacesmightoverlapeachother, thoughwestillonlythinkofthefacesasbeinggluedtoeachotheralongtheiredges.
SecondDemonstrationofEuler’sFormula(Proposition 3.5.4). Suppose that P is a convexpolyhedron. Ourgoalistoshowthat χ(P) = 2.Choosea faceof P; call this face C. We thenstartbyexpanding C in suchaway that it
becomeswiderthantherestof P. SeeFigure 3.6.2 (i)and(ii)foranexampleofsuchstretching.(Notethatthissortofstretchingispossiblepreciselybecause P isconvex.)
Next, wecollapseallof P ontothefaceC, making P completelyflat. Bytheoriginalconvexityof P, weseethatinthecollapsedversionof P therearetwolayersoffaces: wehave C onthebottom, andthentherestof P ontopinasinglelayer. Thecollapsedversionof P isnolongerapolyhedronaswehavediscussedup tillnow, but itdoessatisfycriteria (1)and (2) in thedefinitionofpolyhedra. SeeFigure 3.6.2 (iii) for theresultofcollapsing theexampleshowninFigure 3.6.2 (ii). WhatweseeinFigure 3.6.2 (iii)looksverymuchliketheprojectionswesaw inFigure 3.5.2 (ii)and (iii), althoughweare thinkingof ithere inaverydifferentway.InFigure 3.5.2 (ii)and(iii)wethoughtofthepolyhedronasawireframewithnofaces, andthendrewtheshadowoftheframe; bycontrast, inFigure 3.6.2 (iii)wewanttothinkofthepolyhedronashavingitsfaces, andsimplycollapsingthepolyhedron, facesandall.
3.6CurvatureofPolyhedra 103
(i) (ii) (iii)
C
Figure3.6.2
Observethattheabovestretchingandcollapsingprocesschangesthegeometryof P, butitdoesnotchange V , E or F. Hence, ifwewanttoshowthat χ(P) = 2, itsufficestoworkwiththecollapsedversionof P, insteadoftheoriginal. So, fromnowon, whenwesay“P,” wewillrefertothecollapsedversion.Letusnowcalculatethesumoftheangledefectsin P. Foreachvertexof P thatisintheinterior
of C, weseethat P isflatnear P, andthereforetheangledefectiszero. Hence, theonlyangledefectsthatarenotzeroareontheboundaryofC. Let a1, a2, . . . , an denotetheverticesoftheboundaryof C. Let α1, α2, . . . , αn denotetheinterioranglesat a1, a2, . . . , an respectively,andlet β1, β2, . . . , βn denotethecorrespondingexteriorangles. (SeeFigure 2.3.6 (i) foranexampleoftheseangles.) Considervertex a1. Giventhewaythat P iscollapsedonto C, weseethatthesumoftheanglesatvertex a1 isprecisely 2α1. Hencetheangledefectat a1 is360◦− 2α1. However, wecansimplifythisexpressionas 360◦− 2α1 = 2(180◦−α1) = 2β1,usingwhatweknowaboutexterioranglesfromSection 2.3. Thesameargumentshowsthattheangledefectat a2 is 2β2, andsimilarlyfortherestoftheverticesof C. Finally, weseethatthesumofalltheangledefectsof P equals
2β1 + 2β2 + · · ·+ 2βn = 2(β1 + β2 + · · ·+ βn).
However, wenotethatProposition 2.3.3 (2)tellsusthat β1+β2+ · · ·+βn = 360◦. Itfollowsthatthesumofalltheangledefectsof P equals 720◦.On theother hand, by theGeneralizedDescartes’Theorem (Proposition 3.6.1), which as
mentionedpriortothisdemonstrationcanbeappliedto P evenwhencollapsed, weknowthatthesumoftheangledefectsof P is 360◦ · χ(P). Comparingourtwocalculationsofthesumoftheangledefects, weseethat 360◦ · χ(P) = 720◦. Itfollowsthat χ(P) = 2, whichisthesameas V − E+ F = 2.
104 3. Polyhedra
Part IISYMMETRY &PATTERNS
105
4Isometries
4.1 Introduction
Anexcellentplacetostartthestudyofsymmetry isthebook[Wey52], byHermanWeyl, oneof thegreatmathematiciansof the20thcentury. I recommendacareful readingof thefirstchapter(BilateralSymmetry)only; afterthattheauthorlapsesintosometechnicalitiesthatarebest skippedover, though inbetween the technicalparts therearephilosophical ideas (andpictures)thatarewellworthreading. WebeginwithalengthyquotefromWeyl(pp.3–6); theitalicsareintheoriginal.
“IfI amnotmistakentheword symmetry isusedinoureverydaylanguageintwomeanings. Intheonesensesymmetricmeanssomethinglikewell-proportioned,well-balanced, andsymmetrydenotesthatsortofconcordanceofseveralpartsbywhichtheyintegrateintoawhole. Beauty isboundupwithsymmetry. . . Inthissensetheideaisbynomeansrestrictedtospatialobjects; thesynonym“harmony”pointsmoretowardsitsacousticalandmusicalthanitsgeometricapplications . . .
“Theimageofthebalanceprovidesanaturallinktothesecondsenseinwhichthewordsymmetryisusedinmoderntimes: bilateralsymmetry , thesymmetryofleftandright, whichissoconspicuousinthestructureofthehigheranimals, especiallythehumanbody. Nowthisbilateralsymmetryisastrictlygeometricnotionand, incontrasttothevaguenotionofsymmetrydiscussedbefore, anabsolutelypreciseconcept . . .
“ . . . Symmetry, aswideorasnarrowasyoumaydefineitsmeaning, isoneideabywhichman[sic]throughtheageshastriedtocomprehendandcreateorder,beauty, andperfection.
108 4. Isometries
“. . . FirstI willdiscussbilateralsymmetryinsomedetail . . . Thenweshallgener-alizethisconceptgradually . . . firststayingwithintheconfinesofgeometry, butthengoingbeyondtheselimits throughtheprocessofmathematicalabstractionalongaroadthatwillfinallyleadustoamathematical ideaofgreatgenerality,thePlatonicideaasitwerebehindallthespecialappearancesandapplicationsofsymmetry. Toacertaindegreethisschemeistypicalforalltheoreticknowledge:Webeginwithsomegeneralbutvagueprinciple(symmetryinthefirstsense), thenfindanimportantcasewherewecangivethatnotionaconcreteprecisemean-ing(bilateralsymmetry), andfromthatcasewegraduallyriseagaintogenerality,guidedmorebymathematicalconstructionandabstractionthanbythemiragesofphilosophy; andifweareluckyweendupwithanideanolessuniversalthantheonefromwhichwestarted. Gonemaybemuchofitsemotionalappeal, butithasthesameorevengreaterunifyingpowerintherealmofthoughtandisexactinsteadofvague.”
Thatnicelysumsupwhereweareheading, thoughwewillnotquitemakeittothefullestpossiblelevelofabstraction, thatrequiringagreaterdegreeofmathematicalbackgroundthanweareassuminginthistext. Wewill, nonetheless, getquiteclosetoWeyl’svision.Whentheword“symmetry”isusedcolloquially, itismostofteninreferencetobilateralsym-
metry, alsoknownas left-rightsymmetry. Theroleofbilateralsymmetry inartandnature iswithoutquestionquitelarge, evenifsymmetryisnotverymuchinfavorinthecontemporaryartworld. Indeed, somanyancientculturesusedbilateralsymmetryintheirartandornamen-tationthatitishardnottowonderwhy. Isitbecausethehumanbodyisessentiallybilaterallysymmetric(atleastexternally—theinternalorgansarenotsymmetricallyplaced)? Isitbecausesymmetryhassomearchetypalsymbolism?Biologically, whyisthehumanbodyexternallybilaterallysymmetric(andwhyisitnotsym-
metricinternally)? A relatedissueisthatofleftvs.right. Isthereanyinherentdifferencebetweenthetwo(nottomentionsuperiorityofoneovertheother—recalltheoriginoftheword“sinis-ter”)? Orareleftandrightdistinguishableonlyinthattheyareoppositesofoneanother? Suchphilosophicalquestionsarefascinating, butadiscussionofthemwouldtakeusabitfarafield.Readthefirstchapterof[Wey52]forextremelythoughtfulremarksontheseissues.Theconceptofbilateralsymmetryappliestoplanarobjects, thatis, two-dimensionalobjects
(forexample, adrawing)aswellas tospatialobjects, that is, threedimensionalobjects (forexample, asculpture). Forthesakeofrelativesimplicity, wewillrestrictourattentiontoplanarobjects(exceptinSection 5.7). Notetheword“relative”intheprevioussentence. Aswewillsee, therearemoresubtletiestothestudyofsymmetry—evenofplanarobjects—thanmeetstheeye; planarobjectsareeasiertoworkwiththanspatialones, buteventheyarenottrivial.Thoughtheconceptofbilateralsymmetryisaveryfamiliarone, andmostofuswouldhaveno
troubleidentifyingwhetheranygivenobjectisbilaterallysymmetricornot, aprecisedefinitionofbilateralsymmetrytakessomethought. Imaginethatanintelligentalienlandedinyourbackyard, and, becauseitjusthappenstospeakalanguagethatyouknow, youstartexplainingtoitallsortsofthingsaboutourculture; atsomepointyouusetheword“symmetric”(referring
4.1Introduction 109
tosomethingthatisbilaterallysymmetric), andthealienasksyouformoredetails. Howwouldyouexplain it? Thereareanumberofwaysyoumightexplain theword“symmetric” to thealien(thoughonlyoneofthemwillturnouttobeusefulwhenweconsidersymmetryatitsmostgeneral).
BEFORE YOU READ FURTHER:
Thinkof variousways inwhichyoucouldexplainwhat itmeans for anobject tobebilaterallysymmetric.
ConsiderthethreeobjectsinFigure 4.1.1; twoofthemarebilaterallysymmetric, andoneisnot(thoughithasothersymmetry). SupposeyouwishtoexplaintoouralienthattheheartinFigure 4.1.1 (i)hasbilateralsymmetry. First, youcouldsaythatitlooksthesamewhenreflectedinamirror (unlike thewritingon thispage, forexample, whichwouldbebackwardswhenreflectedinamirror—unlessyouhappentobeLeonardodaVinci). Ofcourse, thisexplanationwouldnothelpyouralienifithadneverseenamirror. Second, youcouldtakethepieceofpaperwiththeheartonitandfolditinhalfalongtheverticallinethroughthecenteroftheheart,noticingthatthetwohalvesarethusseentobethesame. Thisapproachseemssatisfactory, andwouldprobablymakethingsquitecleartothealien.
(i) (ii) (iii)
Figure4.1.1
A thirdexplanationforthesymmetryoftheheartisthatifyoudrewitonapieceofverythinglass, andthenflippedtheglassoverabouttheverticallinethroughthecenteroftheheart, thedrawingoftheheartwouldlookthesameaftertheflipasbeforeit. Consideredanotherway,supposeyouplayedthefollowinggamewithsomeone. Youdrawafigureonapieceofverythinglass, andyouasktheotherpersontoclosehisorhereyes. Youtheneitherfliptheglassoverabouttheverticallineonthepieceofglassoryoudonot. Next, youaskthesecondpersontoopenhisorhereyes, andtellyouwhetherornotyouflippedtheglass. Ifyouhaddrawnanon-symmetricfigureontheglass, thesecondpersonwouldonlyhavetonotewhetherthefigurelookeddifferenttoseewhetheryouhadflippedtheglassornot. Ontheotherhand, hadyoudrawnaheart(oranyotherbilaterallysymmetricfigure, drawnsothatitslineofsymmetryistheverticallineinwhichyouflipped), thepersoncouldnottellwhetheryouhadflippedthe
110 4. Isometries
glassornot, becausetheappearanceoftheheartdoesnotchangeasaresultofaflipabouttheverticallinethroughitscenter.Thisthirdmethodofexplainingthebilateralsymmetryoftheheartmayseemthemostcom-
plicated, butit isthemostusefulforourpurposes. Wesayingeneralthataplanarobjectisbilaterallysymmetricifthereisalinesothatiftheplaneisflippedaboutthisline, theobjectappearsunchanged. (Startingtheinnextsection, wewillcallsuchaflipbythemorestandardmathematicalterm“reflection.”) InFigure 4.1.2 (i)thelineusedtodetectbilateralsymmetryfortheheartisvertical. AsseeninFigure 4.1.2 (ii), thelineusedtodetectbilateralsymmetryofanobjectneednotbevertical. Anobjectmayalsobebilaterallysymmetricwithrespecttomorethanoneline, asinFigure 4.1.2 (iii), oreveninfinitelymaysuchlines, asisthecaseofthecircle(Figure 4.1.2 (iv)).
(i) (ii) (iii) (iv)
Figure4.1.2
Thereareothertypesofsymmetrythanjustbilateralsymmetry. Withhindsight, mathemati-cianscametounderstandthatthecommonfeatureofalltypesofsymmetryisthattheycanbedetectedthroughcertaintypesof“motionsoftheplane,” ofwhichflippingisonespecialcase.Anotherexampleofthetypeof“motionsoftheplane”usefultothestudyofsymmetryisrota-tion. Notall“motionoftheplane”areuseful, however. Forexample, stretchingortearingtheplane, whileinterestinginothercontexts, isnotofuseinthestudyofsymmetry. InChapter 5wewillhaveadetaileddiscussionofthesymmetryofvariouscategoriesofplanarobjects. InthischapterwelaythegroundworkforChapter 5 bygivingadetaileddiscussionoftherelevanttypesof“motionsoftheplane.”
4.2 Isometries–TheBasics
Intheprevioussectionwesawthatonewaytodetectthebilateralsymmetryofaplanarobjectisbyflippingtheobjectinaline, andseeingwhethertheobjectappearsunchanged. Flippingtheplaneinalineisoneexampleofacertaintypeof“motionoftheplane”thatiscrucialinthestudyofsymmetry. Twootherexamplesof“motionsoftheplane”wouldberotatingtheplane90◦ clockwiseaboutsomepoint, andshiftingtheentireplane 5 inchestotheright. Moremessy
4.2Isometries–TheBasics 111
transformationssuchassquashingtheplanedowntoasinglepoint, stretchingtheplane, etc.,canalsobeimagined.Itisveryimportanttonote, however, thatweareinterestedonlyintheneteffectofa“motion,”
not“howitgetsthere.” Forexample, rotationby 90◦ clockwiseaboutsomepointhasthesameneteffectasrotationby 270◦ counterclockwiseaboutthesamepoint, andweconsiderthesetworotationstobethesame“motion.” Similarly, shiftingtheplane 10 inchestotherighthasthesameneteffectasfirstshiftingtheplane 15 inchestotheright, andthenshiftingit 5 inchestotheleft.Theword“motionof theplane” isactuallyanunfortunate term, in that itsusemightgive
thefalseimpressionthatweareinterestedinhowwedothe“motion,” ratherthanjusttheneteffect. Hence, wewillnotusetheinformalterm“motionoftheplane”anymore, andinsteadwilladheretothemorestandardmathematicalterm transformation oftheplane. Byatransformationoftheplane, wemeanaruleofassignmentthattakeseachpointintheplane, andassignsitsomepointwhereitwillendup. Forexample, shiftingtheplane 10 inchestotherighttakeseachpointintheplaneandassignsitanewlocation, namely 10 inchestotherightofitsinitialposition; firstshiftingtheplane 15 inchestotheright, andthenshiftingit 5 inchestotheleft,againtakeseachpointintheplaneandassignsitanewlocationthatis 10 inchestotherightofitsinitialposition. Althoughweashumanbeingsmightthinkofshiftingtheplane 10 inchestotherightasadifferent process thanfirstshiftingtheplane 15 inchestotherightandthenshiftingit 5 inchestotheleft, fromthepointofviewoftransformations, theprocessisirrelevant, andonlytheassignmentofpointstotheirfinallocationsisofinterestinthestudyofsymmetry. Wedenotetransformationswiththesametypeofnotationasusedforfunctions. Thatis, supposeT isatransformationoftheplane. Then, foreachpoint A intheplane, welet T(A) denotetheresultofapplyingthetransformationto A. Forexample, suppose T isthetransformationobtainedbyshiftingtheplane 10 inchestotheright. IfA isapointintheplane, then T(A) willbethepointthatis 10 inchestotherightof A. Notethat T moveseverypointintheplane 10inchestotheright, notjustsomeofthepoints.Tohelpavoidfurtherconfusionovertheterm“transformationoftheplane,” wesummarize
twoimportantpointsasfollows:
1. A transformationoftheplanetakeseachpointintheplane, andassignsittoapointwhereitendsup. Whatcountsistheneteffectofthetransformation, thatis, whereeachpointendsupinrelationtoitsinitialposition. Wemightthinkgeometricallyoftransformationsasprocesses, butthatprocessisjustforourpersonalintuitivebenefit, andhasnomathe-maticalsignificance. Twotransformationsarethesameiftheyhavethesameneteffects,eveniftheyseemdifferentasprocesses.
2. A transformationofplanetransformsthewholeplane, notjustsomepartoftheplane. Evenwhenwewillbelookingatsymmetriesofspecificobjects, andapplyingtransformationsoftheplanetothem, wealwaysneedtothinkofeachtransformationasbeingappliedtothewholeplane.
112 4. Isometries
Forgettingtheabovetwopointsisacommonmistakeamongstudentsfirstlearningaboutsym-metry, andoftenleadstoalotofconfusion. So, pleasekeepthesetwopointsinmind.Wearenotinterestedinalltransformationsoftheplane, butonlythoserelevanttothestudyof
symmetry. Ifwethinkofflippingtheplaneaboutaline, orrotatingtheplaneaboutsomepoint,weseethatbothofthesetransformationshavethenicepropertythattheydonotstretch, shrinkordistort anything. Bothof these transformationspreserve lengths, angles, sizesand shapesprecisely. Itisthispropertyofnon-distortionthatiscrucialtothestudyofsymmetry. RecallfromSection 1.3 thattheconceptofdistancebetweenpointscanbeusedtodescribeobjectssuchaslinesandcircles, andcanbeusedtodetermineangles. Hence, itshouldnotbesurprisingthatpreservingdistancesbetweenpointsisthekeytodescribingthosetransformationsoftheplanethatdonotstretch, shrinkordistort. Moreprecisely, wesaythatatransformationoftheplaneisan isometry if, foranytwopointsA and B intheplane, theirdistancebeforethetransformationequalstheirdistanceafterthetransformation. UsingournotationfordistancebetweenpointsfromSection 1.3, wesay thata transformation T of theplane isan isometry if, forany twopoints A and B intheplane, wehave d(T(A), T(B)) = d(A,B); equivalently, wecansaythat |T(A) T(B)| = |AB|. (Sometextsusetheterm“rigidmotion” tomeanwhatwecallanisometry, thoughwewillnotusethatterm.)Itisalsopossibletodefinethenotionofisometryforthreedimensional(andhigher)space,
thoughwewillbestickingtoisometriesoftheplane(exceptinSection 5.7); theword“isom-etry”willthereforealwaysrefertoanisometryoftheplane, exceptwhereotherwisenoted. Acompletelythoroughandrigoroustreatmentofisometrieswouldbeverylengthy. Inthischapterwewilldiscusssomeofthebasicideasinvolvingisometries, asmuchasisneededforourstudyofsymmetry. Isometriesare, withoutquestion, thefundamental—andunifying—conceptinthemathematicalstudyofsymmetry, andourtimelookingatisometrieswillbewellspent.Therearemanythingstobesaidaboutisometries, butthemostbasicquestionsis: canwe
figureoutallthetypesoftransformationsoftheplanethatareisometries? ThecompleteanswertothisquestionwillbegiveninSection 4.6. Inthemeantime, wecandescribeindetailthreefamiliartypesofisometries.Thesimplesttypeofisometryiscalled translation. A translationistheresultof“sliding”the
planerigidlyinagivendirection, andbyagivendistance. InFigure 4.2.1 weseetheeffectoftranslatingtheplane 3 inchestotheright. Inthisfigure, asinmanyotherfigurestocome, inordertoseetheeffectoftheisometry, wedrawsomethingontheplane, tobeabletocompareitsinitialpositionwithitsfinalposition. Theplaneitselfisblank, andifwedonotdrawanythingonit, wecannotseeanydifferenceofhowtheplanelooksbeforeandafteranisometry. Imaginetheplaneasaninfinite, verythin, perfectlysmooth, sheetofglass—ifthesheetofglassismoved,itdoesnotlookanydifferent. Instead, wetaketwosheetsofglass, oneontopoftheother. Then,wedrawthesame non-symmetric objectonbothsheetsofglass, directlyontopofeachother.Wewilltypicallydrawtheletter F, becauseitissimple, thoughanynon-symmetricobjectwoulddo. (Weusetheterm“non-symmetric”intuitivelyrightnow; wewilldiscusstheterminmoredetailinSection 5.1.) Wethenperformtheisometryononeofthesheetsofglass. Comparingtheunmovedcopyoftheobject(calledthe initialobject) withthemovedone(calledthe terminal
4.2Isometries–TheBasics 113
object), wecanobtainapictureoftheeffectoftheisometry. InFigure 4.2.1 weseelabeledtheinitial F andtheterminal F, illustratingtranslationby 3 inchestotheright. Westress, however,thatitisthewholeplanethatisbeingtranslated, notjusttheletter F.
Finitial
Fterminal
Figure4.2.1
A translationoftheplanecanbeinanydirection, andbyanyamount. Oneusefulwaytodescribeatranslationisasfollows. InFigure 4.2.2 weseetheresultoftranslatingtheplane. Inthisfigure, ratherthandrawinganinitialobjectsuchastheletter F, wesimplydrewonepoint,labeled A; thepoint A wastakentoanewpoint, labeled A ′, bythetranslation. Toseehowtheplanewastranslated, wedrewanarrowfrom A to A ′. Thearrowislabeled v. Thisarrowcompletelycharacterizesthetranslation, inthefollowingsense. Supposewehadstartedwithsomeotherpoint intheplane, say B, insteadof A; let B ′ denotethepoint towhich B wasmovedbythetranslation. Itwouldturnoutthatdrawinganarrowfrom B to B ′ wouldyieldanarrowthatisparalleltothearrowfrom A to A ′. Forourpurposeshere, twoparallelarrowsofthesamelengthareconsideredidentical. Sucharrows, whereweconsiderparallelarrowsofthesamelengthtobeidentical, arecalled vectors. Anytranslationcorrespondstoavector, calledits translationvector obtainedasjustdescribed, andanyvectordeterminesatranslation. Givenavector v, wedenoteby Tv thetranslationcorrespondingtothevector v; thatis, thetranslationobtainedbytakingeverypointintheplane, andmovingitbytheamountanddirectionofthevector v. If A isanypointintheplane, then Tv(A) istheresultoftaking A, andmovingitinthelengthanddirectionof v.
A
A’v
Figure4.2.2
A particularlynoteworthytranslationistranslationbyzerolength(itdoesnotmatterwhichdirection), calledthe trivialtranslation. Thetrivialtranslationdoesnot“moveanything,” thoughitisstillatransformationoftheplane, and, inparticular, anisometry. Recallthatatransformationoftheplaneisaruleofassignmentthattakeseachpointintheplane, andassignsitsomepointwhereitwillendup. Inthecaseofthetrivialtranslation, thetransformationtakeseachpoint
114 4. Isometries
intheplane, andassignsitthefinallocationexactlywhereitstarted. Thisisometrythat“doesnotdoanything”isextremelyimportant(justasthenumberzeroisimportantinthestudyofnumbers). Anothernameforthetrivialtranslationisthe identityisometry, anditisdenoted I.If A isanypointintheplane, then I(A) = A. A translationthatisnotthetrivialtranslationiscalleda non-trivial translation.Anotherfamiliartypeofisometryiscalled rotation. InFigure 4.2.3 weseetheeffectofrotating
theplane 90◦ clockwiseaboutthepoint A; asusual, inordertoseetheeffectofthisisometry,wedrewaletter F ontheplane, andwethenseewherethisletter F endsup. Westress, onceagain, thatitisthewholeplanethatisbeingrotated, notjusttheletter F.
Finitial Fterminal
A
Figure4.2.3
Anyrotationischaracterizedbyknowingtwothings, namelythepointaboutwhichwerotate,calledthe centerofrotation, andtheanglebywhichwerotate. Whenwerotatetheplaneaboutapoint, wecanthinkofourrotationsaseitherclockwiseorcounterclockwise. Becauseweareonly interested in thenet effectof a rotation, not theprocessof rotation, it reallydoesnotmatterwhetherweuseclockwiseorcounterclockwiserotations. Forexample, rotationby 90◦
clockwiseaboutsomepointhasthesameneteffectasrotationby 270◦ counterclockwiseaboutthesamepoint, andsoweconsider these tworotations tobe thesame isometry. Moreover,bothoftheserotationshavethesameneteffectasrotationby 450◦ clockwise. Inordertoavoidredundancy, wewillusuallymakeuseof clockwise rotations; all rotationswillbeassumedclockwise, unlessotherwiseindicated. (Sometextsusecounterclockwiserotations, andsoinanytextyouread, itisimportanttomakesurewhichdirectionofrotationisbeingused.)Thenotationfortherotationbyangle α clockwisewithcenterofrotation A isdenoted RA
α ;ifitisnotimportanttodenotethecenterofrotation, wesometimeswrite Rα. Forexample, therotationshowninFigure 4.2.3 canbewrittenas RA
90◦ . Ifwewanttospecifyacounterclockwiserotation, wewillusenegativeangles. Forexample, therotationshowninFigure 4.2.3 couldalsobewrittenas RA
−270◦ . Insteadofusingdegreestodescribeangles, weoftendescriberotationintermsoffractionsofawhole 360◦ rotation. Forexample, arotationby 90◦ isthesameas 1/4ofawhole 360◦ rotation. Hence, wecanalsowritetherotationshowninFigure 4.2.3 as RA
1/4.
4.2Isometries–TheBasics 115
Ingeneral, weusethenotation RA1/n, where n isawholenumber, tomeanaclockwiserotation
by 360◦/n, that is, a rotationby 1/n ofawhole 360◦ turn. Wecommonlycall rotationby180◦ a halfturnrotation (orjust halfturn). Rotationbyangle 0◦, orequivalentlybyanywholenumbermultipleof 360◦, isreferredtoasthe trivialrotation, andit is thesameisometryastheidentityisometry I mentionedpreviously. Notethat 0◦ isamultipleof 360◦, because 0◦ =0·360◦, sowecansimplystatethatatrivialrotationisonewheretheangleisamultipleof 360◦,where“multiple”inthiscontextwillalwaysmeanbyawholenumber. (Itmayseemstrange,buttranslationbyzeroisindeedthesameisometryasrotationbyzero.) A rotationthatisnotthetrivialrotationiscalleda non-trivial rotation.
Exercise 4.2.1. Drawtheeffectontheletter R showninFigure 4.2.4 astheresultofrotatingtheplaneby 60◦ clockwiseabouteachofthepointsshown. (Therewillbefouranswers,oneforeachpoint.)
BD
C
A R
Figure4.2.4
Thereisaveryinterestingdifferencebetweentranslationsandrotations. Whenwetranslatetheplanebyanyamountotherthanzero, nopointendsupwhereitstarted. Bycontrast, whenwerotatebyanyangleotherthanamultipleof 360◦, thereisalwaysone(andonlyone)pointthatdoesendupwhereitstarted, namelythecenterofrotation. A pointthatendsupwhereitstartedafterwedoanisometryoftheplaneiscalleda fixedpoint oftheisometry. If R isanisometry, andif X isapointintheplane, then X isafixedpointof R preciselyif R(X) = X.Usingthisterminology, weseethatanon-trivialtranslationhasnofixedpoints; thatanon-trivialrotationhasonefixedpoint; andthattheidentityisometryhaseverypointasafixedpoint.Althoughanon-trivial translationhasnofixedpoints, observethatanylinethat isparallel
tothedirectionoftranslationistakentoitselfbythetranslation. Forexample, iftheplaneistranslated 5 inchestotheright, thenanyhorizontallineistakenontoitselfbythetranslation;eachnon-horizontallineisnottakenontoitselfbythistranslation. A linethatistakenontoitselfbyanisometryoftheplaneiscalleda fixedline oftheisometry. A fixedlineisalinethatistakenontoitself; theindividualpointsofthefixedlineneednoteachbetakenontothemselves.
116 4. Isometries
Theidentityisometryhaseverylineintheplaneasafixedline. Donon-trivialrotationshavefixedlines? Theanswerdependsupontheangleofrotation. A rotationby 180◦ (oranymultipleof 180◦)takeseverylinecontainingthecenterofrotationontoitself, soalltheselinesarefixedlinesoftherotation; linesthatdonotcontainthecenterofrotationarenotfixed. Ontheotherhand, arotationthatisnotbyamultipleof 180◦ hasnofixedlines.Thethirdtypeofisometrywewishtoexamineiscalled reflection. Reflectionisthemathemat-
icaltermforflippingtheplaneinaline. InFigure 4.2.5 weseetheeffectofflippingtheplaneinthelinelabeled n; asusual, wedrewaletter F ontheplaneasaninitialobject, andwethenseewherethis F endsupasaresultofthereflection. Westress, asalways, thatitisthewholeplanethatisbeingreflected, notjusttheletter F.
initial
terminalF
Fn
Figure4.2.5
A reflectionischaracterizedbythelineinwhichtheplaneisflipped, calledthe lineofreflec-tion (alsoknownasthe lineofsymmetry or mirrorline). Thenotationforareflectioninline nisMn; ifwehaveanumberoflines, forexample L1, . . . , Ls, thenwewillwritethereflectionsintheselinesas M1, . . . ,Ms whenthemeaningisclear.Althoughwethinkofreflectioninalineasflippingtheplaneaboutthatline, thereisanother
wayofthinkingaboutreflectionthatcapturestheideabetter(especiallyifwewanttocomparereflectionoftheplaneinalinewithreflectionofthreedimensionalspaceinaplane—whichisreflectioninamirror). Recallthatwhatcountsinanisometryisonlyitsneteffect, thatis,whereeachpointoftheplaneendsupinrelationtoitsinitialposition, andnotthegeometricprocessusedtovisualizetheisometry(forexample, flippingtheplaneinthecaseofreflection).LetuslookatFigure 4.2.5. Picksomepointintheinitial F, andthenfinditscorrespondingpointintheterminal F; forexample, wewillpickthepointattheverybottomrightoftheinitial F.Whatistherelationofthispointintheinitial F andthecorrespondingpointintheterminal F?InFigure 4.2.6 welabelthechosenpointontheinitial F by A, anditscorrespondingpointontheterminal F by A ′. Wecanthendrawthelinesegment AA ′, asshowninthefigure. Thecrucialobservationisthat AA ′ isperpendiculartothelineofreflectionm, andthatthepointsA and A ′ areeachthesamedistancefromtheline m, thoughonoppositesidesofit. Thatis,ifwelabelthepointofintersectionof AA ′ and m by O, thenthelengthsof AO and A ′ O
4.2Isometries–TheBasics 117
areequal. Whatwehavesaidaboutthepoint A alsoholdsforanyotherpointintheinitialF. Therefore, giventheline m andtheinitial F, insteadofobtainingtheterminal F byflippingtheplaneabout m, wecouldhavefoundtheterminal F bytakingeachpointintheinitial F,drawingaperpendicularlineto m fromthepoint, andthenlocatingthecorrespondingpointontheterminal F bycontinuingtheperpendicularpastm thesamedistancewewentfromthestartingpointto m. Bydoingthisprocesstosufficientlymanypointsontheinitial F (oranyotherinitialobject), wecouldconstructtheterminalobject. Thismethodalsoholdsinthreedimensionalspace. Whenyoulookatyourselfinthemirror, yourimageisasfarbehindthemirrorasyouareinfrontofit.
initial
terminalF
Fm
A
A’
O
Figure4.2.6
Exercise 4.2.2. Drawtheeffectontheletter R showninFigure 4.2.7 astheresultofre-flectingtheplaneineachofthelinesshown. (Therewillbethreeanswers, oneforeachline.)
Tocomparereflectionswithtranslationsandrotations, recallthenotionoffixedpoints. Inareflection, everypointonthelineofreflectionisfixed, butnootherpoint isfixed. This factcontrastswithnon-trivial translations, whichhavenofixedpoints, andnon-trivial rotations,whichhaveonefixedpointeach. Theidentityisometryhasallpointsfixed. Observethatunliketranslationsandrotations, whichcanbetrivialornot(thatis, therearetranslationsandrotationsthatequaltheidentityisometry), thereisnotrivialreflection; thatis, noreflectioncanequaltheidentityisometry.
Exercise 4.2.3. Suppose m isalineintheplane. Describeallfixedlines ofthereflectionMm.
118 4. Isometries
Rm
p
n
Figure4.2.7
Exercise 4.2.4. [UsedinSections 4.5, 4.3 and4.6, andAppendixA] Supposethat A andB aredistinctpointsintheplane. Let m betheperpendicularbisectorof AB.
(1) Showthat Mm takes A to B andtakes B to A, andthat Mm istheonlyreflectionoftheplanetodoso.
(2) Showthat Mm fixesanypointthatisequidistantto A and B.
Thereisanotherimportantdistinctionbetweenthetranslationsandrotationsontheonehand,andreflectionsontheotherhand. InFigure 4.2.1 weseetheeffectofatranslationontheletterF; inFigure 4.2.3 weseetheeffectofarotationontheletter F. Certainly, inFigure 4.2.1 theterminal F stilllooksjustlikealetter F. InFigure 4.2.3 theterminal F doesnotlookpreciselyliketheletter F usuallydoes, butifyouturnyourhead(orthepage)justtherightamount, theterminal F doeslooklikeastandardletter F. Bycontrast, inFigure 4.2.5, whichshowstheeffectofareflectionontheletter F, nomatterhowyouturnyourhead, theterminal F doesnotlookright. Thereflection“reverses”theletter F (andanyotherinitialobject), andthusitdoesnotlooklikeastandardletter F. Theterminal F looks like themirror imageof the initial F. Theformalterminologyweuseisthattranslationsandrotationsare orientationpreserving, whereasreflectionsare orientationreversing.Wehavesofardiscussedthreeparticulartypesofisometries: translations, rotationsandre-
flections. Arethereanyothertypesofisometries, ordothesethreetypesincludeallisometries?Intuitively, itishardtoimagineanyothertypeofisometry, butthatisnotarigorousargumentthatwoulddemonstratethatthethreetypesofisometriesaretheonlytypesthatexist. Toobtainabetterfeelforthisquestion, weneedtolookatisometriesfromaslightlydifferentpointofview, asdiscussedinSection 4.3.
4.3RecognizingIsometries 119
4.3 RecognizingIsometries
InSection 4.2, wediscussedisometriesastransformationsoftheplane. Toseewhateffectaparticularisometryhas, wewoulddrawanobjectintheplane, suchastheletter F, andseewhathappenedtotheobjectasaresultoftheisometrybycomparingtheinitialobjectwiththeterminalobject. Wenowwanttotakea“backwards”lookatisometries. Supposetwopeopleweretoplaythefollowinggame. Onepersondrawsaletter F onablackboard, tobeusedasaninitialobject. Thesecondpersonthencloseshereyes. Thefirstpersonchoosessomeparticularisometry, performstheisometry, anddrawstheterminal F ontheboardthatresultsfromapplyingtheisometry. Thefirstpersontheneraseseverythingontheboardotherthantheinitial F andtheterminal F. Thesecondpersonnowopenshereyes, andlooksatthetwoletters F ontheboard. Canthesecondpersonfigureoutwhatisometrythefirstpersonused? Inotherwords,insteadoftakinganisometryandseeingwhatitseffectis, thesecondpersonseestheeffectoftheisometry, andtriestofigureoutwhattheisometryis.Itwillturnoutthatthesecondpersoncanalwaysfigureoutwhatisometrythefirstperson
used; howthesecondpersondoessoisthesubjectofthepresentsection, thoughthecompleteanswerwillbegivenonlyinSection 4.5. Thereis, however, onecaveat. Supposethefirstpersondoesthefollowing. First, shetranslatestheplane 15 inchestotheright. Then, beforethesecondpersonopenshereyes, shetranslatestheplane 5 inchestotheleft, anddrawstheterminal Fafterdoingbothtranslations. Theneteffectwillbethattheterminal F liesexactly 10 inchestotherightoftheinitial F. Thefirstpersontheneraseseverythingontheblackboardotherthantheinitialandterminalletters F. Whenthesecondpersonopenshereyesandtriestofigureoutwhatisometrywasused, shewouldmostlikelythinkthattheisometryusedwastranslationoftheplaneby 10 inchestotheright. Thereisnowaythatthesecondpersoncouldguessthatthefirstpersonstartedbytranslatingtheplane 15 inchestotheright, andthentranslatingtheplane5 inchestotheleft. Theonlythingthatthesecondpersoncanfigureoutistheneteffectofwhatthefirstpersondid, nottheparticularprocess. However, asmentionedinSection 4.2, itisonlyneteffectthatisofinterestinourdiscussionofisometries.Letusnow rephraseourquestion. Supposewehave two identical letters F on theplane,
onelabeledastheinitialobject, andonelabeledastheterminalobject. Canwefindasingleisometryoftheplanethatwouldhavetheeffectoftakingtheinitial F totheterminal F? (Itmightwellbeaskedwhetherwecanfindmorethanoneanswer, butitwillturnoutthatthereisnevermorethanone; thisfactisprovedrigorouslyinAppendix A,butwewillnotgointothedetailshere.)Letusstartbylookingatsomeparticularcases. First, considertheinitialandterminalletters
F showninFigure 4.3.1.
Here it seems fairlyclear thatwecanfindasingle isometryof theplane thatwouldhavetheeffectof taking the initial F to the terminal F, namelya translation. Tofigureoutwhichtranslation, labelsomepointontheinitial F bytheletter A; thenlabeltheexactcorrespondingpointontheterminal F bytheletterA ′. InFigure 4.3.2 wehavetakentheleft-mostpointonthebottomofeachletter F asthepointbeinglabeled. Wecanthentake v tobethevectorfrom A
120 4. Isometries
Finitial F
terminal
Figure4.3.1
to A ′. Thetranslation Tv istheisometrywearelookingfor; thatis, ithastheeffectoftakingtheinitial F totheterminal F. Observethatitwouldnothavemadeanydifferencehadwechosenadifferentpoint A intheinitial F, aslongas A ′ isthepointintheterminal F thatcorrespondstoourchoiceof A.
Finitial
A’
A
vFterminal
Figure4.3.2
Next, considertheinitialandterminalletters F showninFigure 4.3.3. Isthereasingleisometrythatwouldhavetheeffectoftakingtheinitial F totheterminal F?
F Finitial terminal
Figure4.3.3
Letusconsiderthethreetypesofsymmetrieswithwhichwearefamiliar, namelytranslations,rotationsandreflections. A translationcouldnotpossiblytaketheinitial F to theterminal FinFigure 4.3.3, becausethelatterisatananglewiththeformer. A reflectionalsocouldnotpossiblytaketheinitial F totheterminal F, becauseareflectionisorientationreversing, and
4.3RecognizingIsometries 121
yettheterminal F isnotreversed. Hence, theonlytypeofisometrywithwhichwearefamiliarthatcouldpossiblyworkisarotation. Itturnsoutthatarotationdoesindeedwork. Tofindtherotationthatworks, weneedtofinditscenterofrotationandangleofrotation.Westartbyfindingthecenterofrotation. Aswasthecaseinthepreviousexample, weproceed
bylabelingsomepointontheinitial F bytheletter A, andlabelingthecorrespondingpointontheterminal F bytheletterA ′. Whereverthecenterofrotationweareseekingislocated, itmustbeequidistanttothetwopoints A and A ′; thatis, ifapoint X intheplaneisthesoughtaftercenterofrotation, then d(X,A) = d(X,A ′). WenowmakeuseofProposition 1.3.1, whichsaysthatapoint X intheplanehas d(X,A) = d(X,A ′) ifandonlyifitisontheperpendicularbisectorof AA ′. Hence, wecannarrowoursearchforthecenterofrotationtothosepointsontheperpendicularbisectorofAA ′. SeeFigure 4.3.4. Howdoweknowwhichpointonthislineisthecenterofrotation? Thetrickistorepeatourprocedurewithanotherpairofcorrespondingpoints, say B and B ′. Thecenterofrotationisalsoontheperpendicularbisectorof BB ′. ThesetwoperpendicularbisectorsareshowninFigure 4.3.5. Noticethatthesetwolinesintersectinpreciselyonepoint. Becausethecenterofrotationmustbeonthetwoperpendicularbisectors,itmustbepreciselythepointwherethetwoperpendicularbisectorsintersect. Wehavethereforefoundthecenterofrotation. ItturnsoutthatitdoesnotmatterwhichpointsA and B wechooseon the initial F; wewillalwaysobtain thesamecenterof rotation, labeled O in thefigure.Finally, tofindtheangleofrotation, justmeasuretheanglebetweenfromthelinesegmentOA
tothelinesegment OA ′; itwouldworkjustaswelltouse OB and OB ′. Alltold, wehavenowdeterminedtherotationthattakestheinitial F totheterminal F.
FFinitial terminal
AA’
Figure4.3.4
Asourthirdexample, considertheinitialandterminalletters F showninFigure 4.3.6. Onceagainweaskwhetherthereisasingleisometrythatwouldhavetheeffectoftakingtheinitial Ftotheterminal F.
122 4. Isometries
FFinitial terminal
AA’
BB’
O
Figure4.3.5
F
Finitial
terminal
Figure4.3.6
Inthiscase, weseethattheterminal F hasreversedorientationwhencomparedtotheinitialF. Hence, theonlytypeofisometrywithwhichwearefamiliarthatcouldpossiblyworkisareflection. Itturnsoutthatareflectiondoesindeedwork. Howdowefindthelineofreflection?Asbefore, westartby labelingsomepointon the initial F by the letter A, and labeling thecorrespondingpointontheterminal F bytheletter A ′. UsingExercise 4.2.4 (1), weknowthatthelineofreflection, ifthereisone, mustbetheperpendicularbisectorofAA ′. SeeFigure 4.3.7.
4.3RecognizingIsometries 123
Ifanyotherpairofcorrespondingpointsontheinitial F andterminal F ischosen, say B and B ′,thentheperpendicularbisectorof BB ′ isseen(inthecaseofFigure 4.3.6)tobethesamelineastheperpendicularbisectorof AA ′. Wehavethereforefoundthedesiredlineofreflection.
F
Finitial
terminal
A
A’
B
B’
Figure4.3.7
Exercise 4.3.1. IneachofthethreepartsofFigure 4.3.8 areshowninitialandterminalletters F, obtainedbyusingan isometry. Foreachof the threecases, statewhat typeofisometrywasused. Moreover, iftheisometryisarotation, indicateitscenterofrotation;if the isometry isa translation, indicate the translationbyanarrow; if the isometry isareflection, indicatethelineofreflection.
Asourfinalexample, considertheinitialandterminalletters F showninFigure 4.3.9. Isthereasingleisometrythatwouldhavetheeffectoftakingtheinitial F totheterminal F?
Asinthepreviousexample, weseethattheterminal F hasreversedorientationwhencom-paredtotheinitial F. Hence, notranslationorrotationcouldwork. A reflectionmightseemlikeagoodbet, butthattoodoesnotwork. InFigure 4.3.10 weseetwopairsofcorrespondingpointsontheinitial F andterminal F, labeledA andA ′, and B and B ′. Weseeinthiscasethattheperpendicularbisectorof AA ′ isnotthesamelineastheperpendicularbisectorof BB ′. Ifareflectiontooktheinitial F totheterminal F, thenitwouldneedtohavebothperpendicularbisectorsasitslineofreflection, whichmakesnosense. Hence, thereisnoreflectionthattakestheinitial F totheterminal F.
Whatcanwesayasaresultofthislastexample? Wewouldhavetodrawoneoftwopos-sibleconclusions: eitherthattwoidenticalcopiesoftheletter F canbedrawnintheplanein
124 4. Isometries
(iii)
(i) (ii)
FterminalF
initial
FFinitial
Finitial
Fterminal
terminal
Figure4.3.8
Finitial
Fterminal
Figure4.3.9
suchaway thatno isometryof theplane takesone to theother, or, alternatively, that thereareisometriesotherthanthethreetypeswehavediscussedsofar(translations, rotationsandreflections). Whichofthesepossibilitiesisthecorrectone? Wewillseetheanswertothisques-tioninSection 4.5. First, however, wewillneedaveryimportanttool, whichisdescribedinSection 4.4.
4.4CombiningIsometries 125
FF
initial
terminalA
A’
B
B’
Figure4.3.10
4.4 CombiningIsometries
Justasnumbersbecometrulyusefulonlywhenwecanadd, subtract, multiplyanddividethem,thetransformationsoftheplanediscussedabovebecomemuchmoreinterestingwhenweseehowtocombinethem. Ifwethinkofatransformationoftheplaneasawayofmovingthepointsintheplane, thenifwearegiventwotransformations, wecancombinethembyfirstdoingoneandthendoingthesecond. Suppose P and Q aretransformationsoftheplane; welet Q ◦ P
denotethecombinedtransformationobtainedbyfirstdoing P andthendoing Q. WerefertoQ ◦ P asthe composition of P and Q. (Thenotation Q ◦ P mayseem“backwards”atfirstglance, inthatitmeansdoing P firstratherthanQ first, butthisnotationisverystandardinthemathematicalliterature, andisalsoquiteconvenientinsomesituations, sowewillstickwithit.) Thephrases“P followedby Q”and“Q following P”meanthesamethingas Q ◦ P. Onewayofobtainingafeelforthenotation Q ◦ P isbylookingatwhat Q ◦ P doestoapointA intheplane. Theresultofapplyingtheisometry Q ◦ P tothepoint A wouldbedenoted(Q ◦ P)(A). However, weknowthatthepoint (Q ◦ P)(A) isobtainedbyfirstdoing P to A,resultinginthepoint P(A), andthendoing Q tothat, resultinginthepoint Q(P(A)). Hence,weseethat (Q ◦ P)(A) = Q(P(A)).Beforewelookatexamplesofcompositionsofisometries, weneedtoaskthefollowingques-
tion: thecompositionoftwoisometriesisatransformationoftheplane; isitalsoanisometry?Fortunately, theanswerisyes. Intuitively, thisassertionseemsreasonable, becausedoingeachoftwoisometriesindividuallydoesnotchangedistancesbetweenpointsintheplane, sodoingthemconsecutivelyshouldnotchangedistancesbetweenpoints. UsingournotationfordistancebetweenpointsfromSection 1.3, wecanwriteoutourdemonstrationformallyasfollows.
Proposition 4.4.1. Supposethat P and Q areisometries. Then Q ◦ P isanisometry.
Demonstration. Supposethat A and B arepointsintheplane. Weneedtoshowthattheirdis-tancebeforethetransformationQ ◦ P equalstheirdistanceafterthetransformation. Because Pisanisometry, weknowthat d(P(A), P(B)) = d(A,B). Because Q isanisometry, weknow
126 4. Isometries
that d(Q(P(A)), Q(P(B))) = d(P(A), P(B)). Itthenfollowsthat d(Q(P(A)), Q(P(B))) =d(A,B). However, we know that Q(P(A)) can be rewritten as (Q ◦ P)(A), and thatQ(P(B)) canberewrittenas (Q ◦ P)(B). Puttingourobservationstogether, wededucethatd((Q ◦ P)(A), (Q ◦ P)(B)) = d(A,B). Wehavethereforeshownpreciselywhatisneededtoseethat Q ◦ P isanisometry.
Letuslookatsomeexamplesofcompositionofisometries. InFigure 4.4.1 weseefourlineslabeledm, n, k and l, withthelinesallintersectinginapointA. Letusstartwithaverysimplecomposition, namely RA
1/4 ◦ RA1/2. Thiscompositionmeansfirstrotatingtheplaneclockwiseby
1/2 ofawholerotationcenteredatA, andthenrotatingtheplaneclockwiseby 1/4 ofawholerotationcenteredatA. Theneteffectofthiscompositionisclearlyrotatingtheplaneclockwiseby 3/4 ofawholerotationcenteredat A. Insymbols, wehave RA
1/4 ◦ RA1/2 = RA
3/4.
A
n
m
k
l
Figure4.4.1
Next, letustryaslightlymorecomplicatedcomposition, namely Mm ◦ RA1/2. Hereitwould
behardtoguesstheanswer, aswedidinthepreviouscomposition. Rather, wewillcalculatethecompositionbydrawinganobjectontheplane, forexampletheletter F, andthenseeingwhathappenstotheletter F asaresultofdoingeachofthetwoisometriesinthegivenorder.Intheleft-mostpartofFigure 4.4.2, weseethataletter F hasbeendrawn. (Wecouldjustaswellhavedrawntheletter F anywhereelseintheplane.) InthemiddlepartofFigure 4.4.2,weseetheresultofdoing RA
1/2 totheplane. Itisimportanttoobservethatwhiletheletter F
hasbeenmovedasaresultofdoing RA1/2, thelinesofreflectionhavenotmoved; theyarefixed
linesofreference, andarenotpartoftheplanethatistransformedwhenwedoanisometry.Hence, whenwereferto Mm, forexample, wewillalwaysbereferringtothesameline, nomatterwhatotherisometrieswemighthavedonepreviously. (Ifthemeaningofsymbolssuchas“Mm”weretodependuponwhatcamebeforeit, thenthesamesymbolswouldmeandifferentthingsindifferentsituations, whichwouldleadtomuchconfusion.) Intheright-mostpartofFigure 4.4.2, weseetheresultofdoing Mm totheletter F inthemiddlepartofthefigure. Thecomposition Mm ◦ RA
1/2, whichwearetryingtocompute, thereforetakestheletter F shownintheleft-mostpartofthefigure, andmovesittowhereitisshownintheright-mostpartof
4.4CombiningIsometries 127
thefigure. Now, byProposition 4.4.1, weknowthatthecomposition Mm ◦ RA1/2 isitselfan
isometry. So, weneedtofinda single isometrythatwouldtaketheletter F intheleft-mostpartofthefiguretotheletter F intheright-mostpartofthefigure. TheanswerisseentobeMk. Wethereforeseethat Mm ◦ RA
1/2 = Mk.
l l
k
Am
n
l
k
F
Am
n
Fk
Am
nF
R1/2
AMm
R1/2
AMm Mk° =
Figure4.4.2
Inpractice, it ispossible todo theabovecompositionabitquickerbydoing itall inonedrawing, butlabelingeachstep, asshowninFigure 4.4.3. Ifyoudocompositionsbythisquickermethod, makesureyoulabeleachstep; otherwise, itwillbeimpossibleforyoutogobackandfigureoutwhatyoudid(orforanyoneelsetounderstandwhatyoudid).
Weareallfamiliarwithsomeofthebasicpropertiesofadditionandmultiplicationofnumbers,forexamplethattheorderofadditionormultiplicationdoesnotmatter; thatis, wealwayshavea+b = b+a foranynumbers a and b, andsimilarlyformultiplication. WewilldiscusssuchpropertiesinmoredetailinChapter 6. Doescompositionofisometriessatisfyallthesamebasicpropertiesasadditionandmultiplicationofnumbers? Unfortunately, theanswerisno.ReferringonceagaintoFigure 4.4.1, letusnowcomparethetwoexpressions RA
1/4 ◦ Mm
and Mm ◦ RA1/4. We leave it to the reader to verify (usingdrawings similar to that shown
inFigure 4.4.2) that RA1/4 ◦ Mm = Ml andthat Mm ◦ RA
1/4 = Mn. Hence, theorderofcompositionofisometriesdoesmatter, incontrasttoadditionofnumbers. Itisnotthecasethatordermatterswiththecompositionofeverytwopossibleisometries, forexampleMm ◦ Mk =RA1/2 and Mk ◦ Mm = RA
1/2, butitisthecasethatordersometimesmatters. Hencewedonot
128 4. Isometries
Am
n
l
k
F F
F
1
2
3
Figure4.4.3
haveageneralruleforcompositionofisometriesanalogoustothegeneralrule a+ b = b+ a
fornumbers.Ontheotherhand, somepropertiesofadditionandmultiplicationofnumbersdoholdfor
compositionofisometries. Forexample, weknowthat a + (b + c) = (a + b) + c foranynumbers a, b and c. Letus lookatanexampleofa similarcalculation forcompositionofisometries. StillreferringtoFigure 4.4.1, considerthetwoexpressions Mk ◦ (RA
1/4 ◦ Mn) and
(Mk ◦ RA1/4) ◦ Mn. Computingeachoftheseexpression, wesee
Mk ◦ (RA1/4 ◦ Mn) = Mk ◦ Mm = RA
1/2
and(Mk ◦ RA
1/4) ◦ Mn = Ml ◦ Mn = RA1/2.
(Weleaveittothereadertoverifyeachstepofthesecalculations.) ThegeneralversionofthispropertyisgivenasPart (2)ofthefollowingproposition.
Proposition 4.4.2. Supposethat P, Q and R areisometries.
1. P ◦ I = P and I ◦ P = P.
2. (P ◦ Q) ◦ R = P ◦ (Q ◦ R).
Demonstration.
(1). Wewillshowthat P ◦ I = P; thefactthat I ◦ P = P canbeshownsimilarly, thedetailsbeinglefttothereader. Let A beapointintheplane. Wewillshowthat (P ◦ I)(A) = P(A).BecauseA waschosenarbitrarily, itwillthenfollowthattheisometry P ◦ I equalstheisometryP, becausetheydothesamethingtoeverypointintheplane. Letusexamine (P ◦ I)(A). Aswehaveseenpreviously, wemayrewritethisexpressionas P(I(A)). Wealsoknowthat I(A) = A.Puttingtheselasttwoobservationstogether, weseethat (P ◦ I)(A) = P(I(A)) = P(A), whichiswhatwewantedtoshow.
4.4CombiningIsometries 129
(2). Let A beapointintheplane. Wewillshowthat ((P ◦ Q) ◦ R)(A) = (P ◦ (Q ◦R))(A). Because A waschosenarbitrarily, itwillthenfollowthattheisometry (P ◦ Q) ◦ R
equalstheisometry P ◦ (Q ◦ R), becausetheydothesamethingtoeverypointintheplane.Startingwiththelefthandsideoftheequationwewishtoprove, wecompute
((P ◦ Q) ◦ R)(A) = (P ◦ Q)(R(A)) = P(Q(R(A)))
= P((Q ◦ R)(A)) = (P ◦ (Q ◦ R))(A).
Proposition 4.4.2 mightnotseemveryimpressive, butitwillbeused(sometimesexplicitly,andmoreoftenimplicitly)throughoutourdiscussionofisometriesandsymmetry. OneuseofPart (2)ofthepropositionthatwementionnowisthatitallowsustowriteexpressionssuchasP ◦ Q ◦ R unambiguously. Thatis, supposewehavethreeisometries, namely P, Q and R,andtwopeopleareaskedtocomposethem, inthegivenorder(whichiswhattheexpressionP ◦ Q ◦ R wouldmean). Giventhatwecanonlycomposetwoisometriesatatime, onepersonmightcompute P ◦ Q ◦ R bydoing (P ◦ Q) ◦ R, and theotherpersonmightcomputeP ◦ Q ◦ R bydoing P ◦ (Q ◦ R). BecauseofPart (2)oftheproposition, weareassuredthatbothpeoplewillobtainthesameanswer, andthereforeitisnotambiguoustowrite P ◦ Q ◦ R
Weend this sectionwith a comment. This section is ratherbrief, andcontainsone fairlysimple idea, namely forming the compositionof two isometriesbyfirst doingone isometryandthendoingtheother. Donotletthesimplicityofthisideafoolyou. Theideaofformingthecompositionof isometries is thecrucialstepthatallowsforamathematical treatmentofsymmetry, and forvarious substantial resultsabout the symmetryofornamentalpatterns, asdiscussedinChapter 5. Indeed, itmightbesaidthatwechoosetostudysymmetryintermsofisometriespreciselybecauseisometriescanbecomposed. Bycomposingisometries, wewillbeabletoviewthevarioussymmetriesofanobjectnotasisolatedthings, butasthingsthatinteractwitheachother, anditispreciselythisinteractionthatwillleadtointerestingresults.
Exercise 4.4.1. ReferringtoFigure 4.4.1, computethefollowingcompositions; thatis, foreachofthefollowingexpressions, findasingleisometrythatisequaltoit.
(1) RA1/4 ◦ RA
1/3.
(2) RA1/4 ◦ Mn.
(3) Mn ◦ Mm.
(4) Ml ◦ RA3/4 ◦ Mn.
(5) RA1/4 ◦ Mm ◦ RA
1/2.
For futureuse, wegive the followingproposition, whichsummarizes theeffectofcompo-sitionofisometriesonthepreservationorreversaloforientation. Thisproposition, Part (3)of
130 4. Isometries
whichintuitivelysays that twonegativesmakeapositive, certainlyseemsplausible, andweomitadetaileddemonstration(whichwouldrequireamoretechnicallysophisticateddefinitionoforientationpreservingandreversingthanwehavegiven).
Proposition 4.4.3. Supposethat P and Q areisometriesoftheplane.
1. If P and Q arebothorientationpreserving, then P ◦ Q isorientationpreserving.
2. If P isorientationpreservingandQ isorientationreversing, orvice-versa, then P ◦ Q isorientationreversing.
3. If P and Q arebothorientationreversing, then P ◦ Q isorientationpreserving.
4.5 GlideReflections
Section 4.3 endedwithaquery: theexampleshowninFigure 4.3.9 demonstratedthateitherwehadtoconcludethattwoidenticalcopiesoftheletter F canbedrawnintheplaneinsuchawaythatnoisometryoftheplanetakesonetotheother, orthatthereareisometriesotherthanthethreetypeswehavediscussedsofar, namelytranslations, rotationsandreflections. Wenowusethenotionofcompositionofisometries, asdiscussedinSection 4.4, toresolvethisquestion.Aspreviouslystated, itisnotpossibletotaketheinitial F totheterminal F inFigure 4.3.9
byatranslation, arotationorareflection. However, drawtheline m halfwaybetweenthetwoletters F, asshowninFigure 4.5.1. Itcanthenbeseenthatfirstreflectingtheplanein m, andthentranslatingtheplanedownwardbytheappropriatedistance, willtaketheinitial F totheterminal F. Itwouldalsoworktotranslatetheplanedownward, andthenreflectinm. Therefore,althoughnosinglereflectionortranslationoftheplanewilltaketheinitial F totheterminal F,itisthecasethatthecompositionofareflectionandatranslationdoestaketheinitial F totheterminal F. Weknowthatreflectionandtranslationareisometries, andsobyProposition 4.4.1,wealsoknowthatthecompositionofareflectionandatranslationisanisometry. Hence, thereisanisometryoftheplanethattakestheinitial F totheterminal F inFigure 4.3.9. Letuscallthisisometry G. Thecrucialpointisthis: eventhough G wasmadeupoutofareflectionandatranslation, itisasingleisometryinitsownright. Recallthatwhatcountsinanisometryisonlyitsneteffect, thatis, whereeachpointoftheplaneendsupinrelationtoitsinitialposition,andnotthegeometricprocessusedtovisualizetheisometry(inthepresentcasefirstreflectingandthentranslating). Giventhatwecannottaketheinitial F totheterminal F inFigure 4.3.9byasingletranslation, rotationorreflection, wededucethattheisometryG isnotatranslation,rotationorreflection. Hence, thereareisometriesthatarenotofthethreefamiliartypes.
Theisometry G discussedaboveisanexampleofnewtypeofisometry, calleda glidereflec-tion. A glidereflectionisanisometrythatisobtainedbyfirstreflectingtheplaneinaline, andthentranslatinginadirectionparalleltothelineofreflection. (Pleasenotetheword“parallel”here.) Thelineofreflectionusedinformingaglidereflectioniscalledthe lineofglidereflection.Ifthetranslationusedinaglidereflectionhaszerolength, theglidereflectioniscalleda trivialglidereflection. Thatis, atrivialglidereflectionisaglidereflectionthatisjustareflection. A
4.5GlideReflections 131
FF
initial
terminal
m
Figure4.5.1
glidereflectionthatisnotjustareflectioniscalleda non-trivial glidereflection. Wenotethatanon-trivialglidereflectionhasnofixedpoints, butthelineofglidereflectionisafixedline.Also, wenotethataglidereflectionreversesorientation.Thenotationforaglidereflection, wherethereflectionisinline m, andthetranslationisby
vector v (whichisparallelto m), is Gm,v. Insymbolswecanwritethat Gm,v = Tv ◦ Mm. Wereiteratetheimportantpointthataglidereflectionisasingleisometry, whichtakeseachpointoftheplane, andassignsitsomepointwhereitwillendup. Thefactthatweconstructaglidereflectionasthecompositionoftwootherisometriesdoesnotdetractfromthefactthattheglidereflectioncanbethoughtofasanisometryinitsownright.Animmediatequestionthatcomestomindconcerningglidereflectionsiswhetheritmatters
ifwefirstreflectandthentranslate, orvice-versa. Althoughingeneraltheorderdoesmatterwhenwecomposetwoisometries, inthecaseofconstructingglidereflections, itturnsoutnottomatter.
Proposition 4.5.1. Supposem isalineintheplane, and v isavectorthatisparalleltom. ThenMm ◦ Tv = Tv ◦ Mm.
Demonstration. Supposem and v arerespectivelyalineandvectorasshowninFigure 4.5.2 (i).Let A beanypointintheplane. Weneedtoshowthatboth Mm ◦ Tv and Tv ◦ Mm take A tothesamepointintheplane. Itwillfollowthat Mm ◦ Tv = Tv ◦ Mm. Ifthepoint A isontheline m, thenitiseasytoseethat Mm ◦ Tv and Tv ◦ Mm take A tothesamepoint; weleavethedetailstothereader. Nowsupposethat A isnotontheline m.IfwefirstdoMm, thepointA istakentothepoint B, asshowninFigure 4.5.2 (ii). Ifwethen
do Tv, thepoint B istakentopoint C, asshowninthefigure. Ontheotherhand, ifwefirstdo Tv, thepoint A istakentothepoint D, asshowninFigure 4.5.2 (ii). Ifwecouldshowthatdoing Mm takes D to C, thenitwouldfollowthatboth Mm ◦ Tv and Tv ◦ Mm take A to D,whichimpliesthat Mm ◦ Tv and Tv ◦ Mm bothtake A tothesamepointintheplane, whichiswhatweneededtoshow.
132 4. Isometries
Considerthequadrilateral ABCD. Asafirststep, wewillshowthatthisquadrilateralisarectangle. First, weuseExercise 4.2.4 (1)toseethat m istheperpendicularbisectorof AB.Next, becausethevector v isparallelto m, itfollowsthateachof AD and BC isparalleltom. Because m isperpendicularto AB, itfollowsbyProposition 1.2.3 that AD and BC arebothperpendicularto AB. Moreover, because Tv takes A to D, and B to C, weseethat AD
hasthesamelengthas BC. WecannowapplyExercise 2.3.5 tothequadrilateral ABCD, andsothisquadrilateralisinfactarectangle.Let P thethepointwherethelinem intersectsthelinesegmentAB; weknowfromprevious
commentsthat P isthemidpointofAB. LetQ bethepointwherethelinem intersectsthelinesegment CD. SeeFigure 4.5.2 (iii). Weknowthat ABCD isarectangle. ByExercise 2.3.2, wededucethat ABCD isaparallelogram. Inparticular, itfollowsthat CD isparallelto AB. Be-causeAB isperpendiculartom, itfollowsfromProposition 1.2.3 that CD isperpendiculartom. ItissimpletoseethatthequadrilateralAPQD isarectangle(itiscertainlyaparallelogram,anduseExercise 2.1.1). Hence AP hasthesamelengthas DQ. Because ABCD isaparal-lelogram, itfollowsfromProposition 2.2.5 (1)that CD hasthesamelengthas AB. BecauseAP hashalfthelengthof AB, andbecause AB and CD havethesamelengths, itfollowsthat DQ hashalfthelengthof CD. Fromalltheabovereasoning, wededucethat m istheperpendicularbisectorof CD. ItfollowsfromExercise 4.2.4 (1)that Mm takes C to D, andthatiswhatweneededtoshow.
D C
v v
(i) (ii) (iii)
m m
D
BA BA P
Q C
m
Figure4.5.2
4.5GlideReflections 133
Exercise 4.5.1. Drawtheeffectontheletter R showninFigure 4.5.3 astheresultofglidereflectingtheplaneusingeachofthepairsoflineofreflectionandtranslationvectorshown.
Rm
p
n
v
u
w
Figure4.5.3
Exercise 4.5.2. Supposethat G isaglidereflection. Whatkindofisometryis G ◦ G?
Exercise 4.5.3. Suppose m isalineintheplane, and v isavectorthatisnotparalleltom. Isitalwaysthecasethat Mm ◦ Tv = Tv ◦ Mm? Explainyouranswer.
Anotherimportantquestionconcerningglidereflectionisthefollowing. Weconstructaglidereflectionbyfirstreflectinginaline, andthentranslatinginadirectionparalleltothelineofreflection. Whatwouldhappen ifwewere to reflect theplane ina line, and then translateinadirectionnotparalleltothelineofreflection. Wouldweobtainyetanothernewtypeofisometry? Itturnsoutthatthecompositionofanyreflectionandanytranslationequalseitherareflectionoraglidereflection(dependingupontherelationshipbetweenthelineofreflectionandthetranslationvector), thoughwewillomitthedetails. Thereforewedonotobtainanynewtypeofisometrybyfirstreflectingandthentranslatinginadirectionnotparalleltothelineofreflection.RecallFigure 4.3.9. Whenwefirstencountered thatexample, wesaw thatno translation,
rotationorreflectionalonecouldtaketheinitial F totheterminal F. Atthetimewewerenotfamiliarwithanyothertypesofisometries, butnowweknowaboutglidereflections, andwe
134 4. Isometries
haveseeninthecurrentsectionthattheinitial F inFigure 4.3.9 canbetakentotheterminal Fbyaglidereflection. Howwasthisglidereflectionfound?Tofindtheisometrythattakestheinitial F totheterminal F inFigure 4.3.9, westartjustas
wedidforreflections, namelybylabelingtwopointsontheinitial F bytheletters A and B,andlabelingthecorrespondingpointontheterminal F bytheletters A ′ and B ′. Wecanthenfindthemidpointsofthetwolinesegments AA ′ and BB ′, anddrawalinethroughthesetwomidpoints. Callthisline m. SeeFigure 4.5.4.
FF
initial
terminal
A
A’
B
B’
m
Figure4.5.4
It is seen that simply reflecting theplane in the line m doesnot take the initial F to theterminal F. However, ifwefirstreflecttheplaneintheline m, andthenfollowthisreflectionbyatranslationparalleltom bytheverticaldistancebetweentheinitial F andterminal F, thenthecompositionofthereflectionandtranslationwillindeedtaketheinitial F totheterminal F.Hence, thereisaglidereflectionthattakestheinitial F totheterminal F.Wecannowstate thecompleteanswer to theproblemof recognizing isometriesby their
effects, whichweleftunfinishedinSection 4.3. ThatouransweriscompletereliesuponPropo-sition 4.6.1 inthenextsection, aswellassomemathematicaldetailsweskipovertoavoidadigression, butwecanstatethecompleteanswerrightnow. Supposewearegivetwoidenti-calletters F intheplane. Wecanthenfindwhichsingleisometrytakesone F totheother, asfollows. (Moreover, thisisometryturnsouttobeunique.) Therearetwocases.
Case1: thetwoletters F havethesameorientation. Connecttwopairsofcorrespondingpointswithlinesegments, andformtheperpendicularbisectors. Ifthetwoperpendicularbisectorsin-tersect, thentheisometryisarotation, withthepointofintersectionbeingthecenterofrotation;theanglecanbefoundbydrawinglinesfromthecenterofrotationtocorrespondingpointsonthetwoletters F, andmeasuringtheanglebetweenthesetwolines. Ifthetwoperpendicularbisectorsareparallel, theisometryisatranslation; simplydrawanarrowfromapointonone Ftoitscorrespondingpointonanother F.
4.5GlideReflections 135
Case2: thetwo Fshaveoppositeorientations. Connecttwopairsofcorrespondingpointswithlinesegments, andfindthemidpointsofthesetwolines. Drawalinethroughthetwomidpoints.If the line through themidpoints isperpendicular to theconnecting linesegments, then theisometryisareflectioninthelinethroughthemidpoints; ifthelinethroughthemidpointsisnotperpendiculartotheconnectinglinesegments, thentheisometryisaglidereflection, obtainedbyfirstreflectinginthelinethroughthemidpoints, andthentranslatingasnecessary.
Exercise 4.5.4. IneachofthethreepartsofFigure 4.5.5 areshowninitialandterminalletters F, obtainedbyusingan isometry. Foreachof the threecases, statewhat typeofisometrywasused. Moreover, iftheisometryisarotation, indicateitscenterofrotation;if the isometry isa translation, indicate the translationbyanarrow; if the isometry isareflection, indicatethelineofreflection; iftheisometryisaglidereflection, indicatethelineofreflectionusedintheglidereflection.
(iii)
(i) (ii)
FterminalFinitial
Finitial
F Finitial
Fterminal
terminal
Figure4.5.5
136 4. Isometries
4.6 Isometries—TheWholeStory
HavinglearnedaboutglidereflectionsinSection 4.5, wearenowfamiliarwithfourtypesofisometries: translations, rotations, reflectionsandglidereflections. Are thereanyother typesofisometries? Asstatedinthefollowingproposition, theanswerisno. Thispropositionisthemostfundamentalfactaboutisometriesoftheplane, anditwillformthebasisformuchofoursubsequentdiscussionofisometriesandofsymmetry(giveninthischapterandthenext). Thedemonstrationofthispropositionissomewhatlengthy, andisgiveninAppendix A.
Proposition 4.6.1. Anyisometryoftheplaneiseitheratranslation, arotation, areflectionoraglidereflection.
WediscussedinSection 4.2 thenotionsoffixedpoints, fixedlines, andorientationpreservingandreversing. Nowthatweknowallthetypesofisometries, wesummarizethesepropertiesforalltypesofisometriesinTable 4.6.1.
IsometryType FixedPoints FixedLines OrientationIdentityisometry I allpoints alllines preservingNon-triv.trans. Tv nopoints linesparallelto v preservingNon-triv.rot. RA
α thepoint A none, oralllinesthrough A preservingRefl. Mm pointson m m, andalllinesperp.to m reversingNon-triv.gl.refl. Gm,v nopoints theline m reversing
.
Table4.6.1
Dowereallyneedallfourtypesofisometries? Theanswerisbothyesandno. Wecouldobtainallisometriesbyusingjusttranslations, rotationsandreflections, ifweallowforcombinationsofthem(becauseglidereflectionsareobtainedbycomposingreflectionsandtranslations). Ac-tually, ifwewanttobethemost“economical”intermsofusingthefewesttypesofisometriestoobtainallothers, wecoulduseonlyreflections. Itturnsout, thoughthisfactisnotatallobvious,thatwecangetalltheotherthreetypesofisometries(andthereforeallisometries)bycombininguptothreereflectionsatatime. Indeed, thebulkoftheproofofProposition 4.6.1, asgiveninAppendix A,involvesshowingthattheneteffectofanyisometryoftheplanecanbeobtainedbyeithertheidentity, orthecompositionofone, twoorthreereflections. If, however, wewanteachisometry(meaningeachpossibleneteffect)tobedescribedbyasingletransformationoftheplane, ratherthanacompositionofothertransformations, thenweneedtranslations, rotations,reflectionsandglidereflections. Ourgoalbeingnotefficiencybutobtaininganunderstandingofisometriesandsymmetry, wewilluseallfourtypesofisometries.Ifwecombinetwoisometries, howdoweknowwhattheresultis? Oneapproachwouldbeto
drawaninitialobjectintheplane, performthetwoisometriesoneaftertheother, andthenlookattheneteffecttakingtheinitialobjecttotheterminalobject. WecouldthenapplythemethodofrecognizingisometriesdiscussedinSections 4.3 and4.5 tofigureoutwhattheresultingsingleisometrywas. However, itwillbemoreusefultobeabletoknowtheresultofcombiningtwo
4.6Isometries—TheWholeStory 137
isometriessimplyfromknowingthetwoisometriesthatarecombined. (Asaroughanalogy, itissimilarlybettertofigureout 547× 23 withpencilandpaper, ratherthanmaking 23 rowsof547 stoneseach, andthencountingthetotalnumberofstones.)Recallingthateveryisometryisoneofourfourtypes, wecanseehowtocombineanytwo
isometriesbybreakingupourdiscussionintovariouscases, dependinguponwhichtwotypesofisometriesarebeingcombined. Insomecaseswecanobtainveryspecificresultsabouttheresultofcomposingtwoisometriesofagiventype; inothercases, itisverydifficulttowriteaformulafortheresultingisometry(unlessweusesomemoreadvancedmathematics), butwecanatleaststateitstype. Insomeofthecases, wewillrestrictourattentiontonon-trivialtranslations,rotationsandglidereflections, toavoidvariousspecialcases. Wewillstateherethethreemostimportantresults, namelythoseconcerningcompositionsof twotranslations, tworeflectionsandtworotations. Itisalsopossibletodiscussthecompositionsoftwoglidereflections, andcompositionsoftwodifferenttypesofisometries; toavoidalengthydigression, wewilldiscusstheseothercasesinAppendix B.Inorder todiscuss the compositionof two translations, weneed todiscuss thenotionof
additionofvectorsintheplane. Supposewehavetwovectors v andw intheplane, representedbyarrows, asshowninFigure 4.6.1 (i). Wecanaddthesetwovectorsasfollows. First, positionthetwoarrowssothattheyhaveacommonstartingpoint, asshownFigure 4.6.1 (ii); thereisnoproblemmovingarrowsthatrepresentvectors, aslongasthearrowsarenotstretchedorshrunk, orrotated. Next, wecanformtheparallelogramwiththetwoarrowsassides, asshownFigure 4.6.1 (iii). Finally, wetakethediagonalintheparallelogramtobeanarrowforthesumof v and w, denoted v +w; seeFigure 4.6.1 (iv). Thistypeofvectoradditionisveryusefulinbothmathematicsandthesciences(forexample, theadditionofforcesinphysics).
(i) (ii)
(iii) (iv)
v
w
vv
w
w
vv
w
w
v
w
v+w
Figure4.6.1
138 4. Isometries
Wecannowuseadditionofvectorstounderstandthecompositionoftranslations.
Proposition 4.6.2. Supposethat Tv and Tw aretranslationsoftheplane. Then Tw ◦ Tv = Tv+w.
Demonstration. Chooseanypoint A intheplane. Thentheresultofdoing Tw ◦ Tv to A isthesameasfirstdoing Tv to A, yielding Tv(A), andthendoing Tw tothat, yielding Tw(Tv(A)).SeeFigure 4.6.2. However, thetriangleshowninthisfigureisthesameasthelowerhalfoftheparallelogramshowninFigure 4.6.1 (iv). Hence Tw(Tv(A)) isthesameas Tv+w(A). Becausethisreasoningholdsforanypoint A intheplane, wededucethattheisometry Tw ◦ Tv hasthesameneteffectastheisometry Tv+w. Itfollowsthat Tw ◦ Tv = Tv+w.
w
v
v+w
Tv(A)A
Tw(Tv(A))
Figure4.6.2
Itisseenthatforanytwovectors v andw intheplane, wehave v+w = w+ v. Combiningthisobservationwiththeaboveproposition, wededucethatforanytwovectors v and w, wehave Tw ◦ Tv = Tv ◦ Tw. Inotherwords, orderdoesnotmatterwhentwotranslationsarecomposed. Bycontrast, orderdoesmatterwhenmostotherisometriesarecomposed.Wenowturntothecompositionoftworeflections; therearethreesubcases, dependingupon
whetherthetwolinesofreflectionareequal, areparallel, orareneitherparallelnorequal.
Proposition 4.6.3. Supposethat Mm and Mn arereflectionsoftheplane.
1. If m = n, then Mn ◦ Mm = I.
2. If m and n areparallel, then Mn ◦ Mm = Tv, where v is thevector that is in thedirectionperpendicularto m and n, andthathaslengthtwicethedistancefrom m ton.
3. If m and n areneitherparallelnorequal, then Mn ◦ Mm = RAα , where A isthepoint
ofintersectionof m and n, andwhere α istwicetheanglefrom m to n.
Demonstration. WeknowthateachofMm andMn areorientationreversing, andtherefore, byProposition 4.4.3 (3)weknowthatMn ◦ Mm isorientationpreserving. GiventhatMn ◦ Mm isanisometrybyProposition 4.4.1, thenitiseitheratranslation, arotation, areflectionoraglidereflectionbyProposition 4.6.1. UsingTable 4.6.1, it followsthat Mn ◦ Mm mustbeeitheratranslationorarotation(ortheidentityisometry, whichisbothatranslationorarotation),becausereflectionsandglidereflectionsareorientationreversing.
4.6Isometries—TheWholeStory 139
(1). Thisisclear, becauseMm istheresultofflippingtheplaneaboutthelinem, andflippingtheplaneaboutthesamelinetwiceleaveseverypointintheplanewhereitstarted.
(2). Weclaimthat Mn ◦ Mm doesnotfixanypointintheplane. Suppose, tothecontrary,that Mn ◦ Mm didfixsomepoint, say B. Thatwouldmeanthat Mn(Mm(B)) = B. WethendeducethatMn(Mn(Mm(B))) = Mn(B). ByapplyingPart (1)ofthispropositiontoMn, wethereforeseethat Mm(B) = Mn(B).Howcould thispossiblyhappen? Thereare threecases, dependinguponwhether B ison
m, ison n orisnotoneitherline. If B ison m, then Mm(B) = B, but Mn(B) = B, butthiscannotpossiblybethecase, giventhatMm(B) = Mn(B). So, weconcludethat B isnotonm.A similarargumentshowsthat B isnoton n. Nowsupposethat B isnotoneitherm or n. ThenMm(B) = B and Mn(B) = B. Inthatcase, weuseExercise 4.2.4 (1)todeducethat m istheperpendicularbisectorofthelinesegmentfrom B to Mm(B), andthat n istheperpendicularbisectorofthelinesegmentfrom B toMn(B). However, giventhatMm(B) = Mn(B), weseethat m mustbethesamelineas n, whichcannotbe, giventhat m and n areparallel, whichmeansthattheyaredistinctlines. Thefinalconclusionofthisargumentisthat Mn ◦ Mm hasnofixedpoints, becauseourassumptiontothecontraryledtoalogicalimpossibility.WeknowthatMn ◦ Mm iseithertheidentityisometry, atranslationorarotation. Theidentity
isometryfixesallpoints, andanon-trivialrotationfixespreciselyonepoint. Hence, weseethatMn ◦ Mm mustbeatranslation. Tofindoutwhichtranslation, wecansimplyseewhathappenstoonepointintheplane. Forconvenience, supposethatboththelines m and n arevertical,asinFigure 4.6.3 (ifnot, simplychangeyourvantagepoint). Consideranypoint X onthelinem. Then Mm fixes X, and Mn takes X toapoint Y thatisdirectlytotherightof X, andatadistancefrom X thatistwicethedistancefromm to n. ItfollowsthatMn ◦ Mm = Tv, wherev isthevectorthatisinthedirectionperpendicularto m and n, andthathaslengthtwicethedistancefrom m to n.
(3). ObservethatthepointA, whichistheintersectionofthelinesm and n, isfixedbybothMm andMn. HencethepointA isfixedbyMn ◦ Mm. SupposeZ isapointonthelinem thatisdifferentfromthepointA. SeeFigure 4.6.4. ThenMm fixesZ, andMn doesnot(becauseZ isnoton n). Hencethepoint Z isnotfixedbyMn ◦ Mm. WethereforeseethatMn ◦ Mm fixessomepointsanddoesnotfixothers. WeknowthatMn ◦ Mm iseithertheidentityisometry, atranslationorarotation. Theidentityisometryfixesallpoints, andanon-trivialtranslationfixesnopoints. ItfollowsthatMn ◦ Mm isanon-trivialrotation. Sucharotationfixespreciselyonepoint, namelyitscenterofrotation. ItthereforemustbethecasethatthepointA isthecenterofrotation. Tofindtheangleofrotation, wecansimplyseewhathappenstoonepointotherthanA. Takethepoint Z chosenabove. SupposethatMn ◦ Mm moves Z tothepointlabeledW in
Figure 4.6.4. Usingthedefinitionofreflection, itcanbeseenthattheanglefromtheline←→AW
totheline n isequaltotheanglefromtheline m (whichisthesameastheline←→AZ)tothe
line n. (A rigorousdemonstrationoftheequalityofthesetwoanglescanbeobtainedbyusingcongruenttriangles; thereaderisaskedtoprovidedetailsinExercise 4.6.1.) Itfollowsthatthe
140 4. Isometries
rotationwithcenterofrotation A thattakes Z to W mustberotationbytheangle α, whichistwicetheanglefrom m to n. Wethereforeseethat Mn ◦ Mm = RA
α .
nm
X Y
Figure4.6.3
Z
W
A
n m
Figure4.6.4
4.6Isometries—TheWholeStory 141
Exercise 4.6.1. [UsedinThisSection] InthedemonstrationofProposition 4.6.3 (3), we
assertedthattheanglefromtheline←→AY totheline n isequaltotheanglefromthelinem
totheline n; seeFigure 4.6.4. Usecongruenttrianglestodemonstratethisclaim.
Weturnnexttothecompositionoftworotations. Onceagainwehavetwomaincases, thistimedependinguponwhetherthetwocentersofrotationarethesameornot. However, inthecasewherethetwocentersofrotationarenot thesame, therearetwosubcases, dependinguponwhether the twoanglesadduptoamultipleof 360◦ ornot. Moreover, whenthe twoanglesdonotadduptoamultipleof 360◦, thenwewillnotbeabletogiveasimpledescriptionoftheresultingisometry, whichisdefinitelyarotation, butforwhichitistrickytodescribethecenterofrotation. A pictorialapproachtofindingthedesiredcenterofrotationisfoundinthedemonstrationofProposition 4.6.4 (thispictorialapproachwillbeusefulinAppendix E).
Proposition 4.6.4. Supposethat RAα and RB
β arerotationsoftheplane.
1. If A = B, then RBβ ◦ RA
α = RAα+β.
2. If A = B, andif α + β isnotamultipleof 360◦, then RBβ ◦ RA
α = RCα+β, where C isa
pointintheplaneuniquelydeterminedbyA, B, α and β (andwhichisdescribedmorepreciselyinthedemonstration).
3. If A = B, andif α+ β isamultipleof 360◦, then RBβ ◦ RA
α = Tv, where v isthevectorfrom A to RB
β(A).
Demonstration. Weknowthateachof RAα and RB
β areorientationpreserving, andtherefore,byProposition 4.4.3 (1)weknowthat RB
β ◦ RAα isorientationpreserving. Giventhat RB
β ◦ RAα
isanisometrybyProposition 4.4.1, thenitiseitheratranslation, arotation, areflectionoraglidereflectionbyProposition 4.6.1. UsingTable 4.6.1, itfollowsthat RB
β ◦ RAα mustbeeither
atranslationorarotation(ortheidentityisometry, whichisbothatranslationorarotation).Considerthefixedpointsof RB
β ◦ RAα , ifthereareany. If R
Bβ ◦ RA
α turnsouttohavenofixedpoints, thenitmustbeanon-trivialtranslation; ifithasatleastonefixedpointandatleastonepointthatisnotfixed, thenitmustbeanon-trivialrotation, andthefixedpointmustbethecenterofrotation; ifithasmorethanonefixedpoint, thenitmustbetheidentityisometry.
(1). Supposethat A = B. Then RBβ ◦ RA
α = RAβ ◦ RA
α . Becauseboth RAβ and RA
α fixthepointA, then RA
β ◦ RAα fixesA aswell. Hence RA
β ◦ RAα hasafixedpoint, andcannotbeanon-trivial
translation; itmustthereforebearotationwithcenterofrotation A, ortheidentityisometry(whichcanalsobethoughtofasarotationwithcenterofrotation A).Nowthatweknowthat RA
β ◦ RAα isarotationwithcenterofrotation A, thequestionisby
whatangle. Drawaninitialobjectintheplane, forexampletheletter F. Ifweperform RAα , then
theresultingimageoftheletter Fwillmakeangle αwiththeoriginalletter F. IfwethenperformRAβ , theresultingimageoftheletter F willnowmakeangle α + β withtheoriginalletter F.
Hence, theonlypossiblerotationthatequals RAβ ◦ RA
α is RAα+β.
142 4. Isometries
(2). Supposethat A = B, andthat α+ β isnotamultipleof 360◦. Byaddingorsubtractingmultiplesof 360◦ toeachofα andβ asnecessary, wecanensurethatα andβ arebothbetween0◦ and 360◦; addingandsubtractingmultiplesof 360◦ doesnotchangetheeffectof RB
β or RAα .
Wenowmakethefollowingconstruction. First, wedrawalinesegmentfrom A to B. Next, wedrawtheangle α atthepointA, andtheangle β atthepoint B, sothatbothanglesarebisectedby AB. SeeFigure 4.6.5. Thetwoanglesintersectinpoints C and D asshown. Inthefigure,forconvenience, wehavepositionedthepoints A and B sothat AB ishorizontal; ifthislinesegmentisnothorizontal, thentheconstructionwouldlookjustlikewhatwehaveshown, butrotatedappropriately. Thereare, infact, anumberofdifferentcasesthatwouldlooksomewhatdifferentfromwhatisshowninthefigure, dependinguponwhethereachoftheangles α andβ islessormorethan 180◦; wehavedrawnthecasewherebothanglesarelessthan 180◦. Thekeypointtonoteisthat, because α + β isnotamultipleof 360◦, thetwoangleswillindeedintersectintwopoints.
A α β B
C
D
Figure4.6.5
Wenowmakethefollowingobservation. Ifweapplytheisometry RAα totheplane, thepoint
labeled C willbetakentothepoint D. Ifwethenapply RBβ, thepoint D willbetakenbackto
C. Hence, thecomposition RBβ ◦ RA
α fixesthepoint C. Ontheotherhand, itcanbeseenthatRBβ ◦ RA
α doesnotfixthepoint D. Therefore, theisometry RBβ ◦ RA
α fixesatleastonepoint,butdoesnotfixallpoints. Wealreadysawthat RB
β ◦ RAα iseitherarotation, atranslationor
theidentityisometry. Because RBβ ◦ RA
α cannotbeanon-trivialtranslation(whichwouldfixnopoints), northeidentityisometry(whichfixesallpoints), itmustbearotation. A rotationfixespreciselyonepoint, namelyitscenterofrotation. Wededucethat RB
β ◦ RAα isarotationwith
centerofrotation C.Nowthatweknowthat RB
β ◦ RAα isarotation, thequestionisbywhatangle. Usingthesame
ideaasinthedemonstrationofPart (1)ofthisproposition, itisseenthattheanglemustbeα+β.Hence, theonlypossiblerotationthatequals RB
β ◦ RAα is RC
α+β.
(3). Supposethat A = B, andthat α+ β isamultipleof 360◦. Because α+ β isamultipleof 360◦, theneither α and β arebothmultiplesof 360◦, orneitheraremultiplesof 360◦. Wehavetolookateachofthesetwocasesseparately.
4.6Isometries—TheWholeStory 143
First, supposethat α and β arebothmultiplesof 360◦. Then RBβ and RA
α areboththeidentityisometry. Itfollowsthat RB
β ◦ RAα isalsotheidentityisometry. Becausetheidentityisometryisa
trivialtranslation, then RBβ ◦ RA
α = Tv, where v haslengthzero.Next, suppose thatneither α nor β is amultipleof 360◦. Thenneither RB
β nor RAα is the
identityisometry. Wewanttoshowthat RBβ ◦ RA
α isatranslation. Wewillshowthisresultbyeliminatingtheotherpossibilities. First, notethat RA
α fixesthepoint A, but RBβ doesnotfixthe
point A, because RBβ isanon-trivialrotation. Itfollowsthat RB
β ◦ RAα doesnotfixthepoint A,
andtherefore RBβ ◦ RA
α isnottheidentityisometry.Now, supposethat RB
β ◦ RAα werearotation. Thenitwouldhavesomecenterofrotation,
sayapoint D. If RBβ ◦ RA
α werearotation, bywhatanglewouldtherotationbe? UsingthesameideaasinthedemonstrationofPart (1)ofthisproposition, itisseenthattheanglemustbeα+ β. Hence, theonlypossiblerotationthatcouldequal RB
β ◦ RAα wouldbe RD
α+β. However,weareassumingthat α+β isamultipleof 360◦. Itwouldthenfollowthat RD
α+β istheidentityisometry, andhencethat RB
β ◦ RAα istheidentityisometry. However, wehaveseenthat RB
β ◦ RAα
doesnotfixthepoint A, soitcouldnotpossiblybetheidentityisometry. Wethereforehavealogicalcontradiction. Theonlyresolutionofthisdilemmaisthat RB
β ◦ RAα cannotbearotation.
Wededucethat RBβ ◦ RA
α mustbeatranslation.Because RB
β ◦ RAα is a translation, it equals Tv, where v is somevector in theplane. To
determinethisvector, wecantakeanypointintheplane, seewhereitendsafterperformingRBβ ◦ RA
α , andthentakethearrowfromthepoint’soriginalpositiontoitsfinalposition. Forconvenience, wechoosethepoint A. Because RA
α fixesthepoint A, thenweseethattheresultofapplying RB
β ◦ RAα to thepoint A results in thepoint RB
β(A). Therefore, wededuce thatRBβ ◦ RA
α = Tv, where v isthevectorfrom A to RBβ(A).
Havingjustgivenpropositionsdescribingthedetailsofsomeofthepossiblecompositionsofisometries, weleavethedetailsoftheothertypesofcompositionstoAppendix B.Whatwestaterightnowisasummaryofallthewaysofcombiningthefourdifferenttypesofisometriesinachart, asseeninTable 4.6.2. Noticethatthistablecanbebrokenintofoursub-boxes, basedonorientationpreservingororientationreversing. Inthistable, weincludetrivialtranslations,rotationsandglidereflections, toavoidspecialcases.
◦ translation rotation reflection glidereflectiontranslation trans. trans.orrot. refl.orgl.refl. refl.orgl.refl.rotation trans.orrot. trans.orrot. refl.orgl.refl. refl.orgl.refl.reflection refl.orgl.refl. refl.orgl.refl. trans.orrot. trans.orrot.
glidereflection refl.orgl.refl. refl.orgl.refl. trans.orrot. trans.orrot.
.
Table4.6.2
144 4. Isometries
Exercise 4.6.2. Supposearotationisfollowedbyaglidereflection, whichisthenfollowedbyareflection. Whattypeofisometry(orisometries)couldbeobtainedasaresultofthiscomposition?
Exercise 4.6.3. Describetheisometrythatresultsfromahalfturnfollowedbyanotherhalf-turn? (Theresultdependsuponwhetherthetwohalfturnshavethesamecenterofrotationornot.)
Exercise 4.6.4. [UsedinSection 5.5] A halfturnisfollowedbyareflection. Supposethatthecenterofrotationof thehalfturnisonthelineofreflection. Showthat theresultingisometryisareflectioninthelinethroughthecenterofrotationandperpendiculartotheoriginallineofreflection.
Exercise 4.6.5. [UsedinSection 5.5] A halfturnisfollowedbyareflection. Supposethatthecenterofrotationofthehalfturnisnotonthelineofreflection. Showthattheresult-ingisometryisaglidereflection, whichhaslineofglidereflectionthroughthecenterofrotation, andperpendiculartotheoriginallineofreflection.
Exercise 4.6.6. [UsedinSection 5.5] A halfturnisfollowedbyaglidereflection. Supposethatthecenterofrotationofthehalfturnisonthelineofglidereflection. Showthattheresulting isometry is a reflection, whichhas lineof reflectionperpendicular to the lineofglidereflection, andatadistancefromthecenterofrotationhalf thedistanceof thetranslationintheglidereflection.
Exercise 4.6.7. [UsedinSection 5.5] A reflectionisfollowedbyaglidereflection. Supposethat the lineof reflection isperpendicular to the lineofglide reflection. Show that theresultingisometryisahalfturn, withthecenterofrotationonthelineofglidereflection.
Finally, we turn to thequestionof inversesof isometries. Bywayofanalogy, consider theintegersandtheoperationofaddition. Giventhenumber 5, isthereanumberthat“cancelsit
4.6Isometries—TheWholeStory 145
out”withrespecttoaddition? Theanswerisyes, namelythenumber−5, because 5+(−5) = 0,and (−5) + 5 = 0. WewilldiscussthisconceptatitsmostgeneralinSection 6.4, butfornow,wewanttoknowwhetherthereisananalogof“cancelingout”intherealmofisometriesandcomposition. Supposewearegivenanisometry P. Canwefindsomeotherisometry Q that“cancels” P out? Thatis, canwefindanisometryQ suchthat P ◦ Q = I andQ ◦ P = I? (Weneedtospecifyboththeseequations, becauseingeneraltheorderofcompositionofisometriesdoesmatter, andwecannotautomaticallydeduceoneoftheseequationsfromtheother.) IfwecanfindsuchanisometryQ, isiscalledan inverseisometry of P; weoftensimplysay inverse ofP forshort. Forexample, considertheisometry P thatistranslationtotherightby 4 inches. IfweletQ betranslationtotheleftby 4 inches, thenclearlyQ cancelsout P, andvice-versa, inthatdoingone, andthentheother, leavesuswiththeidentityisometry. Inotherwords, translationtotheleftby 4 inchesistheinverseoftranslationby 4 inchestotheright.Doeseveryisometryhaveaninverse? Ifso, istheinverseunique? Cantheinverse, ifitexists,
befoundeasily? AsseeninProposition 4.6.5, theanswertoallthesequestionsisyes. Weusethefollowingcommonnotation. Suppose P isanisometry. Theinverseisometry, of P isdenotedP−1. Wethereforehave P ◦ P−1 = I and P−1 ◦ P = I. Theselasttwoequationsmeanthatforanypoint X intheplane, wehave P(P−1(X)) = X and P−1(P(X)) = X.Inorder todiscuss inversesof translations, wewill need the followingnotionconcerning
vectorsintheplane. Supposewehaveavector v intheplane, representedbyanarrow. Wedefinethe inversevector of v tobethevector, denoted−v, representedbythearrowthathasthesamelengthasthearrowrepresenting v, buthavingtheoppositedirection. Clearly v+(−v) = 0and (−v) + v = 0, where 0 heremeansavectorwithlengthzero.
Proposition 4.6.5. Supposethat P isanisometry. Then P hasauniqueinverse. Further, wecanfindtheinverseof P asfollows.
1. I−1 = I.
2. If Tv isatranslation, then (Tv)−1 = T−v.
3. If RAα isarotation, then
(RAα
)−1= RA
−α.
4. If Mn isareflection, then (Mn)−1 = Mn.
5. If Gn,v isaglidereflection, then (Gn,v)−1 = Gn,−v.
Demonstration. WeknowfromProposition 4.6.1 thateveryisometryoftheplaneiseitheratranslation, arotationareflectionoraglidereflection. Hence, oncewedemonstrateParts (2)–(5)ofthisproposition, aswillbedoneshortly, thenitwillbeverifiedthateveryisometryhasaninverse.Toshowthattheinverseofeachisometryisunique, supposetothecontrarythatthereisan
isometry P thathastwodistinctinverses Q and R. Then, bythedefinitionofwhatitmeanstobeaninverse, weknowthat P ◦ Q = I and Q ◦ P = I, andthat P ◦ R = I and R ◦ P = I.Wethenseethat
Q = Q ◦ I = Q ◦ (P ◦ R) = (Q ◦ P) ◦ R = I ◦ R = R,
146 4. Isometries
wherewemakerepeateduseofProposition 4.4.2. Wehavenowderivedthat Q = R, whichisalogicalimpossibility, becauseweassumedthatQ and R weredistinct. Theonlywayoutofthisproblemistodeducethat P cannothavetwodistinctinverses, whichmeansthattheinverseof P isunique.WenowdemonstrateParts (1)–(5)oftheproposition. Thesedemonstrationsareallbasedon
thesameidea, whichisthatinordertoshowthattwoisometriesareinverses, weshowthatthey“canceleachotherout.”
(1). ByusingProposition 4.4.2 (1), weseethat I ◦ I = I. Itfollowsthat I−1 = I.
(2). Supposethat Tv isatranslation. ThenbyProposition 4.6.2 itfollowsthat
T−v ◦ Tv = Tv+(−v) = T0 = I.
A similarargumentshowsthat Tv ◦ T−v = I. Wededucethat (Tv)−1 = T−v.
(3). Supposethat RAα isarotation. ThenbyProposition 4.6.4 (1)itfollowsthat
RA−α ◦ RA
α = RAα+(−α) = RA
0 = I.
A similarargumentshowsthat RAα ◦ RA
−α = I. Wededucethat(RAα
)−1= RA
−α.
(4). SupposethatMn isareflection. ThenbyProposition 4.6.3 (1)itfollowsthatMn ◦ Mn =I. Wededucethat (Mn)
−1 = Mn.
(5). Suppose that Gn,v isaglidereflection. WeknowfromProposition 4.5.1 that Gn,v =Mn ◦ Tv and Gn,v = Tv ◦ Mn, andsimilarlyfor Gn,−v. Wenowcomputethat
Gn,−v ◦ Gn,v = (Mn ◦ T−v) ◦ (Tv ◦ Mn) = Mn ◦ (T−v ◦ Tv) ◦ Mn
= Mn ◦ I ◦ Mn = Mn ◦ Mn = I,
where the thirdequalityholdsbyPart (2)of thisproposition, and thefifthequalityholdsbyPart (4). A similarargumentshowsthat Gn,v ◦ Gn,−v = I. Wededucethat (Gn,v)
−1 = Gn,−v.
5SymmetryofPlanarObjectsandOrnamentalPatterns
5.1 BasicIdeas
Ourgoalinthischapteristoapplythegeneralconceptofmathematicalsymmetrytothestudyofornamentalpatterns. Suchpatternscanbe 1-dimensional, 2-dimensionalor 3-dimensional. Anexampleofa 1-dimensionalornamentalpatternisastringofbeads; anexampleofa 2-dimen-sionalpatternisapieceofwallpaper; anexampleofa 3-dimensionalpatternisapileofcannonballs. Wewillconcentrateon 2-dimensionalornamentalpatterns, that is, planarornamentalpatterns. Suchpatternsarequitecommoninart, crafts, designandarchitecture, andarefoundinmanyculturesaroundtheworld. Forexample, wallpaperisquitecommoninEuropeancul-ture; theMuslimtraditionusescomplicatedgeometricdesignsintheirbuildings; variousAfricanpeoplesuserepeatingpatternsintheirart. (Symmetriesofthreedimensionalobjectshavealsobeenstudied, butaremuchmorecomplicatedthanintheplanarcase, andwewillonlydiscusssuchsymmetryverybrieflyinSection 5.7.) A completestudyofplanarornamentalpatternsin-volvessomefairlyadvancedmathematics, namelygrouptheory(whichwewilldiscussbrieflyinChapter 6, thoughwewillnotbeabletodiscussthefullsetoftechnicalitiesneededforacompletelyrigoroustreatmentofornamentalpatterns). Evenwithoutallthetechnicaltoolsofgrouptheory, however, wecanexamineandanalyzevarioustypesofornamentalsymmetry.Ourmaintoolinthestudyofsymmetryistheconceptofisometry. InChapter 4 wediscussed
someofthefundamentalpropertiesofisometriesoftheplane. AsweviewedthingsinChapter 4,theplaneitselfwasblank, thoughwesometimesdrewafigure(suchastheletter F)onit, inordertoseewhathappenedwhenweperformedanisometry. Now, bycontrast, wewanttostartwithanobjectdrawnontheplane, andthenanalyzethesymmetryofthisobjectbyusingisometries.Bytheterm“object” drawnontheplanewesimplymeanacollectionofpointsintheplane.Thesepointscouldbeisolated(asinFigure 5.1.1 (i)), orcouldformageometricfigure(asinFigure 5.1.1 (ii)), orcouldformapicture(asinFigure 5.1.1 (iii)), orcouldformanythingelse.
148 5. SymmetryofPlanarObjectsandOrnamentalPatterns
Wewilloftenrefertoanobjectdrawnintheplaneasa“planarobject” ora“planarpattern;”wewilloftendroptheadjective“planar,” becauseitwillalwaysbeassumed(exceptforafewcases, wherewewillexplicitlysaythatwearelookingatnon-planarobjects). Wewillignoreissuesofcolor inourdiscussionofornamentalpatterns, andwillconsiderallpatterns tobe“black”pointsonatransparentplane.
(i) (ii) (iii)
Figure5.1.1
RecallourinformaldiscussionofsymmetryinSection 4.1, inwhichwerelatedsymmetrytothenotionoftransformationsoftheplane. Givenaplanarobject, a symmetry oftheobjectisanyisometryoftheplanethatcarriestheobjectontoitself. Thatis, afterperformingtheisometry,theobjectlooksjustasitdidbeforetheisometrywasperformed. Insymbols, if K isanobjectand P isanisometryoftheplane, then P isasymmetryof K preciselyif P(K) = K; wedonotrequirethat P fixeseverypointof K, butonlythat P takesallof K ontoitself. (Althoughweareusingisometriestofindthesymmetryofagivenobject, westressthattheisometriesare, asalways, transformationsofthewholeplane, notjusttheobject.) Notethatwiththisdefinitionof symmetry, weuse theword“symmetry”asanoun (beingan isometry). Weare thereforeinterestedinwhetheranobjecthassymmetries, notinwhetheritis“symmetric,” aswouldbeusedmorecolloquially.Suppose, forexample, thatourobjectistherectangleshowninFigure 5.1.2 (i). Reflectionin
averticallinethroughthecenteroftherectangleisanisometrythathastherectanglelandonitself; wedenotethisverticallineby L1, andthereflectioninthatlinebyM1. Similarly, reflectioninahorizontallinethroughthecenteroftherectangleisanisometrythathastherectanglelandonitself; wedenotethislineby L2, andthereflectioninthislineby M2. SeeFigure 5.1.2 (ii).Rotationby 180◦ aboutthecenteroftherectangleisanisometrythathastherectanglelandonitself; wedenotethisrotationby R1/2. Arethesetheonlysymmetriesoftherectangle? Notquite. Thereisonemoresymmetry, namelytheidentityisometry. Thesefourisometries, namelyI, R1/2, M1 and M2 areallthesymmetriesoftherectangle.
Letuslookatthesymmetriesofsomeotherobjects, startingwiththeobjectshowninFig-ure 5.1.3 (i). Thesymmetriesofthisfigureare I, R1/3 and R2/3. Hence, comparingthisobjectwiththerectangle, weseethatdifferentobjectscanhavedifferentcollectionsofsymmetries.Ontheotherhand, considertheletterH showninFigure 5.1.3 (ii). ObservethattheletterH haspreciselythesamefoursymmetriesastherectangle, namely I, R1/2, M1 and M2. Wethere-
5.1BasicIdeas 149
(i) (ii)
L1
L2
Figure5.1.2
foreseethattwodifferentobjectscanhavethesamesymmetries. Fromadesignpointofview,clearlytherectangleandtheletter H aredifferent. Fromasymmetrypointofview, however,theyarethesame. Bothpointsofviewareworthconsidering, thoughourconcerninthistextis,needlesstosay, symmetry, notdesign. Finally, considertheobjectshowninFigure 5.1.3 (iii).Colloquially, wemightsaythatthisobject“isnotsymmetric.” Whileexpressingavalidsenti-ment, suchastatementisnotaccuratefromourmathematicalperspective. TheobjectshowninFigure 5.1.3 (iii)doesinfacthaveasymmetry, namelytheidentityisometry I (everyobjecthasI asasymmetry). Thattheobjectappearstobe“notsymmetric” isexpressedpreciselybythefactthat I istheonlysymmetryoftheobject.
(i) (ii) (iii)
HFigure5.1.3
Wehavejustlistedallthesymmetriesofafewdifferentobjects. Inprincipleitispossibletolistall thesymmetriesofanyobject. Someobjectshaveinfinitelymanysymmetries(wewillseeexamplesshortly), sowecannotinpracticemakealistofallthesymmetriesofeveryob-ject. However, wecanstillcollectallthesymmetriesofanobjectintheory, evenifwecannotexplicitlylistthem. Anyobjecthasatleastonesymmetry, namely I, andsothecollectionofsymmetriesofanyobjectdoesindeedexist. Wecallthecollectionofallsymmetriesofanobject
150 5. SymmetryofPlanarObjectsandOrnamentalPatterns
the symmetrygroup oftheobject. Theword“group” usedheredoesnotsimplymeanacol-lectionofthings, butisusedinitstechnicalmathematicalmeaning. Themathematicalconceptofa“group,” discussedinChapter 6, ispartofthemathematicalfieldofabstractalgebra, andhasmanyusesbeyondjustthestudyofsymmetry. Groupshavebeenwidelystudiedbymath-ematicians, andsomefactsfromthetheoryofgroupscanbeusedtoobtainrathersurprisingresultsaboutthesymmetriesofobjectssuchasfriezepatternsandwallpaperpatterns(discussedinSections 5.5 and5.6 respectively).Usingourpreviousexamples, weseethatthesymmetrygroupoftherectangleshowninFig-
ure 5.1.2 (i)is {I, R1/2,M1,M2}, andthesymmetrygroupoftheobjectshowninFigure 5.1.3 (i)is {I, R1/3, R2/3}. Anysymmetrygroupcontainsatleastonesymmetryinit, namelytheidentityisometry. Anobjectthatwouldbedescribedcolloquiallyashaving“nosymmetry,” suchastheobjectshowninFigure 5.1.3 (iii), hasasymmetrygroupthatcontainspreciselyonesymmetry,namelytheidentityisometry.
Exercise 5.1.1. ForeachoftheobjectsshowninFigure 5.1.4, listallsymmetries.
(i) (ii) (iii)
(iv) (v)
Figure5.1.4
WeknowbyProposition 4.6.1 thatanyisometryoftheplaneisatranslation, arotation, areflectionoraglidereflection. Hence, anysymmetryofanobjectisoneofthesefourtypes.Tosaveverbiage, whenanobjecthasasymmetrythatisanon-trivialtranslation, wewillreferto itasa translationsymmetry of theobject; whenanobjecthasasymmetry that isanon-trivialrotation, wewillrefertoitasa rotationsymmetry oftheobject; whenanobjecthasa
5.1BasicIdeas 151
symmetrythatisareflection, wewillrefertoitasa reflectionsymmetry oftheobject; whenanobjecthas a symmetry that is anon-trivial glide reflection, wewill refer to it as a glidereflectionsymmetry oftheobject. Whenwearelookingatthesymmetriesofanobject, wewillrefertotheidentityisometryasthe identitysymmetry oftheobject. Forexample, theobjectinFigure 5.1.3 (i)hasrotationsymmetrybutnoreflectionsymmetry; theobjectinFigure 5.1.3 (ii)hasbothrotationandreflectionsymmetry. Wenotethatobjectsthathavetranslationsymmetryorglidereflectionsymmetryhaveto“goonforever.” InFigure 5.1.5 (i)weseeanobjectthat, ifweassumeitcontinuesindefinitelyinbothdirections, hastranslationsymmetry. Ontheotherhand, justbecauseanobject“goesonforever”doesnotmeanitautomaticallyhastranslationsymmetry; seeFigure 5.1.5 (ii). WewillseefurtherexamplesoftranslationsymmetryandglidereflectionsymmetryinSections 5.5 and5.6.
. . . F F F F F F . . . . . . 1 2 3 4 5 6 . . . (i) (ii)
Figure5.1.5
Wewillneedsomeadditionalterminology. Ifanobjecthasatranslationsymmetry, thenwecanlookatthetranslationvectorofthistranslationsymmetry, andwerefertothistranslationvectorasa translationvectoroftheobject. Ifanobjecthasarotationsymmetry, thenwecanlookatthecenterofrotationofthisrotationsymmetry, andwerefertothiscenterofrotationasa centerofrotationoftheobject. A centerofrotationofaplanarobjectmightbethecenterofrotationforrotationsoftheobjectbyvariousangles. Ifanobjecthasareflectionsymmetry, thenwecanlookatthelineofreflection, andwerefertothislineofreflectionasa lineofreflectionoftheobject. Ifanobjecthasaglidereflectionsymmetry, thenwecanlookatthelineofglidereflection, andwerefertothislineofglidereflectionasa lineofglidereflectionoftheobject.Wesee, therefore, thatagivenplanarobjectmightormightnothavecertaindistinguished
pointsthatarecentersofrotations, anditmightormightnothavecertaindistinguishedlines,someofwhichmightbelinesofreflection, andsomelinesofglidereflection.
Exercise 5.1.2. ForeachoftheobjectsshowninFigure 5.1.6, findandindicatethecentersofrotationandthelinesofreflection.
152 5. SymmetryofPlanarObjectsandOrnamentalPatterns
(i) (ii) (iii)
(iv) (v)
Figure5.1.6
Exercise 5.1.3. ThefollowingquestionsinvolvewordsinEnglishwrittenincapitalletters.Assumethatalllettersareassymmetricaspossible, andthatW isobtainedfromM by 180◦
rotation.
(1) Findfourwordsthathaveahorizontallineofreflection. Findthelongestsuchwordyoucanthinkof.
(2) Findoneormorewordsthathaveaverticallineofreflection. Findthelongestsuchwordyoucanthinkof.
(3) Findoneormorewordsthathavea 180◦ rotationsymmetry. Findthelongestsuchwordyoucanthinkof.
Nowthatwehavethenotionoftranslationvectors, centersofrotations, linesofreflectionandlinesofglidereflectionofanobject, wecanstatethefollowingtechnicalresultabouthowthesedistinguishedvectors, pointsandlinesaretreatedbysymmetriesoftheobject. Wewillusethisresultlater, whenwestudythesymmetriesofcertaintypesofplanarobjects. Thisresultshouldnotbesurprising, becausewhatitsaysintuitivelyisthatasymmetryofanobjecttakesspecialkindsofvectors, pointsandlinestothesamekindsofvectors, pointsandlines, whichisreasonablegiventhatasymmetryofanobjectleavestheobjectlookingunchanged. Weomittheproofofthisproposition.
Proposition 5.1.1. Supposethat P isaplanarobject, andlet S beasymmetryof P.
5.1BasicIdeas 153
1. If X isacenterofrotationof P, then S(X) isacenterofrotationof P.
2. If m isalineofreflectionof P, then S(m) isalineofreflectionof P.
3. If m isalineofglidereflectionof P, then S(m) isalineofglidereflectionof P.
4. If v isatranslationvectorof P, then S(v) isatranslationvectorof P.
A keyideainthestudyofsymmetryisthatwecandomorewithsymmetriesthansimplylistthesymmetriesofeachobject. Giventwosymmetriesofanobject, whicharebothisometriesoftheplane, wecanformthecompositionofthesetwosymmetries(asdiscussedinSection 4.4),toobtainanewisometryoftheplane. Whatmakesthiswholestudyofsymmetriesworkfromamathematicalperspectiveisthatthecompositionoftwosymmetriesofanobjectisinfactalsoasymmetryoftheobject. Wenowformulatethisfactmoreprecisely.
Proposition 5.1.2. Supposethat P and Q aresymmetriesofagivenobject.
1. Q ◦ P isasymmetryoftheobject.
2. P−1 isasymmetryoftheobject.
Demonstration. Suppose that theobject forwhich P and Q aresymmetries iscalled K. Bydefinitionofwhatitmeanstobeasymmetryofanobject, weknowthat P andQ areisometries,andthat P(K) = K and Q(K) = K.
(1). Because P and Q arebothisometries, weknowfromProposition 4.4.1 that Q ◦ P isanisometry. Wealsoobservethat (Q ◦ P)(K) = Q(P(K)) = Q(K) = K. Itfollowsthat Q ◦ P
isasymmetryof K.
(2). Because P isanisometry, weknowfromProposition 4.6.5 that P hasaninverseisometryP−1. Additionally, weknowthat P−1 ◦ P = I, whichmeansthat P−1(P(K)) = I(K). BecauseP(K) = K and I(K) = K (thelatterbecause I istheidentityisometry, whichtakeseveryobjectontoitself), itfollowsthat P−1(K) = K. Itfollowsthat P−1 isasymmetryof K.
Wenotethatitisthisabilitytocombinesymmetriesthatmakestheapproachtosymmetryviaisometriestheonethatisparticularlysuitedtoamathematicaltreatment, anditisthemathe-maticalanalysisofsymmetrythatleadstotheinterestingresultsaboutsymmetrythatwewillseelaterinthischapter. Ifwethinkoftheword“symmetry”inthecolloquialusageasanattributeofanobject, thereforebeinganadjectiveratherthananoun, thenwecouldnotmeaningfullycombinesymmetries.Ourgoalinthischapteristoexplorethesymmetriesofplanarobjects. Themostbasicthing
todoistolookatallpossiblesymmetriesofeachgivenobject, thatis, thesymmetrygroupoftheobject. Becausesymmetriesareisometries, wecanuseourknowledgeofisometriestolearnmoreabouttheobject. Inthissectionwehavediscussedsomegeneralideasaboutsymmetriesofobjects. Insubsequentsectionsinthischapter, wewillrestrictourattentiontovariousspecialtypesofplanarobjects, andineachrestrictedcase, wewillbeabletosaymoredefinitiveresults.Actually, ifallwecoulddowouldbetotakeaplanarobject, andfinditssymmetrygroup,
thatwouldbenice, butnotveryinteresting. Whatwouldbemoreinterestingwouldbetoknow
154 5. SymmetryofPlanarObjectsandOrnamentalPatterns
whetherwecouldfindallpossibletypesofsymmetrygroupsthataplanarobjectcouldhave.Ananalogymightbewithbirdwatching. Wecannotlistallpossibleindividualbirdsfoundinagivenregion, butbirdwatchingguideslistallpossibletypesofbirdsthatcanbefoundintheregion, anddescribevariouscharacteristics(forexample, color, shapeofbeak, etc.) thatcanbeusedtoidentifythetypeofanybirdspottedinthewild. Similarly, wecertainlycannotlistallpossibleobjectsthatcouldeverbedrawnintheplane, becausethereareinfinitelymanydifferentthingsthatcanbepictoriallyrepresented. However, andthisisratherremarkable, inthreeimportantcategoriesofplanarornamentalpatterns(whichbetweenthemencompassmanyoftheornamentalpatternsofinterest), wecanlistallpossibletypesofsymmetrygroupsthatcanariseforeacheachcategory. Thethreecategoriesofplanarpatternswewilldiscussarerosettepatterns, friezepatterns, andwallpaperpatterns, whichwillbetreatedindetailinSections 5.4,5.5 and5.6 respectively. OurdiscussionofisometriesinChapter 4, andourgeneraldiscussionofsymmetrygroupsinthissection, isessentiallyaimedatprovidingusthetoolstounderstandtheclassificationofsymmetrygroupsofrosettepatterns, friezepatternsandwallpaperpatterns.(Itwouldbebeyondthescopeofthisbooktoprovideallthetechnicalmathematicaldetailsforvariousproofsneededfortheanalysisoffriezepatternsandwallpaperpatterns, butwewillbeabletogiveallthedetailsforrosettepatterns, andmanyofthekeyideasfortheothertwocases.)Inordertomakeheadwaywiththeideaofclassifyingobjectsbytheirsymmetries, weneed
toaskwhatitwouldtakeinordertobeabletosaythattwoobjects“havethesametypeofsymmetry”? Wesawanexampleearlierinthissection, namelytherectangleandtheletter H,wheretwodifferentobjectshavethesamesymmetrygroups. Ingeneral, wewillsaythattwoobjectshavethesame symmetrytype iftheyhavethesamesymmetrygroups. Thatis, ifwecanmatchupthetranslationsinonesymmetrygroupwiththetranslationsoftheother, therotationsinonesymmetrygroupwiththerotationsoftheother, andsimilarlyforreflectionsandforglidereflections. (Forthosefamiliarwiththetheoryofgroups, itisnotsufficientsimplytorequirethatthetwosymmetrygroupsbeisomorphic; itisnecessarytohaveanisomorphismbetweenthetwogroupsthattakestranslationstotranslations, rotationstorotations, reflectionstoreflectionsandglidereflectionstoglidereflections. Forthosenotfamiliarwiththetheoryofgroups, donotworryaboutthesetechnicalities.)
5.2 SymmetryofRegularPolygonsI
Inordertogetafeelforsymmetrygroups, westartwiththesymmetrygroupsofregularpolygons.Letusexaminethesymmetriesofanequilateraltriangle, asshowninFigure 5.2.1. NoticeinthefigurethatwelabeledtheverticesofthetrianglebyA, B and C. Theselabelsarenotpartofthetriangle, butaretheresimplytohelpusseetheeffectsofvarioussymmetriesonthetriangle.
Thefirst thingwewanttodois tolistallsymmetriesof theequilateral triangle, that is, allisometriesoftheplanethathavethetrianglelandonitself. Itispermissiblethattheseisometriesinterchangethelettersusedtolabelthevertices, becausetheselettersarenotactuallypartof
5.2SymmetryofRegularPolygonsI 155
A
C B
Figure5.2.1
the triangle, andareonlyusedtohelpuskeeptrackofwhat isgoingon. Forexample, onepermissible isometry isreflectionin thevertical linethroughthemiddleof thetriangle. Thisreflectionleavesthetrianglelookingthesame, thoughitinterchangesvertices B and C (leavingA unmoved). Youmightfindithelpfulatthispointtocutanequilateraltriangleoutofpaper,labeltheverticesasinFigure 5.2.1, andperformtheisometrieswewilldiscuss. Unlessyouhavetransparentpaper, ithelpstowriteonbothsidesofthetrianglewhenyoufirstlabelthevertices.Thetrianglecannothaveanytranslationsymmetryorglidereflectionsymmetry, becauseany
translationorglidereflectionoftheplanewouldmovethetriangleoffitself. Thetrianglecanthereforehaveonlyrotationsymmetryandreflectionsymmetry. Whatweneedtofindarethevariouslinesofreflectionofthetriangle, andthevariouscentersofrotationandanglesofrotationof the triangle. InFigure 5.2.2 are indicated the three linesaboutwhich the trianglecanbereflectedwithoutchangingitsappearance; thesethreelinesaredenoted L1, L2 and L3. Thereflectionsthroughtheselinesaredenoted M1, M2 and M3 respectively. Forexample, ifweapplyM2 tothetriangleaspicturedinFigure 5.2.1, weseethatM2 leavesthevertexlabeled Bunmoved, andinterchangestheverticeslabeled A and C; seeFigure 5.2.3. NotethatthelinesL1, L2 and L3 arenotpartofthetriangle, butratherarefixedreferencelines; theynevermove.
L1
L3
L2
Figure5.2.2
156 5. SymmetryofPlanarObjectsandOrnamentalPatterns
A
C B
C
A B
M2
Figure5.2.3
Theidentityisometry, denoted I, iscertainlyasymmetryofthetriangle. Thereareonlytwonon-identity rotations that leave the triangle lookingunchanged: rotationby 120◦ clockwiseaboutthecenterofthetriangle, androtationby 240◦ clockwiseaboutthecenterofthetriangle.Whataboutrotationby 120◦ or 240◦ counterclockwiseaboutthecenterofthetriangle? Thesearecertainlysymmetriesofthetriangle, butrotationby 120◦ counterclockwisehasthesameneteffectasrotationby 240◦ clockwise, andsimilarlyrotationby 240◦ counterclockwisehasthesameneteffectasrotationby 120◦ clockwise. Asalways, weareonlyinterestedintheneteffectofanisometry, andsoitwouldberedundanttousebothcounterclockwiseandclockwiserotations; wewillsticktotheclockwiseones. Itiseasiertothinkofrotationsbyfractionsofwholeturns, rather thandegrees, sowewillwrite R1/3 and R2/3 todenote theclockwiserotationsby 120◦ and 240◦ respectively. Forexample, ifweapply R1/3 to the triangleaspictured inFigure 5.2.1, weseethat R1/3 takesthevertexlabeledA towherevertex Bwas, takesthevertexlabeled B towherevertex C was, andtakesthevertexlabeled C towherevertex A was; seeFigure 5.2.4.
Wenowhaveacompletelistofallsymmetriesoftheequilateraltriangle, namely I, R1/3, R2/3,M1,M2 andM3. Thislististhesymmetrygroupoftheequilateraltriangle; weletG denotethislist. Ournextstepistoseehowthesymmetriesinthislistcanbecombinedviacomposition.UsingProposition 5.1.2 (1), weknowinprinciplethatifwetakeanytwosymmetriesinG, thentheircompositionwillalsobein G. Considerthefollowingexample. Weknowthat M1 andR1/3 aresymmetriesofthetriangle, andsoM1 ◦ R1/3 mustalsobeasymmetryofthetriangle.Hence M1 ◦ R1/3 mustbein G. Whichofthesixmembersof G isitequalto? Thekeypointisthat M1 ◦ R1/3, thoughformedintwostages, hasasingleneteffect, anditisthisneteffectthatequalstheneteffectofpreciselyoneofthesixmembersof G. RecallthatthecompositionM1 ◦ R1/3 meansfirstdoing R1/3 andthendoing M1. InFigure 5.2.5 weseetheneteffectofperforming the two isometries in thespecifiedorder. It is important to recognize that thetransformation M1 alwaysreferstoareflectionintheline L1 exactlyasshowninFigure 5.2.2,nomatterwhathadbeendonetothetrianglepreviously. (Thelines L1, L2 and L3 arenotparts
5.2SymmetryofRegularPolygonsI 157
A
C B
C
B A
R1/3
Figure5.2.4
ofthetriangle, anddonotmovewhenwerotatetheplane; wealwayswantM1,M2 andM3 tomeanthesamethingsatalltimes.) AnexaminationofFigure 5.2.5 revealsthatM1 ◦ R1/3 leavesthevertexoriginallylabeled B unmoved, anditinterchangestheverticesoriginallylabeled A
and C. Hence, theneteffectof M1 ◦ R1/3 isexactlythesameastheneteffectof M2. Wethereforecanwritetheequation M1 ◦ R1/3 = M2.
M1R1/3
A
C B
C
A B
C
B A
M1 ° R1/3 = M2
Figure5.2.5
WecancomposeanymemberofGwithanymemberofG, andtheresultwillbeamemberofG. WecansummarizeallpossiblecompositionsofmembersofG byconstructinga“multiplica-
158 5. SymmetryofPlanarObjectsandOrnamentalPatterns
tion”table forG, whichistheanalogofthemultiplicationtableswelearninelementaryschool;wewillcallsuchatablea compositiontable. ThecompositiontablefortheequilateraltriangleisshowninTable 5.2.1. If P and Q aremembersof G, wefind Q ◦ P inthetablebylookingattheentrylocatedintherowcontainingQ andthecolumncontaining P. Forexample, tofindR2/3 ◦ M3, welookattheentrylocatedintherowcontaining R2/3 andthecolumncontainingM3; thissquarecontains M1. Hence R2/3 ◦ M3 = M1. Thewayweobtainedthe 36 entriesinthetablewassimplybydirectlycalculatingeachone; thesecalculationscanbedoneeitherbymakingdrawingssimilartowhatisshowninFigure 5.2.5, orbyusingacut-outequilateraltriangle. (WewilllearnamoreefficientwaytoconstructthisoperationtableinSection 5.3.)
◦ I R1/3 R2/3 M1 M2 M3
I I R1/3 R2/3 M1 M2 M3
R1/3 R1/3 R2/3 I M3 M1 M2
R2/3 R2/3 I R1/3 M2 M3 M1
M1 M1 M2 M3 I R1/3 R2/3
M2 M2 M3 M1 R2/3 I R1/3
M3 M3 M1 M2 R1/3 R2/3 I
.
Table5.2.1
A numberofthingscanbeseeninTable 5.2.1. First, thetableisclearlysubdividedintofour3× 3 squares, twoofwhichonlyhaverotations(thinkingof I asarotation), andtheothertwohavingonlyreflections. Thereisanicediagonalpatternineachofthefoursquares. Also, notethateachof thesixmembersof G appearsonceandonlyonceineachrow, andonceandonlyonceineachcolumn. Wealsoseeinthetablethattheorderofcompositionofsymmetriesmatters. Forexample, itisseeninthetablethatM1 ◦ R1/3 = M2 , whereas R1/3 ◦ M1 = M3.Hence M1 ◦ R1/3 = R1/3 ◦ M1.Onefinalpointregardingthesymmetriesoftheequilateraltriangle. InSection 4.6 wedis-
cussedthenotionofaninverseisometry. ByProposition 5.1.2 (2), weknowthattheinverseisometryofeachsymmetryof theequilateral triangle isalsoasymmetryof the triangle. WecanstateveryexplicitlywhattheinverseofeachsymmetryofthetriangleisbyusingProposi-tion 4.6.5. Moreprecisely, wehave I−1 = I, R1/3
−1 = R2/3, R2/3−1 = R1/3 M1
−1 = M1,
M2−1 = M2 and M3
−1 = M3.Everythingthatwehavediscussedconcerningtheequilateraltriangleworkssimilarlyforany
regular n-gon. Justaswehavethreerotations(includingtheidentity)andthreereflectionsforthetriangle, yieldingatotalofsixsymmetries, similarlyaregular n-gonhas n rotationsand n
reflections, yieldingatotalof 2n symmetries. Thesmallestrotationofaregular n-gonis R1/n,andallotherrotationsaremultiplesofthisrotation. Forexample, thesquarewillhavesymmetrygroupwithmembers I, R1/4, R1/2, R3/4,M1,M2,M3 andM4, wherethefourreflectionshavelinesofreflectionasshowninFigure 5.2.6. Thesymmetriesofaregular n-gonare:
I, R1/n, R2/n, R3/n, . . . , R(n−1)/n,M1,M2,M3,M4, . . . ,Mn.
5.2SymmetryofRegularPolygonsI 159
Itdoesnotmakeanysubstantialdifferencehowthelinesofreflectionforaregularpolygonarearranged, but, foruniformity(andlaterconvenience), wewillalwaysassumethattheyarearrangedasinFigure 5.2.6, namelywith L1 vertical, andtheothersincounterclockwiseorder(thisorderwillturnouttobeusefulinSection 5.3). Wecanformthecompositiontableforthesymmetrygroupofeachregular n-gon, analogouslytoTable 5.2.1. Thepatternsthatwesawinthetablefortheequilateraltrianglealsoholdforotherregular n-gons.
L1
L3
L4
L2
Figure5.2.6
Exercise 5.2.1. Listallthesymmetriesofaregularpentagon, andofaregularhexagon.
Exercise 5.2.2. Construct thecomposition table for thesymmetrygroupof thesquare.UselinesofreflectionlabeledasinFigure 5.2.6. Calculateeachentryinthetabledirectly;donotsimplycopy thepatternofTable 5.2.1. (Thepointof thisexerciseis toverifybyactualcalculationthatthecompositiontableforthesquarehasthesamepatternasfortheequilateraltriangle.)
160 5. SymmetryofPlanarObjectsandOrnamentalPatterns
Exercise 5.2.3. Fortheregularoctagon, computethefollowingsymmetries(thatis, expresseachasasinglesymmetry).
(1) R1/8 ◦ R3/8;
(2) R1/4 ◦ R5/8;
(3) R1/8 ◦ M3;
(4) M1 ◦ M5;
(5) (M6)−1;
(6) (R3/8)−1.
5.3 SymmetryofRegularPolygonsII
Supposethatwewantedtocompute R1/3 ◦ M3 ◦ M2 ◦ R1/3 ◦ M1 foranequilateraltriangle.UsingthemethodofSection 5.2, wecouldeitherdoitdirectlybydrawingatriangleanddoingeachisometryoneata time, orwecoulduse thecomposition tablegiven inTable 5.2.1 tocomputethecompositionofthefirsttwoisometries, thenthecompositionoftheresultwiththenextisometry, andsoon. Eitherway, itwouldbeaslightlytediouscalculation, thoughwecoulddoit.Nowsupposewewantedtocompute R1/20 ◦ M17 ◦ M53 ◦ R3/100 ◦ M1 foraregular
100-gon. InprinciplewecouldusethemethodofSection 5.2, butinpracticeitwouldbesotediousthatnoonewouldwanttodoit, becausedrawinga 100-gonwouldbeverydifficult,andmakingacompositiontablefora 100-gonwouldtakealongtime.InthissectionwepresentanalternativeapproachtothematerialwediscussedinSection 5.2,
andthisalternativeapproachwillallowustodocalculationsforaregular 100-gonjustaseasilyaswedoforanequilateraltriangle. Insteadoffiguringoutcompositionsofsymmetriesdirectly,wedevelopan“algebra”ofsymmetriesofregularpolygons. NotonlywillitbeeasiertofillintablessimilartoTable 5.2.1 oncewehavethealgebraicapproach, butwewillseethatessentiallythesamerulesworkforallregular n-gons.Tostart, wewant toexpressall thesymmetriesofaregular n-gonintermsofa fewbasic
symmetries. Thesebasicsymmetrieswillbesortoflikeatoms, outofwhichallothersymmetriesarebuilt. Nomatterwhatthevalueof n is, wewillalwaysusethesamethreesymmetriesasourbuildingblocks. InordertodistinguishthealgebraicapproachofthissectionfromthegeometricapproachofSection 5.2, wewilladoptadifferentnotation, which looksmore likealgebra.(However, whatwearedoinghereisnotthesameasthealgebrawelearninschool—thatdealswithnumbers, whereasherewearedealingwithsymmetries. Numbersandsymmetriesdonotbehaveexactlythesame.)
5.3SymmetryofRegularPolygonsII 161
Supposewehavearegular n-gonforsome n. First, welet 1 denotetheidentitysymmetry,previouslydenotedby I. Next, let r denotethesmallestpossiblenon-trivialclockwiserotationsymmetry, previouslydenoted R1/n. Forexample, in thecaseof theequilateral triangle, wewillhave r = R1/3; inthecaseofthesquare, wewillhave r = R1/4. Hence, thesymbol rdenotesrotationbyadifferentangleforeachdifferentregularpolygon. Inallcases, however,weknowthat r isthesmallestpossiblenon-trivialclockwiserotationsymmetry. Third, let mdenotereflectioninaverticalline, previouslydenoted M1. (Itwouldworkjustaswelltoletm beanyotherreflectionsymmetry, butwewillalwayschoosereflectioninavertical line,sothattherewillbenoambiguityaboutwhichreflectionisreferredtoby m.) Last, insteadofwritingcompositionusingthesymbol ◦, wewillsimplyusethestandardalgebraicnotationformultiplicationtodenotecomposition. Hence, whatweusedtowriteas M1 ◦ R1/n wenowdenote mr.Wecanusesome further standardalgebraicnotationaswell. Tostart, if k isanypositive
integer, wewill let rk meantheproductof r withitself k times. Notethat r1 = r. Wewilllet r−1 denote R−1/n, that is, the smallest possible counterclockwise rotationof the n-gon.
Forconvenience, againfollowingstandardalgebraicpractice, wewilllet r0 = 1, andforanypositiveinteger k, wewilllet r−k = (r−1)k. Thesamesortofnotationappliestoexpressionsoftheform mk.Itturnsoutthatwecanrewritealltheothersymmetriesofaregular n-gonusingthethree
symmetries 1, r and m. Letusstartwiththeequilateraltriangle, beforestatingtheresultmoregenerally. InthenotationweusedinSection 5.2, thesymmetriesoftheequilateraltriangleareI, R1/3, R2/3, M1, M2 and M3. (ItisimportantforwhatfollowsthatthelinesforthesethreereflectionsbeaspicturedinFigure 5.2.2.) Wehavealreadyseenthat I = 1, that R1/3 = r,
andthat M1 = m. Itisstraightforwardtoverifythat R2/3 = R1/32 = r2. Whatabout M2 and
M3? Weproceedasfollows. InTable 5.2.1 wesawthat M1 ◦ M2 = R1/3. HenceweobtainM1 ◦ M1 ◦ M2 = M1 ◦ R1/3. Weknowthat M1 ◦ M1 = I (byProposition 4.6.5 (4)),anditthereforefollowsthat M2 = M1 ◦ R1/3. Switchingtoournewnotation, weseethat
M2 = mr. A similarcalculationshowsthat M3 = mr2; thereaderisaskedtosupplythedetails. (Alternatively, wecouldhavetakenanequilateraltriangle, anddirectlyverifiedaswedidinSection 5.2 thatthecomposition mr hasthesameneteffectas M2, andsimilarlyformr2 and M3.) Wecouldalsohaveexpressed M2 as r2m, and M3 as rm, butforthesakeofuniformitywewillalwayskeeptheletterm ontheleftandtheletter r ontheright. Alltold, thesixsymmetriesoftheequilateraltrianglecanbewrittenas 1, r, r2, m,mr and mr2.A similarcalculationwouldshowthattheeightsymmetriesofthesquarecanbewrittenas
1, r, r2, r3, m, mr, mr2 and mr3. Thesamepatternholds fora regular n-gon, where thecompletelistofsymmetriesinournewnotationisasfollows:
Geometricnotation I R1/n R2/n · · · R(n−1)/n M1 M2 M3 · · · Mn
Algebraicnotation 1 r r2 · · · rn−1 m mr mr2 · · · mrn−1
162 5. SymmetryofPlanarObjectsandOrnamentalPatterns
Onceagain, westressthattheaboveequalitiesholdexactlyaswrittenonlyifweassumethatthelinesorreflectionarearrangedasinFigure 5.2.6, namelywiththelinesofreflectionarrangedconsecutivelyincounterclockwiseorder.Now thatweknowhow towriteour symmetries in termsof 1, r and m, we turn to the
compositionofsymmetries. InSection 5.2, wecomputedthecompositionoftwosymmetriesgeometrically, byseeingtheeffectonthen-gonofeachofthetwoisometriesperformedoneaftertheother. Forexample, inthecaseoftheequilateraltriangle, wecomputed M1 ◦ R1/3 = M2
asshowninFigure 5.2.5, whereweseethetwoisometriesperformedinthespecifiedorder.Thoughstraightforward, suchgeometriccomputationsarequitetedious, andarealsopronetoerror. However, becauseallsymmetriesoftheregularn-gonhavenowbeenrewrittenintermsof1, r andm, oncewecanfigureouthowtocomposethesethreebasicsymmetries, wewillthenhaveaquickmethodforcomposinganytwosymmetriesofthen-gon. Thefollowingpropositionlistssomeofthemostofthebasicrulesforcombining 1, r and m.
Proposition 5.3.1. Let 1, r and m bedefinedasaboveforaregular n-gon.
1. r · 1 = r and r = 1 · r;2. m · 1 = m and m = 1 ·m;
3. rarb = ra+b;
4. mamb = ma+b;
5. meven = 1 and modd = m, where even denotesanyevennumber, and odd denotesanyoddnumber;
6. rn = 1.
Demonstration.
(1). Thesetwoequalitiesareclear, because 1 issimplyanothernotationfortheidentityisom-etry I, andwecanapplyProposition 4.4.2 (1).
(2). ThisissimilartoPart (1).
(3). Recall that ra means theproductof r with itself a times, andsimilarly rb means theproductof r withitself b times. Hence, weseethat rarb meanstheproductof r withitselfb+ a times, whichisthesameas a+ b times. Because ra+b alsomeanstheproductof r withitself a+ b times, wededucethat rarb = ra+b.
(4). ThisissimilartoPart (3).
(5). Recallthat m isareflectionoftheplane. Itthenmustbethecasethat m2 = 1, whichisjustarestatementinourcurrentnotationofProposition 4.6.3 (1). If even denotesapositiveevennumber, then meven = m2m2m2 · · ·m2 = 1. If odd denotesapositiveoddnumber,then modd = m2m2m2 · · ·m2m = m. A similarargumentholdsfornegativeevenandoddnumbers.
5.3SymmetryofRegularPolygonsII 163
(6). Recallthat r isjustanothernotationfor R1/n. ItfollowsfromProposition 4.6.4 (1)thatcomposing R1/n withitself n timesisthesameasa 360◦ rotation, whichequalstheidentityisometry.
NoticethatRules (1)–(4)intheabovepropositionarejustlikestandardalgebraicrulesfornum-bers, whereasRules (5)–(6)arenotatalllikethestandardrulesfornumbers. Hence, althoughournotationusing 1, r and m isreminiscentofstandardalgebra, itisnotthesameasit. Weshouldalwayskeepinmindthatthesymbols 1, r andm asusedhereareshort-handnotationsforvariousisometriesoftheplane, anddonotdenotenumbers. NoticealsothatRules (1)–(5)holdidenticallyforallregularpolygons, whereasRule (6)variesfordifferentvaluesof n. Thatis, forasquareRule (6)is r4 = 1, whereasforaregularpentagonthesameruleis r5 = 1.
Exercise 5.3.1. [UsedinThisSection] Foraregular n-gon, showthat rn−a = r−a foranyinteger a. Inparticular, deducethat rn−1 = r−1.
InProposition 5.3.1 wesawtherulesforcombiningexpressionsinvolvingonly r oronly m.Wehavenotyetseentherulesforcombiningexpressionsinvolvingbothm and r together; wenowturntothismissingcase. Asbefore, letusstartwiththecaseoftheequilateraltriangle.Considerthecomposition rm. Becausethisisthecompositionoftwosymmetriesoftheequi-lateraltriangle, weknowitmustbeequaltoasinglesymmetryoftheequilateraltriangle. Inotherwords, itmustbethecasethat rm equalsoneof 1, r, r2,m,mr ormr2. Ifwecalculatetheneteffectof rm fortheequilateraltriangle, usingadrawingoracut-outtriangle(justaswedidinSection 5.2), wewillseethat rm = mr, andthatinfact rm = mr2 (thereadershouldverifythisequality). Ifwetrythesameresultforthesquare, itwillturnoutthat rm = mr3.Itappears, unfortunately, asifwedonothavethesameresultforthetwodifferentpolygons.However, everythingworksoutnicelyifwerewriteourformulas. Forthecaseoftheequilateraltriangle, observethat r2 = r−1 (useExercise 5.3.1 (2)with n = 3), andtherefore rm = mr−1.Inthecaseofthesquare, wehave r3 = r−1, andthereforewealsohave rm = mr−1. Wenowseeageneralpattern, asstatedinthefirstpartofthefollowingproposition; thesecondpartofthepropositiongeneralizestheresultevenfurther.
Proposition 5.3.2. Let 1, r and m bedefinedasaboveforaregular n-gon.
1. rm = mr−1;
2. ram = mr−a foranyinteger a.
Demonstration. Thefirstpartoftheproofisgeometric, whereasthesecondpartisalgebraic.
(1). Wewillshowthat rm = mr−1 byapplyingeachof rm and mr−1 toaregular n-gon,andwewillcompare the results. InFigure 5.3.1 weseea regular n-gon, with someof theverticeslabeled. InFigure 5.3.2 weseetheresultofapplyingfirst m andthen r, yieldingtheneteffectofdoing rm tothe n-gon. InFigure 5.3.3 weseetheresultofapplyingfirst r−1 and
164 5. SymmetryofPlanarObjectsandOrnamentalPatterns
thenm, yieldingtheneteffectofdoingmr−1 tothe n-gon. Wethereforeseethattheneteffectofdoing rm and mr−1 isthesame, andthereforethesetwoisometriesareequal.
(2). If a isapositiveinteger, thenwecanapplyPart (1)ofthispropositiontocompute
ram = rr · · · r︸ ︷︷ ︸a times
m = rr · · · r︸ ︷︷ ︸a−1 times
mr−1 = rr · · · r︸ ︷︷ ︸a−2 times
mr−1r−1 = · · · = mr−1r−1 · · · r−1︸ ︷︷ ︸a times
= m(r−1)a = mr−a.
If a isnotpositive, thentheaboveargumentdoesn’twork, sowetakethefollowingapproach.Let a beanyinteger. Thenwenotethat mra isthecompositionof m and ra. Whatever a is,weknowthat ra issomerotation, andm isareflection. Hence ra isorientationpreserving, andm isorientationreversing. ByProposition 4.4.3 (2)weseethat mra isorientationreversing.Becauseallthesymmetriesofaregularpolygonarerotationsandreflections, wededucethatmra mustbeareflection. Hence, byProposition 4.6.3 (1)weknowthat (mra)(mra) = 1.Hence mramra = 1. Multiplyingbothsidesontheleftby m andontherightby r−a, weobtain mmramrar−a = m1r−a. Cancellingtheadjacentm’s, andcancelling ra and r−a, weobtain ram = mr−a. (ThisdemonstrationactuallymakesunnecessarythedemonstrationofPart(1), andthedemonstrationofPart(2)when a isapositiveinteger, butthosedemonstrationsareintuitivelymorestraightforward, andsowereworthkeeping.)
DA
B C
Figure5.3.1
Wenowhaveallthealgebraicrulesneededforworkingwith 1, r andm. Usingthesealgebraicrules, wecannoweasilyconstructcompositiontables forthesymmetriesofregular n-gons.Insteadoffiguringouteachentryinthesetablesgeometrically(thatis, bydrawingeachcase, orusingacut-out), aswedidinSection 5.2, wecansimplyuseouralgebraicrules, andessentiallyforgetaboutthegeometry. (Wearenotreallyforgettingthegeometry—thealgebraicrulesforcombining 1, r and m summarize thegeometryof the regularpolygons. Havingdevelopedtheserules, wenolongerneedtokeepgoingbacktothegeometry.) Asanexample, letuslookat thecomposition table for theequilateral triangle. Wehavealreadyseen thiscompositiontableinTable 5.2.1, butletusstartfromscratch. WestartoffwithTable 5.3.1, wherewesee
5.3SymmetryofRegularPolygonsII 165
rm
rm
DA
B C
AD
C B
D
C
B A
Figure5.3.2
mr-1
mr-1
DA
B C
C
A B
DD
C
B A
Figure5.3.3
theoperationtablewithnoentriesfilledin(weputinextralinestomakeiteasiertoview). Wewishtocomputefoursampleentriesinthetable, whichwehavelabeled A, B, C and D.WecomputethefourdesiredentriesintheabovetableusingvariouspartsofProposition 5.3.1
andProposition 5.3.2. (Recallthat, asinSection 5.2, theentrylocatedintherowcontainingQ andthecolumncontaining P is Q ◦ P.) Letusstartwithentry A. Thisentryis theresultofthecomposition r · r, whichclearlyequals r2. Theentry B istheresultofthecompositionm ·mr, whichequals m2r = 1 · r = r. Theentry C istheresultofthecomposition mr · r2,whichequals mrr2 = mr1+2 = mr3 = m · 1 = m. Finally, theentry D istheresultofthecomposition mr2 ·mr, whichequals mr2mr; usingProposition 5.3.2 (2), thislastexpressionequals mmr−2r = m2r−2+1 = r−1, andbecauseweareworkingwithanequilateraltriangle,
166 5. SymmetryofPlanarObjectsandOrnamentalPatterns
· 1 r r2 m mr mr2
1r A
r2
m B
mr C
mr2 D
Table5.3.1
weseethatentry D equals r2. WecanthereforestarttofillinourcompositiontableasshowninTable 5.3.2.
· 1 r r2 m mr mr2
1r r2
r2
m r
mr m
mr2 r2
Table5.3.2
Usingthesamesortsofcalculations, wecaneasilycompletetheentiretable, asshowninTable 5.3.3.
· 1 r r2 m mr mr2
1 1 r r2 m mr mr2
r r r2 1 mr2 m mr
r2 r2 1 r mr mr2 m
m m mr mr2 1 r r2
mr mr mr2 m r2 1 r
mr2 mr2 m mr r r2 1
Table5.3.3
WecannowcompareTable 5.3.3 withTable 5.2.1. Asexpected, thetwotablesareentirelyidentical, exceptforthechangeofnotation. Inotherwords, ifwetakeTable 5.3.3, andreplaceeveryinstanceof 1 with I, everyinstanceof r with R1/3, etc., wewillobtainTable 5.2.1 pre-cisely. Hence, thereisnoessentialdifferencebetweentheresultsofcomputingcompositionsofsymmetriesgeometricallyvs.algebraically, althougheachmethodismoreconvenientorintu-itivelyappealingindifferentsituations. Finally, wementionthatwhatwehavejustdoneforthe
5.3SymmetryofRegularPolygonsII 167
equilateraltrianglecanalsobedoneforanyregular n-gon. Oneoftheadvantagesofthealge-braicapproachisthatthealgebraicrulesarethesameforallregular n-gons(withtheexceptionofProposition 5.3.1 (6)).
Exercise 5.3.2. Construct thecomposition table for thesymmetrygroupof thesquare,analogouslytoTable 5.3.3. Useonlyouralgebraicrules, withoutactuallydrawingasquare.(Calculateeachentryinthetabledirectly; donotsimplycopythepatternofTable 5.3.3.)
Oneuseofthealgebraicapproachtosymmetriesofregularpolygonsisthatitallowsustosimplifycomplicatedexpressionsinvolvingsuchsymmetries. Forexample, supposewearegiventheexpressionmr5m6r3mr forsomeregularpolygon. WecansimplifyitusingtherulesgiveninProposition 5.3.1 andProposition 5.3.2. Recallthat m and r arenotregularnumbers, andthereforewecanuseonlytherulesdiscussedinthissection, andnottheregularrulesforalgebra.Theideaistodothesimplificationonestepatatime. Weproceedasfollows, underscoring,andjustifying, eachsteptaken:
mr3m7r9mr = mr3mr9mr byProposition 5.3.1 (5)
mr3mr9mr = mmr−3r9mr byProposition 5.3.2 (2)
mmr−3r9mr = r−3r9mr byProposition 5.3.1 (5)
r−3r9mr = r6mr byProposition 5.3.1 (3)
r6mr = mr−6r byProposition 5.3.2 (2)
mr−6r = mr−6+1 byProposition 5.3.1 (3)
mr−6+1 = mr−5
Wecanobserveintheabovecalculationsomeideasthatcanbeusedinanysimilarsituation.First, wheneverwehavean m toapower, wecansimplifytheexpression. Second, ifwehaveadjacent letters m, or ifwehaveadjacentpowersof r, wecan simplify. Third, ourgeneralstrategyistomoveallthelettersm totheleft, andalltheletters r totheright, sothateventuallywewillhavearesultthatiseitheroftheform ra forsomeinteger a, oroftheform mra forsomeinteger a. Thewaywemovetheletters m totheleftandtheletters r totherightisbyusingProposition 5.3.2.Intheaboveexample, westoppedwhenweobtainedmr−5. Thereisnopossibilityofsimplify-
ingfurther, giventhatwedonotknowhowmanyedgestheregularpolygonhas. Wenowturntoanotherexample, thistimeforaspecifictypeofregularpolygon. Supposearegiventheexpres-sion mr10m7rm forthesquare. Wecansimplifythisexpressionasfollows, thistimeomittingthejustificationforeachstep(whichthereadershouldsupply), thoughwestillunderscoreeachsteptaken.
mr10m7rm = mr10mrm = mmr−10rm = r−9m = mr9 = mr4r4r = mr.
168 5. SymmetryofPlanarObjectsandOrnamentalPatterns
Observethatitwasonlyintheverylaststepthatweusedthefactthatthepolygonwasasquare;uptillthatpoint, weproceededexactlyaswehaddoneforanarbitrarypolygon.
Exercise 5.3.3. Simplifyeachofthefollowingexpressionsforarbritraryregularpolygons.Ineachcase, theanswershouldbeoftheform 1 or ra or mra forsomeinteger a.
(1) mrmr2.
(2) r5mmrm.
(3) r7m3r2mr.
(4) mr3mr3mr4mr4.
(5) m4rm3rm2rm.
Exercise 5.3.4. Simplifyeachofthefollowingexpressionsfortheregularpolygonindi-cated. Ineachcase, theanswershouldbeoftheform 1 or ra ormra forsomenon-negativeinteger a, where a islessthanthenumberofedgesofthepolygon.
(1) rmr2m fortheequilateraltriangle.
(2) m3r6mrm fortheequilateraltriangle.
(3) mr9m2r forthesquare.
(4) mr4mr3mr2mr fortheregularpentagon.
(5) mr4mr3mr2mr fortheregularhexagon.
Intheaboveexamplesofsimplifyingexpressions, westartedwithanexpressioninalgebraicnotation, andusedouralgebraic rules inorder to simplify. Wecanalsouse thismethod tosimplifyanexpressionwritteninthegeometricnotationofSection 5.2, byfirstconvertingtoalgebraicnotation, thensimplifying, andthenconvertingback. Weconsidertheexamplegivenattheverybeginningofthissection, namelysimplifying R1/20 ◦ M17 ◦ M53 ◦ R3/100 ◦ M29
foraregular 100-gon.
R1/20 ◦ M17 ◦ M53 ◦ R3/100 ◦ M1 = r5 ·mr16 ·mr52 · r3 ·m = r5mr16mr52r3m
= r5mr16mr55m = mr−5r16mr55m = mr11mr55m
= mmr−11r55m = r44m = mr−44 = mr−44r100
= mr56 = M57.
ItwouldbeveryunpleasanttotrytosimplifytheaboveexpressiongeometricallyaswedidinSection 5.2.
5.3SymmetryofRegularPolygonsII 169
Exercise 5.3.5. Simplifyeachofthefollowingexpressionsfortheregularpolygonindi-cated. Ineachcase, theanswershouldbeinthegeometricnotation.
(1) R1/3 ◦ M1 ◦ R2/3 ◦ M3 fortheequilateraltriangle.
(2) M2 ◦ R1/3 ◦ M3 ◦M1 fortheequilateraltriangle.
(3) R1/2 ◦ M1 ◦ M3 ◦ R3/4 forthesquare.
(4) R3/5 ◦ M1 ◦ M5 ◦ R2/5 ◦ M3 fortheregularpentagon.
(5) R1/2 ◦ M1 ◦ M3 ◦ R3/4 fortheregular 60-gon.
(6) R1/2 ◦ M1 ◦ M50 ◦ M3 ◦ R3/4 fortheregular 80-gon.
Wefinishthissectionbymentioningthealgebraicapproachtofindinginversesofsymmetriesofregularpolygons, assummarizedinthefollowingproposition. Part (4)ofthepropositionmightlookasifitwerebackwardsatfirstglance, butitiscorrect, andistheresultofthefactthatordermatterswhencombiningsymmetries.
Proposition 5.3.3. Let 1, r and m bedefinedasaboveforaregular n-gon.
1. (ra)−1 = r−a = (r−1)a foranyinteger a;
2. m−1 = m;
3. (mra)−1 = mra foranyinteger a.
4. (xy)−1 = y−1x−1 foranysymmetries x and y.
Demonstration.
(1). Recallthat r isanothernotationfor R1/n. Hence ra isanotherwayofwriting Ra/n. Using
Proposition 4.6.5 (3)weseethat(Ra/n
)−1= R−a/n, andthislastexpressionisseentobethe
sameas r−a, whichinturnisthesameas (r−1)a.
(2). Because m is a reflection, the equationweneed to show is simply a restatementofProposition 4.6.5 (4).
(3). AsdiscussedinthedemonstrationofProposition 5.3.2 (2), weknowthatmra isareflec-tion. ThisequationweneedtoshowissimplyarestatementofProposition 4.6.5 (4).
(4). Wecompute
(xy)(y−1x−1) = x(yy−1)x−1 = x · 1 · x−1 = 1.
Itfollowsthat (xy)−1 = y−1x−1.
WenotethatPart (4)of theabovepropositioncanbeextendedtoanynumberofsymme-tries, notjusttwo. Forexample, inthecaseoffoursymmetrieswewouldhave (xyzw)−1 =w−1z−1y−1x−1.
170 5. SymmetryofPlanarObjectsandOrnamentalPatterns
5.4 RosettePatterns
Regularpolygons, thesymmetriesofwhichwestudiedinSections 5.2 and5.3, maybeinter-estingmathematically, buttheyarenotpatternsofgreataestheticinterest. Wewishtoturnourattentiontoaestheticallymoreinteresting—andmathematicallymorecomplicated—objects, of-tenreferredtoasornamentalpatterns. Westartwithrosettepatterns(orsimply, rosettes) whicharethesimplestofthethreetypesofornamentalpatternsthatwewilltreat. A rosettepatternisdefinedtobeanyplanarobjectthathasonlyfinitelymanysymmetries. SeeFigure 5.4.1 forsomeexamplesofrosettepatterns. (Weleaveittothereadertolistallthesymmetriesforeachoftheobjectsinthefigure, thusverifyingthattheyareindeedrosettepatterns.)Thename“rosette”comesfromrosewindowsincathedrals, althoughinfactthehumanfigures
portrayedinarosewindowoftenpreventthewindowfromhavingnon-trivialsymmetry.
(i) (ii) (iii) (iv)
Figure5.4.1
Someauthorsuse the term“finitefigure” insteadof rosettepattern, butwe feel thisnameissomewhatmisleading, becausewhatisfiniteaboutarosettepatternisonlythenumberofsymmetries, not thegeometricnatureof thefigure. Forexample, the infinitecrossshowninFigure 5.4.2 (i)isarosettepattern(ithaseightsymmetries), eventhoughitisnotgeometricallyfinite. Moreover, notallplanarfiguresthatarefiniteinsizearerosettepatterns, forexamplethecircleshowinFigure 5.4.2 (ii). Thecirclecanberotatedaboutitscenterbyanyangle, andsoithasinfinitelymanysymmetries.
Ourultimategoalforrosettepatternsistoclassifythemaccordingtotheirsymmetrygroups.Thatis, wewishtolistallsymmetrygroupsthatariseasthesymmetrygroupsofrosettepatterns,andtobeabletotakeanygivenrosettepattern, andidentifywhichsymmetrygrouponourlistcorrespondstoit—analogoustoacompletefieldguidetothebirdsofNorthAmericathatlistsalltypesofbirdsthatcanbefound, andgivesidentifyingcharacteristicsforeachtypeofbird. Remarkably, wecanmakeacompletefieldguideforrosettepatterns. (Ofcourse, justasthefieldguideforbirdslistsonlyeachtypeofbird, noteachindividualbird, sotooourlistofsymmetrygroupsofrosettepatternsdescribesonlythesymmetriesofrosettepatterns, nottheparticulardesignelements.) IncontrasttoourdiscussionoffriezepatternsandwallpaperpatternsinSections 5.5 and5.6, whereitwouldbebeyondthescopeofthisbooktogiveallthe
5.4RosettePatterns 171
(i) (ii)
Figure5.4.2
mathematicaldetailsofthedemonstrationsofthemainresults, inthecaseofrosettepatternsweareabletodemonstrateallthepropositionsinthissection; thesedemonstrationsaresomewhatlengthy, andtheyarefoundinAppendices CandD.A symmetrygroupisacollectionofsymmetries. Tounderstandwhatcollectionsofsymmetries
ariseassymmetrygroupsofrosettepatterns, wewilllookateachofthefourtypesofisometriesasappliedtorosettepatterns. Westartwithtranslationsandglidereflections. Intuitively, ifarosettepatternhada translationsymmetry, thendoingthetranslationtwicewouldalsobeasymmetry, andthreetimes, fourtimes, etc. wouldallbesymmetries. Itwouldfollowthattheobjecthadinfinitelymanysymmetries, whichcannotbethecaseforarosettepattern. Hence,arosettepatterncannothavetranslationsymmetry. Similarlyforglidereflectionsymmetry. Wethereforehavethefollowingproposition.
Proposition 5.4.1. A rosettepatternhasnotranslationsymmetryandnoglidereflectionsym-metry.
A rosettepatterncanhavereflectionand/orrotationsymmetry. ThepatterninFigure 5.4.1 (i)hasrotationsymmetrybutnoreflectionsymmetry; thepatterninPart (ii)ofthefigurehasrotationandreflectionsymmetry; thepatterninPart (iv)ofthefigurehasnosymmetryotherthantheidentitysymmetry. Westartourdiscussionbyexaminingrotationsymmetryofrosettepatterns.
BEFORE YOU READ FURTHER:
Whatcanbesaidaboutcentersofrotationofrosettepatterns. Specificially, trytodecidewhetherarosettepatterncanhavemorethanonecenterofrotation. Ifyes, trytodrawsucharosettepattern; ifno, saywhynot.
172 5. SymmetryofPlanarObjectsandOrnamentalPatterns
Ourfirstresultaboutrotationsymmetryofrosettepatternsisthefollowingproposition, whichanswerstheabovequestion.
Proposition 5.4.2. Ifarosettepatternhasrotationsymmetries, allsuchsymmetrieshavethesamecenterofrotation.
Wenowknowthatanyrosettepatternhasatmostonecenterofrotation. Ifthereisacenterofrotationinarosettepattern, canwesaysomethingaboutthepossibleanglesofrotationfortherotationsymmetriesaboutthiscenterofrotation. Itturnsoutthatwecansayagooddealaboutsuchangles. Becausewewillneedasimilaranalysis inour treatmentofwallpaperpatternsinSection 5.6, westateournextpropositioninageneralformthatisnotrestrictedtorosettepatterns.Supposewehaveaplanarobject(notnecessarilyarosettepattern)withacenterofrotation.
Hence, there is at least one rotation symmetryof theobjectwith thispoint as its centerofrotation. Theremightbemorethanonesuchrotationsymmetry; thatis, theremightberotationsymmetriesaboutthiscenterofrotationbyvariousangles. Amongalltheserotationsymmetriesaboutthiscenterofrotation, theremightormightnotbeasmallestclockwiserotationsymmetry(recallthattheterm“rotationsymmetry”alwaysmeansanon-trivialrotation). InFigure 5.4.2 (i)there is a smallest clockwise rotation symmetry, namelyby 90◦; in Figure 5.4.2 (ii) there isnosmallestclockwiserotationsymmetry(rotationsymmetriescanbeusedwitharbitrarilysmallangles). Wenotethatifthereisasmallestclockwiserotationsymmetry, thenrotationbynegativeoftheangleisthesmallestcounterclockwiserotationsymmetry, andvice-versa, soweneedonlyconsiderclockwiserotations.Thesituationswhereacenterofrotationdoesnothaveasmallestrotationsymmetryarecom-
plicatedmathematically, andarenotusefulforus. Bycontrast, thesituationwherecentersofrotationhavesmallestrotationsymmetriesisofgreatinterest. Thecrucialfactisthefollowingproposition, thedemonstrationofwhichisfoundinAppendix C.
Proposition 5.4.3. Let P beaplanarobject. Supposethat A isacenterofrotationof P, andsupposethatthereisasmallestclockwiserotationsymmetryabout A. Thenthereisapositiveinteger n suchthatthefollowingpropertieshold.
1. Thesmallestclockwiserotationsymmetryof P about A is RA1/n.
2. Anyrotationsymmetryof P about A isoftheform[RA1/n
]k= RA
k/n forsomeinteger k.
3. Thecollectionofalltherotationsymmetriesof P is {I, RA1/n, R
A2/n, R
A3/n, . . . R
A(n−1)/n}.
Wecannowdefinesomeveryusefulterminology. Supposethat P beaplanarobject, andsupposethat A isacenterofrotationof P. IfthereisasmallestclockwiserotationsymmetryaboutA, thenbytheabovepropositionweknowthatthesmallestclockwiserotationsymmetryabout A isbyanangleoftheform 360◦/n forsomewholenumber n. Thatis, thesmallestclockwiserotationsymmetryis RA
1/n. Wethensaythatthecenterofrotation A isof order n.(Itismoreconvenienttorefertothenumber n thantothefraction 1/n.) Additionally, if A is
5.4RosettePatterns 173
acenterofrotationoforder n, andifthereisanotherrotationsymmetry RAβ about A forsome
angle β, thenProposition 5.4.3 (2) implies that β isan integermultipleof 360◦/n. That is,wehave β = (k · 360◦)/n forsomewholenumber k, whichmeansthat RA
β istheresultofcomposing RA
1/n withitself k times.Wenowreturntoourdiscussionofrosettepatterns. Supposethatarosettepatternhasrotation
symmetries. ByProposition 5.4.2, allsuchsymmetrieshavethesamecenterofrotation, say A.Becausetherosettepatternhasonlyfinitelymanysymmetries, thentheremustbeasmallestclockwiserotationsymmetryabout A. Therefore, asjustdiscussedinthepreviousparagraph,thecenterofrotation A hasorder n forsomewholenumber n. Wethensaythattherosettepatternisof order n. Ifarosettepatternhasnorotationsymmetrywesaythatitisof order 1.Forexample, therosettepatternsinFigure 5.4.1 areoforders 4, 3, 5 and 1 respectively. Everyrosettepatternhasanorder(whichisoneofthenumbers 1, 2, 3, 4, . . .).Wenowturntoreflectionsymmetryofrosettepatterns.
BEFORE YOU READ FURTHER:
Thinkaboutwhatcanbesaidabouttherelationbetweenlinesofreflectionandcentersofrotationofrosettepatterns.
Thefollowingpropositioncompletelycharacterizestherelationbetweenlinesofreflectionandcentersof rotationof rosettepatterns. Thedemonstrationof thisproposition is found inAppendix D.
Proposition 5.4.4.
1. Ifarosettepatternhasbothreflectionsymmetryandrotationsymmetry, thenalllinesofreflectiongothroughthesinglecenterofrotation.
2. Ifarosettepatternhasmorethanonereflectionsymmetry, thenalllinesofreflectionoftherosettepatterngothroughasinglepoint, andanyrotationsymmetryoftherosettepatternhasthispointasitscenterofrotation.
Putting togetherwhatwehaveseenso far, weknowthat thesymmetrygroupofa rosettepatternhasnotranslationsymmetryorglidereflectionsymmetry; ifithasrotationsymmetries,theyallhavethesamecenterofrotation; ifithasreflectionsymmetryandrotationsymmetry,then all lines of reflection go through the single center of rotation; if it hasmore thanonereflectionsymmetry, alllinesofreflectiongothroughasinglepoint, andthispointisalsothecenterofrotationforallrotationsymmetries. Weknowfurtherthateveryrosettepatternhasanorder, whichisoneof 1, 2, 3, 4, . . .. Onceweknowtheorderofarosettepattern, weknowallthereistoknowaboutitsrotations. Forexample, arosettepatternoforder 5 hasrotations I,R1/5, R2/5, R3/5 and R4/5. Theonlyquestionthatremainsis, therefore, whattypesofreflectionsymmetriesarosettepatterncanhave, onceweknowitsorder.
174 5. SymmetryofPlanarObjectsandOrnamentalPatterns
BEFORE YOU READ FURTHER:
Supposearosettepatternhasorder n. Thinkabout thepossiblenumbersofreflectionsymmetriestherosettepatterncanhave.
Letuslookatsomeexamples. TherosettepatterninFigure 5.4.3 (i)hassymmetries I, R1/5,R2/5, R3/5 and R4/5; ithasnoreflections. TherosettepatterninFigure 5.4.3 (ii)hassymmetriesI, R1/4, R1/2, R3/4,M1,M2,M3 andM4; thefourlinesofreflectioncorrespondingtothefourreflectionsM1,M2,M3 andM4 aresimilartothefourlinesofreflectionshowninFigure 5.2.6.Notice that in thefirst case thereareno reflections, and in the secondcase thenumberofreflectionsisthesameastheorderoftherosettepattern. Itturnsout(aswillbemadepreciseinProposition 5.4.5 below), thateveryrosettepatternfallsintooneofthesetwopatterns.
(i) (ii)
Figure5.4.3
Tomakeourresultprecise, foreachpositiveinteger n, wedefinethesymmetrygroup Cn tobethecollectionofsymmetries
Cn ={I, R1/n, R2/n, R3/n, . . . R(n−1)/n
}.
Foreachpositiveintegern, wedefinethesymmetrygroupDn tobethecollectionofsymmetries
Dn ={I, R1/n, R2/n, R3/n, . . . R(n−1)/n,M1,M2,M3, . . . ,Mn
}.
(Theletters C andD standfor“cyclic”and“dihedral”respectively, thoughwewillnotbeusingtheseterms. Also, wenotethatthereisnocompletelystandardnotationforthesegroups, andsomeauthorsusenotationthatisdifferentfromours—thoughthenamescyclicanddihedralarequitestandard.) Forexample, wehave
C4 ={I, R1/4, R1/2, R3/4
},
andD3 =
{I, R1/3, R2/3,M1,M2,M3
}.
UsingthealgebraicnotationofSection 5.3, wecanalsowrite Cn as
Cn ={1, r, r2, r3, . . . rn−1
},
5.4RosettePatterns 175
and Dn asDn =
{1, r, r2, r3, . . . rn−1,m,mr,mr2,mr3, . . .mrn−1
}.
Wenotethatallthegroups Cn aredifferentfromoneanother, allthegroupsDn aredifferentfromoneanother, andallthegroups Cn aredifferentfromallthegroups Dn.Usingournewnotation, weseethattherosettepatternsinFigure 5.4.3 havesymmetrygroups
oftype C5 and D4 respectively. Itcanbeseen, usingtheideasofSections 5.2 and5.3, thataregular n-gonhassymmetrygroupoftypeDn, andthatDn hassametypeofcompositiontablewesawforaregular n-gon. Thecompositiontablefor Cn issimplytheupperlefthandquarterofthemultiplicationtableforaregular n-gon. Forexample, thecompositiontableforC8, usingthenotationofSections 5.3, isgiveninTable 5.4.1.
· 1 r r2 r3 r4 r5 r6 r7
1 1 r r2 r3 r4 r5 r6 r7
r r r2 r3 r4 r5 r6 r7 1r2 r2 r3 r4 r5 r6 r7 1 r
r3 r3 r4 r5 r6 r7 1 r r2
r4 r4 r5 r6 r7 1 r r2 r3
r5 r5 r6 r7 1 r r2 r3 r4
r6 r6 r7 1 r r2 r3 r4 r5
r7 r7 1 r r2 r3 r4 r5 r6
Table5.4.1
Wecannowstate thecompleteclassificationof symmetrygroupsof rosettepatterns. ThedemonstrationofthispropositionisgiveninAppendix D.
Proposition 5.4.5 (Leonardo’sTheorem). Thesymmetrygroupofarosettepatterniseither Cn
forsomepositiveinteger n, or Dn forsomepositiveinteger n.
Itfollowsthatthecompletelistofallpossiblecollectionsofsymmetriesofrosettespatternsare
C1, C2, C3, C4, . . . , Cn, . . .
D1, D2, D3, D4, . . . , Dn, . . . .
Thereareinfinitelymanysuchgroups, a Cn andaDn foreachinteger n ≥ 1. Thesesymmetrygroupsareknownasthe rosettegroups.TheabovepropositioniscommonlyreferredtoasLeonardo’sTheoreminhonorofLeonardo
daVinci, whoappearstohaveknownthisfact, thoughhedidnothavethemathematicaltoolstoexpressthisknowledgerigorously, nortoprovethattheseareindeedtheonlypossiblecollec-tionsofsymmetriesofrosettepatterns. Leonardo’sinterestinrosettepatternsmayhaveoriginatedinhisinterestinthepossiblesymmetriesofchurcheswithcircularfloorplans. SeeFigure 5.4.4forsomepicturesfromLeonardo’snotebooks.
176 5. SymmetryofPlanarObjectsandOrnamentalPatterns
Figure5.4.4
Leonardo’sTheoremisreallyquiteremarkable, inthatitrulesoutallsortsofcombinationsofsymmetriesasarisingfromrosettepatterns. Forexample, thereisnorosettepatternthesymmetrygroupofwhichhas 5 rotations(includingtheidentity)and 3 reflections; thereisalsonorosettepatternthathasrotationby 1/3 ofawholeturnandby 1/4 ofawholeturn, butbynosmallerrotation.Supposeyouaregivenarosettepattern, forexampletheoneshowninFigure 5.4.5. Howdo
youdetermineitssymmetrygroup? First, figureoutitsorder(order 4 inthecaseofFigure 5.4.5).Thendetermineifithasreflectionsymmetryornot(thereisreflectionsymmetryinthecaseofFigure 5.4.5). Ifthereisreflectionsymmetry, younecessarilyhavethegroup Dn, where n istheorder; ifthereisnoreflectionsymmetry, younecessarilyhavegroup Cn. HencetherosettepatternshowninFigure 5.4.5 hasgroup D4.
Figure5.4.5
Exercise 5.4.1. ForeachoftherosettepatternsshowninFigure 5.4.6, listthesymmetries,andstatewhattypeofsymmetrygroupithas.
5.4RosettePatterns 177
(i) (ii) (iii)
(iv) (v) (vi)
(vii) (viii) (ix)
Figure5.4.6
Exercise 5.4.2. Foreachofthefollowingcollectionsofsymmetries, statewhetherornotitisthesymmetrygroupofsomeplanarobject. Ifyes, giveanexampleofanobjectwiththatsymmetrygroup; ifno, explainwhynot.
(1) {I, R1/3, R2/3,M1,M2}.
(2) {I, R1/3, R2/3}.
(3) {I, R1/2, R3/4}.
(4) {I,M1,M2}.
(5) {I,M1}.
(6) {I, R1/2,M1,M2}.
178 5. SymmetryofPlanarObjectsandOrnamentalPatterns
5.5 FriezePatterns
Wenowturntofriezepatterns(orsimply, friezes), anothertypeofornamentalpattern, whichareslightlymorecomplicatedthanrosettepatterns, butwhicharecorrespondinglymoreinterestingaswell. Aswasthecaseforrosettepatterns, inthecaseoffriezepatternswewillalsobeabletostateacompleteclassificationofthesymmetrygroups, analogoustoLeonardo’sTheoremforrosettepatterns, thoughinthepresentcaseitwouldbebeyondthescopeofthisbooktoincludeallthedetailsofthedemonstrationoftheclassificationforfriezepatterns.A friezepattern (alsoknownasa strippattern) isanyplanarobjectthathastranslationsymme-
try, butsuchthatitstranslationsymmetrysatisfiestwoconditions: (1)alltranslationsymmetriesareinparalleldirections; and(2)thereisasmallesttranslationsymmetry(recallthattheterm“translationsymmetry”alwaysmeansanon-trivialtranslation). InFigure 5.5.1, Parts (i)and(ii)arefriezepatterns(notethatinPart (ii)thebasicunitoftranslationistwopeople). Part (iii)isnotafriezepatternbecauseithastranslationsymmetryinnon-paralleldirections(forexample,horizontalandvertical), andPart (iv)isnotafriezepatternbecauseithasnosmallesttranslationsymmetry.
Figure5.5.1
Itisimportanttorecognizethatanyfriezepatternwill“goonforever.” Forexample, thepattern. . . TTTTT . . ., whichweassumeisgoingonforeverinbothdirections, isafriezepattern. Thepattern TTTTT, whichconsistsofpreciselyfive letters T, isnot a friezepattern (though it isarosettepattern, withsymmetrygroup D1). Ifaplanarobjectdoesnotgoonforever, thenitcannotpossiblyhavetranslationsymmetry, andthereforeitcannotbeafriezepattern. However,noteverythingthatgoesonforeverisafriezepattern. Forexample, anon-repeatinginfinitestripofletters, orastraightline, bothgoonforever, butneitherisafriezepattern. Ofcourse, we
5.5FriezePatterns 179
cannotphysicallydrawsomethingthatgoesonforever. Wewillunderstand, however, thateventhoughourpicturesoffriezepatternsdonotgoonforever, weshouldthinkoffriezepatternsasextendingbeyondjustwhatisdrawn, andgoingonforever. Anobjectthatgoesonforeverisnecessarilyamentalconstruct—butsoaremanyotherthingsinbothmathematicsandoutsideofit. Beingamentalconstructisnoliability, atleastfromamathematicalviewpoint. Indeed, itwouldbeapitytolimitourimaginationtoonlythosethingswecanphysicallyconstruct.Foreaseofdiscussion, wewillassumethatallfriezepatternshavebeenpositionedsothatthe
directioninwhichtheycanbetranslatedishorizontal. (ThisisthecaseinFigure 5.5.1.) Anyfriezepattern, nomatterhowitisoriginallydrawn, canberotatedtomakeit“horizontal,” sowearenotlosinganythingbyourassumption.Ourultimategoalforfriezepatternsissimilartoourgoalforrosettepatterns, namelytoclas-
sifyfriezepatternsaccordingtotheirsymmetrygroups. Whatwemeanbythis, attheriskofrepetition, is tolistallsymmetrygroupsthatariseasthesymmetrygroupsof friezepatterns,andtobeabletotakeanygivenfriezepattern, andidentifywhichsymmetrygrouponourlistcorrespondstoit. Asforrosettepatterns, wewillbeginbylookingateachofthefourtypesofisometriesasappliedtofriezepatterns.Westartbylookingat translationsymmetryof friezepatterns. Actually, thereisnothingto
sayhere. Bydefinition, everyfriezepatternmusthavetranslationsymmetry, subjecttocertainrestrictions. Hence, wecannotdistinguishbetweenvarioussymmetrygroups thatarise fromfriezepatternsbyaskingwhetherornottheyhavetranslationsymmetry—theyalldo.Wenextturntorotationsymmetryoffriezepatterns. ThefriezepatterninFigure 5.5.2 (i)hasno
rotationsymmetry. ThefriezepatterninFigure 5.5.2 (ii)hasrotationsymmetryby 180◦ aboutthepointlabeled A, andalsoaboutallpointspointsthatarehalfwaybetweentwoadjacentletters Z inthepattern, andallpointsthatareatthecenterofaletter Z. (AsinSection 4.2, wewillrefertoa 180◦ asahalfturnrotation, orsimplyhalfturn.) Itisnothardtoseethatafriezepatterncannothaverotationsymmetrybyanyangleotherthan 180◦ (oranintegermultipleof180◦), becausethefriezepatternwouldnotlandonitselfifitwererotatedbyanyotherangle.ConsiderthefriezepatternshowninfriezepatterninFigure 5.5.3. Althoughitistruethateachsquareinthefriezepatterncanberotatedby 90◦ aboutitscenter, anditwilllandonitself, sucharotationisnotasymmetryofthefriezepattern, becausewealwaysrotatethewholeplane,notjustonelittlepieceoftheplane, andifwerotatethewholeplaneby 90◦ thenthefriezepatternwillnotlandonitself. Whenitcomestorotationsymmetryforafriezepattern, itiseitherhalfturnsymmetryornothing.Wenotethatifafriezepatternhashalfturnsymmetryaboutonecenterofrotation, thenithas
infinitelymanycentersofhalfturnrotation, obtainedbyapplyingthetranslationsymmetrytotheoriginalcenterofrotation. Further, allcentersofhalfturnrotationmustbeverticallyinthemiddleofthefriezepattern(assumingthefriezepatternishorizontal).
Next, weturntoreflectionsymmetry. ThefriezepatterninFigure 5.5.4 (i)hasnoreflectionsymmetry; thefriezepatterninFigure 5.5.4 (ii)hasreflectionsymmetryintheverticallinein-dicated(andinotherverticallinesaswell); thefriezepatterninFigure 5.5.4 (iii)hasreflection
180 5. SymmetryofPlanarObjectsandOrnamentalPatterns
(i) (ii)
A
Z Z Z Z Z Z
Figure5.5.2
Figure5.5.3
symmetryinthehorizontallineindicated; thefriezepatterninFigure 5.5.4 (iv)hasreflectionsymmetryinbothverticalandhorizontallines. Itisnothardtoseethatafriezepatterncannothavereflectionsymmetryinalinethatisneitherverticalnorhorizontal, becausethefriezepat-ternwouldnotlandonitselfifitwerereflectedinalinethatisneitherverticalnorhorizontal.Wenotethat ifafriezepatternhasreflectionsymmetryinavertical line, thenitnecessarilyhasreflectionsymmetryininfinitelymanyverticallines, obtainedbyapplyingthetranslationsymmetrytotheoriginalverticallineofreflection. Ontheotherhand, ifafriezepatternhasreflectionsymmetry inahorizontal line, thenthere isonlyonehorizontal lineof reflection,namelythehorizontallinethatisverticallyinthemiddleofthefriezepattern(assumingthefriezepatternishorizontal).
(i) (ii)
H H H H H HD D D D D D(iii) (iv)
F F F F F F A A A A A A
Figure5.5.4
Finally, weconsiderglidereflectionsymmetry. ThefriezepatterninFigure 5.5.5 (i)hasnoglidereflectionsymmetry; thefriezepatterninFigure 5.5.5 (ii)hasglidereflectionsymmetry, wherethelineofglidereflectionishorizontal, andthetranslationinvolvedtakesa ∪ andmovesitontoanadjacent ∩. Ifafriezepatternhasglidereflectionsymmetry, thenthelineofglidereflection
5.5FriezePatterns 181
mustbethehorizontallinethatisverticallyinthemiddleofthefriezepattern(assumingthefriezepatternishorizontal).
F F F F F F ∩ ∪ ∩ ∪ ∩ ∪ (i) (ii)
H H H H H H(iii)
Figure5.5.5
ThefriezepatterninFigure 5.5.5 (iii)hasglidereflectionsymmetry, butitisfundamentallydifferentfromtheglidereflectionsymmetryofthefriezepatterninFigure 5.5.5 (ii). Anyglidereflectionistheresultofcombiningatranslationandareflection. ForthefriezepatterninFig-ure 5.5.5 (iii), weseethateachofthetranslationandthereflection, thattogetherconstitutetheglidereflectionsymmetry, isitselfasymmetryofthefriezepattern. Bycontrast, thefriezepatterninFigure 5.5.5 (ii)hasnoreflectionsymmetryinahorizontalline, andneitherthetranslationnorthereflection, thattogetherconstitutetheglidereflectionsymmetry, isaloneasymmetryofthefriezepattern. Wecallaglidereflectionsymmetry non-trivial ifneitherthetranslationnorthereflectionthattogetherconstitutetheglidereflectionsymmetry, isaloneasymmetryofthefriezepattern. Wenotethatifafriezepatternhasglidereflectionsymmetry, andhasreflectionsymmetryinahorizontalline, thentheglidereflectionsymmetrymustbetrivial; thereaderisaskedtosupplythedetailsinExercise 5.5.1. Inotherwords, ifafriezepatternhasnon-trivialglidereflectionsymmetry, thenitcannothavereflectionsymmetryinahorizontal line; con-versely, ifafriezepatternhasreflectionsymmetryinahorizontalline, thenithasonlytrivialglidereflectionsymmetry. So, theonlytimetolookfornon-trivialglidereflectionsymmetryiswhenthereisnoreflectionsymmetryinahorizontalline.
Exercise 5.5.1. [UsedinThisSection] Supposethatafriezepatternhasglidereflectionsymmetry, andhasreflectionsymmetryinahorizontalline. Showthattheglidereflectionsymmetrymustbetrivial.
Inthecaseofrosettepatterns, wecouldlistthesymmetrieseachpatternhad, becauseeachlistofsymmetrieswasfinite(bydefinitionofwhatitmeanstobearosettepattern). Wecannotmakesuchlistseasilyforfriezepatterns, becausefriezepatternshaveinfinitelymanysymmetrieseach.However, eventhoughwecannotconvenientlylistsymmetriesinthecaseoffriezepatterns, we
182 5. SymmetryofPlanarObjectsandOrnamentalPatterns
canstillaskwhichtypesofsymmetriescanbecombinedwitheachother. Todoso, weaskthefollowingfourquestionsaboutanygivenfriezepattern.
QuestionA: Istherehalfturnsymmetry?
QuestionB: Istherereflectionsymmetryinaverticalline?
QuestionC: Istherereflectionsymmetryinahorizontalline?
QuestionD: Istherenon-trivialglidereflectionsymmetry?
Giventhateachoftheabovequestionshaseitheryesornoastheanswer, thereare 2·2·2·2 =16 possiblecombinationsofanswerstothesequestions. These 16 casesarelistedinTable 5.5.1.
QuestionsA B C D
1 N N N N2 N N N Y3 N N Y N4 N N Y Y5 N Y N N6 N Y N Y7 N Y Y N8 N Y Y Y9 Y N N N10 Y N N Y11 Y N Y N12 Y N Y Y13 Y Y N N14 Y Y N Y15 Y Y Y N16 Y Y Y Y
Table5.5.1
WhatisinterestingisthatnoteverycombinationlistedinTable 5.5.1 canactuallyoccur. Inotherwords, noteverypossibletypeofsymmetryofafriezepatterncanexistincombinationwitheveryothertypeofsymmetry. Usingsomeofthefactsaboutisometriesthatwehavealreadydiscussed, wewillinfacteliminatethemajorityofthelistedcombinationsofanswerstothefourquestions. Ineachcasethatiseliminated, wewillseethattheanswerstothefourquestionscontradicteachother.
5.5FriezePatterns 183
BEFORE YOU READ FURTHER:
TrytoeliminateasmanyofthecasesinTable 5.5.1 asyoucan; foreachcasethatyoueliminate, statewhynofriezepatterncansatisfythatcombinationofsymmetries. Foreachcasethatyoudonoteliminate, trytofindafriezepatternthathasthatcombinationofsymmetries.
ThecasesthatcanbeeliminatedfromTable 5.5.1 arethefollowing:
(4)NNYY. Thereisreflectionsymmetryinahorizontalline, whichimpliesthattheonlyglidereflectionsymmetryistrivial(asmentionedinourdiscussionofglidereflectionsymmetriesoffriezepatterns). Becausethiscasedoeshaveanon-trivialglidereflectionsymmetry, wehaveacontradiction.
(6)NYNY. There is reflectionsymmetry inavertical lineandaglidereflectionsymmetry. Ifthefirstof these isometries is followedby the second, thenbyExercise 4.6.7 weknow theresultingisometryisahalfturnsymmetry. Becausethiscasehasnohalfturnsymmetry, wehaveacontradiction.
(7)NYYN. Thereisreflectionsymmetryverticallineandreflectionsymmetryinahorizontalline. Ifthefirstofthesereflectionsisfollowedbythesecond, thenbyProposition 4.6.3 (3)weknowtheresultingisometryisahalfturnsymmetry. Becausethiscasehasnohalfturnsymmetry,wehaveacontradiction.
(8)NYYY. ThiscaseisjustlikeCase (7).
(10)YNNY. Thereisahalfturnsymmetryandaglidereflectionsymmetry. Thecenterofrotationofthehalfturnsymmetrymustbeonthelineofglidereflection. Ifthehalfturnisfollowedbytheglidereflection, thenbyExercise 4.6.6 wededucethattheresultingisometryisareflectionsymmetryinaverticalline. Becausethiscasehasnoreflectionsymmetryinaverticalline, wehaveacontradiction.
(11)YNYN. There isahalfturn symmetryand reflection symmetry inahorizontal line. Thecenterofrotationofthehalfturnsymmetrymustbeonthehorizontallineofreflection. Ifthehalfturnisfollowedbythereflectioninthehorizontalline, thenbyExercise 4.6.4 wededucethattheresultingisometryisreflectionsymmetryinaverticalline. Becausethiscasehasnoreflectionsymmetryinaverticalline, wehaveacontradiction.
(12)YNYY. ThiscaseisjustlikeCase (4).
(13)YYNN. Thereisahalfturnsymmetryandreflectionsymmetryinaverticalline. Ifthehalfturnisfollowedbythereflectioninaverticalline, thenbyExercises 4.6.4 and4.6.5 wededucethattheresultingisometryiseitherareflectionsymmetryinahorizontallineoraglidereflectionsymmetry. Becausethiscasehasneitherofthesesymmetries, wehaveacontradiction.
(16)YYYY. ThiscaseisjustlikeCase (4).
184 5. SymmetryofPlanarObjectsandOrnamentalPatterns
Therearesevencombinationsofanswersthatwehavenoteliminated, namelyCases (1), (2),(3), (5), (9), (14)and(15). Infact, eachofthesecombinationsdoesarisefromafriezepattern,aswillbeseenveryshortly. Moreover, all friezepatterns thathave thesameanswers to thefourquestionshavethesamesymmetrygroups, andeachofthesesevencombinationsofan-swerscorrespondstoadifferentsymmetrygroup. (Theproofofthesefactsusessomeadvancedmathematicsthatisbeyondthescopeofthisbook.) Insum, therearepreciselysevensymmetrygroupsoffriezepatterns. Thesesevensymmetrygroups, knownasthe friezegroups, areoftendenotedwiththesymbols f11, f12, f1m, f1g, fm1, fmm and fmg. (Therationaleforthesesymbolsisasfollows: the f standsforfrieze; thefirstsymbolafterthe f is 1 ifthereisnore-flectionsymmetryinaverticalline, andis m ifthereis; thesecondsymbolafterthe f is 1 ifthereisnoothersymmetry, ism ifthereisreflectionsymmetryinahorizontalline, is g ifthereisnon-trivialglidereflectionsymmetry, andis 2 ifthereishalfturnsymmetry.) Wesummarizetheclassificationofthesymmetrygroupsoffriezepatterns, andgiveanexampleofeachoftheseventypes, inthefollowingproposition.
Proposition 5.5.1 (ClassificationofFriezePatterns). ThesymmetrygroupofanyfriezepatternisoneofthesevengroupslistedinTable 5.5.2.
QuestionsName A B C D Example
f11 N N N N F F F F F F F F
f1g N N N Y D ∪ D ∩ D ∪ D ∩f1m N N Y N D D D D D D D D
fm1 N Y N N T T T T T T T T
f12 Y N N N S S S S S S S S
fmg Y Y N Y ∪ ∩ ∪ ∩ ∪ ∩ ∪ ∩fmm Y Y Y N O O O O O O O O
Table5.5.2
InSection 5.4 wenotonlystatedthetypesofsymmetrygroupsthatcouldariseforrosettepatterns, namelythe Cn and Dn groups, butweexplicitlylistedallthemembersofeachofthesegroups; forexample, westatedthat
Cn ={1, r, r2, r3, . . . rn−1
}.
Canwegiveasimilarexplicitdescriptionofeachofthesevenfriezegroups? Intheorywecoulddoso, thoughitismorecomplicatedthaninthecaseoftherosettegroups, becauseeachrosettegroupisfinite, whereaseachfriezegroupisinfinite. Consider, forexample, thefriezegroup f11,whichisthesymmetrygroupoffriezepatternsthathavenosymmetryotherthantranslation, forexample · · · FFFFF · · · . Let t denotethesmallestpossibletranslationsymmetrytotherightof
5.5FriezePatterns 185
thisfriezepattern. Thenthecollection f11 ofallsymmetriesofthisfriezepatternis
f11 ={· · · t−3, t−2, t−1, 1, t, t2, t3, · · ·
}.
Wecanthinkof t as t1, and 1 as t0. Althoughwewillnotwriteacompositiontablefor f11,becausesuchatablewouldbeinfinite, wecanexplicitlydescribehowtocombineanytwosymmetriesin f11 bytherule tatb = ta+b.
Exercise 5.5.2. Listthesymmetriesinthefriezegroup f1m, similarlytothewaywelistedthesymmetriesin f11. Onceagainlet t denotethesmallestpossibletranslationsymmetrytotherightofthisfriezepattern, andlet h denotereflectioninahorizontalline.
LetusnowuseTable 5.5.2 toanalyzethesymmetriesofthefriezepatterninFigure 5.5.6.Wefirstaskifthefriezepatternhashalfturnsymmetry. Inthiscasetheanswerisyes; thereadershouldfindacenterofrotationforahalfturnsymmetry. Thenextquestioniswhetherthereisreflectionsymmetryinverticallines. Theanswerisyes; thereadershouldfindaverticallineofreflection. Next, weaskwhetherthereisreflectionsymmetryinahorizontalline. Theanswerisno. Finally, weaskifthefriezepatternhasnon-trivialglidereflectionsymmetry. Theanswerisyes; thereadershouldfindthenon-trivialglidereflection. WethereforehaveanswersYYNYtoQuestionsA,B,C,D.Itfollowsthatthefriezepatternhassymmetrygroup fmg.
Figure5.5.6
Observethatthesituationforfriezepatternsisverydifferentfromrosettepatternsinthefollow-ingcrucialway: thereareinfinitelymanydistinctrosettegroups, butonlysevenfriezegroups.Thisisanamazingfact. Inasense, thegreatergeometriccomplexityoffriezepatternsrestrictshowtheycanbeconstructed. Theredoesnotseemtobeanysimpleintuitivereasonforthenumberoffriezegroups, namelyseven; itsimplycomesoutofthemathematicaldetails.
Exercise 5.5.3. ForeachofthefriezepatternsshowninFigure 5.5.7, statetheanswerstoQuestionsA–D,andstatewhatsymmetrygroupithas.
186 5. SymmetryofPlanarObjectsandOrnamentalPatterns
Figure5.5.7
Exercise 5.5.4. Findandphotocopy 7 friezepatterns, allwithdifferentsymmetrygroups.Foreachofthefriezepatternsyoufind, statetheanswerstoQuestionsA–E,andstatewhatsymmetrygroupithas.
5.6WallpaperPatterns 187
Exercise 5.5.5. ThemathematicianJohnH.Conwayhascomeupwiththefollowingde-scriptivenamesforthesevenfriezegroups, eachonebasedonaformofbodilymotion:hop, jump, step, sidle, spinninghop, spinningjumpandspinningsidle. Performeachtypeofmotion(partoftheproblemisfiguringoutwhateachmotionis), andlookatyourfoot-prints. Thefootprintsfromeachmotionformafriezepattern. MatchupthesefootprintfriezepatternswiththesevenlistedinTable 5.5.2.
5.6 WallpaperPatterns
Thelast, andmostinteresting, ofourthreetypesofornamentalpatternsarewallpaperpatterns.Thoughwallpaperpatternsaremorecomplicatedtechnicallythanfriezepatterns, heretoowewillbeabletostateacompleteclassificationofthesymmetrygroupsthatarise. Onceagainitwouldbebeyondthescopeofthisbooktoincludeallthedetailsofthedemonstrations.A wallpaperpattern isanyplanarobjectthathastranslationsymmetrysubjecttotwocondi-
tions: (1)thetranslationsymmetriesarenotallinparalleldirections; and(2)thereisasmallesttranslationsymmetryinanypossibledirectionforwhichthereistranslationsymmetry. Addi-tionally, weassumethatateverycenterofrotationofthewallpaperpatternthereisasmallestclockwiserotationsymmetry. InFigure 5.6.1, Parts (i)and(ii)arewallpaperpatterns. Part (iii)isnotawallpaperpatternbecausealltranslationsymmetriesareinparalleldirections, andPart (iv)isnotawallpaperpatternbecauseithasnosmallesttranslationsymmetryintheverticaldirec-tion. Thislastexampleshowsthatnoteverythingyoumightputonawalliscalleda“wallpaperpattern”inthetechnicalsense.
Justasafriezepatternhadto“goonforever”inordertohavetranslationsymmetry, thesameholdsforawallpaperpattern, exceptthatwallpaperpatternsgoonforeverinalldirections, notjustone. Ofcourse, anypicturewedrawofawallpaperpatternwillnotgoonforever, butthatissimplytheresultofourhumanlimitations. Wewillunderstand, however, thateventhoughourpicturesofwallpaperpatternsdonotgoonforever, weshouldthinkofwallpaperpatternsasextendingbeyondjustwhatisdrawn, andgoingonforever.Onecontrastbetweenhowwedrawwallpaperpatternsandfriezepatternsisthatfriezepat-
ternswerealwaysdrawnhorizontally(forconvenience), whereasforawallpaperpatternthereisnooneparticulardirectionthatcanbesingledoutandmadehorizontal.Ourultimategoal forwallpaperpatterns is just likeourgoal for friezepatterns, namelyto
classifywallpaperpatternsaccordingtotheirsymmetrygroups. Weproceedverymuchaswedidwithfriezepatterns, namelyfirstexaminingeachofthefourtypesofisometriesasappliedtowallpaperpatterns. Aswithfriezepatterns, wedonotneedtosayanythingabouttransla-tionsymmetryofwallpaperpatterns, becauseeverywallpaperpatternmusthave translationsymmetry, subjecttocertainrestrictions.
188 5. SymmetryofPlanarObjectsandOrnamentalPatterns
Figure5.6.1
Letusstartbyexaminingrotationsymmetryofwallpaperpatterns. Aswith friezepatterns,awallpaperpatternmightormightnothaverotationsymmetry. ThewallpaperpatterninFig-ure 5.6.2 (i)hasnorotationsymmetry(otherthantheidentity); thewallpaperpatterninFig-ure 5.6.2 (ii)hasrotationsymmetryby 120◦ orby 240◦ aboutthepointslabeled A, B and C
(andaboutallsimilarpoints); thewallpaperpatterninFigure 5.6.2 (iii)hasrotationsymmetryby 90◦, 180◦ or 270◦ aboutthepointslabeled X and Y (andallsimilarpoints), androtationsymmetryby 180◦ aboutthepointlabeled Z (andallsimilarpoints). Wethereforeseethatincontrasttofriezepatterns, whererotationsymmetrycanonlybeby 180◦, forwallpaperpatternsrotationsymmetrycanbebyavarietyofangles; moreover, differentcentersofrotationinthesamewallpaperpatterncanhavedifferentanglesofrotation.
IfwelookatthewallpaperpatternshowninFigure 5.6.2 (iii), weseethreecentersofrotation,labeled X, Y and Z respectively. Ofcourse, thewallpaperpatternhasothercentersof rota-tion, besidesthethreethatarelabeled. Indeed, becausewallpaperpatternsrepeatthemselvesinfinitely, ifawallpaperpatternhasonecenterofrotation, thenithasinfinitelymanycentersofrotation. Itwould, therefore, besillyforustoattempttofindliterallyallthecentersofrotation
5.6WallpaperPatterns 189
AB
C
X
Y Z
(i) (ii)
(iii)
Figure5.6.2
ofagivenwallpaperpattern. Whatwecanhopetofindareallthe“genericallydifferent”typesofcentersofrotationofawallpaperpattern. Wemakethisconceptpreciseasfollows.Supposewearegivenawallpaperthathascentersofrotation. Wesaythat twocentersof
rotationofthewallpaperpatternare equivalent ifthereisasymmetryofthewallpaperpatternthattakesonecenterofrotationtotheother(thesymmetrycouldbeanyofthefourtypesofisometries). InFigure 5.6.3, weseefourcentersofrotationlabeledA,B,C, andD. ThepointsAand B areequivalentcentersofrotation, becausereflectionintheverticallinehalfwaybetweenthemisasymmetryofthewallpaperpatternthattakes A to B. Ontheotherhand, notwoofthepointsA, C andD areequivalent, becausenosymmetryofthewallpaperpatterntakesoneofthemtoanother.
Ingeneral, foranycenterofrotationofawallpaperpattern, wecanlookforallthecentersofrotationthatareequivalenttoit; allsuchcentersofrotationwillinfactbeequivalenttoeachotheraswell. Wecallsuchacollectionofequivalentcentersofrotationan equivalenceclass of
190 5. SymmetryofPlanarObjectsandOrnamentalPatterns
A B
DC
Figure5.6.3
centersofrotation. Forexample, inFigure 5.6.3, theequivalenceclassofthecenterofrotationC consistsofallpointsthatareinthemiddlesofallthe“bricks”outofwhichthepatternisbuilt. Foranywallpaperpattern, itcanbeshownthatthecollectionofallitscentersofrotationcanbebrokenupintoafinitenumberofequivalenceclasses, whichwillbedisjointfromeachother. Now, withthisnotionofequivalenceclasses, wecanstatemorepreciselywhatitmeanstofindallthe“genericallydifferent”typesofcentersofrotationofawallpaperpattern. Givenawallpaperpattern, whatwewanttofindispreciselyonecenterofrotationperequivalenceclass. Forexample, thecentersofrotationlabeled X, Y and Z inFigure 5.6.2 (iii)areexactlyonerepresentativefromeachequivalenceclassofcentersofrotationforthiswallpaperpattern.Assuch, wecansayinformallythatwehavefound“allthecentersofrotation”ofthepattern.
Exercise 5.6.1. Foreachwallpaperpatterns shown inFigure 5.6.4, findand labelonecenterofrotationperequivalenceclass(ifthereareany).
Tomakesenseofrotationsymmetryofwallpaperpatterns, weneedtorecallfromSection 5.4thenotionofacenterofrotationofanobjecthavingorder n. Becauseweareassumingthatateverycenterofrotationofawallpaperpatternthereisasmallestclockwiserotationsymmetry,theneverycenterofrotationofawallpaperpatternhassomeorder n, thoughdifferentcentersofrotationofagivenwallpaperpatterncanhavedifferentorders. Forexample, thepoints A,B and C inFigure 5.6.2 (ii)areallcentersofrotationoforder 3, andthepoints X, Y and Z inFigure 5.6.2 (iii)arecentersofrotationoforders 4, 4 and 2 respectively.Foragivenwallpaperpattern, wecanfinditscentersof rotation(that is, wecanfindone
centerofrotationperequivalenceclass). Eachofthesecentersofrotationhasanorder.
5.6WallpaperPatterns 191
(i) (ii)
(iii) (iv)
Figure5.6.4
BEFORE YOU READ FURTHER:
Trytofigureoutwhichnumberscanoccurasordersofcentersofrotationofwallpaperpatterns. Areallnumberspossible, oronlysome? Ifthelatter, whichnumbersoccur?
WeknowfromtheexamplesinFigure 5.6.2 that 2, 3 and 4 canoccurasordersofcentersofrotationofwallpaperpatterns. Arethereanyothernumberspossible? Forinstance, isitpossibletohaveawallpaperpatternwithacenterofrotationoforder 5? Howabout 7, or 583? Itturnsout, quiteremarkably, thatthereareveryfewpossibleordersforcentersofrotationofwallpaperpatterns, aswenowstate.
Proposition 5.6.1. Everycenterofrotationofawallpaperpatternhasorder 2, 3, 4 or 6.
A rigorousproofoftheabovepropositionusesgrouptheory, andisbeyondthescopeofthisbook. Aninformaldiscussionofwhythisresultistruecanbefoundin[Wey52, pp.101–103]. ToseehowremarkableProposition 5.6.1 is, noteinparticularthatitmeansthattherecannotbeawallpaperpatternwithacenterofrotationoforder 5. Simplyarrangingpentagonsinaninfinite
192 5. SymmetryofPlanarObjectsandOrnamentalPatterns
grid, asshowninFigure 5.6.5 (i)willnotyieldawallpaperpatternwithcenterofrotationoforder5, becauseweneedtorotatetheentireplane, notjustonepentagonatatime. InFigure 5.6.5 (ii)weseeanexampleofacleverIslamicdesignthatincorporatespentagons, andmightthereforegiveanillusionoforder 5 centersofrotation, butitisonlyanillusion.
Figure5.6.5
Givenawallpaperpattern, wecanlookforthedifferentcentersofrotationthatithas. Eachcenterofrotation, ifthereareany, hasanorderthatis 2, 3, 4 or 6. Wenowwanttodefinetheorderforthewholewallpaperpattern. Ifthewallpaperpatternhasnocentersofrotation, thenwesaythatthewallpaperpatternisof order 1. Ifthewallpaperpatternhascentersofrotation, wesaythatthewallpaperpatternisof order n if n isthehighestorderfoundamongthecentersofrotationofthewallpaperpattern. Forexample, thewallpaperpatterninFigure 5.6.2 (i)isoforder1; thewallpaperpatterninFigure 5.6.2 (ii)isoforder 3; thewallpaperpatterninFigure 5.6.2 (iii)isoforder 4.
5.6WallpaperPatterns 193
Exercise 5.6.2. FindtheorderofeachwallpaperpatternshowninFigure 5.6.4.
Exercise 5.6.3. Supposewearegivenawallpaperpattern. Supposefurtherthat, amongitscentersofrotation, thewallpaperpatternhasanorder 2 centerofrotationandanor-der 3 centerofrotation. Fromthispartialinformation, canyoudeterminetheorderofthewallpaperpattern? Ifyes, whatistheorder, andwhy?
Weturnnexttoreflectionsymmetry. Aspreviouslymentioned, incontrasttofriezepatterns,wherewedistinguishedbetweenverticalandhorizontallinesofreflection, forwallpaperpat-ternsthereisnosuchdistinction, becauseawallpaperpatterngoesonforeverinalldirections,sowecannotisolateonedirectionas“horizontal.” ThewallpaperpatterninFigure 5.6.6 (i)hasnoreflectionsymmetry; thewallpaperpatternsinFigure 5.6.6 (ii)and(iii)bothhavereflectionsymmetry, inthelinesindicated(andinallsimilarlines). Ifawallpaperpatternhasreflectionsymmetryinsomelineofreflection, thenitwillnecessarilyhaveinfinitelymanylines, obtainedbyapplyingthetranslationsymmetryofthewallpaperpatterntotheoriginallineofreflection.AlthoughthewallpaperpatternsinFigure 5.6.6 (ii)and(iii)bothhavereflectionsymmetry, thereisonemajordifferencebetweenthereflectionsymmetryofthesetwopatterns, namelythatallthelinesofreflectionforPart (ii)areparallel, whereasthelinesofreflectionarenotallparallelforPart (iii).
Ifawallpaperpatternhasonelineofreflection, thenithasinfinitelymanylinesofreflection. Aswasthecasewithcentersofrotation, wewouldliketofindallthe“genericallydifferent”typesoflinesofreflectionofawallpaperpattern; onceagain, weusetheconceptofequivalence.Supposewearegivenawallpaperthathaslinesofreflection. Wesaythattwolinesofreflectionofthewallpaperpatternare equivalent ifthereisasymmetryofthewallpaperpatternthattakesonelineofreflectiontotheother. InFigure 5.6.7, weseethreelinesofreflectionlabeledm, nand k. Thelines m and n areequivalentlinesofreflection, becauserotationinthecenterofrotation A showninFigure 5.6.3 isasymmetryofthewallpaperpatternthattakes m to n. Ontheotherhand, thelines m and k arenotequivalent, becausenosymmetryofthewallpaperpatterntakes m to k.
Givenawallpaperpattern, andalineofreflectionforthiswallpaperpatterns, wecanlookforallthelinesofreflectionthatareequivalenttoit. Wecallsuchacollectionofequivalentlinesofreflectionan equivalenceclass oflinesofreflection. Forexample, inFigure 5.6.7, theequivalenceclassof the lineof reflection m consistsofallvertical linesof reflectionof thewallpaperpattern(thatis, allverticallinesthatareboundariesbetweenthe“bricks”). Givenawallpaperpattern, whatwewanttofindispreciselyonelineofreflectionperequivalenceclass.Forexample, thelinesofreflectionshowninFigure 5.6.6 (iii)areexactlyonerepresentative
194 5. SymmetryofPlanarObjectsandOrnamentalPatterns
(i) (ii)
(iii)
Figure5.6.6
m n
k
Figure5.6.7
fromeachequivalenceclassoflinesofreflectionforthiswallpaperpattern. Assuch, wecansayinformallythatwehavefound“allthelinesofreflection”ofthepattern.
5.6WallpaperPatterns 195
Exercise 5.6.4. ForeachwallpaperpatternshowninFigure 5.6.4, findandlabelonelineofreflectionperequivalenceclass(ifthereareany).
Wenowlookatglidereflectionsymmetryforwallpaperpatterns. Asforfriezepatterns, weareinterestedonlyinnon-trivialglidereflectionsymmetry, that is, glidereflectionsymmetrysuchthatneitherthetranslationnorthereflection, thattogetherconstitutetheglidereflectionsymmetry, isaloneasymmetryofthewallpaperpattern. Justaswasthecaseforfriezepatterns,ifawallpaperpatternhasalineofglidereflectionthatisalsoalineofreflection, thentheglidereflectionsymmetryinthatlineistrivial. Moreover, itturnsoutthatinawallpaperpattern, anylineofreflectionisautomaticallyalineofglidereflection(foratrivialglidereflectionsymmetry);thereaderisaskedtosupplythedetailsinExercise 5.6.5. Puttingtheseobservationstogether,weseethattofindanon-trivialglidereflectionsymmetry, weneedtofindalineofglidere-flectionthatisnotalineofreflection. Wecallsuchlinesofglidereflection non-triviallinesofglidereflection. ThewallpaperpatterninFigure 5.6.8 (i)hasnoglidereflectionsymmetry; thewallpaperpatterninFigure 5.6.8 (ii)hasnon-triviallinesofglidereflectionasindicated(notethattheverticallinesthroughthemiddleoftheletters M aretriviallinesofglidereflection).
(i) (ii)
Figure5.6.8
Exercise 5.6.5. [Used inThis Section] Suppose that awallpaperpatternhas a lineofreflection. Show that this lineof reflectionmustalsobea lineofglide reflection (foratrivialglidereflectionsymmetry). ThisexerciseusesideasfromAppendix C.
Itissometimestrickyinpracticetofindnon-triviallinesofglidereflectioninwallpaperpat-terns, certainlytrickierthanitistofindcentersofrotationandlinesofreflection. Linesofglidereflectionoftentendtobe“inbetween”featuresof thewallpaperpattern, forexampleasin
196 5. SymmetryofPlanarObjectsandOrnamentalPatterns
Figure 5.6.8 (ii). Moreprecisely, ifalineofglidereflectionisparalleltolinesofreflection, thenitmustbehalfwaybetweenthelinesofreflection. SeeExercise 5.6.6 fordetails. Exercise 5.6.7discussestherelationbetweenalineofglidereflectionandcentersofrotationnotonit. Note,however, thatlinesofglidereflectionneednotbeparalleltoanylinesofreflection, andcaninfactintersectlinesofreflection; thereaderisaskedtosupplyanexampleinExercise 5.6.8.
Exercise 5.6.6. [UsedinThisSection] Supposethatawallpaperpatternhasanon-triviallineofglidereflectionthatisparalleltolinesofreflection. Showthatthelineofglidereflec-tionishalfwaybetweenthelinesofreflection. ThisexerciseusesideasfromAppendix C.
Exercise 5.6.7. [UsedinThisSection] Supposethatawallpaperpatternshasanon-triviallineofglidereflection, andithasacenterofrotationthatisnotonthelineofglidereflection.Showthatthereisanothercenterofrotationatthesamedistancefromthelineofglidereflection, buton theother side (thoughnotdirectly across from theoriginal centerofrotation). ThisexerciseusesideasfromAppendix C.
Exercise 5.6.8. [UsedinThisSection] Findanexampleofawallpaperpatternthathasnon-triviallinesofglidereflectionandhaslinesofreflection, andsuchthatthenon-triviallinesofreflectionintersectsomelinesofreflection. Thereissuchanexampleamongthewallpaperpatternsshownsofarinthissection, thoughitslinesofreflectionandlinesofglidereflectionarenotshown.
Justaswehavethenotionofequivalentcentersofrotation, andequivalentlinesofreflec-tion, wehavethesamenotionfornon-triviallinesofglidereflection. Supposewearegivenawallpaperthathasnon-triviallinesofglidereflection. Wesaythattwonon-triviallinesofglidereflectionofthewallpaperpatternare equivalent ifthereisasymmetryofthewallpaperpatternthattakesonelineofglidereflectiontotheother. InFigure 5.6.9, weseethreenon-triviallinesofglidereflectionlabeled a, b and c. Thelines a and b areequivalentlinesofglidereflection,becausereflectionintheverticallinehalfwaybetween a and b isasymmetryofthewallpaperpatternthattakes a to b. Ontheotherhand, thelines a and c arenotequivalent, becausenosymmetryofthewallpaperpatterntakes a to c.
Givenawallpaperpattern, andanon-triviallineofglidereflectionforthiswallpaperpatterns,wecanlookforallthenon-triviallinesofglidereflectionthatareequivalenttoit. Wecallsuchacollectionofequivalentnon-triviallinesofglidereflectionan equivalenceclass ofnon-triviallinesofglidereflection. Forexample, inFigure 5.6.9, theequivalenceclassofthelineofglidereflection a consistsofallverticalnon-triviallinesofglidereflectionofthewallpaperpattern
5.6WallpaperPatterns 197
a b
c
Figure5.6.9
(thatis, allverticallinesthatarehalfwaybetweentheverticallinesthroughtheedgesofthe“bricks”). Foragivenwallpaperpattern, wewanttofindpreciselyonenon-triviallineofglidereflectionperequivalenceclass.
Exercise 5.6.9. For eachwallpaper pattern shown in Figure 5.6.4, find and label onenon-triviallineofglidereflectionperequivalenceclass(ifthereareany).
Inthecaseoffriezepatterns, afterdiscussingthedifferenttypesofsymmetriesthatcouldoccur,wewereledtofourquestionsconcerningthesetypesofsymmetries. Wefollowasimilarplanforwallpaperpatterns, thoughwithoneadditionalquestionthatdoesnothaveananalogamongthefourquestionswehadforfriezepatterns. Uptillnowwehavelookedseparatelyateachofthefourtypesofisometriesastheycanoccurassymmetriesofwallpaperpatterns. Wenowneedtoaskonequestionconcerninghowthesedifferenttypesofsymmetriesinteract. Supposeawallpaperpatternhasbothrotationsymmetryandreflectionsymmetry. Thewallpaperpatternmustthereforehavebothcentersofrotationandlinesofreflection.
BEFORE YOU READ FURTHER:
Supposethatawallpaperpatternhasbothcentersofrotationandlinesofreflection. Mustallthehighestordercentersofrotationbeonlinesofreflection?
Theanswertotheabovequestionisthatinsomewallpaperpatternsthehighestordercentersofrotationareallonlinesofreflection, andinotherwallpaperpatternstheyarenot. (Centersofrotationthatarenotthehighestorderarenotofusetousforourpresentpurpose.) Forexample,inthewallpaperpatterninFigure 5.6.2 (iii), thehighestordercentersofrotationarethepoints
198 5. SymmetryofPlanarObjectsandOrnamentalPatterns
labeledX and Y (theyareorder 4), andboththesepointsareonlinesofreflection. Bycontrast, inthewallpaperpatterninFigure 5.6.3, thehighestordercentersofrotationarethepointslabeledA, C and D (theyareallorder 2); wedonotneedthepoint B, becauseitisequivalentto A.Thepoints C and D areonlinesofreflection, butthepoint A isnotonalineofreflection.Itturnsoutthatwenowhaveeverythingweneedtoclassifywallpaperpatternsaccordingto
theirsymmetries. Aswasthecasewithfriezepatterns, wecannotconvenientlylistallthesym-metriesofwallpaperpatterns, butwecanstillaskwhichtypesofsymmetriescanbecombinedwitheachother. Inparticular, weaskthefollowingfivequestionsaboutanygivenwallpaperpattern.
QuestionA: Whatistheorderofthewallpaperpattern? (Answer: 1, 2, 3, 4, or6.)
QuestionB: Istherereflectionsymmetry? (Answer: YesorNo.)
QuestionC: Istherereflectionsymmetryinnon-parallellines? (Answer: YesorNo.)
QuestionD: Areallhighestordercentersofrotationonlinesofreflection? (Answer: YesorNo.)
QuestionE: Isthereglidereflectioninnon-triviallinesofglidereflection? (Answer: YesorNo.)
Itcanbeseenthatthereare 5 · 2 · 2 · 2 · 2 = 80 possiblecombinationsofanswerstothesequestions. Wewillnotlistall 80 here. Aswithfriezepatterns, itturnsoutthatmostofthese 80casescannotactuallyoccur. Hence, noteverypossibletypeofsymmetryofawallpaperpatterncanexistincombinationwitheveryothertypeofsymmetry. Wewillnotgooverthedetailsofhowtoeliminatethecasesthatcannotoccur; todosowouldbebeyondthescopeofthisbook.Somecasesaresimpletoeliminate, however, andarelefttothereaderinExercises 5.6.10 and5.6.11.
Exercise 5.6.10. [UsedinThisSection] Showthatnowallpaperpatterncanhaveanswers1, yesandyestoQuestionsA–C,regardlessofwhattheanswerstoQuestionsD andE are.(Wecanthereforeeliminatethecombinationsofanswers 1YYNNN, 1YYNNY, 1YYNYN,1YYNYY, 1YYYNN, 1YYYNY, 1YYYYN, 1YYYYY.)
Exercise 5.6.11. [UsedinThisSection] Showthatnowallpaperpatterncanhaveanswers3, yesandnotoQuestionsA–C,regardlessofwhattheanswerstoQuestionsD andE are.Similarly, showthatnowallpaperpatterncanhaveanswers 4, yesandno, oranswers 6,yesandno, toQuestionsA–C.ListallthecombinationsofanswerstoQuestionsA–E thatcanthereforebeeliminated.
AfteralltheimpossiblecombinationsofanswerstoQuestionsA–E areeliminated, itturnsoutthatthereare 17 combinationsofanswersthatdooccur. Moreover, allwallpaperpatternsthat
5.6WallpaperPatterns 199
havethesameanswershavethesamesymmetrygroups, andeachofthese 17 combinationsofanswerscorrespondstoadifferentsymmetrygroup. (Theproofofthesefactsusessomeadvancedmathematicsthatisbeyondthescopeofthisbook.) Insum, thereareprecisely 17 symmetrygroupsofwallpaperpatterns, knownasthe wallpapergroups. Thewallpapergroupsareoftendenotedwiththesymbols p1, pg, pm, cm, p2, pgg, pmg, cmm, pmm, p3, p31m, p3m1,p4, p4g, p4m, p6 and p6m. (Thereareothersetsofsymbolsthatvariousauthorsuse, butthesymbolswehaveusedseemtobethemostcommon; asforthesymbolsusedtodenotethefriezegroups, thereisarationaleforthewallpapergroupsymbols, butitisnotworthdwellingupon.) Thereasonforthenumber 17 isnomoreintuitivelyobviousthanthereasonthattherearepreciselysevenfriezegroups; inbothcasesitcomesoutofthemathematicalanalysis. Wesummarizetheclassificationofthesymmetrygroupsofwallpaperpatternsasfollows.
Proposition 5.6.2 (ClassificationofWallpaperPatterns). Thesymmetrygroupofanywallpaperpatternisoneofthe 17 groupslistedinTable 5.6.1.
QuestionsName A B C D Ep1 1 N N N Npg 1 N N N Ypm 1 Y N N Ncm 1 Y N N Yp2 2 N N N Npgg 2 N N N Ypmg 2 Y N N Ycmm 2 Y Y N Ypmm 2 Y Y Y Np3 3 N N N Np31m 3 Y Y N Yp3m1 3 Y Y Y Yp4 4 N N N Np4g 4 Y Y N Yp4m 4 Y Y Y Yp6 6 N N N Np6m 6 Y Y Y Y
Table5.6.1
Anexampleofeachofthe 17 typesofwallpaperpatternsisgiveninFigures 5.6.10 and5.6.11(thefirstofthesefiguresshowsallthewallpaperpatternsoforders 1 and 2, andthesecondofthefiguresshowsallthewallpaperpatternsoforders 3, 4 and 6).
200 5. SymmetryofPlanarObjectsandOrnamentalPatterns
p1 pg pm
cm p2 pgg
pmg cmm pmm
Figure5.6.10
Thoughthe 17 wallpapergroupsweretreatedmathematicallyonlyinthelate19thcentury,theyseemtohavebeenknownonsomeintuitivelevelearlier. Forexample, wallpaperpatternsforallthesegroupscanbefoundintheAlhambra inGranada, Spain, whichwasbuiltduringthe9th–14thcenturies. SeeFigure 5.6.12 foronesuchpattern. TheAlhambrawasbuiltbytheArabrulerswhocontrolledpartofSpainatthetime. BecauseIslam forbidstheuseofrepresen-tationalpictures, Muslimartistsexcelledatgeometricdesigns. (ItshouldbenotedthatArabicculture wasgenerallymoreadvancedmathematicallythantheEuropeancultureduringtheMid-dleAges. Moreover, duringtheRenaissance, theEuropeanslearnedmuchGreekmathematicsthroughArabictranslations. TheArabicculturehasnotalwaysbeengiventhecredititdeservesinthesematters. Itisnotclear(totheauthor, anyway)whetherthedesignersoftheAlhambraac-tuallyknewexplicitlythattherewereseventeendifferenttypesofsymmetryconfigurationsthatawallpaperpatterncouldhave, orwhethertheyweresimplysogoodatdesigninggeometricpatternsthattheymanagedtofindallofthembyaccident. Asasidenote, theDutchartistM.C.Escher wasinspiredtomakehisownrepeating, interlockingfiguresaftervisitingtheAlhambra,ashisnotebooksshow. Itisclaimedin[Wey52]thattheancientEgyptianshadfoundwallpaperpatternsofall 17 types; manyothercultures, includingChinaandvariouspeoplesinAfrica,alsoexcelatgeometricdesign.
LetusnowuseTable 5.6.1 toanalyzethesymmetriesofthewallpaperpatterninFigure 5.6.13.First, wefindthecentersofrotation. ThepointsA, B and C inFigure 5.6.13 arethreecentersof
5.6WallpaperPatterns 201
p3 p31m p3m1
p4 p4g p4m
p6 p6m
Figure5.6.11
rotation, andallothersareequivalenttothese. Allthesecentersofrotationhaveorder 2, sothewallpaperhasorder 2. Next, weaskifthepatternhasreflectionsymmetry. Theanswerisyes.Thenextquestioniswhetherthereisreflectionsymmetryinnon-parallellines. Becausetherearebothhorizontalandverticallinesofreflection, theanswerisyes. Fourth, weaskifallhighestordercentersofrotationareonlinesofreflection. Allthreeof A, B and C arehighestorder(inthiscaseorder 2), butbecause C isnotonalineofreflection, theanswertothisquestionisno.Finally, weaskwhetherthewallpaperpatternhasglidereflectionsymmetryinnon-triviallinesofglidereflection. Theanswerisyes, asthereadershouldverify. LookingatTable 5.6.1 leadsustoconcludethatthewallpaperpatternhassymmetrygroup cmm.
Exercise 5.6.12. ForeachofthewallpaperpatternsshowninFigure 5.6.14, statethean-swerstoQuestionsA–E,andstatewhatsymmetrygroupithas.
202 5. SymmetryofPlanarObjectsandOrnamentalPatterns
Figure5.6.12
Figure5.6.13
5.6WallpaperPatterns 203
Figure5.6.14
204 5. SymmetryofPlanarObjectsandOrnamentalPatterns
Exercise 5.6.13. Findandphotocopy 4 wallpaperpatterns, allwithdifferent symmetrygroups. Foreachofthewallpaperpatternsyoufind, statetheanswerstoQuestionsA–E,andstatewhatsymmetrygroupithas.
Exercise 5.6.14. Draw 4 wallpaperpatterns, allwithdifferentsymmetrygroups. (IfyouarealsodoingExercise 5.6.13, thenmakesurethewallpaperpatternsyoudrawhavedifferentsymmetrygroupsthantheonesyoufoundandphotocopied.) Foreachofthewallpaperpatternsyoudraw, statetheanswerstoQuestionsA–E,andstatewhatsymmetrygroupithas.
5.7 ThreeDimensionalSymmetry
Havingsofardiscussedthesymmetryofplanarobjectsinthischapter, weturnbrieflytoalookatsymmetryofthreedimensional objects(thatis, spatialobjects). Thestudyofsymmetryofthreedimensionalobjectsisinmanywayssimilartothestudyofsymmetrywehaveseenforplanarobjects, thoughitismorecomplicated, andwewillmentiononlyafewideas, andwillnotgiveathoroughtreatmentaswedidforplanarobjects. Forsomeinterestingissuesconcerningspatialobjects, see[Wey52].Justas thestudyofsymmetryofplanarobjects isbasedonthenotionof isometriesof the
plane, thestudyofthesymmetryofthreedimensionalobjects(whichwewillrefertoas“threedimensionalsymmetry”)isbasedonisometriesofthreedimensionalspace. Assuch, athoroughtreatmentof threedimensionalsymmetrywouldcommencewithanexaminationofallpos-sibletypesofisometriesofthreedimensionalspace. Ratherthangivingacompletetreatmentofisometriesinthreedimensionalspace, whichwouldbeverylengthy, wewilllookatafewexamplesofsymmetriesofthreedimensionalobjects, startingwiththesymmetriesofthecube,analogouslytowhatwedidinSection 5.2, wherewelookedatthesymmetriesoftheregularpolygons. Wepointout, withoutgoingintothedetails, thatallthebasicideasaboutisometriesandsymmetriesthatholdfor theplanehaveanalogsforthreedimensionalspace; forexam-ple, thecompositionoftwosymmetriesofathreedimensionalobjectisstillasymmetryoftheobject, etc.InFigure 5.7.1 weseeacube, withitsverticeslabeled(justaswelabeledtheverticesofregular
polygonsinSection 5.2). Aswithregularpolygons, therearenotranslationorglidereflectionssymmetriesof thecube(thoughother threedimensionalobjectscanhavesuchsymmetries).Clearlytheidentityisometryofthreedimensionalspace, denoted I asintheplanarcase, isasymmetryofthecube. Letusnowtrytofindallthenon-trivialrotationsymmetriesofthecube.Intheplane, eachrotationisperformedaboutapoint, calledthecenterofrotation, whichisfixedbytherotation. Inthreedimensionalspace, bycontrast, eachrotationisperformedaround
5.7ThreeDimensionalSymmetry 205
aline, calledthe axisofrotation. Thereisoneslightcomplicationinvolvingrotationinthreedimensionalspace, however. Intheplane, weusedtheconventionthatrotationbyapositiveangleistakentobeclockwise. Wecouldadoptthisconventionbecausewecanalldistinguishbetweenclockwiseandcounterclockwiserotations. Inthreedimensionalspace, supposewewanttorotateaboutagivenaxisofrotationbyagivenpositiveangle. Inwhichdirectionshouldwerotate? Therearetwopossibilitiesforrotatingbythegivenpositiveangleaboutthegivenaxisofrotation, andweneedtofindawaytospecifywhichoneistobeused. Themethodforsolvingthisproblemisthateveryaxisofrotationwillbegivenadirection, specifiedbyanarrowhead, asseenforexampleonline a inFigure 5.7.2. Wethenadopttheconventionthatwewillconsiderclockwiserotationaboutthelinetobethedirectionofrotationthatappearsclockwisewhenwelookfromthetailofthearrowtowardtheheadofthearrow. Wesaythatsuchrotationfollowstherighthandrule. Thatis, weconsiderclockwiserotationaboutalinewithanarrowheadtobethedirectiongivenbycurlingthefingersofyourrighthand, whenyouplaceyourthumbparalleltotheaxisofrotation, andinthedirectionofthearrowhead. Thisrighthandruleisusedregularlyinphysics.Using theaboveconsiderations in thecaseof thecube, wesee inFigure 5.7.2 a rotation
symmetryof thecube, namelyrotationby 1/4 turnaround the line labeled a, which is theverticallinethroughthecenterofthecube. Noticethattherotationisclockwisewhenviewedfromabove thecube, which is looking in thedirectionof thearrowhead shownon line a.Rotationby 1/2 and 3/4 around line a are also symmetriesof the cube. Wedenote thesesymmetriesby Ra
1/4, Ra1/2 and R
a3/4 respectively. Theseareallthenon-trivialrotationsymmetries
aroundline a.
B C
DAF G
HE
Figure5.7.1
BEFORE YOU READ FURTHER:
Trytofindasmanyrotationsymmetriesofthecubeaspossible.
Whataretheotheraxesofrotationofthecube? Therearetwomorethatareverysimilarto a,namelythe“front-to-back”horizontalline b throughthecenterofthecubethatisperpendiculartothesquare ADHE, andthe“left-to-right”horizontalline c throughthecenterofthecube
206 5. SymmetryofPlanarObjectsandOrnamentalPatterns
B C
DA
a
F G
HE
A B
CDE F
GH
R1/4a
Figure5.7.2
thatisperpendiculartothesquare ABFE. Then Rb1/4, R
b1/2, R
b3/4, R
c1/4, R
c1/2 and Rc
3/4 areallnon-trivialrotationsymmetriesofthecube.Therearealsoothertypesofaxesofrotationofthecube. Thelines a, b and cwerethroughthe
centersofopposingsquarefaces. Therearealsoaxesofrotationthroughmidpointsofopposingedges. Forexample, inFigure 5.7.3 (i)weseethelinethatgoesthroughthemidpointsof AB
and HG, pointinginthedirectionofthemidpointof HG; thislineisdenoted d. Then Rd1/2
isasymmetryofthecube. (Observethat Rd1/4 and Rd
3/4 arenotsymmetriesofthecube.) There
arefiveothersimilaraxesofrotation: theline e thatgoesthroughthemidpointsof BC andEH; theline f thatgoesthroughthemidpointsof CD and E F; theline g thatgoesthroughthemidpointsofAD and FG; theline h thatgoesthroughthemidpointsofAE and CG; andtheline i thatgoesthroughthemidpointsof DH and BF; inallcasesthelinespointinthedirectionofthemidpointofthesecondlistededge. Hence Re
1/2, Rf1/2, R
g
1/2, Rh1/2 and Ri
1/2 aresymmetriesofthecube.
B C
DA
d j
F G
HE
B C
DAF G
HE
(i) (ii)
Figure5.7.3
5.7ThreeDimensionalSymmetry 207
Therearealsoaxesofrotationthroughopposingverticesofthecube. Forexample, inFig-ure 5.7.3 (ii)weseethelinethatgoesthroughtheverticesA andG, pointinginthedirectionofG; thislineisdenoted j. Then Rj
1/3 and Rj
2/3 aresymmetriesofthecube. Therearethreeothersimilaraxesofrotation: theline k thatgoesthrough B and H; theline l thatgoesthrough C
and E; andtheline m thatgoesthrough D and F; inallcasesthelinespointinthedirectionofthesecondlistedvertex. Hence Rk
1/3, Rk2/3, R
l1/3, R
l2/3, R
m1/3 and Rm
2/3 aresymmetriesofthecube. Wenowhaveacompletelistofrotationsymmetriesofthecube.Next, weturntoreflectionsymmetriesofthecube. Intheplane, wereflectedinaline, called
thelineofreflection. Inthreedimensionalspace, wereflectinaplane, calledthe planeofre-flection. Thatreflectionofthreedimensionalspaceisinaplaneisquitereasonableintuitively—mirrorsareplanes!
BEFORE YOU READ FURTHER:
Trytofindasmanyreflectionsymmetriesofthecubeaspossible.
ReferringtothecubeshowninFigure 5.7.1, itisevidentthatthecubehasreflectionsymmetryintheplanethatgoesthroughthecenterofthecubeandisparalleltothetop(ABCD)andthebottom(EFGH). Callthisplane p, anddenotereflectioninthisplaneby Mp. Therearetwoothersimilarplanesofreflection: theplane q thatgoesthroughthecenterofthecubeandisparalleltotheleftside(ABFE)andtherightside(DCGH); andtheplane r thatgoesthroughthecenterofthecubeandisparalleltothefront(ADHE)andtheback(BCGF). ThenMq andMr aresymmetriesofthecube.Thereisanothercollectionofplanesofreflectionofthecube. Forexample, let s denotethe
planecontainingtheedges AB and GH. ThenMs isareflectionsymmetryofthecube. Therearefiveothersimilarplanesofreflection: theplane t containingtheedges BC and EH; theplane u containingtheedges CD and E F; theplane v containingtheedges AD and FG;theplane w containingtheedges AE and CG; andtheplane x containingtheedges BF andDH. HenceMt,Mu,Mv,Mw andMx aresymmetriesofthecube. Wenowhaveacompletelistofreflectionsymmetriesofthecube.Havewenowfoundallthesymmetriesofthecube? Itmightatfirstappearasifwedoknow
allthesymmetriesofthecube, giventhatweknowallthereflectionandrotationsymmetriesofthecube, andweknowthattherearenotranslationorglidereflectionsymmetries. However, inthreedimensionalspace, thecompletelistoftypesofisometriesisnotjusttranslations, rotations,reflectionsandglidereflections. Itturnsoutthattherearetwoadditionaltypesofisometriesinthreedimensionalspace, called rotaryreflections and screws. Bothofthesetypesofisometriesaresimilartoglidereflections, inthattheyaresingleisometriesthataredescribedintermsoftwo-stepprocesses. A rotaryreflectionistheresultoffirstrotatingaroundanaxisofrotation,andthenreflectinginaplanethatisperpendiculartotheaxisofrotation; ascrewistheresultoffirst rotatingaroundanaxisof rotation, and then translating inadirectionparallel to theaxisofrotation. (See[Mar82, Section16.1]forathoroughdiscussionoftheisometriesofthreedimensionalspace, includingthethreedimensionalanalogofProposition 4.6.1.)
208 5. SymmetryofPlanarObjectsandOrnamentalPatterns
Thecubedoesnothaveanyscrewsymmetries, butitdoeshaverotaryreflectionsymmetries.Forexample, considerthecomposition Mp ◦ Ra
1/4. Thiscompositioniscertainlyasymmetryof thecube, being thecompositionof twosymmetries. However, byusingdrawings similartoFigure 5.7.2, itcanbeverifiedthatthiscompositionisnotequaltoanyoftherotationorreflectionsymmetrieswehavelistedforthecube(suchaverificationwouldentailcomparingtheneteffectofMp ◦ Ra
1/4 withtheneteffectsofeachoftherotationsandreflectionsthatwehavefound; weleavethedetailstothereader). Hence, thecomposition Mp ◦ Ra
1/4 isasymmetryofthecube, andisnotequaltoanyothertheothersymmetriesthatwehaveseensofar. Forconvenience, weusethefollowingnotation: If α isanangle, if a isalineinthreedimensionalspace, andif p isaplanethatisperpendicularspace, welet Ca
α,m denotetherotaryreflectionthatconsistsoffirstdoingtherotation Ra
α, andthendoingthereflection Mm. Hence, wewriteCa
1/4,p asanabbreviationforMp ◦ Ra1/4. Therearesixothersimilarrotaryreflectionsymmetries
ofthecubethatcanbeobtainedbycompositionsofrotationandreflectionssymmetriesofthecube, andtheseare Ca
1/2,p, Ca3/4,p, C
b1/4,r, C
b3/4,r, C
c1/4,q, C
c3/4,q.
Thereadermighthavenoticedthatwedidnot include Cb1/2,r and Cc
1/2,q in theabovelistof rotary reflectionsymmetriesof thecube. These twocompositionsare indeedvalid rotaryreflectionsymmetriesofthecube, butitturnsoutthattheyarebothequalto Ca
1/2,p. (Again, thereadercanverifythatthesethreecompositionshavethesameneteffects.) Actually, theneteffectofthesethreecompositionsisaparticularlynicesymmetryofthecube. SeeFigure 5.7.4 forthecomposition Ca
1/2,p. Observethattheneteffecttakeseachvertex, andmovesittothelocationdiametricallyoppositeitwithrespecttothecenterofthecube. Let O denotethecenterofthecube. TheisometrythattakeseverypointinthreedimensionalspaceandsendsittothepointdiametricallyoppositeitwithrespecttoO iscalled inversion inO. Let JO denotethisisometry.Fromnowon, insteadofwriting Ca
1/2,p wewillwrite JO. Itturnsoutthat JO canbeobtained
in sixadditionalwaysas rotary reflections; eachof Mu ◦ Rd1/2, Mv ◦ Re
1/2, Ms ◦ Rf1/2,
Mt ◦ Rg
1/2, Mx ◦ Rh1/2 and Mw ◦ Ri
1/2 isequalto JO.
Wearestillnotfinishedlookingforrotaryreflectionsymmetriesofthecube. Certainly, onecanobtainarotaryreflectionsymmetryofthecubebycomposingarotationsymmetryofthecubewithareflectionsymmetryofthecube(aslongastheplaneofreflectionisperpendiculartotheaxisofrotation). However, notallrotaryreflectionsymmetriesofthecubeareobtainedthatway. Itisalsopossibletoformarotaryreflectionsymmetryofthecubewheretherotaryre-flectionisthecompositionofarotationandareflection, neitherofwhichaloneisasymmetryofthecube, buttheircompositionis. (A similarphenomemonoccurredwhenwestudiedglidere-flectionsymmetryoffriezepatternsandwallpaperpatterns.) Forexample, let j denotetheplanecontainingthecenterofthecubethatisperpendiculartotheline j (showninFigure 5.7.3 (ii)).Thenneither Rj
1/6 nor Mj isasymmetryofthecube, butthecomposition Mj ◦ Rj
1/6, abbre-
viatedasbeforeby Cj
1/6,j, isinfactasymmetryofthecube. Theneteffectofthissymmetryis
showninFigure 5.7.5. Therearesevenothersimilarrotaryreflectionsymmetriesofthecube,andtheseare Cj
5/6,j, Ck
1/6,k, Ck
5/6,k, Cl
1/6,l, Cl
5/6,l, Cm
1/6,m, Cm5/6,m.
5.7ThreeDimensionalSymmetry 209
B C
DAF G
HE
D A
BCH E
FG
H E
FGD A
BC
R1/2
Mp
R1/2
aMp C
1/2, p° =
a
a
Figure5.7.4
B C
DAF G
HE
F B
CGE A
DH
R1/6
jM j C
1/6, j° =j
Figure5.7.5
Wenow, finally, haveacompletelistofsymmetriesofthecube:
I, Ra1/4, R
a1/2, R
a3/4, R
b1/4, R
b1/2, R
b3/4, R
c1/4, R
c1/2, R
c3/4, R
d1/2, R
e1/2,
Rf1/2, R
g
1/2, Rh1/2, R
i1/2, R
j
1/3Rj
2/3, Rk1/3, R
k2/3, R
l1/3, R
l2/3, R
m1/3, R
m2/3,
Mp,Mq,Mr,Ms,Mt,Mu,Mv,Mw,Mx, JO, Ca1/4,p, C
a3/4,p,
Cb1/4,r, C
b3/4,r, C
c1/4,q, C
c3/4,q, C
j
1/6,j, C
j
5/6,j, Ck
1/6,k, Ck5/6,k, C
l1/6,l, C
l5/6,l, C
m1/6,m, C
m5/6,m.
210 5. SymmetryofPlanarObjectsandOrnamentalPatterns
Clearly, thislistofsymmetriesismuchlarger, andmorecomplicated, thanthelistofsymmetriesofthesquare, whichisthetwodimensionalanalogofthecube(thecubehas 48 symmetries,versus 8 forthesquare). Nonetheless, weseethatforthreedimensionalobjectsitispossibletoformcompletelistsofsymmetries; inotherwords, wecanformthesymmetrygroups ofthreedimensionalobjectsjustaswedidforplanarobjects. Moreover, wecanformthecompositionsofsymmetriesofanobjectinthreedimensionalspace, andinprinciplewecouldformcompositiontables forthreedimensionalobjectsjustaswedidforregularpolygonsinSections 5.2 and5.3. Inpracticeformingsuchanoperationtablewouldbeverytimeconsuming—forthecubewewouldhavea 48×48 table, whichwouldhave 2304 entries—andsowewillnotactuallyconstructsuchtables. InExercise 5.7.1 thereaderisaskedtocomputethecompositionsofvarioussymmetriesofthecube; thesecalculationscomputesomeoftheentriesofthecompositiontableforthecube. Thebottomlineisthatsymmetryofthreedimensionalobjectscanbestudiedsimilarlytothestudyofplanarobjects, butthreedimensionalobjectsareagoodbitmorecomplicated.
Exercise 5.7.1. Forthecube, computethefollowingsymmetries(thatis, expresseachasasinglesymmetry).
(1) Rj
1/3 ◦ Ra1/4.
(2) Mp ◦ Rc1/4.
(3) Ms ◦ Mp.
(4) Mq ◦ Ca1/4,p.
Exercise 5.7.2. Foreachofthefollowingpolyhedra, listallofitssymmetries. Usepicturesorwordstodescribetheaxesofrotationandplanesofsymmetry.
(1) A pyramidoverasquare.
(2) A prismoveranequilateraltriangle, wherethesidesarerectangles, butnotsquares.
(3) A regulartetrahedron.
(4) A regularoctahedron.
Exercise 5.7.3. Howmanysymmetriesdoestheprismoveraregular n-gonhave? Assumethat the sidesof theprismare rectangles, butnot squares. (Youdonotneed to list thesymmetries, justcountthem.)
5.7ThreeDimensionalSymmetry 211
Exercise 5.7.4. Whatistherelationbetweenthesymmetriesofaconvexpolygonandthesymmetriesofitsdual?
Intheplane, westudiedthesymmetryofvariousclassesofobjects: regularpolygons, rosettepatterns, friezepatternsandwallpaperpatterns. Arethereanalogsforsuchclassesofobjectsinthreedimensionalspace? Theanswerisdefinitelyyes. A particularlyinterestingclassofob-jectsinthreedimensionalspaceistheanalogofwallpaperpatterns, thatis, patternsinthreedimensionalspacethathavetranslationsymmetryinatleastthreedifferentdirections(wherenotalldirectionsareinasingleplane). Suchpatternsarecalled crystals, becauseofthefactthatthemoleculesinchemicalcrystals, suchassalt(NaCl), alignthemselvesinalattice-likeformthatcorrespondsexactlytothenotionofhavingtranslationsymmetryinthreedifferentdirectionsinthreedimensionalspace. Thestudyofchemicalcrystalsiscalledcrystallography,andthesymmetrygroupsofmathematicalcrystalsarecalledthe crystallographicgroups (alsoknownasthe spacegroups). Thecrystallographicgroupscanbeclassifiedanalogouslytotheclassificationoffriezegroupsandwallpapergroups, althoughtheclassificationismuchmorecomplicated. Thereare 230 crystallographicgroups(incontrastto 17 wallpapergroups). Thecrystallographicgroupswerefirstcompletelyclassifiedin1891byEvgrafStepanovichFedorov(1853-1919)andArthurSchoenflies (1853-1928), eachworkingindependentlyoftheother. See[Sen90]formoredetailsaboutthecrystallographicgroups.Although the symmetrygroupof thecube ismuch largerandmorecomplicated than the
symmetrygroupofthesquare, thereisonesimilaritybetweenthesetwosymmetrygroupsthatwecanobserve. Inthesymmetrygroupof thesquare, half thesymmetriesarerotations(weconsider I tobeatrivialrotation), andhalfarereflections. AswesawinSection 5.4, thisequalsplitbetweenrotationsandreflectionsholdsforallrosettegroupsthathavereflectionsymme-try. Noticeinparticularthatrotationspreserveorientation andreflectionsreverseorientation.Hence, foranyrosettegroup, halfthesymmetriesareorientationpreservingandhalfareorien-tationreversing. Now, inthecaseofthecube, thesymmetrygroupcontainsnotonlyrotationsandreflections, butalsorotaryreflections. However, itstillisthecaseforthecubethathalfofitssymmetriesareorientationpreserving(therotations), andhalfareorientationreversing(thereflectionsandtherotaryreflections). Infact, itturnsoutthatanyfinitesymmetrygroupforanobjectinthreedimensionalspace(andactuallyinanydimensionalspace), eitherconsistsofallorientationpreservingsymmetries, orhalfitssymmetriesareorientationpreservingandhalfareorientationreversing. ThedemonstrationofthisfactisoutlinedinExercise 5.7.5.
212 5. SymmetryofPlanarObjectsandOrnamentalPatterns
Exercise 5.7.5. [UsedinThisSection] Ourgoalistoshowthatforanyfinitesymmetrygroupforanobjectinthreedimensionalspace, preciselyoneofthefollowingsituationsholds:eitherallthesymmetriesareorientationpreserving, orhalfthesymmetriesareorientationpreservingandhalfareorientationreversing. Wewillmakeuseofthefollowingtwofactsaboutisometriesthatwehaveseenfortheplane, andwhichinfactholdtrueinthree(orhigher)dimensionalspace; wewillnotbeabletodemonstratethesetwofacts—thatwouldrequiremoretechnicalitiesthanweareusing. First, theanalogofProposition 4.4.3 holdsinthreedimensions. Second, everyisometryhasaninverse.Suppose G isafinitesymmetrygroupforanobjectinthreedimensionalspace. Theargu-menthasanumberofsteps, mostofwhichhavesomethingforthereadertodo.
(1) Supposethat A, B and C aresymmetriesin G, andthat A = B. Showthat C ◦A = C ◦ B.
(2) If G hasallorientationpreservingsymmetries, thenthereisnothingtodemonstrate,soassumefromnowonthatnotallsymmetriesinG areorientationpreserving. Showthat G hasbothorientationpreservingandorientationreversingsymmetries.
(3) Let {A1, A2, . . . , An} denotetheorientationpreservingsymmetries in G, andlet{B1, B2, . . . , Bm} denotetheorientationreversingsymmetriesin G, where n andm aresomepositiveintegers. Ourgoalistoshowthat n = m, whichwillimplythat G hasthesamenumberoforientationpreservingsymmetriesandorientationreversingsymmetries.
(4) Considerthecollectionofsymmetries {B1 ◦ A1, B1 ◦ A2, . . . , B1 ◦ An}. Showthatthesesymmetriesarealldistinct.
(5) Showthatallthesymmetries {B1 ◦ A1, B1 ◦ A2, . . . , B1 ◦ An} areorientationreversing.
(6) Deduce that every oneof {B1 ◦ A1, B1 ◦ A2, . . . , B1 ◦ An} is contained in{B1, B2, . . . , Bm}.
(7) Deducethat n ≤ m.
(8) Usesimilarideastoshowthatallthesymmetries {B1 ◦ B1, B1 ◦ B2, . . . , B1 ◦ Bm}
aredistinct, andallarecontainedin {A1, A2, . . . , An}. Deducethat m ≤ n.
(9) Becausewehaveseenthat n ≤ m andthat m ≤ n, itfollowsthat n = m, whichiswhatweneededtoshow.
Finally, wearenowinapositiontoclarifysomethingleftunfinishedinSection 3.3, wherewediscussedthesemi-regularpolyhedra. Inparticular, welistedallsuchpolyhedra(inPropo-sition 3.3.1), andwementionedthatallexceptoneofthem(thepseudorhombicuboctahedron)satisfiedastrongerpropertycalledvertextransitivity. WecouldnotdefinethispropertyinSec-
5.7ThreeDimensionalSymmetry 213
tion 3.3, butwenowhavethenecessarytoolforthedefinition, namelysymmetry. A polyhedronissaidtobe vertextransitive if, givenanytwovertices v and w ofthepolyhedron, thereisasymmetryofthepolyhedronthattakes v to w.For example, we claim that theprismover a regular pentagon is vertex transitive. In Fig-
ure 5.7.6 weseeaprismoveraregularpentagon. Toshowthatthisprismisvertextransitive, weneedtoshowthatforanytwoverticesoftheprism, thereisasymmetryoftheprismtakingonevertextotheother. ConsidertheverticeslabeledA and B, asseeninFigure 5.7.6. Observethatrotationby 1/5 ofaturnaroundtheverticallinethroughthecenteroftheprismisasymmetryoftheprism, andthissymmetrytakesvertex A tovertex B. Rotationby −1/5 ofaturntakes B toA. TotakevertexA tovertex C, asseeninthefigure, weneedtherotaryreflectionobtainedbyfirstrotatingby 2/5 ofaturnaroundtheverticallinethroughthecenteroftheprism, andthenreflectingintheplanethatisparalleltothetopandbottompentagons, andishalfwaybetweenthem. Usingtheseideas, itisseenthatforanytwoverticesoftheprism, thereisasymmetryoftheprismtakingonevertextotheother. Hencethisprismisvertextransitive.
A B
C
Figure5.7.6
Itcanbeshownthatallofthesemi-regularpolyhedraotherthanthepseudorhombicubocta-hedronarevertextransitive; weomitthedetails. Bycontrast, thepseudorhombicuboctahedronisnotvertextransitive. InFigure 5.7.7 (i)weseethepseudorhombicuboctahedron, withthreeofitsverticeslabeled. Thereis, forexample, nosymmetryofthepseudorhombicuboctahedronthat takesvertex A tovertex B. Hence, thepseudorhombicuboctahedron isnotvertex tran-sitive. (Thatdoesnot, however, meanthatnovertexofthepseudorhombicuboctahedroncanbetakenbyasymmetrytoanothervertex; forexample, rotationby 1/4 turnaroundtheverti-callinethroughthecenterofthepseudorhombicuboctahedroninFigure 5.7.7 (i)takesvertexA tovertex C.) Bywayofcomparison, observethat fortherhombicuboctahedronshowninFigure 5.7.7 (ii), reflectionin thehorizontalplanethroughthecenterof thepolyhedronisasymmetryoftherhombicuboctahedronthattakesvertex A tovertex B.
Some textsadd thepropertyof vertex transitivity to thedefinitionof semi-regular (thoughwedonot), andiftheydo, thentheydonotconsiderthepseudorhombicuboctahedrontobesemi-regular, andtheyhaveonly13Archimedeansolids.
214 5. SymmetryofPlanarObjectsandOrnamentalPatterns
AA
B B
CC
(i) (ii)
Figure5.7.7
6Groups
6.1 Thebasicidea
AtthestartofChapter 4 wereadaquotebyHermanWeyl whichended:
Toacertaindegreethisschemeistypicalforalltheoreticknowledge: Webeginwithsomegeneralbutvagueprinciple(symmetryinthefirstsense), thenfindanimportantcasewherewecangivethatnotionaconcreteprecisemeaning(bilateralsymmetry), andfromthatcasewegraduallyriseagaintogenerality, guidedmorebymathematicalconstructionandabstractionthanbythemiragesofphilosophy; andifweareluckyweendupwithanideanolessuniversalthantheonefromwhichwestarted. Gonemaybemuchofitsemotionalappeal, butithasthesameorevengreaterunifyingpowerintherealmofthoughtandisexactinsteadofvague.”
Inthepresentchapter, thelastinourbook, wenowindeedrisetothelevelofmathematicalgenerality, andunifyingpower, towhichWeylwasreferring. Atfirstitmightnotbeapparentwhatthematerialinthischapterhastodowithsymmetry, butwewillmaketheconnectionclearinourverylastsection, Section 6.6.Inthischapterwewilldiscussthemathematicalconceptofagroup. Unlikethecolloquialus-
ageofthisword, toamathematicianagroupisaverypreciselydefinedconcept, aswillbeseenbelow. Thoughatfirstglancegroupsappeartobeveryabstract, likeallworthwhileabstractiontheyarebasedonconcreteexamples. Indeed, itistheextremelybroadrangeofexamplesofgroupsthathaveledthegroupconcepttobeconsideredverycentraltomodernmathematics.Thetheoryofgroups, thoughlessthan200yearsold, ishighlydeveloped, withnewdiscoveriesbeingmadeallthetime. Groupsareextremelyusefulineverythingfromgeometrytochemistry;inparticular, groupsarevital to thestudyof symmetry, and it is for this reason thatwedis-cussthemhere. Further, themethodologyofgrouptheoryepitomizestheabstractapproachof
216 6. Groups
modernmathematics(asspearheadedearlierinthiscenturybythegreatmathematicianEmmyNoether), andthischapter’sexcursionintotheabstractshouldbeseenasatasteofwhatmanymathematiciansdotoday.Considertheintegers −3,−2,−1, 0, 1, 2, 3 . . .. Wewillusethestandardabbreviation Z to
denote the setof integers. Theword“set” is simply thecommonlyusedmathematical termtomeana“collection”of things, in thiscasenumbers, thoughasetcouldcontainanytypeofobject, not justnumbers. (The letterZ,by theway, stands for theGermanwordZahlen,whichmeansnumbers.) Ifallwecoulddowith the integerswouldbe towrite themdown,theywouldbeentirelyuseless. Whatmakestheintegerssousefulisthatwecancombinethem,viaaddition, subtraction, multiplicationanddivision. Actually, subtractionisjustdoingaddition“backwards,” anddivisionisjustmultiplication“backwards,” sowereallyneedtoconsideronlyadditionandmultiplication. (Whatdoes 5 − 3 mean? Itmeansthenumberthatyouaddto 3toget 5, namely 2.) Wewillconsidereachofthetwooperations, additionandmultiplication,separately. Eachoftheseoperationsisreferredtoasa binaryoperation,inthatittakestwothings(inthiscasenumbers)asinputs, andgivesoneoutput(inthiscase
alsoanumber).Whatpropertiescanweascribetotheoperationofadditionasappliedtotheintegers? Some
ofthesepropertiesmayseemsoobviousastobehardlyworthmention, buttheirvaluewillbeapparentlateron. (Itmightbethesimplicityofthesepropertiesthatcausedmathematicianstotakesolongtofocusinonthem.) First, wenotethatifwetakeanytwointegersandaddthem,wegetanotherinteger. Wecallthispropertythe closure propertyoftheintegerswithrespecttoaddition. Toappreciatetheworthofthisproperty, notethatifyoutakeanytwointegersanddivideonebytheother, youwillmostlikelynotgetaninteger, forexample 3 dividedby 2.Next, supposeyouwanttoaddanythreeintegers, forexample 2, 3 and 7. Becausewecan
formallyaddonlytwointegersatatime, wehavetogrouptheintegers 2, 3 and 7 withparen-thesestoprescribetheorderofaddition. Ifwekeepthesethreeintegersinthegivenorder, weseethattherearetwowaysofgroupingthem, namely (2 + 3)+ 7 and 2+ (3+ 7). Theformersaystoadd 2 and 3 first, andthentoadd 7 totheresult; thelattersaystoadd 3 and 7 first,andthentoadd 2 totheresult. Ofcourse, wegetthesamefinalanswerinbothcases, inthat(2 + 3) + 7 = 5 + 7 = 12 and 2 + (3 + 7) = 2 + 10 = 12. Indeed, becausewegetthesameanswerbothways, itissafetodroptheparenthesesandsimplywrite 2 + 3 + 7, lettingeachpersondotheadditionanywayshechooses. Wecanstatethispropertymoregenerallybysayingthatforanythreeintegers a, b and c, wealwayshave (a+ b) + c = a+ (b+ c).Wecallthispropertythe associative propertyfortheintegerswithrespecttoaddition.Ifyouwereaskedtochosethesinglemostimportantinteger, whichwouldyouchoose? Al-
though each personmay have a personal favorite number, mathematically the uncontestedleader of the pack is the number zero. Of itsmany properties, 0 is the only number that,whenaddedtoanyothernumber, leavestheothernumberunchanged. Forexample, wehave5 + 0 = 5. Toputthismoregenerally, foranyinteger a, wehave a + 0 = a and 0 + a = a.Wecallthispropertyof 0 the identity propertyfortheintegerswithrespecttoaddition.
6.1Thebasicidea 217
Onewayofobtaining 0 isbyaddingany integer and itsnegative. For example, wehave5 + (−5) = 0. Moregenerally, foranyinteger a, wehave a+ (−a) = 0 and (−a) + a = 0.Notethattheseequationsholdwhether a ispositive, negativeor 0. Theessentialpointhereisthatforanyinteger a, thereisanotherinteger, namely −a, that“cancels a out.” Wecallthispropertythe inverses propertyoftheintegerswithrespecttoaddition.Althoughtheabovefourpropertiesoftheintegersandadditionarethemostcrucialonesfor
ourpurpose, there isonemorepropertywewillmention, which, thoughwellknown, turnsouttobelesscentralthanthepropertiesmentionedsofar. Thisadditionalproperty, calledthecommutative property, saysthattheorderofadditiondoesnotmatter. Forexample, wehave5 + 3 = 3 + 5. Ingeneral, foranytwointegers a and b, wealwayshave a+ b = b+ a.To summarize, we see that the integers togetherwith the operation addition, symbolized
(Z,+), satisfythefourfundamentalpropertiesofclosure, associativity, identityandinverses, aswellastheadditionalpropertyofcommutativity. Theintegerswithadditionsatisfyanumberofotherpropertiesaswell, butafterlookingatmanyothermathematicalsystems, mathematiciansfoundthatthesefourpropertiesareextremelyprevalentinmanyseeminglyunrelatedfields, fromgeometrytoquantummechanics, andhavethereforechosentofocusonthesefourproperties.Letuslookatsomeothermathematicalsystems, toseeiftheysatisfythesamepropertiesas
theintegerswithaddition. Thenextmostobviousexampletoconsideristheintegerswiththeoperationmultiplication, abbreviated (Z, ·). Weneedtocheckwhetherthefourpropertiesofclosure, associativity, identityandinverses, aswellasthecommutativeproperty, holdfor (Z, ·).Letusstartwithclosure. Itiscertainlythecasethatifwemultiplyanytwointegerswegetaninteger, so theclosurepropertyholds. It isalsonothard tosee that theassociativepropertyholds, thatis, foranythreeintegers a, b and c, itisalwaystruethat (a · b) · c = a · (b · c).Whatabouttheidentityproperty? Weneedtofindaspecialmemberoftheintegersthatplaysthesamerolewithrespecttomultiplicationthat 0 doesforaddition; inotherwords, weneedanumberso thatmultiplyingby itdoesnotchangeanything. Certainly thenumber 1 is theintegerwewant. If a isanyinteger, then 1 · a = a and a · 1 = a. Hence 1 istheidentityfortheintegerswithmultiplication, andsotheidentitypropertyholdsfor (Z, ·). Next, weneedtoverifywhethertheinversespropertyholdsfor (Z, ·). Thismeansthatforeveryinteger, weneedtofindanotherintegerthatcancelsitoutbymultiplication, yielding 1. Letustrythisfortheinteger 2. Thereiscertainlyanumberthatcancels 2 outwithrespecttomultiplication, namely1/2, because 2 ·(1/2) = 1 and (1/2) ·2 = 1. Thereisamajorproblemhere, however, becausewearedealingwiththeintegers, and 1/2 isnotaninteger. Thereiscertainlynoothernumberthatcancels 2 outwithrespecttomultiplication, sowehavetoconcludethat 2 doesnothaveamultiplicativeinverseintheintegers. (Thenumber 2 doeshaveanadditiveinverse, namely−2,butthatdoesnothelpushere.) Hence, weseethat (Z, ·) doesnothavetheinversesproperty.Therefore, eventhough (Z, ·) satisfiesthefirstthreepropertiesthat (Z,+) satisfies, itdoesnotsatisfythefourthproperty. Itisnothardtoseethatthecommutativepropertyholdsfor (Z, ·),thatis, foranytwointegers a and b, itisalwaystruethat a · b = b · a.Theproblem thatoccuredwith (Z, ·) might suggest toyouwherewecanfindsomething
thatdoessatisfyallfiveproperieswithrespecttomultiplication. Thenumber 1/2 isnotinthe
218 6. Groups
integers, butitisafraction, sowhydon’twelookatthesetofallfractions, denoted Q. (Theletter Q standsforquotient.)
BEFORE YOU READ FURTHER:
Trytofigureoutwhether (Q, ·) satisfytheclosure, associative, identity, inversesandcom-mutativeproperties.
Justaswith (Z, ·), itisnothardtoseethat (Q, ·) satisfiestheclosure, associativityandidentityproperties(onceagain 1 istheidentitywithrespecttomultiplication). Butthistime, unliketheintegers, itseemsthattherearemultiplicativeinverses. Foranyfraction, thefractionthatcancelsitoutbymultiplicationisjustthereciprocaloftheoriginalfraction. Forexample, thereciprocalof 5/3 is 3/5, andsureenough (5/3) · (3/5) = 1 and (3/5) · (5/3) = 1. Soitappearsasif(Q, ·) hastheinversesproperty. Almost, butthereisstillonelittleglitch. Thenumber 0 canbeconsideredasafraction, say 0/1. Unfortunately, thefraction 0/1 hasnoreciprocal, becausewewouldwanttouse 1/0, butthatisnotallowedbecausewecannotdivideby 0. Itfollowsthatthefraction 0/1 doesnothaveamultiplicativeinverse. However, thenumber 0 istheonlyproblem, becauseanyfractionthatdoesnotequal 0 doeshaveareciprocal. Wewillbypasssthisproblemcausedby 0 asfollows. LetususethesymbolQ∗ todenotethesetoffractionswiththenumber 0 removed. Then, ifweputalltheabovereasoningtogether, weseethat (Q∗, ·) doessatisfythefourpropertiesofclosure, associativity, identityandinverses. Moreover, becausetheorderofmultiplicationoftwonumbersdoesnotmatter, forexample 4 · 7 = 7 · 4, weseethat(Q∗, ·) alsosatisfiesthecommutativeproperty.
Exercise 6.1.1. Determinewhichofthefivepropertiesofclosure, associativity, identity,inversesandcommutativityaresatisfiedbyeachofthefollowingsystems.
(1) Theevenintegerswithaddition.
(2) Theoddintegerswithaddition.
(3) Allrealnumberswithaddition.
(4) Allrealnumberswithmultiplication.
6.2 Clockarithmetic
Sofarwehavebeenconcernedwithvarioussetsofnumbers, suchasintegersandfractions. Allthesesetshavebeeninfinite. Wenowwishtoexamineamathematicalsystemthatisfiniteinsize. Thismathematicalsystemisbasedontheideaof“clockarithmetic,” whichyoumayhaveseen; ifyouhavenot, itwillbesufficientthatyouhaveseenaclock. Allourreferencestotime
6.2Clockarithmetic 219
willbebasedontheAmerican 12 hoursystem(thoughwewillignorea.m.vs.p.m.), asopposedtothe 24 hoursystemusedmanyotherplacesaroundtheworld(andtheU.S.military); eithertimesystemwouldworkforourpurpose, andwehavesimplychosenoneofthemonceandforalltoavoidanyambiguity.Sayitis 2 o’clock, andyouwanttoknowwhattimeitwillbein 3 hours. A sillyquestion, you
maybethinking, becausethetimeinthreehoursissimply 2 + 3 = 5 o’clock. Right, butnowsupposeitis 7 o’clock, andyouwanttoknowwhatthetimewillbein 6 hours. Youcouldgo7 + 6 = 13, butyouwouldn’tsay 13 o’clock, becausethereisnosuchthing; youwouldsay1 o’clock, ofcourse, andyouwouldberight. Howdidyouget 1 o’clock? Yousubtracted 12from 13, because 13wasgreaterthan 12, andthereforetoolarge. Now, supposeitis 11 o’clock,andyouwanttoknowwhattimeitwillbeafter 30 hours(again, ignoringa.m.andp.m.). Youwouldstartbygoing 11 + 30 = 41, butonceagainthisistoolarge, becauseyoucannothave41 o’clock. Theonly“o’clocks”youcanhavearefrom 1 to 12 (roundingofftowholehours,aswearedoing). Therefore, youwanttotake 41 and“bringitdown”tobetween 1 and 12. Todothis, youwanttosubtractfrom 41 asmanycopiesof 12 asyoucan. Thebestyoucandoissubtract 3 times 12, whichis 36. Now, wecompute 41 − 36 = 5, soifyoustartat 11 o’clockandgoanother 30 hours, youendupat 5 o’clock.
Exercise 6.2.1. Ifyoustartat 7 o’clock, andgoanother 20 hours, whattimewillitbe?
Letusnowlookmorecarefullyatwhatwejustdid; wewilldropthe“o’clocks”forconve-nience. Thereweretwothingswewantedtoaccomplish, whichseemedsomewhatatoddswitheachother: ontheonehand, wewantedtorestrictourselvestotheintegers 1 through 12. Ontheotherhand, wewantedtobeabletoaddnumbers, whichtookusoutsideofthe 1 to 12range. Toresolvetheproblem, wetookanynumberthatwasoutsideofthe 1 to 12 range, andreduceditrepeatedlyby 12 untilwewerebackinthedesiredrange. Forexample, wereduced13 to 1 bysubtracting 12, andwereduced 41 to 5 bysubtracting 3 times 12. Inotherwords,weareessentiallyconsidering 13 and 1 asequivalent(fromthepointofviewofclocks), and 41and 5 areconsideredequivalent.Wearethereforeledtoanewnotion, called congruencemod 12. Wesaythattwointegers
are congruentmod 12 iftheydifferbyanintegermultipleof 12. Therefore, wesaythat 13 and1 arecongruentmod 12, andthat 41 and 5 arecongruentmod 12. Ontheotherhand, thenumbers 17 and 3 arenotcongruentmod 12 becausetheirdifferenceis 14, whichisnotanintegermultipleof 12. Thenumbers 5 and 29 arecongruentmod 12, becausetheirdifferenceis 5− 29 = −24 = (−2) · 12.
220 6. Groups
Exercise 6.2.2. Whichofthefollowingpairsofnumbersarecongruentmod 12?
(1) 15 and 3;
(2) 9 and 57;
(3) 7 and −5;
(4) 11 and 1;
(5) 0 and 12.
Forthesakeofbrevity, weintroducethefollowingnotation. Ifintegers a and b arecongruentmod 12, wewritethis a ≡ b (mod 12). Forexample, wehave 41 ≡ 5 (mod 12). If a and b
arenotcongruentmod 12, wewrite a ≡ b (mod 12). Forexample, wehave 3 ≡ 7 (mod 12).Fromtheclockexample, wenoticedthatanyintegerwhatsoevercouldbereducedbymulti-
plesof 12 untilwhatisleftissomewherefrom 1 through 12. Hence, ifweareinterestedonlyinintegersmod 12, thenweneedtoconsideronly 1 through 12, becauseanythingelsecanbereducedtooneofthesenumbers. Foreaseofuselateron, wewillmakeonesmallchangeatthispoint. Insteadofconsideringtheintegersfrom 1 to 12, wewillswitchtotheintegersfrom 0 to11. Thischangehasnosubstantialeffect, because 0 ≡ 12 (mod 12). Anyintegerwhatsoevercanbereducedbymultiplesof 12 untilwhatisleftisanintegerfrom 0 through 11. Inotherwords, forany integerwhatsoever, there isanother integer, this time from 0 to 11, which iscongruentmod 12 totheoriginalinteger; moreover, thereisonlyonesuchintegerfrom 0 to 11.Insymbols, foranyinteger a, thereisauniqueinteger x from 0 to 11 sothat x ≡ a (mod 12).Forexample, ifwelet a = 13, then x = 1, because 1 ≡ 13 (mod 12). If a = 35, then x = 11,because 11 ≡ 35 (mod 12). If a = 12, then x = 0, because 0 ≡ 12 (mod 12). Notethatthenumber a neednotbepositive. If a = −4, then x = 8, because 8 ≡ (−4) (mod 12).Additionally, notethatif a = 7, then x = 7 aswell, because 7 isalreadybetween 0 to 11.
Exercise 6.2.3. Foreachinteger a givenbelow, findtheinteger x from 0 to 11 sothatx ≡ a (mod 12).
(1) a = 18;
(2) a = 41;
(3) a = −17;
(4) a = 3.
Wearenowledtothefollowingmethodforconstructinganewmathematicalsystem, whichsimplyencapsulateswhatwedowhenwetelltime. Oursystemwillhavetwelveobjects, de-
6.2Clockarithmetic 221
noted 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11. ThecollectionofthesetwelveobjectswillbedenotedZ12. Weputthe“hat”ontheseobjectstoindicatethat, althoughtheycorrespondtotheintegersfrom 0 through 11, theydonotbehaveexactlyliketheintegerstowhichtheycorrespond. Thedifferenceisinhowweaddtheelementsin Z12. Letusstartwithsomeexamples. Toadd 3 and4 iseasy; itissimply 3+4 = 7. Ontheotherhand, wecannotsaythat 6+8 is 14, becausethereisnosuchthingas 14 in Z12. So, asonaclock, whatwedoistoreduce 14 byintegermultiplesof 12. Moreconcisely, wewanttofindanintegerfrom 0 to 11 thatiscongruentmod 12 to 14.Thenumberisclearly 2, andsowesay 6+8 = 2. Ingeneral, if a and b aretwonumbersin Z12,tofind a+ b wefirstfind a+b asusual; if a+b isfrom 0 to 11, weputahatoverit, andthatisouranswer; if a+ b islargerthan 11, wefindaninteger x from 0 to 11 sothat x ≡ (a+ b)
(mod 12), andthenlet a + b = x. Forexample, wehave 3 + 7 = 10, and 7 + 9 = 4, and2 + 0 = 2. Itshouldbeclearthatalthoughweusetheusual“+”signtodenote“addition”inZ12, thisoperationisnotthesameasstandardadditionofintegers, becausewereducemod 12.(Itwouldbesensibletoputa“hat”onthe + signthatweusefor Z12, similarlytothehatweputon 0, 1, 2, . . ., 11, butitisnotstandardtodoso, andwewillstickwithstandardnotation.)
Exercise 6.2.4. Calculatethefollowing.
(1) 4 + 5;
(2) 7 + 8;
(3) 5 + 11;
(4) 0 + 3.
Weareinterestedinthesystem (Z12,+). Onehelpfultoolforunderstandingthissystemisadevicethathelpeduslearnmultiplicationaschildren, namelymultiplicationtables, suchasTable 6.2.1, whichshowsmultiplicationupto 10.Thistablesummarizesexplicitlyallpossiblemultiplicationsbetweenintegersfrom 1 to 10.
Forexample, tofind 3 · 7, lookintherowlabeled 3, andthecolumnlabeled 7, andintheintersectionofthisrowandthiscolumnwefind 21, justasexpected.Wecan, similarly, makeanadditiontablefor (Z12,+), showninTable 6.2.2. Forexample,
tofind 4 + 9, lookintherowlabeled 4, andthecolumnlabeled 9, andintheintersectionofthisrowandthiscolumnwefind 1, whichis 4 + 9. Noticethecyclicpatterninthetable.Wenowaskwhether (Z12,+) thesamefiveproperties(discussedinSection 6.1)that (Z,+)
satisfies. Theclosurepropertyholdsfor (Z12,+), becausetheway+wasdefinedforZ12 insuresthataddinganytwoelementsinZ12 yieldsanotherelementinZ12. Asforassociativity, withabitofthoughtitisnothardtoseethatbecausethestandardadditionfortheintegersisassociative,theadditionof Z12 isalsoassociative; wewillomitthedetails.Toseethattheidentitypropertyholds, wenotethat 0 playstheroleofanidentityelement,
becauseforany a in Z12, itisseenthat a+ 0 = a and 0 + a = a.
222 6. Groups
· 1 2 3 4 5 6 7 8 9 101 1 2 3 4 5 6 7 8 9 102 2 4 6 8 10 12 14 16 18 203 3 6 9 12 15 18 21 24 27 304 4 8 12 16 20 24 28 32 36 405 5 10 15 20 25 30 35 40 45 506 6 12 18 24 30 36 42 48 54 607 7 14 21 28 35 42 49 56 63 708 8 16 24 32 40 48 56 64 72 809 9 18 27 36 45 54 63 72 81 9010 10 20 30 40 50 60 70 80 90 100
Table6.2.1
Whatabout inverses? Youmight thinkatfirst that therecannotbe inverses, because therearenonegativenumbersin Z12. Butnegativeisarelativeterm(dependingonyourset, youroperationandyourzero), andinfactthereareinversesin (Z12,+). Letusstartwith 1. What,ifanything, is its inverse in (Z12,+)? Inotherwords, is thereanelement in Z12 that, whenaddedto 1, yields 0. Recallingthat 0 ≡ 12 (mod 12), weseethatthenumberthatcancels1 outisprecisely 11, because 1 + 11 = 12, andtherefore 1 + 11 = 0. Itisnothardtoseethateveryelementin Z12 hasaninversewithrespecttoaddition. Forexample, theinverseof5 is 7, because 5 + 7 = 12, andtherefore 5 + 7 = 0. Hence, theinversespropertyholdsfor(Z12,+). Itisnothardtoseethatthecommutativepropertyalsoholdsfor (Z12,+), forexample5 + 6 = 6 + 5. Wehavethereforeverifiedthat (Z12,+) satisfiesthesamefivepropertiesas(Z,+).
Exercise 6.2.5. Findtheinverseswithrespecttoadditionof 3, 6, 8 and 0 in Z12.
Wecanalsomakeamultiplicationtablefor Z12, showninTable 6.2.3.Notice that themultiplication table for Z12 doesnot have the same simplepattern along
upward-slopinglinesasdidtheadditiontablefor Z12. Moreover, notethatintheadditiontable,eachof 0, 1, . . . , 11 appearsonceandonlyonceineachrowandineachcolumn; thispropertydoesnotholdforthemultiplicationtable. Alltold, multiplicationfor Z12 isnotaswellbehavedasaddition. SeeExercise 6.2.6 fordetails.
Exercise 6.2.6. Whichofthefiveproperties(closure, associativity, identity, inverses, com-mutativity)holdsfor (Z12, ·)? Forthoseelementsof Z12 thathaveinverseswithrespecttomultiplication, statewhattheirinversesare.
6.3TheIntegersMod n 223
+ 0 1 2 3 4 5 6 7 8 9 10 11
0 0 1 2 3 4 5 6 7 8 9 10 11
1 1 2 3 4 5 6 7 8 9 10 11 0
2 2 3 4 5 6 7 8 9 10 11 0 1
3 3 4 5 6 7 8 9 10 11 0 1 2
4 4 5 6 7 8 9 10 11 0 1 2 3
5 5 6 7 8 9 10 11 0 1 2 3 4
6 6 7 8 9 10 11 0 1 2 3 4 5
7 7 8 9 10 11 0 1 2 3 4 5 6
8 8 9 10 11 0 1 2 3 4 5 6 7
9 9 10 11 0 1 2 3 4 5 6 7 8
10 10 11 0 1 2 3 4 5 6 7 8 9
11 11 0 1 2 3 4 5 6 7 8 9 10
Table6.2.2
6.3 TheIntegersMod n
InSection 6.2 webasedourdiscussiononthenumbertwelvebecauseofourfamiliaritywithclocks. Wecan, however, repeatthewholeprocedurewithanyotherpositiveintegerreplacing12. Chooseanypositiveinteger n. Wesaythatanytwointegersare congruentmod n iftheydifferbysomeintegermultipleof n. Insymbols, suppose a and b areintegers. Wesaythat aand b arecongruentmod n, written a ≡ b (mod n), if a−b = kn forsomeinteger k (whichcouldbepositive, negativeorzero). If a and b arenotcongruentmod n, wewrite a ≡ b
(mod n). Forexample, saywechoose n = 5. Then 17 and 2 arecongruentmod 5, written17 ≡ 2 (mod 5), because 17 − 2 = 15 = 3 · 5, whichisanintegermultipleof 5. Also, wehave 3 ≡ 11 (mod 4) because 3−11 = −8 = (−2) ·4. However, wehave 13 ≡ 2 (mod 9),because 13− 2 = 11, whichisnotamultipleof 9.
Exercise 6.3.1. Whichofthefollowingaretrue, andwhicharefalse?
(1) 3 ≡ 9 (mod 2);
(2) 7 ≡ (−1) (mod 8);
(3) 4 ≡ 11 (mod 3);
(4) 0 ≡ 24 (mod 6).
(5) 9 ≡ 9 (mod 5).
224 6. Groups
· 0 1 2 3 4 5 6 7 8 9 10 11
0 0 0 0 0 0 0 0 0 0 0 0 0
1 0 1 2 3 4 5 6 7 8 9 10 11
2 0 2 4 6 8 10 0 2 4 6 8 10
3 0 3 6 9 0 3 6 9 0 3 6 9
4 0 4 8 0 4 8 0 4 8 0 4 8
5 0 5 10 3 8 1 6 11 4 9 2 7
6 0 6 0 6 0 6 0 6 0 6 0 6
7 0 7 2 9 4 11 6 1 8 3 10 5
8 0 8 4 0 8 4 0 8 4 0 8 4
9 0 9 6 3 0 9 6 3 0 9 6 3
10 0 10 8 6 4 2 0 10 8 6 4 2
11 0 11 10 9 8 7 6 5 4 3 2 1
Table6.2.3
Foreachpositiveinteger n greaterthanorequalto 2, wecanformasystemcalled Zn com-pletelyanalogouslytothewayweformed Z12. Wewillobtainonesuchsystemforeachinteger2, 3, 4, . . . (Weskipover thecase n = 1, becausethat turnsout tobeuseless.) Letusstartwiththeexampleof n = 8. Analogouslytowhatwedidwith 12, weseethatfor n = 8, anyintegercanbereducedbymultiplesof 8 untilwhatisleftissomewherefrom 0 through 7. Inotherwords, foranyinteger, thereisauniqueintegerfrom 0 to 7 thatiscongruentmod 8 totheoriginalinteger. Insymbols, foranyinteger a, thereisauniqueintegerfrom 0 to 7, denoted x,sothat x ≡ a (mod 8). Forexample, ifwelet a = 10, then x = 2, because 2 ≡ 10 (mod 8).Nowsupposewestartwith a = 1950. Inthiscase, wecouldproceedbysubtracting 8, andthenanother 8, andthenagainandagain, asmanytimesasareneeded, untilweareleftwithsomenumberfrom 0 to 7. Thatwouldwork, butwouldbeverytedious. A bettermethodwouldbetodivide 1950 by 8. Wewouldthenseethatthequotientis 243, andtheremainderis 6;thatis, weseethat 1950/8 = 243 + (6/8). Itfollowsthat 1950 = 243 · 8 + 6, andhencethat6 − 1950 = (−243) · 8. Wethereforeseethat 6 ≡ 1950 (mod 8), andthereforewecanusex = 6. Wenote, asbefore, thatifwestartwithanumber a thatisalreadyfrom 0 to 7, then x
isjust a itself.
6.3TheIntegersMod n 225
Exercise 6.3.2. Foreachinteger a givenbelow, findtheinteger x from 0 to 7 sothat x ≡ a
(mod 8).
(1) a = 15;
(2) a = 54;
(3) a = 1381;
(4) a = −2;
(5) a = 3;
(6) a = 8.
Asbefore, theset Z8 andwillhave 8 members, denoted 0, 1, 2, 3, 4, 5, 6 and 7. Weaddelementsof Z8 asbefore, exceptthatthistimewereducebymultiplesof 8. Forexample, in Z8
wehave 2 + 3 = 5 andwehave 5 + 4 = 1, wherethelatterholdsbecause 5 + 4 = 9, and1 ≡ 9 (mod 8).
Exercise 6.3.3. Calculatethefollowingin Z8.
(1) 4 + 1;
(2) 3 + 7;
(3) 0 + 3.
Justaswedidfor (Z12,+), wecanformanadditiontablefor (Z8,+), showninTable 6.3.1.Noticethesamediagonalpatternasbefore.
+ 0 1 2 3 4 5 6 7
0 0 1 2 3 4 5 6 7
1 1 2 3 4 5 6 7 0
2 2 3 4 5 6 7 0 1
3 3 4 5 6 7 0 1 2
4 4 5 6 7 0 1 2 3
5 5 6 7 0 1 2 3 4
6 6 7 0 1 2 3 4 5
7 7 0 1 2 3 4 5 6
Table6.3.1
226 6. Groups
Weaskwhether (Z8,+) satisfiesthesamefivepropertiesas (Z,+), andtheanswerisyes.Theclosure, associativeandidentitypropertiesholdfor (Z8,+) justastheydidfor (Z12,+).Weobservethat 0 isonceagaintheidentityelement. Theinversepropertyalsoholds, althoughalittlecautionmustbetaken, becausetheinversesin Z8 arenotthesameasin Z12. Forexample,theinverseof 1 in Z8 is 7, because 1+7 = 8, andso 1 + 7 = 0. Thiscontrastswiththeinverseof 1 in Z12, whichis 11. Thecommutativepropertyalsoholdsfor (Z8,+).
Exercise 6.3.4. Findtheinverseswithrespecttoadditionof 2, 4, 5 and 0 in Z8.
Justas (Z8,+) satisfiesthefivepropertiesofclosure, associativity, identity, inversesandcom-mutativity, sodoes (Zn,+) foranypositiveinteger n, where n ≥ 2. Thesystem (Zn,+) iscalledthe groupofintegersmod n withtheoperationaddition. Noticethat (Zn,+) haspre-cisely n members.
Exercise 6.3.5. Considerthesystem (Z6,+).
(1) Listtheelementsofthissystem.
(2) In (Z6,+), whatare 5 + 2 and 4 + 1?
(3) Constructtheadditiontablefor (Z6,+).
(4) Findtheinverseswithrespecttoadditionof 2, 4, 5 and 0 in Z6.
Exercise 6.3.6. Observethatintheadditiontablefor (Z8,+), showninTable 6.3.1, alltheentriesonthedownwardsslopingdiagonalareevennumbers. Willthesamefactholdintheadditiontableforany (Zn,+)? Ifyes, explainwhy. Ifnot, describewhatdoeshappenonthedownwardsslopingdiagonalfor (Zn,+) ingeneral, andexplainyouranswer.
Havinglookedat (Zn,+), letusnowturnto (Zn, ·). Considerthecaseof (Z5, ·). Themulti-plicationtablefor (Z5, ·) isshowninTable 6.3.2.NoticethatTable 6.3.2 doesnotsatisfythesimplepatternalongupward-slopinglinesasin
Tables 6.2.2 and6.3.1.Itisseenthat (Z5, ·) doessatisfytheclosure, associative, andidentityproperties(with 1 as
identity), andthecommutativepropertyaswell. However, theinversespropertyisnotsatisfied,because 0 hasnoinversewithrespecttomultiplication. Toseethisfact, wenotethataninversefor 0 wouldbesome x in Z5 suchthat 0 · x = 1 and x · 0 = 1. A lookatthetableshowsthatnosuch x exists. Wecanremedythisproblemjustaswedidfor Q inSection 6.1, bydropping
6.3TheIntegersMod n 227
· 0 1 2 3 4
0 0 0 0 0 0
1 0 1 2 3 4
2 0 2 4 1 3
3 0 3 1 4 2
4 0 4 3 2 1
Table6.3.2
theproblematic 0. Letususethesymbol Z∗5 todenote Z5 with 0 removed. Wethenobtainthe
multiplicationtablefor (Z∗5, ·), shown inTable 6.3.3. Weleaveittothereadertoverifythat
(Z∗5, ·) satisfiesallofourproperties.
· 1 2 3 4
1 1 2 3 4
2 2 4 1 3
3 3 1 4 2
4 4 3 2 1
Table6.3.3
Unfortunately, whatworksfor (Z∗5, ·) doesnotworkforall (Z∗
n, ·). Forexample, weseeinTable 6.3.4 themultiplication table for Z∗
6. Notallfivepropertieshold for (Z∗6, ·). First, the
closurepropertydoesnothold; forexample, weseethat 2 · 3 = 0, but 0 isnotin Z∗6. The
associativepropertyholds, asdoestheidentityproperty(withidentity 1), andthecommutativepropertyholds. Theinversespropertydoesnothold; forexample, itisseenthat 2 doesnothaveaninversewithrespecttomultiplication.
· 1 2 3 4 5
1 1 2 3 4 5
2 2 4 0 2 4
3 3 0 3 0 3
4 4 2 0 4 2
5 5 4 3 2 1
Table6.3.4
228 6. Groups
BEFORE YOU READ FURTHER:
Istheresomethingaboutthenumber 5 thatmakes (Z∗5, ·) satisfyallourproperties, and
aboutthenumber 6 thatmakes (Z∗6, ·) failtosatisfytheclosureandidentityproperties?
Ingeneral, what is itaboutan integer n thatwoulddeterminewhetherornot (Z∗n, ·)
satisifiesallfiveproperties? (Youwillmostlikelynotbeabletodemonstrateyouranswerrigorously, unlessyouknowmoreaboutnumbersthanweareassuming, buttrytomakeaneducatedguess.)
Itturnsoutthattherelevantdifferencebetween 5 and 6 thatleadsto (Z∗5, ·) satisfyingourfive
propertiesbut (Z∗6, ·) notsatisfyingallfiveistheissueofprimenumbers vs.compositenumbers.
A positiveintegerisa primenumber ifitsonlypositivefactorsare 1 anditself. A positiveintegerthatisnotprimeiscalledcomposite. Forexample, thenumbers 2, 3, 5 and 7 areprime, whereas6 iscomposite, havingfactors 1, 2, 3, and 6. Itturnsout, thoughthisisfarfromobvious, that(Z∗
n, ·) satisfiesallfivepropertiesifandonlyif n isaprimenumber. Theproofusesfactsaboutprimenumbers.
Exercise 6.3.7. Constructthemultiplicationtablefor (Z8, ·).
6.4 Groups
Intheprevioussectionsofthischapter, wesawanumberofmathematicalsystemsthatsatisfiedthesamepropertiesofclosure, associativity, identityandinverses. (Allthesystemsdiscusseduptillnowalsosatisfiedthecommutativityproperty, butwewillseesystemsthatdonotsatisfythispropertyinalittlewhile.) Mathematicianshaveinfactfoundsomanysystemsthatsatisfythesesamefourfundamentalproperties(thoughnotnecessarilycommutativity), thattheydecidedtogiveallsuchsystemsanamesothattheycouldbestudiedtogether, andcommonpropertiescouldbefound. Wewillcallanysystemsatisfyingthesefourpropertiesagroup.Letusphrasethisconceptmoreprecisely. Firstofall, agroupisasetofobjects, whichmay
benumbers(asinthecaseoftheintegers), butwhichcouldbeotherthingsaswell. Supposethat G isasetofobjects. Themembersof G willbereferredtoas elements of G. Next, thissetofobjects G needsa binaryoperation, whichcombineselementsof G bytakinganytwoelementsof G asinputs, andfortheseinputsgivesauniqueoutput. Now, supposethat ∗ isabinaryoperation. Thatmeansthatforanytwoelements g and h inG, wecancombinethemtogetasinglethingdenoted g ∗ h. Thenotation g ∗ h ismeanttobesimilartothenotationsforadditionandmultiplication, ourtwomostfamiliarbinaryoperations. Wewilldenoteaset Gtogetherwithanoperation ∗ bythepair (G, ∗). Oneexampleofsuchapair (G, ∗) isthepair(Z,+). A pair (G, ∗) iscalleda group ifitsatisfiestheabovementionedfourproperties, whichwenowstateintheirmostgeneralform:
6.4Groups 229
Closureproperty: If g and h arein G, then g ∗ h isin G.
Associativeproperty: If g, h and k arein G, then (g ∗ h) ∗ k = g ∗ (h ∗ k).Identityproperty: ThereisadistinguishedelementinG, calledan identityelement anddenotede, sothatif g isin G, then e ∗ g = g and g ∗ e = g.
Inversesproperty: If g is in G, there isanelement g ′ in G, calledan inverse of g, so thatg ∗ g ′ = e and g ′ ∗ g = e.
Some, thoughnotall, groupsalsosatisfythefollowingproperty:
Commutativeproperty: If g and h arein G, then g ∗ h = h ∗ g.A groupthatalsosatisfiesthecommutativepropertyiscalledan abelian group. (Itwouldbe
entirelyreasonabletocallsuchagroupa“commutativegroup;” however, thatisnotstandardterminology. ThetermabeliangroupisinhonoroftheNorwegianmathematicianNielsAbel.Thischoiceofnamehasgivenrisetothefollowingwellknownjoke(wellknownamongmath-ematicians, atleast). Question: whatispurpleandcommutative? Answer: anabeliangrape.)
Ifwegobackandconsidertheexampleswehaveseensofar, nowusingtheterminologyofgroups, weseethat (Z,+) isanabeliangroup, asis (Zn,+) foreachpositiveinteger n, wheren ≥ 2. Ontheotherhand, thesystem (Z, ·) isnotagroup, becausewesawthatitdidnotsatisfytheinversesproperty. Wealsosawthat (Z∗
n, ·) isagrouppreciselyif n isaprimenumber.
Exercise 6.4.1. Whichofthefollowingsystemsisagroup? Whichisanabeliangroup?
(1) Theevenintegerswithaddition.
(2) Theoddintegerswithaddition.
(3) Allrealnumberswithaddition.
(4) Allrealnumberswithmultiplication.
Thereisonematterweneedtoclarifyrightawayaboutthedefinitionofgroups. Inthestate-mentoftheidentityproperty, wementioned“anidentityelement,” andinthestatementoftheinversesproperty, wementioned“aninverse.” Coulditbethecasethatagrouphasmorethanoneidentityelement, orthatanelementinagrouphasmorethanoneinverse? Intuitivelythatsoundsunlikely, andthefollowingpropositionshowsthatourintuitioniscorrect.
Proposition 6.4.1. Supposethat (G, ∗) isagroup.
1. Thegroup G hasauniqueidentityelement.
2. If g isanelementof G, then g hasauniqueinverse.
Demonstration. WeshowPart (1), leavingPart (2)tothereaderinExercise 6.4.2.
230 6. Groups
(1). Wefollowthestandardmathematicalapproach toshowing thatsomething isunique,whichistosupposethattherearetwoofthething, andthenshowthatthetwothingsareinfactequal. Inparticular, supposethat e and c arebothidentityelementsof G. Then
e = e ∗ c = c,
whereinthefirstequalitywearethinkingof c asanidentityelement, andinthesecondequalitywearethinkingof e asanidentityelement. Itfollowsthat e = c, andthereforetheidentityelementof (G, ∗) isunique.
Bytheaboveproposition, wecannowreferto“theidentityelement”ofagroup, and“theinverse”ofeachelementofthegroup.
Exercise 6.4.2. [UsedinThisSection] DemonstrateProposition 6.4.1 (2).
Groupscomeintwobasicvarieties, infinite orfinite, dependingonhowmanyelementsareinthegroup. Forexample, thegroup (Z,+) isaninfinitegroup, andeach (Zn,+) isafinitegroup.Althoughtheinfinitegroupswehavedealtwithsofarmayseemmorenatural(forexample, theintegers), finitegroupsareofteneasiertoworkwithmathematically. Letuslookatsomefurtherexamplesoffinitegroups.Recall that thegroup (Zn,+) hasprecisely n elements in it. Because thisworks foreach
positiveinteger n with n ≥ 2, weseethatthereisagroupofeachpossiblefinitesize. (Thereisalsoagroupwithoneelement, namelythesetwiththesingleelement 0, andwiththeoperationaddition.) However, therearemanyotherfinitegroupsbesidesthegroups (Zn,+), thoughmanyofthemaremorecomplicatedtoconstruct. Someofthefollowingexamplesoffinitegroupsmayappearsomewhatarbitrary. Wheredidthesegroupscomefrom? Sometimestrialanderrorwasused, thoughifthatweretheonlymethod, notonlywouldthatberathertedious, butitwouldberatherunappealing. Therearevarioussystematicwaysofconstructingfinitegroups,someofwhichareextremelycomplex, yieldinghugegroupswithnicknamessuchas“monster”(seriously).To see someexamplesoffinitegroups, recall thatagroup ingeneral is a setof elements
G, togetherwithanoperation ∗, subject to fourproperties. Whendealingwith the familiaroperationsofadditionandmultiplication, allwehadtodowasnametheseoperations, andeveryoneknewexactlywhatweweretalkingabout. Inunfamiliarsituation, whenwecannotrefertoanoperationbysimplygivinganamewithwhicheveryonewouldbefamiliar, wewillreturntotheideaofthemultiplicationtablementionedpreviously. Todescribeagroup, wewillfirstdescribeasetG, andthendescribeanoperation ∗ usinga“multiplication”table, whichwewillcallan operationtable fromnowon.Weuseoperationtablesjustasweusedadditiontablesfor (Z,+) and (Z,+) previously. Whatwillbenewnowisthatinsteadofusinganoperationtabletogiveapictorial representation ofabinaryoperationwithwhichwearealreadyfamiliar,nowwewill define newbinaryoperationsbygivingoperation tables for them. Ifwedefine
6.4Groups 231
defineabinaryoperation ∗ bygivingitsoperationtable, thentofind a ∗ b, wesimplylookattheentryintheoperationtableintherowcontaining a andthecolumncontaining b.Letusconstructatwo-elementgroupviaanoperationtable. Westartwithaset, labeled T ,
whichcontainstwoelements, called r and s; wecanabbreviatethisbywriting T = {r, s}. Wethenspecifyabinaryoperation ∗ bygivingTable 6.4.1.
∗ r s
r r s
s s r
Table6.4.1
FromTable 6.4.1 wesee, forexample, that r ∗ s = s and s ∗ s = r. Wewant toverifywhether (T, ∗) isagroup. Now, wedonotknowwhether r and s aremeanttodenotenumbers,orperhapshouses, orsomethingelse; wealsodonotknowthat ∗ “means,” inthewaythatweknowwhatadditionandmultiplicationofnumbersmeans. So, isitpossibletoverifywhether(T, ∗) isagroupusingonlytheoperationtablegivenfor ∗? Theanswerisyes—everythingthatcanbeknownabout ∗ iscontainedinitsoperationtable.Theclosurepropertycertainlyholdsfor (T, ∗), becauseanytwoelementsintheset T yieldan
elementof T whencombinedby ∗; thisfactisevidentbecausealltheentriesinTable 6.4.1 arethemselvesin T . Ingeneral, aslongasallentriesinanoperationtablearethemselveselementsoftheoriginalset, thentheclosurepropertywillnecessarilyhold.Tochecktheassociativityof ∗ weneedtocheckmanycases. Inprinciple, onewouldhave
tocheckeverypossiblewaytocombine threeelementsof T (repeatsallowed), tosee if theassociativityruleholds. Forexample, does (r ∗ s) ∗ s equal r ∗ (s ∗ s)? Usingtheoperationtable, weseethat (r ∗ s) ∗ s = s ∗ s = r, and r ∗ (s ∗ s) = r ∗ r = r, whichiswhatwehadhopedfor. Tocheckwhetherassociativityholdsfor (T, ∗), wehavetodoallotherpossiblecases, whichwouldbequitetedious. A casebycasecheckdoesshowthatassociativityholdsinthisexample; wewillnotgothroughthedetails. Becausecheckingforassociativityissotedious(andnotveryenlightening), thereadercanassumetheassociativepropertyforanyoperationtablesweusefromnowon(noteverypossibleoperationtablesatisfiestheassociativeproperty,butwearenotinterestedinthosethatdonot).Wedefinitelycannotassumethetwootherpropertiesofgroups, however, andsoweneedto
verifywhethertheidentityandinversespropertiesholdfor (T, ∗). Toverifywhethertheidentitypropertyholds, weneedtofindanelementof T thatisanidentityelementwithrespectto ∗.Thatis, wewanttofindanelementof T thatbehaveswithrespectto ∗ justasthenumber 0behaveswithrespecttoadditionofnumbers. Lookingatthetable, weseethat r ispreciselysuchanelement, because r ∗ r = r, and r ∗ s = s and s ∗ r = s. Hence, theidentitypropertyholdswithidentityelement r.Toverifytheinversesproperty, wehavetofindaninverseforeachelementof T , wherean
inverseofanelementissomethingthatcancelsitout, whichmeansthattheelementanditsinversecombinetoyieldtheidentityelement. (Ofcourse, ifasetandbinaryoperationdonot
232 6. Groups
haveanidentityelement, thequestionofinversesismoot.) Bylookingatthetable, weseethatr ∗ r = r and s ∗ s = r. Inotherwords, theelements r and s areeachtheirowninverses.Itmayseemsomewhatstrangethatsomethingisitsowninverse, thatis, itcancelsitselfout,butthereisnothinginvalidhere. Consequently, theinversespropertyholdsfor (T, ∗). Puttingthisalltogether, weseethat (T, ∗) isagroup. Doesthecommutativepropertyhold? Notethatr ∗ s = s and s ∗ r = s, andso r ∗ s = s ∗ r. Becausethisistheonlypossiblepairofelementstocheckforcommutativity, anditworks, weseethat (T, ∗) satisfiesthecommutativeproperty,andhenceisanabeliangroup.Thatwasafairbitofefforttoverifythat (T, ∗) wasagroup, butifwewanttobesurethat ∗
asgiveninTable 6.4.1 yieldsagroup, wecannotavoidthateffort, becausenoteveryoperationtableyieldsagroup. Infact, ifyourandomlywritedownanoperationtable, itishighlyunlikelythatitwillyieldagroup. Asanexample, letustakethesameset T = {r, s} asbefore, butletusdefineadifferentbinaryoperation, denoted • thistime, givenbyTable 6.4.2.
• r s
r r s
s s s
Table6.4.2
Wewanttoverifywhetherornot (T, •) isagroup. Thereadermayverifythat (T, •) onceagainsatisfiestheclosure, associativityandidentityproperties, with r againplayingtheroleoftheidentity. Butwhataboutinverses? Fromthetableweseethat r•r = r, so r isitsowninverse.Ontheotherhand, lookingatthetabledoesnotyieldaninversefor s. Weseethat r • s = s
and s • s = s, soneither r nor s canbetheinverseof s. Itfollowsthat (T, •) doesnotsatisfytheinversesproperty, andisthereforenotagroup. So, notalloperationtableswork.Letustryafewmoreoperationtables. ConsiderthesetV = {x, y, z,w}withbinaryoperation
⊕ givenbyTable 6.4.3.
⊕ x y z w
x w z y x
y z w x y
z y x w z
w x y z w
Table6.4.3
Is (V,⊕) agroup? Becausealltheentriesinthetableareintheset V , theclosurepropertyholds. Asmentioned, wecanassume theassociativityproperty. As for the identityproperty,noticethat w playstheroleofanidentityelement, because w combinedwithanythingisthatsamething. Observethatthecolumnunderw intheoperationtableisthesameasthecolumnattheleftendofthetable; similarly, therowtotherightof w isthesameastherowatthetopofthetable. Thisphenomenonholdspreciselybecause w istheidentityelement, andthis
6.4Groups 233
methodcanbeusedtofindidentityelements(iftheyexist)quicklyinanyoperationtable. Forinverses, itisseenthateachelementisitsowninverse(thiswillnotbethecaseforallfinitegroups, sodon’tjumptoanyconclusionshere). Hence, allthepropertiesofagrouphold, and(V,⊕) isindeedagroup. Itisalsothecasethatthecommutativepropertyholdsfor (V,⊕), Toseethis, youcouldtryallpossibilities. Forexample, weseethat x ⊕ y = z and y ⊕ x = z,sothat x ⊕ y = y ⊕ x; similarlyfortheothercases. Hence (V,⊕) isanabeliangroup. Thereisaneasierwaytoseethatcommutativityholdsfor (V,⊕). Noticethattheoperationtablefor(V,⊕) issymmetricaboutitsdownwardslopingdiagonal. Ifyouthinkaboutit, youwillseethatingeneral, foranygroup, thistypeofsymmetryoftheoperationtablewillholdpreciselyifagroupsatisfiesthecommutativeproperty.
Exercise 6.4.3. Foreachcollectionofobjectsandoperationtableindicatedbelow, answerthefollowingquestion:
(a) Istheclosurepropertysatisfied?
(b) Isthereanidentityelement? Ifso, whatisit?
(c) Whichelementshaveinverses? Forthosethathaveinverses, statetheirinverses? (Ifthereisnoidentityelement, thisquestionismoot.)
(d) Isthecommutativepropertysatisfied?
(e) Assumingthattheassociativepropertyholds, dothecollectionofobjectsandgivenoperationformagroup? Iftheyareagroup, isitanabeliangroup?
(1) Theset V = {x, y, z,w} withbinaryoperation ⋄ givenbyTable 6.4.4.
(2) Theset K = {m,n, p, q, r} withbinaryoperation ⋆ givenbyTable 6.4.5.
(3) Theset M = {1, s, t, a, b, c} withbinaryoperation ⊙ givenbyTable 6.4.6.
(4) Theset W = {e, f, g, h,w, x, y, z} withbinaryoperation ∗ givenbyTable 6.4.7.
⋄ x y z w
x z w y x
y x y z w
z y z w x
w z w x z
Table6.4.4
Wehaveseensomeexamplesoffinitegroupsgivenbyoperationtables, andotherexamples(namelythegroups (Zn,+))thatwerenotgivenbyoperationtables. However, anyfinitegroup
234 6. Groups
⋆ m n p q r
m n p q m r
n p r m n p
p q m r p n
q m n p q r
r r p n r q
Table6.4.5
⊙ 1 s t a b c
1 1 s t a b c
s s t 1 b c a
t t 1 s c a b
a a c b 1 t s
b b a c s 1 t
c c b a t s 1
Table6.4.6
hasanoperationtable, evenifthatisnothowthegroupwasinitiallydescribed; nomatterhowabinaryoperationisdefined, wecanalwayswriteoutanoperationtablesimplybyseeingwhatthebinaryoperationdoestoeachpairofelementsofthegroup.
BEFORE YOU READ FURTHER:
Lookattheexamplesofoperationtablesthatwehaveseensofarthatyieldgroups. Canyouseeanynicefeaturesofthewaythattheelementsarearrangedinthesetables?
Thefollowingpropositionstatesaverynicefeatureofoperationtablesoffinitegroups.
Proposition 6.4.2. Supposethat (G, ∗) isafinitegroup. Intheoperationtableforthegroup,eachelementofthegroupappearsexactlyonceineachrow, andonceineachcolumn.
Demonstration. SupposetothecontrarythatasingleelementofG appearstwiceinonerow. Inparticular, supposethatthissameelementappearstwiceintherowcorrespondingtotheelementa; supposethatitappearsinthecolumnscorrespondingtoelements b and c. Itfollowsthata∗b = a∗c. Because (G, ∗) isagroup, weknowthat a hasaninverse, say a ′. Wededucethata ′ ∗ (a ∗b) = a ′ ∗ (a ∗ c), andhencebyassociativityweseethat (a ′ ∗a) ∗b = (a ′ ∗a) ∗ c.If e istheidentityelementofthegroup, thenitfollowsthemeaningofinverseelementsthate ∗ b = e ∗ c. By themeaningof the identityelement, wededuce that b = c. However,weassumedthatthecolumnscorrespondingto b and c aredistinctcolumns, andsowehavearrivedata logical impossibility. Theonlywayoutof thissituationis toconcludethateachelementof G appearsatmostonceineachrow. Inordertofillupeachrowintheoperation
6.4Groups 235
∗ e f g h w x y z
e e f g h w x y z
f f g h e x y z w
g g h e f y z w x
h h e f g z w x y
w w x y z e f g h
x x y z w f g h e
y y z w x g h e f
z z w x y h e f g
Table6.4.7
tablewithelementsof G, itmustbethecasethateveryelementof G appearsatleastonceineachrow. ThefinalconclusionisthateachelementofG appearsexactlyonceineachrow. Thesameideawillworkforcolumnsinsteadofrows, andwewillskipthedetails.
Wenotethateventhougheachelementofthegroupappearsexactlyonceineachrow, andonce in each column, of its operation table, it is definitely not the case that anyoperationtablethatsatisfiesthispropertyyieldsagroup. ThereaderisaskedtofurnishanexampleinExercise 6.4.4.
Exercise 6.4.4. [UsedinThisSection] Findanexampleofafinitesetwithabinaryoperationgivenbyanoperationtable, suchthateachelementofthesetappearsexactlyonceineachrow, andonceineachcolumn, andyetthesetwiththisbinaryoperationisnotagroup.
Exercise 6.4.5. Let C betheset C = {k, l,m}. Constructanoperationon C, bymakinganoperationtable, whichturns C intoagroup.
Whatdoesitmeantosaythattwogroupsaredifferent? Certainly, thegroup (T, ∗) givenbyTable 6.4.1 isdifferentfromthegroup (V,⊕) givenbyTable 6.4.3, becausetheformerhastwoelements(namely r and s), whereasthelatterhasfourelements(namely x, y, z and w).Nowconsidertheset Q = {E, F,G,H} withbinaryoperation ⊞ givenbyTable 6.4.8.Thereadercanverifythat (Q,⊞) isindeedanabeliangroup. Wenowcomparethegroups
(V,⊕) and (Q,⊞). Technicallytheyaredistinctgroups, havingdifferentelementsandoperationtables, butintuitivelytheyappeartobe“essentiallythesame.” Moreprecisely, observethatwe
236 6. Groups
⊞ F G H E
F E H G F
G H E F G
H G F E H
E F G H E
Table6.4.8
canobtainTable 6.4.8 fromTable 6.4.3 bythefollowingsubstitutions:
x 7−→ F
y 7−→ G
z 7−→ H
w 7−→ E.
ThefactthatTable 6.4.8 isobtainedfromTable 6.4.3 bythissubstitutionmeansthatnotonlydotheelementsof V correspondtotheelementsofQ, buttheoperation ⊕ correspondstotheoperation⊞. Hence, wesaythat (V,⊕) and (Q,⊞) areessentiallythesameinthatthesecondgroupisobtainedfromthefirstsimplybyrenamingtheelementsoffirstgroupandrenamingthebinaryoperation. Wenote, moreover, that it isacceptable foroneoperation table tobeobtainedfromanotherbysubstitution, evenifonetablehastoberearranged. Forexample, if⊞ hadinitiallybeengivenbyTable 6.4.9, thattoowouldbeessentiallythesameas ⊕. Ifonegroupcanbeobtainedfromanotherbyrenamingtheelementsandthebinaryoperation, andpossiblyrearrangingtheoperationtable, thenwesaythetwogroupsare isomorphic.
⊞ E F G H
E E F G H
F F E H G
G G H E F
H H G F E
Table6.4.9
Clearly, twogroupswithdifferentnumbersofelementscannotbeisomorphic. Ontheotherhand, not all groups of the same size are isomorphic. For example, consider the set Z ={a, b, c, d} withbinaryoperation ⋆ givenbyTable 6.4.10.Again, thereadercanverifythat (Z, ⋆) isanabeliangroup. However, weclaimthat (Z, ⋆)
isnot isomorphic to (Q,⊞) (andhencenot to (V,⊕) either). Themostdirectway toshowthat (Q,⊞) and (Z, ⋆) arenotisomorphicwouldbetotryeverypossiblewayofrenamingtheelementsofQ as a, b, c and d, andthenobservingthatweneverobtainTable 6.4.10 for ⋆ fromTable 6.4.8 for ⊞, evenafterrearranging. Suchaverificationwouldbequitetedious. Themoreappealingwaytoverifythattwogroupsarenotisomorphicistofindsomepropertyofoneof
6.5Subgroups 237
⋆ a b c d
a a b c d
b b c d a
c c d a b
d d a b c
Table6.4.10
thegroupsthatdoesnotholdfortheothergroup, butsuchthatthepropertywouldbepreservedbyrenamingandrearranginganoperationtable. Forexample, weobservethatinTable 6.4.10,whichhasidentityelement a, eachof a and c isitsowninverses, but b and d arenottheirowninverses(theyareinversesofeachother). Bycontrast, inTable 6.4.8, whichhasidentityelement E, weobservethateachof F, G, H and E isitsowninverses. Hence, in (Z, ⋆) twoelementsaretheirowninverses, whereasin (Q,⊞) allfourelementsaretheirowninverses. Itfollowsthatthesetwogroupscouldnotpossiblybeisomorphic.
Exercise 6.4.6. Isthegroup (Z4,+) isomorphictoeitherof (Z, ⋆) or (Q,⊞)? If (Z4,+)isisomorphictooneofthesetwogroups, demonstratethisfactbyshowinghowtorenametheelementsof (Z4,+) appropriately.
6.5 Subgroups
Oneinterestingphenomenoninthetheoryofgroupsistheideaofasubgroup. Considerthegroup (Z,+). Insidethesetofintegers Z isthesetofevenintegers
E = {. . . ,−6,−4,−2, 0, 2, 4, 6, . . .}.
Thesystem (E,+), isitselfagroup. (YouwereaskedtoverifythisfactinExercise 6.4.1 (1); thepointisthataddingtwoevennumbersgivesanevennumber, sotheclosurepropertyholds,andtheotherpropertiescanbeverifiedsimilarly.) Hence, weseethat (E,+) isbothagroupinitsownright, andit isalsocontainedinthelargergroup (Z,+). Wesaythat (E,+) isasubgroupof (Z,+). Ingeneral, acollectionofelementsofagroup forma subgroup if thiscollection, togetherwiththeoperationoftheoriginalgroup, formagroupintheirownright.Noteverycollectionofelementsofagroupsformsasubgroup. Forexample, considerthesetofoddnumbers, denoted O. Itturnsoutthat (O,+) isnotagroup, becausetheclosurepropertydoesnothold; toseethis, notethatthesumoftwooddnumbersisanevennumber, notanoddnumber.
238 6. Groups
Exercise 6.5.1.
(1) Let T bethesetofallintegermultiplesof 3, thatis, theset
T = {. . . ,−9,−6,−3, 0, 3, 6, 9, . . .}.
Is (T,+) subgroupof (Z,+)?
(2) Let V bethesetofallperfectsquareintegersandtheirnegatives, thatis, theset
V = {. . . ,−16,−9,−4,−1, 0, 1, 4, 9, 16 . . .}.
Is (V,+) subgroupof (Z,+)?
It turnsout thatagroupmayhavemanysubgroups, or itmayhavevery few. Everygroupcontainswhatiscalledthe trivialsubgroup, whichisthesubgroupconsistingofnothingbuttheidentityelement. Thistrivialsubgrouphasonlyoneelementinit, whichmaymakeitseemlessthanexciting, butitisreallyavalidgroup. Forexample, thetrivialsubgroupof (Z,+) isjusttheoneelementset {0}, togetherwiththeoperationofaddition. Notethat 0 + 0 = 0, sotheclosurepropertyholdsforthistrivialgroup; theotherpropertiesofgroupscanalsobeverifiedfor {0}. Everygroupalsocontainsatleastoneothersubgroup, namelyitself. Wedidnotrequirethatasubgrouphavefewerelements thantheoriginalgroup. Ofcourse, whatwearereallyinterestedinaresubgroupsthatarenottheentireoriginalgroup. A subgroupthatisnotequaltotheoriginalgroupiscalleda propersubgroup. Thequestionnowbecomeswhetherthereareanyproper, non-trivialsubgroupsinagivengroup.Letusstartwiththeexampleof (Z8,+), theoperationtableforwhichisgiveninTable 6.3.1.
Wewanttofindasubcollectionofelementsof (Z8,+) thatformagroupbythemselves. Be-cause theoperation + isassociative forall theelementsof Z8, it iscertainlyassociative foranysubcollectionofelements. Asaresult, wewillnothavetoworryaboutassociativitywhenlookingforsubgroupsof (Z8,+), orsubgroupsofanythingelseforthatmatter. Ontheotherhand, wedohavetoworryaboutclosure, identityandinverses. Becausethesubcollectionsof(Z8,+) thatwearelookingformustsatisfytheidentityproperty, theymustcontain 0. So, weneedtofindasubcollectionof Z8 = {0, 1, 2, 3, 4, 5, 6, 7} thatcontains 0, andthatsatisfiestheclosureproperty, andsuchthatforanyelementofthesubcollection, itsinversewillbeinthesubcollection. A goodwaytotrytofindsuchasubcollectionistochoosesomeelements, andconstructtheoperationtable.Letustrythesubcollection A = {0, 1, 3, 7} of Z8, chosenrandomly. Theoperationtablefor
theseelements, showninTable 6.5.1, wasobtainedbydeletingalltheunnecessaryrowsandcolumnsoftheoperationtablefor (Z8,+).Aninspectionoftheoperationtablefor A revealsthat (A,+) isnotasubgroupof (Z8,+).
First, itdoesnotsatisfytheclosureproperty, because, forexample, weseethat 1 + 3 = 4, and
6.5Subgroups 239
+ 0 1 3 7
0 0 1 3 7
1 1 2 4 0
3 3 4 6 2
7 7 0 2 6
Table6.5.1
yet 4 isnotinthesubcollection A. Ingeneral, ifalltheentriesintheoperationtableforthesubcollectionarethemselvesinthesubcollection, thentheclosurepropertyholds; conversely,ifsomeoftheentriesintheoperationtableforthesubcollectionarenotinthesubcollection,thentheclosurepropertydoesnothold. Further, notethat 3 doesnothaveaninversein A,becausethereisnothingin A which, whenaddedto 3, yields 0. (Theelement 3 doeshaveaninversein (Z8,+), namely 5, but 5 isnotin A.) So, ifwewanttofindsubgroups, weneedtochooseoursubcollectionsmorecarefully.Letusnowtrythesubcollection B = {0, 2, 4, 6}. Theoperationtablefor (B,+) isshownin
Table 6.5.2.
+ 0 2 4 6
0 0 2 4 6
2 2 4 6 0
4 4 6 0 2
6 6 0 2 4
Table6.5.2
Thistimethingslookmorepromising. Noticethatalltheelementsofthetablearefromthesubcollection B, sothattheclosurepropertyholds. Theelement 0 isinthecollection, sotheidentitypropertyholds. Asforinverses, notethat 0 isitsowninverse, that 4 isitsowninverse,andthat 2+ 6 = 0 and 6+ 2 = 0, sothat 2 and 6 areinversesofeachother. Hencetheinversespropertyholds. Becausetheassociativitypropertyisautomatic, asmentionedabove, weseethat (B,+) isindeedasubgroup.Arethereanyotherpropersubgroupsof (Z8,+)? Twothatareeasytofindare C = {0} (the
trivialsubgroup)and D = {0, 4}. Itisnothardtoverifythat C and D areindeedsubgroupsbyexaminingtheiroperationtables. Arethereanyothersubgroups? Wecouldexamineeachpossiblesubcollectionof Z8 aswedidsubcollection A above, butthatwouldbeverytedious.Ingeneral, itisnoteasytofindallsubgroupsofagivengroup, butthereisonefactthatis
veryusefulinreducingtheworkincheckingthevarioussubcollections. Thisresult, knownasLagrange’sTheorem, isasfollows. A proofofthistheoremisbeyondthescopeofthistext.
Proposition 6.5.1 (LaGrange’sTheorem). Supposethat (G, ∗) isafinitegroup. Supposethat Hisasubgroupof G. Thenthenumberofelementsin H dividesthenumberofelementsin G.
240 6. Groups
Inotherwords, ifyouhaveafinitegroupandyouarelookingforsubgroups, youcanruleoutanysubcollectionwherethenumberofelementsdoesnotdividethenumberofelementsoftheoriginalgroup. However, justbecauseasubcolletiondoeshaveanacceptablenumberofelementsdoesnotmeanthatthesubcollectionisnecessarilyasubgroup.LetusapplyLagrange’sTheoremtothegroup (Z8,+). Thisgrouphas 8 elements. Theonly
numbersthatdivide 8 are 1, 2, 4 and 8. Wecanignore 1 and 8, becausetheonlysubgroupwithoneelementisthetrivialsubgroup {0}, andtheonlysubgroupwith 8 elementsisthewholeof(Z8,+). Hence, byLagrange’sTheorem, alltheproper, non-trivialsubgroupsof (Z8,+) musthave 2 or 4 elements. Inparticular, therecanbenosubgroupsof (Z8,+) witheither 3, 5, 6 or7 elements. Ontheotherhand, noteverysubcollectionof (Z8,+) with 2 or 4 elementsisasubgroup. Forexample, thesubcollection A discussedabovehad 4 elements, andyetwasnotasubgroup. Infact, itturnsoutthattherearenoothersubgroupsof (Z8,+) otherthan B, CandD givenabove(weomitthedetails, thoughitcouldbeverifieddirectly, albeittediously, bycheckingallsubcollectionsof Z8 witheither 2 or 4 elements).Next, letusfindallsubgroupsofthegroup (Z7,+). Thisgrouphas 7 elements. ByLagrange’s
Theoremanysubgroupwouldhave tohaveanumberofelements thatdivides 7. But 7 isaprimenumber; thatis, therearenonumbersthatdivide 7 exceptitselfand 1. A subgroupwith7 elementswouldjustbethewholegroup (Z7,+), andasubgroupwith 1 elementwouldhavetobethetrivialsubgroup. Inotherwords, weseefromLagrange’sTheoremthat (Z7,+) hasnoproper, non-trivialsubgroup. Thissamereasoningappliestoanyfinitegroupthathasaprimenumberofelements.
Exercise 6.5.2. Thegroup (Z36,+) has 36 elements. Howmanyelementscouldasubgroupof (Z36,+) possiblyhave?
Exercise 6.5.3. Which, if any, of the following subcollections of Z6 are subgroups of(Z6,+)? UseLagrange’sTheorem, andconstructoperationtablesaswedidforsubgroupsof (Z8,+).
(1) A = {0, 3};
(2) B = {0, 2};
(3) C = {0, 1, 4};
(4) D = {0, 2, 4}.
(5) E = {0, 1, 2, 3}.
6.6SymmetryandGroups 241
Exercise 6.5.4. Let (M,⊙) beas inExercise 6.4.3 (3). Which, ifany, of the followingsubcollectionsof M aresubgroupsof (M,⊙)?
(1) E = {1, s};
(2) F = {1, a};
(3) C = {1, s, t};
(4) D = {1, a, b, c}.
Exercise 6.5.5. Findasmanypropersubgroupsasyoucanof (Z12,+). Theoperationtablefor (Z12,+) isgiveninTable 6.2.2.
6.6 SymmetryandGroups
Althoughthestudyofsymmetryappearstobe“geometric”innature, andthestudyofgroupsappearstobe“algebraic,” infactsomeofthesameideasappearinbothfields. Forexample,recallLeonardo’sTheorem(Proposition 5.4.5)aboutrosettepatterns, whichstatedthatthesym-metrygroupofarosettepatterniseitherCn forsomepositiveinteger n, orDn forsomepositiveinteger n. The Cn groupsshouldlookveryfamiliarafterourdiscussionoftheintegersmod n
inSection 6.3. Forexample, wesaw inTable 5.4.1 theoperationtablefor (C8, ·). Comparethatoperationtablewiththeoperationtablefor (Z8,+), showninTable 6.3.1. Itisseenfromthesetwooperationtablesthat (C8, ·) and (Z8,+) areisomorphicgroups; simplyreplace 1by 0, replace r by 1, replace r2 by 2, replace r3 by 3, etc. Thesameideashowsthatforeachpositiveinteger n, thegroup (Cn, ·) isisomorphictothegroup (Zn,+). Wethereforeseethatthesamebasicobjectcanariseinthestudyofgeometryandthestudyofalgebra. Geometryandalgebraare, wesee, notasunrelatedasonemightthinkafterseeingthetwofieldsstudiedratherseparatelyintypicalhighschoolcourses.Tounderstand the relationbetweensymmetryandalgebramoreexplicitly, recall the term
“symmetrygroup” thatwestartedusinginSection 5.1, thoughatthetimewesimplyusedthistermtorefertothecollectionofallsymmetriesofagivenobject. Nowthatwehavediscussedthegeneralconceptofgroups(whichisinherentlyanalgebraicconcept), weneedtoaskwhethera“symmetrygroup”aspreviouslydefinedisindeedagroupaswehavenowdefinedit. Theanswer, notsurprisinglygivenourchoiceofterminology, isyes.
Proposition 6.6.1. Supposethat K isaplanarobject. LetG denotethecollectionofallsymme-triesof K. Then (G, ◦) isagroup.
242 6. Groups
Demonstration. TheclosurepropertiesfollowsfromProposition 5.1.2 (1). Theassociativeprop-ertyfollowsfromProposition 4.4.2 (2). Theidentityof (G, ◦) istheidentitysymmetry I. TheinversespropertiesfollowsfromProposition 5.1.2 (2).
Anotherexampleofasymmetrygroupthatalsoarisesnaturallyinalgebraisthefriezegroupf11, whichwasdiscussedinSection 5.5. Thefriezegroup f11 isthesymmetrygroupoffriezepatternsthathavenosymmetryotherthantranslation, forexample · · · FFFFF · · · . AsstatedinSection 5.5, wehave
f11 ={· · · t−3, t−2, t−1, 1, t, t2, t3, · · ·
},
where t denotesthesmallestpossibletranslationsymmetrytotherightofthefriezepattern. Wecanthinkof t as t1, and 1 as t0, andwecancombineanytwosymmetriesin f11 bytheruletatb = ta+b. Itcannowbeobservedthatthegroup (f11, ◦) isisomorphictothegroup (Z,+),where 1 in f11 correspondsto 0 in Z, where t correspondsto 1, where t2 correspondsto 2,where t3 correspondsto 3, etc.Observethatsymmetrygroupsarenotnecessarilyabelian, forexamplethesymmetrygroup
ofanequilateraltriangle. WementionthattheanalogofProposition 6.6.1 forthreedimensional(andhigher)objectsalsoholds(andforthesamereasons), butwewillnotgointodetailshere.Also, weshouldnotethatalthougheverysymmetrygroupofaplanarobjectisagroup, noteverygrouparisesasthesymmetrygroupofaplanarobject(seeExercise 6.6.1).
Exercise 6.6.1. [Used in This Section] Show that the group (W, ∗) given in Exer-cise 6.4.3 (4)isnotthesymmetrygroupofanyplanarobject. Theideaisasfollows. Giventhat W isfinite, if itwere the symmetrygroupofaplanarobject, itwouldhave tobethesymmetrygroupofa rosettepattern (because thoseareprecisely theplanarobjectswithfinitesymmetrygroups). ByLeonardo’sTheorem (Proposition 5.4.5), weknowthatanyrosettepatternhassymmetrygroupeither Cn or Dn forsomepositiveinteger n. Findreasonstoshowwhy (W, ∗) isnotisomorphictoanyofthe Cn or Dn groups.
Nowthatweknowthatsymmetrygroupsareindeedgroups, variousideasaboutgroupscanbeusedtogainabetterunderstandingofsymmetry. Indeed, mathematicallycompleteproofsofProposition 5.5.1 andProposition 5.6.2, inwhichwestatedtheclassificationoffriezepatternsandwallpaperpatternsrespectively, arebasedonsomeideasfromgrouptheorythatarebeyondthescopeofthistext.Oneconceptfromthetheoryofgroupsthatcanbeappliedtosymmetrygroupsisthenotion
ofsubgroups. (Indeed, subgroupsplayanimportantroleontheproofsreferredtointhepreviousparagraph.) Giventhesymmetrygroupofanobject, wecanaskwhichcollectionsofsymmetriesoftheobjectformsubgroups. Forexample, letusexaminethesymmetrygroupofthesquare,theoperationtableforwhichweseeinTable 6.6.1.
6.6SymmetryandGroups 243
· 1 r r2 r3 m mr mr2 mr3
1 1 r r2 r3 m mr mr2 mr3
r r r2 r3 1 mr3 m mr mr2
r2 r2 r3 1 r mr2 mr3 m mr
r3 r3 1 r r2 mr mr2 mr3 m
m m mr mr2 mr3 1 r r2 r3
mr mr mr2 mr3 m r3 1 r r2
mr2 mr2 mr3 m mr r2 r3 1 r
mr3 mr3 m mr mr2 r r2 r3 1
Table6.6.1
AnexaminationofTable 6.6.1 showsthatthisgrouphasninepropersubgroups, asfollows:
{1},
{1, r2},
{1,m},
{1,mr},
{1,mr2},
{1,mr3},
{1, r, r2, r3},
{1, r2,m,mr2},
{1, r2,mr,mr3}.
Wefoundthesesubgroupsbytrialanderror, thoughwemadeuseofLaGrange’sTheorem, whichsaidthatweonlyneededtolookforsubgroupswith 1, 2 or 4 elements.
Exercise 6.6.2. Foreachofthefollowingobjects, findallpropersubgroupsofitssymmetrygroup.
(1) Theequilateraltriangle.
(2) Theregularpentagon.
Are thereanygeneral rules forfinding subgroupsof symmetrygroups? Forexample, doesthecollectionofalltranslationsymmetriesformasubgroup? Whataboutthecollectionofallrotationsymmetries? Whataboutthecollectionofallrotationandallreflectionsymmetries?Thesecondandthirdofthesecollectionsofsymmetriesarenotalwayssubgroups(andexamplewillbegivenshortly), butthefirstalwaysis, asshownbythefollowingproposition.
244 6. Groups
Proposition 6.6.2. Supposethat K isaplanarobject. LetG denotethecollectionofallsymme-triesof K. Thenthefollowingsubcollectionsof G aresubgroupsof (G, ◦):
1. Alltranslationsymmetriesof A;
2. Alltranslationandrotationsymmetriesof A.
Demonstration.
(1). Let T denote thecollectionofall translationsymmetriesof K. WeknowfromPropo-sition 4.6.2 that thecompositionofanytwotranslations isa translation, andweknowfromProposition 5.1.2 (1)thatthecompositionofanytwosymmetriesofanobjectisasymmetry.Puttingthesetwofactstogether, wededucethat (T, ◦) satisfiestheclosureproperty. Theasso-ciativepropertyfor (T, ◦) isautomaticallytrue, becauseitistruefor ◦ ingeneral(seePropo-sition 4.4.2 (2)). Next, wecanthinkoftheidentityisometry I astranslationby 0, andso I isin T . Therefore (T, ◦) satisfiestheidentityproperty. WededucefromProposition 4.6.5 (2)thattheinverseofanytranslationisatranslation, andweknowfromProposition 5.1.2 (2)thattheinverseofanysymmetryofanobjectisasymmetry. Puttingthesetwofactstogether, wededucethat (T, ◦) satisfiestheinversesproperty. Alltold, weseethat (T, ◦) isagroupinitsownright,andhenceitisasubgroupof (G, ◦).
(2). ThispartisverysimilartoPart (1), andthedetailsarelefttothereader.
Theabovepropositiongivestwoverysimpletypesofsubgroupsofsymmetrygroups, thoughthereareother subgroupsaswell. Thecollectionof all rotation symmetries isnot alwaysasubgroup—itdependsupontheobject. Forarosettepattern, forexample, thecollectionofallrotationsymmetriesisasubgroup; thereaderisaskedtosupplythedetailsinExercise 6.6.4.Bycontrast, forafriezepatternthathashalfturnrotationsymmetry, thecollectionofallrotationsymmetriesisnotasubgroup, becausethecompositionoftwohalfturnrotationsaboutdifferentcentersofrotationisatranslation, andthereforetheclosurepropertyisnotsatisfied.
Exercise 6.6.3. Iseachofthefollowingcollectionofsymmetriesalwaysasubgroupofthesymmetrygroupofaplanarobject. Explainyouranswers.
(1) Thecollectionofallreflectionsymmetries.
(2) Thecollectionofalltranslationandallhalfturnrotationsymmetries.
(3) Thecollectionofallrotationandallreflectionsymmetries.
Exercise 6.6.4. [UsedinThisSection] Showthatforarosettepattern, thecollectionofallrotationsymmetriesisasubgroupofthesymmetrygroup.
SuggestionsforFurtherReading
Therearemanyexcellenttextsthatyoumightwishtoreadtofurtheryourstudyofthematerialdiscussedinthisbook(thoughsomeofthesetextsshouldbeapproachedwithanunderstandingof their strengthsandweaknesses). What follows isavery idiosyncraticallyannotated listofvariousbooksyoumightconsiderforfurtherreading, arrangedbythechaptersinthistext.
GeometryBasics
• Euclid, “TheElements”(3vols.), Dover, 1956.
Oneof thegreatestworksofWesternCivilization, andoneof themore tediousaswell. It’sunquestionablytruethatourliveswouldbeverydifferenttodayifthisbookhadnotbeenwritten,butthat’snoreasontoattempttoreadthewholething. ItiswellworthknowingwhatEuclidwastryingtodo, howhedidit, andwhetherornothesucceeded, butitdoesn’ttakeallthreevolumestogetthat. Lookitover, inanycase. Thisusedtoberequiredreadingforeverypersonclaimingtobeeducated. Unfortunately, lessEuclidinschoolshasnotbeenreplacedbyotherkindsofgeometry.
• RobinHartshorne, “Geometry: EuclidandBeyond,” Springer-Verlag, NewYork, 2000.
OneofthemostimpressivemathematicstextbooksI haverecentlyseen. Thistext, meantasacompaniontoEuclid’s“TheElements,” doesnotsummarizeEuclid, butratherexplainswhathisconceptualunderstandingwasandhowitdiffersfromourcontemporaryapproach, andshowshowEuclidcanbebroughtmathematicallyuptodate. Thoughmostofthebookisaimedatanaudienceofjuniororseniorlevelcollegemathematicsmajors(and, inparticular, makesuseofabstractalgebra), muchofthediscussionofEuclideangeometryinthefirsttwochaptersis
246 SuggestionsforFurtherReading
accessibletoabroaderaudience, andwellworththepriceofhavingtoskipoversometechni-calities. HartshornehasdoneanastonishingjoboffiguringEuclidout, makingthisasubstantialbookwithequallysubstantialrewards.
Polygons
• MarthaBoles&RochelleNewman, “TheGolden-Relationship, Book1,” PythagoreanPress, 1987.
A workbook that actually has you get your hands dirtywith geometric constructions usingstraightedgeandcompass. InbetweentheproblemsandprojectsareveryreadablediscussionsoftheGoldenRatio, Fibonaccinumbersandthelike. Somereadersmayfindthephilosophicalexpositionabitflaky, butit’sworthwadingthroughitforthesakeofthehands-onapproach.Besides, howcanyougowrongwithabookthathasarecipefor“FibonacciFudge”?
• TheodoreA.Cook, “TheCurvesofLife,” Dover, 1914.
Morethanyoueverwantedtoknowaboutspirals, fromrams’hornstospiralstaircases. Thefirstandlastfewchaptersareworthreading; thestuffinbetween(whichisafairbit)makesforfunbrowsing.
• MatilaGhyka, “TheGeometryofArtandLife,” Dover, 1977.
Inspiteof thebroadtitle, mostof thebookfocusesontheGoldenRatioandrelatedtopics.Someofthematerialisgood, thoughabittechnical; otherpartsofthebookarespeculative(toputitpolitely), concerningvariousesoterictheoriestheauthorappearstobelieve. Interestingreadingifyoucandealwithit.
• H.E.Huntley, “TheDivineProportion,” Dover, 1970.
A rhapsodyabouttheGoldenRatio(a.k.a.theDivineProportion), andbeautyinmathematicsingeneral. Someofthematerialisphilosophical, somefairlytechnical. It’sworthpickingbitsandpiecesoutofthisbook.
• RobertLawlor, “SacredGeometry,” Crossroad, 1982.
Greatpictures, andallkindsofesoterictheories—withlotsofgeometricalconstructionsthrownin. Youwillhavetodecideforyourselfwhat’sgoingonhere, becauseI amnotsure.
SuggestionsforFurtherReading 247
Polyhedra
• PeterR.Cromwell, “Polyhedra,” CambridgeUniversityPress, Cambridge, 1997.
A lovelytextforthenon-specialist. Thereisawealthofhistoricalinformationonthestudyofpolyhedra, wonderfulillustrations, andanexcellentchoiceoftopics, rangingfromsuchstan-dardsasthePlatonicsolidstolesswellknown(toapopularaudience)gemssuchasDescartes’TheoremonangledefectsandConnelly’sflexiblesphere. Theonerealdrawbackisthelackofexercisesforthereader, devaluingthisbookasatextbook, butwellworthreadingnonetheless.
• MarjorieSenechalandGeorgeFleck, “ShapingSpace,” Birkhäuser, Boston, 1988.
Thisbookistheproceedingsfromaconferenceonvariousaspectsofpolyhedraandrelatedtopics, whichmightsounddulluntilyoutakealookatit—lookingthroughthisbookmakesmewishthatI hadbeenatthatconference! Thoughafewofthearticlesarequitetechnical, manyareaimedatageneralaudience, includinganicehistoryofthestudyofpolyhedra. Thebookisverywellillustrated. I wouldn’tnecessarilyrecommendbuyingthisoneunlessyouareahardcorepolyhedrafan, butitiswellworthabrowse.
HigherDimensions
• EdwinA.Abbott, “Flatland,” Dover, 1952(orothereditions; alsoavailableontheweb).
A minorclassic, withheavyemphasisonbothwords. “Flatland”recountstheadventuresofASQUARE,wholivesina2dimensionalworld. Thefirstpartofthebook, asatireoftheVictoriansocietyinwhichAbbottlived, describestheracistandsexistsocialorderinwhichourherolives.ThesecondpartofthebookdescribesA SQUARE’S encounterwithlowerandhigherdimen-sionalbeings, thusintroducingthereadertosomeimportantideasaboutthefourthdimensionandhigher. Neithergreatwritingnorbrilliantmathematics, “Flatland”straddlesthefencesowellthatitsplaceinthecanonisassured. (Becarefulwiththeintroductionstovariouseditionsof“Flatland”—theonebyBaneshHoffmannintheDoveredition, andtheonebyIsaacAsimovintheHarperCollinsedition, bothentirelymissthepointofthebook.)
• DionysBurger, “Sphereland,” PerennialLibrary(Harper&Row), 1965.
A modern sequel to “Flatland,” introducingmanymathematical ideas recognizedas impor-tantsincetheadventofEinstein’stheoryofrelativity(whichpost-dates“Flatland”by25years).“Sphereland”waswrittenbyamathematician, whichshowsinboththewellchosenmathemat-icaltopics, andthelessthangrippingnarrativestyle. Thoughmathematicallymoresubstantialthan“Flatland,” itlacksthelatter’ssatiricalbite.
• RudyRucker, “TheFourthDimension: aGuidedTouroftheHigherUniverses,” HoughtonMifflin, 1984.
248 SuggestionsforFurtherReading
A funbookcoveringalotofseriousmaterial, andsomeratheresotericstuff toboot. Ruckermakeshigherdimensions, relativityandgeometryenjoyableandsurprisinginawaynooneelsecan. Lotsofgoodproblemsandpuzzles, andgreatquotesandillustrations. Cometoyourownconclusionsaboutthemorespeculativestuff—I’msureRudywouldn’thaveitanyotherway.
• Thomas F. Banchoff, “Beyond theThird Dimension,” ScientificAmerican Library, NY1990.
I wishI hadwrittenthisone, thoughitisjustaswellthatI didn’t, becauseitishardtoimaginethatanyoneelsewouldhavecomeclose todoing itaswellasBanchoff (aserious researchmathematicianwithagenuineinterestinreachingabroadaudience). Thisbookissuchacare-fullythoughtoutandbeautifullyillustratedtreatmentofhigherdimensionsthatitcouldmakeafinecoffeetablebook, thoughdon’tletthatfoolyou—thisbookdiscussesseriousstuff. Theexcellentchoiceof topics range fromunfoldingand slicinghigherdimensionalcubes (withgreatcomputergraphics)toperspectiveandscaling. Afterreadingtheclassics“Flatland”and“Sphereland,” thiswouldbeanexcellentnextplacetowhichtoturnifyouwanttoknowmoreabouthigherdimensions.
• A.K.Dewdney, “ThePlaniverse,” PoseidonPress, 1984.
A verydetailedexplorationofwhata2-dimensionalworldcouldreallybelike, wrappedinasomewhatsillynarrative. Theemphasisisnotonmathematics(asinFlatland), butonphysics,biologyandtechnologyin2dimensions. Whatwoulda2-dimensionalsailboatlooklike? Howwould2-dimensionalintestineskeepfromsplittingacreatureintwo? It’sallquitefun, thoughabitmorethanyoumightwanttoknow.
• MichioKaku, “Hyperspace,” Anchor, 1994.
A rhapsodyaboutthelatesttheoriesofphysics(forexample, stringtheory, paralleluniverses,wormholesandthelike), andtheirrelationtomathematics, especiallythestudyofhigherdi-mensions. Writtenbyaphysicist, ithastheadvantageofaninsider’sviewofthelatestphysicaltheories, andthedisadvantageofaphysicistsviewofmathematics—whichtothismathemati-cianseemsabitdistorted. Thefirstfewchaptersonhigherdimensionscontainsomeinterestinghistoricaldiscussionoftheriseofpopularinterestinthesubject, butthemathematicalideascanbefoundtreatedbetterelsewhere. Ifyouwanttolearnaboutphysics, thenbyallmeansreadthisbook.
• CharlesH.Hinton, “SpeculationsontheFourthDimension,” Dover, 1980.
Probablyamustforhard-core4thdimensionfans, butnotnecessarilyforanyoneelse. Hinton,amathematicianobsessedwiththe4thdimension, wroteavarietyofessaysand“Flatland”stylestoriesthathavebeenexcerptedandcollectedbyRudyRuckerinthisvolume. Thefictionat-temptstobemorescientificthan“Flatland”(havingadifferentsortof2dimensionalworld, andanticipatingthelaterbook“ThePlaniverse”), butthenarrativeistedious, andimbuedwithHin-ton’smysticalideas. Theessaysarefineinpart, but, aswiththefiction, thereisbetterelsewhere.
SuggestionsforFurtherReading 249
SymmetryofPlanarObjectsandOrnamentalPatterns
• HermanWeyl, “Symmetry,” Princeton, 1952.
A classicbyoneofthegreatmathematiciansofthe20thcentury. Thecaliberofthephilosophicalandhistoricaldiscussionsreflectthestatureoftheauthor. Weyldoeslapseintosomeoverlytechnicalpassages, buttheyarewellworthwadingthroughfortherest. Greatillustrationsaswell.
• Farmer, David, “GroupsandSymmetry,” AMS,1996
Theideaisofthisbookisgreat: anexpositionoflovelymathematicaltopicsincludingsymmetry,ornamentalpatternsandgroups, aimedatnon-mathematicians, donenotbylecturingbutbybriefdiscussioncombinedwithlotsof‘tasks’forthereadertoexplore. Unfortunately, thewritingisattimesawkward, thechoiceofterminologyisonoccasionunfortunate, theorganizationispoorandthe‘tasks’varyfromtrivialtoextremelyhardwithnowarning. A fewextrarevisionswouldhavehelped. A well-meaningbookthatdoesnotquiteliveuptoitspromise.
• GeorgeE.Martin, “TransformationGeometry,” SpringerVerlag, 1982.
A verytechnicalbookappropriateforpeoplewithatminimumsomeCalculuslevelmathemat-ics(thoughCalculusperseisnotrequired). Thebookhasaverynicetreatmentoffriezeandwallpapergroups, tilings, andprojectivegeometry. Thisonedemandsseriousstudy.
Tilings
• BrankoGrünbaum&G.C.Shephard, “TilingsandPatterns,” W.H.Freeman, NY,1987.
Theultimatereferenceonthemathematicaltheoryoftilingsandotherplanarornamentalpat-terns, thismassivebookwillsurelybethedefinitivesourceintheforeseeablefuture. Thoughmostofthetextismathematicallysophisticated, thelovelyintroductionisaccessibletoall, andthepicturesandfiguresthroughoutthetextaregreat. I wouldnotrecommendbuyingthisoneunlessyouareplanningaseriousstudyofthesubject, butitiswellworthlookingthrough.
250 SuggestionsforFurtherReading
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[Gre93] MarvinJ. Greenberg, EuclideanandNon-EuclideanGeometries, 3rded., W. H. Free-man, NewYork, 1993.
[Har00] RobinHartshorne, Geometry: EuclidandBeyond, Springer-Verlag, NewYork, 2000.
[Joh66] NormanW. Johnson, Convexpolyhedrawithregularfaces, Canad. J. Math. 18 (1966),169–200.
[Jus61] NortonJuster, ThePhantomTollbooth, Epstein&Carroll, NewYork, 1961.
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[Sen90] MarjorieSenechal, CrystallineSymmetries: AnInformalMathematicalIntroduction,AdamHilger, Bristol, 1990.
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[WW98] EdwardWallaceandStephenWest, RoadstoGeometry, 2nded., PrenticeHall, UpperSaddleRiver, NJ, 1998.
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