the escher problem

The Escher Problem

Frieze Groups

• Frieze = embroidery from Friez, horizotal ornamented band (architecture).

• We are interested in symmetry groups of such bends. There are 7 frieze groups.

• We start with a rectangular stamp.

Transformations

• Translation

• Halfturn

• Vertical Reflection

• Glide-reflection

Seven Frieze Types

• Groups (notation):• 11 (translations only)• 12 (translations and halfturns)• m1 (translations and vertical reflections)• 1g (translations and glidereflections)• mg (translations, halfturns, vertical reflections and

glide reflections)• 1m (translations and horizontal reflections)• mm (translations, halfturns,vertical reflections,

glide reflections, horizontal reflections)

11

12

m1

1g

mg

1m

mm

The Groups

• Group Elements• Identity I

• Translation T

• Halfturn R

• Glidereflection G

• Vertical mirror V

• Horizontal mirror H.

• Some relations: R2 = V2 = H2 = I, RV = VR = H,..

Subgroups

• (1) T P C1

• (2) G P,R(P) C1

• (3) T,R B C1 £ D1

• (4) T,V A D1

• (5) T,H S D1

• (6) G,V A,R(A) D1

• (7) all(T,G,R,V,H) H D1 £ D1

Exercise

• Explain how the Frieze groups can be described by the four letters (aspects of the pattern):

• b, p, q, d.

b

p

q

d

Discrete Isometries

• Each metric space M determines the group of distance preserving maps, isometries Iso(M).

• A subgroup of Iso(M) is discrete, if any isometry in it either fixes an element of M or moves is far enough.

• Discrete subgroups of I(R2) fall into three classes:• 17 crystallographic groups

• 7 frieze groups

• finite groups (grups of rozettes).

Theorem of Leonardo da Vinci

• The only finite groups of isometries in the plane are the group of rosettes (cyclic groups Cn and dihedral groups Dn).

The Escher problem

• There is a square stamp with asymmetric motif.• By 90 degree rotations we obtain 4 different aspects.• By combining 4 aspects in a square 2 x 2 block, a

translational unit is obtained that is used for plane tilnig. Such a tiling is called a pattern.

• Question: What is the number of different patterns ?• Answer: 23.

Example

Recall Burnside Lemma.

• Let G be a group, acting on space S.

• For g 2 G let fix(g) denote the number of points form S fixed by g.

• Let N denote the number of orbits of G on S.

• Then:

Application

• Determine the group (G) and the sapce (S).

• Pattern can be translated and rotated.

• Basic observatrion: • Instead of pattern consired the block (signature).

• Group operations:• H – horizontal translation

• V – vertical translation

• R – 90 degrees rotation.

(Abstract) group G

• h2 = v2 = r4 = 1.• hv = vh• hr = rv

v vr vr2 vr3

1 r r2 r3

h hr hr2 hr3

hv hvr hvr2 hvr3

Space S

• Space S consists of 4 4 4 4 = 256 signatures.

• Count fix(g) for g 2 G.

• For instance:• fix(1) = 256.

• fix(r) = fix(r3) = 4.

• fix(h) = fix(v) = 16.

• By Burnside Lemma we obtain N = 23.

Homework

• Consider the Escher problem with the motiff on the left.

• H1. Determine the abstract group and its Cayley graph.

• H2. What is the number of different patterns?

• H3. What is the number of different patterns if we reflections are allowed?

• H4. What is the number of different patterns in the original Escher problem if reflections are allowed?

1-dimensional Escher problem

• Rectangular Asymmetric Motiff

• Only Two Aspects.

• 1 x n block (signature)

• Determine the number of patterns:

• Two more variations:• II two motiffs(mirror

images)

• III reflections are allowed.

Space S

• Space S consists of 2 2 .... 2 = 2n signatures.

• Count fix(g) for g 2 G.

• For instance:• fix(1) = 256.

• fix(r) = fix(r3) = 4.

• fix(h) = fix(v) = 16.

• By Burnside Lemma we obtain N = 23.

Solution for the basic case

• where g(n) = 0 for odd n and for even n:

nk

kn ngkn

nf|

/ ))(2)((2

1)(

2/2)2/()( nnng

Program in Mathematica

• f[n_] := (Apply[Plus,Map[EulerPhi[#] 2^(n/#)&,Divisors[n]]] + If[OddQ[n],0,(n/2) 2^(n/2)])/(2n)

• f[n_,m_] := (Apply[Plus,Map[EulerPhi[#] (2 m)^(n/#)&,Divisors[n]]] + If[OddQ[n],0,(n/2) (2 m)^(n/2)])/(2n)

• g[n_] := (Apply[Plus,Map[If[OddQ[#],1,2] EulerPhi[#] 4^(n/#)&,Divisors[n]]] + If[OddQ[n],0,(n) 4^(n/2)])/(4n)

Results for a tape

• n I II III• 1 1 2 1• 2 2 6 4• 3 2 12 6• 4 4 39 23• 5 4 104 52• 6 9 366 194• 7 10 1172 586• 8 22 4179 2131• 9 30 14572 7286• 10 62 52740 26524

Exercise

• Determine the Cayley graph of each of the Frieze groups.

• Determine the crystallographic groups that may arise from the classical Escher problem

17 CRYSTALLOGRAPHIC GROUPS6-števna os?

zrcaljenje? 4-števna os?

p6mm p6

Zrcala v 4 smereh?

zrcaljenje?

p4

3-števna os?

p4mm p4gm

zrcaljenje?2-števna os?

zrcaljenje?

glide?Rombska mreža?glide?Drugo zrcalo?

Rombska mreža?

zrcaljenje?

p1pgpmcmp2c2ggp2mg

3-osi na zrcalih?

p31mp3m1

p3

c2mm p2mm

p1

• p1 = <a,b|ab=ba>

p2

• p2 = <a,b,c| b2=c2=(ab)2=(ac)2=1>

pm

• pm = <a,b,c| b2=c2=1, ab=ba, ac=ca>

pg

• pg = <a,b|ab=ba-1>

cm

• cm = <a,b,c| b2=c2=1, ab=ca>

p2mm

• p2mm = <a,b,c,d| a2=b2= c2=d2= 1, (ab)2=(ad)2=(bc)2=(cd)2=1>

p2mg

• p2mg = <a,b,c| b2 = c2 = 1, (ab)2=(ac)2=1>

p2gg

• p2gg = <a,b| (ab)2=1>