Math 3181 Name:Dr. Franz RotheJanuary 24, 2014All3181\3181_spr14h1.tex

Homework has to be turned in this handout.For extra space, use the back pages, or blank pages between.

due January 22/23

1 Solution of Homework

10 Problem 1.1. As far as two-dimensional geometry is concerned, Hilbert’sProposition 1 reduces to one simple statement: any two different lines either intersectin one point, or are parallel. This statement can be rephrased in many formulations.Encircle the statements below which are such equivalent reformulations; and scratchthrough the statements which are not.

• X Any two different lines which are not parallel, have a unique point of intersec-tion.

• There exists at most one parallel to a given line through a given point.

• On any given line lie two or more points.

• X If two lines have two or more points in common, they are equal.

• Two lines can but need not have some points in common.

• X Two different lines have at most one point in common.

• X Two different points determine a line going through them uniquely.

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10 Problem 1.2 (The five-point incidence geometries). Find all (four)non-isomorphic incidence geometries with five points. For each of them

• Provide a drawing.

• Tell how many lines exist.

• Tell how many lines have three or more points, and describe the properties of theselines.

• Tell which parallel property (elliptic, Euclidean, hyperbolic, or neither) does hold.

Figure 1: There are four five-point incidence geometries.

Answer. There exist four different five-point geometries.

(a) Ten lines have each two points. This model has the hyperbolic parallel property.

(b) Exactly one line with three points. Altogether, there are eight lines. Neither parallelproperty holds.

(c) Two intersecting lines with three points. Altogether, there are six lines. Neitherparallel property holds.

(d) There are five lines, one of which has four points. This model has the ellipticparallel property.

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10 Problem 1.3. How many lines does the hand-shake incidence geometry withan arbitrary number n of points have. How many lines does the straight fan with npoints have.

Answer. The hand-shake incidence geometry with n points has

(n− 1) + (n− 2) + · · ·+ 1 =(n− 1)n

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lines. We see this as follows. We can connect the first point to the other (n− 1) points.Disregarding this point, we connect the second point to (n− 2) different points, and soon. The last line to be drawn is between the (n− 1)-th and the n-th point.

The straight fan with n points has n lines. There is one long line with n− 1 points,and only one point P does not lie on this line. There are n − 1 lines with two pointseach of which connects P to a different point on the long line.

10 Problem 1.4. Which parallel property holds for the hand-shake model with4 points. Which parallel property holds for the hand-shake model with n ≥ 5 points.Which parallel property holds for a straight fan with any numbers n ≥ 3 of points..

Answer. The hand-shake model with 4 points has the Euclidean parallel property. Allthe hand-shake models with n ≥ 5 points have the hyperbolic parallel property. For allstraight fans, the elliptic parallel property holds.

10 Problem 1.5. From the axioms of incidence (I.1) (I.2) (I.3), one shows easilythat there exist two different lines through every point. Complete the following proof forthis statement.

Proof. By axiom (I.3b), there exist at least !! three points that do not lie on a line.We call them A,B and C. Let any point P be given. We now distinguish two cases:

1. In the case that point P is !! different from all three points A,B,C, we draw thethree lines !! PA, PB and PC . At least two of them are different since A,B,Cdo not lie on a line. Thus we are ready.

2. In the case that point P is one of !! the three points A,B,C , we draw the threelines AB,BC and CA. These are three different lines, and two of them go throughthe given point P .

In !! both cases we have obtained two different lines through the arbitrary pointP .

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Definition 1 (Isomorphism of incidence planes). Two incidence planes are calledisomorphic if and only if there exists a bijection between the points of the two planes,and a bijection between the lines of the two planes such that incidence is preserved.

Figure 2: Two isomorphic six-point incidence geometries

10 Problem 1.6. Given two incidence geometries, it is not obvious whetherthey are isomorphic. By corresponding labelling of the points in both geometries, showan isomorphism between the two six-point incidence geometries in the figure on page 4.Color each pair of corresponding sides with the same color, using altogether seven colors.

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Figure 3: Labels of the points and colors of the lines show the isomorphism.

Answer. Such an isomorphism is easy to find as follows. One chooses in both geometriesan arbitrary line with three points, and names the points of these two lines randomlyby A,B,C. The correspondence of the remaining points is now uniquely determined.There exists exactly one line with two points, and point A one of them. The secondpoint on this line is named D. Similarly, there is a line with two points, and point Bamong them. Now we name the second point E. Similarly we get a line with exactlytwo points C and F . The bijection among the lines is now given uniquely, and can beshown by the corresponding colors.

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Definition 2 (Affine plane). An affine plane is a set of points, and a set of lines,satisfying the axioms:

A.1 Every two different points lie on exactly one line.

A.2 If point P does not lie on a line l, there exists exactly one line m through the pointP that does not intersect l.

A.3 There exist three points that do not lie on a line.

Axiom (A.2) is just the Euclidean parallel property. Because of the result of theproblem below, an affine plane is just an incidence plane with the Euclidean parallelproperty.

10 Problem 1.7. The reader should convince himself that in an affine plane

1. on every line lie at least two points;

2. in every point intersect at least two lines.

Complete the proof for these statements given below and provide drawings for the differ-ent cases explained below.

Proof. Let A,B,C be the three points not on a line which exist according to axiom(A.3).

1. Given is any line l. At most !! one of the three lines AB, BC and AC is parallelor equal to the given line l. We may assume that neither line AB nor line ACis parallel or equal to the given line l. Hence they intersect the given line in thepoints P = AB ∩ l and !! Q = AC ∩ l .

• If these two points P 6= Q are different, we are ready.

• We get a more involved argument in the special case that P = Q. In thiscase, we get A = P = Q. If the line l intersects the line BC, we get a secondintersection point R = BC ∩ l on the line l and !! are ready again.

• Still open is only the case that lines BC and l are parallel and the pointA = P lies on the line l. Let m be the parallel to line c = AB through pointC. Since m 6= BC ‖ l, the line m is not parallel to l. Hence these two linesintersect at point !! R = m ∩ l 6= A , which is a second point on the line l.

2. Given is any point P . Assume that P 6= A,B,C. Since these three points do notlie on a line, we get from !! lines PA, PB and PC at least two different onesthrough point P .

In the special case that P = A, we get two different lines AB and AC throughpoint P . We argue similarly in the special cases where !! P = B or P = C.

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Figure 4: In an affine plane, we find a second point on any given line.

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10 Problem 1.8. An illustrative affine incidence geometry is the Cartesian planeZ3 × Z3. In other word, we use analytic geometry and arithmetic modulo 3 for thecoordinates. The ”points” are the ordered pairs (a, b) with a, b ∈ Z3. As in analyticgeometry, lines are given by linear equations ax + by + c = 0 with a, b, c ∈ Z3 and a, bare not both zero. One defines the slope of a line in the usual way. Lines are parallel ifand only if they have the same slope. To find the parallel to a given line through a givenpoint, one uses the point slope equation of a straight line. This procedure shows that theEuclidean parallel property holds.

To better understand this model, provide a drawing. How many lines does the modelhave?

Use clearly different colors for lines with different slopes, but give each set of threeparallel lines different shades of nearby the same color.

Figure 5: A nine-point incidence geometry

Answer.

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Lemma 1 (Proclus’ Lemma).. In any affine plane,

• a third line intersecting one of two parallel lines intersects the other one, too.

• a third line parallel to one of two parallel lines, is parallel to the other one, too.

10 Problem 1.9. Explain why Proclus’ Lemma is an easy consequence of theuniqueness of parallels. Convince yourself that, conversely, Proclus’ Lemma implies theuniqueness of parallels.

Answer. Suppose towards a contradiction that the transversal t intersects one of theparallel lines l and m, but not the other one.

We may assume that P is the intersection point of lines t and m. If lines t and lwould not intersect, then t and m would be two different parallels of line l through pointP . This contradicts the uniqueness of parallels.

Conversely, we now assume Proclus’ Lemma to be true and check the uniqueness ofthe parallel to a given line l through a given point P not on line l. Let m and t betwo parallels to line l through point P—these may be equal or different lines. The linet is a transversal intersecting one of the two parallel lines m ‖ l at point P . Hence itintersects the second line l, too, contrary to the assumption. The only possibility left isthat m = t. Hence the parallel to a given line through a given point is unique.

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10 Problem 1.10. In the figure on page 10 are shown two illustrations of theFano plane: A highly symmetric illustration based on an equilateral triangle, and a morearbitrary chosen one.

Denote the seven points with the same names in both drawings, consistently in a wayto show the isomorphism. After you have obtained the isomorphism, color the sevenlines with seven different colors. Give the corresponding lines in the other illustrationthe same colors.

Figure 6: Fano’s seven-point projective plane

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Figure 7: The symmetric drawing of the Fano-plane is really isomorphic to the projectivecompletion of the affine plane of order 2.

Answer. The figure on page 11 give an illustration based on an equilateral triangle.To check that this symmetric illustration is isomorphic to the illustration I have givenon the left side, one needs to names the points in both illustrations in a way that theincidence relations hold for the same names. Thus the isomorphism is given by thecorrespondence of names.

To find such an isomorphism, the key observation is that a triangle can be mappedto any triangle, but afterwards the correspondence of the remaining points is uniquelydetermined.

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Proposition 1 (Intersections of lines and planes)..

(ii) Any two different planes have either no point in common, or they have one line andno further points in common.

(iii) A plane and a line not lying in this plane have either no or exactly one point incommon.

10 Problem 1.11. Complete the proof of statement (ii) from Proposition 1 above.We use definite names for the objects, explain the steps, and state which axioms areinvolved.

Proof of item (ii). Take any two !! different planes α 6= β. If they do not intersect,we are ready. Assume that the two planes have !! point of intersection A in common.We now use !! (I.7) from Hilbert’s axioms of incidence. Hence, the two planes have a!! further point B 6= A in common.

By axioms (I.1) and (I.2), there exists a !! unique line a through points A and B.Next we use !! (I.6) from Hilbert’s axioms of incidence. Hence every point of line alies !! in the plane α.

Reasoning about the same points A and B and line a, axiom (I.6) yields that everypoint of line a lies in the second plane β, too. In other words, we get the inclusion

a ⊆ α ∩ β

We need still to check that the two planes α and β do not have any point !! outside theline a in common.

Take any point C lying in both planes α and β. We assume towards a contradic-tion that point C does not lie on the line a. Both planes α and β would contain the!! three points A,B and C.

Now we use !! (I.5) from Hilbert’s axioms of incidence. By this axiom, there exists!! at most one plane through the three points A,B and C. Hence these two planes αand β would be equal, contradicting the assumption α 6= β.

Hence the two planes α and β have no point outside the line a in common. Togetherwe get the equality

a = α ∩ βWe have shown that two planes intersecting at one point have a line through this pointas their intersection.

10 Problem 1.12. Write a short clear proof for statement (iii) from Proposition 1above.

Answer. Proof of item (iii). Take a plane α and a line a not lying in α. We assumetowards a contradiction they intersect at two different points A 6= B. By axiom (I.6),every point of a lies in the plane α, contradicting the assumption that the line a is notlying in α.

Hence the plane α and line a have at most one point in common.

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