Math 3181 Name: Dr. Franz Rothe February 9, 2015 All3181\3181_spr15hs1.tex Homework has to be turned in this handout. For extra space, use the back pages, or blank pages between. The homework can be done in groups up to four students due January 27/28

1

Solution of Homework

10 Problem 1.1 (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 these lines. • 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 parallel property holds. (c) Two intersecting lines with three points. Altogether, there are six lines. Neither parallel property holds. (d) There are five lines, one of which has four points. This model has the elliptic parallel property. 1

10 Problem 1.2 (The three,four,five point incidence geometries). There are seven non-isomorphic incidence geometries on a set of three or four or five points. • Find all among them which are an affine plane. Provide a drawing, or drawings. • Find all among them which are straight fans. Provide drawings. • Find all among them which have the hyperbolic parallel property. Provide a drawing, or drawings.

Figure 2: Properties of small incidence geometries. Answer. Among the incidence geometries with 3, 4 or 5 points, • there is one affine plane. It has four points and six lines. • There are three different straight fans, with 3, 4, 5 points. • There is one model with the hyperbolic parallel property. It is the handshake model with five points and ten lines.

2

10 Problem 1.3. Among all finite incidence planes with n points, which are the extreme ones, in the sense that they have the maximal number of lines possible. How many lines does the hand-shake incidence geometry with an arbitrary number n of points have. Which are the extremes ones, in the sense that they have as few lines as possible. How many lines does the straight fan with n points have. Answer. Among the finite incidence planes with n points, the handshake model has the maximum number of lines possible. Every line has exactly two points, and thus corresponding bijectively to a ”handshake” among the n points. The hand-shake incidence geometry with n points has (n − 1) + (n − 2) + · · · + 1 =

(n − 1)n 2

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 so on. The last line to be drawn is between the (n − 1)-th and the n-th point. Among the finite incidence planes with n points, the straight fan has the minimal possible number of lines, which is n lines. 1 The straight fan has 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 points each 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 with 4 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. All the hand-shake models with n ≥ 5 points have the hyperbolic parallel property. For all straight fans, the elliptic parallel property holds. 10 Problem 1.5 (Optional). Show that every finite incidence geometry has at least as many lines as points. http://en.wikipedia.org/wiki/De_Bruijn %E2%80%93Erd%C5%91s_theorem_%28incidence_geometry%29 http://en.wikipedia.org/wiki/Sylvester%E2%80%93Gallai_theorem

1

It is not as easy to prove that n is the minimal number of lines.

3

10 Problem 1.6. As far as two-dimensional geometry is concerned, Hilbert’s Proposition 1 reduces to one simple statement: any two different lines either intersect in one point, or they are parallel. This statement can be rephrased in many formulations. Encircle the statements below which are such equivalent reformulations; and scratch through the statements which are not. • X Any two distinct lines lying in one plane but not parallel to each other, have a unique point of intersection. • 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. 10 Problem 1.7. From the axioms of incidence (I.1) (I.2) (I.3), one shows easily that there exist two different lines through every point. Complete the following proof for this 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 !! the three lines P A, P B and !! P C . At least !! two of them are different since A, B, C do 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 three lines AB, BC and CA. These are three different lines, and two of them go through the given point P . In !! both cases P.

we have obtained two different lines through the arbitrary point

4

Definition 1 (Isomorphism of incidence planes). Two incidence planes are called isomorphic 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 3: Two isomorphic six-point incidence geometries 10 Problem 1.8. Given two incidence geometries, it is not obvious whether they are isomorphic. By corresponding labelling of the points in both geometries, show an isomorphism between the two six-point incidence geometries in the figure on page 5. Color each pair of corresponding sides with the same color, using altogether seven colors.

5

Figure 4: 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 geometries an arbitrary line with three points, and names the points of these two lines randomly by 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 second point on this line is named D. Similarly, there is a line with two points, and point B among them. Now we name the second point E. Similarly we get a line with exactly two points C and F . The bijection among the lines is now given uniquely, and can be shown by the corresponding colors.

6

10 Problem 1.9. Give a highly symmetric illustration for the Fano plane based on an equilateral triangle. This symmetric illustration is really isomorphic to the projective plane obtained by completion from the affine plane with four points. • Provide both drawings, as was already done in the lecture. • Denote the seven points with the same names in both drawings, consistently in a way to show the isomorphism. • After you have obtained the isomorphism, color the lines with seven different colors. Give the corresponding lines in the other model the same colors.

Figure 5: The symmetric drawing of the Fano-plane is really isomorphic to the projective completion of the affine plane of order 2.

Answer. The figure on page 7 give an illustration based on an equilateral triangle. To check that this symmetric illustration is isomorphic to the illustration I have given on the left side, one needs to names the points in both illustrations in a way that the incidence relations hold for the same names. Thus the isomorphism is given by the correspondence of names. To find such an isomorphism, the key observation is that a triangle can be mapped to any triangle, but afterwards the correspondence of the remaining points is uniquely determined.

7

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 point P that does not intersect l. A.3 There exist three points that do not lie on a line. Remark. Axiom (A.1) repeats Hilbert’s incidence axioms (I.1) and (I.2). Axiom (A.2) is the Euclidean parallel property. In problem ?? from the notes, it is proved that on every line of an affine plane lie at least two points. Together with (A.3), we see that Hilbert’s incidence axiom (I.3) holds for an affine plane, too. Hence an affine plane is just an incidence plane for which the Euclidean parallel property holds. 10 Problem 1.10. Convince yourself that in an affine plane, the relation that two lines are either equal or parallel is an equivalence relation. Answer. For any two lines, let l ∼ m mean l = m or l k m. The relation ∼ has the three defining properties of an equivalence relation. Indeed this relation is reflexive: since l = l is a logical axiom. symmetric: since l = m implies m = l, and l k m implies m k l. transitive: assume that m ∼ l and l ∼ k. If two of the three lines are equal, we substitute equals to get m ∼ k. We now assume that all three lines are different. The two lines m and k are either parallel or they intersect at one point, by Hilbert’s first Proposition. If m and k would intersect at point P , the line l would have two different parallels through P , which is impossible in an affine plane. Therefore m and k are parallel, as to be shown. A relation that is reflexive, symmetric and transitive is called an equivalence relation.

8

Because the construction of the affine plane from the real numbers uses only addition, subtraction, multiplication and division, replacing the real numbers by any finite or infinite field F and doing the same construction once more, leads to an affine plane, too. Thus one gets the following general definition. Definition 3 (The Cartesian plane over a field F). The "points" of the Cartesian plane are ordered pairs (x, y) of elements x, y ∈ F. The "lines" of the Cartesian plane of the field F are equations ax + by + c = 0 with coefficients a, b, c from the field, of which a and b are not both zero. A "point lies on a line" if and only if the coordinate pair (x, y) satisfies the equation of the line. Two lines are equal if and only if they contain the same points. Theorem 1 (The Cartesian plane over an arbitrary field). In a Cartesian plane over any field, there exist a unique line between any two points. There exists a unique parallel to a line through a given point. Hilbert’s axioms (I.1)(I.2)(I.3a)(I.3b) and the Euclidean parallel postulate (IV*) hold. Hence a Cartesian plane over a field is an affine plane. 10 Problem 1.11. Use your knowledge of College algebra, and give in a few words an explanation why Theorem 1 is true. Answer. We begin by checking postulates (I.1) and (I.2). Given are two distinct points (x1 , y1 ) and (x2 , y2 ) through which we need a unique line. In the ususal way, one defines the slope y2 − y1 m= x2 − x1 and puts m = ∞ in the special case that x1 = x2 . The line through point (x1 , y1 ) with slope m has the equation (1.1)

y = y1 + m(x − x1 )

in the usual case of finite slope, and x = x1 = x2 for m = ∞. This equation of a line in this form is uniquely determined by two points (x1 , y1 ) 6= (x2 , y2 ), as the reader should check. Hence two lines through two distinct points contain the same points and are equal. To check axiom (I.3a), we use that each field contains the two different elements 0 6= 1. On any given line with finite slope, we get two points on a line by putting x = 0, x = 1. We put y = 0, y = 1 for a vertical line. To check axiom (I.3b), we may for example use the three points (0, 0), (1, 0) and (1, 1). Hence we see that the Cartesian plane over any field is an incidence plane. To get an affine plane, we need still to check that the Euclidean parallel postulate (IV*) does hold. We see that two distinct lines y = mx + k and y = m0 x + k 0 9

with different slopes m 6= m0 do intersect. But two distinct lines with the same slope have m = m0 and k 6= k 0 ,— they do not intersect. Too, one may include vertical line in this statement. Given any line y = mx + k and point (x1 , y1 ), the parallel through the point is obtained by the point-slope formula (1.1). An easy reasoning shows that Euclidean parallel postulate does hold in the case of a vertical line, too. 10 Problem 1.12. An illustrative affine incidence geometry is the Cartesian plane Z3 × Z3 . In other word, we use analytic geometry and arithmetic modulo 3 for the coordinates. The ”points” are the ordered pairs (a, b) with a, b ∈ Z3 . As in analytic geometry, lines are given by linear equations ax + by + c = 0 with a, b, c ∈ Z3 and a, b are not both zero. One defines the slope of a line in the usual way. Lines are parallel if and only if they have the same slope. To find the parallel to a given line through a given point, one uses the point slope equation of a straight line. This procedure shows that the Euclidean parallel property holds. To better understand this model, provide a drawing. How many lines does the model have? Use clearly different colors for lines with different slopes, but give each set of three parallel lines different shades of nearby the same color. Answer.

10

Figure 6: A nine-point incidence geometry Definition 4 (Projective plane). A projective plane is a class of points, and a class of lines satisfying the axioms: P.1 Every two points lie on exactly one line. P.2 Every two lines intersect at exactly one point. P.3 There exist four points of which no three lie on a line. Definition 5 (Pencil, improper elements of an affine plane). An equivalence class of (equal or) parallel lines is called a pencil. To each pencil corresponds an improper point. 2 The line through all improper points is called the improper line. 3 2

This name is given for visual reasons. We want to catch the vague idea of a ”point very far away” in the direction pointed to by a pencil, and make it precise in terms of rudimental set theory. 3 I reserve the names points at infinity and line at infinity to hyperbolic geometry.

11

Definition 6 (The projective completion of an affine plane). By adding the improper points and the improper line, one can construct from any affine plane a projective plane, called the projective completion. Both the proper as well as the improper points (lines) of an affine plane are the points (lines) of a projective plane. 10 Problem 1.13. Convince yourself that the projective completion of an affine plane is indeed a projective plane. Starting from the axioms (A.1)(A.2)(A.3) for the affine plane, one needs to check the validity of axioms (P.1)(P.2)(P.3). Complete the reason why (P.1)(P.2) are valid. Finally, to confirm (P.3), we show existence of four points no three on which lie on a line. Make an effort to check (P.3) really carefully. Check of validity of axiom (P.1). Given are two distinct points A 6= B of the projective plane. We distinguish three cases, depending on whether these are proper or improper points. • In case that both A and B are proper points, existence and uniqueness of the line through A and B is postulated by axiom (A.1). • Assume now that only one of the two points is proper. We may assume that point A is proper, and B∞ is an improper point. By definition, the point B∞ is a pencil of parallel lines. Take any line b in that pencil. By axiom (A.2) there exists a unique parallel m to line b through point A. In the projective completion, the line m is the unique line through points A and B∞ . • Finally, we assume that both A∞ and B∞ are improper points. In the projective completion, the improper line l∞ is a line through points A∞ and B∞ . Any other line cannot go through both A∞ and B∞ , since the extension of each proper line contains only one improper point. We see that in all possible cases, there exists a unique line through points A and B. Hence axiom (P.1) is valid. Check of validity of axiom (P.2). Given are two distinct lines a 6= b of the projective plane. We distinguish three cases, depending on whether these are proper or improper lines. • Let both a and b be distinct proper lines, and assume they intersect at point P in the affine plane. They cannot have at second proper intersection point since by axiom (A.1) the line through two distinct points is unique. They cannot have any further improper intersection point, otherwise lines a and b would have to be parallel to two distinct pencils, which is impossible. • Let both a and b be distinct proper lines, and assume they do not intersect in the affine plane. Hence they belong to the same pencil P∞ of parallel lines. Extended to the projective plane, the two lines intersect at the improper point P∞ . They cannot have any further improper intersection point, otherwise lines a and b could not be parallel. 12

• Assume now that only line a is proper, and b∞ is the improper line. The line a is contained in exactly one pencil A∞ of parallel lines. This pencil is the unique intersection of lines a and b∞ in the projective completion. We see that in all possible cases, there exists a unique intersection point of lines a and b. Hence axiom (P.2) is valid. Check of validity of axiom (P.3). Finally, to confirm (A.3), we obtain four points no three on which lie on a line. By axiom (A.3), there exist, in the affine plane, three points A, B, C not lying on a line. We draw the parallel m to line AB through point C. Let C∞ be the improper point for that direction. We draw the parallel n to line AC through point B. Let B∞ be the improper point for that direction. The lines m and n are not parallel since the respectively parallel lines AB k m and AC k n intersect at A. Hence the lines m and n intersect, in the affine plane, at a point D = m ∩ n. Of the four points A, D, B∞ , C∞ no three lie on a (projective) line, confirming axiom (P.3). Remark. Here is more simple of axiom (P.3): Finally, to confirm (P.3), we show existence of four points no three on which lie on a line. By axiom (A.3), there exist, in the affine plane, three points A, B, C not lying on a line. We draw the parallel m to line AB through point C, and the parallel n to line AC through point B. The lines m and n are not parallel since the respectively parallel lines AB and AC intersect at A. Hence the lines m and n intersect, in the affine plane, at a point D = m ∩ n. Of the four points A, B, C, D no three lie on a (projective) line, thus confirming axiom (P.3).

13

Proposition 1 (Cutdown of a projective plane to an affine plane, and reconstruction of the projective plane). Given is a projective plane. Any line—and all points on it—are marked as improper. The points and lines from the original affine plane are called proper points and proper lines. (a) The proper points and lines yield an affine plane. (b) Concatination of improper elements leads back to the original projective plane. Check of validity of axiom (A.1). We begin to check the validity of the axioms for an affine plane. Existence and uniqueness of a line through any two distinct points is postulated by axiom (P.1), confirming axiom (A.1). Check of validity of axiom (A.2). Next we check existence and uniqueness of a parallel to a given proper line l through the proper point P not lying on l. Let l∞ be the line marked as improper. Since l∞ 6= l, axiom (P.2) yields, in the projective plane, existence of a unique intersection point Q∞ = l∞ ∩ l. The line m = P Q∞ is retained in the affine plane. Since Q∞ = l ∩ m is an improper point, as seen in the affine plane, the lines l and m are parallel. Hence we have obtained the parallel m to the given line l through point P , as required. Any other parallel m0 k l through point P extended, goes through point Q∞ , too. By the uniqueness of the line through P and Q∞ , we conclude m = m0 . Hence uniqueness of parallels holds, too. Check of validity of axiom (I.3a). Let l∞ be the line marked as improper. The set of points P of the original projective plane is cut down to the set of points A = P \ l∞ of the affine plane. Each line of the projective plane, at least !! three points , as has been shown in the solution of Problem ??. For a line l 6= l∞ , only one of them, the point l ∩ l∞ , is improper. Hence there are at least two proper points on the line l. Check of validity of axiom (I.3b). Finally, to confirm (A.3), we obtain three points not lying on a line. By axiom (P.3), there exist, in the projective plane, four points A, B, C, D of which no three lie on a line. In the cases that none, or only one of them lies on the improper line l∞ , we are ready. Now assume that two of them, say C and D lie on the improper line. Hence A and B are proper points. The intersection P = AC ∩ BD exists. • Moreover, point P is a proper point. Indeed if P 6= C would be two improper points, then point A would be improper. If P 6= D would be two improper points, then point B would be improper. • Points A 6= P are distinct, otherwise A, B, D would lie on a line. Points B 6= P are distinct, otherwise A, B, C would lie on a line. • Finally, point P cannot lie on the line AB. Otherwise all points A, B, P, C, D would lie on this single line.

14

Hence we have found three proper points A, B and P not lying on a line, thus confirming axiom (A.3). Proof of item (b). The set of points P of the original projective plane is cut down to the set of points A = P \l∞ of the affine plane. All the parallel lines in one pencil, extended, have the same intersection point with the improper line l∞ . There is a bijection between these pencils and the improper points on l∞ . Hence the reconstruction of the improper points via pencils brings back the original projective plane.

15