Technische Universität München Zentrum Mathematik

Question 1. Incidence matrix with gaps Technische Universität München Zentrum Mathematik Prof. Dr. Dr. Jürgen Richter-Gebert, Bernhard Werner Projective Geometry SS 2016 www-m10.ma.tum.de/projektivegeometriess16 Solutions for Worksheet 1 (2016-04-14) Classwork a) Fill in the gaps in the following matrix in such a way that it represents the cincidence structure of a finite projective plane. The symbol denotes an icidence, the symbol the absence of an incidence. A B C D E F G H I K L M N a b c d e f g h i k l m n b) How can one derive a Cayley table (i.e. a table with operands as labels of the rows and columns, and the corresponding result as the corresponding table entry) for the operations (join) and (meet)? Outline such a table and fill in some entries as examples. c) What s the order of this finite plane? d) Create an (unlabeled) draft of this plane. Hint: All projective planes of this order are isomorphic. e) Label the created draft in accordance with the incidence matrix. How unique is this labeling? f) How many different ways are there to label the rows and columns of the matrix using homogeneous coordinates? g) Give one such labeling explicitely. 1

Solution: a) The table below gives the solution to the problem. The number after the entries hint at the order in which the matrix could be filled. For that we use the following rules: Step 1: A column or row with 4 can only contain else, since every line contains 4 points and through every point run 4 lines. Even Steps: No complete 4 4 minor is allowed. So, when the entries (x, X), (x, Y ) and (y, X) are incident, (y, Y ) cannot be an incidence. The reason is, that both the Join X Y = x and the Meet x y = X have to be unique. So one can deduce a at the fourth corner of a rectangulat built by. Odd steps: A column or row with 9 must contain in the remaining entries. Again, because every line contains 4 points and through every point run 4 lines. A B C D E F G H I K L M N a 1 18 2 19 18 18 14 18 17 14 17 b 1 10 11 11 11 2 c 1 10 8 18 19 8 7 14 17 8 2 d 1 19 20 12 12 18 18 17 21 12 e 2 4 2 12 12 4 2 18 4 19 f 1 10 5 21 22 23 6 6 22 6 6 21 10 g 1 17 17 16 16 15 17 16 16 16 16 h 4 2 20 21 4 2 2 19 4 20 2 i 1 10 4 8 12 12 7 8 13 4 10 k 2 3 2 3 2 2 3 2 l 2 4 21 20 4 2 14 4 8 2 m 1 10 20 23 22 4 6 23 4 14 10 n 1 9 5 2 6 8 6 6 8 9 2

b) To get the Join, one has to find the (unique) line (i.e. row) to two points (i.e. columns) being incident to both i.e. having a at both. An analogue argument holds for the Meet. This results in the following tables for both operations. A B C D E F G H I K L M N A h k l l h k l e h k e e B h n g d h g n g h d d n C k n f m f k n m m k f n D l g f l f g l g a a f a E l d m l b b l m m d d b F h h f f b b c c h c f b G k g k g b b i g i k i b H l n n l l c i c i c i n I e g m g m c g c m c e e K h h m a m h i i m a i a L k d k a d c k c c a d a M e d f f d f i i e i d e N e n n a b b b n e a a e a b c d e f g h i k l m n a N L L N D D K K L D K N b N F E N F G F G G E E N c L F L I F I F H L H I H d L E L M M B B M L E E B e N N I M M I A M A A I N f D F F M M D F M C D C C g D G I B I D B G G D I B h K F F B A F B K A A K B i K G H M M M G K G H K H k L G L L A C G A G A C C l D E H E A D D A H A E H m K E I E I C I K K C E C n N N H B N C B B H C H C c) The plane has 13 = 3 2 + 3 + 1 points, thus order 3. 3

d) As all planes of this order are isomporphic (by the hint), one can use the projective plane over F 3 as a starting point for the drawing. First, one can use {0, 1, 2} R to imagine the points as a subset of RP 2 and depict them acordingly. The modulo arithmetic results in additional incidences, which are best depicted as bent lines. One gets to the additional points of the diagonal by steps in modulo arithmetic. So one gets with steps in the direction (1, 1) starting at (0, 1) first to (1, 2), then to (2, 3). That is, modulo 3, the point (2, 0). Whence, those three points have to be collinear. The order of the points on the line is irrelevant and should be choosen in a way that results in a nice picture. e) 4 points, of which no 3 are collinear, can be assigned arbitrarily (cf. part e). If one places the points A to D in the bottom left corner of the plane, all other labels have to as following: G b N n e a k g i m l d E L I H c C D M f F A B K h f) A projective map is determined by 4 points, of which no 3 are collinear. It maps the projective plane on itself. This means, we can choose 4 points arbitrarily and the rest of the plane unfolds. This unfolding is covered in the next part. 13 possibilities, to choose the point A. 12 possibilities, to choose B distinct from A. 9 possibilities, to choose C off the line connecting A and B. 4 possibilities, to choose D off the three lines A B, B C and C A. This results in 13 12 9 4 = 5616 different ways, to label the rows and columns with equivalence classes. For each single vector there are obviously 2 different scalar multiples which can serve as a representative. With 13 points and lines, this yields a factor of 2 2 13 and a total of 376 883 380 224 of possible labels with concrete coordinates. 4

g) The easy way is to choose 4 points again, with no 3 of them on a common line, and to assign them appropriate coordinates. E.g. A = 0 0 B = 1 1 0 1 C = 0 1 D = 1 From that one can deduce all other points and all lines with the following computation being one of many. 0 1 0 0 0 1 0 1 0 1 1 2 0 0 = 1 L 0 1 = 0 L 1 1 = 2 L 0 1 = 0 L 1 1 0 1 1 0 1 1 1 1 1 1 0 0 1 1 2 0 1 0 0 1 0 2 1 2 = 0 P 0 0 = 1 P 0 1 = 0 L 0 1 = 2 L 0 1 0 0 1 0 0 0 1 1 1 1 0 2 1 1 1 1 1 0 2 1 1 0 0 2 = 2 P 2 1 = 1 L 1 1 = 0 P 1 0 = 2 P 1 1 0 0 1 1 1 0 1 1 0 1 2 0 1 0 1 0 1 0 2 0 2 1 0 1 = 0 L 2 0 = 1 L 0 2 = 1 P 1 0 = 2 P 1 0 1 1 0 1 1 1 1 1 1 1 1 0 2 0 2 1 1 0 1 0 1 = 2 P 0 2 = 2 L 2 0 = 1 P 1 1 1 1 1 0 0 1 0 2 1 1 1 2 2 0 1 1 1 2 = 1 L 0 1 = 1 L 1 2 = 2 L 1 1 0 1 1 1 1 1 1 1 1 1 With the starting points choosen above, one gets the following homogenous coordinates: A = (0, 0, 1) T B = (1, 0, 1) T C = (0, 1, 1) T D = (1, 1, 1) T l b = E = (1, 1, 0) T h f = F = (1, 0, 0) T k g = G = (0, 1, 0) T i c = H = (2, 2, 1) T c g = I = (1, 2, 1) T a h = K = (2, 0, 1) T a k = L = (0, 2, 1) T i f = M = (2, 1, 1) T b n = N = (1, 2, 0) T N D = a = (1, 1, 1) T F G = b = (0, 0, 1) T L F = c = (0, 1, 1) T B M = d = (2, 1, 1) T M I = e = (1, 1, 0) T C D = f = (0, 2, 1) T B D = g = (2, 0, 1) T A B = h = (0, 1, 0) T K G = i = (1, 0, 1) T A C = k = (1, 0, 0) T A H = l = (1, 2, 0) T C I = m = (1, 2, 1) T B C = n = (2, 2, 1) T E l A B C D h k n g f G F b N a K L i c H M I d e m Question 2. (n r, m k ) configurations Given an incidence structure with points P, lines L and incidence relation I satisfying the following conditions: (i) For two points p, q P with p q there exists at most one line l L with pil and qil. (ii) For two lines l, m L with l m there exists at most one point p P with pil and pim. 5

An (n r, m k ) configuration is an incidence structure in the sense just stated which has n points and m lines, with k points on every line and r lines through every point. a) How do the axioms stated above differ from those of a projective plane? b) Show that for every (n r, m k ) configuration the equation n r = m k holds. c) Consider m lines in the real plane in general position, together with all the points of intersection between these lines. Which configuration (i.e. which n, r, k) does one obtain from this? d) Consider n points in the real plane in general position, together with all the lines spanned by these points. Which configuration does one obtain from this? e) Complete the following sketch to form an (8 3, 8 3 ) configuration. f) Find a (13 4, 13 4 ) configuration. g) Find additional (n r, m k ) configurations. Solution: a) First, the axioms of a (n r, m k )-configuration at most one line connecting points, not exactly one. The same holds for intersection of lines. So, Join and Meet are not well-defined operations in these configuration. Second, for general (n r, m k )-configurations there is no axiom assuring the richness of the structure. I.e. one can have a (0 3, 0 5 )-configuration. Recall, that all universal statements concerning the empty set are true. So, if there are 0 points, all of the them lies on 3 lines. b) We have n r = I = m k. Both sides describe the number of incidences, i.e. the pairs of points and incident lines. As each of the n points is incident to r lines, there are n r incidences. Analogically, there are k pairs (i.e. points) for each of the m lines. When one uses incidence matrices (columns for points and rows for lines), then this matrix has n columns with r entries each and m rows with k entries each. As the total number of entries has to be unique, the claim follows. c) When the lines are in general position, no 3 of them run through a single point. Thus, every intersection is incident to exactly r = 2 lines. Since parallel lines are just a special case of intersecting lines, all lines intersect. And k = m 1 intersection points lie on every line. From part b) we infer that there are n = m k r = m (m 1) 2 = ( ) m 2 intersections in total. d) In general position no 3 points lie on a common line. So every line connecting two points is only incident to those k = 2. So every points is connected to every other point by r = n 1 different lines. We have m = n r k = n (n 1) 2 = ( n 2) such lines. e) One obvious extension of the drawing is to draw 4 (actually straight) lines connecting the outer points to form a square. On each of those lines are already 3 points as demanded. The given lines have one point less than needed and through the corners of the outer square run one too few lines. So we must augment all given lines by one corner point. As every given line already contains middle points of two sides and those are adjacent to 3 corner points, every line has to connected to the opposite corner. Otherwise, there would be more than one line connecting two points. 6

f) The projective plane of order 3, i.e. F 3 P 2, fulfills this combinatorial datum. g) There are many ways to obtain more projective planes. Every finite projective plane is an (n r, n r )-configuration. The homework exercises show some possibilities to generate (n r, m k )-configurations. E.g. by the projection of higher-dimensional configurations. One can spread n points evenly on a circle. A line connecting k such points could than be represented by a k-gon (in general non-regular). Given a set of such polygons for a fixed k and drawing all other polygons obtained by a rotation by 2π n, symmetry infers that all points are incident to the same number of lines. It is possible to use multiples of 2π n, as long as every point still lies on the same number of copies. It is important that every connection between points is part of at most 1 polygon to fulfill the axioms of (n r, m k )-configurations. Question 3. Counting Homework The number of points and lines in a projective plane over a field with n elements is equal to a) Show that the above equation is true for any n. n 2 + n + 1 = n3 1 n 1 b) Give different arguments explaining that count, one for each side of the above equation. c) Generalize this equation for arbitrary dimension d, namely for the projective space KP d over some field K with K = n. 7

Solution: a) Multiplying both sides of the equation with (n 1) results in the easy computation (n 2 + n + 1)(n 1) = (n 3 + n 2 + n) (n 2 + n + 1) = n 3 + (n 2 n 2 ) + (n n) 1 = n 3 1 b) The right hand side follows directly from the quotient group describing P: P = {( K 3 0 \ 0 0 K \ {0} For a field K with n elements, the vector space K 3 has n 3 elements. Minus the zero vector, we get n 3 1 possible representatives for our points. Modulo the n 1 non-zero field elements we get the desired number of points. The left hand side results from splitting the projective structures in an affine part and a boundary. The projective plane is an affine plane plus the line at infinity. A projective line is a an affine line plus a point at infinity. An affine plane over a field with n elements has n 2 points. An affine line has n points. And a single point is just 1 point. Together we get n 2 + n + 1 points. This is equivalent to split the representatives of the points in the ones of the form (x, y, 1) T, (x, 1, 0) T and (1, 0, 0) T, respectively: )} x y 1 x, y K = n2 x 1 0 x K = n 1 0 0 = 1 An alternative to obtain the left hand side is to count the n + 1 lines through every point p multiplied by the n points on each of these lines distinct from p. This method, however, is harder to generalise to higher dimensions. c) For the right hand side, the representatives have d + 1 elements, determining the exponent of the numerator. On the left hand side, we have d + 1 affine objects starting with the d-dimensional affine space down to a 0-dimensional point. Whence, the general equation is This can be proven the same way as part a). d r=0 n r = nd+1 1 n 1 Question 4. Subset configurations Given a set A with a elements. Consider all subsets with three elements as points, and all subsets with two elements as lines. Incidence shall be defined using containment of sets. P := {p A p = 3} L := {l A l = 2} pil := l p a) What configuration will result for a = 3? Create a sketch of this configuration and label it. b) Also give the (n r, m k ) name which results from a = 4, and create a labeled sketch of that as well. c) What configuration will result from a = 5? What is special about this situation? d) Can you draw the configuration for a = 5 in such a way that the lines and points of that configuration are indeed lines and points of the Euclidean plane? e) What is the connection between the configuration for a = 5 and Desargues theorem? f) complete and label the following drawing in such a way that it reflects the configuration for A = {1, 2, 3, 4, 5, 6}. 8

123 124 456 134 234 156 145 146 Hint: The points of this drawing can be transferred to 5mm graph paper. g) Give the name of the configuration which results from the general case, i.e. for arbitrary a. h) Describe these configurations as projections of objects in higher dimensions. Solution: a) We get the following (1 3, 3 1 )-configuration. 12 123 13 23 b) We get the following (4 3, 6 2 )-configuration. 24 23 14 13 234 134 34 123 124 12 9

c) We get a (10 3, 10 3 )-configuartion. Remark, that here we have the same number of points and lines, resulting in an (n r, n r )-configuration (abbreviated as an n r -configuartion). It is even self-dual. This becomes obvious when we replace every point p P by its complement A \ p L an vice versa. This transition preserves incidences, leading to the duality of the incidence relation. d) The configuartion is easily embedded into the plane. By Desargue s Theorem we know that when all incidents except for one hold, this last one holds, too. A possible picture is: 24 23 123 124 134 234 345 35 34 125 235 245 25 12 135 13 45 145 15 14 e) The given 10 3 -configuartion is just the configuration depicting Desargue s Theorem. So, when all but one incidents hold, the last one holds, too. This is true in all projective planes in which Desargue s Theorem holds. f) The comlete configuration is 10

23 123 14 24 124 456 16 26 126 34 134 234 25 12 125 36 35 136 135 236 235 356 256 345 245 346 246 13 156 145 56 15 45 146 46 g) For an arbitrary a we get an (n r, m k )-configuration with ( ) a n = 3 m = ( ) a 2 r = 3 k = a 2 To be sure, one can check whether the number of incidents is correct: ( ) a I = n r = 3 = 3 a! 3!(a 3)! 3 = a! 2!(a 3)! = a! (a 2) = 2!(a 2)! ( ) a (a 2) = m k 2 h) One can think of these configurations as intersections of planes in the three-dimensional affine space, such that they are in general position w.r.t. each other and to the drawing plane. In that case, three planes intersect in a single point and two planes in a line. Projecting these into the drawing plane gives points and lines again. Subsets describing points and lines label just the planes used to generate these object by intersections. Question 5. Axiomatics An incidence structure (P, L, I) is a projective plane if and only if the following axioms hold: (i) For p, q P, p q there exists exactly one l L with pil and qil (ii) For l, m L, l m there exists exactly one p P with pil and pim (iii) There exists a, b, c, d P such that no three of these are incident to the same line. Show using these axioms exclusively that the following statements hold in any projective plane: a) For every point p P there exists a line l L which is not incident to p. 11

b) For two lines l, m L the sets of incident points have equal cardinality: {p P pil} = {q P qim} Hint: The lecture in this case showed two inequalities, with the proof of the second one only suggested by a claim that it is analogous to the first. Together these two inequalities result in equality only for finite sets. To properly handle infinite sets, one should define a bijection between them to compare their cardinality. c) For finite projective planes, the number of points and the number of lines is always equal. Hint: Several of these statements have been shown in this or a similar form in the lecture. Simply copying the proofs from the lecture is not the idea behind this task. Instead you should recall all the relevant arguments and formalize the proofs in such a way that you can best follow them yourself. Solution: The following proofs don t use statements from the lecture on purpose, since it is explicitly stated to use axioms only. Some results accord to lemmas from the lecture, but may use a different proof to show a different ansatz. a) Beweis: 1. One starts with the four points a to d, which exists by axiom (iii). 2. With axiom (i) we can look at their Joins, e.g. g = a b and h = b c. 3. The Meet of those is by axiom (ii) unique. As b is incident to both, the Meet has to be b. I.e. g h = b. 4. If p b, at most one of the lines g and h can run through p. So at least one ist not incident to p and can be uesed as l. 5. If p = b, we can choose l = c d. Assuming p = b is incident to this l, the points b, c and d would be on the same line contradicting axiom (iii). b) To prove that two sets have the same cardinality, we have to find a bijection between them. In the lecture, it was enough to find injections, since we condidered only finite sets. 1. There is a point p, neither on l nor on m. 1.1. By axiom (iii) the points a to d exist. Since no 3 are on a common line, at most 2 of them can lie on ech of the lines l and m. 1.2. If one of these lines is not incident to 2 of the points above, one of the points is not incident to both lines. Call it p and proceed with step 2. 1.3. If both lines are incident to 2 of the points above, label them w.l.o.g. in a way, that l = a b and m = c d. Describing the lines in this way is unique by (i). 1.4. the Joins g = a c and h = b d exist, again, by (i). 1.5. g, h, l and m are distinct. Each of these lines is defined by 2 of the points a to d. If two lines were equal, it would contain 3 of these points. That violates (iii). 1.6. The point p = g h exists by (ii). 1.7. p lies not on l or m. We can show this by contradiction: 1.7.1. W.l.o.g. assume that p lies on l. By its definition it also lies on g 1.7.2. l und g have the common point a. Hence, by (ii) we have a = p. 1.7.3. By definition, p lies on h, too. And from h = b d we can infer that a, b, d are collinear. That contradicts (iii). So, in any case we found a point p outside the lines l and m. 2. Projecting one line onto the other through the point p is a bijection f between the lines to be precise between the sets of points on the lines. 2.1. For each point r with ril exists a unique line h = r p connecting it to p, by (i). 2.2. As p is not on l and thus r/neqp, (i) is indeed applicable. Moreover, this implies h l and h m. 2.3. The Join h intersects m in the unique point s = h m, by (ii). 2.4. Define f(r) as s. 2.5. By axiom (i), h is also the unique line incident to both s and p. Thus, f is injective. 12

2.6. By axiom (i), there exists a line h = s p connecting p with every point s on l. 2.7. By axiom (ii), each of these connections intersect l in an appropriate point r = h l, with f(r) = s by the above definition. So, f is surjective. 2.8. We deduce that f is bijective. 3. This means that the sets of points incident to both lines have the same cardinality. c) The proof uses the results from the previous parts and gives the concrete number of points and lines. 1. By part b) we have the same number of points on every line. Denote this number by n + 1 so that the order of the plane is n as usual. Caution: That there is something called order and how the number of elements can be derived from it is not part of the axioms. It cannot be used directly. 2. The number of points in the plane is n 2 + n + 1. 2.1. By axiom (iii) we have 4 points a, b, c, which do not lie on a common line. 2.2. By (i), the Join l = b c exists. 2.3. By (iii), a does not lie on l. 2.4. By b) and the definition above, there are exactly n + 1 points on l. Denote them by p i. 2.5. By (i), for each of these there is a line h i = p i a connecting it to a. 2.6. The lines h i are distinct. Otherwise, 2 points p i and p j would lie on a line together with a. But l = p i p j is unique by (i) and we saw above, that a is not on l. 2.7. On each of these n + 1 lines h i lie, by b), n + 1 points. 2.8. Apart from the point a, lying on all of them, any two of the lines h i do not share another point. Because then, two differnet lines would connect this point and a, violating (i). 2.9. By (ii), every line through a intersects l in the unique point p i. So, besides the n + 1 lines h i there are no other lines through a. 2.10. For every point in the plane distinct from a there is a line connecting it with a by (i). 2.11. As only the n + 1 lines h i contain a, every point in the plane has to lie on one of them. 2.12. Thus, in the plane we have the point a and n(n + 1) further points n points distinct from a for each of the n + 1 lines h i. 2.13. So the total number of points in the plane is n 2 + n + 1. 3. The number of lines is n 2 + n + 1. The proof for this is just the dual of the proof above. 3.1. Every line distinct from l has to intersect l in one of its points p i, by (ii). 3.2. Through every point p i on l run n + 1 line, which we will show below. These are l and n additional lines. 3.3. Except for l, no line contains more than one p i by (i). 3.4. Because of that, in the plane we have l and n(n + 1) further lines n lines distinct from l for each of the n + 1 points p i. 3.5. So the total numberof lines is n 2 + n + 1. 4. What is left to show is, that exactly n + 1 lines go through every ppint of the plane. 4.1. For every point p exists a line l not incident to p by part a). 4.2. By part b), n + 1 points lie on l. 4.3. To each of these n + 1 points we have a line connecting it to p, by (i). 4.4. On the other hand, no line through p does not intersect l. 4.5. This gives a bijection between the n + 1 points on l and the lines through p. 5. As both the number of points and lines is n 2 + n + 1, both number are equal. 13