Bruck nets and partial Sherk planes John Bamberg, Joanna B. Fawcett, Jesse Lansdown 6 1 Abstract. In Bachmann’s Aufbau der Geometrie aus dem Spiegelungsbegriff (1959), it was 0 shown that a finite metric plane is a Desarguesian affine plane of odd order equipped with 2 a perpendicularity relation on lines, and conversely. Sherk (1967) generalised this result to n characterisethefiniteaffineplanesofoddorderbyremovingthe‘threereflectionsaxioms’from a J a metric plane. We show that one can obtain a larger class of natural finite geometries, the 7 so-calledBruck nets ofevendegree,byweakeningSherk’saxiomstoallownon-collinearpoints. 2 1. Introduction ] O Bruck nets generalise the notion of parallelism in affine planes by extending parallelism C to partial linear spaces (a definition is given in Section 2), and they are of great interest due . h to their application to the study of mutually orthogonal latin squares. Sherk characterised t a affine planes of odd order as finite linear spaces with perpendicularity [4], generalising the m work of Bachmann who characterised finite Desarguesian affine planes of odd order as finite [ metric planes [1]. In this paper we further generalise the work of Sherk and Bachmann by 1 characterising Bruck nets of even degree as finite partial linear spaces with perpendicularity. v 1 Thus there is a sense in which the concepts of parallelism and perpendicularity are equivalent 3 in finite partial linear spaces. Indeed, we prove that two lines are parallel precisely when they 2 have a common perpendicular. 7 0 Let G be a geometry consisting of a set of points, a set of lines, a binary relation incidence . 1 between points and lines, denoted by I, and a binary relation perpendicularity between lines, 0 denoted by ⊥. Two lines are said to intersect if they are each incident with a common point. 6 1 We call G a Sherk plane after Sherk [4] if it satisfies the following six axioms. : v A. Two distinct points are incident with a unique line. i B1. For lines ℓ and m, if ℓ ⊥ m then m ⊥ ℓ. X B2. Perpendicular lines intersect in at least one point. r a B3. Given a point P and line ℓ, there exists at least one line m such that mIP and m ⊥ ℓ. B4. Given a point P and line ℓ such that ℓIP, there exists a unique line m such that mIP and m ⊥ ℓ. B5. There exist lines x,y and z such that x ⊥ y, x 6⊥ z, y 6⊥ z and x,y and z do not intersect at a common point. These axioms are equivalent to the incidence and perpendicularity axioms of a metric plane [1]. Thus a Sherk plane generalises the notion of a metric plane by removing the powerful three 2010 Mathematics Subject Classification. primary 51E14;secondary 51E05, 51E15,51F05. Key words and phrases. Bruck net, affine plane, finite geometry, metric plane. The first author acknowledges the support of the Australian Research Council (ARC) Future Fellowship FT120100036. The second author acknowledges the support of the ARC Discovery Grant DP130100106. The third author acknowledges the support of the ARC Discovery Grant DP0984540. 1 2 JOHN BAMBERG, JOANNAB. FAWCETT, JESSELANSDOWN reflections axioms. Any geometry (with a non-empty point set) satisfying Axiom A is a linear space. This concept can be generalised to that of a partial linear space by weakening Axiom A to the following. A*. Two distinct points are incident with at most one line. We can similarly generalise Sherk planes froma linear space equipped with perpendicularity to a partial linear space equipped withperpendicularity. We call a geometry G satisfying Axiom A* and Axioms B1 - B5 a partial Sherk plane. Such a geometry is finite if it possesses only finitely many points. We now state the main result of this paper. Theorem 1.1. The following are equivalent. (1) A finite partial Sherk plane in which the number of lines incident with any point is constant. (2) A finite Bruck net in which the number of lines incident with any point is even and greater than 2. In fact, we prove in Lemmas 2.3 and 3.24 that perpendicularity can only have the form of Definition 2.2. The format of this paper is as follows. In Section 2, we define Bruck nets and give some of their fundamental properties. In Section 3, we prove Theorem 1.1. 2. Bruck nets A (Bruck) net is a partial linear space satisfying the following three axioms. N1: For a line ℓ and point P not incident with ℓ, there exists a unique line m such that mIP and m and ℓ do not intersect. N2: For every line, there exist two points not incident with it. N3: For every point, there exist two lines not incident with it. The line m in Axiom N1 is the parallel to ℓ at the point P. Two lines are parallel to each other if they do not intersect. Parallelism is an equivalence relation, and the equivalence classes of parallel lines are called parallel classes [3, p.141]. We denote the parallel class of ℓ by [ℓ]. A finite net has the following properties [2], [3, p.141]. Lemma 2.1. The following are true in a finite Bruck net. (1) There are r lines incident with every point. (2) There are r parallel classes. (3) There are n points on every line. (4) There are n lines in each parallel class. 2 (5) There are n points in total. (6) There are rn lines in total. (7) Two lines from distinct parallel classes intersect in a unique point. (8) Every point is incident with exactly one line of each parallel class. We call a finite Bruck net an (n,r)-net and say it has order n and degree r. We define perpendicularity as follows. Definition 2.2. Let N be an (n,r)-net where r is even and r > 2. Since N has an even number of parallel classes by Lemma 2.1(2), we may choose an involution τ that acts fixed-point-freely on the set of parallel classes. For lines ℓ and m, define ℓ ⊥ m if and only if [ℓ]τ = [m]. BRUCK NETS AND PARTIAL SHERK PLANES 3 Lemma 2.3. Let N be an (n,r)-net with r even, r > 2 and perpendicularity as in Definition 2.2. Then N is a partial Sherk plane in which every point is incident with r lines. Proof. Axiom A* holds by definition. Axiom B1 holds since τ is aninvolution. By Lemma 2.1(7), any two lines in different parallel classes meet in a point, so Axiom B2 holds. Let ℓ be a line and P a point. By Lemma 2.1(8), there exists m in [ℓ]τ such that mIP. Now [ℓ]τ = [m], so ℓ ⊥ m and Axiom B3 holds. Moreover, any such m is unique by Lemma 2.1(8), so Axiom B4 holds. There are at least four parallel classes by Lemma 2.1(2). Let ℓ be a line and choose m in [ℓ]τ. Now ℓ and m are perpendicular by definition, and they intersect in a unique point P by Lemma 2.1(7). Furthermore, n > 1 by Lemma 2.1(5) and Axiom N2, so by Lemma 2.1(4) and (8), there exists a line g not in [ℓ]∪[m] such that g is not incident with P. Thus B5 holds. By Lemma 2.1(1), there are exactly r lines incident with any point. (cid:3) 3. Proof of Theorem 1.1 If Theorem 1.1(2) holds, then Theorem 1.1(1) holds by Lemma 2.3. Weassumefortheremainder ofthesectionthatTheorem1.1(1)holdsanddenotethepartial Sherk plane by G. In particular, Axiom A* holds. By assumption, every point is incident with the same number of lines, and we denote this number by r. Let v be the total number of points, and b the total number of lines. Since v > 0 and b > 0 by Axioms B5 and B2, it follows from Axiom B3 that r > 0 and from Axioms B3 and B2 that every line is incident with some point. Note that Lemmas 3.2, 3.4, 3.7 and 3.10 are similar to results in [4] but their proofs are included for completeness. Lemma 3.1. There exist three non-collinear points, and some line has at least three points. Proof. Let x, y and z be lines as in Axiom B5. Then x and y intersect at a point P by 1 Axiom B2. By Axiom B3, there exists a perpendicular g from P to z . Clearly g is not x or y as z is not perpendicular to either of these lines. Let Q be the point of intersection of z and g. Now Q 6= P since z is not incident with P by Axiom B5. Let h be a perpendicular from Q to y. Note that h 6= z as z is not perpendicular to y, nor is h equal to x as this would imply that QIx and therefore that g = x by Axiom A*, a contradiction. Furthermore h 6= g or else P Ih, in which case h = x by Axiom B4. Denote the intersection of h and y by R. Now R and P are distinct since g and h are distinct, and R and Q are distinct since g and y are distinct. Moreover, P, Q and R are non-collinear by Axiom A* since g 6= y. Thus there exist three non-collinear points. Let k be a perpendicular from R to g, and let S be the intersection of k and g. Now S 6= P or else k = y by Axiom A*, which would imply that g = x by Axiom B4, a contradiction. Furthermore, S 6= Q or else k = z since g is perpendicular to z at Q and perpendicular to k at S, but then R and Q are on z, so z = h, a contradiction. Thus P, Q and S are pairwise distinct points on the line g. (cid:3) Lemma 3.2. No line is perpendicular to itself. Proof. Let ℓ bealine. ByLemma 3.1, there exists apoint P not onℓ. By AxiomB3, there exists a perpendicular g from P to ℓ. By Axiom B2, there exists a point Q at the intersection of ℓ and g. Now g is the unique line incident with Q and perpendicular to ℓ by Axiom B4. Clearly g 6= ℓ, as g is incident with P and ℓ is not. Thus ℓ cannot be perpendicular to ℓ. (cid:3) Lemma 3.3. r > 2. Proof. Consider the lines x, y and z as in Axiom B5. By Axiom B2, x and y meet at some point P. By Axiom B3, there exists a perpendicular ℓ from P to z. Now ℓ is distinct 1For a point P and line ℓ, a perpendicular from P to ℓ is a line incident with P and perpendicular to ℓ. 4 JOHN BAMBERG, JOANNAB. FAWCETT, JESSELANSDOWN from x and y since z is perpendicular to ℓ but not x or y. Hence x, y and ℓ are three pairwise distinct lines incident with P. There are r lines incident with every point and thus r > 2. (cid:3) Lemma 3.4. r is even. Proof. Let P be a point. There are r lines incident with P. By Axiom B4, every line incident with P has a unique perpendicular at P. It then follows from Lemma 3.2 that r is even. (cid:3) Lemma 3.5. Let ℓ and m be lines that are not perpendicular. Let P and Q be distinct points that are incident with ℓ, and let g and h be perpendiculars from P and Q respectively to m. Then g 6= h, and g and h intersect m in distinct points. Proof. If g = h, then both P and Q are incident with g. Thus g = ℓ by Axiom A*, so ℓ ⊥ m, a contradiction. Hence g and h are distinct. It then follows from Axioms B2 and B4 that g and h meet m in distinct points. (cid:3) Lemma 3.6. Given any line ℓ, the number of lines that are perpendicular to ℓ is equal to the number of points on ℓ. Proof. Denote the number of points on ℓ by n. By Axiom B4, each of the n points of ℓ has a unique perpendicular to ℓ, and these perpendiculars are pairwise distinct by Axiom A* and Lemma 3.2. By Axiom B2, no other lines are perpendicular to ℓ. Thus there are exactly n lines that are perpendicular to ℓ. (cid:3) Lemma 3.7. The number of points incident with any line is constant. Proof. Let ℓ be a line with n points such that every other line has at most n points. Let ′ m be any other line, and let n be the number of points of m. If m is not perpendicular to ℓ, then there are at least n lines that are incident with ℓ and perpendicular to m by Axiom B3 ′ and Lemma 3.5, and there are exactly n lines that are perpendicular to m by Lemma 3.6, so ′ n = n. Otherwise, ℓ is perpendicular to m. Then ℓ and m intersect at a point P. By Lemma 3.3, there exists a line g incident with P that is distinct from ℓ and m. Moreover, g is not perpendicular to ℓ by B4, so it has exactly n points. Similarly, g is not perpendicular to m, so m must also have exactly n points. Thus all lines have exactly n points. (cid:3) We shall denote the number of points on a line by n. We have the following consequence of Lemmas 3.1 and 3.7. Corollary 3.8. n > 3. Lemma 3.9. Axioms N2 and N3 hold. Proof. Let ℓ be a line. Recall that there exists a point P on ℓ. Since r is even, there exists a line g incident with P and not equal to ℓ. By Corollary 3.8, there are at least two additional ′ points on g, and by Axiom A* they are not on ℓ. Thus Axiom N2 holds. Let P be a point. ′ ′ Recall that there exists a line ℓ incident with P . By Corollary 3.8, there exists a point Q on ′ ′ ′ ℓ not equal to P , and since r > 2 by Lemma 3.3, there are two lines other than ℓ incident with Q, which by Axiom A* are not incident with P′. Thus Axiom N3 holds. (cid:3) Lemma 3.10. Let ℓ and m be perpendicular lines. Then ℓ intersects all lines of G except possibly the lines that are perpendicular to m. Proof. Say g is a line that is not perpendicular to m. By Axiom B3 and Lemmas 3.5 and 3.6, there are exactly n perpendiculars from the n points of g to m, and every perpendicular to m arises in this way. Since ℓ is perpendicular to m, the line ℓ intersects g. (cid:3) BRUCK NETS AND PARTIAL SHERK PLANES 5 Lemma 3.11. For a point P and line ℓ such that P is not incident with ℓ, either every line incident with P is perpendicular to ℓ, or there is a unique perpendicular from P to ℓ. Proof. Suppose that some line m is incident with P but not perpendicular to ℓ. Let P1,...,Pn be the points on m, and let hi be a perpendicular from Pi to ℓ for 1 6 i 6 n. Then for all i 6= j, the lines h and h are distinct and meet ℓ in distinct points by Lemma 3.5. Say i j P = P . If h is the unique perpendicular from P to ℓ, then we are done, so suppose that g i i is another perpendicular from P to ℓ. Since ℓ has n points, Axiom B4 implies that g = h for j some j 6= i. Then P and P are both incident with g and m, so g = m by Axiom A*, in which i j case m is perpendicular to ℓ, a contradiction. (cid:3) In the case where a point P is not incident with a line ℓ and there is more than one perpendicular from P to ℓ, we say that P is a pole of ℓ and ℓ is a polar of P. Lemma 3.12. Every line has exactly (n2 −v)/(r−1) poles. Proof. Let ℓ be a line, and let N be the number of poles of ℓ. Let t be the number of pairs (P,m) such that P is a point not on ℓ, and m is a line incident with P and perpendicular to ℓ. Recall that v is the total number of points in G. There are N poles of ℓ, each incident with r lines, and all such lines are perpendicular to ℓ by Lemma 3.11. Moreover, there are v−n−N points not incident with ℓ that are not poles of ℓ, each with exactly one perpendicular to ℓ by Lemma 3.11. Thus t = Nr +(v−n−N). On the other hand, there are n lines perpendicular to ℓ, each with n−1 points not on ℓ, so t = n(n−1). Hence N = (n2 −v)/(r −1). (cid:3) We shall denote the number of poles of a line by N. Lemma 3.13. Every point has exactly (nr −b)/(r −1) polars. Proof. Let P be a point, and let M be the number of polars of P. Let t be the number of pairs (Q,ℓ) such that ℓ is a line not incident with P, and Q is a point incident with ℓ such that the perpendicular from Q to ℓ is incident with P. Recall that b is the total number of lines in G. There are M polars of P, and each such polar ℓ is incident with r points Q such that the perpendicular from Q to ℓ is incident with P by Lemma 3.11. Moreover, there are b−r −M lines not incident with P that are not polars of P, and for each such polar ℓ, there is a unique point Q incident with ℓ such that the perpendicular from Q to ℓ is incident with P by Lemma 3.11. Thus t = Mr+(b−r−M). On the other hand, there are r lines incident with P, each of which is incident with n−1 points besides P, and each such line and point determines a unique perpendicular line not incident with P, so t = r(n−1). Hence M = (nr −b)/(r −1). (cid:3) We shall denote the number of polars of a point by M. Lemma 3.14. Nr = Mn. Proof. Every point is incident with r lines and every line is incident with n points, so vr = bn. Similarly, every line has N poles and every point has M polars, so bN = vM. Thus Nr = Mn. (cid:3) Lemma 3.15. There are exactly b−n(r −1)−1 lines that do not intersect any given line. Proof. There are n points on a line and r lines on a point, so there are n(r − 1) lines intersecting a line ℓ (besides ℓ). Since there are b lines in total, the result follows. (cid:3) Lemma 3.16. Let ℓ and h be perpendicular lines. Then the number of poles of ℓ on h is M. 6 JOHN BAMBERG, JOANNAB. FAWCETT, JESSELANSDOWN Proof. Let Z be the number of lines that do not intersect h. By Lemma 3.10, each of these Z lines is perpendicular to ℓ. Hence there are n−Z−1 perpendiculars to ℓ that intersect h (besides h), each of which intersects h at a pole of ℓ. On the other hand, every pole of ℓ on h is incident with r−1 lines besides h, and all such lines are perpendicular to ℓ by Lemma 3.11. Thus the number of poles of ℓ on h is (n−Z−1)/(r−1). Since Z = b−n(r−1)−1 by Lemma 3.15, the number of poles of ℓ on h is (nr −b)/(r −1), which equals M by Lemma 3.13. (cid:3) Lemma 3.17. Suppose that a line ℓ has a pole P. Then r < n. Proof. Suppose (by way of contradiction) that r > n. By Lemma 3.11, the r lines incident with P are all perpendicular to ℓ, so r = n. By Lemma 3.6, every line that is perpendicular to ℓ is incident with P. By Axiom B3, every point of G lies on a perpendicular to ℓ. Each of the n perpendiculars to ℓ is incident with n−1 points other than P. Thus G has v = n(n−1)+1 points. Let t be the number of triples (Q,R,m) where Q and R are distinct points and m is incident with Q and R. There are v(v −1) pairs of distinct points, so t 6 v(v −1) by Axiom A*, with equality if and only if Axiom A holds. On the other hand, for each point Q, there exist n lines incident with Q, and for each such line m, there exist n−1 points incident with m besides Q. Thus t = vn(n−1) = v(v−1) since v = n(n−1)+1, so G is a Sherk plane, but n is even by Lemma 3.4, which contradicts [4, Theorem 1]. Hence r < n. (cid:3) Lemma 3.18. Suppose that a line ℓ has a pole P. Then n−r +1 6 N. Proof. By Lemma 3.17, we have r < n. Then there are pairwise distinct lines h1,...,hm where m = n−r such that h isnot incident with P and h is perpendicular to ℓ. For1 6 i 6 m, i i there exists a perpendicular a from P to h . Let Q be the intersection of a and h . Then i i i i i P,Q1,...,Qm are pairwise distinct points, none of which are incident with ℓ (by Axiom B4). Moreover, a and h are distinct lines that are incident with Q and perpendicular to ℓ for i i i 1 6 i 6 m, so Q1,...,Qm are poles of ℓ. Hence n−r+1 6 N. (cid:3) Theorem 3.19. No line has a pole. Proof. Suppose for a contradiction that some line ℓ has a pole P. Then r < n by Lemma 3.17. Let W be the number of poles of ℓ that are not equal to or collinear with P. Each of the r lines on P is perpendicular to ℓ by Lemma 3.11, so by Lemma 3.16, W = N −1−(M −1)r. Let t be the number of pairs (Q,h) where Q is a pole of ℓ not equal to or collinear with P, and h is a line perpendicular to ℓ that is incident with Q but not P. Each of the n−r points of ℓ not collinear with P has a unique perpendicular to ℓ, and each such perpendicular is incident with M poles of ℓ by Lemma 3.16, so t = M(n−r). By another count for t, Wr = M(n−r). The two equations above yield (N −1−(M −1)r)r = M(n−r). Now Nr = Mn by Lemma 3.14, so Mr(r −1) = r(r −1). Since r > 1, it follows that M = 1 and N = n/r. By Lemma 3.18, n−r +1 6 N = n/r. Thus (n−r)(r−1) 6 0, so n 6 r, a contradiction. (cid:3) The following extension of Axiom B3 is a consequence of Theorem 3.19. Corollary 3.20. For a line ℓ and point P, there exists a unique line that is perpendicular to ℓ and incident with P. Lemma 3.21. Two lines are parallel if and only if they have a common perpendicular. BRUCK NETS AND PARTIAL SHERK PLANES 7 Proof. If two lines do not have a common perpendicular, then by Lemma 3.10 they must intersect. Conversely, if two lines do have a common perpendicular, then they cannot intersect by Theorem 3.19. (cid:3) Lemma 3.22. Axiom N1 holds. Proof. Let ℓ be a line and P a point not incident with ℓ. By Corollary 3.20, there exists a unique perpendicular g from P to ℓ. By Axiom B4, there exists a unique line m incident with P and perpendicular to g. Clearly m 6= ℓ as m is incident with P but ℓ is not, so m is a line ′ that is incident with P and parallel to ℓ by Lemma 3.21. Say m is another line that is incident ′ with P and parallel to ℓ. Since g and ℓ are perpendicular, so are m and g by Lemma 3.10. But now m and m′ are both perpendicular to ℓ at P, so m = m′ by Axiom B4. (cid:3) The result of Lemmas 3.3, 3.4, 3.9 and 3.22 is the following. Corollary 3.23. G is a Bruck net in which there are r lines incident with any point, where r is even and r > 2. Now that Theorem 1.1(2) holds, we finish by proving that perpendicularity has the form of Definition 2.2. Lemma 3.24. There exists a fixed-point-free involution τ on the set of parallel classes such that ℓ ⊥ m if and only if [ℓ]τ = [m]. Proof. Let ℓ be a line. Denote the set of perpendiculars to ℓ by X . Now X has size n, ℓ ℓ and any two lines in X are parallel by Lemma 3.21, so X is a parallel class by Lemma 2.1(4). ℓ ℓ Define a map τ on the set of parallel classes of G by mapping [ℓ] to X for all lines ℓ. If [ℓ] = [ℓ′], ℓ ′ then ℓ and ℓ have a common perpendicular, say m, by Lemma 3.21, so Xℓ = [m] = Xℓ′. Hence ′ ′ τ is well defined. If ℓ and ℓ are lines such that Xℓ = Xℓ′, then ℓ and ℓ have a common ′ perpendicular, so [ℓ] = [ℓ] by Lemma 3.21. Thus τ is a bijection. Moreover, τ has no fixed points since no line is perpendicular to itself by Lemma 3.2, and τ is clearly an involution. Lastly, two lines ℓ and m are perpendicular if and only if m ∈ X . This occurs precisely when ℓ [m] = [ℓ]τ, as desired. (cid:3) References [1] F. Bachmann. Aufbau der Geometrie aus dem Spiegelungsbegriff. Die Grundlehren der mathematischen Wissenschaften, Bd. XCVI. Springer-Verlag, Berlin-G¨ottingen-Heidelberg, 1959. [2] R. H. Bruck. Finite nets. II. Uniqueness and imbedding. Pacific J. Math., 13:421–457,1963. [3] P. Dembowski. Finite geometries. Classics in Mathematics. Springer-Verlag, Berlin, 1997. Reprint of the 1968 original. [4] F. A. Sherk. Finite incidence structures with orthogonality. Canad. J. Math., 19:1078–1083,1967. Centre for the Mathematics of Symmetry and Computation, School of Mathematics and Statistics, The University of Western Australia, Crawley, W.A. 6009, Australia. E-mail address: [email protected] E-mail address: [email protected] E-mail address: [email protected]