Simplicial Cells in Arrangements of Hyperplanes

Transcription

1 Simplicial Cells in Arrangements of Hyperplanes Christoph Dätwyler This paper is a report written due to the authors presentation of a paper written by Shannon [1] in The presentation was part of a seminar in combinatorics, organized by Prof. Dr. K. Fukuda at ETH Zürich in Shannon proved in his paper the generalization of two older theorems to higher dimensions. The report covers the rst few pages of Shannons paper, in which a proof of one of these generalized theorems is given. 1 Introduction and Denition In a rst part two theorems going back to Roberts, Eberhard and Levi are stated. The goal of the second part then is to arrive at a proof of a generalization of the theorem of Eberhard and Levi. For a proof of a generalization of the theorem of Roberts, the reader is referred to the paper of Shannon. We start with the denition of some basic terms and concepts we will need. Denition. We destinguish the following two types of arrangements of hyperplanes: A projective d-arrangement H is a nite collection of n = n(h) hyperplanes in real d-dimensional projective space with H =. H partitions P d into k-cells, subsets homeomorphic to an open ball in R k, for 0 k d, forming the associated cell complex C(H). A Euclidean d-arrangement H is required to have the additional property that K for each K H with n(k) = d. Then H partitions R d into k-cells, convex sets, some of which are unbounded. If we simply talk about a d-arrangement, we mean by it a projective one. Otherwise the word Euclidean is explicitely written. But let us just note here that, it actually hardly makes any dierence whether a statement is understood in the projective or Euclidean sense, as there is sort of a duality between them. For a projective d-arrangement H and a hyperplane L P d containing no vertices of C(H) we can regard H as a Euclidean d-arrangement in the ane space A d (L) = P d L. In the case where d = 2, we take the following construction as model for a projective arrangement: Consider S 2 the unit 2-sphere in R 3. Intersecting the 1

2 sphere with planes through the origin yields an arrangement of circles on it. By identify antipodal points, we can represent the arrangement as semicircles on a hemisphere. These semicircles are our hyperplanes. To get an intuitive imagination of A 2 (L) one can think of an arrangement of circles on the 2-sphere. Now one adds one more circle, not passing through any of the intersection points of the circles already on the sphere, and call it L. This circle L devides the sphere into two hemispheres with boundary L and pointsymmetric (with respect to the origin) arrangements of hyperplanes. Due to this symmetry one can pick any of these two. The circle L lies in a plane, say P. Consider the (unique) plane P parallel to P, which intersects the chosen hemisphere, say H, at exactly one point. For every point p on H, one can associate the line l p through the origin and p itself, which intersects the plane P at exactly one point i p. Thus one obtains the corresponding Euclidean arrangement (on P ) by sending every point p of H along l p to its according point i p on P. Note that the points on L are sent to innity. The condition that L does not pass through any vertex is necessary in order to end up with a valid Euclidean arrangement. To see this, let L pass through a vertex of the arrangement on S 2. In the corresponding Euclidean arrangement the vertex is sent to innity and the semicircles passing through the vertex become parallel lines. But by the denition of a Euclidean arrangement, parallel lines are not allowed. Denition. A projective or Euclidean d-arrangement is called simple if the intersection of any d + 1 hyperplanes is empty. A projective or Euclidean d-arrangement H is called near trivial (or a near pencil) if there is a point in P d or R d contained in all hyperplanes of H but one. 2 The Theorems of Roberts, Eberhard and Levi Roberts (1889) stated and tried to prove the following Theorem 1 (Roberts). If His a simple 2-arrangement and H is a line of H, then there are at least n(h) 3 triangles of C(H) which have no edge on H. Grünbaum (1972) conjectured that Theorem 1 is true for all 2-arrangements which are not near trivial. This indeed holds true. It directly follows from Theorem 2, the generalization of Theorem 1, proven in Shannons paper, but not discussed in this report. Theorem 2. Let H be a d-arrangement which is not near trivial and let H be a hyperplane of H. Then there are at least n(h) d 1 simplicial d-cells of C(H), each having no facet in H. Observing that the assumption for an arrangement to be simple is stronger than not to be near trivial and that in the case d = 2 simplicial 2-cells are triangles and facets correspond to edges, we get exactly Theorem 1 as special case of Theorem 2 with d = 2. 2

3 One year after Roberts Eberhard proved the following result for simple planar arrangements and Levi in 1926 proved it for all planar arrangements: Theorem 3. Let H be a 2-arrangement and H a line of H. There are at least three triangles of C(H), each having an edge on H. C(H) contains at least n(h) triangles altogether. Remark. We can construct a simple 2-arrangement of n 4 lines having a cell complex containing exactly n triangles by letting the arrangement consist of n tangent lines to a circle in R 2 (considered as part of P 2 ). 3 A Generalization of the Theorems of Eberhard and Levi Similar to Theorem 1, Theorem 3 can be generalized. Theorem 4. Let H be a d-arrangement and H a hyperplane of H. Then 1. there are at least d + 1 simplicial d-cells of C(H), each having a facet on H 2. C(H) contains at least n(h) simplicial d-cells altogether. The topic of this section is to discuss, how Theorem 4 can be proven. The way we will go about it is as follows: Use Lemma 5 to prove Lemma 7. Use Lemma 8, a restatement of Lemma 7 in a dierent context, and Lemma 9 to prove Theorem 4. Let us start with introducing the notion of an induced arrangement. Denition. Let U = K 1 and V = K 2, K 1, K 2 H, be two non-intersecting ats of H. The induced (projective or Euclidean) arrangement on U with respect to V is dened as the set {U} {U H : H H, U H, V H} and denoted by L(H, U, V ). If V = we write L(H, U) instead of L(H, U, ). Now we can formulate and prove Lemma 1 Lemma 5. Let H be a Euclidean d-arrangement. bounded d-simplex. Then C(H) contains a Proof. Induction on d: d = 1: Since H =, C(H) has k 2 vertices and k 1 1 bounded 1-cells. d 2: Let H be a hyperplane of H and p a vertex of C(H) such that dist(p, H) is positive and minimal. Since L(H, H, {p}) is Euclidean of dim d 1 it has by the induction hypothesis a bounded d 1-simplex p 1 p 2...p d. Since p was chosen to be the point nearest to H, each of the segments pp i will be an edge of C(H). Hence pp 1...p d will be a bounded d-simplex of 3

4 C(H). One has to justify why such a point p of minimal positive distance always exists. This essentially follows from how a Euclidean d-arrangement was dened. Because not all hyperplanes intersect in a single point, there exists a vertex p L 0, L 0 a hyperplane, and a hyperplane L such that p / L. Because the intersection of any d hyperplanes is non-empty, we can write p = d 1 i=0 L i. Then l = d 1 i=1 L i is a line and l L 0 because otherwise p would not be a vertex (note that l L 0 = p). Because p / L it follows that l L / L 0 and because any d hyperplanes have to intersect, the intersection point of l L is non-empty and hence a candidate for the point with positive minimal distance. The following Lemma 6 is just Lemma 5 adapted to the projective setting. Lemma 6. Let H be a d-arrangement and let L P d be a hyperplane containing no vertices of C(H). Then there is a d-simplex D C(H) with L D =. The next step uses Lemma 5 to prove Lemma 7. Denition. If H is a Euclidean d-arrangement, then an unbounded d-cell D C(H) is a d-cone if the closure of D has only one extreme point. D is simplicial if the closure of D is the convex hull of d rays. Lemma 7. If H is a Euclidean d-arrangement, then C(H) contains at least d + 1 simplicial d-cones. Proof. Let C be the convex hull of the vertices C(H) (green in the picture). Lemma 5 implies that C is a d-dimensional convex polytope in R d and therefore has at least d + 1 extreme points (red). Let p be one such extreme point and let L be a hyperplane not of H such that L C = and such that the distance between L and C is minimized only at p. 4

5 Then L(H, L, {p}) is a Euclidean d 1-arrangement and by Lemma 5 contains a bounded d 1-simplex p 1...p d C(L(H, L, {p})) (in the picture this would correspond to p 1 p 2 ). The convex hull of the rays pp 1,..., pp d will be the closure of a d-cone with vertex p (grey) and since C has at least d + 1 extreme points, the claimed statement follows. 8. Again we can formulate Lemma 7 in the projective context. This is Lemma Lemma 8. Let H be an arrangement in P d and H a hyperplane of H. Suppose the convex hull (in A d (H)) of the vertices C(H) not contained in H is d-dimensional. Then there are at least d + 1 d-simplices of C(H), each having a facet contained in H. Suppose H is an arrangement in P d and U and V are ats of H satisfying U V = and the dimension of U + the dimension of V = d 1 U H or V H for each H H then H is the join of the two induced arrangements L(H, U) and L(H, V ) and we write H = L(H, U) L(H, V ). Lemma 9. Let H be an arrangement in P d. If there is a hyperplane H H containing some face of each d-cell of C(H), then there exist ats U and V of H such that H = L(H, U) L(H, V ). Proof. The proof uses the following result due to Zaslavsky: 5

6 Theorem. Let H be an arrangement in P d. Then the number of d-cells which do not contain some face of a hyperplane H of H is independent of the choice of H. This allows us to make the following assumption: (*) Each hyperplane H of H contains some face of every d-cell of C(H). This is actually only possible for near pencils (follows directly from Theorem 2). Now use induction on d and n = n(h) d + 1: d = 1: The above assuption (*) implies n = 2 and the conclusion follows with U and V the two points of the arrangement. d 2, n = d + 1: Take U = K 1, V = K 2, where K 1 and K 1 are any collections with K 1 K 2 = H and U V =. d 2, n d + 2: Consider H = H {H 0 }, H 0 a hyperplane of H. Case 1: H is trivial. Then take U = H 0, V = H. Case 2: Otherwise, since n(h ) < n and H satises assuption (*), there are ats X and Y of H (and hence of H) such that H = L(H, X) L(H, Y ). Case 2.1: X H 0 or Y H 0. Then we are done, as H 0 can be included in one of the arrangements. Case 2.2: Otherwise consider X H 0 L(H, X) and Y H 0 L(H, Y ). Case 2.2.1: E.g. X H 0 is incident with each maximal cell of C(L(H, X)). Then, since dim(x) < d there exist ats U and V of L(H, X) such that L(H, X) = L(H, U ) L(H, V ). If e.g. U X H 0, we let U = U and V = V Y giving H = L(H, U) L(H, V ). Case 2.2.2: There exist maximal cells A and B, not incident with X H 0 in L(H, X) and Y H 0 in L(H, Y ) respectively. Then A B contains two d-cells of C(H) one of which 6

7 cannot be incident with H 0 contrary to assuption (*). This concludes the proof. Now we are ready to prove Theorem 4. Proof of Theorem Induction on d: d = 1: Clear. d 2: Distinguish two cases: Case 1: H is not the join of two induced arrangements. Then the result follows from Lemma 8 and 9 because Lemma 9 asserts that there exists a hyperplane H of H and a d-cell of C(H) such that this d-cell does not have a face on H. Hence Lemma 8 applies since the convex hull of the vertices C(H) not contained in H is d-dimensional. Case 2: H = L(H, U) L(H, V ) where dim(u) = r, dim(v ) = s and r + s = d 1. Let H be a hyperplane of H containing e.g. U. By the induction hypothesis there are at least r + 1 simplicial r-cells in C(L(H, U)) and s + 1 simplicial s-cells in C(L(H, V )) each of the latter having a facet in H V. The join of one such cell from C(L(H, U)) and another from C(L(H, V )) contains two simplicial d-cells of C(H), each with a facet in H. There are at least (r + 1)(s + 1) possible joins of maximal simplicial cells and hence at least 2(r + 1)(s + 1) = 2(rs + r + s + 1) r+s=d 1 = 2(rs + d) 2d d + 1 simplicial d-cells of C(H) each having a facet in H. 2. Since the total number of incidences between hyperplanes and facets of d- simplices is at least n(h)(d+1) (by 1. of Theorem 4), it follows that there 7

8 must be at least n(h) simplicial d-cells altogehter in C(H) as a d-simplex has d + 1 facets and hence n(h)(d+1) d+1 = n(h). References [1] Shannon, R.W., Simplicial Cells in Arrangements of Hyperplanes, Geom. Dedicata, 8(2): ,