On the Connected Components of the Space of Line Transversals to a Family of Convex Sets Jacob E. Goodman, Richard Pollacky, and Rephael Wengerz March 23, 1997
Abstract
Let L be the space of line transversals to a nite family of pairwise disjoint compact convex sets in R3 . We prove that each connected component of L can itself be represented as the space of transversals to some nite family of pairwise disjoint compact convex sets.
Introduction
Let A be a family of convex sets in Rd . A k-transversal for A is a k- at, an ane subspace of dimension k, that intersects every member of A. (For basic facts about transversal theory, see [1] or [3].) Let A be the set of all 0ktransversals of A, considered as a subspace of the ane Grassmannian Gk;d of all k- ats in Rd . We are interested in studying the connected components of this space A and, in particular, in the question of whether each of these connected components can be represented as the space of k-transversals of some other family B of convex sets. City College, City University of New York, New York, NY 10031, U.S.A. (
[email protected]). Supported in part by NSF grant DMS91-22065 and by NSA grant MDA904-92-H-3069. y Courant Institute of Mathematical Sciences, New York University, New York, NY 10012, U.S.A. (
[email protected]). Supported in part by NSF grant CCR9122103 and by NSA grant MDA904-92-H-3075. z Ohio State University, Columbus, OH 43210, U.S.A. (
[email protected]). Supported in part by NSA grant MDA904-93-H-3026 and by the NSF Regional Geometry Institute (Smith College, July 1993) grant DMS90-13220.
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This question was posed in [2], where it was shown that a convexity structure can be de ned on G0k;d in which the convex sets are precisely the sets of k-transversals to families of convex point sets in Rd . This convexity structure coincides with the usual one on Rd if k = 0, and retains many of its properties for k > 0. In particular, the convex hull of a family F of k- ats can be de ned by means of a duality operator, , between sets of k- ats and families of convex point sets, as follows. If F is a family of k- ats and A is a family of convex point sets in Rd , let F be the family of convex point sets meeting every member of F and let A be the family of k- ats meeting every member of A. Then conv(F ), the convex hull of F , is simply F . A family F of k- ats is convex if conv(F ) = F . It is easily seen that A is convex for any family A of convex point sets. One can equally well obtain the convex hull as follows. A family F of k- ats is said to surround a k- at f if there is some j - at g containing f such that every (j ? 1)- at h containing f and lying in g strictly separates two k- ats of F lying in g ; i.e., h cannot \escape to in nity" within g without containing a member of F after moving some positive distance. (If j = k we interpret this condition as meaning that f 2 F .) A family F of lines in R3 , for example, surrounds a line f if either (i) f 2 F , (ii) f lies strictly between two lines of F with all three contained in a plane g , or (iii) each plane h through f lies strictly between two lines of F parallel to h. The family of all k- ats surrounded by F turns out to be exactly the convex hull of F [2]. Notice that this characterization of the convex hull applies equally well to 0- ats (or points) in Rd , where it agrees with the standard one for point sets. (For a complete discussion of convexity on the ane Grassmannian, see [2].) If F is a convex set of k- ats in Rd , then F = A for some family A of convex point sets in Rd ; F is then said to be presented by A. A convex set F may be presented by many di erent families of convex point sets. A convex set of k- ats need not be connected for k > 0. For instance, a nite family of mutually skew lines in R3 is convex, by the \surrounding" criterion. Problem 7.2 of [2] asks whether the connected components of a convex set of k- ats are necessarily convex. In general the answer turns out to be negative, as we will show by exhibiting an example of a convex family of lines in R3 that has a non-convex connected component. If, however, a convex set of k- ats in Rd is presented by some nite family of suitably separated compact convex point sets, then we conjecture that the connected components of F will themselves be convex. We prove this conjecture for 2
the case of a convex set of lines in R3 presented by a nite family of pairwise disjoint compact convex point sets; this is the main result of the present paper.
A Non-Convex Connected Component
In this section we exhibit a connected set L of lines in R3 whose convex hull has two arcwise connected components. One of these components must contain L, so it cannot be convex. Start with the set of all the tangents to a unit circle centered at O in the (x; y )-plane. Let be an angle irrational with respect to and for each n = 1; 2; . . . translate the pair of lines tangent at (cos n ; sin n ) and (? cos n ; ? sin n ) toward O until they are each at distance 1=n from it. For each direction , 0 < , we now have a pair of parallel lines around O in direction ; ll in all the lines between these two parallels in each direction , getting a strip S . It follows from the irrationality of that arbitrarily close to each direction there are arbitrarily narrow strips. Finally, for each , 0 < , translate this strip S upward to a strip T at height z = . This gives a \spiral staircase" L having the z -axis as an axis of symmetry, closed at the height z = 0 and open at the height z = . Each line in a strip T is connected to the member of T passing through the z -axis. Since the latter form a connected family for 2 [0; ), L is connected. We claim that the convex hull of L is L [ z -axis, so that conv(L) has two components, L and the z -axis. We will show this by using the \surrounding" criterion above. The fact that the z -axis 2 conv(L) is seen by considering any plane through the z -axis; since each T has positive width this plane will be trapped by lines in L strictly parallel to it, so that the z -axis is surrounded by L. It remains only to show that no other line l 2= L lies in conv(L). If l is parallel to the (x; y )-plane, any plane through l not itself parallel to the (x; y )-plane has lines of only a single strip T parallel to itself, with T depending only on the line l, so that by rotating around l we can nd a position in which lies entirely on one side of T , hence escapes in the opposite direction. Suppose, then, that l is not horizontal. Let be any plane through l. meets the (x; y )-plane in a line having some direction . Look at the plane z = . By varying slightly, if necessary, we may assume that the intersection l1 of with that plane misses the z -axis, say by some distance > 0. Since l1 varies continuously with the choice of through l, 3
we can choose 0 suciently close to so that the corresponding line l0 (at height 0 equal to its horizontal direction) lies at distance > =2 from the z -axis, but so that the strip in direction parallel to l0 lies entirely within =2 of the z -axis (i.e., has width < ). Since the only lines that can \block" 0 in either direction are those parallel to 0 , hence in the strip in question, it follows that 0 can escape, i.e., that l is not in the convex hull of L.
Convex Connected Components
In this section, it will be more convenient to work in the ane oriented Grassmannian, G10 ;d, consisting of all oriented lines in Rd . If A is a family of convex sets in Rd , let A~ represent the family of oriented k- ats meeting all the members of A. Clearly, each unoriented k-transversal of A gives rise to a pair of oriented k-transversals of A. Let A be a nite family of pairwise disjoint compact convex point sets in d R and let A~ be its set of oriented line transversals. Each line in A~ induces a linear ordering on the sets in A corresponding to the order in which the line meets the members of A. Each transversal within a given connected component of A~ induces the same linear ordering on A. (This follows easily from the pairwise disjointness; for more details, and a generalization to k-transversals for k > 1, see [3].) Thus each connected component of A~ is associated with some linear ordering of A. Similarly, each connected component of the space A of unoriented line transversals is associated with a pair of linear orderings of A (usually referred to as a geometric permutation of A [4].) Two or more connected components of A~ may be associated with the same linear ordering of A. (For an example, see [3], p. 174.) The exception is in R2 , where components of A~ are associated with distinct linear orderings, and (likewise) components of A with distinct pairs of linear orderings. More generally, if A is suitably separated, distinct components of the space of hyperplane transversals are associated with distinct order types (see [3]). A proof for unoriented hyperplane transversals in Rd , which can easily be adapted to the oriented case, can be found in [5]. We include a proof here for the case of oriented lines in R2 for the convenience of the reader.
Lemma 1 Let A be a nite family of pairwise disjoint compact convex point sets in R2 . Each connected component of A~ determines a distinct linear ordering of A. Proof: Suppose directed lines l; l0 2 A~ meet the members of A in the 4
l
2
k
1 l0
0k
p
3
4
Figure 1: Directed lines inducing the same linear ordering. same order a1; . . . ; an . We will show that l can be moved continuously to l0 through A~ . If A consists of a single set, then A~ is connected and the theorem is trivially true. Assume A contains two or more sets. If l is parallel to l0 , then it must have the same orientation as l0. Then all the parallel lines between l and l0 with the same orientation as l lie in A~ and connect l to l0. Suppose l intersects l0 at a point p. Lines l and l0 divide the plane into four quadrants labeled 1,2,3,4 in clockwise order as in Figure 1. For each set ai 2 A, choose points i 2 l \ ai and 0i 2 l0 \ ai . Each line segment i 0i lies in a single quadrant. (If a segment lies on one of the quadrant boundaries, assign it arbitrarily to one of the adjacent quadrants.) Two or more line segments cannot lie in quadrants 2 and 4, since l and l0 would meet these segments in di erent orders. If a single segment, k 0k , is contained in these two quadrants, no endpoint i or 0i lies between k and p or 0k and p. Rotate l around k until it passes through 0k and then rotate it around 0k to l0. If no segments are contained in quadrants 2 and 4, rotate l around p to l0. 2 Lemma 1 has a simple corollary for line transversals in any dimension.
Corollary 1 Let A be a nite family of pairwise disjoint compact convex
point sets in Rd . If l; l0 2 A~ meet the members of A in the same order and are coplanar, then l and l0 lie in the same connected component of A~ .
Proof: Assume that l; l0 2 A~ generate the same ordering on A and that l and l0 both lie in a plane . Let A0 = f \ a j a 2
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Ag, considered as a
family of subsets of . By Lemma 1, l and l0 are connected through the space of transversals of A0 . Since every transversal of A0 is a transversal of A, it follows that l and l0 lie in the same connected component of A~. 2 We will refer to a vector v as a positive combination of vectors v1; v2; . . . ; vn if v = 1 v1 + v2 + + n vn for real numbers i 0. The set of all positive combinations of nitely many vectors belonging to a set V is the positive hull of V ; it is a convex cone centered at the origin O. As before, let A be a nite family of pairwise disjoint compact convex point sets in Rd . For each directed line l, let vl be the unit vector in the direction of l. For each connected component L A~ , de ne C (L) to be the positive hull of the vectors vl ; l 2 L.
Lemma 2 Let A be a nite family of at least two pairwise disjoint compact
convex point sets in R3 and let L be a connected component of A~ . A directed line l 2 A~ belongs to L if and only if vl belongs to C (L).
Proof: If l is in L, then by de nition C (L) contains vl. Assume that l lies
in A~ but not in L. We wish to show that vl does not lie in C (L). Let ~l be some line in L, and suppose rst that l and ~l meet A in di erent orders. There is some pair of sets a; b 2 A such that l meets a before b and ~l meets b before a. Let h be a plane separating a from b. Every line in L must point toward the half-space bounded by h containing a, while line l points toward the half-space bounded by h containing b. Thus vl cannot be a positive combination of the vectors vl0 , l0 2 L. Now suppose l meets the members of A in the same order that l~ does. Let be a plane perpendicular to l, and denote by (a), (l0), and (v ) the orthogonal projection of any set a 2 A, any line l0 2 R3 , and any vector v 2 R3 onto . If vl is a positive combination of vectors vl0 , l0 2 L, then the origin, O, is in the convex hull of (vl0 ), l0 2 L. T Let = a2A (a). (See Figure 2.) Notice that if a line l0 is parallel to l, then l0 is in A~ if and only if l0 \ lies in . Each directed line of A~ parallel to l and oriented in the same direction is clearly in the same connected component of A~ as l. By Corollary 1, no such line intersects a line of L. Thus (l0), for each line l0 2 L, misses . Without loss of generality, assume that lies to the left of (l~). Since 0 (l ) is a directed line that misses for every l0 2 L and L is connected, must be to the left of (l0) for every l0 2 L. Let t be the translate of (~l) supporting on its left. Line t does not support any (a); a 2 A, or else 6
ta
t
l\
(~l)
tb
p
Figure 2: Projection of sets onto plane . l~ would not intersect a. Thus t \ consists of only a single point p, since otherwise some line strictly between ~l and t would already support . Let t+ be the open half-plane in bounded by t not containing . By Helly's theorem, there are two distinct point sets (a); (b), with a; b 2 A, such that (a) \ (b) \ t+ = ;. Let ta and tb be supporting lines to (a) and (b) at p. Orient ta and tb in directions va and vb , respectively, so that (v~l ) lies in the open cone between va and vb . For each l0 2 L, the vector (vl0 ) is not a positive scalar multiple of va or vb or else l0 would not intersect a or b. Since L is connected, all the vectors (vl0 ), l0 2 L, also lie in the open cone between va and vb . Thus 0 is not a convex combination of (vl0 ), l0 2 L, and so vl is not in the 2 positive hull of vl0 , l0 2 L. The following simple lemma holds for k-transversals in Rd for any k and d.
Lemma 3 Let A be a non-empty family of convex point sets in Rd and let
A and A~ represent the sets of unoriented and oriented k-transversals of A, respectively. If every point set in A is compact, then so are A and A~ . 7
Proof: For each compact convex set a 2 A, the set of (oriented) k-transversals of a is compact. A (A~ ) is the intersection of these compact sets of (oriented) k-transversals and so is also compact. 2 We return to lines in R3 where A~ is the set of oriented line transversals of A. Given a set L of directed lines, let us denote the set of lines with reverse orientation by Lrev . If L is a connected component of A~ , then clearly Lrev is also.
Theorem 1 Let A be a nite family of pairwise disjoint compact convex
point sets in R3 . If L is a connected component of A~ , then L [ Lrev is convex. Moreover, L[Lrev is itself the space of oriented line transversals of some nite family of pairwise disjoint compact convex point sets.
Proof: If A consists of a single convex set, then A~ has just one connected
component and the theorem is trivially true. We will therefore assume that A consists of more than one convex set. Let L be a connected component of A~ . Let h be some plane strictly separating a pair of sets in A translated to pass through O. The vectors vl , l 2 L, all lie in one of the open half-spaces, say h+ , bounded by h. Thus C (L) lies in h+ as well. By Lemma 3, since A is compact, so is A~ . Hence so are the sets L and ~ A n L, and the convex cone C (L) is closed. By Lemma 2, if l 2 A~ n L, then vl does not lie in C (L). Hence, by compactness, there is an angle > 0 such that the angle between vl , for any l 2 A~ nL, and any vector v 2 C (L) is always greater than . We can therefore enlarge C (L) slightly to form a new closed convex cone C 0 with vertex O such that C (L) lies in the interior of C 0, C 0 h , and vl does not lie in C 0 for any l 2 A~ n L. Consider a directed line l 2 A~ . Line l belongs to L [ Lrev if and only if vl or ?vl lies in the interior of C 0 . Hence l 2 L [ Lrev if and only if l \ C 0 is unbounded. By compactness, there is a ball B centered at O such that C \ l is contained in B for every l in A~ n (L [ Lrev). Similarly by compactness, there is a ball B 0 centered at O such that C \ l meets B 0 for every l 2 L. Finally, let B 00 be a ball centered at O containing all the sets a 2 A. Call the largest of these three balls B 000. Translate the plane h to a plane h0 that intersects the cone C 0 but not the ball B 000. Let c = C 0 \ h0 , and notice that since C 0 h+ was closed, c is compact. Then every line in L [ Lrev intersects c, while no line in A~ n (L [ Lrev ) does. Thus A [ fcg consists of pairwise disjoint compact convex sets, and we have L [ Lrev = (A [ fcg)~ . 2 8
Our desired result then follows as a corollary.
Corollary 2 Let A be a nite family of pairwise disjoint compact convex
point sets in R3 . If L is a connected component of A , then L is convex. Moreover, L is itself the space of oriented line transversals of some nite family of pairwise disjoint compact convex point sets.
Proof: Again, we may assume without loss of generality that A consists of
at least two convex sets. Since the sets in A are disjoint, L gives rise to two connected components of A~ coming from the two orientations of lines in L. By Theorem 1, there is a compact convex set c disjoint from each of the sets in A such that the two components together constitute the space of directed line transversals of A [ fcg. Thus L is the space of undirected line transversals of A [ fcg. 2
References [1] Eckhoff, J. Helly, Radon, and Caratheodory type theorems. In Handbook of Convex Geometry, P. M. Gruber and J. M. Wills, Eds. North Holland, Amsterdam, 1993, pp. 389{448. [2] Goodman, J. E., and Pollack, R. Foundations of a theory of convexity on ane Grassmann manifolds. Tech. rep., Courant Institute, 1994. [3] Goodman, J. E., Pollack, R., and Wenger, R. Geometric transversal theory. In New Trends in Discrete and Computational Geometry, J. Pach, Ed. Springer-Verlag, Heidelberg, 1993, pp. 163{198. [4] Katchalski, M., Lewis, T., and Zaks, J. Geometric permutations for convex sets. Discrete Math. 54 (1985), 271{284. [5] Wenger, R. Geometric permutations and connected components. Tech. Rep. TR-90-50, DIMACS, 1990.
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