The complexity of the envelope of line and plane arrangements
TThe complexity of the envelopeof line and plane arrangements
David Bremner, Antoine Deza and Feng XieSeptember 15, 2007
Abstract
A facet of an hyperplane arrangement is called external if it belongs to exactly one boundedcell. The set of all external facets forms the envelope of the arrangement. The number ofexternal facets of a simple arrangement defined by n hyperplanes in dimension d is hypothesizedto be at least d ` n − d − ´ . In this note we show that, for simple arrangements of 4 lines or more, theminimum number of external facets is equal to 2( n − n ( n − and ( n − n −
3) + 5.
Let A d,n be a simple arrangement formed by n hyperplanes in dimension d . We recall that anarrangement is called simple if n ≥ d + 1 and any d hyperplanes intersect at a distinct point. Theclosures of connected components of the complement of the hyperplanes forming A d,n are calledthe cells, or d -faces, of the arrangement. For k = 0 , . . . , d −
1, the k -faces of A d,n are the k -facesof its cells. A facet is a ( d − A d,n , and a facet belonging to exactly one bounded cellis called an external facet. Equivalently, an external facet is a bounded facet which belongs to anunbounded cell. For k = 0 , . . . , d −
2, an external k -face is a k -face belonging to an external facet.Let f k ( A d,n ) denote the number of external k -faces of A d,n . The set of all external facets formsthe envelope of the arrangement. It was hypothesized in [1] that any simple arrangement A d,n hasat least d (cid:0) n − d − (cid:1) external facets. In Section 2, we show that a simple arrangement of n lines has atleast 2( n −
1) external facets for n ≥
4, and that this bound is tight. In section 3, we show that asimple arrangement of n planes has at least n ( n − external facets for n ≥
5, and exhibit a simpleplane arrangement with ( n − n −
3) + 5 external facets. For polytopes and arrangements, werefer to the books of Edelsbrunner [3], Gr¨unbaum [6] and Ziegler [7] and the references therein.
Proposition 2.1.
For n ≥ , a simple line arrangement has at least n − external facets.Proof. The external vertices of a line arrangement can be divided into three types, namely v , v and v , corresponding to external vertices respectively incident to 2, 3, and 4 bounded edges. Let usassign to each external vertex v a weight of 1 and redistribute it to the 2 lines intersecting at v thefollowing way: If v is incident to exactly 1 unbounded edge, then give weight 1 to the line containing1 a r X i v : . [ m a t h . M G ] S e p avid Bremner, Antoine Deza and Feng Xie v ; if v is incident to 2 or 0 unbounded edges,then give weight 0 . v . See Figure 1 for an illustration of theweight distribution. A total of f ( A ,n ) weights is distributed and we can also count this quantityline-wise. The end vertices of a line being of type v or v , we have three types of lines, h , , h , and h , , according to the possible types of their end-vertices. As a line of type h , contains 2vertices of type v , its weight is at least 2. Similarly the weight of a line of type h , weight is atleast 1. Remarking that a line of type h , contains at least one vertex of type v yields that theweight of a line of type h , is at least 1 + 0 . . n ≥ h , is at most 2 as otherwise the envelope would be convex which is impossible, see for example [4].Therefore, counting the total distributed weight line-wise, we have f ( A ,n ) ≥ n −
2. Since for aline arrangement the number of external facets f ( A ,n ) is equal to the number of external vertices f ( A ,n ), we have f ( A ,n ) ≥ n − v v v
12 1212 12
Figure 1: The weight distribution for the lines of an arrangement (the shaded area corresponds tothe bounded cells).
For n ≥
4, consider the following simple line arrangement: A o ,n is made of the 2 lines h and h forming, respectively, the x and x axis, and ( n −
2) lines defined by their intersections with h and h . We have h k ∩ h = { k − ε, } and h k ∩ h = { , − ( k − ε } for k = 3 , , . . . , n − h n ∩ h = { , } and h n ∩ h = { , ε } where ε is a constant satisfying 0 < ε < / ( n − A o , . One can easily check that A o , has 2( n −
1) external facets and therefore the lower bound given in Proposition 2.1 is tight.
Proposition 2.2.
For n ≥ , the minimum possible number of external facets of a simple linearrangement is n − . Proposition 3.1.
For n ≥ , a simple plane arrangement has at least n ( n − external facets.Proof. Let h i for i = 1 , , . . . , n be the planes forming the arrangement A ,n . For i = 1 , , . . . , n ,the external vertices of the line arrangement A ,n ∩ h i are external vertices of the plane arrange-ment A ,n . For n ≥
5, the line arrangement A ,n ∩ h i has at least 2( n −
2) external facets byProposition 2.1, i.e., at least 2( n −
2) external vertices. Since an external vertex of A ,n belongs to3 planes, it is counted three times. In other words, the number of external vertices of A ,n satisfies avid Bremner, Antoine Deza and Feng Xie h h Figure 2: An arrangement combinatorially equivalent to A o , f ( A ,n ) ≥ n ( n − for n ≥
5. As the union of all of the bounded cells is a piecewise linear ball,see [2], the Euler characteristic of the boundary gives f ( A ,n ) − f ( A ,n ) + f ( A ,n ) = 2. Sincean external vertex belong to at least 3 external edges, we have 2 f ( A ,n ) ≥ f ( A ,n ). Thus, wehave 2 f ( A ,n ) ≥ f ( A ,n ) + 4. As f ( A ,n ) ≥ n ( n − , it gives f ( A ,n ) ≥ n ( n − For n ≥
5, we consider following simple plane arrangement: A o ,n is made of the 3 planes h , h and h corresponding, respectively, to x = 0, x = 0 and x = 0, and ( n −
3) planes definedby their intersections with the x , x and x axis. We have h k ∩ h ∩ h = { k − ε, , } , h k ∩ h ∩ h = { , k − ε, } and h k ∩ h ∩ h = { , , − ( k − ε } for k = 4 , , . . . , n −
1, and h n ∩ h ∩ h = { , , } , h n ∩ h ∩ h = { , , } and h n ∩ h ∩ h = { , , ε } where ε is a constantsatisfying 0 < ε < / ( n − A o , where, for clarity, only the bounded cells belonging to the positive orthant aredrawn.We first check by induction that the arrangement A ∗ ,n formed by the first n planes of A on +1 , has 2( n − n −
3) external facets. The arrangement A ∗ ,n is combinatorially equivalent to theplane cyclic arrangement which is dual to the cyclic polytope, see [5] for combinatorial propertiesof the (projective) cyclic arrangement in general dimension. See Figure 4 for an illustration of A ∗ , . Let H +3 denote the half-space defined by h and containing the positive orthant, and H − the avid Bremner, Antoine Deza and Feng Xie h h h Figure 3: An arrangement combinatorially equivalent to A o , other half-space defined by h . The union of the bounded cells of A ∗ ,n in H − is combinatoriallyequivalent to the bounded cells of A ∗ ,n − and therefore has 2( n − n −
4) facets on its boundary byinduction hypothesis, including (cid:0) n − (cid:1) bounded facets contained in h . These (cid:0) n − (cid:1) bounded facetsalso belong to a bounded cell of A ∗ ,n in H − and therefore are not external facets of A ∗ ,n . Thus,the number of external facets of A ∗ ,n belonging to a bounded cell in H − is 2( n − n − − (cid:0) n − (cid:1) .The union of the bounded cells of A ∗ ,n in H +3 can be viewed as a simplex cut by n − (cid:0) n − (cid:1) + 2( n −
3) = n ( n −
3) facets on its boundary, including the (cid:0) n − (cid:1) bounded facets contained in h belonging to a bounded cell of A ∗ ,n in H − . Thus, the number ofexternal facets of A ∗ ,n belonging to a bounded cell in H +3 is n ( n − − (cid:0) n − (cid:1) . Therefore, A ∗ ,n has n ( n −
3) + 2( n − n − − (cid:0) n − (cid:1) = 2( n − n −
3) external facets. We now consider how theaddition of h n to A ∗ ,n − impacts the number of external facets. This impact is similar in nature to avid Bremner, Antoine Deza and Feng Xie h n to the first n − A o ,n . The addition of h n creates (cid:0) n (cid:1) new bounded cells:one above h that we call the n -shell , and the other ones being below h . The n -shell turns n − A ∗ ,n − above h into internal facets of A o ,n , and adds 3 external facets. For eachexternal facet of A ∗ ,n − belonging to h which is turned into an internal facet of A o ,n , one externalfacet of A o ,n on h n and not incident to h is added. Below h , the addition of h n creates 3( n − A o ,n with an edge on h . Finally, n − h and bounded by h n are created from unbounded facets of A ∗ ,n − . Thus, the total number ofexternal facets of A o ,n is 2( n − n − − ( n −
4) + 3 + (3( n −
4) + 2) + ( n −
4) = ( n − n −
3) + 5. h h h Figure 4: An arrangement combinatorially equivalent to A ∗ , Remark 3.1.
We do not believe that A o ,n minimizes the number of external facets. Among the 43simple combinatorial types of arrangements formed by 6 planes, the minimum number of externalfacets is while A o , has 23 external facets. See Figure 5 for an illustration of the combinatorialtype of one of the two simple arrangements with planes having 22 external facets. The far awayvertex on the right and 3 bounded edges incident to it are cut off (same for the far away vertex onthe left) so the 10 bounded cells of the arrangement appear not too small. avid Bremner, Antoine Deza and Feng Xie Acknowledgments
Research supported by NSERC Discovery grants, by MITACS grants, by theCanada Research Chair program, and by the Alexander von Humboldt Foundation.
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21 (1996) 111–148.[5] D. Forge and J. L. Ram´ırez Alfons´ın: On counting the k -face cells of cyclic arrangements. European Journal of Combinatorics
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Convex Polytopes . V. Kaibel, V. Klee and G. Ziegler (eds.), Graduate Texts inMathematics 221, Springer-Verlag (2003).[7] G. Ziegler:
Lectures on Polytopes . Graduate Texts in Mathematics 152, Springer-Verlag (1995). avid Bremner, Antoine Deza and Feng Xie David Bremner
Faculty of Computer Science,University of New Brunswick, New Brunswick, Canada . Email : bremner @ unb.ca Antoine Deza, Feng Xie
Department of Computing and Software,McMaster University, Hamilton, Ontario, Canada . Email : deza, xief @@