It All Depends on How You Slice It: An Introduction to Hyperplane Arrangements

Transcription

1 It All Depends on How You Slice It: An Introduction to Hyperplane Arrangements Paul Renteln California State University San Bernardino and Caltech April, 2008

2 Outline Hyperplane Arrangements Counting Regions The Intersection Poset Finite Field Method

3 Hyperplane Arrangements Hyperplane Arrangements Begin with R n = {(x 1, x 2,..., x n ) : x i R}. An affine hyperplane is the set of points in R n satisfying an equation of the form a 1 x a n x n = b. A hyperplane arrangement is simply a collection of affine hyperplanes.

4 Hyperplane Arrangements A Hyperplane Arrangement An arrangement of four lines in the plane

5 Hyperplane Arrangements Another Hyperplane Arrangement An arrangement of three planes in three space

6 Counting Regions Regions A region of the arrangement A is a connected component of the complement R n H A H Regions can be bounded or unbounded. The total number of regions is r(a), and the number of bounded regions is b(a).

7 Counting Regions Region Counting r(a) = 10 b(a) = 2 (shaded)

8 Counting Regions Deletion and Restriction Is there a better way to count regions? Yes! Definition Let A be an arrangement and H A a hyperplane. A = A\H is called the deleted arrangement. A = {K H : K A } is called the restricted arrangement. (A, A, A ) is called a triple of arrangements.

9 Counting Regions A Triple (A, A, A ) of Arrangements A and H A A

10 Counting Regions Region Counting Recurrence Theorem (Zaslavsky, 1975) r(a) = r(a ) + r(a )

11 Counting Regions Proof by Example (Don t try this at home!) A and H r(a) = 11 = A r(a ) = 7 + A r(a ) = 4

12 Counting Regions Arrangements in General Position Let s apply this result. An arrangement A is in general position if you can move the hyperplanes slightly and not change the number of regions.

13 Counting Regions Two Arrangements in general position not in general position

14 Counting Regions Counting Regions of General Position Line Arrangements Start with an arrangement A of k lines in general position in the plane, and choose a particular line H. By hypothesis, H meets A in k 1 points, which divide H into k regions. So r(a ) = k. Hence r(a) = r(a ) + k, where A contains k 1 lines. By continuing to delete lines in this way, we get r(a) = r( ) (k 1) + k. When no lines are present there is one region, so r( ) = 1. Hence, for a general position line arrangement we have r(a) = 1 + k i = 1 + i=1 k(k + 1) 2 = 1 + k + ( ) k. 2

15 Counting Regions Counting Regions of a General Position Line Arrangement r(a) = 1 + k + ( ) k 2 = ( ) 4 2 = 11

16 The Intersection Poset The Number of Regions in an Arbitrary Arrangement We can use the recurrence (and induction) to show that ( ) ( ) ( ) k k k r(a) = 1 + k n for k hyperplanes in general position in n dimensional space. (Ludwig Schläfli, 1901) But what if the hyperplanes are not in general position? Zaslavsky developed a powerful tool to compute r(a) in general. To describe this, we must take a long detour...

17 The Intersection Poset Partially Ordered Sets A partially ordered set (poset) is a set P and a relation satisfying the following axioms (for all x, y, and z in P ): 1. (reflexivity) x x. 2. (antisymmetry) x y and y x implies x = y. 3. (transitivity) x y and y z implies x z. 4. Posets are represented by their (Hasse) diagrams.

18 The Intersection Poset Hasse Diagrams Some posets

19 The Intersection Poset The Intersection Poset of an Arrangement The intersection poset L(A) of the arrangement A has as its elements all the intersections of all the hyperplanes. It is ordered (for good reason) by reverse inclusion, so A B A B. The minimum element is the ambient space R n.

20 The Intersection Poset An Arrangement and Its Intersection Poset A labeled arrangement A R 2 Its intersection poset L(A)

21 The Intersection Poset The Möbius Function for Posets The (closed) interval [x, y] of a poset is the set of all points between x and y, including endpoints: [x, y] = {z : x z y}. The Möbius function µ(x, y) is defined (recursively) on the interval [x, y] by the two properties: 1. µ(x, x) = x < y z [x,y] µ(x, z) = 0

22 The Intersection Poset Some Möbius Function Values The values of the Möbius functions µ(ˆ0, x)

23 The Intersection Poset The Characteristic Polynomial We define the characteristic polynomial associated to the arrangement A by. χ(a, q) = x L(A) µ(ˆ0, x)q dim(x)

24 The Intersection Poset The Characteristic Polynomial of an Arrangement χ(a, q) = x L(A) µ(ˆ0, x)q dim(x) = q 2 4q + 5.

25 The Intersection Poset Zaslavsky s Theorem Theorem (Zaslavsky, 1975) With the definitions above r(a) = µ(ˆ0, x) = χ(a, 1) b(a) = x L(A) x L(A) µ(ˆ0, x) = χ(a, 1)

26 The Intersection Poset Zaslavsky s Theorem continued Idea of Proof. One can show that, for any triple (A, A, A ) of arrangements, χ(a, q) = χ(a, q) χ(a, q), from which it follows that χ(a, 1) and r(a) satisfy the same recurrence. As they agree on the empty set, the claim follows. A similar argument works for b(a).

27 The Intersection Poset Counting Regions via Zaslavsky s Theorem χ(a, q) = q 2 4q r(a) = χ(a, 1) = 10 b(a) = χ(a, 1) = 2

28 Finite Field Method Finite Field Method Recall that the defining equation of a hyperplane can be written a 1 x 1 + a n x n = b for some real numbers {a 1, a 2,..., a n, b}. In many cases of interest the numbers a 1, a 2,..., a n, b are integers. When this holds there is a particularly nice way to compute the characteristic polynomial. For any positive integer q let A q denote the hyperplane arrangement A with defining equations reduced mod q. Then we have the following result.

29 Finite Field Method The Characteristic Polynomial for Integral Arrangements Theorem (Crapo and Rota (1971), Orlik and Terao (1992), Athanasiadis (1996), Björner and Ekedahl (1996)) For q a sufficiently large prime χ(a, q) = # F n q H A q H where F n q denotes the vector space of dimension n over the finite field with q elements.

30 Finite Field Method Remarks Identifying F n q with {0, 1,..., q 1} n = [0, q 1] n, χ(a, q) is the number of points in [0, q 1] n that do not satisfy modulo q the defining equations of any of the hyperplanes in A. We need large q to avoid lowering the rank of the defining matrix, but as both sides are polynomials in q, the two sides will agree for all q.

31 Finite Field Method Reflection Arrangements Consider the following families of arrangements: A n = {x i x j = 0 1 i j n} D n = A n {x i + x j = 0 1 i j n} B n = D n {x i = 0 1 i n} These are examples of reflection arrangements associated to finite Coxeter groups of types A n 1, D n, and B n, respectively.

32 Finite Field Method Computing the Characteristic Polynomial I What is χ(a n )? According to the finite field method, we want the number of points in [0, q 1] n satisfying x i x j for all 1 i j n. This is the same thing as asking for vectors (x 1, x 2,..., x n ) all of whose entries are distinct mod q. Well, we can pick x 1 in q ways, then x 2 in q 1 ways, and so on. Thus χ(a n, q) = q(q 1)(q 2) (q n + 1). It follows that r(a n ) = n! (and b(a n ) = 0).

33 Finite Field Method Computing the Characteristic Polynomial II What is χ(b n )? Now we want to count the points satisfying x i x j, x i x j, and x i 0. Since we do not allow 0, there are only q 1 (nonzero) choices for the first entry, q 3 nonzero choices for the second entry (because we must avoid the first entry and its negative), etc.. Thus χ(b n, q) = (q 1)(q 3) (q 2n + 1). It follows that r(b n ) = 2 n n! (and b(b n ) = 0).

34 Finite Field Method The Catalan Arrangement C n = {x i x j = 1, 0, +1 1 i j n} C 3 (projected)

35 Finite Field Method The Catalan Arrangement Using the finite field method you can show that χ(c n, q) = q(q n 1)(q n 2) (q 2n + 1) Hence where r(c n ) = n!c n and b(c n ) = n!c n 1 is the n th Catalan number. C n = 1 n + 1 ( ) 2n As of April 1, 2008, Richard Stanley has listed 164 combinatorial interpretations of C n on his website. n

36 Finite Field Method Fundamental Open Question When does the characteristic polynomial factor completely over the integers?