Enumerative geometry of hyperplane arrangements

Transcription

1 Enumerative geometry of hyperplane Max Department of Mathematics US Naval Academy Joint work with Will Traves and Thomas Paul Partially supported by the Simons Foundation and the Office of Naval Research. June 29, 2013

2 Outline Multivariate of Counting

3 Classical enumerative geometry Counting some algebraic varieties that satisfy certain geometric conditions. Typical problems: How many conic sections are tangent to five given lines in the projective plane? How many lines in R 3 pass through 4 general lines? Note: Usually the varieties in these problems do not have much more structure than their dimension and degree.

4 Arrangement setting Choose a matroid or geometric lattice L with rank r + 1. M(L)= the set of hyp arr s in P r with lattice L Main question: What is the degree N L of M(L)? Classical Enumerative Geometry view: Let D = dim M(L) Fix D general position points in P r. How many N L with intersection lattice = L contain these D points?

5 Let L = Easy example Then r = 2 and D = 4 but view this in P 2. Question: How many different pairs of lines in P 2 contain 4 points? Answer: N L = ( 4 2) /2! = 3

6 Let L = Easy example Then r = 2 and D = 4 but view this in P 2. Question: How many different pairs of lines in P 2 contain 4 points? Answer: N L = ( 4 2) /2! = 3

7 Let L = Easy example Then r = 2 and D = 4 but view this in P 2. Question: How many different pairs of lines in P 2 contain 4 points? Answer: N L = ( 4 2) /2! = 3

8 Let L = Easy example Then r = 2 and D = 4 but view this in P 2. Question: How many different pairs of lines in P 2 contain 4 points? Answer: N L = ( 4 2) /2! = 3

9 Generic Arrangements An arrangement G n,k = {H 1,..., H k } in P n is if the intersection of any n + 1 hyperplanes dim M(G n,k ) = nk H i1 H in+1 = 0 Theorem (Carlini) The number of of size k in P n through nk points is N Gn,k = 1 k! ( kn n )( (k 1)n n ) ( ) n = (kn)! n k!(n!) k. This came up when studying the Chow variety of zero dimensional degree k cycles in P n.

10 Star in P 2 : A star arrangement S k = {H 1,..., H k } in P 2 has dim M(S k ) = k + 2 k H i = pt. I=1 Proposition: The number of star S k that contain k + 2 points is ( ) ( ) k + 2 k + 2 N Sk = /2 = 3 2, 2, k 2 4

11 Multivariate polynomial The multivariate polynomial of an arrangement A = {H 1,..., H k } is Z A (q, v 1,..., v k ) = B A q rk(b) H i B v i G 2,k a arrangement in P 2 Fact: N G2,k = Z G2,k (1, 0, 2, 4,..., 2(k 1)) = (2k 1)!!

12 numbers For an arrangement A in P n and integers p, l such that p + l = dim M(A) the characteristic numbers are N A (p, l) = the number of combinatorially equivalent to A that contain p points and are tangent to l lines N A = N A (dim M(A), 0) N A (p, l) are in general very difficult to compute Usually if you can compute all the characteristic numbers for your object then you can compute all enumerative problems with that object. To compute this we will need the class of a curve is the number of lines passing through a given general point and tangent to the curve at a simple point. For example, the class of a smooth curve of degree d is d(d 1).

13 polynomial Adapting a Fulton-MacPherson theorem to line in P 2 we get: Theorem The number of line with intersection lattice isomorphic to L A through p points and tangent to D p smooth curves of degrees n 1,..., n D p and classes m 1,..., m D p in general position is write down D p C = µ p (m i µ + n i ν) i=1 expand the polynomial C plug in the characteristic numbers for each term in the expansion N A (k, D k) = µ k ν D k sum all terms.

14 3 and 4 lines in P 2 Theorem (Paul, Traves, W.) p N G2,3 (p, 6 p) Theorem (Paul, Traves, W.) p N G2,4 (p, 8 p) Do each example separately. Examine the Chow ring of A = A[(P 2 ) k (P 2 ) s ] where s = L(A) 2 =the number of intersection points of lines in A. A = A[(P 2 ) k (P 2 ) s ] = Z[x 1,...,x k,y 1,...,y s] (x 3 1,...,x 3 k,y 3 1,...,y 3 s ) Form a class [M(A)] A that represents the moduli space and the tangency conditions. Expand this class in A. The coefficient of this class is N A (p, l) WARNING: Many of these cases have excess intersection and multiplicities that must be accounted for.

15 N G2,4 (0, 8) The projective dual of G 2,4 is the braid arrangement A 3

16 N G2,4 (0, 8) The projective dual of G 2,4 is the braid arrangement A 3

17 N G2,4 (0, 8) The projective dual of G 2,4 is the braid arrangement A 3 The dual of the 8 line conditions for G 2,4 are 8 point conditions for A 3.

18 N G2,4 (0, 8) The projective dual of G 2,4 is the braid arrangement A 3 The dual of the 8 line conditions for G 2,4 are 8 point conditions for A 3. Hence N G2,4 (0, 8) = = number of braid that contain 8 general points

19 An arrangement A of k n hyperplanes in P n is called a d-cone if there is a linear space X of dimension d common to all the hyperplanes in A and if no point outside of X lies on more than n of the hyperplanes. Any d-cone A is a cone over the arrangement in P n d 1, obtained by replacing each hyperplane in P n d 1 by the linear span of the hyperplane and X.

20 Generic d-cones in P n Let A be a d-cone arrangement of k hyperplanes in P n. Then A is determined by 1 X G(d, n)=the Grassmanian of d-dimensional linear subspaces of P n 2 k points in P(C n+1 /X) = P n d 1 D = dim M(A) = G(d, n) (P n d ) k = (d + 1)(n d) + k(n d 1) In order to get N A we will need to know how many ways there are to choose X and satisfy our point conditions. This is exactly the subject of Schubert calculus.

21 Schubert Calculus H (G(d, n), Z) is generated by σ α where α is a d + 2 tuple of non-increasing non-negative integers α i d n. The products of these classes are given by the Pieri and Giambelli formulas. If α α t = dim G(d, n) = (n d)(d + 1) then the product has well defined degree denoted G(d,n) σ α 1 σ αt which is the number of d planes in the intersection of the corresponding Schubert varieties. Let (1,..., 1, 0,..., 0) =: 1 i where there are i 1 s. For s = (s 0,..., s d+1 ) N d+2 let σ s = d+1 i=0 σ s i 1 i

22 Main theorem Theorem (Paul, Traves, W.) If A is a d-cone in P n consisting of k hyperplanes then the the number of d-cones that pass through D = (d + 1)(n d) + k(n d 1) points in general position is N A = σ s( ) k D Γ where Γ = { )( s 0,s 1,...,s d+1 (n) s d+1,(n 1) s d,...,(n (d+1)) s 0 d+1 (s 0,..., s d+1 ) N d+2 : i=0 k! d+1 is i = dim G(d, n), i=0, s i = k }

23 Mulţumesc!!!

24 0-cones from P 1 to P 2 Theorem (Paul, Traves, W.) If A is a 0-cone of k lines in P 2 then dim M(A) = k + 2 and the characteristic numbers are N A (k + 2, 0) = 3 ( ) k+2 4, NA (k + 1, 1) = ( ) k+1 2, NA (k, 2) = 1. All other characteristic numbers are 0. Note: To be tangent to a line an intersection point of the arrangement must be on the line. For a 0-cone to be tangent to 2 lines then the unique intersection point of the arrangement must be on the intersection point of the 2 lines. Then the k-points uniquely determine the arrangement.