Counting Hyperplane Arrangements Will Traves Department of Mathematics United States Naval Academy CAGE Seminar Drexel/Penn, Philadelphia 8 OCT 2013 Traves (USNA) Counting Arrangements 8 OCT 2013 1 / 29 Warm-up Question Joint work with Max Wakefield and Tom Paul Question How many generic arrangements of k lines pass through d points in general position in P2 ? Traves (USNA) Counting Arrangements 8 OCT 2013 2 / 29 Warm-up Question Joint work with Max Wakefield and Tom Paul Question How many generic arrangements of k lines pass through d points in general position in P2 ? If d is too small there are infinitely many arrangements. If d is too large there are no arrangements. If d = 2k then there is a finite number of arrangements through the d points. Traves (USNA) Counting Arrangements 8 OCT 2013 2 / 29 Warm-up Question Joint work with Max Wakefield and Tom Paul Question How many generic arrangements of k lines pass through d points in general position in P2 ? If d is too small there are infinitely many arrangements. If d is too large there are no arrangements. If d = 2k then there is a finite number of arrangements through the d points. No line can pass through 3 points (by general position assumption); PHP shows that each line goes through 2 points. 2k 2k − 2 2 Count = ··· /k! = (2k − 1)!! 2 2 2 Traves (USNA) Counting Arrangements 8 OCT 2013 2 / 29 Tutte Polynomial of A 2k 2k − 2 2 Count = ··· /k! = (2k − 1)!! 2 2 2 Multivariate Tutte polynomial of a generic arrangement of k lines is T (q = 1, x1 , x2 , . . . , xk ) = (1 + x1 )(1 + x2 ) · · · (1 + xk ). Count = T (1, 0, 2, 4, . . . , 2k − 2). The Tutte polynomial is an invariant of the intersection lattice of A. This suggests a generalization our warm-up question. Traves (USNA) Counting Arrangements 8 OCT 2013 3 / 29 The Intersection Lattice of A Problem Given a hyperplane arrangement A in Pn , count the number of hyperplane arrangements with intersection lattice isomorphic to L(A) that pass through d points in general position. The intersection lattice (or poset) of A encodes how the hyperplanes intersect: Traves (USNA) Counting Arrangements 8 OCT 2013 4 / 29 The Moduli Space MA Definition If A is an arrangement of k hyperplanes in Pn then set MA := {arrangements with intersection lattice ∼ = L(A)} ⊂ (Pn∗ )k /Sk with dimension d = dA . Hyperplane H : a0 x0 + · · · + an xn = 0 corr. to (a0 : . . . : an ) ∈ Pn∗ A passes through P ⇐⇒ (H1 · · · Hk )(P) = 0 – a codim 1 condition dimension d = # point conditions giving a finite counting problem Traves (USNA) Counting Arrangements 8 OCT 2013 5 / 29 Degree NA of MA The degree NA of MA is the number of arrangements in MA through d points in general position. The map (Pn∗ )k ,→ P(C[x0 , . . . , xn ]k ) (H1 , . . . , Hk ) 7−→ H1 · · · Hk factors through (Pn∗ )k /Sk . The condition that A passes through a point, (H1 · · · Hk )(P) = 0, is a linear condition on P(C[x0 , . . . , xn ]k ) and pulls back to a multilinear condition on (Pn∗ )k . Traves (USNA) Counting Arrangements 8 OCT 2013 6 / 29 Finding d the Easy Way... Dim MA = # degrees of freedom you have in drawing A and represent each degree by a point constraint: Virtual dimension: vdim(MA ) = 2k − X (np − 2) with np =| {H ∈ A : p ∈ H} | Traves (USNA) Counting Arrangements 8 OCT 2013 7 / 29 Finding d the Easy Way...Fails! Traves (USNA) Counting Arrangements 8 OCT 2013 8 / 29 An (Intractible?) Open Problem Problem Find a combinatorial algorithm to compute dim MA from L(A). Mn̈ev’s Universality Theorem ⇒ MA arbitrarily complicated Need to “see” syzygies among defining equations for MA from L(A). Traves (USNA) Counting Arrangements 8 OCT 2013 9 / 29 Cohomological Approach Goal: count arrangements without precise location of constraint points. Move points −→ arrangements vary −→ number remains constant Codim-d constraint −→ degree-d class ∈ (graded) cohomology ring. Deformed constraints have the same class. Traves (USNA) Counting Arrangements 8 OCT 2013 10 / 29 Cohomology ring structure The structure of the cohomology ring depends on the structure of the underlying moduli space X . The cells of a cell-decomposition of X determine a Z-module basis for H ∗ (X , Z). Cohomology class with degree d represented by dual homology class (via Poincaré duality). [V ∩ W ] = [V ] ∗ [W ] if V and W intersect transversely. [V ∪ W ] = [V ] + [W ]. Traves (USNA) Counting Arrangements 8 OCT 2013 11 / 29 Deformation Deformed constraints have the same class. Example: Line ax + by + cz = 0 in P2 represented by [a : b : c]. Line through P[1 : 0 : 0] ⇔ a = 0. Line through Q[0 : 1 : 0] ⇔ b = 0. Deformation F (t): a(1 − t) + bt = 0 so [P-pencil] = [Q-pencil]. Q[0 : 1 : 0] P [1 : 0 : 0] t=0 Traves (USNA) t=1 Counting Arrangements 8 OCT 2013 12 / 29 Example: Pn and Bézout’s Theorem Pn has one cell in each dimension d = 0, . . . , n H ∗ (Pn , Z) = Zc0 ⊕ Zc1 ⊕ · · · ⊕ Zcn w/ cd codim-d class c0 = [Pn ] = Identity element c12 = c2 = [codim-2 plane] c1 = [hyperplane] c1n = cn = [point] H ∗ (Pn , Z) = Z[h]/(hn+1 ) w/ h = c1 = [hyperplane]. Deformation: [degree d hypersurface] = dh. Bézout: [d1 -hyp ∩ d2 -hyp ∩ · · · ∩ dn -hyp] = (d1 h) · · · (dn h) = d1 · · · dn hn . Traves (USNA) Counting Arrangements 8 OCT 2013 13 / 29 Example: Generic Arrangements in Pn Kunneth formula: H ∗ ((Pn )k , Z) = Z[h1 , h2 , . . . , hk ]/(h1n+1 , . . . , hkn+1 ). Question How many generic arrangements of k hyperplanes go through d = kn points? A goes through P ⇔ H1 (P)H2 (P) · · · Hk (P) = 0. [H1 (P)H2 (P) · · · Hk (P) = 0] = h1 + · · · + hk [{arrs satisfying kn point conditions }] kn kn = (h1 + · · · + hk ) = n,n,...,n (h1 h2 · · · hk )n = kn so answer is n,n,...,n /k !. Traves (USNA) Counting Arrangements kn n,n,...,n points/arrs 8 OCT 2013 14 / 29 Characteristic Numbers Definition Given a family of plane curves, the characteristic number N(p, `) counts the number of such curve that pass through p points and are tangent to ` lines in general position. Let Nk (p, 2k − p) be the nontrivial characteristic numbers for the family of generic arrangements of k lines in P2 . Traves (USNA) Counting Arrangements 8 OCT 2013 15 / 29 Characteristic Numbers Definition Given a family of plane curves, the characteristic number N(p, `) counts the number of such curve that pass through p points and are tangent to ` lines in general position. Let Nk (p, 2k − p) be the nontrivial characteristic numbers for the family of generic arrangements of k lines in P2 . We can compute these for k = 3 and k = 4 and use them to solve more general enumerative problems. Problem How many arrangements of 3 lines passing through 3 points are tangent to 3 elliptic curves? Traves (USNA) Counting Arrangements 8 OCT 2013 15 / 29 The Heinz Problem Question How many generic arrangements of 3 lines pass through 3 points and are tangent to 3 lines? M = {(L1 , L2 , L3 , P12 , P13 , P23 ) : Pij ∈ Li ∩ Lj } ⊂ (P2∗ )3 × (P2 )3 Traves (USNA) Counting Arrangements 8 OCT 2013 16 / 29 A Comhology Calculation M = {(L1 , L2 , L3 , P12 , P13 , P23 ) : Pij ∈ Li ∩ Lj } ⊂ (P2∗ )3 × (P2 )3 3 ) H ∗ ((P2∗ )3 × (P2 )3 , Z) = Z[x1 , x2 , x3 , y12 , y13 , y23 ]/(x13 , . . . , y23 xi = [hyperplane (point) condition on Li ] yij = [hyperplane (line) condition on Pij ] [Pij ∈ Li ] = [{ax0 + by0 + cz0 = 0}] = [{ax0 }] = xi + yij [M] = (x1 + y12 )(x2 + y12 )(x1 + y13 )(x3 + y13 )(x2 + y23 )(x3 + y23 ) Labeled Count = [M](x1 + x2 + x3 )3 (y12 + y13 + y23 )3 = 342[pt] Count = N3 (3, 3) = 342/3! = 57. Traves (USNA) Counting Arrangements 8 OCT 2013 17 / 29 Why is this called the Heinz Problem? 57 Varieties! Traves (USNA) Counting Arrangements 8 OCT 2013 18 / 29 Characteristic Numbers for k = 3 Theorem (Paul, Wakefield, T-) The characteristic numbers for the family of generic arrangements with 3 lines are: Points p Lines ` N3 (p, `) 0 6 15 1 5 30 2 4 48 3 3 57 4 2 48 5 1 30 6 0 15 Why is the table symmetric? → Duality! Why is the table unimodal? Traves (USNA) Counting Arrangements 8 OCT 2013 19 / 29 Enumerative Problem Problem How many arrangements of 3 lines passing through 3 points are tangent to 3 elliptic curves? Class of a smooth curve of degree d is d(d − 1). Fulton-MacPherson: x 3 (6x + 3y )3 = 63 x 6 Count Traves (USNA) +62 3 3 2 x 5y +6 · 32 3 2 x 4 y 2 +33 x 3 y 3 = 216N3 (6, 0) +324N3 (5, 1) +162N3 (4, 2) = 216(15) +324(30) +162(48) = 22, 275 Counting Arrangements +27N3 (3, 3) +27(57) 8 OCT 2013 20 / 29 4 Lines Consider N4 (0, 8). Then 82 62 quadruple lines each count as 6! labeled arrangements in M = MA ∪ M 0 . N4 (0, 8) = ([M](y12 + · · · + y34 )8 − = 16, 695. 8 2 6 2 6!)/4! By duality there are 16,695 braid arrangements through 8 points. Traves (USNA) Counting Arrangements 8 OCT 2013 21 / 29 Table of Characteristic Numbers Theorem (Paul, Wakefield, T-) The characteristic numbers for the family of generic arrangements with 4 lines are: Points p Lines ` N4 (p, `) 0 8 16695 1 7 17955 2 6 13185 3 5 8190 4 4 4410 5 3 2070 6 2 855 7 1 315 8 0 105 What happens with 5 lines? → Excess Intersection – higher dimensional collections of arrangements that contribute a finite number to the degree computation. Traves (USNA) Counting Arrangements 8 OCT 2013 22 / 29 d-coned Generic Arrangements We consider arrangements where each hyperplane contains a common linear space, but the hyperplanes are otherwise generic. d-coned generic arrangement A in Pn : H∼ = Pn−(d+1) containing generic arr. B Ω linear space of dim. d, Ω ∩ H = ∅ each H in A is span of hyp in B with Ω. Traves (USNA) Counting Arrangements 8 OCT 2013 23 / 29 Running Example Question How many 1-coned arrangements in P5 of 6 hyperplanes pass through D points in general position? Each hyperplane is determined by its normal vector modulo line Ω, i.e. by a point in P(C6 /Ω) ∼ = P3 . D = dim MA = dim G(1, 5) × (P3 )6 = 2 × 4 + 6 × 3 = 26. G(1, 5) = Grassmannian of lines in P5 . Traves (USNA) Counting Arrangements 8 OCT 2013 24 / 29 Cohomology of the Grassmannian Each line in P5 is the span of 2 points in C6 . Cellular decomposition of G(1, 5): cells determined by position of leadingones in RREF of spanning matrix: 1 ∗ ∗ 0 ∗ ∗ σ20 = M ∈ M2×6 : M ∼ → codimσ20 = 2 0 0 0 1 ∗ ∗ Schubert calculus tells us how to multiply cohomology classes Giambelli formula → special classes Pieri formula → multiplication of special classes Duality: σα σα0 = [single Ω] Traves (USNA) Counting Arrangements 8 OCT 2013 25 / 29 A cohomology class Z := {(Ω, H) ∈ G(1, 5) × P5 : Ω ⊂ H}. H must contain 2 independent points on Ω: so codim Z = 2. [Z ] = aσ20 + bσ11 + cσ10 h + dσ00 h2 Use duality to find coefficients: [Z ] = σ11 + σ10 h + σ00 h2 Traves (USNA) Counting Arrangements 8 OCT 2013 26 / 29 # 1-coned arr. of 6 hyperplanes through 26 points in P5 Pull-back to G(1, 5) × (P5 )6 : [Zi ] = σ11 + σ10 hi + σ00 hi2 ANS = (h1 + h2 + h3 + h4 + h5 + h6 )26 6 Y [Zi ] i=1 Schubert Calculus gives: 2 4 2 3 σ00 σ11 = σ00 σ10 σ11 = 1σ44 = 1[arr] h 6 26 ANS = 1 2,4 3,3,5,5,5,5 + 1 = 10270301132391300 Traves (USNA) 6 1,2,3 4 2 σ10 σ11 = 2[arr]. i 6 26 + 2 4,2 /6! 4,4,4,4,5,5 Catalan numbers 26 3,4,4,5,5,5 Counting Arrangements 8 OCT 2013 27 / 29 General Result We have a general expression for the dimension D of MA for d-coned arrangements of k hyperplanes in Pn . We can also express the number of such arrangements passing through D points using Schubert Calculus. Traves (USNA) Counting Arrangements 8 OCT 2013 28 / 29 Parting thoughts 1 Need a better sense of how the intersection lattice of A affects the dimension of MA and the enumerative counts. 2 Is there a recursive relationship between the enumerative counts for d-coned arrangements of k hyperplanes in Pn passing through D = dim MA points? 3 Is there a role for quantum cohomology here? If so, are the cohomology rings associatitve and can we leverage this to get nice generating functions for some class of enumerative problems (c.f. moduli of curves)? 4 This project is in its infancy and there’s room for many others to contribute. Traves (USNA) Counting Arrangements 8 OCT 2013 29 / 29