Counting the points in the Hilbert scheme Anna Brosowsky - PowerPoint PPT Presentation

Counting the points in the Hilbert scheme Counting the points in the Hilbert scheme Anna Brosowsky [Collaborators: Murray Pendergrass, Nathanial Gillman] [Mentors: Dr. Amin Gholampour, Rebecca Black, Tao Zhang] Department of Mathematics Cornell University MAPS REU Research Fair, 2016 1 / 16

Counting the points in the Hilbert scheme Outline Background 1 Modules Hilbert schemes Recursion 2 Motivation Going down Coming up Formula 2 / 16

Counting the points in the Hilbert scheme Background Modules Definition Definition For a ring R , an R -module M is an additive abelian group with an operation · : R × M → M such that for all r 1 , r 2 ∈ R , m 1 , m 2 ∈ M , we have r · ( m 1 + m 2 ) = r · m 1 + r · m 2 ( r 1 + r 2 ) · m = r 1 · m + r 2 · m 1 R · m = m r 1 · ( r 2 · m ) = ( r 1 r 2 ) · m . Examples: R n and Z n are Z -modules using usual multiplication. Any ring R is an R -module over itself. 3 / 16

Counting the points in the Hilbert scheme Background Modules Torsion Definition Let M be an R -module, for R a ring. Then m � = 0 ∈ M is torsion if there exists some r � = 0 ∈ R such that rm = 0 . M is called a torsion module if every m ∈ M is torsion. If no m ∈ M is torsion, then M is torsion-free . Examples: R n is a torsion-free R -module, since a · � v = � 0 implies a = 0 or � b = � 0 for any a ∈ R and � b ∈ R n . Z / Z n is a torsion Z -module since for any a ∈ Z / Z n , n · a = na = 0 ∈ Z / Z n . 4 / 16

Counting the points in the Hilbert scheme Background Hilbert schemes Definition Definition Let k = F q be a finite field with q elements, and R = k [ y ] . The punctual Hilbert scheme of type ( m 0 , m 1 ) is defined as k 2 = { I ⊆ k [ x, y ] | k [ x, y ] /I ≃ m 0 ρ 0 + m 1 ρ 1 , Hilb ( m 0 ,m 1 ) 0 V ( I ) = 0 } . The stratified version is defined as k 2 = { I ∈ Hilb ( m 0 ,m 1 ) k 2 | I | l ≃ F I ⊕ T I , Hilb ( m 0 ,m 1 )( d 0 ,d 1 ) 0 0 T I ≃ d 0 ρ 0 + d 1 ρ 1 } . where F I is a torsion-free R -module, T I is a torsion R -module, and I | l = I/x · I . 5 / 16

Counting the points in the Hilbert scheme Background Hilbert schemes Example y ✻ y ✻ y ✻ ✻ ✻ ♣ ♣ ♣ ♣ ♣ ♣ ♣ ♣ ♣ ♣ ♣ ♣ ♣ ♣ ♣ ♣ ♣ ♣ ♣ ♣ ♣ ♣ ♣ ♣ ♣ ♣ ♣ ♣ ♣ ♣ ♣ ♣ ♣ ♣ 1 1 ♣ ♣ ♣ ♣ ♣ ♣ ♣ ♣ ♣ ♣ ♣ ♣ ♣ ♣ ♣ ♣ ♣ 0 0 ♣ ♣ ♣ ♣ ♣ ♣ ♣ ♣ ♣ ♣ ♣ ♣ ♣ ♣ ♣ ♣ ♣ 1 1 0 1 0 ♣ ♣ ♣ ♣ ♣ ♣ ♣ ♣ ♣ ♣ ♣ ♣ ♣ ♣ ♣ ♣ ♣ 0 1 0 1 0 0 1 0 ♣ ♣ ♣ ♣ ♣ ♣ ♣ ♣ ♣ ♣ ♣ ♣ ♣ ♣ ♣ ♣ ♣ 1 0 1 0 1 1 0 1 ♣ ♣ ♣ ♣ ♣ ♣ ♣ ♣ ♣ ♣ ♣ ♣ ♣ ♣ ♣ ♣ ♣ 0 1 0 1 0 1 0 1 0 0 1 0 1 0 ✲ ✲ ✲ ♣ ♣ ♣ ♣ ♣ ♣ ♣ ♣ ♣ ♣ ♣ ♣ ♣ ♣ ♣ ♣ ♣ x x x I x · I I/ ( x · I ) I = � y 4 , xy 3 , x 2 y, x 4 � ∈ Hilb (4 , 5) k 2 0 x · I = � xy 4 , x 2 y 3 , x 3 y, x 5 � I/ ( x · I ) ≃ Ry 5 ⊕ kxy 3 ⊕ kx 2 y 2 ⊕ kx 2 y ⊕ kx 4 T I = kxy 3 ⊕ kx 2 y 2 ⊕ kx 2 y ⊕ kx 4 ≃ 3 ρ 0 + ρ 1 F I = Ry 5 6 / 16

Counting the points in the Hilbert scheme Recursion Motivation Outline of our goal Find the generating function for the Hilbert scheme of points, which has the form � #Hilb ( m 0 ,m 1 ) k 2 � � · t m 0 0 t m 1 1 m 0 ,m 1 ≥ 0 where k = F q . Need to count the number of points in Hilb ( m 0 ,m 1 ) k 2 . 0 Do this by counting points in the stratified version. k 2 = � Hilb ( m 0 ,m 1 ) d 0 ,d 1 ≥ 0 Hilb ( m 0 ,m 1 )( d 0 ,d 1 ) k 2 , so 0 0 k 2 = � #Hilb ( m 0 ,m 1 ) d 0 ,d 1 ≥ 0 #Hilb ( m 0 ,m 1 )( d 0 ,d 1 ) k 2 0 0 Specifically, want a recursion giving the number of points in stratified Hilbert scheme in terms of number of points in smaller Hilbert scheme 7 / 16

Counting the points in the Hilbert scheme Recursion Going down Getting I ′ For any ideal I , define x · I ′ to be the kernel of the map I → I | l → F I . Exact commutative diagram shows uniqueness. 0 0 0 T I I | l F I 0 x · I ′ 0 I F I 0 x · I x · I 0 0 For monomial ideals, get I ′ by deleting the first column of the Young diagram, so if I ∈ Hilb ( m 0 ,m 1 )( d 0 ,d 1 ) k 2 , then 0 I ′ ∈ Hilb ( m 0 − d 1 ,m 1 − d 0 )( d ′ 0 ,d ′ 1 ) k 2 . 0 8 / 16

Counting the points in the Hilbert scheme Recursion Going down Why ( m 0 − d 1 , m 1 − d 0 ) ? Recurse by “chopping off” first column and sliding diagram over, then counting which ideals give same diagram. Same as removing last block in each row 0 in torsion part ⇒ 1 in last box. 1 in torsion part ⇒ 0 in last box. Also requires d ′ 0 ≤ d 1 and d ′ 1 ≤ d 0 in smaller scheme. 1 0 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 − → 9 / 16

Counting the points in the Hilbert scheme Recursion Coming up Recover I from I ′ Fix a torsion module T and map ϕ : I ′ → T , and define I = ker ϕ . Exact commutative diagram shows if F = ker I ′ | l → T is torsion-free, then I | l ≃ F ⊕ x · T . 0 0 I ′ | l 0 F T 0 I ′ 0 I T 0 x · I ′ x · I ′ 0 0 10 / 16

Counting the points in the Hilbert scheme Recursion Coming up Choosing torsion-free For any I ′ ∈ Hilb ( m 0 − d 1 ,m 1 − d 0 )( d ′ 0 ,d ′ 1 ) k 2 , the number of possible I 0 it came from is the number of F such that F is a rank 1 torsion-free submodule of I ′ | l with I ′ | l /F ≃ T ≃ d 1 ρ 0 + d 0 ρ 1 . Rank 1 since the torsion-free part is always the first column above the Young diagram, generated by single element y a . d 1 ρ 0 + d 0 ρ 1 since we require x · T ≃ d 0 ρ 0 + d 1 ρ 1 and multiplying by x switches the parity of basis elements. Or, number of F ⊆ F I ′ , rank 1 and torsion free, with F I ′ /F ≃ ( d 1 − d ′ 0 ) ρ 0 + ( d 0 − d ′ 1 ) ρ 1 times number of ways to embed into I ′ | l . 11 / 16

Counting the points in the Hilbert scheme Recursion Coming up How many F ? From [1], at most one F ⊆ F I ′ which works. If I ′ is a monomial ideal, then F I ′ = Ry d ′ 0 + d ′ 1 and F = Ry d 0 + d 1 . Basis for F I ′ /F is { y j | d ′ 0 + d ′ 1 ≤ j < d 0 + d 1 } . Since F I ′ /F ≃ ( d 1 − d ′ 0 ) ρ 0 + ( d 0 − d ′ 1 ) ρ 1 , must have d 1 − d ′ 0 even degree basis elements and d 0 − d ′ 1 odd degree ones. Three cases to check: 1 If d ′ 0 + d ′ 1 + d 0 + d 1 ≡ 0 mod 2 , then d 0 − d 1 = d ′ 1 − d ′ 0 . 2 If d ′ 0 + d ′ 1 + d 0 + d 1 ≡ 1 mod 2 and d ′ 0 + d ′ 1 ≡ 0 mod 2 , then 1 + d 0 − d 1 = d ′ 1 − d ′ 0 . 3 If d ′ 0 + d ′ 1 + d 0 + d 1 ≡ 1 mod 2 and d ′ 0 + d ′ 1 ≡ 1 mod 2 , then d 0 − d 1 − 1 = d ′ 1 − d ′ 0 . 0 = d 0 − d 1 + ( − 1) d ′ 0 + d ′ so d ′ 1 − d ′ 1 (( d ′ 0 + d ′ 1 + d 0 + d 1 )%2) . 12 / 16

Counting the points in the Hilbert scheme Recursion Coming up How many ways to embed? Suppose F = Ry d 0 + d 1 , b 1 , . . . , b d ′ 0 are basis for trivial torsion elements, and c 1 , . . . , c d ′ 1 basis for non-trivial torsion elements. If we don’t care about type, then can embed F as d ′ d ′ � � 0 1 F := R ( y d 0 + d 1 , � β i b i + γ j c j ) i =1 j =1 1 possible � for any β i , γ j ∈ k . q choices for each ⇒ q d ′ 0 + d ′ F . We do care about type, so can only use torsion elements of same type as y a . Therefore q r possible � F , where � d ′ if d 0 + d 1 ≡ 0 mod 2 0 r = d ′ if d 0 + d 1 ≡ 1 mod 2 1 13 / 16

Counting the points in the Hilbert scheme Recursion Formula Re-CURSE-ion � k 2 = q r · #Hilb ( m 0 − d 1 ,m 1 − d 0 )( d ′ 0 ,d ′ 1 ) #Hilb ( m 0 ,m 1 )( d 0 ,d 1 ) k 2 0 0 0 ≤ d ′ 0 ≤ d 1 0 ≤ d ′ 1 ≤ d 0 0 = d 0 − d 1 +( − 1) ( d ′ 0+ d ′ 1) (( d ′ d ′ 1 − d ′ 0 + d ′ 1 + d 0 + d 1 )%2) where � d ′ if d 0 + d 1 ≡ 0 mod 2 0 r = d ′ if d 0 + d 1 ≡ 1 mod 2 1 Let a, b, c, d ∈ Z ≥ 0 . The base cases are k 2 = 0 k 2 = 0 #Hilb (0 ,b> 0) , ( c,d ) #Hilb ( a,b ) , ( c>b,d ) 0 0 k 2 = 0 k 2 = 0 #Hilb ( a,b ) , ( c,d>a ) #Hilb ( a � =0 ,b ) , (0 , 0) 0 0 k 2 = 1 #Hilb (0 , 0) , (0 , 0) 0 14 / 16

Counting the points in the Hilbert scheme Summary Summary Found recursion for number of points in stratified Hilbert scheme! Hard to work with, so unsuccessful in finding a closed form with this method. Still interesting, especially because of special case closed formulas. In Future Look at more special cases. Pursue abacus method. 15 / 16

Counting the points in the Hilbert scheme Appendix For Further Reading References K. Yoshioka. The Betti numbers of the moduli space of stable sheaves of rank 2 on P 2 . J. reine angew. Math. , 453:193–220, 1994. S.M. Gusein-Zade, I. Luengo, and A. Melle-Hern´ andez. On generating series of classes of equivariant Hilbert schemes of fat points. Moscow Mathematical Journal , 10(3):593–602, 2010. ´ A. Gyenge, A. N´ emethi, and B. Szendr˝ oi. Euler characteristics of Hilbert schemes of points on simple surface singularities. http://arxiv.org/abs/1512.06848 , 2015 16 / 16