Recursions and Colored Hilbert Schemes Anna Brosowsky [Joint Work - PowerPoint PPT Presentation

Recursions and Colored Hilbert Schemes Recursions and Colored Hilbert Schemes Anna Brosowsky [Joint Work With: Nathaniel Gillman, Murray Pendergrass] [Mentor: Dr. Amin Gholampour] Cornell University 24 September 2016 WiMiN Smith College 1/26

Recursions and Colored Hilbert Schemes Outline Background 1 Recursion 2 Future Work 3 2/26

Recursions and Colored Hilbert Schemes Background Outline Background 1 Recursion 2 Future Work 3 3/26

Recursions and Colored Hilbert Schemes Background The Problem Object of Study: The Hilbert scheme of type ( m 0 , m 1 ) . Long Term Goal: Find the Poincar´ e polynomial of the punctual Hilbert scheme of type ( m 0 , m 1 ) . Why? The Poincar` e polynomial is a topological invariant , meaning it doesn’t change with stretching and bending. Short Term Goal: Count the points of the punctual Hilbert scheme of type ( m 0 , m 1 ) . Why? We can use this to find a generating function, which we can then use in the Weil conjectures, to find the Poincar´ e polynomial. 4/26

Recursions and Colored Hilbert Schemes Background What is the punctual Hilbert scheme of type ( m 0 , m 1 ) ? We’ll first introduce some background: 1 Monomial ideals 2 Young diagrams 3 Group actions 4 Colored Young diagrams 5/26

Recursions and Colored Hilbert Schemes Background Ideals Definition An ideal I ⊂ k [ x, y ] for a field k is a set of polynomials, but with a few rules attached: α, β ∈ I implies α + β ∈ I α ∈ I and m ∈ k [ x, y ] implies α · m ∈ I Theorem Every polynomial ideal can be written as � g 1 , . . . , g n � = { f 1 g 1 + · · · + f n g n | f i ∈ k [ x, y ] } , which is the set of all k [ x, y ] linear combinations of the g i . These g i are called the generators. 6/26

Recursions and Colored Hilbert Schemes Background Monomial ideals Definition A monomial ideal is an ideal generated by monomials. Example � x � = { Ax | A ∈ k [ x, y ] } = { Ax 3 + By 3 | A, B ∈ k [ x, y ] } x 3 , y 3 � � are monomial ideals within the polynomial ring k [ x, y ] . � x + y � = { A ( x + y ) | A ∈ k [ x, y ] } is not a monomial ideal. 7/26

Recursions and Colored Hilbert Schemes Background Young diagrams A Young diagram is a visual representation of a monomial ideal. Example We’ll construct a Young diagram for the monomial ideal x 4 , x 2 y, xy 3 , y 4 � � ⊂ k [ x, y ] ✻ y ♣ ♣ ♣ ♣ ♣ ♣ y 4 ♣ ♣ ♣ ♣ ♣ ♣ xy 3 ♣ ♣ ♣ ♣ ♣ ♣ ♣ ♣ ♣ ♣ ♣ ♣ x 2 y ♣ ♣ ♣ ♣ ♣ ♣ x 4 ✲ ♣ ♣ ♣ ♣ ♣ ♣ x 8/26

Recursions and Colored Hilbert Schemes Background Group actions Our Z 2 group action is defined by x �→ − x , y �→ − y Example Under the given transformation: 1 = x 0 y 0 �→ ( − x ) 0 ( − y ) 0 = 1 , ∴ 1 �→ 1 x 3 y 5 �→ ( − x ) 3 ( − y ) 5 = x 3 y 5 , ∴ x 3 y 5 �→ x 3 y 5 y 5 = x 0 y 5 �→ ( − x ) 0 ( − y ) 5 = − y 5 , ∴ y 5 �→ − y 5 and in general, x a y b �→ ( − 1) a + b x a y b 9/26

Recursions and Colored Hilbert Schemes Background Colored Young diagrams We can combine the Young diagram and the group action to “color the Young diagram.” Procedure: 1 Draw the Young diagram as before 2 If the monomial for a box maps to itself, we “color” the box with a 0 3 If the monomial for a box maps to the negative of itself, we “color” the box with a 1 10/26

Recursions and Colored Hilbert Schemes Background Example: coloring a Young diagram Example � x 4 , x 2 y, xy 3 , y 4 � Recall our previous ideal, ⊂ k [ x, y ] . We can see that 1 �→ 1 , x �→ − x , y �→ − y , xy �→ xy , etc. ✻ ✻ y y ♣ ♣ ♣ ♣ ♣ ♣ ♣ ♣ ♣ ♣ ♣ ♣ ♣ ♣ ♣ ♣ ♣ ♣ ♣ ♣ 1 → ♣ ♣ ♣ ♣ ♣ ♣ ♣ ♣ ♣ ♣ 0 1 ♣ ♣ ♣ ♣ ♣ ♣ ♣ ♣ ♣ ♣ 1 0 ♣ ♣ ♣ ♣ ♣ ♣ ♣ ♣ ♣ ♣ ✲ 0 1 0 1 ✲ ♣ ♣ ♣ ♣ ♣ ♣ ♣ ♣ ♣ ♣ x x 11/26

Recursions and Colored Hilbert Schemes Background And finally... the punctual Hilbert scheme The punctual Hilbert scheme of type ( m 0 , m 1 ) is defined as � � k [ x, y ] � k 2 = � Hilb ( m 0 ,m 1 ) I ⊆ k [ x, y ] ≃ m 0 ρ 0 + m 1 ρ 1 , V ( I ) = 0 � 0 I where k = F q is a finite field of order q . But the subset of k 2 made of monomial ideals looks like Hilb ( m 0 ,m 1 ) 0 { Young diagrams with m 0 0’s and m 1 1’s } 12/26

Recursions and Colored Hilbert Schemes Background Example: punctual Hilbert scheme of type (4 , 5) Example x 4 , x 2 y, xy 3 , y 4 � � Recall the monomial ideal ⊂ k [ x, y ] . ✻ y ♣ ♣ ♣ ♣ ♣ ♣ ♣ ♣ ♣ ♣ 1 ♣ ♣ ♣ ♣ ♣ 0 1 ♣ ♣ ♣ ♣ ♣ 1 0 ♣ ♣ ♣ ♣ ♣ 0 1 0 1 ✲ ♣ ♣ ♣ ♣ ♣ x is in Hilb (4 , 5) x 4 , x 2 y, xy 3 , y 4 � k 2 , since � By definition, the ideal 0 the ideal has 4 0’s and 5 1’s when represented as a colored Young diagram under our group action. 13/26

Recursions and Colored Hilbert Schemes Recursion Outline Background 1 Recursion 2 Future Work 3 14/26

Recursions and Colored Hilbert Schemes Recursion Stratified Hilbert schemes The stratified punctual Hilbert scheme of type ( m 0 , m 1 ) , ( d 0 , d 1 ) , written Hilb ( m 0 ,m 1 ) , ( d 0 ,d 1 ) k 2 , is a subset of Hilb ( m 0 ,m 1 ) k 2 . The 0 0 strata is determined by the first box outside the Young diagram in each row. d 0 = number of those boxes containing zero d 1 = number of those boxes containing one. 15/26

Recursions and Colored Hilbert Schemes Recursion Stratified Hilbert scheme example Example ✻ y ♣ ♣ ♣ ♣ ♣ ♣ 1 0 ♣ ♣ ♣ ♣ ♣ ♣ 0 1 0 � y 4 , xy 3 , x 2 y, x 4 � ∈ Hilb (4 , 5) , (3 , 1) k 2 ♣ ♣ ♣ ♣ ♣ ♣ 1 0 1 0 ♣ ♣ ♣ ♣ ♣ ♣ 0 1 0 1 0 ✲ ♣ ♣ ♣ ♣ ♣ ♣ x k 2 = Hilb ( m 0 ,m 1 ) � Hilb ( m 0 ,m 1 ) , ( d 0 ,d 1 ) k 2 0 0 d 0 ,d 1 ≥ 0 k 2 = ⇒ #Hilb ( m 0 ,m 1 ) #Hilb ( m 0 ,m 1 ) , ( d 0 ,d 1 ) � k 2 0 0 d 0 ,d 1 ≥ 0 16/26

Recursions and Colored Hilbert Schemes Recursion Example Example y ✻ ✻ y ♣ ♣ ♣ ♣ ♣ ♣ ♣ ♣ ♣ ♣ ♣ ♣ ♣ ♣ ♣ ♣ ♣ 1 0 ♣ ♣ ♣ ♣ ♣ → 0 1 ♣ ♣ ♣ ♣ ♣ ♣ 0 1 0 ♣ ♣ ♣ ♣ ♣ 1 0 ♣ ♣ ♣ ♣ ♣ ♣ 1 0 1 ♣ ♣ ♣ ♣ ♣ 0 1 0 1 ✲ ♣ ♣ ♣ ♣ ♣ ♣ 0 1 0 1 0 ✲ ♣ ♣ ♣ ♣ ♣ x ♣ ♣ ♣ ♣ ♣ ♣ x y ✻ y ✻ → → ♣ ♣ ♣ ♣ ♣ ♣ ♣ ♣ ♣ ♣ ♣ ♣ ♣ ♣ 0 1 0 0 1 ✲ ✲ ♣ ♣ ♣ ♣ ♣ ♣ ♣ x x 17/26

Recursions and Colored Hilbert Schemes Recursion Recursion 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 18/26

Recursions and Colored Hilbert Schemes Recursion 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 outside diagram ⇒ 1 in last box 1 outside diagram ⇒ 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 − → 19/26

Recursions and Colored Hilbert Schemes Recursion Recursion 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 20/26

Recursions and Colored Hilbert Schemes Recursion Some Special Cases #Hilb ( k,k +1) , ( k +1 ,k − 1) k 2 = 1 0 #Hilb ( k,k +1) , ( k +1 ,k − 1) k 2 = 1 0 k 2 = q k #Hilb ( k,k ) , (1 , 0) 0 #Hilb ( k +1 ,k ) , (0 , 1) k 2 = q k 0 � k #Hilb ( k,k +1) , (2 , 0) k 2 = q k − 1 � 0 2 � k k 2 = #Hilb ( k +1 ,k ) , (0 , 2) � q k 0 2 21/26

Recursions and Colored Hilbert Schemes Future Work Outline Background 1 Recursion 2 Future Work 3 22/26

Recursions and Colored Hilbert Schemes Future Work Generating Function Prove Dr. Gholampour’s conjectured generating function for the number of points in the punctual Hilbert scheme of n points 1 � (#Hilb n 0 ,n 1 ) t n 0 0 t n 1 � 1 = (1 − q j − 1 ( t 0 t 1 ) j )(1 − q j ( t 0 t 1 ) j ) n 0 ,n 1 ≥ 0 j ≥ 1 t m 2 0 t m 2 + m � · 1 m ∈ Z 23/26