The Hilbert Polynomial of the Irreducible Representation of the Rational Cherednik Algebra of Type A n in Characteristic p n

Transcription

1 The Hilbert Polynomial of the Irreducible Representation of the Rational Cherednik Algebra of Type A n in Characteristic p n Merrick Cai Mentor: Daniil Kalinov, MIT Kings Park High School May 19-20, 2018 MIT PRIMES Conference

2 Vector Space A vector space defined over a field k, is a collection of vectors, which may be multiplied by scalars λ k, and added together.

3 Vector Space A vector space defined over a field k, is a collection of vectors, which may be multiplied by scalars λ k, and added together. k n k[x 1, x 2,..., x n ] k[[x 1, x 2,..., x n ]] k[ x, x] Mat n k)

4 Algebra An algebra is a vector space V equipped with a bilinear product; i.e., the vectors can be multiplied while preserving linearity.

5 Algebra An algebra is a vector space V equipped with a bilinear product; i.e., the vectors can be multiplied while preserving linearity. k[x 1,..., x n ] k[[x 1,..., x n ]] k[x 1, x 2,..., x n, x1, x2,..., xn ] Mat n k)

6 Graded Algebra An algebra A is graded if A = n 0 A n for subspaces A n and A i A j A i+j.

7 Graded Algebra An algebra A is graded if A = n 0 A n for subspaces A n and A i A j A i+j. Example The algebra A = k[x 1, x 2,..., x n ] has a grading given by A i the subspace of homogeneous degree i polynomials.

8 Hilbert Series The Hilbert series of a graded algebra A is given by hz) = n 0 dima n )z n.

9 Hilbert Series The Hilbert series of a graded algebra A is given by hz) = n 0 dima n )z n. Example The algebra A = k[x 1,..., x n ] has the usual grading by degree. Then dima i ) = ) n+i 1 i, so ha z) = n+i 1 ) i 0 i z i = 1 1 z) n.

10 Representation A representation of an algebra A is a vector space V equipped with a homomorphism ρ : A EndV ).

11 Representation A representation of an algebra A is a vector space V equipped with a homomorphism ρ : A EndV ). Example Take V = C n and G = S n. Then the group algebra k[s n ] acts on v V by permuting the indices; e.g., [123)] x, y, z) = z, x, y).

12 Representation A representation of an algebra A is a vector space V equipped with a homomorphism ρ : A EndV ). Example Take V = C n and G = S n. Then the group algebra k[s n ] acts on v V by permuting the indices; e.g., [123)] x, y, z) = z, x, y). A subrepresentation is a subspace W V which remains closed under the action of ρa).

13 Irreducible Representation A representation A, V ) is irreducible if there does not exist any proper) subspace W V which is closed under the action of A.

14 Irreducible Representation A representation A, V ) is irreducible if there does not exist any proper) subspace W V which is closed under the action of A. Example Let A = k[s n ] be the group algebra of S n and V = C n be a vector space where S n acts by permutations. Then Span{1, 1, 1,..., 1)} is an irreducible subrepresentation.

15 Differential Operators Let V = k[x]. The differential operator acts by x x k = kx k 1. In characteristic 0 we can define the algebra of differential operators as a subalgebra in Endk[x]) generated by x and x.

16 Differential Operators Let V = k[x]. The differential operator acts by x x k = kx k 1. In characteristic 0 we can define the algebra of differential operators as a subalgebra in Endk[x]) generated by x and x. But in characteristic p, p x acts by 0, this is problematic. So to define k[x, x ], use the fact that [ x, x] = 1. So k[x, x ] = k x, y /[y, x] = 1).

17 Rational Cherednik Algebra of Type A n The Cherednik algebra H t,c n) in characteristic 0 is generated by the following in Endk[x 1,..., x n ]/x x n )):

18 Rational Cherednik Algebra of Type A n The Cherednik algebra H t,c n) in characteristic 0 is generated by the following in Endk[x 1,..., x n ]/x x n )): Polynomials in k[x 1,..., x n ] Acts by multiplication

19 Rational Cherednik Algebra of Type A n The Cherednik algebra H t,c n) in characteristic 0 is generated by the following in Endk[x 1,..., x n ]/x x n )): Polynomials in k[x 1,..., x n ] Acts by multiplication Elements of S n Acts by permuting the x i s

20 Rational Cherednik Algebra of Type A n The Cherednik algebra H t,c n) in characteristic 0 is generated by the following in Endk[x 1,..., x n ]/x x n )): Polynomials in k[x 1,..., x n ] Acts by multiplication Elements of S n Acts by permuting the x i s The Dunkl operators D yi An extension of the partial derivative D yi = t xi c k i [D yi, D yj ] = 0 1 σ ik x i x k

21 Rational Cherednik Algebra of Type A n The Cherednik algebra H t,c n) in characteristic 0 is generated by the following in Endk[x 1,..., x n ]/x x n )): Polynomials in k[x 1,..., x n ] Acts by multiplication Elements of S n Acts by permuting the x i s The Dunkl operators D yi An extension of the partial derivative D yi = t xi c k i [D yi, D yj ] = 0 1 σ ik x i x k The relevant cases are t = 1 and t = 0. We will work with t = 0. We need more abstract definition for characteristic p as for differential operators.

22 Dunkl Operators The Dunkl operator can be described by D yi = t xi c k i 1 σ ik x i x k.

23 Dunkl Operators The Dunkl operator can be described by D yi For t = 1, an example of D y1 x1 x2 2x 3 3 ) : = t xi c k i 1 σ ik x i x k.

24 Dunkl Operators The Dunkl operator can be described by D yi For t = 1, an example of D y1 x1 x2 2x 3 3 ) : = t xi c k i 1 σ ik x i x k. 1 x1 x1 x 2 2 x 3 3 ) = x 2 2 x 3 3

25 Dunkl Operators The Dunkl operator can be described by D yi For t = 1, an example of D y1 x1 x2 2x 3 3 ) : = t xi c k i 1 σ ik x i x k. 1 x1 x1 x2 2x 3 3 ) = x 2 2 x3 3 1 σ 12 x 1 x x1 2 x2 2x 3 3 ) ) = x 3 x1 x2 2 x2 1 x 2 3 x 1 x 2 = x 1 x 2 x3 3

26 Dunkl Operators The Dunkl operator can be described by D yi For t = 1, an example of D y1 x1 x2 2x 3 3 ) : = t xi c k i 1 σ ik x i x k. 1 x1 x1 x 2 2 x 3 3 ) = x 2 2 x σ 12 x 1 x 2 x1 x 2 2 x σ 13 x 1 x 3 x1 x 2 2 x 3 3 ) = x 3 3 x1 x 2 2 x2 1 x 2 x 1 x 2 ) = x 1 x 2 x 3 3 ) = x 2 2 x1 x 3 3 x3 1 x 3 x 1 x 3 ) = x 2 1 x 2 2 x 3 x 1 x 2 2 x 2 3

27 Dunkl Operators The Dunkl operator can be described by D yi For t = 1, an example of D y1 x1 x2 2x 3 3 ) : = t xi c k i 1 σ ik x i x k. 1 x1 x1 x 2 2 x 3 3 ) = x 2 2 x σ 12 x 1 x 2 x1 x 2 2 x σ 13 x 1 x 3 x1 x 2 2 x 3 3 ) = x 3 3 x1 x 2 2 x2 1 x 2 x 1 x 2 ) = x 1 x 2 x 3 3 ) = x 2 2 x1 x 3 3 x3 1 x 3 x 1 x 3 ) = x 2 1 x 2 2 x 3 x 1 x 2 2 x 2 3 For k > 3, then 1 σ 1k x 1 x k x1 x 2 2 x 3 3 ) = x 2 2 x 3 3

28 Dunkl Operators The Dunkl operator can be described by D yi For t = 1, an example of D y1 x1 x2 2x 3 3 ) : = t xi c k i 1 σ ik x i x k. 1 x1 x1 x 2 2 x 3 3 ) = x 2 2 x σ 12 x 1 x 2 x1 x 2 2 x σ 13 x 1 x 3 x1 x 2 2 x 3 3 ) = x 3 3 x1 x 2 2 x2 1 x 2 x 1 x 2 ) = x 1 x 2 x 3 3 ) = x 2 2 x1 x 3 3 x3 1 x 3 x 1 x 3 ) = x 2 1 x 2 2 x 3 x 1 x 2 2 x 2 3 For k > 3, then 1 σ 1k x 1 x x1 k x2 2x 3 3 ) = x 2 2 x3 3 D y1 x1 x2 2x 3 3 ) = x1 x1 x2 2x 3 3 ) 1 σ 1k k 1 x 1 x x1 k x2 2x 3 3 ) = x2 2x c x 1 x 2 x3 3 + x 1 2x 2 2x 3 + x 1 x2 2x 3 2 n 3)x 2 2x 3 3 )

29 Dunkl Operators The Dunkl operator can be described by D yi For t = 1, an example of D y1 x1 x2 2x 3 3 ) : = t xi c k i 1 σ ik x i x k. 1 x1 x1 x 2 2 x 3 3 ) = x 2 2 x σ 12 x 1 x 2 x1 x 2 2 x σ 13 x 1 x 3 x1 x 2 2 x 3 3 ) = x 3 3 x1 x 2 2 x2 1 x 2 x 1 x 2 ) = x 1 x 2 x 3 3 ) = x 2 2 x1 x 3 3 x3 1 x 3 x 1 x 3 ) = x 2 1 x 2 2 x 3 x 1 x 2 2 x 2 3 For k > 3, then 1 σ 1k x 1 x x1 k x2 2x 3 3 ) = x 2 2 x3 3 D y1 x1 x2 2x 3 3 ) = x1 x1 x2 2x 3 3 ) 1 σ 1k k 1 x 1 x x1 k x2 2x 3 3 ) = x2 2x c x 1 x 2 x3 3 + x 1 2x 2 2x 3 + x 1 x2 2x 3 2 n 3)x 2 2x 3 3 ) The singular polynomials are those which are in the kernel of all Dunkl operators D yi y j for all i, j.

30 Baby Verma Modules By M t,c denote the Verma module k[x 1,..., x n ]/x x n ) with a standard structure of H t,c n) representation

31 Baby Verma Modules By M t,c denote the Verma module k[x 1,..., x n ]/x x n ) with a standard structure of H t,c n) representation Ideal of symmetric polynomials is a subrepresentation

32 Baby Verma Modules By M t,c denote the Verma module k[x 1,..., x n ]/x x n ) with a standard structure of H t,c n) representation Ideal of symmetric polynomials is a subrepresentation Denote by N t,c the quotient by this subrepresentation, which is the baby Verma module

33 Contravariant form The contravariant form B : Sh Sh k is defined by B1, 1) = 1 and for y h, x h, g Sh, f Sh, then Byg, f ) = Bg, D y f )) and Bg, xf ) = BD x g), f ). The kernel of B is given by x Sh such that for all y Sh, then By, x) = 0.

34 Contravariant form The contravariant form B : Sh Sh k is defined by B1, 1) = 1 and for y h, x h, g Sh, f Sh, then Byg, f ) = Bg, D y f )) and Bg, xf ) = BD x g), f ). The kernel of B is given by x Sh such that for all y Sh, then By, x) = 0. The kernel is a subrepresentation Define L t,c = M t,c /kerb L t,c = N t,c /kerb L is an irreducible representation of H t,c

35 Purpose To find the Hilbert polynomial of the irreducible quotient L t,c in the polynomial representation of the rational Cherednik algebra of type A n, when the characteristic p n. The singular polynomials generate a subrepresentation so we would like to find them and remove them.

36 Goal To find the smallest d such that degree d polynomials in the simultaneous kernel of the Dunkl operators D yi y j exist, and find the dimension of this kernel.

37 Past Results: Balagovic/Chen Balagovic and Chen showed that if no singular polynomials existed, the Hilbert polynomial of N t,c τ) is as follows:

38 Past Results: Balagovic/Chen Balagovic and Chen showed that if no singular polynomials existed, the Hilbert polynomial of N t,c τ) is as follows: t = 1 = h N1,c τ)z) = 1 z 2p ) 1 z 3p) 1 z np ) 1 z) n 1,

39 Past Results: Balagovic/Chen Balagovic and Chen showed that if no singular polynomials existed, the Hilbert polynomial of N t,c τ) is as follows: t = 1 = h N1,c τ)z) = 1 z 2p ) 1 z 3p) 1 z np ) 1 z) n 1, t = 0 = h N0,c τ)z) = 1 z 2 ) 1 z 3) 1 z n ) 1 z) n 1.

40 Past Results: Balagovic/Chen Balagovic and Chen showed that if no singular polynomials existed, the Hilbert polynomial of N t,c τ) is as follows: t = 1 = h N1,c τ)z) = 1 z 2p ) 1 z 3p) 1 z np ) 1 z) n 1, t = 0 = h N0,c τ)z) = 1 z 2 ) 1 z 3) 1 z n ) 1 z) n 1. They showed that ) 1 z p n 1 h Lt,cτ)z) = h z p ) 1 z for some polynomial h with integer coefficients.

41 Past Results: Balagovic/Chen Balagovic and Chen showed that if no singular polynomials existed, the Hilbert polynomial of N t,c τ) is as follows: t = 1 = h N1,c τ)z) = 1 z 2p ) 1 z 3p) 1 z np ) 1 z) n 1, t = 0 = h N0,c τ)z) = 1 z 2 ) 1 z 3) 1 z n ) 1 z) n 1. They showed that ) 1 z p n 1 h Lt,cτ)z) = h z p ) 1 z for some polynomial h with integer coefficients. They also proved that kerb is a maximal proper submodule of the Verma module M t,c τ), and that L t,c τ) is irreducible.

42 Methods We wrote a program in Sage to compute the dimensions of the subspaces for various p, n in L t,c

43 Methods We wrote a program in Sage to compute the dimensions of the subspaces for various p, n in L t,c We compared the dimension to those predicted by Balagovic/Chen for N t,c to find existence of singular polynomials

44 Methods We wrote a program in Sage to compute the dimensions of the subspaces for various p, n in L t,c We compared the dimension to those predicted by Balagovic/Chen for N t,c to find existence of singular polynomials We computed these singular polynomials

45 Methods We wrote a program in Sage to compute the dimensions of the subspaces for various p, n in L t,c We compared the dimension to those predicted by Balagovic/Chen for N t,c to find existence of singular polynomials We computed these singular polynomials We conjectured a pattern and looked to prove it

46 Current Progress for t = 0 For p n: The singular polynomials are x i for i = 1, 2,..., n The Hilbert polynomial is 1

47 Current Progress for t = 0 For p n: The singular polynomials are x i for i = 1, 2,..., n The Hilbert polynomial is 1 The case t = 1 and p n was done by Devadas and Sun.

48 Progress for p = 2 and t = 0 For p = 2, the following polynomials are singular for distinct i, j, k: x 2 i + x i x j + x 2 j x i x j + x j x k + x k x i

49 Progress for p n 1 and t = 0 For p odd and distinct i, j, k, l, the following polynomials are singular: x j + x k )x i x j x k ) x i x j )x k x l )

50 Conjectures The Hilbert polynomial for p = 2 and t = 0 is h L0,c z) = 1 + n 1)z + n 1)z 2 + z 3 Etingof conjectures that for n = kp + r, then h L0,c z) = [r] z![p] z Q r n, z) and h L1,c z) = [r] z p![p] z p[p] n 1 z Q r n, z p ), for Q r n, z) = ) n 1 r 1 z r+1 + r ) i=0 z i, n r 2+i i [k] z! = [k] z [k 1] z [1] z, and [w] z = 1 zw 1 z

51 Future Work In the future, we would like to find the Hilbert polynomials for L t,c, and the singular polynomials for various p, n. We would like to study more cases in t = 0 and prove irreducibility, then consider the connection between t = 0 and t = 1.

52 Acknowledgements I would like to thank: My parents My mentor, Daniil Kalinov Dr. Pavel Etingof Dr. Tanya Khovanova The MIT Math Department The MIT PRIMES program Sheela Devadas, Yi Sun, Martina Balagovic, Harrison Chen