Outline 1. Background on Symmetric Polynomials 2. Algebraic definition of (modified) Macdonald polynomials 3. New combinatorial definition of Macdonal

Transcription

1 Algebraic and Combinatorial Macdonald Polynomials Nick Loehr AIM Workshop on Generalized Kostka Polynomials July 2005 Reference: A Combinatorial Formula for Macdonald Polynomials" by Haglund, Haiman, and Loehr, JAMS 18 (2005),

2 Outline 1. Background on Symmetric Polynomials 2. Algebraic definition of (modified) Macdonald polynomials 3. New combinatorial definition of Macdonald polynomials 4. Proof that the two definitions agree 5. q; t-kostka polynomials 2

3 Symmetric Polynomials Let n be a positive integer. Suppose: ffl P is a polynomial in x 1 ;:::;x n. ffl P is homogeneous of degree n. ffl Permuting the subscripts of the x i 's always leaves P unchanged. Then P is a symmetric polynomial of order n. Example: (n = 3) P = 5x x x 3 3 (1=2)x 1x 2 x 3 : Fact: V n = fsymm. polys. of order ng is a vector space. 3

4 Partitions A partition of n is a list of positive integers μ = (μ 1 ;:::;μ n ) with μ 1 μ n 0 and μ μ n = n: Notation: μ ` n means μ is a partition of n. Example: There are 5 partitions of n = 4: (4; 0; 0; 0); (3; 1; 0; 0); (2; 2; 0; 0); (2; 1; 1; 0); (1; 1; 1; 1): Diagram of μ: Put μ i boxes in row i. 4

5 Bases for Symmetric Polys. Fact: dim(v n ) = number of partitions of n. We use partitions to index bases of V n. The six classical bases of V n : ffl monomial basis fm μ : μ ` ng ffl elementary basis fe μ : μ ` ng ffl homogeneous basis fh μ : μ ` ng ffl power-sum basis fp μ : μ ` ng ffl Schur basis fs μ : μ ` ng ffl forgotten basis ff μ : μ ` ng 5

6 Modern Bases for V n ffl Zonal symmetric polys. fz μ : μ ` ng ffl Jack's symmetric polys. ffl Hall-Littlewood basis ffl original Macdonald basis fp μ : μ ` ng ffl dual Macdonald basis fq μ : μ ` ng ffl integral Macdonald basis fj μ : μ ` ng ffl transformed Macdonald basis fh μ : μ ` ng ffl modified Macdonald basis f ~ Hμ : μ ` ng 6

7 The Monomial Basis For μ ` n, the monomial symmetric poly. m μ is the sum of all distinct monomials obtained by permuting the subscripts of the monomial x μ 1 1 xμ 2 2 xμ n n : Fact: fm μ : μ ` ng are linearly independent polynomials that span V n. They form the monomial basis for V n. Example. For n = 3, m (3;0;0) = x x x 3 3 m (2;1;0) = x 2 1 x 2 + x 2 1 x 3 + x 2 2 x 1 m (1;1;1) = x 1 x 2 x 3 +x 2 2 x 3 + x 2 3 x 1 + x 2 3 x 2 7

8 The Power-Sum Basis For k > 0, the k'th power-sum is p k (x 1 ;:::;x n ) = x k 1 + x k x k n : Define p 0 = 1. For μ ` n, define the power-sum symmetric poly. p μ to be p μ = n Y p μi (x 1 ;:::;x n ): i=1 Fact: fp μ : μ ` ng is a basis for V n. Example: For n = 3, p (3;0;0) = x x x 3 3 = m (3;0;0) p (2;1;0) = (x x x 2 3)(x 1 + x 2 + x 3 ) = m (3;0;0) + m (2;1;0) p (1;1;1) = (x 1 + x 2 + x 3 ) 3 = m (3;0;0) + 3m (2;1;0) + 6m (1;1;1) 8

9 The Parameters q and t Macdonald polynomials involve variables x 1 ;:::;x n and two extra parameters q and t. Formally, let F = Q (q; t) be the field whose elements are formal quotients of polynomials in two variables q and t. Examples: 4, 3t 1, +5 q 2 (3=7)t lie in F. From now on, view V n as a vector space over this field. Example: (3t 1)x (3t 1)x q 2 (3=7)t x 1x 2 2 V 2. 9

10 Two Special Linear Maps We can define linear maps on the vector space V n by specifying their effect on any basis. Define: A t (p μ ) = A q (p μ ) = Y i:μ i >0 Y i:μ i >0 [t μ i 1] [q μ i 1] 1 A pμ : 1 A pμ ; Note: Terms in parentheses are elements of F (scalars)! Extending by linearity, we get two linear maps A q and A t mapping V n into itself. 10

11 Conjugation; Domination Conjugate of μ: μ 0 is the partition whose parts are the columns in the diagram of μ. conjugate Dominance partial ordering: For ; μ ` n, μ μ iff i» μ μ i for all i. Example: (1; 1; 1) μ (2; 1; 0) μ (3; 0; 0). But (3; 1; 1; 1; 0; 0) 6μ (2; 2; 2; 0; 0; 0), (2; 2; 2; 0; 0; 0) 6μ (3; 1; 1; 1; 0; 0). 11

12 Algebraic Definition of Modified Macdonald Polys. Def./Thm. There exists a unique basis f ~ Hμ : μ ` ng of V n satisfying these axioms: (1) The coefficient of x n 1 in ~ Hμ is 1. (2) Let A t ( ~ Hμ ) = P `n c ;μ m (c ;μ 2 F). Then c ;μ = 0 except when μ μ. (3) Let A q ( ~ Hμ ) = P `n d ;μ m (d ;μ 2 F). Then d ;μ = 0 except when μ μ 0. The ~ Hμ 's are the modified Macdonald polynomials. 12

13 Comments/Complaints The algebraic definition for ~ Hμ just given: ffl requires a hard proof to justify (uniqueness easy, but existence unclear!) ffl is completely non-explicit ffl seems totally unmotivated ffl gives us no intuition about ~ Hμ Yet, this definition was the only one available for the last 16 years! ( ) 13

14 The New Definition We're about to give Haglund's conjectured combinatorial definition for ~ Hμ, which: ffl proves the existence claim in the earlier definition by giving a construction for ~ Hμ ffl is an explicit sum of weighted combinatorial objects ffl shows that ~ Hμ is in N [q; t][x 1 ;:::;x n ], not just in Q (q; t)[x 1 ;:::;x n ]. ffl has intuitive appeal due to its concreteness and simplicity ffl exhibits the combinatorial significance of the cryptic algebraic axioms defining ~ Hμ ffl leads to elegant proofs of results on Jack's polys., Hall-Littlewood polys., etc. 14

15 Combinatorial Definition Haglund's combinatorial formula: ~H μ = X objects T q qwt(t ) t twt(t ) ~x xwt(t ) where the objects and weights depend on μ. The objects: all fillings of the boxes of μ with integers from 1 to n, repeats allowed. The ~x-weight of T : xwt(t) = # of 1's in T x 1 x # of n's in T n : Example: n = 10, μ = (3; 3; 3; 1; 0;:::;0). T = xwt(t) = x 1 1 x4 2 x1 3 x2 4 x1 5 x1 9 : 15

16 Major Index and Inversions Let w = w 1 ;w 2 ;:::;w s be a list of integers. The major index maj(w) is the sum of all i < s such that w i > w i+1. The inversions of w, inv(w), is the number of pairs i < j with w i > w j. Examples: maj(4; 2; 2; 3) = 1; inv(4; 2; 2; 3) = 3 maj(5; 4; 1) = 3; inv(5; 4; 1) = 3 maj(2; 9; 2) = 2; inv(2; 9; 2) = 1 16

17 The t-weight Given an object T, let w (j) be the list of integers in column j of μ, from top to bottom. Define X twt(t) = maj μ (T) = μ 1 maj(w (j) ): j=1 Example: T = maj μ (T) = = 6: 17

18 Inversion Triples Consider a configuration of cells in T like this: y z x These three cells form an inversion triple of T iff x < y» z or y» z < x or z < x < y: 18

19 The q-weight Given an object T, let w (0) be the list of integers in the lowest row of μ, from left to right. Suppose T has K inversion triples. Define qwt(t) = inv μ (T) = inv(w (0) ) + K: Example: T = inv μ (T) = = 5: Full weight of T: q 5 t 6 x 1 1 x4 2 x1 3 x2 4 x1 5 x

20 Example: ~ H(2;1;0) x 1 x 1 x x x 2 x 2 x 3 x 3 x 3 x 1 x 1 x 2 x 1 x 1 x x x 2 x 3 x 1 x 2 x x q x 1 x 2 x 1 x 1 x 3 x 1 x 1 x 3 t x 1 x 1 x 3 x 2 x 2 x x x 2 x 3 x 2 x 2 x 3 x 1 x 2 x 2 x 1 x 2 x 2 x 1 x 2 x x x 3 x 3 x 1 x 3 x 3 x 1 x 3 x 3 x 2 x 3 x 3 x 2 x 3 x x x 3 x 3 x 1 x 2 x 3 x 1 x 2 x 3 x 1 x 2 x 3 x 1 x 2 x 3 ~H (2;1;0) = 1m (3;0;0) + (1 + q + t)m (2;1;0) +(1 + 2q + 2t + )m (1;1;1) : 20

21 Steps in the Proof Let C μ denote Haglund's formula for ~ Hμ. 1. Show the coefficient of x n 1 in C μ is Prove C μ is symmetric (i.e., C μ 2 V n ). 3. Interpret A t (C μ ) and A q (C μ ) as sums of signed, weighted objects. 4. A t (C μ ) = X a ;μ m ; A q (C μ ) = X b ;μ m : `n `n Use cancellation of objects to show a ;μ 6= 0 ) μ μ; b ;μ 6= 0 ) μ μ 0 : 5. C μ satisfies all axioms, so C μ = ~ Hμ. 21

22 Interpreting A t (C μ ) One can prove that A t (C μ ) is a sum of signed, weighted objects: A t (C μ ) = X objects T sgn(t)q qwt(t ) t twt(t ) ~x xwt(t ) The objects: fillings of μ using the alphabet f1; 2;:::;n;1; 2;:::;ng consisting of positive and negative letters. The ~x-weight: Q n # of i's and i's in T i=1 x i : Example: n = 10, μ = (3; 3; 3; 1; 0;:::;0). T = xwt(t) = x 1 1 x4 2 x1 3 x2 4 x1 5 x1 9 : 22

23 Signs and Weights Consider an object T with P positive letters and N negative letters. q-weight: qwt(t) = inv μ (T). t-weight: twt(t) = maj μ (T) + P. Sign: sgn(t) = ( 1) N. Example: T = Full weight of T is ( 1) 5 q 2 t 8+5 x 1 1 x4 2 x1 3 x2 4 x1 5 x

24 Cancelling Pairs of Objects Idea: Cancel pairs of terms in A t (C μ ) = X sgn(t)q qwt(t ) t twt(t ) ~x xwt(t ) objects T with equal weights and opposite signs. Example: T = T contributes the term q 2 t 8+5 x 1 1 x4 2 x1 3 x2 4 x1 5 x1 9. U = U contributes the term +q 2 t 9+4 x 1 1 x4 2 x1 3 x2 4 x1 5 x1 9. Terms for T and U cancel in A t (C μ )!!! 24

25 Finding Matched Pairs To cancel an object T: ffl Choose i minimal y such that i or i appears above the lowest i rows in T. ffl Find the topmost and then leftmost occurrence of i or i in T. ffl Flip the sign of this symbol to get U. ffl Check: sign reverses, but weights are preserved! y If no such i exists, then T contributes an uncancelled term to A t (C μ ). 25

26 Proving Axiom 2 for C μ A t (C μ ) = X `n a ;μ m. Show: a ;μ 6= 0 ) μ μ. 1. a ;μ 6= 0 ) the coefficient of x 1 X A t (C μ ) = 1 x n n sgn(t )q qwt(t ) t twt(t ) ~x xwt(t ) in is nonzero. objects T 2. So, there must be an uncancelled object T with xwt(t) = x 1 1 x n n. 3. For each i, the number of i's and i's in T is exactly i. 4. For each i, letters in f1;:::;i;1;:::;ig occur in the lowest i rows of μ. 5. So i» μ μ i 8i; which says that μ μ! 26

27 Interpreting A q (C μ ) As with A t (C μ ), we can prove that A q (C μ ) = X sgn(t)q qwt(t ) t twt(t ) ~x xwt(t ) objects T Objects: fillings of μ with entries from f1;:::; n; 1; 2;:::; ng. ~x-weight: Q n # of i's and i's in T i=1 x i : Sign: ( 1) # of negative letters in T. q-weight: inv μ (T)+ (# of positive letters in T). t-weight: maj μ (T). 27

28 Cancelling Pairs of Objects Idea: Cancel pairs of terms in A q (C μ ) with equal weights and opposite signs. Example: T = Term for T is ( 1) 5 q 6+5 t 6 x 1 1 x4 2 x1 3 x2 4 x1 5 x1 9. V = Term for V is ( 1) 4 q 5+6 t 6 x 1 1 x4 2 x1 3 x2 4 x1 5 x1 9. Terms for T and V cancel in A q (C μ )!!! As before, not every object T can cancel. 28

29 Proving Axiom 3 for C μ A q (C μ ) = X `n b ;μ m. Show: b ;μ 6= 0 ) μ μ b ;μ 6= 0 ) there is an uncancelled object T with xwt(t) = x 1 1 x n n. 2. For each i, the number of i's and i's in T is exactly i. 3. The new cancellation mechanism implies that, for all i, T never has two letters in fi; ig in the same row. 4. The last condition easily implies i» μ μ 0 i 8i; so that μ μ 0! 29

30 Applications of New Formula ffl best proof of existence, integrality of ~ Hμ ffl explicit combinatorial formulas for all five Macdonald bases (P μ, Q μ, J μ, H μ, ~ Hμ ) ffl explanation and proof of the Lascoux-Schützenberger cocharge statistic for Hall-Littlewood polynomials ffl simple proof of Sahi and Knop's formula for Jack polynomials ffl expansion of ~ Hμ using LLT polynomials ffl insight into Kostka-Macdonald coefficients ~ K ;μ 30

31 Monomial Expansion Expand Macdonald polynomials in terms of monomial symmetric functions: ~H μ = X a ;μ m (a ;μ 2 Q (q; t)) Lemma: The combinatorial formula for ~ Hμ is symmetric in the x i 's. Corollary: If (c 1 ;:::;c N ) is any sequence that rearranges to the partition, then X a ;μ = T :μ![n] jt 1 (fig)j=c i q inv(t ) t maj(t ) : 31

32 Schur Expansion Now expand Macdonald polynomials in terms of Schur functions: ~H μ = X ~K ;μ s The scalars ~ K ;μ 2 Q (q; t) are the (modified) q; t-kostka polynomials. Theorem: [positivity and polynomiality] ~K ;μ 2 N [q; t]: Open Problem: Find a combinatorial formula for the q; t-kostka polynomials. 32