Quadratic transformations and the interpolation kernel

Transcription

1 Quadratic transformations and the interpolation kernel Eric M. Rains Department of Mathematics Caltech Elliptic integrable systems and hypergeometric functions Lorentz Center, Leiden July 15, 2013 Partially supported by NSF

2 Several purposes of talk: (1) Explain a new technique for reducing analytic statements about elliptic special functions to algebraic statements (2) As an application, prove things about the interpolation kernel (an analytic continuation of the interpolation functions) (3) Prove a number of multivariate quadratic transformations. (Original abstract predates (2) and the full version of (3), hope you won t mind!) 1

3 Notation reminder: Γ p,q (x) := 0 i,j 1 pi+1 q j+1 /x 1 p i q j. x Warning: p and q will vary considerably in some of the identities, so no suppression of subscripts! Elliptic Dixon density: (n) D (z 1,..., z n ; u 0,..., u 2n+2m+3 ; p, q) ((p; p)(q; q)/2)n 1 i n 0 r<2n+2m+4 Γ p,q (u r z := i ±1 ) n! 1 i<j n Γ p,q (z i ±1 z j ±1 ) 1 i n Γ p,q (z i ±2 ) dt, where dt is Haar measure on unit torus (but extended to contour integral). Elliptic Selberg density: (n) S (z 1,..., z n ; u 0,..., u 2m+5 ; t; p, q) (n) D (z 1,..., z n ; u 0,..., u 2m+5 ; p, q) = Γ p,q (t) n 1 i<j n Γ p,q (tz ±1 i z ±1 j ) 2

4 Main technical idea: When working with p-elliptic functions (i.e., meromorphic functions on C invariant under multiplication by p), things often remain well-behaved when p 0. So we can expand in a power series there. Two holomorphic functions agree iff their power series agree, and there can be other approaches to understanding the power series. 3

5 Example: The W (E 7 )-symmetry of the order 1 elliptic beta integrals Recall this has 8 parameters multiplying to p 2 q 2. If four parameters remain fixed as p 0, and four are Θ(p 1/2 ), then limit is Askey-Wilson density. Lemma: The normalized integral of a symmetric Laurent polynomial over the Askey-Wilson density is a rational function of the parameters. Proof: Compute the integral by expanding in Askey-Wilson polynomials and taking the constant term! 4

6 Corollary: The power (Puiseux) series expansion of the given elliptic beta integral has (essentially) rational function coefficients. (Ratio of integrand to AW integrand is a Puiseux series with symmetric Laurent polynomial coefficients; apply the Lemma) Rational functions agree if they agree on a Zariski dense set, much easier than a set with a limit point. E.g., cases where both sides become Frenkel-Turaev are Zariski dense. Works for multivariate version as well (replace Askey-Wilson by Koornwinder). 5

7 The interpolation kernel Many multivariate identities (including Littlewood identities) involve an interpolation function or two in the integrand. These are products of p-elliptic and q-elliptic functions, and q-elliptic cases mess up the power series. But it turns out there is an analytic continuation with well-behaved power series! This lets us promote identities for p-elliptic interpolation functions to identities for general functions... Note that interpolation functions depend on a pair of partitions (one for each of p- and q-elliptic pieces). 6

8 Key property (for this purpose) of interpolation functions: an integral representation. Elliptic Dixon transformation ( Type I ) implies that certain integral operators take elliptic functions to elliptic functions. One of these takes interpolation functions in n 1 variables to interpolation functions in n variables, leaving the partition pair alone. Since adding 1 to every part of one of the partitions just multiplies by a nice factor, can combine. Result is an expression of any n-dimensional interpolation function as an n 1-dimensional integral of n 1-dimensional interpolation functions. Iterate: obtain ( ) n 2 -dimensional integral. Here the partition pair only appears via p λ iq µ i. So can analytically continue! If we take y i = p λ iq µ it n i, then have symmetry y i C/y i for some C; can rescale to make C = 1. 7

9 In this way, we obtain function K (n) c (x 1,..., x n ; y 1,..., y n ; t; p, q) as n(n 1)/2-dimensional integral, s.t. K (n) c (x 1,..., x n ;..., p λ iq µ it n i a,... ; t; p, q) is essentially an interpolation function (for any a). (Dependence on x of correction factor is a pair of elliptic Gamma functions.) Recursive formula is: K (n) c ( x; y; t; p, q) = 1 i n Γ p,q (cx ±1 i y n ±1) Γ p,q (c 2 )Γ p,q (t) n 1 i<j n Γ p,q (tx ±1 K (n 1) t 1/2 c ( z; y 1,..., y n 1 ; t; p, q) i x ±1 j ) (n 1) D ( z; (pq/t 1/2 c)y ±1 n, t 1/2 x ±1 1,..., t1/2 x ±1 n ; p, q) 8

10 Key observation: If c p α for 0 < α < 1/2, then for a big open subset of parameters, all contours are unit circles for p 0. Each limiting integral operator takes symmetric Laurent polynomials to symmetric Laurent polynomials. Corollary: If c p α for 0 < α < 1/2, then the coefficients of the Puiseux series expansion of K c (n) ( x; y; t; p, q) are polynomials. Remains true with slight modification for c p 1/2. 9

11 This implies a number of properties: A big one is symmetry: Proof: Check this for K c (n) ( x; y; t; p, q) = K c (n) ( y; x; t; p, q) x = q λ 1t n 1 a,..., q λn a y = q µ 1t n 1 b,..., q µn b where it s a statement about p-elliptic interpolation functions. Implies all Puiseux series coefficients agree! No idea how to prove this except via formal limit. Symmetry is pretty powerful: in particular, most potential poles from integral representation violate symmetry, so go away! 10

12 Also have generalized Cauchy identity: along the lines K c (n) ( x; y; t; p, q) λ an explicit expansion C λ R (n) λ ( x; t 0, c/t n 1 u 0 ; q, t; p) R (n) λ ( y; u 0, c/t n 1 t 0 ; q, t; p) (with both t 0, u 0 freely chosen) Note this is not a terminating sum! How to interpret this: C λ p β λ some β > 0. So the right-hand side converges in the ring of formal power series! This formal sum then gives the Puiseux series expansion of kernel. (Proof: if y i = t n i q µ iu 0, then both sides are the same interpolation function, and this is Zariski dense.) 11

13 Why kernel? Because it s the kernel of an integral operator acting nicely (as generalized eigenoperator) on interpolation functions. Turns out to specialize to several known operators: c 1 : identity c q 1/2 : simplest q-difference operators c p 1/2 : simplest p-difference operators c p l/2 q m/2 : iterated difference operators c = t 1/2 : known integral operator Also c = (pq/t) 1/2 : new integral operator relating to an identity proved by van de Bult. (Can evaluate kernel in this case using elliptic Cauchy identity.) 12

14 Comparing c = t 1/2 and c = (pq/t) 1/2 has a funny (and surprisingly useful!) consequence: it implies that K c (n) ( x, y; pq/t; p, q) = Γ p,q (t) 2n Γ p,q (tx ±1 i x ±1 j, ty i ±1 y j ±1 ) i<j K c (n) ( x, y; t; p, q). Not clear if there s any way to make sense of this symmetry below elliptic level, or even at the formal level... 13

15 Also have analogue of Jackson summation ( braid relation ): ( z; x; t; p, q)k (n) ( z; y; t; p, q) (n) S ( z; t 0, t 1 ; t; p, q) K c (n) = 1 i n d Γ p,q (ct 0 x ±1 i, ct 1 x ±1 i, dt 0 y ±1 i, dt 1 y ±1 i ) K (n) cd ( x; y; t; p, q) where t 0 t 1 = pq/c 2 d 2. (Proof: when y is a partition, this is the integral equation) When d = c 1, this says (ignoring the fact that it diverges!) that K c (n) and K (n) c 1 are inverses. A careful limit gives a sort of Bailey Lemma. 14

16 Why braid relation? Apart from ABA=BAB structure, turns out integral operators fit into a sort of twisted representation of Coxeter group of type W (E n ); this is the only nontrivial braid relation. Interesting consequence: Since difference operators are limits of integral operators, the kernel satisfies difference equations (from braid or other identities). Can show (outside scope of talk) that this extends to a generalized Fourier transform preserving a nice class of operators. (Ask me about noncommutative geometry and elliptic Painlevé! ) Example: This (algebraically) extends work of Ruijsenaars on E 8 symmetries of relativistic Heun equation to full van Diejen integrable system. But not right now! 15

17 Quadratic transformations Have following admittedly bizarre evaluation: K (2n) (pqt) 1/4(..., t±1/2 z i,... ; y; t; p, q) (n) S ( z; t1/2 v ±1 ; t 2 ; p, q) = 1 i 2n Γ p,q ((pqt) 1/4 v ±1 y ±1 i ) Γ p,q ((pq/t) 1/2 ) 2n 1 i<j 2n Γ p,q ((pq/t) 1/2 y ±1 i y ±1 j ) (Proof by induction in n using the integral representation & known symmetries) Note integral is over t 2 Selberg density... Corresponds to formal nonterminating version of elliptic Littlewood identity. 16

18 Since the integrals are Koornwinder in the limit, we not only have polynomial coefficients, but can analytically continue in t n (uses symmetric function version of the formal kernel!), and apply Macdonald involution. This gives analogue of dual Littlewood identity. Can also obtain an elliptic version of Kawanaka identity as follows: Take the integral dual Littlewood identity and swap p and q. This does not behave well formally, but does specialize to an algebraic statement as (the new) p 0. This statement treats p and 1 differently, so is not modular invariant; using this gives another algebraic identity, and the corresponding integral does behave well, so we can extend back to the integral by comparing formal series. 17

19 Taking t pq/t in the dual Littlewood identity and using the braid relation to change c gives quadratic transformation (essentially Conjecture Q3 of my Elliptic Littlewood identities ): K (n) t 1/4 ( x; y; t; p, q) c (n) S ( x; t 1/4 t 1, t 1/4 t 2, ±t 1/4, ±p 1/2 t 1/4, t 1/4 v 0, t 1/4 v 1 ; t 1/2 ; p, q) Γ p,q (t 1/2 ct 1 y i ±1, t 1/2 ct 2 y i ±1 ) = 1 i n K (n) q 1/2 ( x; y; t; p, q) (n) c S ( x; q±1/2 t 1, q ±1/2 t 2, q 1/2 v 0, q 1/2 v 1 ; t; p, q 2 ) (with suitable balancing conditions.) Even nontrivial (but proved by van de Bult) when y is a geometric progression: kernel becomes product of Gamma functions. 18

20 All told, total of 16 quadratic transformations arise using these symmetries (and another 4 follow from known algebraic statements via the symmetries). Several are new even in Macdonald/Koornwinder limit. Favorite example (Q5): if t 2n 1 t 0 t 1 t 2 t 3 = p 1/2 q 2, then (2n) S (; q ±1/2 t 0, q ±1/2 t 1, q ±1/2 t 2, q ±1/2 u 0, pqv ±1 ; t; p, q 2 ) (n) S (; t 0, t 1, t 2, u 0, ±1, ±t 1/2, p 1/2 q 1/2 v ±1 ; t; p 1/2, q) + Γ p,q 2(t 2 0, t2 1, t2 2, u2 0, t, t, pqv±2 )Γ p 1/2 ( 1, t),q (n 1) S (; t 0, t 1, t 2, u 0, ±t, ±t 1/2, p 1/2 q 1/2 v ±1 ; t; p 1/2, q). Limiting case of full kernel version is fact that O O(2n) s λ(o) = δ λi mod 2=0. 1 i 2n 19

21 Why two integrals here? This arose via the Macdonald involution (from Q3), so we needed to analytically continue in the dimension. But trying to go back to a finite integral gives a 0/0 issue; the second integral gives correction term. Another subtlety is that there s an extra symmetry after analytically continuing in t n which lets us deal with case LHS is odd-dimensional as well. 20

22 Other things to note: (1) Given the kernel (and reasonable control over its singularities), it is completely straightforward to write down multivariate elliptic analogue of Askey-Wilson function. (This needs a name!) (2) When t = q, the kernel is a determinant; when c = p/t, it is (essentially) independent of q. (Use the braid relation to write it as a difference operator applied to the pq/t case.) Combining gives a determinantal representation for K (n) p/t. Comparing to the generalized Cauchy expansion shows this generalizes an Cauchy-like identity of Warnaar at the Macdonald level, with RHS the Izergin-Korepin determinant (for the six vertex partition function). (Related to an SOS-type partition function studied by Filali.) 21