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$\DeclareMathOperator\Rep{Rep}$It is well-known that the Macdonald "$P$" polynomials deform the Jack "$J$" polynomials [1]. The latter have profound relations with representation theory. For example, setting $\alpha=1$ on $J$ recovers the Schur polynomials; setting $\alpha=2$ on $J$ recovers the zonal polynomials (related to Gelfand pairs, Selberg $\alpha$-integral…); setting $\alpha=(-1)$ almost recovers the monomial symmetric polynomial. Given such results, it is clear that Macdonald "$P$" polynomials must also have profound meaning in representation theory. Indeed, this has been long pursued using several methods: DAHA, equivariant K-theory, geometric of Hilbert schemes, generalized intertwiners. Some combinatorial interpretations have also been found (c.f. Ram and Yip - A combinatorial formula for Macdonald polynomials, ASEP, etc).

In this thread I'm mainly interested in seeing a more direct relation with the symmetric groups $S_{d}$. In short, how do Macdonald polynomials hint about $(q,t)$-deforming the symmetric groups $S_{d}$ (or their group algebras, or their representation categories…)?

Frobenius Formula and Symmetric Functions

Let us start by recalling the relation between the symmetric groups and the Jack polynomials. Let $V_{\lambda}$ be the irreducible representation of $S_{d}$ corresponding to the partition $\lambda$. This induces a character $\chi^{\lambda}$ whose values are of interest. Since characters are class functions, it is enough to look at its values on the set of conjugacy classes of $S_{d}$, which is parametrized by partitions

$$\mu = (d,d,\dotsc,d,d-1,d-1,\dotsc,2,1,1,\dotsc,1),$$

with $\sum_{i} \mu_{i} = d$. The answer is beautifully given by the Frobenius formula [2]. Here's a concise formulation:

$$p_{\mu} = \sum_{\lambda} \chi^{\lambda}({\mu}) s_{\lambda},$$

where $p$ denotes the power-sum symmetric polynomials and $s$ denotes the Schur symmetric polynomials. In other words, the transformation matrices between the power-sum symmetric functions and the Schur symmetric functions are given by the characters in $\Rep(S_{\infty})$!

Jack Polynomials and $\alpha$-deformation

Now let's recall how Jack polynomials are defined in [1]. Construct an inner product $(,)$ on a suitable ring such that

$$(s_{\lambda}, s_{\mu}) = \delta_{\lambda,\mu}.$$

Then the $p$'s also form an orthogonal basis: $$(p_{\lambda}, p_{\mu}) = \delta_{\lambda,\mu} z_{\lambda}$$ for some scalar $z_{\lambda}$. Moreover, the Frobenius formula says that the character values are given by $(p_{\mu}, s_{\lambda})$.

Now, deforming the inner product to $(,)_{\alpha}$ by requiring

$$(p_{\lambda}, p_{\mu})_{\alpha} = \delta_{\lambda,\mu} z_{\lambda} \alpha^{l(\lambda)},$$

we then have the Jack polynomials as a certain orthogonal basis under this deformed pairing.

Question

  1. By the Frobenius formula and the definition of the Jack's polynomial, it seems that there should be a way to $\alpha$-deform $\Rep(S_{\infty})$ such that the character values are given by $(p_{\mu}, s_{\lambda})_{\alpha}$.

  2. Can one deform $\Rep(S_{\infty})$ further so that the character values are given by $(p_{\mu}, s_{\lambda})_{(q,t)}$, where the pairing (this gives the Macdonald functions) is given by deforming $\alpha^{l(\lambda)}$ to $\prod_{i=1}^{\lambda} \frac{1-q^{\lambda_{i}}}{1-t^{\lambda_{i}}}$. This question is asked essentially in [3].

Reference

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  • $\begingroup$ Given an answer to (1) -- which I don't have -- I would guess that the answer to (2) might be its analogue for the Hecke algebra (with parameter $t$): the Jack limit of a Macdonald polynomial is (e.g.) obtained by setting $q=t^\alpha$ and letting $t\to 1$. [Note that Macdonald uses $P$ for the monic convention of both Jack and Macdonald polynomials, $P_\lamdba^{(\alpha)}(x)$ and $P_\lamdba(x;q,t)$ resp, reserving the symbol $J$ for another (maybe integral?) normalisation.] $\endgroup$
    – Jules Lamers
    Commented Dec 14, 2021 at 23:45
  • $\begingroup$ @JulesLamers What is the Hecke algebra with parameter $t$? I only know one with parameter $q$. But notably the Schur functions do not seem to admit pure $q$ deformation - one needs a nontrivial $\alpha$ for the $q$-deformation to be nondegenerate. Put it in another way: for $q=1$, the Macdonald "$P$" function is Schur immediately. $\endgroup$
    – Student
    Commented Dec 15, 2021 at 3:43
  • $\begingroup$ It's the same, but Macdonald denotes the parameter $q_{Hecke}$ by $t^{1/2}$. For any $q=t$ Macdonald reduces to Schur, but I don't think it does if $q=1$ but $t$ is not. $\endgroup$
    – Jules Lamers
    Commented Dec 15, 2021 at 7:13

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