Sign up ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

I have been recently learning a lot about Macdonald polynomials, which have been shown to have probabilistic interpretations, more precisely the eigenfunctions of certain Markov chains on the symmetric group.

To make this post more educational, I will define these polynomials a bit. Consider the 2-parameter family of Macdonald operators (indexed by powers of the indeterminate $X$) for root system $A_n$, on a symmetric polynomial $f$ with $x = (x_1, \ldots, x_n)$:

$$D(X;t,q) = a_\delta(x)^{-1} \sum_{\sigma \in S_n} \epsilon(\sigma) x^{\sigma \delta}\prod_{i=1}^n (1 + X t^{(\sigma \delta)_i} T_i),$$

(mathoverflow doesn't seem to parse $T_{q,x_i}$ in the formula above, so I had to use the shorter symbol $T_i$, which depends on q).

where $\delta$ is the partition $(n-1,n-2,\ldots, 1,0)$, $a_\delta(x) = \prod_{1 \le i < j \le n} (x_i - x_j)$ is the Vandermonde determinant (in general $a_\lambda(x)$ is the determinant of the matrix $(a_i^{\lambda_j})_{i,j \in [n]}$).

$x^{\sigma \delta}$ means $x_1^{(\sigma \delta)_1} x_2^{(\sigma \delta)_2} \ldots x_n^{(\sigma \delta)_n}$.

Also $(\sigma \delta)_i$ denotes the $\sigma(i)$-th component of $\delta$, namely $n-i$.

Finally the translation operator $T_i = T_{q,x_i}$ is defined as $$ T_{q,x_i}f(x_1, \ldots, x_n) = f(x_1, \ldots, x_{i-1}, q x_i , x_{i+1} ,\ldots, x_n).$$

I like to think of the translation operator as the quantized version of the differential operator $I + \partial_i$, where $q-1$ is analogous to the Planck constant(?).

If we write $D(X;q,t) = \sum_{r=0}^n D_{n-r}(q,t) X^r$, then Macdonald polynomials $p_\lambda(q,t)$ are simply simultaneous eigenfunctions of these operators. When $q=t$ they become Schur polynomials, defined by $s_\lambda = a_{\delta +\lambda} / a_\delta$. When $q= t^\alpha$ and $t \to 1$, we get Jack symmetric polymomials, which are eigenfunctions of a Metropolis random walk on the set of all partitions that converge to the so-called Ewens sampling measure, which assigns probability proportional $\alpha^{\ell(\lambda)} z_\lambda^{-1}$. When $q = 0$, they become the Hall-Littlewood polynomials and when $t=1$ they become the monomial symmetric polynomials etc.

I was told repeatedly by experts that Macdonald polynomials exhaust all previous symmetric polynomial bases in some sense. Does anyone know a theorem that says that every family of symmetric polynomial under some conditions can be obtained from Macdonald polynomials by specializing the $q$ and $t$?

share|cite|improve this question

3 Answers 3

I don't know this story too well, but at least for type $A_1$ there is a 5-parameter family of symmetric polynomials called Askey-Wilson polynomials that specialize to the Macdonald polynomials. (I'm not sure about the story for higher ranks or other types.) These polynomials are controlled by the representation theory of a 5-parameter double affine Hecke algebra $H_{q,a,b,c,d}$ (introduced by Sahi, I believe). When all the parameters are 1, this algebra is $\mathbb C[X^{\pm 1},Y^{\pm 1}]\rtimes (\mathbb C\mathbb Z_2)$. (The Macdonald polynomials are controlled by Cherednik's double affine Hecke algebra of type $A_1$, which is isomorphic to $H_{q,t,1,1,1}$.)

Oblomkov showed that $H_{q,a,b,c,d}$ is a universal deformation of $\mathbb C[X^{\pm 1}, Y^{\pm 1}]\rtimes (\mathbb C \mathbb Z_2)$, which in some sense shows that the Askey-Wilson polynomials are the largest family of polynomials which have similar properties to Macdonald polynomials.

See and references therein for more details.

share|cite|improve this answer

Another way of seeing Macdonald polynomials is via the following definition (I now take it from Macdonald's lecture series "Symmetric functions and orthogonal polynomaials", Section 1.11, but you find it in various places, like for the definition for any root system of finite type.): The $P_\lambda(q,t)$ are characterized by the two conditions

  1. $P_\lambda = m_\lambda + \sum_{\mu < \lambda} c_{\lambda \mu} m_\mu,$
  2. $\langle P_\lambda,P_\mu \rangle_{q,t} = 0 \text{ if } \lambda \neq \mu$,

where "<" is the dominance order on partitions, and where the scalar product is defined by $$ \langle P_\lambda,P_\mu \rangle_{q,t} = \delta_{\lambda \mu} z_\lambda \prod_{i=1}^{l(\lambda)} \frac{1-q^{\lambda_i}}{1-t^{\lambda_i}}.$$ (for details the the lecture notes above.)

Using this definition, it is not too hard to see that one obtains elementary symmetric functions, monomial symmetric functions, Schur functions, Hall-Littlewood functions, and Jack symmetric functtions for appropriate specializations for $q$ and $t$.

Using this definition (which actually needs a proof as it overdetermines these polynomials), one can somewhat derive the condition for any family of symmetric functions by saying that they need to be triangular when written in the monomial basis, and that they need to behave appropriately with respect to a specialization of $q$ and $t$ in the inner product in the second condition.

I hope that's somewhat giving you a recipe to check if your family can be obtained from the Macdonalds, but I have no idea how hard it might be to check how that inner product looks like on your family of symmetric polynomials.

share|cite|improve this answer
Hi Christian, Thank you for your detailed reply. Yes I was indeed aware of this characterization of Macdonald Polynomials. In fact the Markov chain interpretation of Macdonald polynomials uses this inner product in a crucial way. I guess my original question can be rephrased to mean whether this $(q,t)$-family is in some sense universal. I understand that in the literature people have only succeeded in generalizing MacD polynomials to inhomogeneous or asymmetric directions. So somehow this is the end of the story? But why? Another thought I had was that maybe MacD polynomials were –  John Jiang Feb 12 '12 at 18:53
the only ones that have the restriction property. i,e., $P_\lambda(x_1, \ldots, x_{n-1}, 0) = P_\lambda(x_1,\ldots, x_{n-1})$. Does that characterize them? –  John Jiang Feb 12 '12 at 18:54

There is a result by Sergei Kerov (in his book Asymptotic representation theory of the symmetric group and its applications in analysis) which somewhat charaterizes the Macdonald symmetric functions. Namely, in Chapter 2 he proves the following.


  1. $P_\lambda=m_\lambda+\sum_{\mu<\lambda}c_{\lambda\mu}m_\mu$

  2. $P_\lambda$'s are orthogonal with respect to a scalar product $\langle p_\lambda,p_\mu\rangle =\delta_{\lambda\mu}z_\lambda w_\mu$, where $z_\lambda=1^{m_1}m_1!2^{m_2}m_2!...$, $w_\mu=w_{\mu_1}w_{\mu_2}...$ (for some sequence $w_1,w_2,...$)

  3. $P_\lambda(x_1,x_2)=x_1x_2 P_{\lambda-1}(x_1,x_2)$ (index shift property for two variables)

Then essentially the constants $w_1,w_2,...$ correspond to the Macdonald scalar product (or its variants; e.g., $w_n\equiv 1$ is the Schur case, etc.)

So, from this perspective you cannot invent anything else.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.