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Let $J_\lambda^{(\alpha)}(x)$ be the Jack polynomials in $N$ variables, with a normalization such that the coefficient of the monomial polynomial $m_\lambda$ is equal to 1.

They satisfy the identity $$ \det(x)^s J_\lambda^{(\alpha)}(x^{-1})=J_{\widetilde{\lambda}}^{(\alpha)}(x),$$ where $s$ is a large enough integer and $\widetilde{\lambda}=(s-\lambda_N,s-\lambda_{N-1},...,s-\lambda_1)$.

For a single variable this is obvious, but not for many variables. I saw this in a paper. The paper referenced a book, but in the book this was left as an exercise.

What are ways to prove this identity?

(As specified by Jules Lamers in the comments, $x$ is an invertible $N\times N$ matrix whose eigenvalues are the arguments of the Jack polynomials)

[EDIT]: Jack polynomials in $N$ variables are eigenfunctions of the differential operator $$ D(\alpha)=\frac{\alpha}{2}\sum_{i=1}^Nx_i^2\frac{\partial ^2}{\partial x_i^2}+\sum_{j\neq i}\frac{x_i^2}{x_i-x_j}\frac{\partial}{\partial x_i},$$ the eigenvalue of $J_\lambda^{(\alpha)}(x)$ being $$e_\lambda=\alpha b(\lambda')-b(\lambda)+(N-1)|\lambda|,$$ with $b(\lambda)=\sum_i(i-1)\lambda_i$.

So I guess showing that $\det(x)^s J_\lambda^{(\alpha)}(x^{-1})$ is an eigenfunction of $D(\alpha)$ with eigenvalue $e_{\widetilde{\lambda}}$ would do it.

Another possibility is to work with the expression in terms of tableaux. We know that $J_\lambda^{(\alpha)}(x)=\sum_{T\in S(\lambda)}w_\alpha(T)x^T$, where the sum is over tableaux of shape $\lambda$ and $w_\alpha(T)$ is an appropriate weight. Now $\det(x)^sJ_\lambda^{(\alpha)}(x^{-1})=\sum_{T\in S(\lambda)}w_\alpha(T)x^{s^N-T},$ and then take it from here.

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    $\begingroup$ I don't think that would help. But the book is Log-gases and Random Matrices, by Peter Forrester, and the paper is also by him. $\endgroup$
    – Marcel
    Mar 11, 2022 at 16:35
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    $\begingroup$ Here's a sketch: (Working in type $GL_n$): take the transformation $x^{\alpha} \mapsto x^{-w_0\alpha}$ from $\mathbb{C}[P] \to \mathbb{C}[P]$, and show that the Jack polynomial $J_{\lambda} \mapsto J_{-w_0\lambda}$ via this transformation. (You can use the constant term scalar product for showing this since the kernel $\Delta$(in Macdonald's notation) is $S_n$ invariant. Now use the fact that $J_{\lambda+(1^n)} = x_1...x_nJ_{\lambda}$. Does this help? (continued below for more...) $\endgroup$
    – ArB
    Mar 16, 2022 at 14:12
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    $\begingroup$ ... $w_0$ is the longest permutation that takes dominant weights to antidominant weights... in one line notation $[n , n-1,..., 1]$... $P = \mathbb{Z}^n$ is the weight lattice for $GL_n$ and $\mathbb{C}[P]$ is the group algebra spanned by $x^{\alpha} : \alpha \in P$... $\endgroup$
    – ArB
    Mar 16, 2022 at 14:16
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    $\begingroup$ I am using the scalar product $(f,g) = ct(f\bar{g}\Delta)$ where $\bar{g}(x_1,...,x_n)=g(x_1^{-1},...,x_n^{-1})$ and $\Delta$ is given in chapter VI sec 9 of Macdonald's Symmetric Functions and Hall Polynomials book. This is for the Macdonald $P$ polynomials, and you can in-fact prove your statement for the $P$ polynomials and then take the limit for Jack case. $\endgroup$
    – ArB
    Mar 16, 2022 at 15:01
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    $\begingroup$ The Macdonald $P$ polynomials are the unique orthogonal basis of $\mathbb{C}[P]^W$ for this scalar product with unitriangularity wrt monomial basis . So to show image of $P_{\lambda}$ under the transformation $x^{\alpha} \mapsto x^{-w_0\alpha}$ is $P_{-w_0\lambda}$ you may show that the image satisfies orthogonality and triangularity. $\endgroup$
    – ArB
    Mar 16, 2022 at 15:06

1 Answer 1

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I tried to use the eigenfunction approach, but did not get the desired result. I will post the answer anyway in case someone can help.

The Jack polynomial $J_\lambda(z_1,...,z_N)$ satisfies $$ D(\alpha)J_\lambda(z)=\frac{\alpha}{2}\sum_{i=1}^N z_i^2J_\lambda^{ii}(z)+\sum_{i\neq j}^N\frac{ z_i^2}{z_i-z_j}J_\lambda^{i}(z)=e_\lambda J_\lambda(z),$$ where I omited the superscript $(\alpha)$ and denoted $J^{i}$ the derivative with respect to the $i$th argument and by $J^{ii}$ the second derivative. Substituting $z_i=1/x_i$ after the derivatives, we have $$ \frac{\alpha}{2}\sum_{i=1}^N \frac{1}{x_i^2}J^{ii}(x^{-1})+\sum_{i\neq j}^N\frac{x_j}{x_i(x_j-x_i)}J^{i}(x^{-1})=e_\lambda J_\lambda(x^{-1}). \quad (*)$$

Now let $f_\lambda(x)=\det(x)^sJ_\lambda(x^{-1})$ and apply $D(\alpha)$ to this, in the hope of obtaining $e_{\widetilde{\lambda}}f_\lambda$ as a result.

We have $$ D(\alpha)f_\lambda(x)=\frac{\alpha}{2}\sum_{i=1}^N x_i^2f_\lambda^{ii}(x)+\sum_{i\neq j}^N\frac{x_i^2}{x_i-x_j}f_\lambda^{i}(x),$$ which is $$\left(\frac{\alpha}{2}s(s-1)N+\frac{sN(N-1)}{2}\right)\det(x)^sJ_\lambda(x^{-1})+\left(\sum_{i=1}^N\frac{\alpha(1-s)}{x_i}-\sum_j\frac{1}{x_i-x_j}\right)\det(x)^sJ_\lambda^{i}(x^{-1}) +\det(x)^s\frac{\alpha}{2}\sum_i \frac{1}{x_i^2}J_\lambda^{ii}(x^{-1}).$$

Using $(*)$ to replace the second derivatives, we get $$ \frac{N}{2}s(\alpha s-\alpha+N-1)\det(x)^sJ_\lambda(x^{-1})+\left(\sum_{i=1}^N\frac{\alpha(1-s)}{x_i}-\sum_j\frac{1}{x_i-x_j}\right)\det(x)^sJ_\lambda^{i}(x^{-1}) +\det(x)^s\left(e_\lambda J_\lambda(x^{-1})-\sum_{i\neq j}^N\frac{x_j}{x_i(x_j-x_i)}J^{i}(x^{-1})\right),$$ or $$\left(e_\lambda+\frac{N}{2}s(\alpha s-\alpha+N-1)\right)\det(x)^sJ_\lambda(x^{-1})+\left(\sum_{i=1}^N\frac{\alpha(1-s)}{x_i}-\sum_j\frac{1}{x_i-x_j}-\sum_{i\neq j}^N\frac{x_j}{x_i(x_j-x_i)}\right)\det(x)^sJ_\lambda^{i},$$ or $$ D(\alpha)f_\lambda(x)=\left(e_\lambda+\frac{Ns}{2}(\alpha (s-1)+N-1)\right)f_\lambda-\sum_{i=1}^N(\alpha(s-1)+N-1)\frac{1}{x_i}\det(x)^sJ_\lambda^{i}.$$

This is NOT what we would like. I do not know where the mistake is.

As for the eigenvalue, we have $\widetilde{\lambda}=(s-\lambda_N,...,s-\lambda_1)$, and $|\widetilde{\lambda}|=sN-|\lambda|$. We know that $$ e_\lambda=\alpha b(\lambda')-b(\lambda)+(N-1)|\lambda|,$$ with $b(\lambda)=\sum_i(i-1)\lambda_i$ and $b(\lambda')=\sum_i\lambda_i(\lambda_i-1)/2$. So I get $$e_{\widetilde{\lambda}}= e_{\lambda}+\alpha\frac{s(s-1)N}{2}+\frac{sN(N-1)}{2}-(\alpha s+N-1)|\lambda|.$$

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