# Jack function in power symmetric basis

In Macdonald's book, the Jack symmetric function $$J_{\lambda}(x_1,\ldots, x_n)$$ for a partition $$\lambda$$ is defined by three properties (orthogonality, triangularity, and normalization). In the following paper (http://www-math.mit.edu/~rstan/pubs/pubfiles/73.pdf) its existence and uniqueness appear as Theorem 1.1. The Jack symmetric functions can be seen as eigenfunctions for the operators $$D(\alpha)= \alpha/2 \sum_{i=1}^{n}x_i^2\frac{\partial}{\partial_i^2}+\sum_{i\neq j}\frac{x_i^2}{x_i -x_j}\frac{\partial}{\partial x_i}$$ with the eigenvalues as given in Theorem 3.1 of the above paper.

Now a recent paper of Chapuy and Dolega (https://arxiv.org/pdf/2004.07824.pdf) defines the following operator, which is given in the power symmetric basis,

$$D_{\alpha}^{'}= \alpha/2 \sum_{i,j\geq 1}ij p_{i+j}\frac{\partial^2}{\partial p_i \partial p_j} + 1/2 \sum_{i,j\geq 1}(i+j) p_{i}p_{j}\frac{\partial^2}{\partial p_{i+j}}+(\alpha -1)\sum_{i\geq 1}\frac{i(i -1)}{2}p_i\frac{\partial}{\partial p_i}$$ and it defines the Jack symmetric functions to be those function which are eigenfunctions to these operators; they give the eigenvalues in terms of $$\alpha$$, which takes a nice form. In the paper it appears in Proposition 5.1.

My question is how to derive this operator from Stanley's paper and to apply it on the Jack symmetric function. Do we need to express the Jack symmetric function in the power symmetric basis? I cannot do it in the case of a general partition. Also, the operator in Stanley's paper involves finitely many variables, but in the Chaupy and Dolenga paper it involves infinitely many. I hope someone can give me more details.

• Are you claiming that the operators $D\left(\alpha\right)$ and $D'_\alpha$ are identical? In that case, this shouldn't be hard to prove. Indeed, both are differential operators of second order (i.e., operators $f$ satisfying $f\left(abc\right) - f\left(ab\right)c - f\left(ac\right)b - f\left(bc\right)a + f\left(a\right)bc + f\left(b\right)ac + f\left(c\right)ab = 0$ for all $a, b, c$ in the algebra), and thus their equality needs only to be checked on a generating set of the algebra and the pairwise products of generators from this generating set. Commented May 6, 2020 at 10:25
• Regarding the number of variables, you can restrict to n variables, if you are only interested in Jack polynomials of degree n. This is a standard observation when dealing with symmetric functions. Commented May 6, 2020 at 10:35
• I guess they are identical as per the new paper goes. How does one get the operator in power symmetric basis from the monomial basis is the jacobian change of coordinates. I can't get the transformation.
– GGT
Commented May 6, 2020 at 11:56
• @RichardStanley Can you comment on this? I too am confused by equality (11) in your paper; it seems to be tailored to $n$ variables, but $\Lambda$ is a ring of symmetric functions in infinitely many variables. Commented May 11, 2020 at 13:01
• @darijgrinberg Yow! The operator $D(\alpha)$ that I define should be on $\Lambda_n\otimes \mathbb{Q}(\alpha)$. (I use $\Lambda_n$ to denote $n$ variables and $\Lambda^n$ to denote degree $n$, same as Macdonald.) Commented May 17, 2020 at 0:54

Note that Stanley's Laplace-Beltrami operator $$D(\alpha)$$ depends on $$n$$ and acts on the algebra $$\mathbb{Q}(\alpha)\otimes \Lambda^n$$, where $$\Lambda^n$$ denotes the algebra of symmetric polynomials in $$n$$ variables $$x_1,\dots,x_n$$. Because of that I prefer to modify your notation and denote Stanley's operator by $$D_n(\alpha)$$. Let me introduce a modified version of this operator $$D'_n(\alpha)$$, which also acts on $$\mathbb{Q}(\alpha)\otimes \Lambda^n$$ and is defined by setting $$D'_n(\alpha)f := \big(D_n(\alpha)-(n-1)\deg(f)\big)f$$ for homogenous $$f$$ and extended by linearity. Recall that the algebra $$\Lambda$$ of symmetric functions is defined as the projective limit of $$\Lambda_n$$ with respect to the morphism $$\rho_n : \Lambda_{n+1}\to \Lambda_n$$ which kills the last variable: $$\rho_n(f)(x_1,\dots,x_n) := f(x_1,\dots,x_n,0).$$ It is easy to check that $$\rho_n D'_{n+1}(\alpha) = D'_{n}(\alpha)\rho_{n},$$ so you can define an operator $$D'_\alpha := \lim D'_{n}(\alpha)$$ which acts on $$\mathbb{Q}(\alpha)\otimes \Lambda$$. Stanley's computations used in the proof of his Theorem 3.1 give you immediately an expression for $$D'_\alpha$$ as a differential operator in power-sums.