Applying $\sum_i \partial_{x_i}$, $\sum_i x_i \partial_{x_i}$ and $\sum_i x_i^2 \partial_{x_i}$ to Schur polynomials

Question

The operators $L_k=\sum_i x_i^k\frac{\partial}{\partial x_i}$, with integer $k$, take symmetric polynomials into symmetric polynomials.

Is it known how to write the result of the application of $L_0$, $L_1$ or $L_2$ to a Schur polynomial as a sum of Schur polynomials?

Answer 1

Since $x \, \partial_{x} \, x^k = k \, x^k$, the operator $L$ just records the total degree of a polynomial. For Schur (and Jack, Macdonald, $\dots$) polynomials it thus acts by $$L \, s_\lambda(x_1,\dots,x_n) = |\lambda| \, s_\lambda(x_1,\dots,x_n) \,,$$ where the eigenvalue $|\lambda|=\sum_i \lambda_i$ the weight of the partition $\lambda$.

The operator $L$ is well known. At the Jack level, in terms of Dunkl operators we have $L = \sum_i d_i$ (maybe up to a multiplicative factor), which is an element of the center of the degenerate affine Hecke algebra. At the Macdonald level, in terms of Cherednik--Dunkl operators we recognise the difference operator $q^L = Y_1 \cdots Y_n = D_n^n(x;q,t)$ as the $n$th Macdonald operator, which lies in the center of the affine Hecke algebra. Physically, $L = \mathrm{i} \sum_i \partial_{\xi_i}$ is interpreted as the total momentum operator (up to a factor of $\hbar$) in terms of the additive variables (coordinates of particles moving on the unit circle) defined via $x_i = e^{\mathrm{i}\, \xi_i}$.

Answer 2

The answer for $L_1=\sum_ix_i^2\partial_i$ can be derived in a rather straightforward way (I changed your convention a little bit to match the usual formulas for Virasoro algebra). Namely, use the determinantal definition of Schur function $$ s_\lambda=\frac{a_{\lambda+\delta}}{a_{\delta}}=\frac{\det(x_j^{\lambda_i+n-i})}{\det(x_j^{n-i})}. $$ Indeed, since $L_1$ is a derivation, we have $$ L_1\left(\frac{a_{\lambda+\delta}}{a_{\delta}}\right)=\frac{L_1(a_{\lambda+\delta})a_{\delta}-L_1(a_{\delta})a_{\lambda+\delta}}{a_{\delta}^2}= \frac{L_1(a_{\lambda+\delta})}{a_{\delta}}- \frac{L_1(a_{\delta})a_{\lambda+\delta}}{a_{\delta}^2}. $$ Moreover, $L_1(a_{\delta})=(n-1)e_1a_{\delta}$ since we can apply our derivation $L_1$ row by row, and almost each time we get a determinant with two equal rows, so only the derivation of row $1$ counts, and $$ \frac{L_1(a_{\delta})a_{\lambda+\delta}}{a_{\delta}^2}=(n-1)e_1s_\lambda= (n-1)\sum_{\square\in\mathrm{Ins}(\lambda)}s_{\lambda+\square} $$ thanks to the Pieri rule (here $\mathrm{Ins}(\lambda)$ are the places where boxes can be added to $\lambda$ to get a valid diagram, as in the other answer). Similarly, $$ L_1(a_{\lambda+\delta})=\sum_{\square\in\mathrm{Ins}(\lambda)}(\lambda_{i(\square)}+n-i(\square))a_{\lambda+\square} $$ (where, of course, $i(\square)$ is the number of the row of $\lambda$ where $\square\in\mathrm{Ins}(\lambda)$ is located) ,so $$ \frac{L_1(a_{\lambda+\delta})}{a_{\delta}}=\sum_{\square\in\mathrm{Ins}(\lambda)}(\lambda_{i(\square)}+n-i(\square))s_{\lambda+\square}, $$ and finally $$ L_1\left(\frac{a_{\lambda+\delta}}{a_{\delta}}\right)=\sum_{\square\in\mathrm{Ins}(\lambda)}(\lambda_{i(\square)}-i(\square)+1)s_{\lambda+\square} $$

Answer 3

The answer by Jules Lamers considers $k=1$.

In the comments to that answer, people have mentioned that the case $k=0$ has been solved. Namely, for a given partition $\lambda$ let Del($\lambda$) be the set of boxes in its diagram that may be deleted and the result is still a diagram. Then $$ L_0(s_\lambda)=\sum_{\square \in {\rm Del}(\lambda)} [c(\square)+N]s_{\lambda-\square},$$ where $N$ is the number of variables, $ c(\square)$ is the content of a box and $\lambda-\square$ is the diagram obtained from $\lambda$ upon deletion of that box.

I have done some experimentation, and I would like to propose a conjecture for the action of $L_2$ on a Schur polynomial.

Let Ins($\lambda$) be the set of boxes that may be inserted in its diagram and the result is still a diagram. Then I believe that $$ L_2(s_\lambda)=\sum_{\square \in {\rm Ins}(\lambda)} c(\square)s_{\lambda+\square}.$$

For example, in the diagram for $(2,1)$ three boxes can be inserted, leading to either $(3,1)$, $(2,2)$ or $(2,1,1)$. In the first case the content of the inserted box is $2$, in the second case it is $0$ and in the third case it is $-2$. So $L_2(s_{(2,1)})=2s_{(3,1)}+0s_{(2,2)}-2s_{(2,1,1)}.$

I would be grateful if someone could prove this. Such a simple result cannot be too hard.