Since the Laplace operator $$ \Delta=\frac{\partial^2}{\partial x_1^2}+\cdots+\frac{\partial^2}{\partial x_n^2} $$ preserves symmetric polynomials in $n$ variables, its action on the Schur polynomial $s_{\lambda} $ can again be expressed in terms of Schur polynomials $$ \Delta(s_\lambda) =\sum_{\mu} c_{\lambda \mu} s_\mu, $$ for some $\mu $ and $c_{\lambda \mu}.$

For the case $n=2, a>b+1$, I have empirically established the following explicit formula $$ \frac{1}{2} \Delta(s_{a,b})=\frac{b(b-1)}{2}s_{a,b-2}+b s_{a-1,b-1}+\left(\frac{a(a-1)}{2}+b+1 \right)s_{a-2,b} -\sum_{i=b+1}^{\left[\frac{a+b-2}{2}\right]} (a-2i+b-1) s_{a-i+b-2,i}. $$

I tried, considering that the action of $\Delta $ on the power sums $p_k$ is quite simple, to use the well-known decomposition of Schur polynomials through power sums (Murnaghan-Nakayama Rule) but so far without much success.

Question. Is anything known about $\Delta(s_\lambda)$ for $n>2$?

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    $\begingroup$ See mathoverflow.net/questions/464595 for similar investigations. $\endgroup$ Commented May 20 at 19:55
  • $\begingroup$ If I did the calculation right, $\Delta (\frac{f}{g})= \frac{ \Delta f}{g} -2 \frac{\nabla g \cdot \nabla f}{g^2}+ 2 f \frac{\nabla g\cdot \nabla g}{g^3} + \frac{ f \Delta g}{g^2}$. (But surely this formula is standard.) Applying this to the Weyl character formula, with $g$ the Vandermonde determinant, seems promising: last term vanishes since $\Delta g=0$, first term is easy, second and third terms are harder but might be doable. $\endgroup$
    – Will Sawin
    Commented May 20 at 20:32
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    $\begingroup$ @Will Sawin Yes, that was my initial plan but I couldn't implement it even for $n=2$ $\endgroup$
    – Leox
    Commented May 20 at 20:59
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    $\begingroup$ Some more empirical values. I tried looking up some of the coefficients for fixed $b$ in FindStat (with various combinations of dividing by two and working for $n=0$ or $n=1$) with no success. Those values took about an hour of computer time; I'll try to extend with a longer run and update the gist. $\endgroup$ Commented May 24 at 12:51
  • $\begingroup$ For computer calculation it turns out to be much much faster to convert from the Schur to the monomial basis, calculate the Laplacian, and then convert from the monomial to the Schur basis. The action of $\Delta$ on the monomial symmetric polynomials isn't quite as nice as on the power sums, but maybe the Kostka numbers are easier to work with than the Murnaghan-Nakayama rule? $\endgroup$ Commented May 31 at 7:28

1 Answer 1


Just a comment, but 59 characters too long:

Here is an idea that may be useful. Let $x=(x_1,\dots,x_n)$ and $y=(y_1,y_2,\dots)$. We know that $$ C(x,y):=\sum_\lambda s_\lambda(x)s_\lambda(y) =\exp \sum_{k\geq 1}\frac 1kp_k(x)p_k(y). $$ We can compute the Laplacian of the right-hand side (in the $x$ variables). It will be of the form $P\cdot C(x,y)$, where $P$ is a polynomial in the $p_i(x)$'s and $p_i(y)$'s. One can (in principle, at any rate) use the Murnaghan-Nakayama rule to express $P\cdot C(x,y)$ in terms of Schur functions only. Then take the coefficient of $s_\lambda(y)$ (which will require the Littlewood-Richardson rule). This looks pretty messy, but maybe something can be said.


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