11
$\begingroup$

Consider the generating function $$ G_n(x_1,x_2,\ldots,x_n, t_1,t_2,\ldots,t_n) =\sum_{\lambda}s_{\lambda}(x_1,x_2,\ldots, x_n) t_1^{\lambda_1}t_2^{\lambda_2} \cdots t_n^{\lambda_n}, $$ where the sum is over all partitions $\lambda=(\lambda_1, \lambda_2,\ldots, \lambda_n)$ and $s_\lambda$ is a Schur polynomial.

For small $n$ such generating function is easy to find, for example for $n=2$ by direct calculation we have $$ G_2(x_1,x_2, t_1,t_2)=\sum_{\lambda}s_{\lambda}(x_1,x_2) t_1^{\lambda_1}t_2^{\lambda_2}=\frac{1}{(1-x_1 t_1)(1-x_2 t_1)(1-x_1 x_2 t_1 t_2)}. $$ If we put $t_1=t_2=1$ then we come to well-known Littlewood identity

$$ \sum_{\lambda}s_{\lambda}(x_1,x_2)=\frac{1}{(1-x_1)(1-x_2)(1-x_1 x_2)}. $$

Question. Is there any close expression for the generating function $G_n$ for arbitrary $n?$

$\endgroup$
0

1 Answer 1

15
$\begingroup$

This is done in my paper The character generator of SU(n). I believe there was an essentially the same previous MO question, but I am unable to find it.

$\endgroup$
2
  • $\begingroup$ In case the link to the PDF file is broken, the DOI is (almost certainly) doi.org/10.1063/1.524687 $\endgroup$ Commented Dec 20, 2021 at 13:46
  • $\begingroup$ @Richard Stanley. Thank you for the answer! $\endgroup$
    – Leox
    Commented Dec 20, 2021 at 17:26

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .