1
$\begingroup$

I'm trying to prove $s_\lambda(x_1,\ldots,x_n)=\sum_\mu s_\mu(x_1,\ldots,x_j)s_{\lambda/\nu}(x_{j+1},\ldots,x_n)$.

Using Littlewood-Richardson coefficients, I got the right hand side to equal to $\sum_{\mu,\nu,\lambda}C^\lambda_{\mu\nu} C^\lambda_{\mu\nu} s_\lambda$. I don't know how this would equal to $s_\lambda$. I'm new to this and would appreciate any help.

$\endgroup$
1
  • $\begingroup$ $s_\lambda(x_1,\ldots,x_n)=\sum_\mu s_\mu(x_1,\ldots,x_j)s_{\lambda/\mu}(x_{j+1},\ldots,x_n)$ surely? $\endgroup$ Dec 2, 2019 at 7:51

1 Answer 1

1
$\begingroup$

This comes straight from the interpretation of Schur functions as generating functions of semi-standard Young tableaux.

The coefficient of $x_1^{a_1}\cdots x_n^{a_n}$ in $s_\lambda(x_1,\ldots,x_n)$ is the number of SSYTs of shape with $a_i$ $i$s in them. In one of these, the instances of $1,\ldots,j$ occupy a subtableau of shape $\mu$ and the instances of $j+1,\ldots,n$ occupy its complement which has shape $\lambda\setminus\mu$. For a given $\mu$ the number of such tableaux on $\lambda$ is the product of the coefficients of $x_1^{a_1}\cdots x_j^{a_j}$ in $s_\mu(x_1,\ldots,x_j)$ and that of $x_{j+1}^{a_{j+1}}\cdots x_n^{a_n}$ in $s_{\lambda\setminus\mu}(x_{j+1},\ldots,x_n)$. Adding up over all $\mu$ gives the identity in question.

$\endgroup$

You must log in to answer this question.

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