# Skew Schur function identity $s_\lambda(x_1,\ldots,x_n)=\sum_\mu s_\mu(x_1,\ldots,x_j)s_{\lambda/\nu}(x_{j+1},\ldots,x_n)$

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.

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

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.