Integral of Schur functions over $SU(N)$ instead of $U(N)$

Question

Schur functions are irreducible characters of the unitary group $U(N)$. This implies $$ \int_{{U}(N)}s_\lambda(u)\overline{s_\mu(u)}du=\delta_{\lambda\mu},$$ where the overline means complex conjugation. My question is what is the result of the same integral performed over $SU(N)$ instead, $$ \int_{{SU}(N)}s_\lambda(u)\overline{s_\mu(u)}du=?$$

Answer 1

The desired integral is given in equation (13) of arXiv:1812.06069:

$$\int_{{SU}(N)}s_\lambda(u)\overline{s_\mu(u)}du=\sum_{q=-\infty}^\infty\prod_{i=1}^N\delta_{\lambda_i,\mu_i+q},$$ where $\lambda=(\lambda_1,\lambda_2,\ldots\lambda_N)$ and $|\lambda|=\sum_{i}\lambda_i$, with $\lambda_1\geq\lambda_2\cdots\geq 0$. This is still an orthogonality relation, because the Schur functions $s_\lambda$ and $s_\mu$ are identical in $SU(N)$ iff $\lambda=\mu+(q,q,\ldots q)$ for some integer $q$.

The $U(N)$ integral $\int_{{U}(N)}s_\lambda(u)\overline{s_\mu(u)}du=\delta_{\lambda\mu}$ corresponds to the $q=0$ term in the sum over $q$. It follows that the integrals over $SU(N)$ and over $U(N)$ are the same if $|\lambda|,|\mu|<N$, because then only the $q=0$ term contributes.