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, $$ \int_{{SU}(N)}s_\lambda(u)\overline{s_\mu(u)}du=?$$

$\begingroup$ isn't the answer the same for $U(N)$ and $SU(N)$? $\endgroup$– Carlo BeenakkerAug 8 '19 at 18:55

1$\begingroup$ @CarloBeenakker Why should it be the same? (I don't think they are the same) $\endgroup$– thedudeAug 8 '19 at 18:57

1$\begingroup$ At least for $N=2$ it's explictly known how a $U(N)$ irreducible repsentation reduces upon restricting to $SU(N)$ (see e.g. here math.stackexchange.com/questions/2284576/…); the integral counts the number of irreducibles, if I'm not led astray by analogy with finite groups... $\endgroup$– Dima PasechnikAug 8 '19 at 21:38
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.