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,
$$ \int\_{{SU}(N)}s\_\lambda(u)\overline{s\_\mu(u)}du=?$$



Integral of Schur functions over $SU(N)$ instead of $U(N)$
The desired integral is given in equation (13) of [arXiv:1812.06069](https://arxiv.org/abs/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.


