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

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

• isn't the answer the same for $U(N)$ and $SU(N)$? Aug 8, 2019 at 18:55
• @CarloBeenakker Why should it be the same? (I don't think they are the same) Aug 8, 2019 at 18:57
• 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... Aug 8, 2019 at 21:38

$$\int_{{SU}(N)}s_\lambda(u)\overline{s_\mu(u)}du=\sum_{q=-\infty}^\infty\prod_{i=1}^N\delta_{\lambda_i,\mu_i+q},$$ 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 $$|\lambda|,|\mu|, because then only the $$q=0$$ term contributes.