Trace identity for $SU(N)$ matrix integral I would like to know if there's a nice way to compute the following:
$$ \int_{SU(N)} \underbrace{ dU}_{\text{Haar Measure}} \mathrm{tr} \left(U^n \right)~?$$
The following is necessary:
$U \in SU(N)$
$dU$ is the haar measure (left invariant measure).
The integration is carried over the whole group manifold.
What I tried:
If $U \in SU(N)$ it can be written as $U = \exp \left(i \vec{\alpha} \cdot \vec{T} \right)$ for $\alpha \in \mathbb{R}^{N^2 -1 }$ and $\vec{T}$ being the generators of SU(N).  THis implies that
$$\mathrm{tr} (U^n) = \mathrm{tr} (\exp(i n \vec{\alpha} \cdot \vec{T}))$$
For every integration range with some $\alpha$, there is an equivalent value $-\alpha$, so i'm convinced the answer has to be real.  However I'm not sure where to proceed from there.
 A: The result is not always zero. See my two answers to
https://mathoverflow.net/questions/255492/how-to-constructively-combinatorially-prove-schur-weyl-duality?noredirect=1&lq=1
for a formula for such integrals, and my answer to
https://mathoverflow.net/questions/330964/nonnegativity-of-an-integral-over-the-unitary-group?noredirect=1&lq=1
for a hopefully instructive practice example.
A general remark is that if $N$ does not divide $n$, then the integral is zero. For $n=kN$, then the result is
$$
\frac{0!1!\cdots (N-1)!}{k!(k+1)!\cdots (k+N-1)!}
\ ({\rm det}(\partial U))^k\ {\rm tr}(U^{kN})\ .
$$
Here $\partial U$ is the matrix given by the differential operators $\frac{\partial}{\partial U_{ij}}$ instead of the matrix elements $U_{ij}$ themselves.
For $k=1$, it is easy to do the computation and find the value $(-1)^{N-1}$.
In general what one can do, is express the power sum in the Schur basis. This involves the characters of the symetric group. Then the only contributions should come for rectangular partitions with $N$ rows and $k$ columns. I would have to spend some time to check details but, I think the result is basically a character value where one partition is with one part $kN$ and the other is $N^k$, i.e., $N$ parts equal to $k$.

Update:
After looking up the literature a bit here is the result.
Denote by $I(N,k)$ the integral with $n=kN$. The other cases vanish.
One can turn this into an integral over $U(N)$ by writing
$$
I(N,k)=\int_{U(N)}dU\ \overline{({\rm det}\ U)^k}\ {\rm tr}(U^{kN})\ .
$$
When the partitions involved have at most $N$ parts, this group integral inner product is the same as the Hall inner product of symmetric functions of the eigenvalues of $U$.
So
$$
I(N,k)=\langle s_{\lambda},p_{\mu}\rangle=\chi_{\mu}^{\lambda}\ .
$$
Here $s_{\lambda}$ is the Schur function for the rectangular partition $\lambda=(k,k,\ldots,k)$, with $n$ parts of size $k$. The $p_{\mu}$ is the power sum corresponding to the partition $\mu=(kN)$, i.e., with a single part of size $kN$. A particular case of the Murnaghan-Nakayama rule says that the symmetric group character value $\chi_{\mu}^{\lambda}$ is zero unless $\lambda$ is a hook, in which case one has
$$
\chi_{\mu}^{\lambda}=(-1)^{kN-\lambda_1}\ .
$$
But our $\lambda$ is a hook only if $N=1$ or $k=1$. For $k=1$ one recovers my previous formula $(-1)^{N-1}$.
Reference for the stated facts about symmetric functions:
https://www.symmetricfunctions.com/murnaghanNakayama.htm#murnaghanNakayamaRule
