This should be both well-known and probably easy, but I was wondering if the following is known (and, if so, how to easily calculate the thing or where to read about how to calculate it):

what is $$\int_{\mathrm{SU}(n)} \mathrm{tr}(U^k) dU?$$ (Here by "$dU$" I mean normalized Haar measure.)

Of course for $k$ not a multiple of $n$ the integral is zero. (It's always zero on $\mathrm{U}(n)$ by the same argument.) Also by Weyl's integration formula (i.e. averaging over conjugates of $U$) one immediately reduces to performing the integral over diagonal matrices, but I didn't see a way to get a clean answer out.

Apologies if this is easy! I just couldn't find a good reference.


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    $\begingroup$ It's the dimension of the invariant subspace of the $k^{th}$ tensor power of the defining representation of $\text{SU}(n)$. This is at least good enough to figure out the answer when $n = 2$; you get the Catalan numbers. In general maybe you can say something involving Schur-Weyl duality. $\endgroup$ – Qiaochu Yuan Dec 12 '14 at 23:55
  • $\begingroup$ Schur-Weyl duality is the key word! One gets something like $k!\cdot \frac{0!\cdots (n-1)!}{(k/n)!\cdots (k/n+n-1)!}$ via the hook-length formula (I think the only relevant representation of S_k is the one corresponding to a $k/n\times n$ box?), but I have to write this down to check. At least it's OK in the n=2 case. Thanks!! I totally missed this interpretation! $\endgroup$ – alpoge Dec 13 '14 at 0:39
  • $\begingroup$ Such integrals crop up in lattice gauge theory, for which see, e.g. Creutz's book. $\endgroup$ – Steve Huntsman Dec 13 '14 at 0:58
  • $\begingroup$ Wait --- now I'm sure I'm being dumb --- but the above computes the average of $\mathrm{tr}(U)^k$ rather than that of $\mathrm{tr}(U^k)$, no? $\endgroup$ – alpoge Dec 13 '14 at 3:09
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    $\begingroup$ You can view it as the trace of a $k$-cycle in $S_k$ acting on the invariant subspace of $V^{\otimes k}$. $\endgroup$ – Will Sawin Dec 24 '14 at 2:15

I should have stared at the Weyl integration formula longer --- it's clear from that that, once $k\gg_n 1$, the integral is zero. In fact this article shows that the integral vanishes once $k\geq 2n-1$ (it already vanishes when $k$ is not a multiple of $n$). Hence the only case left is $k=n$, and, if I've interpreted their result correctly, the answer is $(-1)^{n-1}$. This could almost certainly have been read off from the formula in this special case, of course.

[To explain the title, note that the element $\int_{\mathrm{SU}(n)} U^n dU$ commutes with all of $\mathrm{SU}(n)$ and must therefore be a scalar, so it suffices to know its trace.]


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