# Integration over special unitary group

It is known that for $SU(N)$ $$\int \chi_{\mu_1}(UV_1)\chi_{\mu_2}(U^{-1}V_2)\, dU = \delta_{\mu_1\mu_2}\frac{\chi_{\mu_1}(V_1V_2)}{\dim(\mu_1)}$$ where $dU$ is Haar measure on $SU(N)$ normalized such that with respect to it $\operatorname{Vol}(SU(N))=1$; and $\chi_{\mu}(U)$ means the trace of $U$ in the irreducible representation $\mu$.

I came across this integral in "On quantum gauge theories in two dimensions", Edward Witten, Commun. Math. Phys. 141, 153-209 (1991).

My question is simple. What else is known about integrals on $SU(N)$ of the form $$\int \chi_{\mu_1}(U^{\pm 1}V_1)\dots\chi_{\mu_n}(U^{\pm 1}V_n)\, dU \ ?$$

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I edited my negative answer now, and gave a rep-theoretic strategy. –  plusepsilon.de Jun 27 '13 at 9:52
@RicardoAndrade what was the point of adding the Haar Measure tag? the question does not really concern the details of that measure, and I find the tag rather gratuitous, representation-theory would be more apt –  Yemon Choi Jun 27 '13 at 10:01
@Yemon: You are correct. I apologize for the tagging error. I have replaced the tag. –  Ricardo Andrade Jun 27 '13 at 10:06

Do the Weingarten formulas help at all? See Theorem 2.5, and the surrounding discussions, in http://arxiv.org/abs/0903.5143

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Thanks for the answer. In Weingarten's formula all group elements in the integrand are in fundamental representation. But I don't think if I can get any more general results:) –  user10001 Jun 27 '13 at 10:36
$SU(1)$ is easy. For $SU(2)$, this involves integrals of the Chebyschev polynomials. If you can't find anything useful in the usual integral tables, there is no hope to find something for any other $n$.
Edit: As a second thought, the character $\chi_{\mu_1}\cdot \chi_{\mu_2}$ equals the character $\chi_{\mu_1 \otimes \mu_2}$, where $\mu_1 \otimes \mu_2$ is the tensor product of $SU(n)$-reps. I think that Fulton-Harris' book about representation theory explains you what you get for tensoring $SU(n)$-reps, i.e., how they decompose into irreducible things. So you have to clue together those $\chi$'s with $U$ and those with $U^{-1}$, and you can use the Schur orthogonality relations.
re your comment about decomposing tensors - what, you get for SU(n), once $n\geq3$, is a mess, although I think the physicists have hacked out SU(3) quite explicitly. I think Benoit Collins may have some recent work related to asymptotic behaviour for large n –  Yemon Choi Jun 27 '13 at 9:54