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When trying to build a dual formulation for lattice gauge theories using Weingarten integration I am getting sums of the kind

$$I^{m, n}_{\mu, \nu} (\sigma, \tau) = \sum_{\pi \in S_n} \chi_\mu (\pi \sigma) \chi_\nu(\pi \tau), $$

where $\sigma$ and $\tau$ are some fixed permutations in $S_m$ ($m \geq n$), $\mu$ and $\nu$ are some irreducible representations of $S_m$ defined by the partitions of $m$, and $S_n$ is a subgroup of $S_m$ leaving the last $m - n$ elements in place.

Now when $m = n$ the sum is trivial using the orthogonality rule for irreducible representations. Also when $\sigma$ and $\tau$ are in $S_n$, I can just use the Littlewood-Richardson rule to expand restrictions of $\mu$ and $\nu$ to $S_n$ into irreducible representations of $S_n$ and get the result. But can something be done in the general case?

I guess the answer cannot be written in terms of $S_m$ characters of $\sigma$ and $\tau$ since fixing the $S_n$ subgroup breaks the permutation symmetry -- in particular $$ I^{m, n}_{\mu, \nu} (\rho^{-1} \sigma \rho, \rho^{-1} \tau \rho) = \Sigma^{m, n}_{\mu, \nu} (\sigma, \tau) $$

does not necessary hold for every $\rho \in S_m$, (just for $\rho$ in $S_n$). Still maybe there is some way to obtain the sum?

One possibility I am thinking of is taking the representation matrices for $\pi$, $\sigma$ and $\tau$ as $\Pi_\mu$, $\Sigma_\mu$, $T_\mu$ and rewrite

$$I^{m, n}_{\mu, \nu} (\sigma, \tau) = \sum_{\pi \in S_n} {\Pi_\mu}_{i j} {\Sigma_\mu}_{j i} {\Pi_\nu}_{k l} {T_\nu}_{l k}, $$ where $i$, $j$ enumerate Young tables of shape $\mu$ and $k$, $l$ enumerate Young tables of shape $\nu$.

After that if I could somehow express ${\Pi_\mu}_{i j}$ in terms of the matrix elements of irreducible representations of $S_n$, I would be able to use the orthogonality property for the matrix elements and get an answer in terms of matrix elements of $\sigma$ and $\tau$. But this would require some analogue of Littlewood-Richardson rule for matrix elements. (There should be a basis in which $\Pi_\mu$ is a block diagonal matrix, built from irreducible representation matrices for $S_n$, but I am not sure how to define such a basis).

Being a physicist, it is hard for me to understand if this is a simple problem, or an unsolvable one, so any advice is appreciated.

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  • $\begingroup$ In the case when the restrictions of the two characters to $S_n$ don't have common irreducible constituents the answer is $0$, it follows from general group theory and is contained in the paper ftp.math.uni-rostock.de/pub/preprint/2005/pre05_10.ps $\endgroup$ – Yellow Pig Feb 10 at 17:54
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    $\begingroup$ Thank you very much. This can be used to simplify the result i had (removing extra terms in the summation). Also intersting is the "only if" part -- so for all the terms that are left the result must be nonzero. Still, it would be very useful to know any kind of simplification for nonzero terms (though maybe it does not exist). $\endgroup$ – Volodymyr Chelnokov Feb 12 at 11:09

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