It is known that every irreducible representation of the unitary group $U(n)$ can be uniquely described by the non-increasing sequence $\lambda=(\lambda_1,\ldots,\lambda_n)$ of integers (denote the corresponding representation as $V_{\lambda}$). Does there exist a formula (like Littlewood-Richardson rule for $GL(n)$) for the decomposition of the tensor product $V_{\lambda}\otimes V_{\mu}$?

In terms of characters it means that we need to decompose the product of 'Schur functions' $s_{\lambda}s_{\mu}$ into the sum of some $s_{\nu}$ (but here $\lambda,\mu,\nu$ might have negative 'parts').

The case when $\lambda_1\ge\ldots\ge\lambda_n\ge 0\ge\mu_1\ge\ldots\ge\mu_n$ is particularly interesting for me.

Any references would be appreciated.

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    $\begingroup$ I am not totally familiar with this, but maybe link.springer.com/content/pdf/10.1007/PL00001288.pdf is what you're looking for? $\endgroup$ Commented Jul 22, 2021 at 11:32
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    $\begingroup$ Aren't the representation theories of $GL(n)$ and $U(n)$ simply the same, by the unitarian trick (en.wikipedia.org/wiki/Unitarian_trick)? I think you can just use the ordinary Littlewood-Richardson rule. $\endgroup$
    – Will Sawin
    Commented Jul 22, 2021 at 22:51
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    $\begingroup$ Oh, is the issue negative weights? There's not much special about them. You can just shift up all the weights by $k$ to make them nonnegative, apply Littlewood-Richardson, then shift down by $k$ the weights of all the summands appearing. This works because shifting weights by $1$ is the same as the tensor product with the determinant. $\endgroup$
    – Will Sawin
    Commented Jul 22, 2021 at 22:53
  • $\begingroup$ @WillSawin Indeed. However, this seems to be a bit artificial because this number $k$ can be chosen arbitrarily (if it's big enough). Is it possible to obtain formula in terms of $\lambda$ and $\mu$? (LR-rule is not the easiest thing to worl with, so I'd rather avoid adding more complications, if possible.) By the way, your observation can be also seen from the character point of view (i. e. via bialternant formula for "Schur functions"). $\endgroup$
    – richrow
    Commented Jul 23, 2021 at 7:14
  • $\begingroup$ I think you can see that the formula can't be much simpler than the Littlewood-Richardson rule from the same reason that it can't be much more complex - it's the same problem, up to shifts. $\endgroup$
    – Will Sawin
    Commented Jul 23, 2021 at 12:40


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