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Irreducible finite dimensional representations of the group $GL_n$ are parameterized by the highest weights, that is by nonincreasing sequences of integers $$ \lambda_1 \ge \lambda_2 \ge \dots \ge \lambda_n. $$ Let us restrict to the case when $\lambda_n \ge 0$. Then one can encode a highest weight by a Young diagram with $\lambda_i$ boxes in the $i$-th row.

The Littlewood-Richardson rule describes the decomposition of a tensor product $V^\lambda \otimes V^\mu$ into a direct sum of irreducibles. It says that the multiplicity of $V^\nu$ in the tensor product $V^\lambda \otimes V^\mu$ is equal to the number of so-called Littlewood-Richardson tableux in the skew-diagram $\nu\setminus\lambda$ of weight $\mu$. See Littlewood-Richardson rule for precise definitions.

Note that the rule is not symmetric with respect to $\lambda$ and $\mu$!

On the other hand, the category of representations of $GL_n$ has a commutativity morphism: it is a bifunctorial isomorphism $$ c_{V,W}:V\otimes W \to W\otimes V, \qquad v\otimes w \mapsto w\otimes v. $$


Is there a possibility to make Littlewood-Richardson rule compatible with the commutativity morphism?

To be more precise, is there a way to associate with every Littlewood-Richardson tableau in a skew diagram $\nu\setminus\lambda$ of weight $\mu$ an embedding $V^\nu \to V^\lambda\otimes V^\mu$ such that the composition $V^\nu \to V^\lambda\otimes V^\mu \stackrel{c}\to V^\mu\otimes V^\lambda$ is the embedding associated to an appropriate Littlewood-Richardson tableau in a skew diagram $\nu\setminus\mu$ of weight $\lambda$?

Let me emphasize that I am asking about the $GL_n$ case, although this question has sense for any reductive group.

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Let $LR(\mu/\lambda;\nu)$ be the set of Littlewood-Richardson tableaux of shape $\mu/\lambda$ and weight $\nu$. Then there is a canonical bijection between $LR(\mu/\lambda;\nu)$ and $LR(\mu/\nu;\lambda)$, presented in a paper by Pak and Vallejo ("Fundamental Symmetry map"), in a paper by Danilov and Koshevoi ("Commutor"), and in a paper by Henriques and Kamnitz.

Is this what you want ?

In the paper by Pak and Vallejo actually two "Fundamental Symmetry maps" are presented. Danilov and Koshevoi show that they coincide, and that they coincide with their "commutor", and with the map defined by Henriques and Kamnitzer.

The references:

Igor Pak and Ernesto Vallejo.
Reductions of Young tableau bijections
SIAM J. Discrete Math. 24 (2010), no. 1, 113--145.
doi: 10.1137/070689784

V.I. Danilov and G.A. Koshevoi
The Robinson-Schensted-Knuth correspondence and the bijections of commutativity and associativity.
2008 Izv. Math. 72 689
doi: 10.1070/IM2008v072n04ABEH002415

A. Henriques and J. Kamnitzer
The octahedron recurrence and $gl_n$-crystals
Adv. Math. 206:1 (2006), 211-249

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You probably also want to read Henriques and Kamnitzer's paper where they show that this commutor (and a similar one defined for any Lie type) do not obey the braid relation, but do obey a variant they call the cactus relation. The analogue of the braid group is now the fundamental group of the moduli space of stable real $n$-pointed genus zero curves. – David Speyer Sep 14 '10 at 12:12
As far as I understand the bijections described in these papers describe the structure of the tensor product in the category of crystals, which is not equivalent to the category of $GL_n$-representations. So, this is NOT what I need. Also note, that I don't ask how the bijection looks, I only ask whether it exists (in the category of $GL_n$-representations). – Sasha Sep 15 '10 at 3:10
The categories aren't equivalent, but their Grothendieck rings are isomorphic. Since LR tableaux are only computing the coefficients in these Grothendieck rings -- not, e.g. giving specific intertwiners between V_a @ V_b and V_c -- I'm not sure what you think you DO need that these don't supply. – Allen Knutson Sep 18 '10 at 2:01
Dear Allen, I don't understand your comment. My principal interest is in the action of the commutativity isomorphism on the tensor product of irreducibles. Certainly, it is not the thing you can see on the level of Grothendieck rings, right? – Sasha Sep 18 '10 at 8:33
Looking back, I completely agree that my comment is not addressing your question; sorry! If your question could be answered, I would expect it would give a nice formula for Sym^k and Alt^k of a representation, which I don't think is known. – Allen Knutson Jan 27 '11 at 4:13

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