Is there any explicit decomposition of tensor product of two finite dimensional irreducible modules of simple lie algebras whose highest weights are same?
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The two modules in the question are actually isomorphic, since highest weights determine irreducibles. Anyway, the answer to your question is that there is usually no explicit decomposition known. Of course, there are various algorithmic approaches to tensor product decomposition, but the process tends to get very long. From the viewpoint of the old Brauer method, it's easiest to handle a tensor product when one highest weight is much bigger than the other, whereas the case you consider is rather hard.
For some perspective on the relatively small highest weight $\rho$ (half-sum of positive roots, or sum of fundamental dominant weights), it's worth looking into one of Kostant's papers: Flag manifold quantum cohomology, the Toda lattice, and the representation with highest weight ρ, Selecta Math. (N.S.) 2 (1996), no. 1, 43–91.
[In a more positive direction, if the given module happens to be self-dual, then the trivial module occurs precisely once as a summand of the module tensored with itself.]
answered Apr 28, 2014 at 16:50
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$\begingroup$ Is it really a summand? Not disputing, just asking. I know it's a composition factor $\endgroup$ Commented Apr 28, 2014 at 17:18
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$\begingroup$ @Geoff: Yes, in this classical finite dimensional situation all representations are completely reducible. On the other hand, finding explicit tensor product decompositions is quite a big problem as the dimensions of the factors go up. (Brauer's method required knowing all weight multiplicities in one factor, besides which there can be lots of "cancellation". But all methods are tricky to implement as dimensions grow.) $\endgroup$ Commented Apr 28, 2014 at 18:46
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$\begingroup$ I agree with everything Jim says, and want to add: I don't think that any of the positive rules (e.g. Littelmann's) for tensor product decompositions let you compute the symmetric and alternating square. $\endgroup$ Commented Apr 28, 2014 at 22:29
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$\begingroup$ @Jim If $\mathfrak{g}$ is a simple lie algebra,what is the multiplicity of the irreducible representation (of $\mathfrak{g}$) $V(\lambda)$ in tensor product of $V(\lambda)$ with $\mathfrak{g}$ ? $\endgroup$ Commented Apr 29, 2014 at 13:04
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$\begingroup$ If it's any help, for an irreducible character (I prefer characters) $\chi\equiv \chi_w$ ($w$ being the highest dominant weight) of a compact connected (simple) Lie group, there exists an integer $N$ (depending ostensibly on the irreducible, but probably only on the group) such that every irreducible whose highest dominant weight in the convex hull of the orbit of $w$ (under the Weyl group) and satisfies the obvious congruence relation wrt the centre) appears in $\chi_w^N$. For small values of $n$, and the group SU($n$), the $N$ depends only on $n$ (but I forget what the values of $N$ are). $\endgroup$ Commented Apr 29, 2014 at 15:23