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I wonder if there is a way to compute the symmetric tensor power of irreducible representations for classical Lie algebras: $\mathfrak{so}(n)$, $\mathfrak{sp}(n)$, $\mathfrak{sl}(n)$.

The question is motivated by reading "Introduction to Quantum Groups and Crystal Bases" by Hong, J. and Kang, S.-J. The book provides an algorithm for computing the tensor product of any two irreducible representations for classical Lie algebras. Could it be generalized to symmetric parts of tensor products? Any references are very much appreciated!

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For different reasons (I think), I asked some quantum group experts whether their ability to decompose tensor products also worked for symmetric powers of a representation. My impression is that it is an open problem to find a crystal basis approach to symmetric powers. – Marty Jan 31 '10 at 18:44
Unfortunately, I was given the same answer that this is an open "crystal problem" – Eugene Starling Jan 31 '10 at 21:41

I assume, since you haven't explicitly stated it, that you're taking these Lie algebras in characteristic 0 -- the question is much harder in positive characteristic (and in particular, the word "compute" will be ambiguous there). With this assumption, one can use theoretical techniques as in David Speyer's answer; also check out Fulton & Harris's Representation Theory book. But I'd also point out that if you have specific irreps in mind, there are computer packages that will do this computation, eg LiE; and there's even a web interface here: that will compute the answer in low rank.

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Yes, you are right, the characteristic I have in mind is zero. Thank you for the reference, the algorithm used there has an open source, which is well commented with one more useful reference at the end. However, I still hope that the problem is solved somewhere. – Eugene Starling Jan 31 '10 at 21:15

It's maybe worth mentioning a reason why the crystal techniques don't immediately solve this problem: If $V$ and $W$ are representations of $\mathfrak g$, then we can find the corresponding representations of the quantum group $V_v$,$W_v$, and take their tensor product. The crystal basis theory shows you how to combinatorially understand the decomposition of $V_v\otimes W_v$ and so this also solves in turn the problem of decomposing $V\otimes W$. However the "flip map" $\sigma\colon V_v\otimes W_v \to W_v\otimes V_v$ is not an isomorphism representations (even though the representations are in fact isomorphic). Thus $\sigma$ doesn't induce an automorphism of $V_v\otimes V_v$, and so the "symmetric part" of the quantum version of a second tensor power doesn't immediately make sense.

This gets reflected on the combinatoral side by the fact that if $B_1$ and $B_2$ are the crystals of highest weight representations, then their tensor product $B_1 \boxtimes B_2$ and $B_2 \boxtimes B_1$ are isomorphic but this isomorphism is not realized by the flip map: one first has to twist by the map which takes a "highest weight" crystal to a "lowest weight" crystal (this has been studied by a number of people: e.g. Berenstein, Henriques-Kamnitzer, and in the context of Littelmann paths by Biane-Bourgerol-O'Connell and probably others too!) So again, at least at first sight it's not immediately clear what the "symmetric part" of a crystal $B_1 \boxtimes B_1$ should be.

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Computing the character of a symmetric power is fairly easy; see this note by Stavros Kousidos. Decomposing that character into irreps is a form of plethysm, and is very hard.

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Could you please be more precise. Is it only hard to solve or is it unsolved? Are there cases for which it is solved? – Eugene Starling Jan 31 '10 at 21:34
What David means is: there is an algorithm for computing which simples appear with which multiplicity. However, it is very computationally intensive, so it is not very practical in large cases. – Ben Webster Feb 1 '10 at 0:24
I'll just note: what it seems likely are open at this point is to find a positive algorithm, one where the multiplicities are presented in a subtract-free way. Crystal bases give such an algorithm for tensor products; I don't think one exists for symmetric powers. – Ben Webster Feb 1 '10 at 0:46
There are known algorithms for this decomposition and, as Ben says, they are not positive. I don't know what their efficiency is, either in practice or in the sense of theoretical complexity. – David Speyer Feb 1 '10 at 12:50
Ok, I see that I should wait a number of years until the positive algorithm will be found!:) – Eugene Starling Feb 1 '10 at 14:19

Here's a very special case for $\mathfrak{gl}_n$ in characteristic 0 (which I have found useful in my work). Let $V$ be the vector representation, and for a partition $\lambda$ with at most $n$ parts, let ${\bf S}\_{\lambda}(V)$ denote the corresponding highest weight representation. Then $Sym^n(Sym^2 V) = \bigoplus\_{\lambda} {\bf S}\_{\lambda}(V)$ where the direct sum is over all partitions $\lambda$ of size $2n$ with at most $n$ parts such that each part of $\lambda$ is even. Similarly, $Sym^n(\bigwedge^2 V) = \bigoplus\_{\mu} {\bf S}\_{\mu}(V)$ where the direct sum is over all partitions $\mu$ of $2n$ with at most $n$ parts such that each part of the conjugate partition $\mu'$ is even. If you want the corresponding result for $\mathfrak{sl}\_n$ we just introduce the equivalence relation $(\lambda\_1, \dots, \lambda\_n) \equiv (\lambda\_1 + r, \dots, \lambda\_n + r)$ where $r$ is an arbitrary integer.

One reference for this is Proposition 2.3.8 of Weyman's book Cohomology of Vector Bundles and Syzygies (note that $L\_\lambda E$ in that book means a highest weight representation with highest weight $\lambda'$ and not $\lambda$).

Another reference is Example I.8.6 of Macdonald's Symmetric Functions and Hall Polynomials, second edition, which proves the corresponding character formulas.

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Thank you for Weynman reference! The answer in the case of rank-two tensors is known to me, it is related to building invariant tensors for $\mathfrak{so}(n)$ and $\mathfrak{sp}(n)$ by taking tensor powers of the invariant tensor with the lowest rank -- the rank two symmetric and rank two antisymmetric, respectively – Eugene Starling Feb 3 '10 at 13:12

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