Symmetric square of the adjoint representation of a semi-simple Lie algebra

Question

General Outline

Assume one has a decomposition of the tensor product of the adjoint representation into irreducible components:

$\mathfrak{g} \otimes \mathfrak{g} = \bigoplus_{\lambda} V_{\lambda}^{\oplus m(\lambda)} $

where $\mathfrak{g}$ is a semi-simple Lie algebra, $\lambda$ denotes the highest weight of the irreducible representation in $V_{\lambda}$ and $m(\lambda)$ is the multiplicity of the irrep $\lambda$ in the decomposition.

We can also decompose $\mathfrak{g} \otimes \mathfrak{g}$ into it's symmetric square and second exterior power (see ref 1)

$\mathfrak{g} \otimes \mathfrak{g} = Sym^2(\mathfrak{g}) \oplus \Lambda^2(\mathfrak{g})$.

I am intersted in computing $Sym^2(\mathfrak{g})$.

Question I

Is there a (computationally efficient) way of checking which irreps $V_{\lambda}$ occurring in the decomposition of $\mathfrak{g} \otimes \mathfrak{g}$ contribute to the symmetric square $Sym^2(\mathfrak{g})$?

Refined Question II

I am particularly interested in the case of $\mathfrak{g} = \mathfrak{su}(n)$. Any generalities applying beyond this case would be welcome.

Attempt At a Solution

The $n$th symmetric tensor power $Sym^n(V_{\lambda})$ of a irrep $V_{\lambda}$ is defined as the subrepresentation of $V_{\lambda}^{\otimes n}$ which is invariant under the permutation group on $n$ elements, $S_n$. For the case at hand, we have $V_{\lambda} = \mathfrak{g}$ and $n=2$ where $S_2$ has two elements being the identity and an order two element $\sigma$ with eigenvalues $\pm 1$. The $+1$ eigenspace is the subrepresentation $Sym^2 (\mathfrak{g})$ while the $-1$ eigenspace is that of the exterior power $\Lambda^2(\mathfrak{g})$. For $n=2$ these are orthogonal subspaces so we have the decomposition given above: $$\mathfrak{g} \otimes \mathfrak{g} \cong Sym^2(\mathfrak{g}) \oplus \Lambda^2(\mathfrak{g}).$$ If this approach is correct, then the answer to I can be given by computing which irreps occurring in the decomposition $\bigoplus_{\lambda} V_{\lambda}^{\oplus m(\lambda)}$ are invariant under $\sigma$. Such irreps are by definition contained in $Sym^2(\mathfrak{g})$.

Resulting Question III

If this approach is correct, my question is then how to actually check which irreps contained in $\bigoplus_{\lambda} V_{\lambda}^{\oplus m(\lambda)}$ are invariant under $\sigma$ in terms of the irrep data. That is, I am wondering if this can be done by simply knowing the highest weight $\lambda$ for $V_{\lambda}$ or if I need to write down the tensor structure of $V_{\lambda}$ in terms of projections onto $\mathfrak{g} \otimes \mathfrak{g}$ elements. I'd prefer an "abstract" approach which only requires knowledge of something like the highest weights $\lambda$ occurring in the decomposition since writing down projectors and deriving the explicit tensor structure can be difficult and tedious. I feel one should also be able to check this simply by knowing the Young tableaux for $V_{\lambda}$, but I am not seeing how to achieve this.

Thank you for any feedback.

Answer 1

You can use LiE (online version here) to compute the decomposition of $\mathrm{Sym}^2(\mathfrak{g})$ into irreducibles using its symmetrised tensor power function. This should give you the answers you seek for low dimensions at least and you can possibly see what the pattern is going into higher dimensions.

Let's assume $\mathfrak{g}$ is simple. A couple of little things to notice: the Killing form is a nondegenerate, invariant bilinear form on $\mathfrak{g}$. This means there will always be a 1-dimensional trivial bit in $\mathrm{Sym}^2(\mathfrak{g})$ (and a copy of $\mathfrak{g}$ in $\bigwedge^2\mathfrak{g}$). Other than that for type $A_n$ the representations seem to show up exactly as David says in his comments above. So we have the trivial module, a copy of $\mathfrak{g}$ (highest weight $\lambda_1 + \lambda_n$), and then modules with highest weights $2\lambda_1 + 2\lambda_n$ and $\lambda_2 + \lambda_{n-1}$.