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

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.

• I think the answer for type $A_n$ is the highest weights $2\lambda_1+2\lambda_n$, $\lambda_2+\lambda_{n-1}$, $\lambda_1+\lambda_n$ and $0$. But someone will hopefully be along who knows for sure. Aug 27, 2021 at 18:24
• For type $A$, I would approach proving that by writing the adjoint as the tensor product of $\lambda_1$ and $\lambda_n$ (minus a trivial), and then taking the tensor square of this. One can write down the factors of this, then compute their dimension. Now there should be a unique way of assembling these into modules of the right dimensions for the exterior and symmetric square. Aug 27, 2021 at 18:27
• I get the weights of the product natural * natural * dual * dual to be 20...02, 20...10, its dual 01...02, 01...10 and 00...00 twice. The dimensions of these are easy, and are $n^2(n+1)^2/4-1$, $n^2(n^2-1)/4$, $n^2(n^2-1)/4$, $n^2(n-1)^2/4-1$ and $1$. There's a unique way of adding a subset of these together to obtain $n^2(n^2+1)/2$. Aug 27, 2021 at 20:45
• @EdRich - "For the case at hand, we have $V_λ=g$ and $n=2$ where $S_2$ has two elements being the identity and..". In your case the Lie algebra is semi simple. If $\mathfrak{g}$ is simple it follows there are no submodules $V\subseteq \mathfrak{g}$ hence $\mathfrak{g}\cong V(\lambda)$ is irreducible. In general if $\mathbb{S}(\mu)\mathbb{S}(\lambda)$ is the composite of two Schur-Weyl modules there is no known formula for its decomposition into irreducibles. In your case you seek $Sym^2(V(\lambda))$. Aug 28, 2021 at 11:51
• @EdRich - You must somehow give a construction of all highest weight vectors $v\in Sym^2(\mathfrak{g})$ and this problem may be formulated as a problem in "invariant theory". There is a group acting on $Sym^2(\mathfrak{g})$ and the highest weight vectors are invariant under this action. Aug 28, 2021 at 15:42

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 $$\lambda_2 + \lambda_{n-1}$$.