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Let $\mathfrak g$ be a simple Lie algebra. Is there any way to find the $\mathfrak g$-invariant subspace of $\mathfrak g \otimes \mathfrak g\ $? I am familiar with the result for $\mathfrak g = sl_2(\mathbb C)$ in which case the subspace is $\mathbb C \Omega,$ where $\Omega$ is a Casimir element corresponding to the Killing form. Can the same result have any generalization for arbitrary simple Lie groups (or for arbitrary Lie groups)?

I am quite curious at this stage. Any comment or suggestions would be appreciated.

Thanks!

Anacardium
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    Yes any simple Lie algebra has a single (up to scale) invariant bilinear form given by the Killing form. This can naturally be associated with an element of $\frak{g}\otimes\frak{g}$ and its span is an invariant subspace (indeed a trivial representation of $\frak{g}$. – Callum Feb 08 '23 at 22:42
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    To be super clear on terminology, this is the only subspace of $\frak{g}$-invariant elements (I.e. it is a trivial representation) but there are other $\frak{g}$-invariant subspaces (in general this tensor product will decompose into irreducible subrepresentations for example) – Callum Feb 08 '23 at 22:45
  • @Callum$:$ What's that association i.e. which element of $\mathfrak g \otimes \mathfrak g$ does the Killing form correspond? – Anacardium Feb 09 '23 at 04:28
  • Any bilinear form is an element of $\frak{g}^* \otimes\frak{g}^$. Since it is nondegenerate it also defines an isomorphism $\frak{g}^ \cong \frak{g}$ (which we can also see is $\frak{g}$-equivariant in this case) and thus $\frak{g}^* \otimes\frak{g}^* \cong \frak{g}\otimes\frak{g}$. So the Killing form is sent to an element of $\frak{g}\otimes\frak{g}$ – Callum Feb 09 '23 at 07:48

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