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For any nontrivial representation $R$ of a Lie group $G$, I think it's true that $$R \subset R \otimes A$$ where $A$ is the adjoint representation. I've checked this for a few simple cases.

In terms of the physicists' notation with tensors, this is true for the following reason. Let the fundamental representation be represented by a rank 1 tensor with an upper index, and let the dual representation have a lower index. The adjoint representation is the traceless part of their product (see here) so it has one upper and one lower index. Now if we tensor this with any representation $R$, we can recover the index structure of $R$ by contracting anything, e.g. $$A^i_j R^{kl}_m \supset A^i_j R^{jl}_m$$ has the index structure of the $R$ representation: two upper, one lower.

How would a mathematician prove this statement? Is there a simple and elegant reason this is true?

knzhou
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1 Answers1

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Is there a simple and elegant reason this is true?

Sure, I think so. I'm going to work at the level of Lie algebras and assume $R$ is irreducible. We identify the adjoint representation, as a vector space, with the Lie algebra itself, and consider the action map $$A \otimes R \longrightarrow R \quad \text{given by} \quad x \otimes v \longmapsto x(v),$$ where $x(v)$ denotes the action of the element $x$ of the Lie algebra $A$ on the vector $v$ in the representation $R$. This is a homomorphism of representations$^{[1]}$, which is non-zero if $R$ is non-trivial. Thus $R$ is canonically a quotient of $A \otimes R$.

It follows (at least for finite dimensional representations of compact Lie groups, or in any situation where the representation $A \otimes R$ is semisimple) that $R$ is (perhaps non-canonically) a subobject of $A \otimes R$.

[1] Proof (in case you are interested): For $x,y \in A$ and $v \in R$ we have $$y(x \otimes v)=\mathrm{ad}(y)(x) \otimes v+ x \otimes y(v) \longmapsto (\mathrm{ad}(y)(x))(v)+x (y (v))=y( x (v)),$$ using the definition of representation of a Lie algebra.

Stephen
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