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Given a semisimple Lie algebra $\frak{g}$ over $\mathbb{C}$, and a finite dimensional irreducible representation $V$, with dual representation $V^*$, we know that the decomposition of $V \otimes V^*$ must contain a copy of the trivial module. Is it true that it will only ever contain a single copy of the trivial representation?

Does this generalise to the setting of suitably qualified semisimple rigid monodial categories? A specific guess:

For a semisimple rigid braided monoidal category $\mathcal{C}$, with $V$ a simple object in $\mathcal{C}$, and $V^*$ its dual, the decomposition of $V \otimes V^*$ into simple objects contains one and only one copy of the unit object of $\mathcal{C}$.

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    $\begingroup$ Identifying $V\otimes V^{*}\simeq \operatorname{Hom}(V,V)$, this follows from the Schur's lemma. $\endgroup$ – Victor Protsak Jun 21 '18 at 14:16
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$\newcommand{\C}{\mathcal{C}}$ To expand on Victor's comment: you don't actually need $\C$ to be semi-simple (let's assume it's abelian to be safe). By definition the multiplicity of $1_C$ in $V\otimes V^*$ is the dimension of $$Hom_\C(1_\C,V\otimes V^*)\cong Hom_\C(V,V).$$

Now Schur Lemma is valid at this level of generality, so if $V$ is simple this space is 1-dimensional.

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