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Let $ G $ be a semisimple Lie group and $ (\pi,V) $ a faithful finite dimensional representation of $ G $. Consider the action of $ G $ on $ V \otimes V $ by $$ g \cdot (v_1 \otimes v_2)= gv_1 \otimes gv_2 $$ Is it always true that one of the irreducible sub representations of $ V \otimes V $ is the adjoint representation of $ G $?

Thoughts so far:

This is true for $ G=SU_n $ and the natural $ n $ dimensional faithful representation. Indeed in this case $ V \otimes V $ decomposes as a direct sum of the adjoint representation (dimension $ n^2-1 $) with one copy of the trivial representation (dimension 1).

Ian Gershon Teixeira
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    This is not true even for $GL$ and the natural module $V$. $V\otimes V^*$ decomposes as the sum of the adjoint and the trivial. $V\otimes V$ always has $\Lambda^2(V)$ and $S^2(V)$ as submodules, or dimensions $n(n-1)/2$ and $n(n+1)/2$. It will rarely hold for small-dimensional modules $V$, is my guess. – David A. Craven Jun 17 '22 at 21:57
  • @DavidA.Craven is it true if we require G to be semisimple? Or are there still obvious counterexamples? – Ian Gershon Teixeira Oct 13 '22 at 01:39
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    @IanGershonTeixeira The claim isn't true for $SL(3, \mathbb R)$ and the standard representation (call it $V$), for example. Indeed, decomposing into irreducible subrepresentations gives $V \otimes V \cong S^2 V \oplus \bigwedge^2 V \cong S^2 V \oplus V^*$, but these subreps have dimension $6, 3$, resp., whereas the adjoint representation $\mathcal{sl}_3(\Bbb R)$ has dimension $3^2 - 1 = 8$. – Travis Willse Oct 13 '22 at 02:06
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    Are you sure you didn't mean to ask about $V \otimes V^{\ast}$? This always contains $\mathfrak{g}$ as a subrepresentation, given by the action map $\mathfrak{g} \to \text{End}(V)$. – Qiaochu Yuan Oct 13 '22 at 03:32
  • @MarianoSuárez-Álvarez Ya I was just confused/didn't do a careful job reading. Other papers referred me to https://arxiv.org/pdf/math/0502080.pdf and the introduction there mentions small tensor powers of $ V $ like $ V^{\otimes 2} $ but in retrospect the actual result relevant for me is Theorem 1.5, which is about $ V \otimes V^* $, as David A. Craven immediately and Qiaochu Yuan yesterday both suspected. – Ian Gershon Teixeira Oct 13 '22 at 23:02

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The claim is false in general; see below for some counterexamples for simple $G$. As mentioned in the comments, by definition any faithful representation $V$ of $\mathfrak{g}$ defines an embedding $\mathfrak{g} \hookrightarrow \operatorname{End} V \cong V \otimes V^*$ of the adjoint representation, $\mathfrak{g}$, that is, $V \otimes V^*$ always contains an isomorphic copy of the adjoint representation.

In the case that $V$ admits a nondegenerate $G$-invariant bilinear form $\beta$, that form defines an isomorphism of $G$-representations $V \to V^*$ by $v \mapsto \beta(v, \,\cdot\,)$, i.e.,"raising an index", and so in that case $V \otimes V \cong V \otimes V^*$, and in particular $V \otimes V$ contains a copy of $\mathfrak{g}$. This condition applies, for example, to all faithful representations of simple Lie groups of type $B, C, F$, and $G$, as well as $E_7$ and $E_8$.

We can find simple counterexamples to the original claim by looking for simple Lie algebras with outer automorphisms satisfying a particular condition.

  • For $\operatorname{SL}_n(\Bbb C)$, $n > 2$, and its standard standard representation $V$, the tensor square decomposes into irreducibles as $$V \otimes V \cong \operatorname{Sym}^2 V \oplus \bigwedge\!{}^2 \,V ,$$ neither summand of which is the adjoint representation. (The decomposition still holds for $n = 2$, but in that case the adjoint representation is $\mathfrak{sl}_2(\mathbb C) \cong \operatorname{Sym}^2 V$.)

  • For $G = \operatorname{SO}_{4 m + 2}(\mathbb C)$, $m > 0$, and $V$ and $S_\pm$ the standard and positive (negative) spin representation, respectively, the tensor square of $S_\pm$ decomposes into irreducibles as $$S_\pm \otimes S_\pm \cong V \oplus \bigwedge\!^3 V \oplus \cdots \oplus \bigwedge\!^{2 m - 3} V \oplus \bigwedge\!^{2 m - 1} V \oplus \bigwedge\!^{2 m + 1}_\pm V,$$ where $\bigwedge\!^{2m + 1}_\pm V$ is the representation of (anti-)self-dual $(2m + 1)$-forms, but the adjoint representation is $\mathfrak{so}_{4 m + 2}(\mathbb C) \cong \bigwedge\!^2 V$ .

  • For a more exotic example, take $G$ to be the exceptional complex simple Lie group $\operatorname{E}_6$, and let $V$ denote either of the $27$-dimensional fundamental representations. Then, the tensor square of $V$ decomposes into irreducibles as: $$V \otimes V = W \oplus V^* \oplus \operatorname{Sym}^2 V ,$$ where $W$ is the (only) $351$-dimensional subrepresentation of $\bigwedge^2 V$. (The representation $W$ is itself also a counterexample, but the decomposition of $W \otimes W$ into irreducibles is more complicated.)

Travis Willse
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