Let $ V=\mathbb{C}^n $ and let $ V^* $ be the dual space of $ V $.

Let $ M(n,\mathbb{C}) $ be the vector space of all $ n \times n $ complex matrices. Let $ U(n) $ be the group of unitary matrices. Let $ G $ be a subgroup of $ U(n) $.

There is a natural action of $ G $ on $ V $ by $ g \cdot v=gv $, and a natural action of $ G $ on $ V^* $ by $ g \cdot v=v^* g^* $ where $ g^*=g^{-1} $ is the conjugate transpose of $ g $. Let $ W $ be a subrepresentation of $ V $, call this representation $ \lambda: G \to GL(W) $. And let $ \lambda^*: G \to GL(W^*) $ be the corresponding subrepresentation of $ V^* $.

There is also a natural action of $ G $ on matrices by conjugation $ g \cdot T = gTg^{-1} $. Let $ R $ be a subrepresentation of $ M(n,\mathbb{C}) $ with respect to this conjugation action. Call this representation $ \omega: G \to GL(R) $.

Take two vectors $ w_1,w_2 \in W $ and a matrix $ T \in R $ and consider the product $ w_1^*Tw_2 $ where $ w_1^* $ denotes the dual vector of $ w_1 $ with respect to the standard Hermitian form.

Is it true that if the decomposition of $ \lambda^* \otimes \omega \otimes \lambda $ into irreducibles has no copies of the trivial representation then it must be the case that $$ w_1^* Tw_2=0 $$ for all $ w_1,w_2 \in W $ and $ T \in R $?

Context: This appears to be a form of/ related to/ generalization of/ what physicists call the Wigner-Eckart theorem. https://en.wikipedia.org/wiki/Wigner%E2%80%93Eckart_theorem

  • $\begingroup$ I can't make heads or tails of what the Wigner–Eckart theorem is saying in non-physics / purely representation-theoretic language. Also, "branching rules"? $\endgroup$
    – coiso
    Dec 6, 2023 at 3:06
  • $\begingroup$ @coiso lol ya me too. Some of the links here to more mathematical physics papers are kind of helpful though physics.stackexchange.com/questions/401013/… and physics.stackexchange.com/questions/721439/… $\endgroup$ Dec 6, 2023 at 3:07
  • $\begingroup$ Oh and branching rules is a term common in physics for taking a tensor product of representations and decomposing it into a direct sum of irreps. So a physicist would say that this is using "branching rules" in the sense that we are decomposing $ \pi^* \otimes Ad \otimes \pi $ into irreps to see if the trivial irrep is a subrep. I forgot that the term "branching rules" isn't really used in math. The term does appear for example here en.wikipedia.org/wiki/…. $\endgroup$ Dec 6, 2023 at 15:46
  • $\begingroup$ I'd say the term is used in math. But I believe "branching rules" refer to actual formulas used to calculate the multiplicity of irreps in tensor products. We're simply using the fact decomposition into irreps exist, we're not using any branching rules. $\endgroup$
    – coiso
    Dec 6, 2023 at 17:12
  • $\begingroup$ ya that's fair! anyway I've removed the term branching rules from the title $\endgroup$ Dec 6, 2023 at 17:46

1 Answer 1


Consider the representation $ (W^* \otimes R \otimes W, \lambda^* \otimes \omega \otimes \lambda) $. The map $$ W^* \otimes R \otimes W \to \mathbb{C} $$ given by $$ w_1 \otimes T \otimes w_2 \mapsto w_1^*Tw_2 $$ is a homomorphism of $ G $ representations. In particular note that the action of $ g $ here is by $$ g \cdot (w_1^* \otimes T \otimes w_2) =(g \cdot w_1^*) \otimes (g \cdot T) \otimes (g \cdot w_2)=(w_1^* g^*) g T g^* (g w_2)=w_1^* T w_2 $$ so the action of $ g $ on $ \mathbb{C} $ is trivial. So the image of this irrep must be either the trivial rep $ \mathbb{C} $ or the zero rep $ 0 $. But a completely reducible representation can only have a quotient which is the identity irrep $ \mathbb{C} $ if it already includes a copy of the identity irrep. Thus if $ \lambda^* \otimes \omega \otimes \lambda $ does not include a copy of the trivial irrep then the quotient cannot be the trivial irrep and thus must be the $ 0 $ representation. In other words we have $$ w_1^*Tw_2=0 $$ for all $ w_1,w_2 \in W $ and $ T \in R $ if there are no copies of the trivial irrep in $ \lambda^* \otimes \omega \otimes \lambda $.

  • $\begingroup$ The universal property is overkill to see $V^\ast\otimes W\otimes V\to\Bbb C$ is a morphism. (Plus, since you wrote $V\times W\times V$ instead of $V^\ast\otimes W\otimes V$, it is not actually a morphism in the first argument. Note $V^\ast$ has conjugate scalar multiplication.) Otherwise argument is good. Note the hypothesis is necessarily false if $W\ne 0$, since $v_1^\ast Tv_2=0$ for all $v_1,v_2\in V$ implies $T=0$. (Set $v_1=Tv_2$ to get $\|Tv_2\|^2=0$, hence $Tv_2=0$, for all $v_2$.) $\endgroup$
    – coiso
    Dec 6, 2023 at 2:57
  • $\begingroup$ @coiso that's fair I'll take that out. Also what do you mean by "Note the hypothesis is necessarily false if $ W \neq 0 $." Which hypothesis? $\endgroup$ Dec 6, 2023 at 3:11
  • $\begingroup$ The hypothesis that $V^\ast\otimes W\otimes V$ contains no copy of the trivial 1D rep. $\endgroup$
    – coiso
    Dec 6, 2023 at 3:14
  • $\begingroup$ Good point! I've updated (including changing some notation) to express more clearly the generality I'm interested in. Of course if we take the subrepresentation of $ V $ to be all of $ V $ we exactly recover the original more specific result which is the one you correctly characterize in your comment. $\endgroup$ Dec 6, 2023 at 15:39

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