For a representation V of a finite group G, when is Hom(W, W⊗V) trivial for all irreps W?

Question

This is probably really easy, but I just need someone to help me get mentally unstuck. As part of a description of the McKay correspondence, I want to show that if $G$ is a finite subgroup of $SU(2)$ and $V$ the corresponding 2-dimensional representation, then $\dim \text{Hom}(W, W \otimes V) = 0$ for any irreducible representation $W$ of $G$. I suspect the result is true in slightly greater generality, but it clearly can't always be true. Since $\dim \text{Hom}(W, W \otimes V) = \dim \text{Hom}(W \otimes W^{\ast}, V)$, the result is false if, for example, $V \simeq W \otimes W^{\ast}$ or is a direct summand thereof for some $W$.

So I am wondering when, for a given $G$ and $V$, it is always true that $\dim \text{Hom}(W, W \otimes V) = 0$ for all irreducible representations $W$. One can easily reduce to the case that $V$ is irreducible.

Answer 1

One necessary condition is that the center of $G$ needs to act trivially in $V$ for $\mathrm{Hom}(W, W \otimes V)$ to ever be non-trivial. The character of the center justs multiplies in a tensor product, and so we can't have a map from $W$ to $V \otimes W$ if $V$ has a non-trivial central character. (This also follows from Ben's observation above.)

I don't know if this is also sufficient. This boils down to looking at groups with trivial center.

In any case, you originally asked about finite subgroups of SU(2) and its fundamental representation. These are all classified, and none of them have trivial center, so all satisfy this property. Proof of the last bit: any subgroup $G$ of SU(2) gives a subgroup $\overline{G}$ of SO(3) by projection. If $\overline{G}$ has even order, it has an element of order 2, which necessarily lifts to an element of order 4 whose square is $-1$, so $-1 \in G$. The only subgroups of SO(3) of odd order are the odd cyclic groups, which lift to Abelian groups. Some references are is these notes by Dolgachev or an earlier MO question.

Answer 2

This is true if and only if $V$ doesn't occur in the permutation representation of $G$ acting on itself by conjugation (since that's the sum over irreps of $W\otimes W^*$). This is probably the cleanest description you're likely to get.

Of course, one can also state this in terms of characters in which case you want $\sum_{g\in G} \chi_v(g)|C_{G}(g)|=0$.

Answer 3

For a compact group $G$ one can define the following equivalence: given two irreps $X$ and $Y$, $X \sim Y$ if they both appear as summands in a finite string of tensor products of irreps $X_1 \otimes X_2 \otimes \dots X_n$. The equivalence classes have the structure of an abelian group which turns out to be the dual of the centre of $G$. This was conjectured in http://arXiv.org/abs/math/0311170 and proven in http://arxiv.org/abs/math/0312257. Thus $Hom(W \otimes W^*,V) \neq 0$ iff V is in the identity class (i.e. the centre acts trivially on $V$).

Answer 4

If you care only about semisimple Lie theory, you can get most of the way just from looking at the weights of the corresponding representations. Let $W$ be simple with highest weight $\mu$, and let $\mu^*$ be the highest weight of $W^*$. Let ${\rm wt}(W)$ be the set of weights of $W$, and let $Q$ be the root lattice for your Lie group. Then ${\rm wt}(W) \subseteq \mu + Q$ and ${\rm wt}(W^*) \subseteq \mu^* + Q$, and so ${\rm wt}(W \otimes W^*) \subseteq \mu + \mu^* + Q$ is again contained within some coset of $Q$ in the weight lattice $P$. But we know that $0$ is a weight of $W \otimes W^*$, and so $\mu + \mu^* + Q = Q$. Which is to say that the weights of $W \otimes W^*$ are all roots.

In particular, if $V$ is semisimple with highest weight that is not a root, then it is certainly the case that $\hom(V, W\otimes W^*) = 0$. This handles e.g. the defining representations of ${\rm SL}(n)$.

When ${\rm wt}(V) \subseteq Q$, I'm not sure of the answer.