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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.

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If you know the representation theory of $\rm SU(2)$, the fact that $\hom(W,W\otimes V) = 0$ is immediate for $W$ a finite-dimensional irrep. Indeed, remember that the finite-dimensional irreps of $\rm SU(2)$ are classified by their dimension — there is one for each positive integer — and we have the decomposition $[n] \otimes 2 = [n-1] \oplus[n+1]$. But I assume that you know this line of reasoning, and are either trying to prove the representation theory or are simply interested in the more general case. –  Theo Johnson-Freyd Apr 25 '10 at 20:27
To be more precise: yes, I know this, but it says nothing about how the representations [n], [n-1], [n+1] decompose as representations of G. –  Qiaochu Yuan Apr 25 '10 at 21:25

4 Answers 4

up vote 5 down vote accepted

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.

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The center of the cyclic group of order n certainly doesn't act trivially, but the result is still true for that case...? –  Qiaochu Yuan Apr 25 '10 at 22:02
Perhaps you were confused about which direction I was proving? I tried to clarify. As a special case of what I wrote (or what Ben explained), for any non-trivial rep $V$ of any abelian group, $\mathrm{Hom}(W, W \otimes V)$ is always trivial. –  Dylan Thurston Apr 25 '10 at 22:13
He's answering the contra-positive. If $Hom(V,W^*\otimes W)$ is non-zero, then Z(G) acts trivially on $V$, you agree? –  David Jordan Apr 25 '10 at 22:14
Ah. Yes; I had also come to this conclusion by playing around with characters. Can it be shown that a finite subgroup of SU(2) has nontrivial center without already knowing what they are? –  Qiaochu Yuan Apr 25 '10 at 22:25
I can prove that a nontrivial finite subgroup of SU(2) has nontrivial center without the classification, which means that your answer is what I need; thanks! The proof is as follows: if V is irreducible, then dim V = 2 divides the order of G, so -I is in G. Otherwise, V is a sum of two one-dimensional representations and G is abelian. –  Qiaochu Yuan Apr 26 '10 at 4:44

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 $\sum_{g\in G} \chi_v(g)|C_{G}(g)|=0$.

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Thanks! I'm going to hold out for an answer more specific to the case I care about, but if one isn't forthcoming I'll accept this answer. –  Qiaochu Yuan Apr 25 '10 at 21:13
Another easy remark is that if $g_1,g_2,\ldots,g_k$ is a full set of representatives for the conjugacy classes of $G$, then Ben's condition is equivalent to $\sum_{i=1}^{k} \chi_{V}(g_i) =0.$ –  Geoff Robinson Jun 2 '11 at 19:56

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 $V$).

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This sounds great, but I'm not sure what the meaning of "compare" is in your first sentence. –  Qiaochu Yuan Apr 25 '10 at 23:21
I'm concerned. He only wants to allow $W^*\otimes W$, and not a string of products of such. If he allows strings of size larger than two, then I think it's well known for $G$ finite that $V$ descends to a rep of $G/Z$, since $G/Z$ acts faithfully on $C[G]$, so that every $G/Z$ rep appears in some tensor power of $C[G]$. –  David Jordan Apr 25 '10 at 23:28
@Qiaochu: I replaced "compare" with "appear" (sorry, I mixed words from different languages). @David: it is enough to find one such W to conclude that V is in the identity class, you don't need to check it for all strings. –  pasquale zito Apr 26 '10 at 0:27
Thank you for clarifying, but I'm still confused. I see that the Hom condition implies that V is in the identity class. But I don't see why every V in the identity class must be contained in W^* ot W for an irreducible W. It seems that V could be contained in W^* ot W ot W^* ot W (here "ot" is shorthand for \otimes). Is there a statement about strings of length 2 in the irreps? Sorry if I'm being dense. –  David Jordan Apr 26 '10 at 0:36
@David: you are absolutely right, silly me! –  pasquale zito Apr 26 '10 at 3:31

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.

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I think you're answering a different question than I'm posing. How is highest weight theory applicable to finite group representations? –  Qiaochu Yuan Apr 25 '10 at 21:13
Yes, when wt(V) is in Q it's automatically a summand of something of the form WW*. For one proof see Claim on the bottom of page 17 of arxiv.org/abs/0810.0084 There's an interesting related result about fusion categories having a "universal grading group" whose trivial part is exactly the summands of WW* see arxiv.org/abs/math/0610726 –  Noah Snyder Apr 25 '10 at 21:17
@QY: I must have misunderstood the question. Oh, I see: I missed the words "if G is a finite subgroup of" and thought you were ust interested in the representation theory of SU(2). My bad. –  Theo Johnson-Freyd Apr 25 '10 at 23:08

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