# Dimension of Hom(U, V)

all. I am reading a paper and I could not understand the dimension of the vector space $Hom_{\Gamma}(\rho_i\otimes \rho, \rho_j)$. The paper is http://pages.uoregon.edu/njp/Lusztig92.pdf, section 1.1, page 113. Here $\rho$ is a 2-dimensional C-vector space. $\Gamma$ is a finite subgroup of the special linear group $SL(\rho)$. Then $\rho$ is a $\Gamma$-module. Let $I$ be the set of isomorphism classes of simple $\Gamma$-modules. For each $i\in I$, $\rho_i$ is a simple $\Gamma$-module in class $i$. It is said that $dim Hom_{\Gamma}(\rho_i\otimes \rho, \rho_j)$ is 0, 1, or 2. I don't know why. Could you explain this? Thank you very much.

I have another question about this paper. On page 114, equation (c), it is said that $(M, M')=|\Gamma|^{-1}\sum_{\gamma\in \Gamma}tr(\gamma, M)tr(\gamma^{-1}, M')(2-tr(\gamma, \rho))$. Here $(M, M')$ is defined on page 113, section 1.2. $(M, M')=(M\otimes C^2: M')-(M\otimes \rho: M')$, $(M, M')=dim Hom_{\Gamma}(M, M')$. Why the equation (c) is true? What does $tr(\gamma, \rho)$ mean? We could define trace of a linear map of a matrix. But here $(\gamma, \rho)$ is not a matrix or a map. Thank you very much.

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This follows from the lemmas proven in this blog post about the McKay correspondence (see the paragraph under the $\tilde{D}_n$ diagrams). The more general result is also true and also follows by the lemmas in the post: if $\rho$ is a faithful self-dual representation of a finite group $G$ and $\rho_i, \rho_j$ are irreducible, then $\dim \text{Hom}_G(\rho_i \otimes \rho, \rho_j) \le \dim \rho$.

Lemma 6 implies something stronger: the inequality is strict unless $G$ has at most two irreducible representations! Unfortunately I don't know a more direct proof. One is tempted to write down an argument using characters and the triangle inequality, but it doesn't seem to work and I don't know if it can be fixed.

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thank you very much. If we use the lemmas, we need to have the condition $\rho$ is a faithful self-dual representation of $\Gamma$. I don't know whether $\rho$ is a faithful self-dual representation. It is only said that $\rho$ is a 2 dimensional space and is a representation of $\Gamma$. – user Mar 12 '11 at 21:19
@Jianrong: any $2$-dimensional representation with image in $\text{SL}$ is self-dual. This is because a $2 \times 2$ matrix of finite order and determinant $1$ must have eigenvalues $\lambda, \lambda^{-1}$ for some root of unity $\lambda$, hence has real trace. And since $\Gamma$ is supposed to be a subgroup, the representation is faithful by definition. – Qiaochu Yuan Mar 12 '11 at 21:25
thank you very much. – user Mar 13 '11 at 0:33

I don't think the first question is a general feature of groups, so it is probably something special about finite subgroups of SL(2,C). They are basically cyclic or dihedral, and so ρi and ρj have dimension at most 2, and the tensor product has dimension at most 4, so you immediately get Tij is at most 4, and I guess you just rule out having too many linear constituents (and the few exceptions).

Actually this is a general feature of finite groups and their irreducible representations. Suppose χ and θ are irreducible characters and ρ is a character with complex conjugate ρ′. Then [ χρ, θ ] = [ χ, ρθ ]. Whenever φ is a character, [ χ, φ ] ≤ φ(1)/χ(1), and equality holds iff φ = (φ(1)/χ(1))⋅χ. In our case we get [ χρ, θ ] ≤ χ(1)/θ(1) ⋅ ρ(1) but also [ χ, ρθ ] ≤ θ(1)/χ(1) ⋅ ρ′(1) = θ(1)/χ(1) ⋅ ρ(1). Since at most one of { θ(1)/χ(1), χ(1)/θ(1) } is bigger than 1, we get that [ χρ, θ ] ≤ ρ(1). In the paper, ρ(1) = 2. (This does not require faithful or self-dual characters.)

For your third question, I believe tr(g,V) means the trace of the group element g acting on the G-module V. Sometimes he uses M to mean the module and sometimes he uses ρ to mean the module.

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The largest exceptional finite subgroup (the binary icosahedral group) has an irreducible representation of dimension $6$, so I don't think your argument works. – Qiaochu Yuan Mar 12 '11 at 20:13
@Jack, thank you very much. – user Mar 12 '11 at 21:19
When you say the finite subgroups of $SL(2,C)$ are «basically cyclic or dihedral» you are leaving out the interesting one! :) – Mariano Suárez-Alvarez Mar 12 '11 at 22:18
@Jack: this is interesting. Do you have a reference / proof of that inequality $\langle \chi, \varphi \rangle \le \frac{\varphi(1)}{\chi(1)}$? I don't think I am familiar with this. – Qiaochu Yuan Mar 16 '11 at 19:25
@Qiaochu: It follows from the orthogonality relations. I think of it as: [χ,φ]χ is a submodule of φ, so [χ,φ]χ(1) ≤ φ(1), since φ(1) is the dimension of the module φ. It does require χ to be (absolutely) irreducible ([χ,χ]=1), but I don't think any other field restrictions are needed (the Hom's involved factor through/land in semisimple modules). – Jack Schmidt Mar 17 '11 at 1:46