5
$\begingroup$

I asked this question yesterday on the setting of an exercise problem (Ex 2.8) from Serre's book "Linear representations of Finite Groups" (I'm teaching myself representation theory...)

Now that that is sorted, I'm still stuck on the actual problem: Let $\rho: G\to GL(V)$ be a linear representation of a finite-dimensional complex vector space $V$, and suppose $V = W_1 \oplus \dotsb \oplus W_1 \oplus W_2 \oplus \dotsb \oplus W_2 \oplus \dotsb \oplus W_l \oplus \dotsb \oplus W_l$ is a decomposition into irrreducible representations, and let $V_i := W_i\oplus \dotsb \oplus W_i$.

Let $H_i$ be the vector space of linear mappings $h: W_i \to V_i$ such that $\rho_s h = h \rho_s$ for all $s\in G$.

Show that $\dim H_i = \dim V_i / \dim W_i$. [Hint: Use Schur's lemma.]

What I've got so far: if $h\neq0$ then $h(W_i)$ must be an irreducible subrepresentation of $\rho$ isomorphic to $W_i$, and such $h$ mapping to the same image form a one-dimensional subspace by Schur's lemma. For the given explicit decomposition $V_i := W_i\oplus \dotsb \oplus W_i$ with $k$ summands, let $h_\alpha$ send the first summand to the $\alpha$-th summand. These $h_\alpha$ are linear independent, but I wasn't able to show that they span...

Please give me some hints!

$\endgroup$

1 Answer 1

Reset to default
2
$\begingroup$

You are asked to compute the dimension of the vector space $\mathrm{Hom}_G(W_i,V_i)$ of $G$-module homomorphisms from $W_i$ to $V_i$; explicitly, this consists of linear maps $\phi$ from $W_i$ to $V_i$ such that for all $g \in G$ and $w \in W_i$ one has $\phi(g w)=g \phi(w)$. The right way to think about this is to use the fact that Hom is additive (in both variables, for any representations---this is the fact that a map to a direct sum is uniquely determined by its compositions with the projections, which can be arbitrary): $$\mathrm{Hom}_G(U,V \oplus W) \cong \mathrm{Hom}_G(U,V) \oplus \mathrm{Hom}_G(U,W),$$ so that $$\mathrm{Hom}_G(W_i,V_i)=\mathrm{Hom}_G(W_i,W_i \oplus \cdots \oplus W_i) \cong \mathrm{Hom}_G(W_i,W_i) \oplus \cdots \oplus \mathrm{Hom}_G(W_i,W_i) .$$ Now Schur's lemma implies that the dimension of $\mathrm{Hom}_G(W_i,W_i)$ is $1$, so the dimension you are after is just the number of summands---in other words, the quotient $\mathrm{dim} (V_i) / \mathrm{dim}(W_i)$.

$\endgroup$
5
  • $\begingroup$ May I ask if $\cong$ in your case means isomorphic as $G$-modules, or just vector spaces? $\endgroup$
    – user71815
    Jun 10, 2013 at 17:24
  • $\begingroup$ @user71815 As vector spaces---both sides of the iso are $G$-modules, but with the trivial action of $G$ (they are precisely the trivial isotypic component of the space of all linear maps, thought of as a $G$-module via "conjugation"). $\endgroup$
    – Stephen
    Jun 10, 2013 at 18:49
  • 1
    $\begingroup$ @Steve : Can you kindly explain why $Hom_G (W_i\oplus W_i)$ is 1-dimensional by Schur's lemma? If the two copies of W_i are not $equal$, just $isomorphic$, then the lemma says that each element in $Hom_G(W_i\oplus W_i)$ is either zero map, or an isomorphism. But I can't see why that implies dimension is 1. $\endgroup$
    – Hajime_Saito
    Jun 15, 2013 at 6:35
  • $\begingroup$ @usersujo, This uses the stronger version of Schur's lemma, depending upon the fact that the base field is algebraically closed. If $W$ is an irrep and $\alpha:W \rightarrow W$ is an automorphism, then (since our ground field is alg. closed) we may find an eigenvector $w \in W$ for $\alpha$ of eigenvalue $\lambda$. Since $\alpha$ commutes with the action of the group, the set of eigenvectors with eigenvalue $\lambda$ is actually a submodule of $W$, and must therefore be everything (by irreducibility). Thus $\alpha$ is just mult. by a scalar, which is what I claimed. Usually [contd] $\endgroup$
    – Stephen
    Jun 16, 2013 at 1:17
  • $\begingroup$ ...this fact is also referred to as "Schur's lemma". It's completely standard. $\endgroup$
    – Stephen
    Jun 16, 2013 at 1:19

You must log in to answer this question.