# Why $\dim V^{G} = \dim\operatorname{Hom}_{G}(\mathbb{C}, V)$?

Let $G$ be a finite group. Let $V^{G} = \{v\in V: \pi(g)v=v, \forall g\in G\}$ be the subspace of invariants, where $(\pi, V)$ is a representation of $G$. Why $\dim V^{G} = \dim \operatorname{Hom}_{G}(\mathbb{C}, V)$? By Schur Lemma, $\dim \operatorname{Hom}_{G}(\mathbb{C}, V)$ is the number of trivial representations of $G$ which occur in $V$. Thank you very much.

• Hint: Consider the map $f\mapsto f(1)$. Show that it lands in $V^G$ and is a bijection on that. – Tobias Kildetoft Jan 30 '14 at 14:46

A map $f$ in $\hom_G(\mathbb{C},V)$ is a function $f:\mathbb{C}\to V$ which is $G$-linear. We can write this explicitly as $f(g.\lambda)=gf(\lambda)=\lambda g f(1)$. Since $f(g.\lambda)=\lambda f(g.1)$ we can cancel out $\lambda$ to get $f(g.1)=g f(1)$ but the action of $G$ on $\mathbb{C}$ is trivial so that $f(g.1)=f(1)$. We conclude that $$f(1)=g.f(1)$$ This implies that $f(1)\in V^G$. It is easy to see that this map is injective. We can show that it is surjective: take $v\in V^G$ and define $f(\lambda)=\lambda v$ (this is inspired in the previous argument). You can check easily that this $f$ is $G$-linear and that this construction is the inverse of the previous one.

• thank you very much. I think that we don't have $f(\lambda)=\lambda f(1)$ $(\lambda \in \mathbb{C})$ since $f$ is $G$-linear but not necessarily $\mathbb{C}$-linear. – LJR Feb 1 '14 at 8:43
• Usually when a group acts on a vector space one assume that the action is linear (that is $\mathbb{C}$-linear) – Quimey Feb 1 '14 at 16:16

If you're already familiar with the fact that representations decompose into irreducibles (in the semisimple case), you can reason as follows:

• If $U_1\oplus\cdots\oplus U_m$ is a sum of reps then $(u_1,\cdots,u_m)$ is $G$-invariant iff each $u_i\in U_i$ is.
• If $u\in U\setminus0$ where $U$ is irreducible then $u$ is $G$-invariant iff $\,U$ is the trivial representation.

Conclude that the space of $G$-invariants of $V$ is precisely the sum of trivial reps inside $V$.${}^\dagger$ As you have stated, the number of trivial reps is equal to $\hom_G(\Bbb C,V)$ by Schur's lemma.

${}^\dagger$Note that the decomposition of $V$ into irreducibles is not unique. For example, if $V=\Bbb C^2$ is the sum of two trivial reps, then $V=\Bbb C(1,0)\oplus\Bbb C(0,1)=\Bbb C(1,1)\oplus\Bbb C(1,-1)$ gives two different decompositions (different because all of the factors are distinct subspaces). However, for each irreducible $U$ occuring in $V$ with exponent $e$, the subspace $U^{\oplus e}$ occurring in $V$ is unique.