2
$\begingroup$

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.

$\endgroup$
  • 2
    $\begingroup$ Hint: Consider the map $f\mapsto f(1)$. Show that it lands in $V^G$ and is a bijection on that. $\endgroup$ – Tobias Kildetoft Jan 30 '14 at 14:46
5
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – LJR Feb 1 '14 at 8:43
  • $\begingroup$ Usually when a group acts on a vector space one assume that the action is linear (that is $\mathbb{C}$-linear) $\endgroup$ – Quimey Feb 1 '14 at 16:16
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.