2nd half of proof of $\dim V^G=\frac{1}{|G|}\sum_{g \in G}\chi(g) $

Question

Lemma. Let $\rho: G \rightarrow GL(V)$ be a representation, character $\chi$. Then

$$\dim V^G=\frac{1}{|G|}\sum_{g \in G}\chi(g) $$

Proof. RHS: $$\frac{1}{|G|}\sum_{g \in G}tr \rho(g)=\frac{1}{|G|}tr \sum_{g \in G} \rho(g)$$.

Let $\displaystyle M:=\sum_{g \in G} \rho(g)$. $M$ is a linear map: $V \rightarrow V$ and image lands in $V^G$ since $\rho(h)\sum_{g \in G}\rho(g)v= \sum_{g \in G}\rho(g)$.

I do not understand how what they are trying to do in the bit "since $\rho(h)\sum_{g \in G}\rho(g)v= \sum_{g \in G}\rho(g)$.". I can see that somehow it is related in to G-invariance. In particular I am struggling to see what the $\rho(h)$ is related to and why and how it acts on $\sum_{g \in G}\rho(g)$.

Answer 1

The statement $$\rho(h)\sum_{g \in G}\rho(g)v = \sum_{g \in G}\rho(g)v$$ should have the $v$ that I've included in the right hand side. Then it is exactly the statement that the image of $M$ lands in $V^G$. This is because if $v$ is any vector and $M = \sum_{g \in G}\rho(g)$ then $\sum_{g \in G}\rho(g)v = Mv$ is what a vector in the image of $M$ looks like. Now the above statement says $\rho(h)Mv = Mv$. In other words, $\rho(h)$ fixes the vector $Mv$ for all $h$ so $Mv \in V^G$.

Answer 2

The set $ \lbrace hg: g \in G \rbrace $ =$G$. (You're making h act by left multipliation on $G$ which defines a bijection on G). (In other words, as g ranges over G exactly once, then so will $hg$ hence the $\sum\limits_{g \in G} \rho(hg)=\sum\limits_{hg \in G} \rho(hg) =\sum\limits_{g \in G} \rho(g)$