# Occurrences of trivial representation is equal to dimension of $\{v\in V:\varphi(g)v=v\}$.

Suppose $\varphi\colon G\to GL(V)$ is a complex representation with character $\psi$. If $W=\{v\in V:\varphi(g)v=v,\ \forall g\in G\}$, why is $\dim W=(\psi,\chi_1)$, where $\chi_1$ is the principal character of the trivial representation?

Since the irreducible characters are a basis, write $\psi=\sum a_i\chi_i$, and then $(\psi,\chi_1)=a_1$ by the orthogonality relations. So $(\psi,\chi_1)$ is the number of times the trivial representation occurs in a direct sum decomposition of $\varphi$.

I think the $\mathbb{C}G$-module $V$ decomposes as $$V=a_1V_1\oplus\cdots\oplus a_rV_r$$ where $V_i$ is the irreducible submodule for the representation $\varphi_i$ with character $\chi_i$. So $a_1V_1\subseteq W$, and $\dim a_1V_1=a_1$ since each $V_1$ is $1$-dimensional. MY question is, how can we be sure that $W$ isn't any bigger than this? Why can't there be elements in the other summands which happen to be fixed by all $\varphi(g)$?

Because any vector which is fixed by all $\rho(g)$ spans a $1$-dimensional subrepresentation.
• Is the idea that if such a nonzero $v\in V_i$, by simplicity $\langle v\rangle =V_i$, so the irreducible representation $\varphi_i$ on $V_i$ is necessarily the trivial representation? May 5, 2015 at 6:30
• Exactly. ${}{}$ May 5, 2015 at 6:31