# Why is $\sum_{g \in G} \rho(g) =0$ for any nontrivial irreducible representation

Let $F$ be an arbitrary field, and $(\rho, V)$ be an irreducible representation of $G$. Then $$\sum_{g \in G} \rho(g) = \begin{cases} 0 & \text{ if } \rho \neq 1_G, \\ |G|1_V & \text{ if } \rho = 1_G. \end{cases}$$

The case when $\rho =1_G$ is clear. But why is the sum $0$ for nontrivial irreducible representations?

Thanks very much.

• You probably want $G$ to be a finite group and $V$ to be a vector space over $F$. – tomasz Jun 17 '13 at 14:13
• You mean $g \neq 1_{G}$ (and $g = 1_{G}$) since $\rho : G \rightarrow \text{GL}(V)$. – Gil Jun 17 '13 at 14:14
• @GilYoungCheong: I think he means what he wrote in this case. $1_G$ is simply the trivial representation. – tomasz Jun 17 '13 at 14:16
• @tomasz Thanks. Then I learn another notation. – Gil Jun 17 '13 at 14:17

## 1 Answer

Here's a hint:

Suppose that $x = \rho(\sum g) v \neq 0$ for some $v$. What happens if you act $g \in G$ on $x$? What does that mean for the subspace spanned by $x$?