## Schur’s Lemma

This is a standard result that any beginner in the study of the representation theory would be aware of. I merely present restate the theorem a little(essentially I don’t) and prove it.

## Theorem:

Let $V,W$ be representations of a group $G$. If $f:V \mapsto W$ is a $G-linear$ map, then the following are true assuming that $V$ and $W$ are not isomorphic:

1)If both $V,W$ are irreducible, then $f$ is either an isomorphism or the zero map.

2)If only $V$ is irreducible, then $f$ is either injective or the zero map.

3)If only $W$ is irreducible, then $f$ is either surjective or the zero map

If $V,W$ are isomorphic as representations, then $f$ is a scalar multiplication map.

#### Proof:

Let $\rho_{V},\rho_{W}$ be the respective representations. Here, $V,W$ are vector spaces over an algebraically closed field.

We tackle the first part of the theorem.

Consider the kernel of $f$, say $Ker(f)$, which is set of all $v \in V$ such that $f(v)=0$. Since $f$ is a G-linear map, $f(g.v)=g.f(v)=0$ implies that if $v \in Ker(f)$, then $g.v \in Ker(f)$. Hence, $Ker(f)$ is stable under the action of $G$ which makes it a sub-representation. Since $V$ is irreducible, this means that either $Ker(f)$ must be zero or $f$ must be the zero map. This proves that $f$ is injective.