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.


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.


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.

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