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.

Using the same technique, we can also prove surjectivity. Consider some image of $f$ , say $f(v) \in W$ . Again, since the map is G-module homomorphism, $f(g.v)=gf(v)$ , which means that $f(g.v)$ is also in the image of $f$ . So the image of $f$ is a subspace of $W$ stable under the action of $G$ . Since $W$ is also irreducible, the image must be all of $W$ .

To tackle the second and third parts of the theorem, you simply use the above the two arguments.

If $V,W$ are isomorphic as representations, then by the Fundamental Thoerm of Algebra, the map $f$ has an eigenvalue $\lambda$ . Let $x \neq 0$ be a corresponding eigenvector.

Now, construct another G-linear homomorphism: $f'=f-\lambda I$ . $f'(x)=\lambda I-\lambda I=0$ . $f$ is also a homomorphism of representations. By the previous results, any homomorphism must be either an isomorphism or zero. No isomorphism would map a non-zero element to zero. So, $f'$ is the zero map.

So, $f(v)=\lambda v$ .

This lemma is key to developing the row orthogonality relations in representation theory which I’ll not discuss here

This entire result can be translated(or rather extended) to the language of modules quite easily.

If $A,B$ are two simple modules over a ring $R$ , then any homomorphism $f:A \mapsto B$ is either invertible or the zero map. In particular, when $A \cong B$ , Schur’s Lemma can be restated as:

The endomorphism ring of a simple module $M$ over a ring $R$ ( $End_{R}(M)$ ) is a division ring.

All this stuff can be generalized a little further by allowing the modules/vector spaces to be over some arbitrary algebraically closed field.

Theorem:

Proof:

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