Schur's lemma

This fact is related to: linear representation theory
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Statement

Let $G$ be a group and $\rho _{1}:G\to GL(V_{1}),\rho _{2}:G\to GL(V_{2})$ be linear representations of $G$ over a field $k$. Then the following hold:

1. If $\rho _{1}$ is irreducible, then any homomorphism of representations from $\rho _{1}$ to $\rho _{2}$ is either injective, or the zero map
2. If $\rho _{2}$ is irreducible, then any homomorphism of representations from $\rho _{1}$ to $\rho _{2}$ is either surjective, or the zero map
3. If both are irreducible, then any homomorphism of representations from $\rho _{1}$ to $\rho _{2}$ is either an isomorphism of representations, or the zero map
4. If $\rho _{1}=\rho _{2}$ is irreducible, and the field is an algebraically closed field, then any homomorphism from $\rho _{1}$ to $\rho _{2}$ is a scalar multiplication map

Proof

Verbal proof

1. For this, we use the fact that the kernel of any homomorphism of representations is an invariant subspace. If $\rho _{1}$ is irreducible, the kernel is either the whole of $V_{1}$ (in which case we have the zero map) or the zero subspace (in which case we have an injective map).

1. For this, we use the fact that the image of any homomorphism of representations is an invariant subspace. If $\rho _{2}$ is irreducible, the image is either the whole of $V_{2}$ (in which case we have a surjective map) or the zero subspace (in which case we have the zero map).

1. This follows from the previous two parts.

1. Any homomorphism from a representation on that field to itself can be represented by a linear operator from the vector space to itself. This linear operator must have an eigenvalue, because the field is algebraically closed. Subtracting that eigenvalue times the identity matrix, we now get a linear operator that has zero as an eigenvalue, which is also a homomorphism of representations.

But part (3) tells us that any homomorphism of representations is either the zero map or an isomorphism. An isomorphism cannot have zero as an eigenvalue (since it is an invertible linear operator) and hence the new linear operator we get must be the zero map.

Thus the original linear operator must have been a scalar times the identity.