Schur's Lemma (Representation Theory)

This proof is about Schur's Lemma (Representation Theory). For other uses, see Schur's Lemma.

Theorem

Let $\struct {G, \cdot}$ be a finite group.

Let $V$ and $V'$ be two irreducible $G$-modules.

Let $f: V \to V'$ be a homomorphism of $G$-modules.

Then either:

$\map f v = 0$ for all $v \in V$

or:

$f$ is an isomorphism.

Corollary

Let $\struct {G, \cdot}$ be a finite group.

Let $\struct {V, \phi}$ be a $G$-module.

Let the underlying field $k$ of $V$ be algebraically closed.

Let:

$\map {\mathrm {End}_G} V := \leftset {f: V \to V: f}$ is a homomorphism of $G$-modules$\rightset {}$

Then:

$\map {\mathrm {End}_G} V$

is a field, with the same structure as $k$.

Proof

From Kernel is G-Module, $\map \ker f$ is a $G$-submodule of $V$.

From Image is G-Module, $\Img f$ is a $G$-submodule of $V'$.

By the definition of irreducible:

$\map \ker f = \set 0$

or:

$\map \ker f = V$

This article, or a section of it, needs explaining. In particular: Link to a result which shows this. While it does indeed follow from the definition, it would be useful to have a page directly demonstrating this. You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by explaining it. To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{Explain}} from the code.

If $\map \ker f = V$ then by definition:

$\map f v = 0$ for all $v \in V$

Let $\map \ker f = \set 0$.

Then from Linear Transformation is Injective iff Kernel Contains Only Zero:

$f$ is injective.

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It also follows that:

$\Img f = V'$

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Thus $f$ is surjective and injective.

Thus by definition $f$ is a bijection and thence an isomorphism.

$\blacksquare$

Source of Name

This entry was named for Issai Schur.