Theorem
Schur's Lemma
Let $(\rho, V)$ and $(\sigma, W)$ be irreducible representations
(Irreducible Representation) of $G$ over an algebraically closed field $k$.
- Any $G$-module homomorphism $\phi : V \to W$ is either zero or an isomorphism.
- $\mathrm{End}_G(V) \cong k$.
Proof
(1) Since $\ker(\phi)$ and $\mathrm{Im}(\phi)$ are subrepresentations of $V$
and $W$ respectively, irreducibility forces each to be $\{0\}$ or the whole
space. The non-trivial cases give an isomorphism.
(2) For any $\lambda \in k$, the map $\phi - \lambda \cdot \mathrm{id}_V$ is
again a $G$-endomorphism. Since $k$ is algebraically closed, $\phi$ has some
eigenvalue $\lambda_0$, and $\phi - \lambda_0 \cdot \mathrm{id}_V$ has a
nontrivial kernel. By (1) it must be identically zero, so $\phi = \lambda_0
\cdot \mathrm{id}_V$.
Depends on
Dependency Graph
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