Schur Lemma

General Version

Let $S_1$ and $S_2$ be simple $R$-modules. Then ${\mathrm{Hom}}_R(S_1,S_2)=0$ unless $S_1\simeq S_2$ in which case the endomorphism ring ${\mathrm{End}}_R(S_1)$ is a division ring.

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Representation Theory Version

Schur lemma

Let $V$ and $W$ be two finite-dimensional irreducible representations of a group $G$ and $\varphi :V\to W$ is $FG$-homomorphism, then either $\varphi$ is invertible, or $\varphi =0$. Moreover, if $V=W$, then $\varphi$ is an isomorphism.

In particular, if $F$ is algebraically closed and $V=W$, then a non-zero homomorphism $\varphi :V\to V$ is a scalar multiple of the identity endomorphism.

a Corollary

Corollary

Any finite-dimensional division $k$-algebra over an algebraically closed field $k$ is isomorphic to $k$.

?- 或更学术地归入 Wedderburn’s little theorem（魏德本小定理） 的更广泛背景中，尽管它原本是关于有限除环的。

just a Schur Lemma Meetup

• If $\theta :V\to V$ is a $\mathbb{C}G$-isomorphism, then $\theta$ is a scalar multiple of the identity endomorphism $1_V$.

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If $k$ is algebraically closed field, $\rho$ is a irreducible representation of $G$ in $V$, $\dim V<\infty$ and $S:V\to V$ morphism of $G$-modules, then there exists $\lambda \in k$ such that $S=\lambda \mathrm{Id}$.

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Let $V,W$ be two irreducible $A$-modules. If $\phi :V\to W$ is a homomorphism, then either $\phi =0$ or $\phi$ is an isomorphism. If $k$ is algebraically closed and $V$ is finite-dimensional irreducible, then for any $\phi :V\to V$ there is $\lambda \in k$ such that $\phi =\lambda \mathrm{Id}$.

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