Let $G$ be a linear algebraic group and let $\rho:G \rightarrow GL(V_{1})$ and $\psi:G \rightarrow GL(V_{2})$ be finite representations. Why is $Hom_{G}(V_{1},V_{2}) \subset Hom_{\mathfrak G}(V_{1},V_{2})$, where $\mathfrak G$ is the Lie algebra of $G$?
algebraic group to the lie algebra and hom
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representation-theory
lie-algebras
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0Your title does not make a lot of sense! – 2012-04-15
1 Answers
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Because a representation of the group induces a representation of the Lie algebra of the group on the same vector space, and an homomorphism of representations of a group is automatically an homomorphism for the associated Lie algebra representations.
This is immediate in the case of Lie groups. If $\rho:G\to\mathrm{GL}(V)$ is a homomorphism of Lie groups, turning $V$ into a representation of $G$, then we can take the differential $d\rho:\mathfrak{g}=T_eG\to T_e\mathrm{GL}(V)=\mathfrak{gl}(V)$, which is automatically a Lie algebra homomorphism, and this turns $V$ into a $\mathfrak{g}$-module.
The algebraic group version of this is exactly the same, mutatis mutandi.
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0@Rkoustach, If $G$ is a Lie group of dimension *zero*, then the Lie algebra does not see anything... More generally, the Lie algebra can only see what happens in the connected component of the identity. – 2012-04-16