# algebraic group to the lie algebra and hom

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 $G$?

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Your title does not make a lot of sense! – Mariano Suárez-Alvarez Apr 15 '12 at 22:48

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.
@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. – Mariano Suárez-Alvarez Apr 16 '12 at 7:20