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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

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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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te llego el mail? –  Pedro Tamaroff Apr 15 '12 at 23:57
    
Sipi. Está en mi lista de cosas para leer :) –  Mariano Suárez-Alvarez Apr 16 '12 at 0:00
    
@Meriano Ok. Tiene una pequeña errata que puedo corregir si queres, pero no cambia mucho. Tengo tiempo? –  Pedro Tamaroff Apr 16 '12 at 0:04
    
thank you! why are the two homs not equal then? perhaps I misunderstood something. –  Rkoustach Apr 16 '12 at 7:09
    
@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

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