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Given a representation of an algebraic group $$\Gamma: G \to GL(V)$$ If we take the differential of $\Gamma$ at $e$, we get $$\Gamma^*: Lie(G) \to gl(V)$$ Suppose that $\Gamma^*$ turns out to be an irreducible and faithful representation. Then does that imply that $\Gamma$ has a finite kernel ? Also, is $\Gamma$ irreducible representation. I would appreciate if a small proof quoting results from a text is given as an answer. Any comments are also welcome !

Thanks !

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Hint: $\Gamma$ does not have necessarily a finite kernel. Consider $f:R\rightarrow Gl(2,R)$ defined by:

$f(t) =\pmatrix{cos(t) & sin(t)\cr -sin(t) & cos(t)}$. Its kernel is $\{2\pi n, n\in Z\}$.

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  • $\begingroup$ Refer Humphreys page 85 for this which tells that this is true. I $\endgroup$ – Jagdeep Singh Apr 1 '16 at 7:42

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