Representations of reductive Lie group

Let $G$ be a reductive algebraic group and $\varrho$ a representation of $G$ in $GL(n)$. Is it true that $\varrho$ is completely reducible? Moreover, how are related the representations of the Lie algebra $\mathfrak{g}$ of $G$ with the one of $G$?Finally, the centre of the identity component of $G$ consists of semisimple transformations, is it true also for $\mathfrak{g}$?

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Are you in positive characteristic? In characteristic zero, a linear algebraic group is reductive if and only if each of its (finite dimensional) representations is semisimple (aka completely reducible). In positive characteristic this fails. –  Xandi Tuni May 12 '10 at 8:34
I'm in characteristic zero, in particular on the complex numbers. –  Michele Torielli May 12 '10 at 9:19

You need to be over a field of zero characteristic and your representation needs to be rational, i.e. matrix entries need to be algebraic functions on $G$. Then it is completely reducible, see any book on algebraic groups, e.g., Jantzen or Humphreys.

You can always differentiate, so a differential of a map $G\rightarrow GL(V)$ is a representation of ${\mathfrak g}$. In the opposite direction, a certain care is required. To integrate a vector field, you need exponential function, which is not, in general, algebraic. However, for a semisimple group in characteristic zero, you have enough nilpotent elements $X\in{\mathfrak g}$, so that the polynomials $e^{\rho (X)}$ define a representation of the group.

Finally, the answer is no. Take ${\mathfrak g}$ to be one-dimensional Lie algebra acting on $K^2$ by the nilpotent nonzero transformation.

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Can you explain a bit more in the case in which I start with a representation of the Lie algebra?in my case, I'm using the adjoint representation of $\mathfrak{g}$ in $\mathfrak{gl}(n)$. –  Michele Torielli May 12 '10 at 9:48
There's a complete proof that representations of reductive groups in characteristic zero are semisimple in II, section 5, of the notes on my website. –  JS Milne May 12 '10 at 11:18
What exactly do you need explaining? –  Bugs Bunny May 12 '10 at 11:58
@milne:thank you. @Bunny:it's not clear to me what happen if I try to lift a representation of $\mathfrak{g}$. –  Michele Torielli May 12 '10 at 12:55
The answer to my last question is then no, but is it true that the element in the centre of $\mathfrak{g} \subset \mathfrak{gl}(n)$ are diagonalizable? –  Michele Torielli May 12 '10 at 13:21

Is it true that $\rho$ is completely reducible?
Certainly not. The counterexamples given to you in your previous question easily adapt for groups. For example, the additive group of the ground field has a 2d representation $x\mapsto\begin{pmatrix}1&x\\\ 0&1\end{pmatrix}$.