## 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 2010 at 8:34 I'm in characteristic zero, in particular on the complex numbers. – Michele Torielli May 12 2010 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 2010 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 2010 at 11:18 What exactly do you need explaining? – Bugs Bunny May 12 2010 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 2010 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 2010 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 $x\mapsto\begin{pmatrix}1&x\\ 0&1\end{pmatrix}$.

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The additive group is unipotent, which is the opposite of reductive. – Victor Protsak May 12 2010 at 10:35
@Victor: thanks! yes, I guess I read the title of this question better than the body :( I edited the answer accordingly. – Vladimir Dotsenko May 12 2010 at 10:51
Sorry for the misunderstanding. The fact is that a reductive algebraic group is reductive also as Lie group. – Michele Torielli May 12 2010 at 10:58
@Michele: that's fine, a good lesson for me to read things carefully. A reductive algebraic group is reductive as a Lie group, sure, but I fail to see how it is relevant: for a reductive Lie group, the statement you are interested in is false, while for a reductive algebraic group it is true, so you'd better be careful too :) – Vladimir Dotsenko May 12 2010 at 11:01
I wonder where a reductive Lie group is defined solely in terms of its Lie algebra: this sounds like folk etymology to me. For connected linear Lie groups, the following definition is common: a subgroup G of GL(n,R) which is fixed under the transpose map is a reductive Lie group, see e.g. Knapp or Wallach, vol 1. Then a reductive Lie group is algebraic and it is also reductive as an algebraic group. – Victor Protsak May 12 2010 at 21:40