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 $\mathfrak{g}$?

$\begingroup$ 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. $\endgroup$ – Xandi Tuni May 12 '10 at 8:34

$\begingroup$ I'm in characteristic zero, in particular on the complex numbers. $\endgroup$ – 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 onedimensional Lie algebra acting on $K^2$ by the nilpotent nonzero transformation.

$\begingroup$ 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 $\mathfrak{gl}(n)$. $\endgroup$ – Michele Torielli May 12 '10 at 9:48

$\begingroup$ 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. $\endgroup$ – JS Milne May 12 '10 at 11:18


$\begingroup$ @milne:thank you. @Bunny:it's not clear to me what happen if I try to lift a representation of $\mathfrak{g}$. $\endgroup$ – Michele Torielli May 12 '10 at 12:55

$\begingroup$ 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? $\endgroup$ – Michele Torielli May 12 '10 at 13:21
Edit: I was too fast with my answer, but I am going to keep it here to possibly prevent others from the misunderstanding that I had. The problem is that you ask about Lie groups in the title of your question, and about algebraic groups in the body of your question! The answers differ: for an algebraic group, reductive = "trivial unipotent radical" (and this, in char=0, gives the complete reducibility), while for a Lie group, it is "Lie group whose Lie algebra is reductive" (so the additivel group is perfectly fine). In my view, this mismatch is terrifying, but not much can be done, and there probably will be misconceptions about it forever.
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 $x\mapsto\begin{pmatrix}1&x\\\ 0&1\end{pmatrix}$.

1$\begingroup$ The additive group is unipotent, which is the opposite of reductive. $\endgroup$ – Victor Protsak May 12 '10 at 10:35

$\begingroup$ @Victor: thanks! yes, I guess I read the title of this question better than the body :( I edited the answer accordingly. $\endgroup$ – Vladimir Dotsenko May 12 '10 at 10:51

$\begingroup$ Sorry for the misunderstanding. The fact is that a reductive algebraic group is reductive also as Lie group. $\endgroup$ – Michele Torielli May 12 '10 at 10:58

1$\begingroup$ @Victor: I believe that at some point one of the existing versions of terminology might have been "reductive" for Lie groups with reductive Lie algebras and "linearly reductive" for what now is more commonly called reductive. Some traces of that mess remained till today, see, e.g. eom.springer.de/R/r080440.htm (which messes up as many things as possible) and eom.springer.de/l/l058500.htm. But it's probably true that this terminology is mostly dormant, and is more or less nonexistent in reliable textbooks. $\endgroup$ – Vladimir Dotsenko May 13 '10 at 8:24

1$\begingroup$ Victor, does not one of the sentences in Popov's article read "A linear algebraic group over a field of characteristic 0 is reductive if and only if its Lie algebra is a reductive Lie algebra"? It is more than an bit misleading, isn't it? $\endgroup$ – Vladimir Dotsenko May 14 '10 at 6:47