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 $\mathfrak{g}$?

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. 


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.


