Is every complex semisimple Lie algebra associated to a complex semisimple algebraic group?

Question

I know that every complex semisimple algebraic group has a complex semisimple Lie algebra. Can we go in the other direction?

Given a complex semisimple Lie algebra $\mathfrak{g}$, can we find a complex semisimple algebraic group $G$ such that $Lie(G)=\mathfrak{g}$? Is the category of complex semisimple algebraic (or Lie) groups equivalent to the category of complex semisimple Lie algebras?

This seems like it would be a fairly standard result if true, but I haven't been able to find an answer to my question anywhere.

Answer 1

The answer to the first question is yes; you can just run through the classification of complex simple Lie algebras and do it case-by-case.

The answer to the second question is no; the correct statement is that the category of finite-dimensional Lie algebras is equivalent to the category of finite-dimensional simply connected Lie groups, over either $\mathbb{R}$ or $\mathbb{C}$, and this equivalence respects restricting to semisimple things on both sides.

Answer 2

Regarding the 1st question: Doing this on "case-by-case" basis is an awful idea (just think about discovering an algebraic group with the Lie algebra $E8$ with bare hands); trying this over reals is even harder since the number of "exceptional cases" is so high. A better way to do so is to observe that given a semisimple (finite dimensional, over real or complex numbers) Lie algebra ${\mathfrak g}$, the automorphism group $Aut({\mathfrak g})$ is an algebraic group (it is given by an obvious set of equations) whose Lie algebra is isomorphic to ${\mathfrak g}$. The latter part does require a proof but does not require a case-by-case analysis and works equally well over real and over complex numbers; the key is to show that all derivations of ${\mathfrak g}$ are inner, see e.g. here for a very short proof (all you need to know is the definition and the fact that the Killing form is nondegenerate in this case).