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

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.

• I presume you already know about the Baker-Campbell-Hausdorff expansion and its consequences? – SZN Jun 13 '17 at 0:30
• @SZN I'm not familiar with that no – leibnewtz Jun 13 '17 at 0:31
• oh. Then check it out. Also see the exponential map. – SZN Jun 13 '17 at 0:32
• @SZN I'm on the Wikipedia page, but I fail to see how this answers the question. I know about the exponential map, but that already assumes we have a Lie group in mind, doesn't it? – leibnewtz Jun 13 '17 at 0:34
• The exponential map maps a Lie algebra back onto its Lie group. The surjectivity of the map has to do with the topology of the Lie group. I read your question as asking about the Lie algebra - Lie group correspondence for the complex semisimple case in particular. Perhaps I have misunderstood. – SZN Jun 13 '17 at 0:39

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.

• Cool thanks. So if the answer to the second question is no, is the correspondence between semisimple Lie algebras and groups not functorial? Is there some weaker sense in which they're equivalent? For example are their associated representation categories equivalent? – leibnewtz Jun 13 '17 at 1:51
• @leibnewtz: there are functors going in both directions, they just don't form an equivalence of categories, but it's very close. For a simply connected Lie group $G$ with Lie algebra $\mathfrak{g}$, $\text{Rep}(G)$ and $\text{Rep}(\mathfrak{g})$ (finite-dimensional representations) are equivalent. – Qiaochu Yuan Jun 13 '17 at 1:53
• Ah, so there's no escaping the simply-connected condition, though I guess it's not too restrictive. – leibnewtz Jun 13 '17 at 1:55
• @leibnewtz: it's not very hard to describe what happens if you relax this condition. If $G$ is connected but not simply connected then we have $\text{Rep}(\mathfrak{g}) \cong \text{Rep}(\widetilde{G})$ where $\widetilde{G}$ is the universal cover, and $\text{Rep}(G)$ is the subcategory of $\text{Rep}(\widetilde{G})$ on which $Z = \pi_1(G)$, regarded as the kernel of the projection $\widetilde{G} \to G$, acts trivially. A nice case to understand here is $G = SO(3)$, or if you want a complex algebraic group, $G = PSL_2(\mathbb{C})$. – Qiaochu Yuan Jun 13 '17 at 1:59
• That's unexpectedly nice. I guess I'll go try to construct some universal covers then :) – leibnewtz Jun 13 '17 at 2:09

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).

• Thank you for your answer. Unfortunately I don't see the obvious set of equations making $Aut(\mathfrak{g})$ an algebraic group. Could you elaborate? – leibnewtz Jun 15 '17 at 2:28
• @leibnewtz: If $e_i$'s denote a basis in the Lie algebra then the equations are $[Ae_i, Ae_j]=[e_i,e_j]$, $\forall i, j$. An element $A\in GL({\mathfrak g})$ satisfies these equations if and only if it defines an automorphism of the Lie algebra. – Moishe Kohan Jun 15 '17 at 3:34