# Complex Lie algebra implies complex Lie group?

I'm not an expert on Lie algebras and Lie groups, but I've learnt a very known theorem which states (someones call it Lie 3rd theorem):

"For every real finite dimensional Lie algebra $\mathfrak{g}$ exist one and only one connected and simply connected real Lie group $G$ such as $\mathfrak{g}=Lie(G)$".

Now, I'm asking if the complex version of the above theorem also holds, thus

"For every complex finite dimensional Lie algebra $\mathfrak{g}$ exist one and only one connected and simply connected complex Lie group $G$ such as $\mathfrak{g}=Lie(G)$".

Or, does exist a quite different, but essentially equivalent version of that theorem?

Thanks for the help,

Diego

• The theorem is true for any nondiscrete, locally compact, complete valued field of characteristic zero. It is done in that generality in Serre's book Lie Algebras and Lie groups. That includes p-adic fields and other weirder examples. – Mariano Suárez-Álvarez Feb 19 '17 at 22:34
• (The result is false if you want the group to be algebraic, and from that fact it is that there are algebraic Lie algebras, those which come from an algebraic Lie group; most Lie algebras, in some sense, are not algebraic) – Mariano Suárez-Álvarez Feb 19 '17 at 22:39
• @MarianoSuárez-Álvarez No: the theorem as stated, true for real/complexes, is not true for $p$-adics as you don't have many connected $p$-adic Lie groups! The statement of the theorem that holds in this setting is that every finite-dim Lie algebra is Lie algebra of a Lie group over the given field, unique up to local isomorphism. (Actually, in the non-Archimedean setting, two Lie groups are locally isomorphic iff have isomorphic compact open subgroups). – YCor Feb 20 '17 at 5:40