# The Simply Connected Subgroups of GLn(C)?

A friend of mine and I were trying to answer a question related to his research and he couldn't remember whether or not the special linear group over the complex numbers, SLn(C),was simply connected. (It IS,of course.)

This got me wondering:What are all the simply connected topological subgroups of the general linear group over C? Is there a simple characterization of all of them up to isomorphism? What about thier fundamental groups as topological spaces?Are THEY simply connected if the subgroup is? I would expect them to have fundamental groups as basepoints should be easy to choose via the identity matrix.

So is there such a characterization for the simply connected subgroups of GLn(C)?

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The way to do this problem is to take all of the simply connected Lie groups, and then classify their faithful complex representations. – Greg Kuperberg Jun 5 '10 at 4:34
"What about thier fundamental groups as topological spaces?" If the subgroup is simply connected, its fundamental group is trivial, no? – Steve D Jun 5 '10 at 5:53
Greg, is a simply connected subgroup of GL_n(C) necessarily a Lie group? – Qiaochu Yuan Jun 5 '10 at 6:26
@Yuan: Cartan's theorem: Every closed subgroup of a Lie group is a Lie subgroup. – Xandi Tuni Jun 5 '10 at 8:05
@Qiaochu: $\textit{Any}$ subgroup of $GL_n(\mathbb{C})$ is a Lie group (in particular, it need not be closed or connected). However, its topology is inherited from the matrix space only if it is closed (in general, one uses the exponential map on the connected component of the identity and makes all connected components open). – Victor Protsak Jun 5 '10 at 10:58

There can be no such classification except for small n because that would imply the classification of nilpotent Lie algebras up to isomorphism, which is a well-known wild problem.

By Lie-Engel's theorem, any nilpotent Lie algebra of $n$ by $n$ matrices is a direct sum of a central ideal and a Lie algebra of strictly upper triangular matrices. Every Lie subalgebra $\mathfrak{g}\subseteq\mathfrak{gl}_n$ consisting of nilpotent matrices is exponential: the exponential map $\exp$ is a diffeomorphism. In particular, $G=\exp(\mathfrak{g})$ is a simply-connected Lie subgroup of $GL_n(\mathbb{C})$ and $G$ and $\mathfrak{g}$ determine each other up to isomorphism.

By Ado's theorem, every $k$-dimensional Lie algebra is linear and, in fact, a subalgebra of $\mathfrak{gl}_n$ for some $n$ depending on $k$ ($n=k^2+1?$). So classifying simply connected subgroups of $GL_n$ includes as a subproblem classifying all $k$-dimensional nilpotent Lie algebras up to isomorphism. For $k\geq 7$ there are continuous parameters in this moduli space and for general $k$ the classification is considered impossible (I forgot precise bounds).

Of course, in practice there is Greg's way.

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