Question

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)?

Answer 1

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.