Question

If G is a group, its abelianization is the abelian group A and the map G → A such that any map G → B with B abelian factors through A. Abelianization is a functor, and in general a very lossy operation. The map G → A is always a surjection/quotient, because we can construct A by dividing G by the minimal normal subgroup that contains all conjugations ghg^-1h^-1 for g,h∈G.

If V is a finite-dimensional (super)vector space over a field K, then the abelianization of GL(V) is isomorphic to the multiplicative group K^* of non-zero numbers in K. Indeed, the determinant exhibits the desired isomorphism.

Here are two questions I'm curious about:

1. What can be said about the abelianizations of other (finite-dimensional) Lie groups?
2. If V is an infinite-dimensional vector space, what can be said about the abelianization of GL(V)? Most infinite-dimensional vector spaces have some analytic structure, e.g. topological vector spaces, and so it's reasonable to ask that the operators in GL(V) should preserve that structure; you are welcome to take your favorite type of infinite-dimensional vector space and your favorite type of GL(V), if you want.

Answer 1

I don't have anything to say about specific examples, but here are some general remarks. A way to construct the abelianization of any compact group is to consider its image under the product of all its 1-dimensional unitary representations. This is because a compact abelian group is characterized by its set of characters by Pontrjagin duality. More generally, you can construct the double Pontrjagin dual of a locally compact group to get its locally compact abelianization as a subgroup of a space of maps to U(1) with the compact-open topology.

Answer 2

(In some sense, this is just a restatement of what Eric said above....)

For compact groups, quite a lot can be said. Every compact group H' has a finite cover H which is Lie group isomorphic to $T^{k} \times G$, where $G$ is compact and simply connected.

Then, one can easily show that [H,H] = {$e$}$\times G$ and hence that the abelianization of $H$ (which, as in the finite case is H/[H,H]) is $T^{k}$.

The same holds true of the original group $H'$: the abelianization is $H'/[H',H']$ and is isomorphic to $T^{k}$.