Abelianization of Lie groups 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:


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*What can be said about the abelianizations of other (finite-dimensional) Lie groups?

*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.

 A: 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.
A: (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 $T^{k}$.
