# Fundamental group of a topological group

It is well known that the fundamental group of a path-connected topological group is abelian. Suppose that $$G$$ is a connected topological group and let $$Ab(G)$$ the abelianization of the topological group $$G$$. Is there a relation between $$\pi_{1}G$$ and $$\pi_{1}(Ab(G))$$ ?

• There's a functorial homomorphism. Have you looked at any examples? Do you have any more precise question?
– YCor
Mar 25 '19 at 19:52
• @YCor sure, how far the functorial homomorphism from being an isomorphism
– lab
Mar 25 '19 at 19:59
• Have you looked at any examples?
– YCor
Mar 25 '19 at 20:06
• @YCor no, do you have a trivial easy example ?
– lab
Mar 25 '19 at 20:09
• SU(2) and SO(3) are good ones... Mar 25 '19 at 21:04

This question is arguably too broad, but I interpret that to mean the OP is somewhat new to this area. So, I'll give some basics, in a CW answer, and I encourage others to add to it if they want to. First, for any group $$G$$, the abelianization $$Ab(G)$$ is defined to be the group $$G/[G,G]$$ where $$[G,G]$$ is the commutator subgroup. There is a natural quotient map $$G \to Ab(G)$$. The assignment of $$G$$ to $$Ab(G)$$ is also a functor from the category of groups to the category of abelian groups, and this functor realizes the latter as a reflective subcategory of the former.

If $$G$$ is a topological group, then $$G/[G,G]$$ can be given the quotient topology. See Section 5 of these notes. A common situation of interest is when $$G$$ is a (compact) Lie group. Taking the fundamental group is a functor $$X\mapsto \pi_1(X)$$ to the category of groups. Note that $$\pi_1(G)$$ does not need to be a topological group. See the thesis of Jeremy Brazas. Because $$\pi_1$$ is a functor, there is a natural homomorphism $$\pi_1(G) \to \pi_1(Ab(G))$$. The comments demonstrate that this morphism need not be injective or surjective in general. Here are some links where examples are computed: here, here, here.

If $$G$$ is a compact connected Lie group, then standard classification results show that there is a diagram $$G\xleftarrow{f}H\xrightarrow{g}T\times K$$, where $$f$$ and $$g$$ are surjective with finite kernel (and so are coverings), and $$T$$ is a torus, and $$K$$ is a finite product of simple Lie groups, which have finite fundamental groups. This gives maps $$\pi_1(G)\xleftarrow{\pi_1(f)}\pi_1(H)\xrightarrow{\pi_1(g)}\pi_1(T)\times\pi_1(K)$$, where $$\pi_1(f)$$ and $$\pi_1(g)$$ are injective with finite cokernel. This shows that $$\pi_1(G)$$, $$\pi_1(H)$$ and $$\pi_1(T)$$ are the same up to a finite error. We also get surjective homomorphisms $$\text{Ab}(G)\xleftarrow{\text{Ab}(f)}\text{Ab}(H)\xrightarrow{\text{Ab}(g)}\text{Ab}(T\times K)=T$$, where $$\text{Ab}(G)$$ and $$\text{Ab}(H)$$ are compact connected abelian Lie groups and therefore tori. There are a few more details to sort out, but I am pretty sure that $$\text{Ab}(f)$$ and $$\text{Ab}(g)$$ are again finite coverings so $$\pi_1(\text{Ab}(G))$$ and $$\pi_1(\text{Ab}(H))$$ are again the same as $$\pi_1(T)$$ up to a finite error. For any given $$G$$ it should not be hard to pin down the details and find $$\text{Ab}(G)$$ and $$\pi_1(\text{Ab}(G))$$ explicitly.

As a consequence of the Peter-Weyl Theorem, an arbitrary compact Hausdorff group can be written as the inverse limit of a filtered diagram of compact Lie groups, and one could use this to transfer some results from the Lie case to the general case.

For any connected Lie group $$G$$, there is a maximal compact subgroup $$G_0\leq G$$ such that the inclusion $$G_0\to G$$ is a homotopy equivalence. One should be able to learn something from this, but I have not thought through the details.

• There's no obvious reduction from connected Lie groups to the easy case of compact connected Lie groups, because taking the abelianization does not commute with "passing to a maximal compact subgroup". To be more concrete, if $G=\mathrm{SL}_2(\mathbf{R})$ then abelianization induces $\mathbf{Z}\to 0$ on $\pi_1$, but for the maximal compact subgroups, it induces the identity of $\mathbf{Z}$.
– YCor
Mar 27 '19 at 10:35