$\pi_1$ Sequence of Topological Groups

Question

Consider a connected topological group $G$ (not necessarily Lie). You have some maps $G\times G\to G$, such as projection to either summand, or multiplication $(g,h)\mapsto gh$. Now let's look at a slightly more complicated but naturally-occuring map: $(g,h)\mapsto ghg^{-1}h^{-1}$, i.e. $G\times G\to [G,G]\hookrightarrow G$. What goes on at the fundamental group level?

In other words, is it true that $\pi_1(G\times G)\to\pi_1([G,G])\to\pi_1(G)$ is exact?

I have a rather ad hoc reason to believe that the first map is trivial (as $\pi_1$ is abelian here, the commutator $[g,h]$ will unwind itself to the constant loop) and so I would want the second map to be injective.

Update: The comments below take care of this when $G$ is a Lie group!
So what can obstruct $\pi_1$ being injective on $[G,G]\hookrightarrow G$ for non-Lie groups?
Update: It has also been pointed out that this works for finite-dimensional topological groups!
That leaves a possible counterexample for the infinite-dimensional case.

Answer 1

This is really more of a comment, but it kind of answers one of the OP's question, so I am indulging myself: It is a result of W. Browder (Annals, 1961) that $\pi_2$ of a finite dimensional $H$-space is trivial, so the result holds in that setting. I learned of this (and also that this is not true without the finite dimensionality assumption) from @Allen Hatcher's answer to this (very relevant) question: Homotopy groups of Lie groups

Answer 2

This was originally supposed to produce a counterexample, but when I got to the end the proof actually showed that you can't produce a certain family of counterexamples. I'll post this, but only selfishly because I don't want to hit "delete" on what's written. Sorry!

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There are no "homotopical" counterexamples (meaning that, if we make "abelianization" homotopy invariant, no counterexample exists). There are two things that help explain this. The first is that the homotopy theory of based, connected spaces is equivalent to the homotopy theory of simplicial groups (and half of this equivalence is the "classifying space" functor); the second is that, for "good" simplicial groups, the homotopy groups of the abelianization are shifts of the group homology.

Let $X$ be any space. There is a simplicial group $G$ so that

• $|BG|$ is homotopy equivalent to $X$, and

• $G$ is levelwise free.

For the first property, you can use (for example) the Kan loop group construction. For the second property, one can replace any simplicial group by a weakly equivalent one which is levelwise free (for example, by a cofibrant replacement in the model category of simplicial groups).

There is then a short exact sequence of simplicial groups $1 \to [G,G] \to G \to G_{ab} \to 1$, which gives rise to a (quasi)fibration sequence on classifying spaces and a long exact sequence on homotopy groups. $$ \cdots \to \pi_2(G) \to \pi_2(G_{ab}) \to \pi_1([G,G]) \to \pi_1(G) \to \pi_1(G_{ab}) \to 0 $$ so it is equivalent to show that $\pi_2(G) \to \pi_2(G_{ab})$ is always surjective. We also have that $$\pi_i(G) = \pi_{i+1}(BG) = \pi_{i+1}(X)$$ and, because $G$ is levelwise free, $$\pi_i(G_{ab}) = H_{i+1}(BG) = H_{i+1}(X).$$ Therefore, to show injectivity it suffices to show that for any simply-connected space $X$, the map $\pi_3(X) \to H_3(X)$ is surjective. The usual Serre spectral sequence for a fibration $F \to X \to K(\pi_2(X), 2)$ allows one to show that there is an exact sequence $$ H_4(X) \to H_4(K(\pi_2(X),2)) \to \pi_3(X) \to H_3(X) \to H_3(K(\pi_2(X),2)) \to 0. $$ Therefore, it is necessary and sufficient that $H_3(K(A,2))$ vanishes for any abelian $A$.

To show this, I think that the shortest method is to use a free resolution $0 \to R \to F \to A \to 0$ and apply the Serre spectral sequence to the resulting fibration $K(R,2) \to K(F,2) \to K(A,2)$. The total space and fiber have trivial homology in even degrees (they're a colimit of products of $\mathbb{CP}^\infty$). If there was something in $H_3(K(A,2))$, it would have to support a differential to $H_2(K(R,2)) = R$ in order to not survive the spectral sequence, but the edge homomorphism to $H_2(K(F,2)) = F$ is an injection.

Answer 3

What is more interesting is to look at the fundamental groupoid $\pi_1 G$ of a topological group, particularly in the non-connected case. This groupoid inherits a group structure, and so becomes what is called a group-groupoid, or a $2$-group, i.e. a group internal to the category of groupoids. Such objects are equivalent to crossed modules, which are objects which model connected, pointed weak homotopy 2-types.

Here are a couple of relevant papers:

R. Brown and C.B. Spencer, $\cal G$-groupoids, crossed modules and the fundamental groupoid of a topological group'', Proc. Kon. Ned. Akad. v. Wet. 7 (1976) 296-302.

R. Brown and O. Mucuk, ``Covering groups of non-connected topological groups revisited'', Math. Proc. Camb. Phil. Soc, 115 (1994) 97-110.

and just now a search on Baez 2-groups gave 82,700 hits, e.g.

Higher-Dimensional Algebra V: 2-Groups. John C. Baez.

The second paper revisits the work of R.L. Taylor which showed that there is in general an obstruction to a non-connected topological group having a universal covering topological group.

Later: The following may be seen as still not answering the original question, but I hope will be interesting to some readers!

I was trying to convey that the fundamental groupoid $\Phi=\pi_1 G$ of a topological group $G$ contains useful information, so I hope it will be useful to set that out in some, but not full, detail.

As said above, $\Phi$ is in fact a group-groupoid, and so has an associated crossed module, from which may be recovered the group-groupoid. This crossed module, say $\delta: C \to G$, is part, see below, of an exact sequence

$$ 0 \to \pi_1(G,e) \to C \to G \to \pi_0 G \to 1$$

which is known as a crossed sequence. This crossed sequence determines a cohomology class $ k \in H^3(\pi_0 G, \pi_1 (G,e)) $, as shown by Mac Lane, which may also be identified with the first Postnikov invariant of the classifying space $BG$. This invariant is trivial if and only if $G$ admits a universal covering group (assuming $G$ is suitably nice locally), by which is meant a topological group $U$ with a covering map $p: U \to G$ which is also a homomorphism of topological groups, and such that $p$ restricts to a universal covering map for each component of $G$.

Now to give more on the crossed module $\delta: S \to G$. According to one convention, $C$ is the costar (or star in another convention) of $\pi_1 G$ at $e$, i.e. the elements of $\pi_1 G$ which end at $e$. The map $\delta$ then is just the source map. The group structure on $C$ is induced by the multiplication in $G$: i.e. on path classes $[a][b]=[c]$ where $c(t)=a(t)b(t)$. The operation of $G$ on $C$ is by conjugation: $[a]^g= [g^{-1}ag]$. The crossed module rules are of the form: CM1) $\delta(\alpha ^g)= g^{-1}(\delta \alpha)g $; CM2) $\alpha^{-1} \beta \alpha = \beta ^{\delta \alpha}$, for all $\alpha, \beta \in C, g \in G$.

Notice that in the crossed sequence displayed, the fundamental group is the kernel of $\delta$, and is an abelian group and in fact a module over $\pi_0 G$. But this abelian group, and even the module, is, in the non-connected case, and in general, but a pale shadow of the $2$-type of the classifying space.

What I have been trying to convey is the idea that throwing away a larger structure, in this case restricting from a groupoid to a group, may throw away needed information. Thus in homotopy theory there is a concentration on homotopy groups, e.g. second homotopy groups. But it has been shown that in some cases the best way to determine this group may be, indeed may only be, by calculating the whole $2$-type, as determined by a crossed module. This seems to me an interesting inversion of a traditional approach.

Maybe this site is not the place for such remarks?