Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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.

share|improve this question
5  
To see that the first map is trivial, take two paths and homotope the first such that it is constant on the identity for the first half and the second such that it is constant on the identity on the second half. Now the commutator with respect to the pointwise product vanishes. This is basically also what you do to prove that $\pi_1(G)$ is abelian. –  Ulrich Pennig Aug 17 '12 at 6:58
1  
The second map can easily seen to be injective for compact Lie groups using the long exact sequences on homotopy groups associated to the abelianization short exact sequences and the fact that $\pi_2$ of a Lie group is $=0$. –  Fernando Muro Aug 17 '12 at 8:56
8  
the commutator map $G\times G\to G$ is a classical instance of a map that is trivial on all $\pi_k$ ($G$ connected Lie group) but is not homotopic to a constant in general (precisely, if the maximal compact subgroup of $G$ are non-abelian). –  YCor Aug 17 '12 at 14:32
    
@Fernando: I see that this works even without compactness of the Lie group, great! So in general if $G$ is not Lie then we might have $\pi_2(G_{ab})\ne 0$, in which case perhaps there is another way to go about this. –  Chris Gerig Aug 17 '12 at 18:17
1  
This questions was asked (and answered) five years ago(!): lehigh.edu/~dmd1/je228 (by answered, I mean, just as now, positively for Lie groups, and with "umm..." for general top. groups: lehigh.edu/~dmd1/je228 –  Igor Rivin Aug 21 '12 at 16:04

3 Answers 3

up vote 4 down vote accepted

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

share|improve this answer
    
Thanks for the link and reference and relevant answer! –  Chris Gerig Aug 22 '12 at 0:09

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!


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.

share|improve this answer

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?

share|improve this answer
18  
@Ronnie: What does your answer have to do with the given question? –  John Klein Aug 17 '12 at 13:14
    
@John: Good question! In answers on this site, there is the specific question, and also the broader aspects, of which young people may need to be told; and in this case those aspects are not in the standard texts. So the abelian nature of the fundamental group of a topological group is a small part of the structure of the fundamental groupoid of a topological group, which leads to the subject of $2$-groups. Since the late 1960s I have publicised that $1$-dimensional homotopy is better expressed in terms of groupoids rather than groups, and I feel need to keep repeating it! –  Ronnie Brown Aug 18 '12 at 9:53
12  
@Ronnie: with all due respect, it is not clear to me how the broader aspects, as you put it, fit in with the particular concrete question that was asked. Maybe you can elaborate better... In some way, what you seem to be doing is presenting comments that belong at best in the "remarks" field (and in my opinion, many of your comments amount to unabashed advertising for your own stuff). But let the people who vote on these matters decide for themselves how useful your answers are... –  John Klein Aug 19 '12 at 2:14
16  
@Ronnie: I think you would draw better attention to your work if you mentioned it in a way that actually answered the question. By posting your work in a way that does not respond to the question at hand, you potentially confuse a lot of people. People could spend a lot of time reading your papers -- assuming (by your reputation) that somehow your work is relevant to this particular question, only to get frustrated later once they come to understand it does not. This could draw more negative attention to your work than positive. –  Ryan Budney Aug 19 '12 at 2:22

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.