The classifying space $BG=|Nerve(G)|$ of an arbitrary topological group $G$ does not necessarily have the homotopy type of a CW-complex but the fundamental group should still be accessible. What is $\pi_{1}(BG)$? A reference on this would be great. My initial guess: $\pi_{1}(BG)$ is the quotient group $\pi_{0}(G)$ for arbitrary $G$

Motivation: There is a natural way to make $\pi_1$ a functor to topological groups. I am interested in relating the topologies of $G$ and $\pi_{1}(BG)$ but the topology on $\pi_{1}(X)$ is boring (discrete) when $X$ is a CW-complex.

  • $\begingroup$ Should that be $\pi_0(G)$? Since the universal bundle $EG$ is contractible, then the long exact sequence of homotopy groups break into [1=\pi_1(EG)\to\pi_1(BG)\to \pi_0{G}\to\pi_0(EG)=1.] $\endgroup$ – Fei YE Oct 9 '10 at 20:19
  • $\begingroup$ Is it obvious that the function $\pi_{1}(BG)\rightarrow \pi_{0}(G)$ from the LES is a homomorphism? $\endgroup$ – Jeremy Brazas Oct 9 '10 at 21:05
  • $\begingroup$ @Jeremy, that it's a homomorphism looks fairly obvious to me --- but don't ask me why... @Fei, following on Jeremy's remark, the proofs (e.g. of long exact sequence) are usually presented for topological groups of the homotopy type of a CW complex, which is a specialization explicitly excluded by the question as posed. Put another way, it's not clear to me even that the sequence is exact. Jeremy, just how arbitrary can these groups be? Is a real algebraic group with the Zariski topology a case you want to consider? $\endgroup$ – some guy on the street Oct 9 '10 at 21:42
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    $\begingroup$ In general, there is no map $\pi_1(BG) \to \pi_0(G)$. See my post below. $\endgroup$ – André Henriques Oct 9 '10 at 22:27

If $G$ is homeomorphic to a Cantor set (e.g. $G=\mathbb Z_p$), then $BG$ contains a copy of the Hawaiian earrings in it. To see this, take a sequence of points of $G$ that converges to the identity element: you'll get a corresponding sequence of 1-cells in $BG$ that converge to the the degenerate 1-cell. The fundamental group of the Hawaiian earrings is a rather wild object, and looks nothing like the free group that you might naively expect.

If, on the other hand, if you agreed to redefine $BG$ to be the fat geometric realization of the simplicial space $NG$, then you would get $\pi_1(BG)\cong\pi_0(G)$, as desired. I would even bet that the above isomorphism respects the natural topologies that are present on both sides.

  • $\begingroup$ Could you explain the subtelty between fat and ordinary geometric realization ? $\endgroup$ – BS. Oct 10 '10 at 9:34
  • $\begingroup$ Thanks very much Andre! The business about the geometric realization makes sense. I am not that familiar with fat geometric realizations but I believe the difference comes from restricting the identifications to only strictly increasing maps in $\Delta$ (correct me if I am wrong). Is there somewhere I can find a proof of the assertion about the group isomorphism? $\endgroup$ – Jeremy Brazas Oct 10 '10 at 15:03
  • $\begingroup$ I don't think that this "known". But I think that it's not too difficult prove it using transversality w.r.t the subset of centers of simplices. $\endgroup$ – André Henriques Oct 11 '10 at 9:41
  • $\begingroup$ @Andre: As you know, the usual geometric realization is given as a certain colimit (coend). I think it is possible to show that the fat geometric realization the same thing as (i.e. weakly equivalent to) taking the homotopy colimit of this same diagram. Right? If so, doesn't your claim follow from the weak homotopy invariance of the homotopy colimit? You simply replace G be a CW approximation. $\endgroup$ – Chris Schommer-Pries Oct 12 '10 at 17:04
  • $\begingroup$ @Chris: I believe that you are correct. But viewing things up to weak homotopy equivalence would tend to forget the topology on $\pi_0(G)$, and I think that Jeremy cares about the latter. $\endgroup$ – André Henriques Nov 7 '10 at 12:58

The first reference in this general area was:

N. E. Steenrod, "Milgram's classifying space of a topological group", Topology 7 (1968) 349–368.

Working in the category of compactly generated Hausdorff spaces, and combining his Theorems 8.1 and 8.3 we have:

Theorem: Let $G$ be a topological group with unit $e$, such that $(G, e)$ is an NDR. Then the canonical map $EG \to BG$ is a quasi-fibration and a principal $G$-bundle.

The NDR-condition asks that the inclusion of $e$ in $G$ is a cofibration. That is of course weaker than asking for $G$ to be a CW-space, but not as general as how the question was posed.

Since $EG$ is contractible, the long exact sequence in homotopy for a quasi-fibration, at the usual base point of $BG$, gives a bijection $$ \partial : \pi_1(BG) \cong \pi_0(G) $$ of $\pi_1(BG)$-sets. This at least gives you an isomorphism between $\pi_1(BG)$ and some group structure on $\pi_0(G)$.


Readers might also like to know a result on the first k-invariant of BG contained in

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.

This shows that the associated crossed module to fundamental groupoid of $G$ has $k$-invariant which is exactly the $k$-invariant of $BG$.

Since this query is about non-connected topological groups, another related ressult is on universal covers of non-connected topological groups

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

which relates the question of the existence of topological group universal covers of $G$ to the theory of ostructions to extensions of abstract groups.

  • $\begingroup$ I should have added that the $k$-invariant referred to is exactly the obstruction to there being a `universal covering' topological group $\tilede G$ of $G$ (in the sense that there is a morphism $p: \tilde{ G} \to G$ of topological groups which on each component is a universal cover of spaces). $\endgroup$ – Ronnie Brown Nov 25 '11 at 19:06
  • $\begingroup$ Misprint: $\tilede{ G}$ should be $\tilde{G}$. $\endgroup$ – Ronnie Brown Nov 25 '11 at 19:07

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