4
$\begingroup$

I've been reading some papers carefully, with their proofs (Notations are given at the end).

The following comes from "Braids, mapping class groups and categorical delooping" by Song & Tillmann.

The group completion of $\coprod_{k \geq 0} BB_{k}$ is given by $\mathcal{G}(\coprod_{k \geq 0} BB_{k}) \simeq \mathbb{Z} \times BB_{\infty}^{+} \simeq \Omega^{2}S^{2}$. So,

$BB_{\infty}^{+} \simeq \Omega_0^{2} S^{2}$, where the subscript 0 indicates that only the 0-component is considered.

Now, the following seems to be a well-known result, but I am not able to see how :

$\Omega_0^{2} S^{2} \simeq \Omega^2S^{3}$.

Notations :

Let $B_n$ denote the braid group with $n$ strands, and let $B_{\infty}$ be its direct limit.

For a CW complex $X$, let $\Omega X$ be its loop space, $X^{+}$ its plus construction, and $S^n$ the $n$-sphere.

Given a (discrete) group $G$, we write $BG$ for its classifying space, and $\mathcal{G}$ for its group completion, i.e. $\mathcal{G} = \Omega B$.

Improve this question
$\endgroup$
6
  • 1
    $\begingroup$ For a simply-connected space $X$, you have $\Omega_0X = \Omega X$. So you can reduce the problem to showing $\Omega_0^2S^2 \simeq \Omega_0^2S^3$. I guess you mean homotopy equivalence here? $\endgroup$
    – David Roberts ♦
    Commented Apr 15 at 7:35
  • 5
    $\begingroup$ I think I'd want to actually write down the specific map, for instance looping the Hopf fibration $S^3 \to S^2$, to get $\Omega_0^2S^3 \to \Omega_0^2S^2$. But then this is a map of pointed connected spaces that induces isomorphisms on all homotopy groups, and you should be able to use Whitehead's theorem to then get that this is a homotopy equivalence. $\endgroup$
    – David Roberts ♦
    Commented Apr 15 at 7:38
  • $\begingroup$ I have a similar question. How is the group completion of the disjoint union and $\mathbb{Z} \times BB_{\infty}^{+}$ homotopy equivalent? $\endgroup$
    – May
    Commented Apr 17 at 6:23
  • $\begingroup$ @David Roberts, thank you so much! I can't believe I missed this part, I immediately got it when you said Hopf fibration :). $\endgroup$
    – wind
    Commented Apr 17 at 6:31
  • $\begingroup$ @May sounds like a good separate question ;-) $\endgroup$
    – David Roberts ♦
    Commented Apr 17 at 11:25

1 Answer 1

Reset to default
3
$\begingroup$

The point is that $S^3$ is the universal $2$-connected cover of $S^2$, via the Hopf fibration $S^3 \to S^2 \to K(\mathbb Z, 2)$. As suggested by David Roberts, this map induces an isomorphism on $\pi_i$ for $i \geq 3$ (as you can see from the long exact sequence in homotopy). So on double loop spaces, it induces an isomorphism on $\pi_i$ for $i \geq 1$, hence an equivalence when restricted to the zeroth connected component.

Improve this answer
$\endgroup$
1
  • $\begingroup$ I got it immediately when David Roberts mentioned Hopf fibrations. Can't believe I missed this step! Thanks :) $\endgroup$
    – wind
    Commented Apr 17 at 6:32

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .