17
$\begingroup$

The Artin braid groups $B_n$ and the symmetric groups $S_n$ are closely related by the maps $1 \to P_n \to B_n \to S_n \to 1$. The infinite symmetric group has interesting interactions with homotopy theory, due to a result of Barratt-Priddy(-Segal(-Quillen(-others))) that "identifies" the sphere spectrum $QS^0$ with $S_{\infty}$.

In light of the short exact sequence above, are there any Barrat-Priddy-esque results on the infinite braid group $B_{\infty}?$ Is there even a loop space structure on $B_{\infty}?$

Ditto for the pure braids $P_{\infty}$, the kernal of $B_{\infty} \to S_{\infty}$.

Improve this question
$\endgroup$

1 Answer 1

Reset to default
19
$\begingroup$

Yes, $B \beta_\infty$ is homology equivalent to $\Omega^2_0 S^2$, the zero component of the double loop space of $S^2$. The map $B_\infty \to S_\infty$ induces the obvious stablisation map $\Omega^2_0 S^2 \to Q_0S^0$.

Improve this answer
$\endgroup$
6
  • $\begingroup$ Do you know a proof (or where one is found)? $\endgroup$ Commented Oct 30, 2010 at 20:49
  • $\begingroup$ I learnt it from G. Segal "Configuration-spaces and iterated loop-spaces" Invent. Math. 21 (1973) for example, special case b) after Theorem 3. I suspect this is not the earliest proof. $\endgroup$ Commented Oct 30, 2010 at 20:56
  • $\begingroup$ Ah, that's a great theorem! Do the pure braids fit in here? Fred Cohen has a result on the simplicial group $P_{\infty}$ containing a simplicial subgroup homotopy equivalent to $\Omega S^2$, but I'm not sure how/if that relates to your statement above. $\endgroup$ Commented Oct 30, 2010 at 21:11
  • 1
    $\begingroup$ The homology of the pure braid groups does not stabilise in the usual sense, so the homology of $BP_\infty$ is going to be very large. However, you should look at the preprint "Representation theory and homological stability" by Church and Farb. $\endgroup$ Commented Oct 30, 2010 at 21:17
  • 6
    $\begingroup$ Closely related to this: there's a virtual vector bundle (classified by a map $B(B_\infty)\to BO$), such that the associated Thom spectrum is equivalent to the mod $2$ Eilenberg MacLane spectrum. That is, "mod 2 homolgy = braid bordism". See F. Cohen, "Braid orientations and bundles with flat connections", Invent. Math. 46 (1978). $\endgroup$ Commented Oct 30, 2010 at 22:40

You must log in to answer this question.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.