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Let $B_n$ be the braid group on $n$ strands, $B_{\infty}$ the direct limit of braid groups. For a discrete group $G$, we let $BG$ to be the classifying space of $G$.

After reading this question, I was curious about the following :

Why is the group completion of $\coprod_{k \geq 0} BB_{k}$ homotopy equivalent to $\mathbb{Z} \times BB_{\infty}^{+}$?

(I think I've seen such things come up fairly often in algebraic topology - where you have symmetric groups $\Sigma_n$, $\operatorname{GL}_n(R)$, mapping class groups, etc instead of braid groups here. Can we somehow generalize all of them?)

PS) Let me edit the last part. After skimming through some papers, let me phrase it this way.

Given a family of groups $\{G_{k}\}$, say $G_{k} \subset G_{k+1}$, (what I have in mind are symmetric groups, braid groups, mapping class groups, general linear group, etc) let $M := \coprod_{k \geq 0} BG_k$, and $G_{\infty} = $ direct limit of $G_k$.

Is it true that the group completion of $M$, is $\mathbb{Z} \times BG_{\infty}^{+}$? Maybe we need some additional structure for $M$ and the groups $G_k$ - like, that they form a monoidal category or something....?

Basically, what are the conditions so that the group completion of $M$ is, $\mathbb{Z} \times BG_{\infty}^{+}$?

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3 Answers 3

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This is explained in my paper "Group-Completion", local coefficient systems, and perfection, Q. J. Math. (Quillen Memorial Issue) 64 (3) (2013) 795-803. A sufficient condition is for the monoid to be homotopy commutative.

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  • $\begingroup$ Thank you so much! :) $\endgroup$
    – May
    Commented Apr 29 at 6:46
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See Proposition 1 in

McDuff, D.; Segal, G.
Homology fibrations and the "group-completion'' theorem.
Invent. Math. 31 (1975/76), no. 3, 279–284.
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  • $\begingroup$ Group Completion Theorem (let's say the paper that you have cited) is about homology equivalence, and not homotopy equivalence. I think some of the examples given in the paper are relevant (along with Proposition 2 somehow), but I can't see it clearly... $\endgroup$
    – May
    Commented Apr 17 at 14:57
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    $\begingroup$ @May: after the plus construction, both sides are simple spaces, so that a homology equivalence implies a homotopy equivalence - this is in fact why we need the plus construction $\endgroup$
    – Maxime Ramzi
    Commented Apr 17 at 15:39
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    $\begingroup$ For another very good paper on the matter of the group completion theorem see dpmms.cam.ac.uk/~or257/GCrem.pdf . $\endgroup$
    – Ulrich Pennig
    Commented Apr 17 at 16:11
  • $\begingroup$ @ Maxime Ramzi) I see... So do we apply the plus construction on the homology equivalence $M_{\infty} \rightarrow \Omega BM$ (Notation as in McDuff & Segal's paper, with $M$ = disjoint union of braid groups)? In that case, how do we get rid of the plus on the right hand side, and where is the $\mathbb{Z}$ coming from? (It must be related with $\pi_0(M)$, I guess) $\endgroup$
    – May
    Commented Apr 17 at 16:42
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    $\begingroup$ @May : the $\mathbb Z$ comes from $\pi_0(M_\infty)$ : if you take $M= \coprod_n BB_n$, then $M_\infty$ is not $BB_\infty$, rather its $\pi_0$ is $\mathbb Z$ $\endgroup$
    – Maxime Ramzi
    Commented Apr 17 at 17:52
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Probably not what you are expecting, but the answer is known to be true for both.

I asked my advisor and he told me that it had been known since the 80s that it was true for monoidal categories (perhaps you need to tweak the conditions a little bit). Further, he remarked that there weren't much references but told me to check out "Higher Algebraic K-theory Part II" by Quillen.

Also, you might want to check out the book "Infinite Loop Spaces" by J.F. Adams, Page 89~91.

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