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When talking about the Eilenberg-Maclane space $K(G,n)$, we usually restrict our attention to the situation where $G$ is abelian. In that case, we get $\Omega K(G,n)=K(G,n-1)$, so we can call $K(G,n)$ a delooping of $K(G,n-1)$.

Since $\pi_n$ is always abelian for $n>1$, it only makes sense to talk about $K(G,1)=BG$ for $G$ nonabelian anyways. So there definitely shouldn't be delooping of this space, because then it would have $\pi_2=G$, which is impossible. From the previous paragraph, it seems like we should therefore be able to say that the nonabelianity of $G$ (i.e., the nontriviality of the commutator $[G,G]$) is the obstruction to delooping $BG$. But this isn't very satisfying, because I can't quite see what's going on with the actual space.

All of which motivates my (slightly open-ended/up-to-interpretation) question:

How should I think about delooping? Is it nothing more than thing like "for the space $X$ that we care about, it just so happens that we've got $Y$ with $\Omega Y\simeq X$", or is there a definite way to measure obstructions? In the cases where a delooping exists, is there an explicit method for its construction?

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This is a bit vague. There's machinery that detects when something is a loop space, I suggest you read up about that first and then ask a more focussed question when you've done that. –  Loop Space Oct 8 '10 at 17:18
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Start with J. P. May's "The geometry of iterated loop spaces", LNM 271, or Boardman-Vogt's "Homotopy invariant algebraic structures on topological spaces", LNM 347. Roughly, the structure of a nice multiplication operation on X provides you with a delooping. Stasheff's joint review: ams.org/mathscinet-getitem?mr=420610 –  Tyler Lawson Oct 8 '10 at 17:26
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The answer to your motivation -- before your bold question -- is that $\pi_1$ of a loop space is always abelian. There's a very cute argument for this, which is a standard intro algebraic-topology homework problem. But your bolded question is different. "Delooping" could be interpreted as the classifying space functor. So one interpretation of your question could be, provided the classifying space functor is defined for $X$, when is $\Omega BX$ naturally equivalent to $X$? Is that the kind of question you're interested in? –  Ryan Budney Oct 8 '10 at 18:14
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@Andrew: I figured as much, I just didn't even know what the machinery is called. Which is what (hopefully) distinguishes my question from a question like "Tell me about group theory." I didn't expect someone to write me a textbook, of course. Maybe the problem is also that I'm just looking for a nice picture I can keep in my head, which I would get from understanding this machinery. –  Aaron Mazel-Gee Oct 8 '10 at 20:36
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(continued) So people sometimes say "classifying space" for delooping, if it's not a group. Note that the following two conditions are equivalent: (1) $\Omega Y$ is equivalent to $X$. (2) There is a fibration with base $Y$, fiber $X$, and contractible total space. –  Tom Goodwillie Oct 10 '10 at 2:16

2 Answers 2

up vote 8 down vote accepted

One possible answer: Stasheff proved that a (connected) space $X$ is (homotopy equivalent to) a loopspace if and only if $X$ is an algebra over the $A_\infty$ operad (or rather I should say an $A_\infty$ operad).

See for instance this article.

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This is very nice. Thanks! –  Aaron Mazel-Gee Oct 8 '10 at 21:38
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That's not exactly right as stated -- to be a loop space X has to not only be an algebra over an $A_\infty$ operad, but must also be grouplike, i.e. $\pi_0$ of it is a group (under the monoid operation induced by the $A_\infty$ structure). Otherwise $\Omega B X$ is only a "group completion" of X. –  Mike Shulman Oct 8 '10 at 22:54
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(The linked article simplifies the situation by mentioning only connected spaces, which are of course automatically grouplike.) –  Mike Shulman Oct 8 '10 at 22:56

I'm not sure whether you'll like this, but my natural response to "how should I think about delooping?" is to invoke (higher) category theory. You may know that a homotopy 1-type, i.e. a space (probably a CW complex) with $\pi_n=0$ for n>1, is uniquely specified up to (weak) homotopy equivalence by its fundamental groupoid. In fact, one has an equivalence of (2- or homotopy-) categories, so we can identify homotopy 1-types with groupoids. Under this identification, the space BG which deloops a discrete group G is identified with the groupoid with one object and G as the automorphism group of that object. So one-step delooping of a discrete group really is just the simple process of considering a group as a one-object groupoid (although in the homotopy theory world it requires a fairly elaborate construction).

At higher levels, the "homotopy hypothesis" in higher category theory (which is a theorem for some definitions of higher category) says that homotopy n-types can be identified with n-groupoids, and arbitrary homotopy types with ∞-groupoids. Moreover, the identification of groups with one-object groupoids is believed to continue to higher categories as well: 2-groups (i.e. groupoids equipped with an extra group structure up to coherent isomorphism) can be identified with one-object 2-groupoids, and similarly an n-group can be identified with a one-object n-groupoid. Thus, deloopability of a space requires that it be equipped with a suitable group structure (up to homotopy, i.e. up to equivalence), and in that case its delooping corresponds on the categorical side to regarding an n-group as a one-object n-groupoid (where possibly n=∞).

Finally, as to the obstruction to delooping BG when G is not abelian, only when G is abelian is BG itself a (2-)group. The reason is the same one that other people have mentioned—Eckmann-Hilton—but I prefer to think about it in these terms.

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I do like this. In my imaginary autobiography, this chapter of my life will be entitled "How I learned to stop worrying and love category theory". Why doesn't the identification with groupoids extend to an equivalence involving non-simply-connected spaces, though? This seems like it's probably answered in your second paragraph, but I can't figure it out. –  Aaron Mazel-Gee Oct 10 '10 at 2:32
    
It does. In fact, only a non-simply-connected space can be modeled by a non-discrete groupoid, since the automorphism groups of the groupoid specify the $\pi_1$s of the space. –  Mike Shulman Oct 10 '10 at 4:18

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