Skip to main content

Timeline for How should I think about delooping?

Current License: CC BY-SA 2.5

18 events
when toggle format what by license comment
Oct 15, 2010 at 6:22 comment added Daniel Pomerleano There are interesting deloopings for BG when G non abelian sometimes for example when G is $U(\infty)$. There is this H-space $BU^{\otimes}$ and BBU tensor comes up in twisted K-theory
Oct 10, 2010 at 20:44 comment added Aaron Mazel-Gee Thanks for the reference. So there's a funny topological monoid $M=\mathbb{N}\cup \{ \infty\}$, and it's valid to say that $\Omega BM=*$?
Oct 10, 2010 at 17:11 comment added Ryan Budney Aaron, there are lots of cases where group-completion "destroys" the homotopy-type of the original monoid. Consider the monoid $\\{0,1,2,\cdots,\infty\\}$ under addition. It's group-completion is trivial. You might like to read this thread: mathoverflow.net/questions/13942/…
Oct 10, 2010 at 2:41 comment added Aaron Mazel-Gee @ Tom, regarding your last point: This isn't an example since $\mathbb{N} \cong \mathbb{Z}$ as spaces, but I'm pretty sure isn't $\Omega B \mathbb{N}$ also just the circle. If so, perhaps we could construct an example where a topological monoid and its groupification already have distinct homotopy types (although off the top of my head, I can't think of any topological monoids that aren't already groups -- maybe it's impossible?). And in any case, what do we want to mean when we're asking for a delooping, then? Is there some sort of universal delooping?
Oct 10, 2010 at 2:18 comment added Tom Goodwillie Finally, may I point out that the same space can have essentially different deloopings. Two spaces may have homotopy equivalent loopspaces without being homotopy equivalent themselves.
Oct 10, 2010 at 2:16 comment added Tom Goodwillie Note that $\Omega Y$ has a multiplication which is not associative but is associative up to homotopy in a certain strong sense ("$A_\infty$"). It is a marvelous fact (pointed out by others here) that this has a converse: If $X$ has an $A_\infty$ multiplication (and if the resulting monoid $\pi_0(X)$ is a group), then $X$ can be delooped. But sometimes you deloop a space by discovering a fibration having that space as fiber and having contractible total space.
Oct 10, 2010 at 2:16 comment added Tom Goodwillie (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.
Oct 10, 2010 at 1:58 comment added Tom Goodwillie "Classifying spaces": I'm just trying to clarify some language here. To a topological group $G$ is associated a space $BG$ which may be called a delooping of $G$ or a classifying space of $G$. It is called a delooping because $\Omega BG$ is (weakly homotopy) equivalent to the space $G$. It is called a classifying space because it is the base of the universal principal $G$-bundle (or you could think of some other type of bundle with structure group $G$, for example real oriented vector bundles of rank $n$ if $G=SO(n)$.
Oct 8, 2010 at 23:09 answer added Mike Shulman timeline score: 16
Oct 8, 2010 at 22:48 comment added Ryan Budney I wasn't aware the result is named the Eckmann-Hilton argument, but yes, it's essentially that argument. On your second point, classifying spaces make sense for a variety of objects, in particular for categories. mathoverflow.net/questions/23857/… So you could try to make sense of your question for quite a variety of $X$'s.
Oct 8, 2010 at 21:38 vote accept Aaron Mazel-Gee
Oct 8, 2010 at 21:16 comment added Aaron Mazel-Gee @Ryan: That's just something like the Eckmann-Hilton argument, right? Even without knowing that, my point was just that a priori it's impossible to shift the homotopy up a notch if $\pi_1$ is nonabelian. I do like the perspective that I'm just looking for a functor $B$, but what do you mean "provided the classifying space functor is defined for $X$"? As far as I know, a classifying space is just the representing space for $X$-bundles. Are there pathological topological spaces where such a thing doesn't exist, or is it deeper than that?
Oct 8, 2010 at 20:36 comment added Aaron Mazel-Gee @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.
Oct 8, 2010 at 18:14 comment added Ryan Budney 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?
Oct 8, 2010 at 17:26 comment added Tyler Lawson 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
Oct 8, 2010 at 17:25 answer added Kevin H. Lin timeline score: 10
Oct 8, 2010 at 17:18 comment added Andrew Stacey 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.
Oct 8, 2010 at 17:12 history asked Aaron Mazel-Gee CC BY-SA 2.5