MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $G$ be a (discrete) group, and $1/G$ the corresponding groupoid with one object. Consider the diagram in (the 2-category) Groupoids with one vertex, labeled $1/G$, the one arrow from that vertex to itself, given by the identity map.
$$ \begin{matrix} 1/G \\ {\huge \circlearrowleft} \\ \scriptstyle \mathrm{id} \end{matrix}$$ (This diagram is equivalent to the pair of parallel arrows $1/G \overset{\rm id}{\underset{\rm id}\rightrightarrows} 1/G$. Note that I am not filling in the loop with a 2-cell.)

A cute fact is that the ("2-") limit of this diagram in Groupoids is the action groupoid $G/G$ of the adjoint action of $G$ on itself. (See e.g. 2 limit in nLab or HTT Chapter 4 for a definition of limits.)

Now, in homotopological terms, the groupoid $1/G$ looks like the classifying space ${\rm B}G$, and the above diagram looks like ${\rm B}G \times S^1$. I have the possibly-mistaken impression that limits are supposed to look like topological cones (but maybe this is because we use words like "cone" when talking about limits).

Question: In terms of homotopy, how should I visualize the limit cone $$ \lim\left( \begin{matrix} 1/G \\ {\huge \circlearrowleft} \\ \scriptstyle \mathrm{id} \end{matrix} \right) \quad \begin{matrix} {\huge \to} \\ {\large \circlearrowleft \!\!\!\!\!\! \circlearrowleft} \end{matrix} \quad \begin{matrix} 1/G \\ {\huge \circlearrowleft} \\ \scriptstyle \mathrm{id} \end{matrix} $$ ?

(Edits: per Quid's request in the comments, I replaced some broken images with diagrams, trying to reconstruct them from memory. $\circlearrowleft \!\!\!\!\! \circlearrowleft$ is my attempt at a doubled circle arrow, i.e. a 2-cell filling in the cone walls.)

share|cite|improve this question
    
The images disappeared. Perhaps you can fix this in some way, whence this comemnt. – quid Dec 25 '15 at 14:07
    
@quid Unfortunately, in various computer moves I worry I've lost those diagrams. They might have been $$\begin{matrix} && 1/G && \\ & \swarrow && \searrow & \\ 1/G &&&& 1/G \\ & \nwarrow && \nearrow & \\ && 1/G && \end{matrix}$$ which is equivalent to $1/G \rightrightarrows 1/G$. Or maybe $1/G \leftrightarrows 1/G$. But I think they were both $$\begin{matrix} 1/G \\ \circlearrowleft \end{matrix} $$ i.e. a self-loop leaving from $1/G$ and then returning to it. – Theo Johnson-Freyd Dec 26 '15 at 18:43
    
Thanks for the reply. Perhaps then you could just replace the missing graphics by something so that the Q&A pair makes sense. – quid Dec 26 '15 at 18:56
    
@quid Sorry for the delay --- is this better? – Theo Johnson-Freyd Jan 7 at 21:13
    
Thanks a lot. To me it looks fine (this is just from the look of it, I have no idea about the math in detail). – quid Jan 8 at 2:02
up vote 10 down vote accepted

You should think of the limit of that diagram as "loops in $BG$." A loop in $BG$ is a principal $G$-bundle on $S^1$. Every principal $G$-bundle on the circle comes from taking the trivial principal $G$-bundle on $[0,1]$ and identifying the fibers over $0$ and $1$ (making this the basepoint). If you like left principal bundles, this gluing map has to be right multiplication by an element $g$ of $G$. Of course, you can still do gauge transformations on the circle, and these will have the effect of conjugating $g$ by the value of the gauge transformation at the basepoint.

Thus, principal bundles on $S^1$ can be thought of as $G/G$.

share|cite|improve this answer
    
So, just to emphasize this (and David Ben-Zvi also emphasizes this), I was wrong to read the first diagram as ${\rm B}G\times S^1$; rather, I should read it as $\operatorname{Maps}(S^1\to{\rm B}G)$. (Why?) Then the limit statement is precisely that $\operatorname{Maps}(S^1\to{\rm B}G)=G/G$ up to homotopy. (Why would this be the limit statement, and not, say, the colimit statement? Come to think of it: what's the colimit of that diagram?) – Theo Johnson-Freyd Jan 16 '11 at 6:38
    
Hrm, on my computer the equation Maps(S^1 -> BG) = G/G seems to print in the wrong place. – Theo Johnson-Freyd Jan 16 '11 at 6:39

To slightly amplify what Ben wrote, the diagram is precisely a presentation of $Map(S^1,BG)=L(BG)$ rather than of $BG\times S^1$. More generally the loop space of a space $X$ can be presented as the homotopy fiber product $LX= X\times_{X\times X} X$, the self-intersection of the diagonal, which is a slightly different way (which I find more convenient) to present self-homotopies of the identity map of $X$. In the case of a groupoid (or a stack) this results in the inertia groupoid, i.e., objects (points) together with automorphisms. Again in the case of $BG$ we have one object (the trivial $G$-torsor on a point, in one presentation) and its automorphisms form a $G$, with automorphisms given by $G$ acting adjointly.

On the level of functions/chains (interpretation depending on your context), rather than points, you get a formula that looks more like what you wrote, i.e. $$F(X) \otimes S^1= F(X) \otimes_{F(X)\otimes F(X)} F(X),$$ aka the Hochschild homology (or chains) of functions on $X$.

share|cite|improve this answer
    
If $F(X)$ stands for $C^{\bullet}(X)$ or for $\mathrm{Map}(X,\mathbb{R})$ -but I may have totally misinterpreted what you wrote- then what does $F(X)\otimes S^1$ denote? Is it just tensor product of abelian groups? – Qfwfq Dec 25 '15 at 19:20

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.