# How to interpret topologically that the equalizer in Groupoids of ${\rm id}, {\rm id}: BG \rightrightarrows BG$ is $G/G$ (adjoint action)?

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.)

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The images disappeared. Perhaps you can fix this in some way, whence this comemnt. – user9072 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. – user9072 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). – user9072 Jan 8 at 2:02

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 $G/G$.

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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 $X$.

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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