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This question ought to have a straightforward (perhaps even glaringly obvious) answer, but so far I've already wasted a few hours trying to untangle this web of inconsistent identifications. I'm sure someone more experienced in this area will be able to quickly point out exactly where things go wrong.

Let $G$ be a Lie group and let $\mathbf{B} G$ denote the moduli stack obtained by fixing a model for the homotopy orbits of $G$ acting on the point (For example, one can take the nerve of the resulting action groupoid).

This smooth stack classifies smooth $G$-bundles. More precisely, if we are given smooth manifold $M$, the mapping space ${\rm Map}(M,\mathbf{B}G)$ is equivalent to the infinity groupoid ${\rm Bun}^{\rm sm}_G(M)$ of smooth principal $G$-bundles on $M$, with smooth isomorphisms and higher isomorphisms between them (To prove this one can, for example, resolve $M$ by its Cech nerve and identify the resulting mapping space directly).

I was under the impression that the map ${\rm Bun}_G^{\rm sm}(M)\to {\rm Bun}_G(M)$, which forgets the smooth structure and returns the underlying (topological) principal $G$-bundle was a natural equivalence. If this is the case, then since the latter is equivalent to ${\rm Map}(M,BG)$ and the projection $M\times \mathbb{R}\to M$ induces an equivalence $${\rm Map}(M,BG)\to {\rm Map}(M\times \mathbb{R},BG)\;,$$ we should conclude that $\mathbf{B}G$ is homotopy invariant.

On the other hand, if $\mathbf{B}G$ were homotopy invariant, then it should be equivalent to the locally constant stack of its global sections. The global sections functor defines a right infinity adjoint and should therefore commute with delooping. But the global sections of $G$ just return the points of $G$, viewed as a discrete simplicial set. We would therefore be forced identify $\mathbf{B}G$ with the locally constant stack of $BG^{\delta}$, where $\delta$ indicates that we have taken $G$ to have the discrete topology. This stack clearly classifies something very different. In particular, if $M$ is a smooth manifold, then we can identify the space of maps to this stack with the space ${\rm Map}( M ,BG^{\delta})$, where we have forgetten the smooth structure on $M$.

Question

Clearly, something is wrong here. Can someone help point me in the right direction?

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  • $\begingroup$ The global sections of $\Omega\operatorname{Bun}_G^\mathrm{sm}$ are not $G^\delta$, they are the homotopy type of $G$. $\endgroup$ – Marc Hoyois Oct 15 '16 at 14:42
  • $\begingroup$ @MarcHoyois Then maybe I have made an error in identifying ${\rm Bun}^{\rm sm}_{G}$ with $\mathbf{B}G$? The latter should have global sections $\mathbf{B}G^{\delta}$ no? $\endgroup$ – Daniel Grady Oct 15 '16 at 14:45
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    $\begingroup$ Indeed, $\mathbf BG$ is a $1$-truncated sheaf on smooth manifolds, and $\operatorname{Bun}_G^\mathrm{sm}\simeq\operatorname{Bun}_G$ is the associated constant sheaf (aka its shape). $\endgroup$ – Marc Hoyois Oct 15 '16 at 14:55
  • $\begingroup$ @MarcHoyois ah,thanks. I thought this might be the case. Then ${\rm Bun}^{\rm sm}_{G}$ is precisely the reflection of $\mathbf{B}G$ through the shape modality. I was confused because the two stacks seemed to provide the same information, but I'll think a bit more about the distinction. $\endgroup$ – Daniel Grady Oct 15 '16 at 14:59
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    $\begingroup$ The homotopy fiber is $\mathbf B\operatorname{fib}(G \to \int G)$. This classifies principal $G$-bundles that are trivialized "up to homotopy". More precisely, an object over $X$ is a principal $G$-bundle with a section, a morphism is a morphism of $G$-bundles respecting the given sections up to specified fiberwise homotopy, a $2$-morphism is an equivalence of such data, etc. $\endgroup$ – Marc Hoyois Oct 15 '16 at 18:39

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