This could as well have been asked in the comments to this question, but I prefer to open a new one for the sake of clarity.

Say $G$ is a reductive group over the complex numbers, with compact real form $K$. On the one hand one can construct

  1. The algebraic stack $[*/G]$ over the complex numbers,

on the other hand one can construct

  1. The classifying space $\mathrm{B} K$, which is a topological space well defined up to homotopy.

What is, exactly, the relation between the two objects?

The classifying space $\mathrm{B} G$ of the underlying topological group $G$ is well known to be homotopy equivalent to $\mathrm{B} K$.

And $[*/G]$ has, I think, an underlying topological stack $[*/G]_{\mathrm{top}}$.

How do $\mathrm{B} G$ and $[*/G]_{\mathrm{top}}$ compare? They live in the same (2-)category, namely possibly "infinite dimensional" topological stacks. Is there a way they are two different "models" of the same "object"?

  • $\begingroup$ What do you mean by "underlying topological stack"? $\endgroup$ – Anton Fetisov Nov 18 '14 at 1:56
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    $\begingroup$ @AntonFetisov you can take the analytic topology on the $\mathbb{C}$-valued points of the groupoid in schemes, and get a topological groupoid: this presents a topological stack. One can take any reductive algebraic group and consider the same comparison. $\endgroup$ – David Roberts Nov 18 '14 at 6:50
  • $\begingroup$ @Qfwfq - BG gives a topological stack by Yoneda, via sheaves on Top. *//G is a topological stack of groupoids. They have the same homotopy type (in the sense that the homotopy colimit of the nerve of *//G is a BG), but *//G lifts to, for instance, a differentiable and even a holomorphic stack. They are definitely not equivalent in the 2-category of topological stacks. $\endgroup$ – David Roberts Nov 18 '14 at 6:53
  • $\begingroup$ @AntonFetisov ...and there's probably also a clever way to see it via Kan extension along the functor Sch^op --> Top^op. $\endgroup$ – David Roberts Nov 18 '14 at 6:57

I'm not sure why you're replacing $G$ with $K$. So let me compare $[*/G]$ and $B(G^{an})$. As you say, for the purposes of homotopy theory, $B(G^{an})$ is as good as $BK$.

One way to understand a stack $\mathcal{X}$ is via a hypercover, which is a simplicial scheme. The easiest way to get a hypercover is to choose an atlas $U\to \mathcal{X}$ and to consider the simplicial scheme $U_\bullet$ with $$U_0=U,$$ $$U_1=U\times_{\mathcal{X}}U,$$ $$U_2=U\times_{\mathcal{X}}U\times_{\mathcal{X}}U,$$ $$U_3=U\times_{\mathcal{X}}U\times_{\mathcal{X}}U\times_{\mathcal{X}}U$$ and so on, with the evident face and degeneracy maps.

Let's try this with $[*/G]$. There is an obvious atlas $*\to [*/G]$. The hypercover we obtain has $0$-th level $*$, first level $*\times_{[*/G]}*=G$, second level $*\times_{[*/G]}*\times_{[*/G]}*=G\times G$, and so on; working out the face and degeneracy maps, we see that this is exactly the usual simplicial "bar construction" model for $BG$.

Let me make the exact relationship explicit, now. There is a natural functor $\text{Var}_\mathbb{C}\to \text{Top}$ given by sending a variety over $\mathbb{C}$ to the topological space underlying it, in the analytic topology. One may extend this functor to simplicial varieties over $\mathbb{C}$ in the obvious way. Now if we apply this functor to the hypercover described above, obtained via the cover $*\to [*/G]$, one gets exactly the usual bar construction of $B(G^{an})$. As a lemma, you should convince yourself that it actually doesn't matter what (smooth) hypercover we take--we get the same homotopy type.

So in terms of the "underlying topological space," (defined as in the previous paragraph) the stack is weakly equivalent to the usual topological object. Of course, the stack knows a lot more than the space $BG$--it has some actual algebraic geometry associated to it as well.

Added Later. I think it might also be worth thinking about another relationship between $[*/G]$ and $B(G^{an})$. This relationship is "adjoint" to the one I described above.

Consider the category of simplicial schemes, with the model structure induced by the Yoneda embedding into the category of simplicial presheaves on $\text{Sch}$, with your favorite model structure (weak equivalences should be given by pointwise weak equivalences). Now, localize at the subcategory generated by smooth hypercovers.

The category of simplicial sets naturally embeds in the category of simplicial schemes, just by sending a set $S$ to the scheme $\sqcup_S \text{Spec}(\mathbb{C})$. Choosing a hypercover of $BG$ (which one doesn't matter, because of the localization) we get a simplicial presheaf on $\text{sSch}$ via Yoneda. Restricting this presheaf to the essential image of $\text{sSet}$, one recovers the usual $BG$ up to weak equivalence.

The sense in which the stack $[*/G]$ "knows more" than the space $BG$ is that the presheaf above is certainly not determined by what it does on the essential image of $\text{sSet}$; the "algebraic geometry" of $[*/G]$ is precisely the data of what it does on the whole category $\text{sSch}$, from a functor-of-points POV.

  • $\begingroup$ It's not obvious to me that the model structure on simplicial presheaves restricts to a model structure on simplicial schemes. How does one verify the factorisation axiom? $\endgroup$ – Zhen Lin Dec 10 '14 at 8:40
  • $\begingroup$ @ZhenLin: Sorry it took me so long to respond to this. You're right; one should work in the category of simplicial presheaves. I was being sloppy. $\endgroup$ – Daniel Litt Jul 17 '17 at 1:13

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