$BG$ the stack, $BG$ the simplicial presheaf

I have a theoretical question about comparing two objects that I have recently come across.

For concreteness, let us work over the category $$C$$ of schemes over $$k$$. Let $$G$$ be an algebraic group over $$k$$. One can construct the stack $$BG$$ as a fibered category in groupoids and as a simplicial scheme $$(BG)_n=G^n$$ (with certain face and degeneracy maps which I don't specify). What is the precise relation between these two versions of $$BG$$? Can I obtain one from the other?

I think I might have heard that "the two constructions are equivalent because the simplicial $$BG$$ has no higher homotopy groups". Does this make sense? How does this implication work?

Any answer that could help me better understand the relation between these two $$BG$$s is very welcome.

• What you heard is true in the case of $G$ being Abelian. In general, $BG$ can be nontrivial in more than one dimension. For instance, $\pi_nU\simeq\mathbb{Z}$ for $n$ even and trivial otherwise. Oct 18, 2018 at 4:33
• Any presentation of an algebraic stack $\mathscr X$ as a quotient of a scheme $U$ by an étale equivalence relation $R \rightrightarrows U$ gives rise to a simplicial scheme by taking its "Čech hypercovering": let $X_i$ be the $i$-fold fibre product of $U$ over $\mathscr X$. Any algebraic stack has such a presentation by [Tag 04T3]. It's not clear to me how well-defined this is, let alone whether this is an equivalence onto some subcategory of simplicial schemes. Oct 18, 2018 at 5:56
• @Qfwfq I think that's supposed to be the n-th unitary group U(n). Oct 18, 2018 at 14:47
• @Horstenson Rather $\text{colim } U(n)$. The individual groups $U(n)$ have well-understood homotopy only up to about degree $2n$ by comparison to this colimit, and after that it becomes more complicated.
– mme
Oct 18, 2018 at 15:40
• @user51223 That has nothing to do with $U$ not being abelian and everything about $G$ being discrete Oct 19, 2018 at 17:37

The two constructions are not quite equivalent. Let me write $$\mathbf BG$$ for the stack and $$B_\bullet G$$ for the simplicial scheme to better distinguish between them. There is a third relevant player, $$BG$$, which is the presheaf of ∞-groupoids on $$C$$ presented by $$B_\bullet G$$.

The precise relation between these three objects is the following:

1. $$\mathbf BG$$ is the fppf sheafification of $$BG$$
2. $$BG$$ is the colimit of the simplicial object $$B_\bullet G$$
3. $$B_\bullet G$$ is the Čech nerve (= $$0$$-coskeleton) of either $$\mathrm{Spec}(k) \to \mathbf BG$$ or $$\mathrm{Spec}(k) \to BG$$

In more detail, algebraic stacks over $$k$$ (in the sense of Artin, say) form a full subcategory of the $$2$$-category of fppf sheaves of groupoids (classically, "stacks in groupoids") on $$C$$, which is itself a full subcategory of the ∞-category of presheaves of ∞-groupoids on $$C$$. Thus, both $$\mathbf BG$$ and $$BG$$ live in this ∞-category (in fact they both belong to the subcategory of presheaves of groupoids), and one is the fppf sheafification of the other. The étale sheafification suffices if $$G$$ is smooth.

The reason $$\mathbf BG$$ and $$BG$$ are not the same is that there is a unique homotopy class of map $$X\to BG$$ from any scheme $$X$$, but homotopy classes of maps $$X\to \mathbf BG$$ are in bijection with isomorphism classes of $$G$$-torsors on $$X$$. In fact $$BG$$ is the full subpresheaf of $$\mathbf BG$$ spanned by the trivial $$G$$-torsors.

(Added details about 2.) $$BG$$ being the colimit of $$B_\bullet G$$ is the manner in which simplicial sets give rise to ∞-groupoids. Of course, one must define ∞-groupoids at some point and one way to do that is to start with simplicial sets and invert weak equivalences, so that we have a localization functor {simplicial sets} → {∞-groupoids}. Once higher category theory is set up however, it turns out that this functor is the composition of the inclusion of simplicial sets into simplicial ∞-groupoids and of the colimit over $$\Delta^{op}$$ functor (this is the standard fact that a simplicial set is canonically the homotopy colimit of itself). In practice this is often a more useful way to think about it.

• Thanks for your answer. I am finding it difficult to understand why $BG$ is the colimit of $B_{\bullet}G$. Could you please detail that a little more? Oct 19, 2018 at 5:01