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.

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    $\begingroup$ 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. $\endgroup$
    – user51223
    Oct 18, 2018 at 4:33
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    $\begingroup$ 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. $\endgroup$ Oct 18, 2018 at 5:56
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    $\begingroup$ @Qfwfq I think that's supposed to be the n-th unitary group U(n). $\endgroup$
    – Horstenson
    Oct 18, 2018 at 14:47
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    $\begingroup$ @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. $\endgroup$
    – mme
    Oct 18, 2018 at 15:40
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    $\begingroup$ @user51223 That has nothing to do with $U$ not being abelian and everything about $G$ being discrete $\endgroup$ Oct 19, 2018 at 17:37

1 Answer 1


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.

  • $\begingroup$ 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? $\endgroup$
    – HuynA
    Oct 19, 2018 at 5:01

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