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I've started to learn about stacks, and a question arose in my attempts of looking at the very definition of a stack by several points of view. First, I recall some background and fix the notation (which is a mix of Fantechi's Stacks for everybody, Edidin's Notes on the construction of the moduli space space of curves, and my personal notation: I'm sorry about that).

Let $\mathfrak S:=\underline S:=Sch/S$ be the "base" category of schemes over a fixed scheme $S$, and let $Gpd$ be the category of groupoids. We say that a groupoid fibration $\pi:\mathfrak X\to \mathfrak S$ is a stack if two things happen: $(i)$ every descent datum is effective, and $(ii)$ isomorphisms are a sheaf for $\mathfrak X$ (with respect to the étale topology).

(It may help rephrase $(ii)$ as follows: for every $S$-scheme $B$ and for every two objects $X,Y$ in the fiber $\mathfrak X_B$, the presheaf $\mathcal I_B^{X,Y}:\underline B\to \textrm{Sets}$ is a sheaf on the (big) étale site associated to $B$. Here $\mathcal I_B^{X,Y}$ takes a $B$-scheme $f:B'\to B$ to the set of isomorphisms $f^\ast X\cong f^\ast Y$ in $\mathfrak X_{B'}$.)

My question: can we say that to give a stack is the same as to give a sheaf of groupoids $\mathcal F:\mathfrak S\to Gpd$ on the étale site associated to $S$?

My attempts. I'll sketch how I began to prove that the answer is yes, and convince myself that the answer is no.

First, given $\mathcal F$, we construct a groupoid fibration $\pi:\mathfrak X\to\mathfrak S$ fiberwise, by attaching $\mathfrak X_B:=\mathcal F(B)$ over $B\in\mathfrak S$. We then need to check $(i)$ and $(ii)$. Ok, let us stop here for the moment.

Conversely, if we have $\pi$, we may define $\mathcal F$ by $\mathcal F(B):=\mathfrak X_B$ on objects and by $\mathcal F(f:B'\to B)=(f^\ast:\mathfrak X_B\to\mathfrak X_{B'})$ on arrows. This seems to be natural. Let us stop here.

In both directions, there is a problem: I have to use (or prove) exactness of the sequence $$ \mathcal F(B)\to\prod_i\mathcal F(B_i)\rightrightarrows \prod_{i,j}\mathcal F(B_i\times_BB_j), $$ for $\{B_i\to B\}$ a covering of $B$. But does exactness make sense in $Gpd$? For instance, is $Gpd$ abelian?

Also, it seems to me that $(i)$ has nothing to do with the sheaf condition: I feel like $(i)$ doesn't help me to prove anything when it is assumed, and cannot be proven when starting with $\mathcal F$.

Any correction/insight is welcome. Thank you!

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1. That definition of the sheaf condition does not require any abelian category structure; the correct way of saying it is to say "it is an equaliser diagram." 2. No, a stack is not merely a sheaf of groupoids, though any sheaf of groupoids does indeed give a stack. Isomorphism is not equality! –  Zhen Lin Feb 11 '13 at 17:45
    
@Zhen Lin: 1. I did not know the equaliser version! Then OK, I'll accept it as a definition of sheaf and forget exactness. In that case, do you know whether the first arrow is a monomorphism? 2. if you have a sheaf of groupoids how do you get a stack? in particular, how do you check $(i)$? –  Brenin Feb 11 '13 at 21:46
    
1. The universal arrow out of an equaliser is always a monomorphism by abstract nonsense. 2. My pronouncement about sheaves of groupoids was premature. Some extra conditions controlling the isomorphisms in the groupoids is required to ensure that all descent data are effective. (The stack condition is essentially a 2-categorical version of the sheaf condition, but when there are non-trivial isomorphisms then in general neither condition implies the other. In essence, the sheaf condition only concerns "strict" descent data where all isomorphisms are identities.) –  Zhen Lin Feb 11 '13 at 22:07
    
I see. Of course I would like to see some examples of this "neither condition implies the other", but perhaps I should work them out myself. If you have any advise in this direction, it is welcome. –  Brenin Feb 11 '13 at 23:28

1 Answer 1

up vote 6 down vote accepted

If we have a sheaf of groupoids, then in particular the presheaf of objects must be a sheaf as well. Thus one way of showing that a stack is not just a sheaf of groupoids is to show that the presheaf of objects it gives is not a sheaf.

Rather than working with complicated sites like the big étale site, let me construct an example for the standard site for a single topological space $X$. Consider the stack $\textbf{Pic}$ of line bundles over $X$: as a fibred category, its objects are pairs $(U, L)$ where $U \subseteq X$ is open and $L$ is a (real) line bundle over $U$, and its morphisms are fibrewise linear isomorphisms. (Warning: The fibre $\textbf{Pic}(U)$ is a groupoid, but it is not the Picard group of $U$ in general!) It is a standard exercise to check that $\textbf{Pic}$ is a stack: this amounts to showing that line bundles can be glued together.

For convenience, we assume $\textbf{Pic}$ is skeletal, so that all isomorphisms in $\textbf{Pic}$ are automorphisms. Let $\textrm{Pic}$ be the presheaf that assigns to each open $U \subseteq X$ the set of isomorphism classes of line bundles over $U$. Obviously, $\textrm{Pic}$ is the presheaf of objects of $\textbf{Pic}$. Now, $\textrm{Pic}$ is not a sheaf in general: by definition, if $L$ is a line bundle over $X$ and $\mathfrak{U}$ is a sufficiently fine open cover of $X$, then $L$ pulls back along $\mathfrak{U}$ to the trivial line bundle; but the Möbius strip is a non-trivial line bundle over $X = S^1$, so in this case we see that $\textrm{Pic}$ fails to even be a separated presheaf.

Now, consider the sheaf $\mathscr{O}_X^\times$ of continuous non-vanishing (real-valued) functions on $X$. This is a sheaf of groups and gives rise to a category $\mathbf{G}$ fibred in groupoids over the standard site of $X$ via the Grothendieck construction. Then $\textbf{G}$ is not a stack in general: again, for $X = S^1 \subseteq \mathbb{C}$, consider the open cover $\mathfrak{U} = \{ X \setminus \{ +1 \}, X \setminus \{ -1 \} \}$ and the descent data induced by the evident trivialisation of the Möbius strip over $\mathfrak{U}$. In this case, the failure of $\textbf{G}$ to be a stack is directly connected to the non-triviality of $\check{H}^1 (X, \mathscr{O}^\times_X)$!


Addendum. In fact, every stack is "weakly" equivalent to a strong stack, i.e. one that comes from a sheaf of groupoids. This is a result of Joyal and Tierney [Strong stacks and classifying spaces]. What this means is a bit subtle: the definition of "weak" equivalence is such that every (pre)sheaf of groupoids is "weakly" equivalent to a strong stack.

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Moreover, if $X$ is a variety, then the sheafification of the Picard stack is the standard relative Picard scheme (and gives us the structure $\mathbf{Pic}\to Pic$ of a $\mathbb{G}_m$-gerbe). –  Matt Feb 14 '13 at 3:42
    
@Zhen Lin: this is a beautiful answer! I still need some time to digest it. For the moment I have this question: why is Pic the presheaf of objects of $\bf{Pic}$? Indeed, you said that ${\bf{Pic}}(U)\neq \textrm{Pic }U$ in general. Or, are they equal by the "skeletal" assumption? –  Brenin Feb 16 '13 at 11:38
    
What I meant was that $\textbf{Pic}(U)$ and $\textrm{Pic}(U)$ are not isomorphic as groupoids. But if $\textbf{Pic}(U)$ is skeletal, then there is an obvious bijection between the objects of $\textbf{Pic}(U)$ and the elements of $\textrm{Pic}(U)$. –  Zhen Lin Feb 16 '13 at 13:07
    
I see. Thank you very much! –  Brenin Feb 16 '13 at 16:38

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