# Stacks and sheaves

I'm a bit confused by the double role which sheaves play in the theory of stacks.

On the one hand, sheaves on a site are the obvious generalization of a sheaf on a topological space. On the other hand a sheaf on a site is (or better its associated category fibered in sets is) a very particular stack itself, so a generalization of a space. This is not completely confusing: more or less it amounts (I believe) to identifying a space X with the sheaf of continuos functions with values in X.

But now my question is the following. An equivalent condition for a fibered category to be a prestack is that for any two objects (over the same base object), the associated functor of arrows should be a sheaf. In particular this is true for a stack, so for any stack and any two objects in it we have a sheaf, and so a stack (over a comma category).

What is the meaning of this geometrically?

For instance take the stack $\mathcal{M}_{g,n}$.

Giving two objects in the stack (over the same base object) means giving two families $X$ and $Y$ of stable pointed curves over the same scheme $S$, and the associated functor of arrows maps every other scheme $f \colon T \rightarrow S$ to the set of morphism between $f^* X$ and $f^* Y$. How should I think of the associated stack as a space?

To avoid misunderstandings I give the defition of the functor of arrows. Let $\mathcal{F}$ be a fibered category over $\mathcal{C}$. Take $U \in \mathcal{C}$ and $\xi, \eta \in \mathcal{F}(U)$. Then there is a functor $F \colon \mathcal{C}/U \rightarrow Set$ defined as follows. For a map $f \colon T \rightarrow U$ we put $F(f) = Hom(f^* \xi, f^* \eta)$. The action on arrows requires some diagrams to be described, but it's really the only possible one.

• Thank you for this question. I had a similar confusion, but did not know enough to form a question on it! Feb 3, 2010 at 1:52
• I would happily drop the relative scheme language, but the fact is that the functor of arrows is only defined in a comma category, so I see no way not to work relatively. Feb 3, 2010 at 2:10
• I'm sorry my phrasing is not clear. I'm just spelling out the functor of arrows construction in the case of a moduli space of curves. I have reworded it a bit, hope now it makes more sense. Feb 3, 2010 at 2:13
• I'm not sure I understand what you mean. The condition of the functor of arrows being a sheaf must be satisfied for every couple of objects in any fiber, not only in the fiber over spec Z. Moreover I don't see how it becomes a functor of points when we take S = Spec Z. There are two objects involved, and maps between their pullbacks. Feb 3, 2010 at 2:28
• Alright, I think I've got what you mean here finally. X and Y are objects of the fiber p^{-1}(S) on the category of schemes. Then what you're verifying with your functor of arrows is that all of the pullbacks exist. This is just the same as checking that a presheaf is a functor. Since $\mathcal{M}_{g,n}$ is a stack, it's not necessary to check any other compatibility conditions. I still don't understand what you mean by "Associated stack". Is that the fibered full subcategory generated by those two objects (and by fullness, we have that it too is a stack)? Feb 3, 2010 at 2:33

Let me see if I understand your example correctly: you are fixing $X$ and $Y$, families of curves over $S$, and now you are considering the functor which maps an $S$-scheme $T$ to the set of $T$-isomorphisms $f^*X \to f^*Y$ (where $f$ is the map from $T$ to $S$).

If I have things straight, then this functor shouldn't be so bad to think about, because it is actually representable, by an Isom scheme. In other words, there is an $S$-scheme $Isom_S(X,Y)$ whose $T$-valued points, for any $f:T \to S$, are precisely the $T$-isomorphisms from $f^*X$ to $f^*Y$. (One can construct the Isom scheme by looking inside a certain well-chosen Hilbert scheme.)

One way to think about this geometrically is as follows: one can imagine that two curves over $k$ (a field) are isomorphic precisely when certain invariants coincide (e.g. for elliptic curves, the $j$-invariant). (Of course this is a simplification, and the whole point of the theory of moduli spaces/schemes/stacks is to make it precise, but it is a helpful intuition.) Now if we have a family $X$ over $S$, these invariants vary over $S$ to give a collections of functions on $S$ (e.g. a function $j$ in the genus $1$ case), and similarly with $Y$. Now $X$ and $Y$ will have isomorphic fibres precisely at those points where the invariants coincide, so if we look at the subscheme $Z$ of $S$ defined by the coincidence of the invariants, we expect that $f^*X$ and $f^*Y$ will be isomorphic precisely if the map $f$ factors through $Z$. Thus $Z$ is a rough approximation to the Isom scheme.

It is not precisely the Isom scheme, because curves sometimes have non-trivial automorphisms, and so even if we know that $X_s$ and $Y_s$ are isomorphic for some $s \in S$, they may be isomorphic in more than one way. So actually the Isom scheme will be some kind of (possibly ramified) finite cover of $Z$.

Of course, if one pursues this line of intuition much more seriously, one will recover the notions of moduli stack, coarse moduli space, and so on.

The families $X$ and $Y$ over $S$ correspond to a map $\phi:S \to {\mathcal M}_g \times {\mathcal M}_g$. The stack which maps a $T$-scheme to $Isom_T(f^*X, f^*Y)$ can then seen to be the fibre product of the map $\phi$ and the diagonal $\Delta:{\mathcal M}_g \to {\mathcal M}_g \times {\mathcal M}_g$.

In the particular case of ${\mathcal M}_g$ the fact that this fibre product is representable is part of the condition that ${\mathcal M}_g$ be an algebraic stack.

But in general, the construction you describe is the construction of a fibre product with the diagonal. This might help with the geometric picture, and make the relationship to Mike's answer clearer. (For the latter:note that the path space into $X$ has a natural projection to $X\times X$ (take the two endpoints), and the loop space is the fibre product of the path space with the diagonal $X\to X\times X$.)

• Thank you, this gives me some intuition on the situation. Still I think there should be some relation between $Z$, or $Isom_S(X, Y)$ and $\mathcal{M}$. The point of view of Mike gives, in a different setting, a kind of complementary intuition. Feb 3, 2010 at 12:47
• Dear Andrea, I added some remarks at the end of my answer that might help connect the intuitions of the two replies. Feb 3, 2010 at 14:25

I'm not a geometer, but here's one way to think of it. Let M be a stack (in groupoids, say) and X an object, and consider just a single point p:X→M. Then $Hom_M(p,p)$ is a sheaf, which is the "isotropy group" of the stack M at the point p, i.e. the space of automorphisms of p in M. If you think of your stacks as like topological spaces, with isomorphisms corresponding to paths (which makes the most sense when you move up to ∞-stacks, where higher morphisms correspond to higher homotopies between paths), then the isotropy-group object of a point p corresponds to the loop space ΩX of a topological space X.

As far as the first part of your question goes, I have exactly the same confusion (and it will probably get worse once I start thinking about sheaves on stacks...).

There are two things that give me the illusion of some comprehension.

1. Thinking of sheaves as (generalized) spaces is probably not terrible, in the same way you think of a bundle just as its total space. It is indeed true that any sheaf (with values in a reasonable category I guess) is the sheaf of sections of its etale space. (Although I must confess I don't particularly like etale spaces and I'm aware that this is probably not the right picture, as we shouldn't think of sheaves on a site in this fashion, but hey).
2. If we think of fibred categories (+cleavage) as presheaves with values in the 2-category Cat, then there is a generalization of the sheaf condition which yields the notion of a stack. Of course, arrows being a sheaf is a consequence of this, so if you find the generalization of the sheaf condition more natural, then arrows-being-a-sheaf might be viewed as a formal consequence I guess.

Anyway it's just a thought.