When we study sheaves of sets (on a space X or a site C) we are often interested in the stalks of the sheaf (at either a point $p:1\to X$ or a left exact, cover-preserving functor $a:C\to Sets$). I wonder how one generalizes this notion to stacks?

Since we can always pass from a stack to an equivalent sheaf, one option might be to say the stalk of a stack is (any category equivalent to) the stalk of the associated sheaf of categories. This seems a little roundabout, though? Is there a more intrinsic definition of these categories?

  • $\begingroup$ I presume by stack you mean a fibred category (satisfying the usual properties), and by a sheaf of categories you mean a presheaf of categories satisfying the appropriate coherence (weaker than that for a sheaf of sets). The main problem is perhaps what you consider to be the 2-category of stacks, and what you want to take for 1-arrows, in order to define a point and the pullback along the inclusion of that point. $\endgroup$ – David Roberts Mar 31 '11 at 6:36
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    $\begingroup$ Related question: Is there a nice construction of an étale space associated to a stack? $\endgroup$ – Martin Brandenburg Mar 31 '11 at 6:46
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    $\begingroup$ I'm confused by the question. Are you trying to understand "the stalk of a sheaf on a stack" or "the stalk of a stack"? It sounds like it's the latter, but then you should consider this question: what is the stalk of a space over $X$? After all, a space over $X$ corresponds to a sheaf over $X$ (in most reasonable categories of spaces). This strikes me as a pretty weird thing to think about. Is this something anybody would want to generalize to stacks? $\endgroup$ – Anton Geraschenko Mar 31 '11 at 9:48
  • $\begingroup$ @David: maybe I'm confused, but I think even in the case of sheaves of modules over a scheme when you take the pullback along the inclusion of a point you don't get the stalk of the sheaf, but its fiber $\endgroup$ – Qfwfq Mar 31 '11 at 19:21
  • $\begingroup$ @Martin: Yes. I explain this in this preprint: arxiv.org/abs/1011.6070 $\endgroup$ – David Carchedi Apr 3 '11 at 20:50

Here is the categorical way to think about the stalk:

Let $X$ be space, and $x \in X$ a point. Regard $x$ as a map $$x:pt \to X$$.

Then $x$ induces a geometric morphism $$x_*:Sh(pt) \to Sh(X)$$ $$Sh(pt) \stackrel{}{\longleftarrow} Sh(X):x^*$$

(so $x^*$ is left-adjoint to $x_*$ and left-exact).

Now notice that $Sh(pt) \cong Set$. Under this identification, the stalk of $F \in Sh(X)$ at $x$ is the set $x^*\left(F\right)$.

This can be done in the $2$-categorical setting with stacks with no adjustment:

$x$ induces an adjoint pair of $2$-functors

$$x_*:St(pt) \to St(X)$$ $$St(pt) \stackrel{}{\longleftarrow} St(X):x^*$$

and under the identification $St(pt) \cong Gpd,$ the stalk of $\mathscr{X} \in St(X)$ at $x$ is the groupoid $x^*\left(\mathscr{X}\right)$.

A way of viewing this in a geometric way is by using the etale-realization of the stack $\mathscr{X}$, (explained in arxiv.org/abs/1011.6070 ) which is a topological stack $$L\left(\mathscr{X}\right)$$ equipped with a local homoemosphism to $X$. The stalk can then be viewed as the fiber of this map over $x$.

  • $\begingroup$ My comment comes too late, but I have the impression that your construction actually gives the fiber, not the stalk, of whatever thing you are considering (sheaf, space, stack,...). $\endgroup$ – Qfwfq Mar 18 '15 at 17:21
  • $\begingroup$ The fiber of the etale-space $L(F) \to X$ of a sheaf $F$ over $X$ at a point $x$ IS precisely the stalk $F_x.$ These are one and the same. $\endgroup$ – David Carchedi Mar 19 '15 at 9:53
  • $\begingroup$ Oh I see what you mean. (On the other hand, if $F$ is a sheaf of modules on a $k$-scheme $X$ and $x:\mathrm{Spec}(k)\to X$ is a (closed) point, the $k$- vector space $x^{*} F$ is the fiber at $x$, not the stalk). $\endgroup$ – Qfwfq Mar 19 '15 at 12:35
  • $\begingroup$ Yes, but the question was about generalizing from stalks of sheaves of sets to stalks of stacks so, although I agree that what I said may not make sense for sheaves of more interesting objects than sets (like modules), it does answer the question asked. $\endgroup$ – David Carchedi Mar 19 '15 at 21:09

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