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For topological spaces and simplicial sets one can consider each pair of parallel morphisms $f,g:X\rightrightarrows Y$ as equipped with a set of 2-morphisms given by homotopies $H:f\simeq g$ (let's ignore the homotopies-between-homotopies).

In the realm of algebraic geometry, spaces and maps are usually too rigid to admit homotopies on the nose.

Which is the "geometric meaning" of 2-morphisms $\alpha$ between a pair of parallel 1-morphisms $f,g:\mathcal{X}\rightrightarrows \mathcal{Y}$ between algebraic stacks?

Of course enlightining theorems and examples are welcome :)

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2 Answers 2

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Whatever a stack is, it should in particular be a functor $\text{CRing} \to \text{Gpd}$ from commutative rings to groupoids (a prestack). Groupoids naturally form a $2$-category and so such functors also naturally form a $2$-category. For example, there are constant prestacks with constant value some groupoid, and then $2$-morphisms between these are $2$-morphisms between the corresponding groupoids. The connection back to the topological story is that groupoids model homotopy $1$-types by the homotopy hypothesis, but at this category level it is possible to be more explicit.

(In other words, my claim is that there are spaces hiding in the story of stacks, namely the values of those stacks on various rings.)

Example. Let $G, H$ be two groups, and consider the constant prestacks $BG, BH$. Then a pair of morphisms $f_1, f_2 : BG \to BH$ is a pair of group homomorphisms $G \to H$, and (exercise) a $2$-morphism between them is a choice of an element $h \in H$ such that $f_1 = h f_2 h^{-1}$.

In particular, let $G = H$ and let $f_1 = f_2 = \text{id}_G$. Then a $2$-morphism $\text{id}_G \to \text{id}_G$ is a central element of $G$. Hence $\text{Aut}(\text{id}_G) \cong Z(G)$. The corresponding topological fact is that the topological monoid of homotopy equivalences $BG \to BG$ has $\pi_1$ naturally isomorphic to $Z(G)$.

Example. Consider the stack $\text{Pic} : \text{CRing} \to \text{Gpd}$ which assigns to a commutative ring $R$ the groupoid whose objects are invertible $R$-modules and whose morphisms are morphisms of $R$-modules, and let $f = g = \text{id}_{\text{Pic}}$. Thinking of $\text{Pic}$ as being represented by $B \mathbb{G}_m$ and keeping the above example in mind, I'm led to suspect that $\text{Aut}(\text{id}_{\text{Pic}})$ can be identified with $\mathbb{G}_m$ itself (and in particular that it makes sense as a group scheme and not just as a scheme, once this is suitably defined).

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  • $\begingroup$ For better or worse, ‘prestack’ is actually defined to be the analogue of separated presheaf, not an arbitrary presheaf. $\endgroup$
    – Zhen Lin
    Feb 4, 2015 at 22:27
  • $\begingroup$ Huh. Well, by prestack I mean an arbitrary thing. $\endgroup$ Feb 4, 2015 at 22:28
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    $\begingroup$ Silly observation: Most naturally occurring stacks do not arise as functors - a choice of pullbacks is extra data (albeit typically unique up to a contractible family of choices). $\endgroup$
    – S. Carnahan
    Feb 8, 2015 at 2:55
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As long as your stacks are attached to moduli problems, the 2-morphisms describe symmetries of the parametrized objects. For example, let $X$ be the spectrum of a field $k$, let $Y$ be the stack of elliptic curves, and let the maps $f,g: X \to Y$ describe elliptic curves over $k$. Then a 2-morphism between $f$ and $g$ is a $k$-isomorphism between the corresponding elliptic curves. In particular, if $f$ and $g$ are the same map, then a 2-morphism is an automorphism of the curve they describe.

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  • $\begingroup$ And if $X$ is any scheme, you get the family version of your example, as for any other moduli stack. But, what if $X$ itself is a (non rigid) stack? $\endgroup$
    – Qfwfq
    Feb 5, 2015 at 11:02
  • $\begingroup$ @Qfwfq Nothing really changes. A 1-morphism $X \to Y$ produces a family of elliptic curves over the stack $X$, and a 2-morphism is an automorphism of the family. This is why I didn't bother to make $X$ complicated. $\endgroup$
    – S. Carnahan
    Feb 7, 2015 at 3:18
  • $\begingroup$ I see. But.. it sounds somehow.. artificial (though correct, of course). What's an automorphism of a family of elliptic curves over $X$ in terms of the geometry of $X$? Perhaps the (more local) question I want to ask is: what's a 2-morphism between $f,g:[U/G]\to [V/H]$ in terms of the geometry of the two actions? (But this I can think by myself I guess) $\endgroup$
    – Qfwfq
    Feb 7, 2015 at 19:28
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    $\begingroup$ If $f$ and $g$ are given by $H$-equivariant maps $\tilde{f}, \tilde{g}: P \to V$, where $P$ is an $H$-torsor over $[U/G]$, then a 2-morphism $f \Rightarrow g$ is a section of $H$ on $V$ whose action takes $\tilde{f}$ to $\tilde{g}$. $\endgroup$
    – S. Carnahan
    Feb 8, 2015 at 2:48

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