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Given a complex manifold, you can `weaken' its structure to give a smooth manifold. Is there an analogous construction that constructs a stack over the category of smooth manifolds from a stack over the category of complex manifolds?

Obviously, this is possible for a stack represented by the quotient of a complex manifold by a group action, so I'd imagine that this should at least be possible for DM stacks, but I can't think of any general construction that doesn't involve some kind of atlas.

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  • $\begingroup$ So I can understand what you're after: how do you explain the case of complex manifolds without using atlases? $\endgroup$
    – Omar Antolín-Camarena
    Apr 12, 2012 at 13:13
  • $\begingroup$ One way to obtain a smooth manifold from a complex manifold is as follows: You can define a smooth structure on a manifold using the sheaf of smooth $\mathbb C$ valued functions. This is a sheaf of $C^\infty$ rings in the sense that as well as the operations of addition and multiplication you have the operations $(x_1,\dotsc,x_n)\mapsto f(x_1,\dotsc,x_n)$ where $f$ is any $C^\infty$ function. The sheaf of smooth $\mathbb C$ valued functions is the sheaf of $C^\infty$ rings generated by the sheaf of holomorphic functions. $\endgroup$
    – Brett Parker
    Apr 12, 2012 at 23:54
  • $\begingroup$ What I was hoping for was something like this: Add in pullback' objects and morphisms to your original stack from the extra objects and morphisms in the category of manifolds, then somehow take the stack generated' by what you get. $\endgroup$
    – Brett Parker
    Apr 12, 2012 at 23:59
  • $\begingroup$ @Brett - what sort of stack are you thinking of? A category fibred over the category of complex manifolds with effective descent? Or a geometric stack, which is a stack with an atlas - that is, an epimorphism from a representable stack? If the former, you want David C's answer. If the latter, then take your pick as to whose answer you like best. $\endgroup$
    – David Roberts ♦
    Apr 13, 2012 at 3:46
  • $\begingroup$ @David R, I'm dealing with moduli stacks, so they don't come with a given atlas. On the other hand, probably all the cases when I need this construction I can prove that an atlas does exist... $\endgroup$
    – Brett Parker
    Apr 13, 2012 at 5:08

2 Answers 2

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Denote by $$u:CxMfd \to Mfd$$ the forgetful functor from complex manifolds to smooth manifolds. Let $$u_!:St\left(CxMfd\right) \to St\left(Mfd\right)$$ denote its 2-categorical prolongation. Explicitly, this is given by the bicategorical Kan extension of $y_{Mfd} \circ u$ along the Yoneda embedding $$y_{CxMfd}:CxMfd \to St\left(CxMfd\right),$$ where $y_{Mfd}$ is similarly defined. $u_!$ is the unique weak colimit preserving functor which agrees with $y_{Mfd} \circ u$ on representables.

I claim that $u_!$ sends holomorphic stacks (stacks coming from groupoid objects in complex manifolds) to differentiable stacks.

Indeed, let $\mathcal{X}$ be a holomorphic stack coming from a groupoid object $X_1 \rightrightarrows X_0.$ Then, $\mathcal{X}$ is the weak colimit of the truncated semi-simplicial diagram $$X_2\mspace{5mu} \{(3\mspace{5mu} parallel \mspace{5mu} arrows)\}\mspace{5mu} X_1 \rightrightarrows X_0,$$ viewing each $X_i$ as a representable presheaf on $CxMfd$. Applying $u_!$ to this diagram, yields that $u_!\left(\mathcal{X}\right)$ is the weak colimit of the same diagram, now viewing each $X_i$ as a representable presheaf in $Mfd$. This in turn implies that $u_!\left(\mathcal{X}\right)$ is the stackification of the weak presheaf of groupoids arising canonically from $X$ viewed as a Lie groupoid. In particular, this implies that $u_!$, when restricted to holomorphic stacks agrees with the answer of David Roberts, only, it makes no explicit reference to atlases.

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  • $\begingroup$ Though for your last paragraph you need to know that every holomorphic stack can be presented by a groupoid object. I was thinking there would be some sort of Kan extension version of this, but didn't know exactly how it would go. $\endgroup$
    – David Roberts ♦
    Apr 13, 2012 at 2:42
  • $\begingroup$ I was just assuming that was the definition of holomorphic stack. Is it not? If not, I'm a bit confused. Or do you mean that I use atlases in proving this? My point just was that that construction itself doesn't need/use an atlas, and actually works for all stacks, not just ones coming from groupoid objects. $\endgroup$
    – David Carchedi
    Apr 13, 2012 at 2:48
  • $\begingroup$ Hmm, I suppose you're right. I would say that any stack arising from a groupoid has a specified atlas, so one doesn't need to choose an atlas, but in the localisation version all one needs to know that an atlas exists, which is true by definition. But I suppose we don't know what the OP needs: general stacks or geometric stacks. $\endgroup$
    – David Roberts ♦
    Apr 13, 2012 at 3:44
  • $\begingroup$ I haven't yet understood the construction of $u_!$ (I'm a beginner in this area) Is it true that all stacks are weak colimits of representable stacks, so stacks should be viewed somehow as a `weak colimit completion' of the original category? $\endgroup$
    – Brett Parker
    Apr 13, 2012 at 6:49
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    $\begingroup$ @Brett & David: Every stack is a weak colimit of representables. Take a stack $\mathcal{X}.$ It corresponds to a fibered category $$p_{\mathcal{X}}:\int \mathcal{X} \to C.$$ Let $y:C \to St(C,J)$ be the Yoneda embedding. Then $\mathcal{X}$ is the weak colimit of $y \circ p_{\mathcal{X}}.$ When a stack $\mathcal{X}$ comes from a groupoid object, one can write it as a much smaller colimit of representables in the way I wrote. I prove this in the appendix of my thesis, but it's widely known. $\endgroup$
    – David Carchedi
    Apr 13, 2012 at 12:12
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If you think of the 2-category of (geometric) stacks $GeomStack(ComplexMfld)$ over the site of complex manifolds as a localisation of (a certain sub-2-category of) the 2-category $Gpd(ComplexMfld)$ of groupoids internal to the site of complex manifolds, then you can use the fact the forgetful functor $ComplexMfld \to SmoothMfld$ gives rise to a 2-functor $Gpd(ComplexMfld) \to Gpd(SmoothMfld)$, and this gives rise to a 2-functor between the localisations aka the 2-categories of stacks by the universal property of localisations.

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  • $\begingroup$ "but I can't think of any general construction that doesn't involve some kind of atlas." Mhh, this construction looks general and involves atlases. $\endgroup$
    – Jan Weidner
    Apr 12, 2012 at 9:46
  • $\begingroup$ How generally will this approach work? (I really want to apply this to some categories that are like complex and smooth manifolds...) Do you know a reference which proves that gemetric stacks are a localization of groupoids in a general setting, or do you know what assumptions are needed on the category you're working over for the proof to work? $\endgroup$
    – Brett Parker
    Apr 13, 2012 at 0:19
  • $\begingroup$ @Brett: I work out in my thesis that differentiable stacks are equivalent to the bicategory of Lie groupoids with principal bundles as morphisms, and also in the topological setting. The proof works in much greater generality. You can take a look if you'd like. There's a link on my webpage. Also, I show it for more general Grothendieck topologies than just the open cover one. $\endgroup$
    – David Carchedi
    Apr 13, 2012 at 0:58
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    $\begingroup$ @Brett, in my thesis I prove a general result about localising 2-categories of internal groupoids using anafunctors. That anafunctors and stacks are related is not too hard to see, but I haven't written it down yet. That the 2-category of stacks is a localisation is covered, in special cases, by Pronk (in a 1996 article) and as David C said, in his thesis. As far as assumptions go, a site with finite products and a subcanonical superextensive pretopology should be enough (see the nLab for the definition of superextensive, but it covers pretty much any category of spaces you might care about) $\endgroup$
    – David Roberts ♦
    Apr 13, 2012 at 2:41
  • $\begingroup$ @David: Re: "That anafunctors and stacks are related is not too hard to see, but I haven't written it down yet." - If you do write it down, let me know. It would nice to have this finally settled once and for all. $\endgroup$
    – David Carchedi
    Apr 13, 2012 at 2:58

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