Not signed in (Sign In)

Start a new discussion

Not signed in

Want to take part in these discussions? Sign in if you have an account, or apply for one below

  • Sign in using OpenID

Discussion Tag Cloud

Vanilla 1.1.10 is a product of Lussumo. More Information: Documentation, Community Support.

Welcome to nForum
If you want to take part in these discussions either sign in now (if you have an account), apply for one now (if you don't).
    • CommentRowNumber1.
    • CommentAuthorUrs
    • CommentTimeApr 2nd 2015
    • (edited Feb 10th 2016)

    am working on putting some genuine detailed content into smooth groupoid. So far there is now discussion of the groupoid-enriched category of groupoid-valued presheaves, Cech nerves, and the stack condition.

    Then it breaks off and some rough old material kicks in which needs to be harmonized. Will continue later, need to go offline now for a little.

    • CommentRowNumber2.
    • CommentAuthorUrs
    • CommentTimeApr 2nd 2015

    have now expanded a good bit further. But there are still some loose ends…

    • CommentRowNumber3.
    • CommentAuthorJohn Dougherty
    • CommentTimeMay 10th 2015

    In Remark 6 it says that one can define smooth groupoids as stacks on the site of all smooth manifolds instead of the site of Cartesian spaces. It says

    Everything discussed so far goes through verbatim for [SmoothMfd\mathsf{SmoothMfd}], too, but the descent condition in def. 4 is a much stronger condition. For instance the presheaves of the form (BG) (\mathbf{B}G)_\bullet from example 3 satisfy descent on CartSp\mathsf{CartSp}, but not all SmoothMfd\mathsf{SmoothMfd}.

    I’m not clear on how to understand the situation. In particular, I’m not sure what the condition in def. 4 is stronger than. Here’s one possibility. On the first approach we define a pre-smooth groupoid to be a (2,1)-presheaf on SmoothMfd\mathsf{SmoothMfd}. We then say that a pre-smooth groupoid is a smooth groupoid if it satisfies the stack condition. On the second approach, we do the same thing on the site CartSp\mathsf{CartSp}. As it happens (because all of the objects in CartSp\mathsf{CartSp} are contractible), the stack condition has a particularly simple expression on this site, and this simplified expression is given as the descent condition in def. 4. These two approaches give equivalent definitions of smooth groupoids. However, if one were to try to use the stack condition as expressed in def. 4 on the first approach (by replacing the n\mathbb{R}^{n} with arbitrary objects of SmoothMfd\mathsf{SmoothMfd}), then one would not obtain the stack condition on SmoothMfd\mathsf{SmoothMfd}. Rather, one would obtain something more restrictive. For example, (BG) (\mathbf{B}G)_{\bullet} does not satisfy this translated condition, though it does satisfy the correct stack condition on SmoothMfd\mathsf{SmoothMfd}. The reason that one can’t just make the obvious replacements in def. 4 and obtain the stack condition on SmoothMfd\mathsf{SmoothMfd} is that def. 4 is a simplified expression, and this simplification relies on properties of CartSp\mathsf{CartSp} that SmoothMfd\mathsf{SmoothMfd} doesn’t have.

    Is something like this right?

    • CommentRowNumber4.
    • CommentAuthorDavid_Corfield
    • CommentTimeMay 10th 2015
    • (edited May 10th 2015)

    I’m not sure what the condition in def. 4 is stronger than.

    Presumably that any pre-smooth groupoid that satisfies descent for SmoothMfd\mathsf{SmoothMfd} will satisfy descent for CartSp\mathsf{CartSp} but this doesn’t hold in general vice versa.

    While we’re at def. 4, the two occurrences before and in it:

    …and precomposes it with the canonical map C({U i})XC(\{U_i\}) \to X


    given by pre-composition with C({U i})XC(\{U_i\}) \to X

    and another in the commutative diagram between these should presumably be C({U i}) nC(\{U_i\}) \to \mathbb{R}^n.

    • CommentRowNumber5.
    • CommentAuthorDavid_Corfield
    • CommentTimeMay 10th 2015

    So I fixed that and other typos.

    Back to #3, presumably the issue is that not all smooth groupoids are pre-smooth groupoids satisfying descent on CartSp (though most are) and fewer satisfy descent on SmoothMfd (including important ones). But then

    So a smooth groupoid is a stack on the site CartSp

    is misleading.

    • CommentRowNumber6.
    • CommentAuthorUrs
    • CommentTimeMay 11th 2015

    David, thanks, these were indeed typos, thanks for fixing them.

    John: it’s simply that CartSpCartSp is a subcategory of SmoothMfdSmoothMfd and hence requiring descent over the former site is only a subset of the conditions of requiring descent over the latter site.

    (In terms of the localization of simplicial presheaves (called “pre-smooth groupoids” in the entry) this is counterbalanced by the fact that over the smaller site the condition for cofibrancy is stronger that that over the larger site (for the projective model structure).)

    David, so it is not misleading to say that smooth groupoids are stacks over CartSpCartSp, indeed there is an equivalence of (2,1)(2,1)-catgeories

    Stacks(CartSp)Stacks(SmoothMfd) Stacks(CartSp) \simeq Stacks(SmoothMfd)

    and either of these (2,1)(2,1)-categories is called Smooth1TypeSmooth1Type in the entry, whose objects may be called smooth stacks.

    • CommentRowNumber7.
    • CommentAuthorDavid_Corfield
    • CommentTimeMay 11th 2015

    Oh, I see. It was this part that confused me:

    For instance the presheaves of the form (\mathbf{B}G)_\bullet = (G \stackrel{\longrightarrow}{\longrightarrow} \ast) from example 3 satisfy descent on CartSp, but not all SmoothMfd. Still, once we have defined the higher category of smooth groupoids, the definition will be equivalent for both choices of sites.

    The choice of the smaller site is the one that is easier to work with, and therefore we stick with that. In fact, most every example of a pre-smooth groupoid that one runs into satisfies descent on CartSp.

    From that last sentence, not every example of a pre-smooth groupoid satisfies descent on CartSp? But then what happens to such pre-smooth groupoids under localization,

    which is really just the identity as a functor. Instead of doing anything to the objects, passing along this functor just means to change the definition of the hom-groupoids from the direct definition of def. 3 to the localized definition.

    • CommentRowNumber8.
    • CommentAuthorUrs
    • CommentTimeMay 12th 2015

    Yes, in general not every pre-sheaf is a sheaf, of course, and this is as true over CartSpCartSp as over any other site.

    Under localization, a pre-sheaf or more generally a pre-stack becomes equivalent to its sheafification or more generally its stackification, which you may think of as being the universal way of approximating “from the right” the object that does not necessarily satisfy descent by one that does.

    Thanks for these comments, I see that the entry leaves some puzzlement that should be clarified. But at the moment I am not sure what could be added to the entry to help make this clearer. But once you and John D. are content with the discussion here, you could maybe make a suggestion.

    • CommentRowNumber9.
    • CommentAuthorJohn Dougherty
    • CommentTimeMay 13th 2015

    One way to make the entry a bit clearer would be to modify Remark 6 to

    There is an alternative formulation of the whole theory where instead of the site CartSpCartSp one uses the site SmoothMfdSmoothMfd of all smooth manifolds. Everything discussed so far goes through verbatim for that site, too, but the descent condition over SmoothMfdSmoothMfd is a much stronger requirement than the descent condition over CartSpCartSp.

    Assuming this is true (and I gather from our discussion that it is), I think I can start to see more of my confusion. Take BG\mathbf{B}G; it satisfies descent over CartSpCartSp, so is a stack over CartSpCartSp. Assuming that the equivalence Stacks(CartSp)Stacks(SmoothMfd)Stacks(CartSp) \simeq Stacks(SmoothMfd) is given by inclusion, BG\mathbf{B}G is a stack over SmoothMfdSmoothMfd. But BG\mathbf{B}G doesn’t satisfy descent over SmoothMfdSmoothMfd, so it’s not a stack over SmoothMfdSmoothMfd.

    I think the problem with that reasoning is that I’m not accounting for localization correctly, but I don’t know where exactly my problem is. Definition 4 gives the stack condition. Remark 5 claims that a smooth groupoid is a stack on CartSpCartSp. Proposition 3 refers to “a smooth groupoid in the sense of def. 4”, but def. 4 doesn’t define smooth groupoids. I assumed that I was supposed to take “smooth groupoid” to mean “pre-smooth groupoid that satisfies descent (as defined in def. 4)”.

    Now I think that was a mistake on my part. Smooth groupoids are defined in def. 8 as objects of the simplicial localization of PreSmooth1TypePreSmooth1Type. The canonical localization functor acts on objects as the identity, and so every pre-smooth groupoid is a smooth groupoid, when considered as an object of L lwePreSmooth1TypeL_{\text{lwe}}PreSmooth1Type via the localization functor. If that’s right, then I was mistaken when I started thinking that (i) “smooth groupoid” meant (ii) “stack on CartSp” meant (iii) “pre-smooth groupoid satisfying the stack condition”. If “stack on CartSp” means “object of the (2,1)(2,1)-category Stacks(CartSp)Stacks(CartSp)”, then I think (i) and (ii) mean the same thing, but I don’t know how (iii) fits in. After all, a pre-smooth groupoid that doesn’t satisfy the stack condition is still an object of the (2,1)(2, 1)-category Stacks(CartSp)Stacks(CartSp) by application of the localization functor. So what does the stack condition do here?

    • CommentRowNumber10.
    • CommentAuthorDavid_Corfield
    • CommentTimeMay 13th 2015

    That’s a very clear way of expressing the confusion, John.

    I think the problem with that reasoning is that I’m not accounting for localization correctly,

    I think that’s right, and hence we need to understand

    Under localization, a pre-sheaf or more generally a pre-stack becomes equivalent to its sheafification or more generally its stackification

    This kind of statement confuses us as poor philosophers brought up on a diet of set theory. There’s something in a collection undergoing a process, and we’re told no member of the collection changes, but that miraculously that thing becomes equivalent to a kind of thing it wasn’t before localisation.

    A colourful comparison might be that it’s like London becoming rail-connected to Paris on building the Channel Tunnel. London hasn’t changed, but it’s now connected.

    Of course, a Bradleian metaphysician would probably want to say that London has changed, since everything is defined by its internal relations. Russell couldn’t seem to fathom that kind of thinking.

    • CommentRowNumber11.
    • CommentAuthorDavid_Corfield
    • CommentTimeMay 13th 2015

    Strange how so often parallel discussions have common themes. Compare #10 with this.

    • CommentRowNumber12.
    • CommentAuthorUrs
    • CommentTimeMay 13th 2015
    • (edited May 13th 2015)


    yes, so just as David says, there is something happening as one sends an object through this equivalence:

    Stacks(CartSp)Stacks(SmoothMfd). Stacks(CartSp) \stackrel{\simeq}{\longrightarrow} Stacks(SmoothMfd) \,.

    The presheaf of groupoids (BG) :CartSpGrpd(\mathbf{B}G)_\bullet \colon CartSp \to Grpd given by n(C ( n,G)*)\mathbb{R}^n \mapsto (C^\infty(\mathbb{R}^n,G) \stackrel{\longrightarrow}{\longrightarrow}\ast) does NOT go, under this equivalence, to the presheaf on the site of manifolds which would be given by sending XSmoothMfdX \in SmoothMfd to the groupoid (C (X,G)*)(C^\infty(X,G) \stackrel{\longrightarrow}{\longrightarrow}\ast), but goes to the stackification of the latter, which is the stack GBund()G\mathbf{Bund}(-) of GG-principal bundles.

    Even if this further stackification seems like a drastic operation, it gives here an equivalence of (2,1)(2,1)-categories.

    • CommentRowNumber13.
    • CommentAuthorMike Shulman
    • CommentTimeMay 14th 2015

    There’s something in a collection undergoing a process, and we’re told no member of the collection changes, but that miraculously that thing becomes equivalent to a kind of thing it wasn’t before localisation.

    Interesting observation! It’s not fundamentally different from the process of forming the quotient of an equivalence relation in set theory: elements of a set which were not equal become equal after quotienting. In that case we usually call the image in the quotient the “equivalence class” rather than the object itself. But even in the case of localization, I don’t think it’s really correct to say that “no member of the collection changes”. From a structural point of view, a thing is a thing only insofar as it is a member of some set/type (this sounds kind of like your Bradleian), so it doesn’t make sense to say that a pre-stack “is the same thing as” its image in the localization. It’s true that when we analyze categories in terms of sets of objects and morphisms, one construction of the localization takes its set of objects to be the same as the set of objects of the original category, but even in set theory that isn’t the only way to construct a localization, and of course there are more things in heaven and earth than set theory. (-:

    • CommentRowNumber14.
    • CommentAuthorJohn Dougherty
    • CommentTimeMay 14th 2015

    Thanks all, my confusion has been pretty much resolved. I was quite happy to say that localization made a pre-stack equivalent to its stackification even though it acted as the identity on the objects. I took it to be just another instance of the importance of keeping track of types. I just didn’t see how localization was related to the Stacks(CartSp)Stacks(SmoothMfd)Stacks(CartSp) \simeq Stacks(SmoothMfd) equivalence, and I wasn’t sure what type to assign to (BG) (\mathbf{B}G)_{\bullet} in different sentences. But now that I get it, I can’t quite see why I was confused, so I unfortunately don’t have a concrete suggestion for an edit to the entry.

Add your comments
  • Please log in or leave your comment as a "guest post". If commenting as a "guest", please include your name in the message as a courtesy. Note: only certain categories allow guest posts.
  • To produce a hyperlink to an nLab entry, simply put double square brackets around its name, e.g. [[category]]. To use (La)TeX mathematics in your post, make sure Markdown+Itex is selected below and put your mathematics between dollar signs as usual. Only a subset of the usual TeX math commands are accepted: see here for a list.

  • (Help)