As will become clear, this is in some sense a follow up on my earlier question Why should I prefer bundles to (surjective) submersions?. As with that one, I hope that it's not too open-ended or discussion-y. If y'all feel it is too discussion-y, I will happily close it.

Let $\rm Man$ be the category of smooth (finite-dimensional) manifolds. I can think of (at least) two natural "smooth structures" on $\rm Man$, which I will outline. My question is whether one of these is the "right" one, or if there is a better one.

I should mention first of all that there many subtly different definitions of "smooth structure" — see e.g. n-Lab: smooth space and n-Lab: generalized smooth space and the many references therein — and I don't know enough to know which to prefer. Moreover, I haven't checked that my proposals match any of those definitions. In any case, the definition of "smooth structure" that I'm happiest with is one where I only have to tell you what all the smooth curves are (and these should satisfy some compatibility condition). So that's what I'll do, but I'm not sure if they do satisfy the compatibility conditions. Without further ado, here are two proposals:

  1. A smooth curve in $\rm Man$ is a fiber bundle $P \to \mathbb R$.
  2. A smooth curve in $\rm Man$ is a submersion $Y \to \mathbb R$.

Then given a manifold $M$, we can make it into a category by declaring that it has only identity morphisms. Then I believe that the smooth functors $M \to {\rm Man}$ under definition 1 are precisely the fiber bundles over $M$, whereas in definition 2 they are precisely the submersions over $M$.

(Each of these claims requires checking. In the first case, it's clear that bundles pull back, so all bundles are smooth functors, and so it suffices to check that if a surjective submersion to the disk is trivializable over any curve, then it is trivializable. In the second case, it's clear that if a smooth map restricts to a submersion over each curve, then it is a submersion, so any smooth functor in a submersion, and so one must check that submersions pull back along curves.)

I can see arguments in support of either of these. On the one hand, bundles are cool, so it would be nice if they were simply "smooth functors". On the other hand, we should not ask for smooth functions (i.e. 0-functors) to be necessarily "locally trivializable", as then they'd necessarily be constant. Maybe the correct answer is definition 2, and that bundles are "locally constant smooth functors", or something.

Anyway, thoughts? Or am I missing some other good definition?


In the comments, folks have asked for applications, which is very reasonable. The answer is that I would really like to have a good grasp of words like "smooth functor", at least in the special case of "smooth functor to $\rm Man$". Of course, Waldorf and Shreiber have explained these words in certain cases in terms of local gluing data (charts), but I expect that a more universal definition would come directly from a good notion of "smooth structure" on a category directly.

Here's an example. Once we have a smooth structure on $\rm Man$, we can presumably talk about smooth structures on subcategories, like the category of $G$-torsors for $G$ your favorite group. Indeed, for the two definitions above, I think the natural smooth structure on $G\text{-tor}$ coincide: either we want fiber bundles where all the fibers are $G$-torsors, or submersions where all the fibers are $G$-torsors, and in either case we should expect that the $G$ action is smooth. So then we could say something like: "A principle $G$-bundle on $M$ is (i.e. there is a natural equivalence of categories) a smooth functor $M \to G\text{-tor}$", where $M \rightrightarrows M$ is the (smooth) category whose objects are $M$ and with only trivial morphisms. (Any category object internal to $\rm Man$ automatically has a smooth structure.) And if I understood the path groupoid mod thin homotopy $\mathcal P^1(M) \rightrightarrows M$ as a smooth category, then I would hope that the smooth functors $\mathcal P^1(M) \to G\text{-tor}$ would be the same as principle $G$-bundles on $M$ with connection. Functors from the groupoid of paths mod "thick" homotopy should of course be bundles with flat connections. Again, Schreiber and Waldorf have already defined these things categorically, but their definition is reasonably long, because they don't have smooth structures on $\rm Man$ that are strong enough to let them take advantage of general smooth-space yoga.

Here's another example. When I draw a bordism between manifolds, what am I actually drawing? I would like to say that I'm drawing something close to a "smooth map $[0,1] \to \rm Man$". I'm not quite, by my definitions — if you look at the pair of pants, for instance, at the "crotch" it is not a submersion to the interval. So I guess there's at least one more possible definition of "smooth curve in $\rm Man$":

  • A smooth curve in $\rm Man$ is a smooth map $X \to \mathbb R$.

But this, I think, won't be as friendly a definition as those above: I bet that it does not satisfy the compatibility axioms that your favorite notion of "smooth space" demands.

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    $\begingroup$ How do you plan to use this category? A potential application would give your question a good point of focus. $\endgroup$ – S. Carnahan Mar 31 '10 at 3:30
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    $\begingroup$ But what are you going to do with this? Are you really going to be working in a setting where the difference between the two really matters? (I mean these to be naive questions from someone uses manifolds only in the simplest ways and is still quite uneducated on why category theory is useful for studying differential geometry) $\endgroup$ – Deane Yang Mar 31 '10 at 3:36
  • $\begingroup$ I like your category. Feels like something potentially useful. $\endgroup$ – Bo Peng Mar 31 '10 at 4:48
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    $\begingroup$ I agree with Carnahan and Yang, there is no correct definition until you have a purpose for the device you want to construct, be it a category or whatever. $\endgroup$ – Ryan Budney Mar 31 '10 at 5:29
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    $\begingroup$ @Bo: what do you mean by "useful"? $\endgroup$ – BCnrd Mar 31 '10 at 5:34

I think that the most interesting part of your question is the part you put in parentheses!

(and these should satisfy some compatibility condition)

What are your compatibility conditions? That is everything here. If you specify the correct conditions, you may find that all your definitions collapse to just one.

I have an issue with Konrad's answer (which I doubt very much that he will be surprised to hear me express!). Whenever I heard words like "Grothendieck topology" or "sheaves" or encoding similar ideas then I feel that something's been lost. I don't like the idea that "smooth" is just "really nice continuity". "Smooth" sits alongside continuity and can be expressed in a different way which is extremely simple: takes smooth curves to smooth curves.

Of course, I would say that, as everyone by now presumably knows that I prefer Frolicher spaces to the other variants (like Chen spaces or diffeological spaces, see generalized spaces for links). It is interesting that Chen's third definition (by my count) was stronger than his eventual sheaf condition and was more along the lines of "a map is smooth if enough tests say that it is smooth".

But Frolicher spaces have a problem, which is that it is extremely difficult to prise them away from being a set-based theory. The compatibility condition is so strong that it forces an underlying set. I'd really like to figure out how to make this extension, and I know that Urs would as well. If I could just encourage you and Konrad over to the nLab to play around with these ideas to see how they could work ...

If you want to study The Smooth, The Whole Smooth, and Nothing But The Smooth, then you should do so and not flirt continually with continuity. The stronger compatibility condition means that more stuff is smooth than you first thought (witness my recent question on this) and that makes it interesting! The unexpected happens, so study it!

This isn't much of an answer so far, it's more of a commentary on your question which (as is usual for me) is too long for an actual comment. So let me end with an actual answer (which I freely confess that I stole from a rabbi):

That is such a great question, why on earth would you want an answer?

  • $\begingroup$ I need to remember that answer. It seems as though it should have many applications! $\endgroup$ – Mike Shulman Apr 1 '10 at 6:09
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    $\begingroup$ Hi Andrew, I am very surprised about your issue! For me, Grothendieck topology has nothing to do with topology, or continuity. I totally agree that "smooth" is not just an improvement of "continuous". A Grothendieck topology is the basis of descent theory, and makes sense in any context, for example diffeological spaces. There, a good Grothendieck topology is formed by "subductions", i.e. maps along which one can locally lift plots. $\endgroup$ – Konrad Waldorf Apr 1 '10 at 18:36
  • $\begingroup$ But the "locally lift" statement implies topology! For example, I've finally realised why you diffeologists like "sitting instants": it's because you don't know how to concatentate curves without them. Even if all the jets agree, you can't concatentate without a sitting instant because otherwise it doesn't "locally lift". I admit that I don't truly understand Grothendieck topologies and that I have a bit of a knee-jerk reaction to the word "topology" when thinking about "smootheology". But then, despite my involvement in the nLab I haven't really grokked the higher stuff yet. $\endgroup$ – Loop Space Apr 2 '10 at 18:22

This is an extremely interesting question. I think what you need is the notion of a stack over smooth manifolds (i.e. a sheaf of groupoids). Such a stack could assign to each smooth manifold $U$

  1. the category of smooth fibre bundles over $U$.
  2. the category of surjective submersions over $U$.

The problem is the gluing axiom for this stack. To state it, you have to decide for a Grothendieck topology on the category of smooth manifolds. Here you have again the two choices 1 and 2, and some more.

According to the calculations I just did (and I should add that it's late and I am tired), all four possible combinations work. So it's again up to you!

Personally, I have a preference for the submersions. If you require fibre bundles, it seems that the manifolds in a "connected component" of the stack are all diffeomorphic, whereas it should be possible to smoothly change the diffeomorphism type.

  • $\begingroup$ Great, and I'm with you on wanting to smoothly change diffeomorphism type. Incidentally, I do not believe that there should be the word "surjective" in #2: the empty manifold is a perfectly good manifold of any dimension, and I see no reason not to allow it to be hit by a smooth map. $\endgroup$ – Theo Johnson-Freyd Mar 31 '10 at 15:52

Perhaps Erhesmann's Fibration Theorem helps to clarify the relation between these two possible definitions of smooth curves. The theorem states that a proper surjective submersion is in fact a locally trivial fibre bundle (with compact fibres).

This means that the two notions of smoothness will coincide on the subcategory ManproperMan consisting of manifolds and proper maps. On the other hand, if I've understood correctly what a smooth functor M to Man is, then a bundle over M with compact fibres should be the same as a smooth functor to the subcategory Manproper.

  • $\begingroup$ Wait, in your last sentence, are you saying that every bundle is proper? Certainly if I have a proper surjective submersion, it is a bundle, but I think I can want bundles without compact fibers. $\endgroup$ – Theo Johnson-Freyd Mar 31 '10 at 16:17
  • $\begingroup$ How can a proper surjective submersion be a fibre bundle? Cover a manifold by finitely many open sets, and define the total space of the submersion to be the disjoint union. Then, the fibres will be disjoint unions of finitely many points, but over some points you find one, while over other points you may find two or three, so this is not a locally trivial fibre bundle. $\endgroup$ – Konrad Waldorf Mar 31 '10 at 16:59
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    $\begingroup$ Theo, apologies - I've edited the answer to clarify that the bundles one gets from functors to the the subcategory of proper maps always have compact fibres. Konrad, your example will not be a proper map, as the inclusion of an open ball in Euclidean space is not proper. $\endgroup$ – Jeffrey Giansiracusa Mar 31 '10 at 18:08
  • $\begingroup$ @Konrad: 'proper' means in this context that preimages of compact sets are compact. This is obviously not the case in your example. It is only for closed maps enough that preimages of points are compact. $\endgroup$ – Lennart Meier Apr 1 '10 at 10:53

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