Timeline for Is there a sheaf theoretical characterization of a differentiable manifold?
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| Oct 4, 2022 at 0:05 | comment | added | Somatic Custard | For reference, in Mac Lane, Moerdijk - Sheaves in Geometry and Logic (bottom of p75), there is a sheaf-theoretic description of smooth manifolds which does require the second-countable condition. | |
| Feb 22, 2022 at 15:07 | comment | added | LSpice | I hope I may say that, while one could speak of schemes as perhaps being tailor-made for some purpose, if one can ever describe anything as Taylor-made then it is surely analytic manifolds. 😄 | |
| Apr 13, 2017 at 12:19 | history | edited | CommunityBot |
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| Feb 18, 2012 at 12:18 | vote | accept | Daniel Moskovich | ||
| Feb 18, 2012 at 12:18 | history | bounty ended | Daniel Moskovich | ||
| Feb 15, 2012 at 7:41 | comment | added | Ryan Budney | I'm still confused. What do you mean by "necessary"? Manifolds were not conceived of in an algebraic context -- I like to say they're solutions to the "flat earth problem" -- many things can be locally flat but not globally. The point of sheaves in some sense is that they allow for the combinatorial assembly of local data. Since manifolds don't have local data, putting them into the sheaf context amounts to opening a peanut with a table saw. It does the job but there are lighter and more direct ways to get to where you want to go. Er, anyhow, enough ideology. :) | |
| Feb 15, 2012 at 4:25 | comment | added | Daniel Moskovich | @Ryan: There's a Grothendieck-Serre idea of quasicoherent sheaves standing in for spaces, which works super-well in algebraic geometry. Vaguely, I wish I understood where these non-algebraic conditions like "paracompact Hausdorff" fit into that picture in the context of differential topology, and how they are necessary. It feels like there should be a bigger, algebraic, category, which is still useful (some souped-up synthetic Frolicher space story?). But the question is much more specific, about how sheaves can't fully stand-in for smooth spaces. | |
| Feb 15, 2012 at 1:07 | comment | added | Ryan Budney | @Daniel, when you say "the right one" what are you talking about? Such talk always makes me feel strange, as if there's some unspoken context everyone is afraid to make explicit. | |
| Feb 14, 2012 at 14:42 | comment | added | Zoran Skoda | Johaness and Martin: what about Dubuc's C-inf schemes and Moerdijk-Reyes models of smooth infinitesimal analysis (synthetic differential geometry) -- they are based on $C^\infty$-rings. ncatlab.org/nlab/show/smooth%20algebra | |
| Feb 14, 2012 at 14:39 | comment | added | Zoran Skoda | lierre: it is not a choice of coordinates but an existence of coordinates. Coordinate free means that there is some low level language which refers to coordinates but does not depend on choices, and then one builds up on this low level language rarely or almost never needing in general considerations to unpack into the lowest level language. | |
| Feb 14, 2012 at 13:28 | comment | added | Lierre | Coordinate free you said ? And what is $\mathbb R^n$, in "is locally isomorphic to $(\mathbb R^n, \mathcal O)$", if not a choice of coordinates ? | |
| Feb 12, 2012 at 14:26 | comment | added | Martin Brandenburg | @Fernando: I give you a downvote, for the following reason: The question by Daniel M. is if $M$ is paracompact provided that $\mathcal{O}_M$ is acyclic. The question is not if there are interesting examples of manifolds which are not assumed to be hausdorff or paracompact. And even if it was the question, you only give a rather incomplete comment instead of an answer. Yes, it is good to be open-minded, but it is not good to dismiss the goal of a question. | |
| Feb 12, 2012 at 12:32 | comment | added | Martin Brandenburg | @Johannes: Well manifolds are not special cases of schemes; but there are attempts to unify them (via derived stacks, I think). There should be some kind of "generalized commutative algebra" which is the local study for both of them. | |
| Feb 12, 2012 at 12:31 | comment | added | Daniel Moskovich | @Johannes Ebert: Surely you must be right... but in any case, I'm wondering whether the sheaf approach gives another way to think about the "second-countable Hausdorff" condition- more generally, I wonder whether sheaf properties can capture significant general topological properties of a space (and in doing so, motivate them in a new way, perhaps). I also wonder in how deep of a sense "because the structure sheaf is acyclic" is the motivation for requiring manifolds to be paracompact... | |
| Feb 12, 2012 at 11:48 | comment | added | Johannes Ebert | @Daniel: yes, manifolds are a special case of schemes, but this viewpoint is not so useful for the study of manifolds, because the scheme formalism is taylor-made to globalize results from commutative algebra. But from the viewpoint of commutative algebra , $C^{\infty}(\mathbb{R})$ is rather pathological. This is why commutative algebra is not the right local tool for differential geometry. | |
| Feb 12, 2012 at 7:02 | history | edited | Daniel Moskovich | CC BY-SA 3.0 |
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| Feb 12, 2012 at 6:54 | history | bounty started | Daniel Moskovich | ||
| Feb 11, 2012 at 23:39 | comment | added | Daniel Moskovich | @Ryan: Conversely, a good sheaf definition would make manifolds look like schemes, which is a more flexible concept, and might conceptually explain why the definition is "the right one". There is certainly some controversy on that count- answers below claim "Hausdorff" is too strong; the natural concept of a useful Frechet space (one in which you have inverse function theorems) isn't 100% clear; and there are Chen spaces, Frolicher spaces, Diffeologies... These are just random thoughts, nothing concrete- but a sheaf definition does satisfy me much more than an "atlas" definition right now. | |
| Feb 11, 2012 at 0:16 | comment | added | Fernando Muro | You dipute it without arguments | |
| Feb 10, 2012 at 23:01 | comment | added | Ryan Budney | It's not clear to me there's any advantage in this formalism for manifolds. From a historical perspective, demanding someone to know what a sheaf is before a manifold seems kind of backwards. And the end result is, you've got a definition that pre-supposes the student is comfortable with a higher-order level of baggage and formalism than the manifold concept, moreover, you haven't really made the subject any easier, you've just brought in an extra layer of definitions to wade through. Cute? Yes. Nice? Maybe not. | |
| Feb 10, 2012 at 19:27 | comment | added | Charles Staats | I seem to recall that Wikipedia deliberately does not include the "second countable Hausdorff" conditions so that they can talk about non-Hausdorff manifolds (and call the long line a manifold). | |
| Feb 10, 2012 at 18:10 | comment | added | Zhen Lin | @Daniel: Separatedness, in the sense of being a Hausdorff topological space, is a property of topological spaces with points, but not a property of locales (pointless topological spaces). The assignment $X \mapsto \textbf{Sh}(X) : \textbf{Top} \to \mathfrak{Topos}$ factors through the category of locales $\textbf{Loc}$. As such there is no hope of finding a purely sheaf-theoretic characterisation without referring to stalks or points somewhere. | |
| Feb 10, 2012 at 15:45 | comment | added | Daniel Litt | @Paul Siegel: No I don't thinks so--the set of germs of continuous functions strictly contains the set of germs of smooth functions. | |
| Feb 10, 2012 at 15:05 | comment | added | Paul Siegel | Oops again! Perhaps what I said might be relevant to germs of continuous functions... In any event, as Will pointed out these matters can't be settled locally. Sorry to all for cluttering the comment area with my dumb mistakes. | |
| Feb 10, 2012 at 14:56 | comment | added | Andreas Blass | @Paul: No. The germ contains much more information than just the value at $x$. To begin with, it determines all the derivatives, of all orders, at $x$. But even that isn't all, since two smooth functions can have the same Taylor series at $x$ without coinciding on any neighborhood of $x$. | |
| Feb 10, 2012 at 12:38 | comment | added | Paul Siegel | I don't dispute that non-Hausdorff manifolds come up naturally - the quotient of any manifold by a bad group action comes to mind. I dispute that "you can develop most of the theory without these conditions". | |
| Feb 10, 2012 at 12:33 | comment | added | Paul Siegel | If we think of a germ at $x$ as an equivalence class of smooth functions defined on neighborhoods of $x$, then isn't the germ completely determined by the common value that all of the functions in the equivalence class take at $x$? | |
| Feb 10, 2012 at 12:04 | comment | added | Fernando Muro | I invite you to do some google search if you think that non-Hausdorff manifolds are completely useless. In my opinion, it's positive to be open-minded in math. | |
| Feb 10, 2012 at 10:28 | answer | added | Stefan Waldmann | timeline score: 4 | |
| Feb 10, 2012 at 9:46 | answer | added | Johannes Ebert | timeline score: 65 | |
| Feb 10, 2012 at 8:28 | comment | added | Qfwfq | @PaulSiegel: and on the usual real line, it's the fiber of $\mathcal{O}$ that is isomorphic to $\mathbb{R}$, not the stalk which is a ring of germs of smooth functions. | |
| Feb 10, 2012 at 7:53 | answer | added | Sebastian | timeline score: 6 | |
| Feb 10, 2012 at 5:07 | answer | added | Tom Goodwillie | timeline score: 25 | |
| Feb 10, 2012 at 4:04 | comment | added | Tom Church | The answers to this question include some more examples of why it is important that your manifolds be Hausdorff: mathoverflow.net/questions/13072/… | |
| Feb 10, 2012 at 2:56 | comment | added | Paul Siegel | @Will Sawin: Oops, that was stupid. The condition on stalks that I gave is equivalent to $T_0$, isn't it. Thanks for the correction... | |
| Feb 10, 2012 at 2:20 | comment | added | Will Sawin | @Paul Siegel: I don't think that statement is true, since either origin has a neighborhood excluding the other origin. Hausdorff is not a local condition, and everything that's locally Euclidean is locally Hausdorff. | |
| Feb 10, 2012 at 2:04 | comment | added | Paul Siegel | It seems to me that hardly any of the theory works without these conditions. Can you really think of a nontrivial theorem about manifolds which uses neither partitions of unity nor a metric? | |
| Feb 10, 2012 at 2:01 | comment | added | Paul Siegel | I don't think about sheaves all that often, but it seems to me that $M$ is Hausdorff if and only if all of the stalks of $\mathcal{O}_M$ are isomorphic to $\mathbb{R}$. For example, the stalk at (either) origin in the line with the double point is $\mathbb{R}^2$. I'm not exactly sure how to make $M$ second countable, but on the other hand the only way I ever see that axiom used is to construct partitions of unity. So I would guess that the "fine" condition should do the job | |
| Feb 10, 2012 at 1:38 | comment | added | Daniel Moskovich | @Tom Goodwillie: phrase edited. | |
| Feb 10, 2012 at 1:30 | history | edited | Daniel Moskovich | CC BY-SA 3.0 |
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| Feb 10, 2012 at 1:24 | comment | added | Tom Goodwillie | I'm all in favor of sheafy points of view, but: You can avoid the unpleasantness of an arbitrary choice of atlas by defining a smooth structure to be a maximal atlas. (Of course in practice you will use the fact that an atlas determines a maximal atlas.) Also, is "looks the same at all points" really the same as "diffeomorphism group acts transitively"? What if $M$ is not connected? | |
| Feb 10, 2012 at 1:13 | comment | added | Daniel Moskovich | Also, I'm not "married" to the textbook definition of a differentiable manifold, but if one were to define anything strictly more general, the usefulness of this extra generality (including extra objects) would need to be strongly justified. | |
| Feb 10, 2012 at 1:06 | history | edited | Daniel Moskovich | CC BY-SA 3.0 |
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| Feb 10, 2012 at 0:45 | comment | added | Daniel Moskovich | I want to be able to do differential topology. In particular, I want approximation theorems which approximate smooth functions on a closed subset extended by continuous functions, by smooth functions. I want to spline diffeomorphisms together. I want to glue along boundaries, I want inverse and implicit function theorems. Maybe I even want Morse Theory. | |
| Feb 10, 2012 at 0:38 | history | edited | Daniel Moskovich | CC BY-SA 3.0 |
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| Feb 10, 2012 at 0:34 | comment | added | Fernando Muro | Why do you want manifolds to be second-countable Hausdorff? You exclude long lines and the line with double origin. You can develop most of the theory without these conditions. You won't have metrizability, though. | |
| Feb 10, 2012 at 0:30 | history | asked | Daniel Moskovich | CC BY-SA 3.0 |