Timeline for Intuition/meaning behind/physical content of the concept of a smooth structure
Current License: CC BY-SA 4.0
19 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 8, 2022 at 20:05 | comment | added | Russ Phelan | @AivazianArshak My point is: visualization is nice, but not always helpful. The kinds of spaces that admit different smooth structures are far, far away from "physical intuition". I really do think the right answer to your question is to show functions that are smooth under one choice, but not another, then "look" at what those functions are like. That will build intuition for how smooth structure can vary. My secondary point is that you may not want to trust the current visualizations you like: they don't necessarily contain useful or accurate information. | |
| Apr 8, 2022 at 20:04 | comment | added | Russ Phelan | @AivazianArshak I think you are putting a bit too much faith in your visualizations. For example, adding a metric to a shape does not make it "hard", it just restricts the ways in which it can bend. For example a cylinder minus a line is isometric to a flat rectangle. You only get "rigidity" in the intuitive sense if you are talking about invariance under rigid motions in R^n. Same with topological spaces: thinking of them of points collected in a "soft form" may be a comforting idea, but topological spaces can be incredibly pathological, and it would be easy to destroy this intuition. | |
| Mar 29, 2022 at 10:06 | comment | added | Tobias Diez | Since you can recover the smooth structure of a manifold from the ring of smooth functions, the question is indeed equivalent to finding a continuous function that is smooth wrt to one differentiable structure but not wrt to the other one. | |
| Mar 28, 2022 at 19:02 | comment | added | Arshak Aivazian | @ThomasRot Analogy: the question "how to think about whether a function is an isometry or not" (it's clear how to think - "preserve distances" is an initially intuitive idea) is not related to the question "how to think about what information a metric stores" (which also seems fine intuitively). However, whether a given set-theoretic function is an isometry or not depends on the metrics on dom-e and cod-e. | |
| Mar 28, 2022 at 19:01 | comment | added | Arshak Aivazian | @ThomasRot I understand it. I don't understand what this has to do with the conversation. Sam Hopkins obviously meant that $\mathbb{R}$ has a fixed standard differential structure and asked to what extent my question differs from the visualization of a continuous non-differentiable function. | |
| Mar 28, 2022 at 13:14 | comment | added | Thomas Rot | The smoothness is not a property of a function on a manifold. It depends on the smooth structure. We typically choose the standard smooth atlas on $\mathbb{R}$,but I chose a different one. The two smooth structures are diffeomorphic in this case. | |
| Mar 24, 2022 at 19:26 | comment | added | Arshak Aivazian | @ThomasRot I don't understand what you want to say. You just renamed the points. The same $|x|$ under the new name (according to the translation by $f$) is also not smooth. I said in my comment that the smoothness of a function is a property of the function (and the physical meaning of the smoothness of a function is perfectly clear; we think of more or less all macrophenomena in nature as smooth). But a smooth structure on a manifold is an additional structure (which has nothing to do with the smoothness of a function). It is not very clear what it corresponds to in nature. | |
| Mar 24, 2022 at 15:19 | comment | added | Thomas Rot | @AivazianArshak: There is a smooth structure on $\mathbb{R}$ for which $g(x)= |x|$ is differentiable. Namely consider the homeomorphism $f:\mathbb{R}\rightarrow \mathbb{R}$ $f(x)=-x^2$ if $x<0$ and $f(x)=x^2$ if $x\geq 0$. The map $f$ induces a smooth structure on $\mathbb{R}$ for which $g$ is differentiable (as g(f(x))=x^2). Of course in this case $\mathbb R$ with the standard smooth structure and $\mathbb R$ with the smooth structure induced by $f$ are diffeomorphic. But I still think that gives insight into what a smooth structure does. This is the comment of Marco Golla above. | |
| Mar 23, 2022 at 19:09 | comment | added | LSpice | Re, the smoothness of a manifold is an additional structure (on the topological manifold, I guess), whereas the smoothness of a function is an additional structure on its domain—there's no way to tell from a bare function, i.e., abstract set of ordered pairs such that …, whether or not it is smooth. | |
| Mar 23, 2022 at 14:47 | history | edited | Overflowian | CC BY-SA 4.0 |
added 1 character in body
|
| Mar 22, 2022 at 23:19 | answer | added | Deane Yang | timeline score: 2 | |
| Mar 22, 2022 at 22:57 | review | Close votes | |||
| Mar 28, 2022 at 13:31 | |||||
| Mar 22, 2022 at 22:53 | comment | added | Maxime Ramzi | @MarcoGolla : wow I had no clue $S^7$ had such a simple exotic smooth structure, thanks ! | |
| Mar 22, 2022 at 22:45 | comment | added | Marco Golla | As for the "simplest example", there are at least two candidates: 1. $3\mathbb{CP}^2\#20\overline{\mathbb{CP}}\vphantom{C}^2$ and $K3\#\overline{\mathbb{CP}}\vphantom{C}^2$; 2. $S^7$ and $\{v^2+w^2+x^2+y^3+z^7 = 0\} \cap S^9 \subset \mathbb{C}^5$. | |
| Mar 22, 2022 at 22:41 | comment | added | Marco Golla | @MaximeRamzi: maybe the two questions are not so separate, since whenever you have two smooth manifolds that are homeomorphic but not diffeomorphic, any homeomorphism gives a continuous, non-differentiable function. | |
| Mar 22, 2022 at 22:30 | comment | added | Maxime Ramzi | I think the question is not about the distinction between manifolds that have or don't have smooth structures, or between maps of manifolds that are or aren't smooth, but given a manifold that does have at least one smooth structure, what do the different ones mean. A related question is "what's the simplest example of two distinct smooth structures on a given topological manifold ?" | |
| Mar 22, 2022 at 22:26 | comment | added | Arshak Aivazian | Probably yes, because the smoothness of a function is a property, and "the smoothness of a manifold" is an additional structure. At least continuous non-smooth functions themselves are easy to visualize, for example $|x|$. | |
| Mar 22, 2022 at 22:23 | comment | added | Sam Hopkins | Is it that different from "visualizing" a function (of one real variable, say) which is continuous but not differentiable? | |
| Mar 22, 2022 at 22:20 | history | asked | Arshak Aivazian | CC BY-SA 4.0 |