Timeline for Manifolds with polynomial transition maps
Current License: CC BY-SA 3.0
19 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jul 19, 2017 at 15:03 | comment | added | Johannes Huisman | @RobertBryant : Yeah, you'e right; a real algebraic variety is locally isomorphic to a closed algebraic subset of $R^n$, and not necessarily locally isomorphic to an open subset of $R^n$. Sorry for the confusion. | |
| Jul 19, 2017 at 14:30 | comment | added | Robert Bryant | @JohannesHuisman: Could you give a reference for this claim? The Nash-Tognoli Theorem implies that every compact $C^\infty$ manifold is diffeomorphic to a real algebraic variety and hence admits an atlas with transition maps that are algebraic, but I don't see how one gets rational from this result. | |
| Jul 19, 2017 at 13:35 | comment | added | Johannes Huisman | @KevinCasto : Any compact $C^\infty$ manifold admits an atlas whose transition maps are rational by the Nash-Tognoli Theorem. | |
| Jul 19, 2017 at 12:12 | comment | added | Tom Goodwillie | @Kevin Casto: Maybe you ask that as a question. | |
| Jul 19, 2017 at 7:06 | comment | added | Ali Taghavi | @RobertBryant Now I understand the details of your interesting answer. Thank you for your patient on my slow understanding. | |
| Jul 19, 2017 at 7:01 | vote | accept | Ali Taghavi | ||
| Jul 18, 2017 at 13:47 | comment | added | Ali Taghavi | @RobertBryant This was the procedure which I was thinking about but now I realize may be it does not work: we fix a base point p on M. For every q we choose a curve $\gamma$ joining p to q. We cover the curve with open sets such that two consecutive open sets have non empty intersection. then there is an obvious rescaling. to prove that this is independent of choosing the curve we use simple connectivity. May be this is my fault? Now I understand Cech cohomology or the connection you constructed are two possible approach. | |
| Jul 18, 2017 at 13:34 | comment | added | Robert Bryant | @AliTaghavi: The question is how do you do the rescaling? You need to produce a decision procedure to determine how to scale each member of the given atlas so as to get a new atlas with unimodular transitions. I am saying that you need to specify exactly how you are using the hypothesis of simple-connectivity to construct this decision procedure before I will believe that you know how to do it. The way I chose (although it could be avoided by an appeal to Cech cohomology) was to construct a flat connection and then appeal to simple-connectivity to provide the global flat section $\mu$. | |
| Jul 18, 2017 at 13:28 | history | edited | Robert Bryant | CC BY-SA 3.0 |
Fixed typos and expanded the explanation of analytic continuation using the polynomial transition maps.
|
| Jul 18, 2017 at 13:16 | comment | added | Ali Taghavi | @RobertBryant In my previous comment I am still assuming simple connectivity for such rescaling. Is this re scalling legal(with simple connectivity)? So we have a global volum form then we extend $\psi$ analytically then we have $\psi^* (dx)=\mu$. So I do not understant the necessity of "connecdtion". I would appreciate if you more explain. | |
| Jul 18, 2017 at 13:11 | comment | added | Robert Bryant | @AliTaghavi: How could you determine how to scale the members of your atlas so as to arrange that the transitions are all unimodular? You need to use simple-connectivity somewhere, otherwise, it may not be possible. For example, consider the quotient $M^2$ of $\mathbb{R}^2\setminus\{(0,0)\}$ by the scaling $(x,y)\mapsto(2x,2y)$, which is a torus. Let the atlas $\mathcal{A}$ consist of the local sections of the quotient mapping $\pi:\mathbb{R}^2\setminus\{(0,0)\}\to M^2$ with small circular domains. You cannot scale the members of $\mathcal{A}$ to obtain a unimodular atlas $\mathcal{A}'$. | |
| Jul 18, 2017 at 12:44 | comment | added | Ali Taghavi | @RobertBryant What would be the mistake of this simplification of your argument: With re scaling, we have an atlas whose transition maps satisfy $Det (f')=1$ (rather than arbitrary constant) So we have easily an analytic volume form $\mu$ globally on $M$. Now we extend $\psi$ to $M$ uniquely. Again we have $\psi^* (dx)=\mu$. | |
| Jul 17, 2017 at 17:42 | comment | added | Robert Bryant | @AliTaghavi: The (flat) connection provides a global volume form with respect to which all the coordinate charts are unimodular. Then, yes, the fact that $\psi$ is analytic on $U$ and has a unique analytic continuation to all of $M$ coupled with the fact that $\mu$ is analytic on the whole of $M$ implies that the analytically continued $\psi$ satisfies $\psi^*(\mathrm{d}x)=\mu$ globally, since these analytic functions are equal on a nonempty open set. | |
| Jul 17, 2017 at 17:35 | comment | added | Ali Taghavi | @RobertBryant Thank you very much for your interesting answer. What is the role of connection in your answer? Am I mistaken to think that $\psi^* (dx)=\mu$ is globally true because both sides are real analytic and they are equal only in a given arbitrary chart?(By analytic continuation). | |
| Jul 17, 2017 at 15:36 | comment | added | Kevin Casto | What about if the transition functions are rational functions? This includes the sphere, for example. | |
| Jul 17, 2017 at 15:33 | comment | added | Robert Bryant | @TomGoodwillie: I could have got away without it, but it was convenient to be able to write that $\psi^*(\mathrm{d}x)=\mu$ globally, which made the contradiction obvious, since, otherwise, it was not immediate (to me) that the analytic continuation of $\psi$ would always be a local diffeomorphism. | |
| Jul 17, 2017 at 15:28 | comment | added | Tom Goodwillie | Where are you using the volume form? | |
| Jul 17, 2017 at 13:21 | history | edited | Robert Bryant | CC BY-SA 3.0 |
added 1943 characters in body
|
| Jul 17, 2017 at 12:32 | history | answered | Robert Bryant | CC BY-SA 3.0 |