Timeline for Manifolds with polynomial transition maps
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jul 19, 2017 at 12:11 | comment | added | Tom Goodwillie | You're welcome! Thanks for your interesting question. | |
| Jul 19, 2017 at 7:10 | comment | added | Ali Taghavi | Thank you very much for your very interesting answer and your supplementary comments on my commented question.I am really sorry that I can not accept both answers simultaneously. | |
| Jul 18, 2017 at 2:30 | comment | added | Tom Goodwillie | For each connected chart in $\mathcal A$ the canonical map $U_\phi\to\tilde M$ maps $U_\phi$ homeomorphically to an open subset. The space $\tilde M$ is Hausdorff. There is a map $\pi:\tilde M\to M$ given by sending each $U_\phi$ to $U$ by "the identity". To see that $\pi$ is a covering projection, argue as follows. Suppose that $(U,\phi)$ is a chart in $\mathcal A$. Then, for every polynomial map $h:\mathbb R^n\to \mathbb R^n$ with polynomial inverse, $(U,h\circ\phi)$ is again a chart in $\mathcal A$. I claim that $\pi^{-1}(U)$ is the disjoint union of the open sets $U_{h\circ\phi}$. | |
| Jul 18, 2017 at 2:05 | comment | added | Tom Goodwillie | For each chart $(\phi,U)$ in the atlas, take a copy of the space $U$; call it $U_\phi$. Form the disjoint union of $U_\phi$ over all charts. Now make a quotient space as follows. Given charts $(U,\phi)$ and $(V,\psi)$, we identify a point $p\in U_\phi$ with a point $q\in V_\psi$ if and only if $p=q\in M$ and $\phi$ and $\psi$ coincide in some neighborhood of $p=q$. This quotient space is $\tilde M$. | |
| Jul 17, 2017 at 19:46 | comment | added | Ali Taghavi | Thank you very much for your answer. My apology if my question is elementary: May I ask you to elaborate the construction of $\tilde{M}$? | |
| Jul 17, 2017 at 19:09 | comment | added | Tom Goodwillie | To explain why $\tilde M$ is a covering space of $M$: If a diffeomorphism between open sets of $\mathbb R^n$ is given by a polynomial formula and the same holds for the inverse diffeomorphism, then the same formulas give diffeomorphisms $\mathbb R^n\to \mathbb R^n$. This fails if we say "rational" or analytic" rather than "polynomial". | |
| Jul 17, 2017 at 15:36 | history | answered | Tom Goodwillie | CC BY-SA 3.0 |