Timeline for Flat metric on compact surface minus a point
Current License: CC BY-SA 4.0
16 events
| when toggle format | what | by | license | comment | |
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| Dec 24, 2019 at 12:41 | history | edited | Robert Bryant | CC BY-SA 4.0 |
Corrected some typos, rearranged the material in a more logical order, and added a section on smoothing.
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| Dec 19, 2019 at 12:37 | comment | added | Robert Bryant | @GiannidelFiore: You are welcome, and best wishes to you for the holidays as well. If you want to know the details of the construction, let me know, and I'll try to find some time to put them into the answer during the holiday break. | |
| Dec 17, 2019 at 17:47 | comment | added | Gianni del Fiore | Fantastic, thanks again for sharing your deep knowledge with this community. Season's Greetings | |
| Dec 17, 2019 at 13:10 | comment | added | Robert Bryant | @GiannidelFiore: Yes, in the sense that there will be a real-analytic atlas on the 2-sphere such that the coefficients of $h$ in any chart of the atlas will be real-analytic. In fact, when the $L_i$ are rational, you will be able to make them algebraic. | |
| Dec 17, 2019 at 12:11 | comment | added | Gianni del Fiore | Interesting! Can we build such an $h$ analytic? | |
| Dec 17, 2019 at 6:14 | comment | added | Robert Bryant | @GiannidelFiore: Well, yes. You can have a smooth $h$ on the $2$-sphere that degenerates at three points, with each degenerate point having $L=1/3$ and the induced metric on the complement being flat. More generally, I think you can do it with any three $L_i>0$ that add up to $1$. | |
| Dec 16, 2019 at 21:44 | comment | added | Gianni del Fiore | Of course, now I realize that the actual behavior of $h$ at $p$ is irrelevant to the construction of the complex coordinate chart $\zeta$. We can then use the same idea to conclude that $\chi(M)\leq0$ even when $h$ degenerates at two points. Do you know any counterexample in the case of three singular points? | |
| Dec 16, 2019 at 20:53 | comment | added | Robert Bryant | @GiannidelFiore: That doesn't matter. As long as $h$ is positive definite and smooth except possibly at $p$, the normal form described above will hold on a punctured neighborhood of $p$, and that is all that is needed for the argument. (Note that I did not claim that $\zeta$ must be smooth at $p$, and, in fact, I do not know that it would be.) | |
| Dec 16, 2019 at 19:47 | comment | added | Gianni del Fiore | Thank you very much for the details, that was a clever use of Gauss-Bonnet. What if $h$ simply degenerates but does not vanish at $p$? | |
| Dec 16, 2019 at 12:31 | history | edited | Robert Bryant | CC BY-SA 4.0 |
Slightly simplified the detail checking for the argument against positive Euler characteristic and fixed a couple of typos.
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| Dec 16, 2019 at 9:17 | comment | added | Robert Bryant | @GiannidelFiore: Yes. See above | |
| Dec 16, 2019 at 9:17 | history | edited | Robert Bryant | CC BY-SA 4.0 |
Added the requested information about the positive Euler characteristic case.
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| Dec 16, 2019 at 0:39 | vote | accept | Gianni del Fiore | ||
| Dec 16, 2019 at 0:38 | comment | added | Gianni del Fiore | @MartindeBorbon and Robert Bryant, interesting construction, many thanks for your explanations. Robert Bryant, could you give more details on how the Gauss Bonnet Theorem can be used to rule out the case $\chi(M)>0$? | |
| Dec 15, 2019 at 5:34 | history | edited | Robert Bryant | CC BY-SA 4.0 |
Added examples covering the unorientable case.
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| Dec 14, 2019 at 21:23 | history | answered | Robert Bryant | CC BY-SA 4.0 |