Timeline for Necessary and sufficient criteria for a surface to cover a surface
Current License: CC BY-SA 2.5
10 events
| when toggle format | what | by | license | comment | |
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| Nov 29 at 9:48 | comment | added | LuckyJollyMoments | If anyone is interested in the theorem mentioned in the first sentence of the question body (the necessary and sufficient condition for the existence of a covering between closed orientable surfaces), you can refer to Example 1.41 and Exercise 23 in Section 2.2 of Hatcher’s Algebraic Topology. | |
| Mar 5, 2010 at 17:08 | vote | accept | Jonah Sinick | ||
| Mar 5, 2010 at 17:08 | answer | added | Jonah Sinick | timeline score: 2 | |
| Jan 23, 2010 at 18:32 | comment | added | Tom Church | For a reference you might check out the literature on dessin d'enfants; I feel like I've seen such criteria discussed in that more combinatorial context. | |
| Jan 23, 2010 at 12:05 | comment | added | Pete L. Clark | Anyway, I realize I am not answering your actual question: i.e., do I know a reference? Unfortunately no, sorry, although I agree that it must have been worked out dozens of times. | |
| Jan 23, 2010 at 12:03 | answer | added | Dmitri Panov | timeline score: 4 | |
| Jan 23, 2010 at 12:00 | comment | added | Pete L. Clark | I had a previous answer which Dmitri Panov pointed out was wrong. Take II (as a comment this time): we start with an arbitrary branched covering of compact Riemann surfaces $S' \rightarrow S$. Then we remove a finite set of points on $S$ together with the complete preimage in $S'$. In order to get an unramified covering we must remove at least the branch locus from $S$; we can also remove any finite number of unbranched points if we wish. I think that every topological unramified cover $S' \rightarrow S$ arises in this way. Now we examine cases using Riemann-Hurwitz... | |
| Jan 23, 2010 at 11:25 | history | edited | Jonah Sinick | CC BY-SA 2.5 |
Added requirement that surfaces be orientable
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| Jan 23, 2010 at 11:22 | comment | added | Pete L. Clark | Are you sure your criterion is correct in the closed case? Let S be the closed, nonorientable surface of Euler characteristic -2 and let S' be the closed orientable surface of Euler characteristic -6. Then $\chi(S')/\chi(S) = 3$, so your divisibility condition is satisfied, but since $3$ is odd, the covering map does not factor through the orientation cover (a degree 2 cover) and therefore passing to an odd degree cover cannot make a nonorientable surface orientable. | |
| Jan 23, 2010 at 10:49 | history | asked | Jonah Sinick | CC BY-SA 2.5 |