Timeline for Basic question about branch points on Riemann surfaces
Current License: CC BY-SA 2.5
11 events
| when toggle format | what | by | license | comment | |
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| Sep 4, 2020 at 11:21 | comment | added | BCLC | @FrancescoPolizzi Re why branch points form discrete set, do you disagree with Federico Fallucca here? (It's my question, and I also refer to Rick Miranda - Algebraic Curves and Riemann Surfaces) | |
| Sep 4, 2020 at 11:21 | comment | added | BCLC | Georges Elencwajg and @rfauffar Re why branch points form discrete set, do you disagree with Federico Fallucca here? (It's my question, and I also refer to Rick Miranda - Algebraic Curves and Riemann Surfaces) | |
| Apr 5, 2011 at 15:52 | history | edited | Georges Elencwajg | CC BY-SA 2.5 |
Added Bibliography and comments
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| Apr 5, 2011 at 14:51 | comment | added | Francesco Polizzi | Dear Georges, I see. I'm still digesting your answer, but it starts to look ok to me. Thank you | |
| Apr 5, 2011 at 14:34 | comment | added | Georges Elencwajg | Dear Francesco, here is the subtle point which (I hope!) will take care of your objection. What you say is true but some points near $q_n$ are branch points of $f$ not because of the behaviour of $f$ near $1/n$, which as you correctly noted is étale at points near to but different from $1/n$, but because of some nasty far away $1/N$ such that $q_N$ happens to be very close to $q_n$. Once again this is hidden by those local coordinates which give us the feeling that $0$ is some determined point whereas all sorts of points get called $0$ the minute you focus your attention on them! | |
| Apr 5, 2011 at 14:10 | comment | added | rfauffar | I find it weird that Miranda "assumes" that branch points are discrete because ramification points are discrete. I guess it's just an error in the text maybe? (page 46 at the top) | |
| Apr 5, 2011 at 13:49 | comment | added | Francesco Polizzi | Dear Georges, I'm still not completely convinced. If the map $f$ can be locally described as $w=\zeta^2$ around $w=0$, then there is a neighborhood of $w=0$, i.e. a neighborhood of $z=q_n$, where $f$ is not branched (since outside $w=0$ we have two distinct preimages). But this seems to contradict your assertion that $f$ has dense branch locus... | |
| Apr 5, 2011 at 13:35 | comment | added | Georges Elencwajg | Dear Francesco, I think it is "the suitable coordinates" which make it a little difficult to follow. Put $w=z-q_n$ at the target and $\zeta=z-1/n$ at the source . Then locally my map is described by $\zeta\mapsto w=\zeta^2$, as it should. The confusion arises because we constantly replace the values of $f$ near $f(b)$ by values near $0$ by surreptitiously changing $w$ to $w-f(b)$ and calling that "suitable coordinates"! | |
| Apr 5, 2011 at 13:13 | comment | added | Francesco Polizzi | Dear Georges, there is something I do not understand. Locally around a branch point, in suitable coordinates a holomorphic map between Riemann surfaces can be written as $z \to z^n$, where $n$ is the ramification number (see for instance Farkas-Kra, p. 12) so it seems to me that the branch locus should be actually discrete. Am I missing something? | |
| Apr 5, 2011 at 13:07 | vote | accept | rfauffar | ||
| Apr 5, 2011 at 13:03 | history | answered | Georges Elencwajg | CC BY-SA 2.5 |