Timeline for Choosing tau for elliptic curves over the rational numbers with prescribed ramification data
Current License: CC BY-SA 2.5
6 events
| when toggle format | what | by | license | comment | |
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| Jan 4, 2011 at 11:07 | comment | added | Ariyan Javanpeykar | For the sake of brevity, I left out the exact set-up of my problem. I now see that this wasn't a good idea. I edited the question. Hopefully it'll be a bit clearer now. | |
| Jan 4, 2011 at 5:55 | comment | added | S. Carnahan♦ | Sorry, I shouldn't have called it the injectivity radius, but I meant the maximum radius in the $q$-disk where $j$ is injective, which is $e^{-2\pi}$. Again, when you say that you know the morphism $f$, it sounds like you have some datum that gives you the isomorphism type of the source $E$. How is the map $f$ described to you otherwise? A set of branch points in the line is not sufficient to characterize $E$. | |
| Jan 3, 2011 at 16:43 | comment | added | Ariyan Javanpeykar | What is the injectivity radius? What does it depend on? I really don't know anything about X as a compact Riemann surface besides its genus and the given morphism f. So for example I "dont know" the j-invariant. So I ask, can one "explicitly" write down the j-invariant in terms of the data given (branch points and degree of f)? | |
| Jan 3, 2011 at 15:24 | comment | added | S. Carnahan♦ | For any orbit, there are infinitely many points within any ball of positive radius around zero. If instead you want $e^{2 \pi i \tau}$ to be small, then you should get a well-defined answer in most cases (namely, inside the injectivity radius). Given the $j$-invariant of $E$, you should be able to approximate $q$ in various absolute values by power series methods (but I haven't tried this). | |
| Jan 3, 2011 at 14:58 | comment | added | Ariyan Javanpeykar | I want to know how this orbit you mention looks like and choose a "small" representative in a way. This could mean something like finding a tau such that its absolute value is bounded by d times r, say. I really hope I'm making sense here. | |
| Jan 3, 2011 at 14:49 | history | answered | S. Carnahan♦ | CC BY-SA 2.5 |