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Apr 13, 2017 at 12:58 history edited CommunityBot
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May 29, 2012 at 21:33 comment added Ariyan Javanpeykar @Abdelmalek. I see what you were getting at right now. It's a shame there isn't a "de Franchis" where the target is an elliptic curve. Anyway, as a final comment (which won't help you again unfortunately) I just came across this "effective" version of de Franchis' theorem by accident: projecteuclid.org/DPubS/Repository/1.0/… You can probably find much more stuff on "effective" versions of de Franchis theorem if you're interested a bit in the situation for genus >1 curves. Again, sorry to not be able to say anything else useful.
May 29, 2012 at 20:03 comment added Abdelmalek Abdesselam @Ariyan: thanks for your comments. For the first point about the "family", I realize the use of that word may have misled you. I simply meant a reasonably explicit denumerable collection of dessins (D_n) such that for every elliptic curve E over Q there exists an n for which the Belyi map encoded by D_n goes from E to P1. As for the second point, thanks for pointing out De Franchis thm. However, in the case I am interested in, Y is of genus 1 (the given elliptic curve) while X is the hypothetical modular curve to be constructed.
May 29, 2012 at 19:52 comment added Ariyan Javanpeykar ....probably know this is called de Franchis' theorem: en.wikipedia.org/wiki/De_Franchis_theorem I think this is again a very (unnecessarily) long comment. My apologies.
May 29, 2012 at 19:52 comment added Ariyan Javanpeykar I see that I slightly misunderstood your last question. You are just asking for a cover $X\to Y$. I was assuming you wanted the cover $X\to \mathbf{P}^1$ to factor through your cover $\mathbf{Y}\to\mathbf{P}^1$. I wrote the above answer with this in mind. (By the way, Will Sawin wrote his answer before me and he also seems to have understood you want your map to factor through $Y\to \mathbf{P}^1$.) Finally, I think I understand what you mean by $H_{X,Y}$. Is $H_{X,Y} $ the number of finite morphisms $X\to Y$? If yes, then $H_{X,Y}$ is finite if the genus of $Y$ is at least $2$. As you....
May 29, 2012 at 19:44 comment added Ariyan Javanpeykar ...family condition?
May 29, 2012 at 19:44 comment added Ariyan Javanpeykar @Abdelmalek. Sorry for the late reply! I didn't see your comment. Let me also apologize for "wanting" a single dessin to capture all elliptic curves over Q. This was a "typo". I really meant to ask: what do you mean by a family of dessins that capture all elliptic curves over Q? I still don't understand what this means because I'm too used to thinking about Belyi maps. What would "a family of dessins that capture all elliptic curves over Q" mean in the Belyi maps terminology? Is it a Belyi map $f_E:E\to \mathbf{P}^1_\mathbf{Q}$ for every elliptic curve $E$ over $\mathbf{Q}$ satisfying some...
May 14, 2012 at 16:16 comment added Abdelmalek Abdesselam @Ariyan: this reformulation is quite different from my original question. I don't understand why you want 1) a single dessin to capture all elliptic curves over Q, 2) the map X->Y to form a commutative triangle with the maps X->P1 and Y->P1?
May 12, 2012 at 11:08 history edited Ariyan Javanpeykar CC BY-SA 3.0
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May 12, 2012 at 11:02 history answered Ariyan Javanpeykar CC BY-SA 3.0