Timeline for Geometric decomposition of J(11)
Current License: CC BY-SA 4.0
14 events
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| May 4, 2023 at 20:25 | history | edited | LSpice | CC BY-SA 4.0 |
Tidying, while this is on the front page
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| Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
replaced http://mathoverflow.net/ with https://mathoverflow.net/
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| Oct 2, 2010 at 19:40 | comment | added | François Brunault | I think the reason is that two distinct isogeny classes of elliptic curves over $\mathbf{Q}$ can be isomorphic over $\mathbf{C}$. Using Pari/GP I found that $11A \cong 121D$ and $121A \cong 121C$ over $\mathbf{C}$. Since $J_0^{\mathrm{new}}(121) \sim 121A \times 121B \times 121C \times 121D$ over $\mathbf{Q}$, your guess is exact! | |
| Oct 2, 2010 at 15:17 | comment | added | François Brunault | Thanks for the details on the fibered product. I agree that $J_1(N)$ is a factor of $J(N)$, but I am a little bit confused about $J_0(N^2)$ because for $N=11$ not all elliptic curves of conductor $121$ appear in $J(11)$. Maybe this is because we are looking things over $\mathbf{C}$ and not over $\mathbf{Q}$ ? | |
| Oct 1, 2010 at 0:54 | history | edited | Soroosh | CC BY-SA 2.5 |
A one line explanation on why the fiber product is biration to X(N)
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| Oct 1, 2010 at 0:49 | vote | accept | Soroosh | ||
| Sep 30, 2010 at 1:32 | comment | added | François Brunault | Could you give some details on how you get $X_{\mu}(N)$ as a fibered product ? | |
| Sep 29, 2010 at 22:09 | answer | added | François Brunault | timeline score: 9 | |
| Sep 29, 2010 at 21:10 | comment | added | Soroosh | Also, I'm mostly interested in this geometrically. So, I want to know the simple isogeny factors over C. | |
| Sep 29, 2010 at 19:39 | comment | added | Soroosh | You are definitely right. I meant X(N) to be the twisted full level N structure, and I changed the wording to reflect that. On the other hand, I think the fiber product is actually smooth at the cusps, since the covering of X_0(N) by X_1(N) is unramified at the cusps. I think for primes N larger than 5, that is an actual isomorphism. I also agree with you that representation theory is probably better than geometry for this type of problems. However, I was trying to figure out what's happening geometrically, and there seems to be something there. | |
| Sep 29, 2010 at 19:29 | history | edited | Soroosh | CC BY-SA 2.5 |
Fixed the minor mistakes pointed out by BConrad
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| Sep 28, 2010 at 14:07 | answer | added | William Stein | timeline score: 5 | |
| Sep 28, 2010 at 1:50 | comment | added | BCnrd | The curve $X(N)$ in its standard definition as a moduli space is not geometrically connected over $\mathbf{Q}$). You must mean to use the variant $X_{\mu}(N)$ classifying "twisted" full level-$N$ structures of type $\mathbf{Z}/(N) \times \mu_N$ as symplectic spaces? And are you interested in the simple isogeny factors over $\mathbf{Q}$, or working geometrically (which might come to the same thing, but only after the fact)? Also, the "fiber product" description looks non-smooth near the cusps. It is at best birational, no? Anyway, representation theory should be better than geometry here. | |
| Sep 28, 2010 at 1:10 | history | asked | Soroosh | CC BY-SA 2.5 |