Timeline for On the algberaicity of the universal elliptic curve associated to a torsion free subgroup
Current License: CC BY-SA 3.0
20 events
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| May 30, 2012 at 15:02 | comment | added | Hugo Chapdelaine | Ok I was being stupid, ma section is not well defined. | |
| May 30, 2012 at 10:44 | comment | added | Hugo Chapdelaine | @Kevin, I agree that the terminology "universal elliptic curve" (away from $0$ and $1728$) confusing but I could not come up with a better terminology. | |
| May 30, 2012 at 10:22 | comment | added | Hugo Chapdelaine | @Donu, note that my non-trivial section $s:P^1(\mathbf{C})\rightarrow Y$ for the Hopf surface $Y=(C^2-(0,0))/<(2,2)>$ is given by $s([x_0,x_1])\mapsto [2x_0,2x_1]$. | |
| May 30, 2012 at 7:42 | comment | added | Kevin Buzzard | Probably $y^2=x^3-tx-t$ is not universal in some functorial sense (because it does not parametrise families with $j$-invariant not 0 or 1728 over arbitrary bases), it's universal in the much weaker sense that it happens to go through each $j$-invariant not 0 or 1728 exactly once. I think calling it "the universal elliptic curve" might cause confusion not least because of the implicit implication of uniqueness. | |
| May 30, 2012 at 2:53 | comment | added | Will Sawin | @Keerthi: For the first question, I'm not sure. For the second, take the Weirstrauss form $y^2=x^3-tx-t$ and compute $j$. The map $t\to j$ is rational linear, thus has an inverse. The bad points $t=0,t=\infty, \Delta=0$ correspond to $j=0$, $j=1728$, and $j=\infty$, thus you can consider it a family over $|mathbb C-\{0,1728\}$. | |
| May 30, 2012 at 2:09 | comment | added | Keerthi Madapusi | Also, what is the universal elliptic curve over $\mathbb{C}-\{0,1728\}$? | |
| May 30, 2012 at 2:08 | comment | added | Keerthi Madapusi | I'm confused. Do semi-continuity and base change not apply for proper maps of analytic spaces? This is all you need to conclude that $f_*\mathcal{O}(3\sigma)$ is trivial. | |
| May 30, 2012 at 1:34 | history | edited | Donu Arapura | CC BY-SA 3.0 |
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| May 30, 2012 at 1:20 | comment | added | Will Sawin | Shouldn't you need the vector bundle to be algebraically trivial? And that is certainly not true. For a curve minus a single point, the line bundle group is the Jacobian, nontrivial unless the curve is rational. | |
| May 30, 2012 at 1:10 | history | edited | Donu Arapura | CC BY-SA 3.0 |
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| May 29, 2012 at 23:29 | comment | added | Filippo Alberto Edoardo | @ Donu: Don't you need projective to apply GAGA? Then how can you be sure the bundle you are starting with is algebraic? | |
| May 29, 2012 at 21:35 | comment | added | Hugo Chapdelaine | Well I don't think you can always extend it in the analytic category when the base has a non-trivial fundamental group. | |
| May 29, 2012 at 21:26 | comment | added | Donu Arapura | It suffices to extend in the analytic category, and then invoke GAGA to conclude the extended bundle is algebraic. | |
| May 29, 2012 at 21:23 | comment | added | Hugo Chapdelaine | @Donu, I'm a bit puzzle by what you said about the existence of a section. Take $Y=(\mathbf{C}^2-(0,0))/<(2,2)>$, then it seems to me that the map $s:\mathbf{P}^1\rightarrow Y$ given by $s(x)=(2,2)$ gives you section of your elliptic fibration. In this case, all fibers are isomorphic to the elliptic curve $\mathbf{C}^{\times}/<2>$. | |
| May 29, 2012 at 21:20 | comment | added | Hugo Chapdelaine | Well holomorphic vector bundles over (connected) non-compact Riemann surfaces are trivial in the analytic category since the construction of sections which trivialize your bundle is transcendental but I don't see why it should trivialize in the algebraic category. | |
| May 29, 2012 at 20:05 | comment | added | Donu Arapura | Hugo, these are separate issues. A vector bundle on a punctured disk is trivial, so it extends. So applying this remark and the above construction, we see $E$ extends to a family of possibly singular cubics over $0$ and $1728$. | |
| May 29, 2012 at 20:02 | comment | added | Hugo Chapdelaine | the 3rd $\mathbf{C}$ should be replaced by $\mathbb{P}^1(\mathbf{C})$ of course. | |
| May 29, 2012 at 19:10 | comment | added | Hugo Chapdelaine | Donu, I think I'm missing something important here. Say that you take the universal elliptic curve $E\rightarrow \mathbf{C}-\{0,1728\}$, then you cannot extend your vector bundle over the smooth projective closure of $\mathbf{C}-\{0,1728\}$ which is $\mathbf{C}$, since there is no universal elliptic curve over $\mathbf{C}$. So may be you could develop a little bit what you mean when you say extending the vector bundle to the smooth projective locus. | |
| May 29, 2012 at 18:12 | history | edited | Donu Arapura | CC BY-SA 3.0 |
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| May 29, 2012 at 17:43 | history | answered | Donu Arapura | CC BY-SA 3.0 |