Analogue of Shafarevich-Ogg's theorem over complex numbers - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T11:31:43Z http://mathoverflow.net/feeds/question/74853 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/74853/analogue-of-shafarevich-oggs-theorem-over-complex-numbers Analogue of Shafarevich-Ogg's theorem over complex numbers shenghao 2011-09-08T10:21:18Z 2011-09-09T08:01:14Z <p>Let <code>$f:E\to D^*$</code> be a family of complex elliptic curves parametrized by the punctured open disk <code>$D^*.$</code> Assume that the monodromy on <code>$H^1$</code> is trivial (i.e. $R^1f_*\mathbb Z$ is a constant sheaf on <code>$D^*$</code>). Does this imply that $f$ extends to a family of elliptic curves over the full disk $D?$ </p> <p>Here's an attempt, which I'm not sure if it works: maybe there is an equivalence (Riemann-Deligne?) between families of elliptic curves over a (smooth) base over $\mathbb C$ and variations of $\mathbb Z$-Hodge structures satisfying Griffiths transversality, of rank 2 and weight 1 on the same base. The constant local system certainly extends to $D.$</p> http://mathoverflow.net/questions/74853/analogue-of-shafarevich-oggs-theorem-over-complex-numbers/74875#74875 Answer by ulrich for Analogue of Shafarevich-Ogg's theorem over complex numbers ulrich 2011-09-08T13:22:32Z 2011-09-08T13:22:32Z <p>The answer is, as you expected, yes:</p> <p>Choosing a symplectic isomorphism of $R^1f_* \mathbb{Z}$ with the constant sheaf $\mathbb{Z}^2$ (i.e. full level structure) gives an analytic map $c:D^* \to \mathfrak{h}$ where $\mathfrak{h}$ is the upper half plane (viewed as the moduli space of elliptic curves with full level structure). Since $\mathfrak{h}$ is analytically isomorphic to a bounded domain it follows from the removable singularity theorem that $c$ extends to an analytic map $D \to \mathfrak{h}$. Pulling back the universal family over $\mathfrak{h}$ -- the fibre over $\tau$ is the elliptic curve $\mathbb{C}/\mathbb{Z} + \mathbb{Z}\tau$ with the level structure given by $1 \mapsto (1,0)$, $\tau \mapsto (0,1)$ -- gives the required extension.</p> <p>The main point in the above is the analyticity of $c$. This is a consequence of the fact that <code>$R^1f_*\mathcal{O}_E$</code> is a holomorphic sub-bundle of $R^1f_*\mathbb{Z} \otimes \mathcal{O}_{D^*}$.</p> http://mathoverflow.net/questions/74853/analogue-of-shafarevich-oggs-theorem-over-complex-numbers/74880#74880 Answer by Francesco Polizzi for Analogue of Shafarevich-Ogg's theorem over complex numbers Francesco Polizzi 2011-09-08T14:02:49Z 2011-09-08T14:02:49Z <p>Let me give another proof using the classical Picard-Lefschetz theory. </p> <p>We first take the flat limit of your family $f \colon E \to D^*$, obtaining a family $g \colon S \to D$. Now we have to understand what is the central fiber $g^{-1}(0)$.</p> <p>I will refer to the paper by Durfee <a href="http://www.kryakin.com/files/Invent_mat_%282_8%29/28/28_03.pdf" rel="nofollow">"The monodromy of a degenerate family of curves"</a>, Inventiones Mathematicae 28 (1975). By Theorem $2$, the monodromy of $g$ is of finite order if and only if there exists a finite base change $D \to D$, $z \to z^n$ such that the pull-back family $g' \colon S' \to D$ has a most nodes as singularities and contains no vanishing cycles in its central fibre. This means that $g' \colon S' \to D$ is a <em>smooth</em> family.</p> <p>If $n \geq 2$ then the central fibre of $g \colon S \to D$ would be of type ${}_nI_0$ in Kodaira's classification, that is a smooth elliptic curve with multiplicity $n$. But then the monodromy of $g \colon S \to D$ would be of order $n$, contradiction.</p> <p>Then $n=1$, that is $g=g'$ and this shows that $g$ is a smooth family. </p>