How to use automorphisms to produce isotrivial non trivial families - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T18:12:18Z http://mathoverflow.net/feeds/question/15884 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/15884/how-to-use-automorphisms-to-produce-isotrivial-non-trivial-families How to use automorphisms to produce isotrivial non trivial families Andrea Ferretti 2010-02-20T10:59:51Z 2010-02-20T20:41:22Z <p>In many places the existence of automorphism is acknowledged as one of the reasons why fine moduli spaces cannot exist. A typical example is the following.</p> <p>Consider a curve $C$ with a nontrivial automorphism, for instance a hyperelliptic curve with its involution $\phi$. Now let $B$ be any scheme with a free action of (in this case) $\mathbb{Z}/(2)$.</p> <p>Let $D$ be the quotient of $B \times C$ by the diagonal action of $\mathbb{Z}/(2)$ and let $B'$ be the quotient of $B$ by the action of $\mathbb{Z}/(2)$.</p> <p>Consider the map $f \colon D \rightarrow B'$. Then the fibers of $f$ are all isomorphic to $C$, but for a suitable choice of $B$ the family $f$ is not isomorphic to the product $C \times B'$.</p> <p>How can I get an explicit example of the last assertion?</p> <blockquote> <p>How can I produce $B$ as above such that $f$ is not isomorphic to the trivial family?</p> </blockquote> http://mathoverflow.net/questions/15884/how-to-use-automorphisms-to-produce-isotrivial-non-trivial-families/15885#15885 Answer by Dmitri for How to use automorphisms to produce isotrivial non trivial families Dmitri 2010-02-20T11:13:23Z 2010-02-20T11:13:23Z <p>Here is an explicit example. Denote by $E$ the elliptic curve $E=\mathbb C/(\mathbb Z+i\mathbb Z)$. Now let $B=C=E$. The involution that we will consider is $$(z,w)\in (E\times E)\to (z+\frac{1}{2},-w).$$</p> <p>Notice now that this quotient is not a direct product anymore. You can see this by calculating $H_1(E\times E)/\mathbb Z_2$ and verifying that it is not $\mathbb Z^4$. Alternatively, you can notice that the quotient does not have a non-vanishing holomorphic volume form. Indeed if such form existed it could be lifted to $E\times E$ to a form $cdz\wedge dw$ ($c\ne 0$), which would give a contradiction since $dz\wedge dw$ is anti-invariant. </p> http://mathoverflow.net/questions/15884/how-to-use-automorphisms-to-produce-isotrivial-non-trivial-families/15928#15928 Answer by Zsolt Patakfalvi for How to use automorphisms to produce isotrivial non trivial families Zsolt Patakfalvi 2010-02-20T20:41:22Z 2010-02-20T20:41:22Z <p>If $C=B$ with the same $Z/2$ action, and $g(C/Z_2) \geq 2$, then you always get a non-trivial family:</p> <p>Suppose it was trivial. Then it had another projection $q : D \to C$. Since $C$ is an \'etale cover of $B'$, and both have genus at least two, $g(C) > g(B')$ by the Hurwitz-formula. That is, also by the Hurwitz formula, all the maps from $B'$ to $C$ are the constant maps. So, all sections $B' \to D$ of $f$ would have to be contained in a fiber of $q$, i. e. they have to be one fiber of $q$. However there are two natural sections $E$ and $F$ of $p$: $[c] \to [(c,c)]$ and $[c] \to [(c,-c)]$. So both of these have to be contracted to a point by $q$. Given any $b \in B$, a choice $c \in C$ such that $[c]=b$ gives an isomorphism $a_c : p^{-1}(d) \to C$, which sends $[(c,c')]$ to $c'$. Now one can construct automorphisms $\varphi_{c'}$ of $C$ for any $c' \in C$ by the following composition:</p> <p>$C \to^{a_c^{-1}} p^{-1}([c]) \to^q C \to^s p^{-1}([c']) \to^{a_{c'}} C$</p> <p>where</p> <p>$S=(q|_{p^{-1}([c'])})^{-1}$</p> <p>One specialty about $\varphi_{c'}$ is that it takes $c$ and $-c$ to $c'$ and $-c'$, respectively. This follows from the fact, that $E$ and $F$ are contracted to a point by $q$. So, $C$ has infinitely many different automorphisms, which is a contradiction by the assumption that $g(C) \geq 2$. That is our assumption is false, therefore $Y$ is not a product family. </p>