Involution on Hyperelliptic curves and their Jacobians - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-20T08:47:24Z http://mathoverflow.net/feeds/question/64839 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/64839/involution-on-hyperelliptic-curves-and-their-jacobians Involution on Hyperelliptic curves and their Jacobians Rex 2011-05-12T21:25:29Z 2011-05-12T22:20:13Z <p>Let $X$ be a hyperelliptic curve and let $i:X\to X$ denote the hyperelliptic involution. Once we fix a point $x_0\in X$ we get the Abel-Jacobi map $AJ:X\to J$ where $J$ denotes the Jacobian variety. Now the Jacobian is also equipped with an involution, namely $x\mapsto x^{-1}$. Is it possible to choose the base point $x_0$ in such a way that the restriction of the involution on the Jacobian is the involution on $X$. </p> http://mathoverflow.net/questions/64839/involution-on-hyperelliptic-curves-and-their-jacobians/64845#64845 Answer by Dan Petersen for Involution on Hyperelliptic curves and their Jacobians Dan Petersen 2011-05-12T21:47:42Z 2011-05-12T21:47:42Z <p>Yes, pick $x_0$ to be a Weierstrass point, i.e. a fixed point of the hyperelliptic involution. </p> <p>Let $\sigma$ denote the hyperelliptic involution on $X$. Under the Abel-Jacobi map we have $x \mapsto [x-x_0]$ and $\sigma(x) \mapsto [\sigma(x) - x_0]$. Now $[x + \sigma(x) - 2x_0]$ is the divisor of a function, since it is the pullback under the hyperelliptic map of a degree zero divisor on $\mathbf P^1$. Hence $AJ(x)$ and $AJ(\sigma(x))$ are inverses.</p> http://mathoverflow.net/questions/64839/involution-on-hyperelliptic-curves-and-their-jacobians/64848#64848 Answer by Simon Rose for Involution on Hyperelliptic curves and their Jacobians Simon Rose 2011-05-12T22:20:13Z 2011-05-12T22:20:13Z <p>Another way to think of this is the following (at least over $\mathbb{C}$).</p> <p>Consider the diagram $$\begin{array}{ccccc} X &amp; \xrightarrow{AJ \times AJ \circ \sigma} &amp; J \times J &amp; &amp; \\ \downarrow &amp; &amp; \downarrow &amp; &amp; \\ \mathbb{P}^1 &amp; \to &amp; Sym^2J &amp; \xrightarrow{+} &amp; J \end{array}.$$ Note that the composition along the bottom row must be constant, as there are no non-trivial maps from $\mathbb{P}^1$ to any Abelian variety.</p> <p>What this tells you is that $f(x) = AJ(x) + AJ\big(\sigma(x)\big)$ is constant. So if you translate it (i.e. choose a different basepoint): $$\tilde{AJ}(x) = AJ(x) - \frac{f(x)}{2}$$ then you find that the corresponding $\tilde{f}(x) = 0$ for all $x$. That is, the two involutions agree with each other as desired.</p>