Are Jacobians principally polarized over non-algebraically closed fields? - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T22:26:51Z http://mathoverflow.net/feeds/question/7505 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/7505/are-jacobians-principally-polarized-over-non-algebraically-closed-fields Are Jacobians principally polarized over non-algebraically closed fields? David Zureick-Brown 2009-12-01T21:00:37Z 2010-02-13T09:14:42Z <p>How does one define the Torelli map $M_g \to A_g$ of moduli <strong>stacks</strong>? On geometric points a curve maps to its principally polarized Jacobian. </p> <p>So what I am asking is: if I have a curve $C$ over a non-algebraically closed field $k$ such that $C(k)$ is empty, is the Jacobian of C still principally polarized? After base change to $\bar{k}$ one has a theta divisor; does it descend? Also, is the relative Jacobian of a family of curves principally polarized?</p> <p>The thing I am confused about is that the theta divisor naturally lives on $Pic^{g-1}$ as the image of the map from the symmetric power $C^{g-1}$; this is a torsor under $Pic_0$, but not itself an abelian variety.</p> <p>Also, the classical Torelli theorem says that this map is an injection on field valued points. Is this actually a locally closed immersion of stacks?</p> http://mathoverflow.net/questions/7505/are-jacobians-principally-polarized-over-non-algebraically-closed-fields/7513#7513 Answer by David Zureick-Brown for Are Jacobians principally polarized over non-algebraically closed fields? David Zureick-Brown 2009-12-01T22:05:41Z 2009-12-01T22:11:19Z <p>Incidentally, as I posted this question someone who knew the answer wandered into my office.</p> <p>The map $M_g \to A_g$ factors through the moduli space $\tau_g$ of pairs (A,P,L) where A is an abelian variety, P is an A torsor, and L is an ample line bundle on P which is geometrically a principal polarization. The map $M_g \to \tau_g$ is given by $C \mapsto (Pic_0, Pic_{g-1}, L(\theta))$, where the theta divisor on $Pic_{g-1}$ is given by the image of $C^{g-1}$. </p> <p>To construct the map $\tau_g \to A_g$, note that $Pic_0(A) \cong Pic_0(P)$, so that L indeed gives a map $A \to A^{\vee}$ given by $a \mapsto t^*_aL \otimes L^{-1}$. </p> <p>The point is one doesn't need to descend the theta divisor. The reference to this is 5.1 of Martin Olsson's book <em>Compactifying moduli spaces of abelian varieties</em>.</p> http://mathoverflow.net/questions/7505/are-jacobians-principally-polarized-over-non-algebraically-closed-fields/15177#15177 Answer by BCnrd for Are Jacobians principally polarized over non-algebraically closed fields? BCnrd 2010-02-13T09:14:42Z 2010-02-13T09:14:42Z <p>There's a more down to earth way to deal with this, which is already explained in Mumford's GIT: make an fppf (or even etale) surjective base change to acquire a section, use that to define the principal polarization, and then show it is independent of the choice. (Short reason: varying the choice amounts to a morphism from the smooth proper curve to a Hom or Isom scheme that is unramified over the base, hence constant.) Thus, by descent theory one gets the polarization over the original base.</p> <p>This is related to the same issue which comes up in explaining why a polarization of an abelian scheme need not arise as the "Mumford construction" $\phi_{\mathcal{L}}$ even though it automatically does so on geometric fibers (due to the special nature of $k$-simple finite commutative $k$-groups when $k = \overline{k}$). That is, a definition of "polarization" which is better-suited to the relative case is not to mimic what one traditionally does over an algebraically closed field (the Mumford construction) but rather something which makes more effective use of the Poincar\'e bundle. The possible lack of $\mathcal{L}$ over the base is analogous to the possible lack of a section of the curve to define the principal polarization. See the Wikipedia page on abelian varieties'' for more on this. :)</p>