This is a follow-up to this question (and my answer thereto). Given an algebraic curve, is there a way to realize its Jacobian explicitly as a zero-set of a bunch of polynomials in $\mathbb{P}^n.$ I care most about the complex case, but all ground fields are of interest...
$\begingroup$
$\endgroup$
asked Aug 12, 2015 at 11:47
Add a comment
|
2 Answers
$\begingroup$
$\endgroup$
3
You might look at Mumford's three articles.
MR0219542 Mumford, D. On the equations defining abelian varieties. III. Invent. Math. 3 1967 215–244.
MR0219541 Mumford, D. On the equations defining abelian varieties. II. Invent. Math. 3 1967 75–135.
MR0204427 Reviewed Mumford, D. On the equations defining abelian varieties. I. Invent. Math. 1 1966 287–354.
There are also articles (David Grant, Victor Flynn, ...) that I think give explicit equations for jacobians of genus 2 curves and explicit equations for the group law.
answered Aug 12, 2015 at 11:53
-
$\begingroup$ Thanks! A particular question is whether these can be hypersurfaces (I think this is not what happens in Greg Anderson's paper...) $\endgroup$ Commented Aug 12, 2015 at 13:36
-
2$\begingroup$ @IgorRivin: do you mean hypersurfaces in projective space? If so, then this only happens for $n=1$ (i.e. elliptic curves in $\mathbf P^2$) by Lefschetz hyperplane theorem. $\endgroup$ Commented Aug 12, 2015 at 14:10
-
2$\begingroup$ @potentiallydense Ah. yes. I was being dense :) $\endgroup$ Commented Aug 12, 2015 at 14:11
$\begingroup$
$\endgroup$
1
This paper by Greg Anderson does just that.
answered Aug 12, 2015 at 12:03