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Let $g\geq 4$. Are there infinitely many compact genus $g$ Riemann surfaces with (mutually) isogenous Jacobians?

Does the situation change in positive characteristic?

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  • $\begingroup$ What exactly do you mean by isogeny? Are you asking for a holomorphic maps between the tori? Or surjective group homomorphisms with finite kernel (in which case you are looking at finite-index subgroups of a lattice). $\endgroup$
    – JHM
    May 6, 2013 at 19:03
  • $\begingroup$ Being principal, jacobians do not embed as finite-index sublattices of other jacobians (i.e. are not finitely-covered by any other jacobian). $\endgroup$
    – JHM
    May 6, 2013 at 19:06
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    $\begingroup$ I thought those two notions of isogeny were equivalent over the complex numbers. Don't elliptic curves provide a counterexample to your second comment? I wanted to prove that given a hyperbolic complete curve $X$, there are only finitely many hyperbolic curves it maps to. By looking at Jacobians, we are led to the above question (with the guess that the answer is no!) $\endgroup$
    – Raju
    May 6, 2013 at 20:26
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    $\begingroup$ @RK I think Martel's second comment is wrong, too. I don't know the answer to the question you asked, but if you want to see a proof of the statement in your comment, this is de Franchis's theorem. $\endgroup$ May 6, 2013 at 23:20
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    $\begingroup$ @R.vanDobbendeBruyn I just don't expect it to be true because $\mathcal{M}_g$ and $\mathcal{A}_{g,1}$ are birational for $g=2,3$. $\endgroup$
    – Raju
    Apr 28, 2020 at 1:19

2 Answers 2

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I believe I have an example in genus 5:

Humbert curves (see either Varley's "Weddle's Surfaces, Humbert's Curves, and a Certain 4-Dimensional Abelian Variety", or exercise batch F in chapter 6 of ACGH) are in 1-1 correspondence with 5-tuples lines in the plane -- the plane being $I_C(2)$ -- up to projective transformations of the plane. These 5 lines are the image of the double cover from $W^1_4(C)$ to rank < 4 quadrics enveloping the canonical image of $C$. Each of the 5 lines is marked with 4 points - its intersection points with the other four lines, and is the 2-quotient (under the map $W^1_4\to I_C(2)$ ) of an elliptic curve which lies inside the Jacobian of $C$, where four points are the ramification points. The Jacobian of C is isogenous to the product of these five elliptic curves.

Giving the plane the projective coordinates $x,y,z$, we now consider the lines: $$x=0,\quad y=0,\quad x=z,\quad y=z,\quad x+y=az.$$ The $y$ (resp $x$) coordinate of the intersection points on the first (resp second) line are $0,1,a,\infty$; whereas the $x$ (resp $y$) coordinate of the intersection points on the third (resp fourth) lines are $0,1, 1-a, \infty$. Since $x\mapsto 1-x$ takes $0\to1,1\to0,\infty\to\infty, a\to1-a$, the four elliptic curves associated with these four lines are isomorphic. Finally, The elliptic curve associated with fifth line has an extra involution, (flipping $x,y$), so it is the same curve regardless of $a$. Hence if we pick any $a_1,a_2$ which give isogenous elliptic curves, we get isogenous Humbert curves.

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(Edit: this doesn't answer the question - I misread it to mean that we were looking for infinitely many isogenies of distinct Jacobians in arbitrary genus.)

There are explicit constructions that provide infinitely many examples of pairs of non-isomorphic curves with isogenous Jacobians in any genus.

Jean-François Mestre has a Richelot-like construction that, for each $g \ge 2$, produces a $g+1$-dimensional family of pairs of hyperelliptic curves with $(2,\ldots,2)$-isogenous Jacobians. For the details, see https://www.ams.org/journals/jag/2013-22-03/S1056-3911-2012-00589-X/ (if you don't have access to that, then there's an earlier version in French at https://arxiv.org/abs/0902.3470 , and a translation of that version into English at http://www.lix.polytechnique.fr/~smith/Mestre---pairs.pdf ).

In characteristic $p$, you can produce examples over finite fields almost trivially using Frobenius. Let $q$ be a power of $p$. For each curve $C$ over $\mathbb{F}_{q^2}$, there is a Galois-conjugate curve $C'$ over $\mathbb{F}_{q^2}$ defined by raising the coefficients of the defining equations of $C$ to the $q$-th power, and then $q$-th powering defines a degree-$q$ morphism $\pi: C \to C'$, which (in genus $g>0$) induces an inseparable isogeny between $J_C$ and $J_{C'}$. If $C$ and $C'$ are not isomorphic (and in general they are not), then you have a pair of distinct isogenous Jacobians.

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    $\begingroup$ Note that this does not answer the question, which is whether there exist infinitely many non-isomorphic curves with Jacobians isogeneous to a fixed abelian variety. $\endgroup$
    – abx
    Apr 27, 2020 at 12:37
  • $\begingroup$ Sorry, I misread the question. I've added a comment to indicate this. $\endgroup$
    – Ben Smith
    Apr 27, 2020 at 13:15
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    $\begingroup$ Your last example works for the intended question over infinite fields of characteristic $p$, instead of finite ones, because you can take the pullback of any curve corresponding to a point in the moduli space with transcendental coordinates by infinitely many different powers of Frobenius, and the Jacobians will be isogenous. $\endgroup$
    – Will Sawin
    Apr 27, 2020 at 13:24
  • $\begingroup$ @WillSawin: by "infinite" you mean "not algebraic over $\mathbf F_p$". $\endgroup$ Apr 27, 2020 at 22:39
  • $\begingroup$ @R.vanDobbendeBruyn That would be accurate although I guess I didn't say "over all infinite fields of characteristic $p$" so maybe I am safe. $\endgroup$
    – Will Sawin
    Apr 27, 2020 at 22:41

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