5
$\begingroup$

The question is kind of self contained, but let me develop a bit further.

Assume K is a CM field of degree $2g$, that is, a quadratic imaginary extension of a totally real field. A CM type of $K$ is a set $\Phi$ consisting of $g$ complex embeddings of $K$ such that $\mathrm{Hom}(K, \mathbb{C})=\Phi \cup \overline{\Phi}$.

Given $(K, \Phi)$, there always exists a $g$-dimensional abelian variety $A$ such that $\mathrm{End}(A) \otimes \mathbb{Q}=K$ and that $K$ acts on $H^0(A, \Omega^1_A)$ through $\Phi$. One easily constructs as a complex torus, starting from the embedding $K \subset \mathbb{C}^g$ given by $\Phi$.

The following question seems much more subtle:

Is there always such an abelian variety of the form $A=\mathrm{Jac}(C)$ for a smooth projective curve $C$?

$\endgroup$
  • $\begingroup$ I think it follows from the paper "The existence of an abelian variety over $\overline{\mathbb{Q}}$ isogenous to no Jacobian" by Tsimerman that this is not always true. $\endgroup$ – naf Dec 12 '13 at 13:28
  • $\begingroup$ Precisely. Actually, Chai and Oort proved that the Andre-Oort conjecture (then only known under GRH) implies the following. For any closed subvariety X of $A_g$, there exists a CM point not isogenous to an abelian variety parametrised by any point of $X$. This, applied with $X$=Torelli locus for $g \geq 4$ implies a negative answer to your question. Tsimerman gave an argument for removing the GRH assumption from the particular case of Adre-oort Chai and oort needed so this result is unconditional. The Andre-Oort conjecture for sub varieties of $A_g$ is now proved unconditionally anyway..... $\endgroup$ – user42721 Oct 2 '16 at 6:48
6
$\begingroup$

``Given $(K,\Phi)$ , there always exists a $g$ -dimensional abelian variety $A$ such that $End(A)\otimes Q=K$ and that $K$ acts on $H^0(A,\Omega^1)$ through $\Phi$. One easily constructs as a complex torus, starting from the embedding $K\subset C^g$ given by $\Phi$."

Actually, this is not always the case. For example, if $K$ is a quartic CM-field containing an imaginary quadratic subfield (i.e., $K$ is a compositum of two imaginary quadratic fields) then it is not isomorphic to the endomorphism algebra of any abelian surface (or a complex torus), see Sect. 5 of arXiv:1312.0377 [math.NT].

However, if $[K;Q] \le 6$ and there exists a simple complex $g$-dimensional CM-abelian variety $B$ (with $g\le 3$) of CM-type $(K,\Phi)$ then there exists a jacobian $J$ isogenous to $B$. (It's true, because the Siegel upper half-plane $H_g$ is the $Sp(2g,R)$-orbit, $Sp(2g,Q)$ is everywhere dense in $Sp(2g,R)$ in the classical topology, and the ``Torelli locus" is open in $H_g$ if $g\le 3$ and therefore meets every (dense) $Sp(2g,Q)$-orbit; compare with Remark 3 at the end of Sect. 2 in arXiv:0912.4325 [math.NT]). This implies $End(J)\otimes Q=K$ and the CM-type of $J$ is $\Phi$.

For big $g$ the situation seems to be murky; however, as far as I understand, it is expected that not every $K$ is isomorphic to the endomorphism algebra of a $g$-dimensional jacobian.

$\endgroup$
1
$\begingroup$

It is certainly not what is expected : a conjecture of Coleman predicts that for $g\geq N$ (see below) there are only finitely many Jacobians of genus $g$ which are CM. Coleman's original conjecture was with $N=4$, but I think there are now counter-examples for $N\leq 7$, so one should state the conjecture with $N$ at least 8.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.