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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 $C$?

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  • $\begingroup$ 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
    Commented Dec 12, 2013 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
    Commented Oct 2, 2016 at 6:48

2 Answers 2

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``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.

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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.

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