# Do all simple factors of jacobians of curves come from correspondences?

For this question I will let the overly ambiguous word curve mean: smooth projective and connected curve over $\mathbb C$ (or equivalently a smooth compact Riemann-Surface).

Let $C$ be a curve over $\mathbb C$. And suppose that $f: D \to C$ and $g:D \to E$ are dominant maps of curves. Then the pair $(f,g)$ forms a correspondence from $C$ to $E$, and one gets a map between Jacobians $f_* \circ g^* : J(E) \to J(C)$. Then the image of $J(E)$ under this map is going to be a sub-abelian variety of $J(C)$.

My questions is, is it possible to get all simple sub-abelian varieties of $J(C)$ from correspondences in this way?

• give me a start here. what makes this seem plausible? Oct 20, 2015 at 3:30
• It does seem pretty plausible. First, any morphism $J(E)\to J(C)$ comes from a line bundle on $C\times E$, which in turns comes from a divisor in $C\times E$... Which is kind of a curve (probably singular, but we can resolve this, also probably reducible, which is more annoying). Secondly, for any abelian variety $A$, we can find a curve $E\subset A$ that generates it (again, the obvious construction gives a reducible curve, so some argument would be required?), and then $A$ becomes a quotient of $J(E)$. Combining these two facts we'd get the required claim. Oct 20, 2015 at 5:14
• P.S. A reference for the second fact: Theorem III.10.1 of Milne's jmilne.org/math/CourseNotes/AV.pdf (the section has the telling title `Abelian varieties are quotients of Jacobian varieties :) And the first fact should follow from Bertini's Theorem... Oct 20, 2015 at 5:22
• @roysmith: I have no idea wether this is something plausible. I just know one important case, namely that it is true for modular curves if one does it over $\mathbb Q$ instead of over $\mathbb C$ (at least if one does not demand that correspondences are given by irreducible curves). So a related question is: how big is the subring of ${\rm End}_{\mathbb C} J(C)$ generated by correspondences from $C$ to itself? Oct 20, 2015 at 21:20
• @Maarten Derickx: No, I don't, but it does not seem particularly hard. Fix base points $e\in E$, $c\in C$, we need to show that for any line bundle $L$ on $E\times C$, there exist $n_1$ and $n_2$ such that the line bundle $L(n_1(\{e\}\times C)+n_2(E\times \{c\}))$ is of the form $O(D)$ for smooth curve $D\subset E\times C$ (which then gives a correspondence between the two curves). However, $O((\{e\}\times C)+(E\times \{c\}))$ is ample, therefore, $L(n(\{e\}\times C)+n(E\times \{c\}))$ is very ample for $n\gg 0$, and the claim follows from Bertini's Theorem. Oct 21, 2015 at 1:50

This argument is mostly contained in t3suji's comments, but with some of the proofs somewhat expanded.

As a convention (consistent with that of the theory of Chow motives), all actions of correspondences $\alpha \in \operatorname{CH}^*(X \times Y)$ on cohomology (or Jacobians) are covariant: $\alpha_* \colon H^*(X) \to H^*(Y)$, etc.

Lemma. If $X$ and $Y$ are smooth projective varieties over $\mathbb C$, then there is an isomorphism $$\operatorname{NS}(X \times Y) \cong \operatorname{NS}(X) \times \operatorname{NS}(Y) \times \operatorname{Hom}(\operatorname{Alb}_X,\operatorname{Pic}^0_Y).\label{Eq NS}\tag{1}$$

Proof. The Künneth formula gives an isomorphism of $\mathbb Z$-Hodge structures $$H^2(X \times Y) = \big( H^2(X) \otimes H^0(Y)\big) \oplus \big( H^0(X) \otimes H^2(Y) \big) \oplus \big( H^1(X) \otimes H^1(Y) \big).\tag{2}\label{Eq Künneth}$$ By Poincaré duality, the last term equals $\operatorname{Hom}(H^{2\dim X-1}(X),H^1(Y))$. Taking $(1,1)$-parts in (\ref{Eq Künneth}) gives the result by Lefschetz's (1,1) theorem and the theory of abelian varieties over $\mathbb C$. $\square$

There is also a version for $\operatorname{Pic}$ for smooth projective varieties over an arbitrary field (or even over a base, under some mild hypotheses to ensure existence of $\operatorname{\textbf{Pic}}_{X/S}$ and $\operatorname{\textbf{Alb}}_{X/S}$). The above analytic proof has the advantage that it shows us how classes act on Jacobians: if $X$ and $Y$ are curves, then $\operatorname{Alb}_X \cong \operatorname{Pic}_X^0 \cong \operatorname{Jac}_X$, and under these identifications, a class $\alpha \in \operatorname{CH}^1(X \times Y)$ induces the map $\alpha_* \colon \operatorname{Jac}_X \to \operatorname{Jac}_Y$ given by the corrresponding component of $\alpha$ under (\ref{Eq NS}).

Construction. Let $A \subseteq \operatorname{Jac}_C$ be any abelian subvariety. Let $E$ be any curve admitting a surjection $\operatorname{Jac}_E \to A$; for example, $E = C$ with the projection $\operatorname{Jac}_C \to A$ given by $n \pi$ for $\pi \in \operatorname{End}^\circ(\operatorname{Jac}_C)$ the projector corresponding to the isogeny factor $A$ of $\operatorname{Jac}_C$, and $n \in \mathbb Z_{> 0}$ such that $n\pi$ is integral. The composition gives a morphism $\phi \colon \operatorname{Jac}_E \to \operatorname{Jac}_C$, corresponding by (\ref{Eq NS}) to a class $\alpha \in \operatorname{NS}(X \times Y)$.

The class $(1,1,0)$ in (\ref{Eq NS}) is ample, hence for some $n \gg 0$ there exists a very ample line bundle $\mathscr L$ on $X \times Y$ mapping to $(n,n,\alpha) \in \operatorname{NS}(X \times Y)$. Let $D \in |\mathscr L|$ be a smooth member. By the discussion above, the cycle $[D] \in \operatorname{CH}^1(E \times C)$ induces the map $\phi$.

The graphs of the maps $g \colon D \to E$ and $f \colon D \to C$ give cycles $[\Gamma_g]^\top \in \operatorname{CH}^1(E \times D)$ and $[\Gamma_f] \in \operatorname{CH}^1(D \times C)$, whose actions on $\operatorname{Jac}$ are given by $g^*$ and $f_*$ respectively. Moreover, we have $$[\Gamma_g]^\top \circ [\Gamma_f] = [D] \in \operatorname{CH}^1(E \times C),$$ which shows that indeed $\phi = f_* \circ g^*$. $\square$