# Can an abelian variety dominate a variety of general type?

Let $X$ be a projective (not necessarily smooth) normal variety of general type over $\mathbb{C}$. Let $A$ be an abelian variety and let $A\to X$ be a surjective morphism.

Is $X$ zero-dimensional?

I have the feeling that one can pull-back differential forms on a resolution $\tilde{X}$ of ${X}$ (by which I mean sections of $\omega_{\tilde{X}}^{\otimes n}$) to obtain differential forms on $A$. But how does one do that?

• One easy solution -- take a resolution of $X$, and blow up $A$ appropriately so it maps to the resolution. But plurigenera are a birational invariant, so the blowup is harmless. May 21 '18 at 23:59
• @DanielLitt Ok. This argument then shows that the Kodaira dimension of $X$ is non-positive, right? May 22 '18 at 0:41
• I posted a general answer that works in more generality, but you can probably get an elementary proof along your argument in some particular cases.... Pulling back and wedging by a general element of $H^0(\Omega ^{\dim A-\dim X}_A)$ gives an injection $H^0(\Omega ^{\dim X}_X)\to H^0(\Omega _A^{\dim A})$ and hence $h^0(\omega _X)\leq 1$...... May 22 '18 at 8:03

Theorem: Let $A$ be an abelian variety and $f:A\to X$ a dominant morphism to a projective variety of general type, then $\dim X=0$.
Proof: Replacing $f$ by a birational model of the Stein factorization we may assume that $f':A'\to X'$ is a projective morphism of smooth varieties with $f'_*\mathcal O _{A'}=\mathcal O _{X'}$, $X'$ is of general type and $A'\to A$ is birational (and hence $\kappa (A')=\kappa (A)=0$). Let $F$ be a general fiber, then $\kappa (F)\geq 0$ (since $K_{A'}|_F\cong K_F$). By Theorem 3 of http://www.numdam.org/article/CM_1981__43_2_253_0.pdf we have $\kappa (A')\geq \kappa (F)+\kappa(X')$. But then $\dim X'=\kappa(X')=0$.