Varieties dominated by products of curves

Question

Let $X$ be an irreducible smooth projective variety of dimension $d$. Do there exist irreducible smooth projective curves $C_1, C_2,\ldots, C_d$, an open subset $U\subset C_1\times C_2\times\ldots\times C_d$ and a dominant morphism $f:U\to X$.

Answer 1

Ulrich observes in the comments that C. Schoen provides counterexamples in his paper "Varieties dominated by product varieties". Beyond Schoen's work, this problem has some interesting history, which I learned from this note of Oort.

Grothendieck, in attempting to prove the Weil conjectures, had hoped that every variety was rationally dominated by a product of curves. He asked Serre if this was true; Serre showed that a sufficiently general surface contained in an explicit Abelian variety of dimension $5$ is a counterexample. (See p. 145 of the Grothendieck-Serre Correspondence, which is a really amazing book.) The counterexample is quite beautiful and very simple, and of a rather different nature than Schoen's.

Essentially Serre observes that of $S\subset A$ is a smooth surface passing through the origin and satisfying the following property:

$(*)$ If $C, C'$ are curves contained in $S$, then $C+C'$ is not contained in $S$

then $S$ cannot be rationally dominated by a product of curves. This is because the rational map must extend to a morphism (as it is a map into an Abelian variety) given by adding two maps $C\to A, C'\to A$. So it suffices to find a surface $S$ satisfying $(*)$. Serre does this by writing down an explicit analytic germ at the origin satisfying $(*)$ (not too hard) and then approximating this germ by an honest surface (which one may take to be a complete intersection, for example).

Oort notes that Schoen seems not to have been aware of Serre's counterexample.

Answer 2

For another family of counter-examples to Grothendieck's question one can take any compact quotient of the unit ball of $\mathbb{C}^n$ (for $n\ge 2$). At least for a topologist/geometer it is a very natural family of examples.

Let me explain this : suppose that we have a surjective holomorphic map $f : S_1 \times \cdots \times S_n\to X= B/\Gamma$ where $\Gamma$ is a torsion free cocompact lattice in ${\rm PU}(n,1)$ ($n\ge 2$) and where the $S_{j}$'s are compact Riemann surfaces. We will derive a contradiction.

At the level of fundamental groups, $f_{\ast}$ has an image of finite index since $f$ is surjective. Replacing $\Gamma$ by a finite index subgroup we assume that $f_{\ast}$ is surjective.

Now we have the following fact:

The group $\Gamma$ is not generated by a family of pairwise commuting, nontrivial, normal subgroups $(H_i)_{1\le i \le N}$ if $N\ge 2$.

This can be seen easily using the fact that the Zariski closure of each such $H_i$ should be ${\rm PU}(n,1)$.

Going back to the morphism $f : S_1 \times \cdots \times S_n \to X$ this says that (after possibly reordering the indices) $f_{\ast}(\pi_{1}(S_{j})$ is trivial for $j\ge 2$.

Hence for a fixed $j\ge 2$ the map $z\mapsto f(p_{1}, \ldots , z, \ldots ,p_{n})$ from $S_{j}$ to $X$ induces the trivial morphism on $\pi_{1}$, hence is constant (since there is no compact Riemann surface in the unit ball of $\mathbb{C}^{n}$).

This implies that $f$ factors through the projection from the product of the $S_{j}$'s onto $S_{1}$ : $f=g\circ pr_{1}$ where $pr_{1}$ is the first projection and $g : S_{1}\to X$ is holomorphic. But this is impossible if $f$ is dominant.

I have not thought on whether we can push this argument to answer the original question (where $f$ is only defined on an open set of the product of the $S_{j}$'s).