# Finite etale covers of products of curves

Probably this question can be phrased in a much greater generality, but I will just state it in the generality I require. I work over $$\mathbb{C}$$.

Let $$C_1, C_2 \subset \mathbb{P}^1$$ be non-empty open subsets and $$f: X \to C_1 \times C_2$$ a non-trivial finite etale cover. Does there exist $$i\in \{1,2\}$$ such that the composition $$X \to C_1 \times C_2 \to C_i$$ has non-connected fibres?

• I do not think so, at least in general. Think of a double cover $\bar{X}$ of $\mathbb{P}^1 \times \mathbb{P}^1$ branched over a curve of type $L_1 + L_2 +M_1 +M_2$ (it gives you an étale double cover with $C_i=\mathbb{P}^1$ minus two points). The general fibres of the composition $\bar{X} \to \mathbb{P}^1 \times \mathbb{P}^1 \to \mathbb{P}^1$ are smooth double cover of $\mathbb{P}^1$ branched at two points, hence they are again isomorphic to $\mathbb{P}^1$, and so the fibres of your composition are isomorphic to $\mathbb{P}^1$ minus the ramification, i.e. $\mathbb{P}^1$ minus two points. – Francesco Polizzi Apr 10 '19 at 9:01
• If you take a branch curve of type $L_1+L_2+L_3+L_4+M_1+M_2+M_3+M_4$, then the fibres of your compositions will be elliptic curves minus four points, and so on... – Francesco Polizzi Apr 10 '19 at 9:08
• @Francesco Polizzi: Are you able to provide an answer with an explicit counter-example? – Daniel Loughran Apr 10 '19 at 9:58
• @DanielLoughran: I will try – Francesco Polizzi Apr 10 '19 at 10:44

The answer is no, at least in general, as shown by the following counterexample.

Take a double cover $$\bar{f} \colon \bar{X} \to \mathbb{P}^1 \times \mathbb{P}^1$$, branched over a reducible curve of the form $$B=L_1 + L_2 + M_1 + M_2$$ (here $$|L|$$ and $$|M|$$ are the two pencil of lines on the quadric).

Such a cover exists because $$B$$ is $$2$$-divisible in $$\mathrm{Pic}(\mathbb{P}^1 \times \mathbb{P}^1)$$, and corresponds to an étale cover $$f \colon X \to C_1 \times C_2$$, where each $$C_i$$ is $$\mathbb{P}^1$$ - {two points}.

If these points are (say) $$0$$ and $$1$$ in both factors, then the equation for $$X \subset \mathbb{C} \times (\mathbb{C}-\{0, \, 1\})^2$$ is
$$z^2 = xy(x-1)(y-1), \quad f(z, (x, \,y)) = (x,\, y).$$

It is clear that the general line in $$|L|$$ and $$|M|$$ intersects the branch locus $$B$$ transversally at two points, hence both compositions $$\bar{X} \to \mathbb{P}^1 \times \mathbb{P}^1 \to \mathbb{P}^1$$ have connected fibres, the general one being isomorphic to $$\mathbb{P}^1$$ (double cover of $$\mathbb{P}^1$$ branched at two points).

Then both compositions $$X \to C_1 \times C_2 \to C_i$$ have connected fibres, the general one being isomorphic to the $$\mathbb{P}^1$$ above minus the ramification, i.e. $$\mathbb{P}^1$$ minus two points, that is clearly connected.

In the same vein, choosing as $$B \subset \mathbb{P}^1 \times \mathbb{P}^1$$ a divisor of type $$B = \sum_{i=1}^{2g+2} L_i + \sum_{i=1}^{2g+2} M_i,$$ both compositions $$\bar{X} \to \mathbb{P}^1 \times \mathbb{P}^1 \to \mathbb{P}^1$$ have connected fibres, the general one being isomorphic to a hyperelliptic curve $$\Sigma_g$$ of genus $$g$$, and so both compositions $$X \to C_1 \times C_2 \to C_i$$ (here each $$C_i$$ is $$\mathbb{P}^1$$ minus $$2g+2$$ points) have connected fibres, the general one being isomorphic to $$\Sigma_g$$ minus $$2g+2$$ distinct points.

• Thanks for the answer. Out of interest, I think that the explicit example you constructed is a quartic del Pezzo surface, and the two projections to $\mathbb{P}^1$ are conic bundles on the surface. – Daniel Loughran Apr 10 '19 at 15:55
• @DanielLoughran: You are welcome. Note that $\bar{X}$ in the first example has four nodal singularities, corresponding to the four nodes of the branch curve $B$. – Francesco Polizzi Apr 10 '19 at 16:02

The question already has a beautiful answer, but here's a different point of view which you may find helpful.

Let $$F_i = \pi_1(C_i, x_i)$$, which is a free group on $$\#(\mathbf{P}^1\setminus C_i) - 1$$ generators (the etale fundamental group will be the profinite completion of this). Then $$F = \pi_1(C_1\times C_2, x_1\times x_2) = F_1\times F_2$$.

A finite etale cover of $$C_i$$ or $$C_1\times C_2$$ corresponds to a finite set (its fiber at the basepoint $$x_i$$ or $$x_1\times x_2$$) with an action of $$F_i$$ or $$F$$. The cover is connected if and only if the action on that set is transitive.

Let $$\sigma_i \colon C_i\to C_1\times C_2$$ be the section $$\sigma_1(x) = (x, x_2)$$, $$\sigma_2(x) = (x_1, x)$$. Then for a finite etale cover $$X\to C_1\times C_2$$, the composition $$X\to C_1\times C_2\to C_i$$ has connected fibres if and only if the pull-back of $$X$$ along $$\sigma_{2-i}$$ is connected.

So now the question is equivalent to: suppose that $$S$$ is a finite set with more than one element with an action of $$F=F_1\times F_2$$. Is it possible that $$F_1$$ and $$F_2$$ both act transitively on $$S$$? It is very easy to construct such examples.

The easiest one could be $$S$$ with two elements, with every generator of each $$F_i$$ acting by a nontrivial involution. If there are only two punctures on each curve, this coincides with Francesco Polizzi's construction, and we see that his example is in some sense minimal.