# Surfaces of general type with $q=1$

Let $$X$$ be a smooth projective connected surface of general type over $$\mathbb{C}$$ with $$q(X) = 1$$, where $$q(X) = \mathrm{h}^1(X,\mathcal{O}_X)$$. Let $$E$$ be the Albanese variety of $$X$$, and let $$X\to E$$ be the Albanese map (having fixed a point). Let $$0$$ be a closed point of $$E$$.

Let $$F$$ be the scheme-theoretic fibre over $$0$$. Does $$F$$ have an irreducible reduced component? That is, does $$F$$ have an irreducible component of multiplicity one?

Of course, for a general $$0$$, the fiber $$F$$ is smooth. I am wondering whether the Albanese map has a multiple fibre or not.

• It may certainly happen. Take an Enriques surface $S$; it admits an elliptic fibration $S\rightarrow \mathbb{P}^1$ with 2 double fibers, say above $0$ and $1$. Now take a double cover $\pi :E\rightarrow \mathbb{P}^1$ branched along 4 points $\neq 0,1$, and let $T$ be the pull back of $S$ by $\pi$; its Albanese map is the projection $p:T\rightarrow E$ with 4 double fibers. Then let $\rho :X\rightarrow T$ be a double covering branched along a smooth, ample curve in $T$, transversal to the double fibers. Its Albanese map $p\circ \rho$ still has 4 double fibers.
– abx
Apr 14 '20 at 7:34
• @abx That's a nice example, thank you! But I am wondering whether this always happens. (I think I phrased the question in a confusing manner. My apologies.) Basically, given a surface of general type $X$ with $q=1$, my question is whether we can prove that $X\to Alb(X)$ has a multiple fibre.
– Pat
Apr 14 '20 at 7:57
• No, of course it does not happen always Apr 14 '20 at 7:57
• How can I find an example of a surface with $q=1$ such that $X\to \mathrm{Alb}(X)$ has no multiple fibres? If I start with one of your examples, I could consider ramified coverings of $X$, but I fear this might increase $q$...
– Pat
Apr 14 '20 at 8:00
• Beautiful...Many thanks!
– Pat
Apr 14 '20 at 9:14

You can find plenty of examples with multiple Albanese fibres by considering surfaces isogenous to a product, namely of the form $$S=(C \times F)/G$$, where $$G$$ is a finite group acting faithfully on the smooth curves $$C$$, $$F$$ and whose diagonal action on the product is free.
On surfaces of general type with $$p_g=q=1$$ isogenous to a product of curves, Communications in Algebra 36 (2008), no. 6, 2023-2053, arXiv:math/0601063.