Timeline for Surfaces of general type with $q=1$
Current License: CC BY-SA 4.0
11 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 14, 2020 at 9:14 | vote | accept | Pat | ||
| Apr 14, 2020 at 9:14 | comment | added | Pat | Beautiful...Many thanks! | |
| Apr 14, 2020 at 8:19 | comment | added | Francesco Polizzi | You can construct examples where the Albanese fibres have multiple components but also contains components of multiplicity one by considering a variation of the isogenous constructions, where $G$ is allowed to have isolated fixed points on the product. Look at my paper arxiv.org/pdf/math/0703066.pdf and at the other related papers of mine on the arXiv. | |
| Apr 14, 2020 at 8:08 | comment | added | Francesco Polizzi | CS surface is rigid. Morally, one expects that, one the surface is not rigid, the "general" element in moduli has reduced Albanese fibres. | |
| Apr 14, 2020 at 8:07 | comment | added | Francesco Polizzi | For instance, Cartwright-Steger surface (which has $p_g=q=1$, $K^2=9$) has the property you want. See the main theorem here: arxiv.org/pdf/1412.4137.pdf | |
| Apr 14, 2020 at 8:00 | comment | added | Pat | 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 $q$... | |
| Apr 14, 2020 at 7:57 | comment | added | Francesco Polizzi | No, of course it does not happen always | |
| Apr 14, 2020 at 7:57 | comment | added | Pat | @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. | |
| Apr 14, 2020 at 7:50 | answer | added | Francesco Polizzi | timeline score: 3 | |
| Apr 14, 2020 at 7:34 | comment | added | abx | 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. | |
| Apr 14, 2020 at 7:06 | history | asked | Pat | CC BY-SA 4.0 |