Timeline for Euler characteristic of isotrivial elliptic surfaces

5 events

when toggle format	what		by	license	comment
Jan 19, 2022 at 21:07	comment	added	Dori Bejleri		Yes that's right. In fact one way to construct such surfaces is to take $\mathbb{P}^1 \times E$ and take a subgroup of $\mathrm{Aut}(E)$ (necessarily cyclic) and consider the action on $\mathbb{P}^1 \times E$ where the cyclic group acts on $\mathbb{P}^1$ with two fixed points $0$ and $\infty$. Then take the quotient, resolve singularities and contract $(-1)$-curves in the fibers. The subgroup of $\mathrm{Aut}(E)$ is exactly the monodromy over $\mathbb{A}^1 \setminus \{0\}$ and it acts by dual weights on the fixed points over $0$ and $\infty$ which leads to the singular fibers coming in pairs.
Jan 19, 2022 at 21:00	comment	added	Enclitic Sarcool		Ahh, I see now that there is monodromy along the generator of the fundamental group of $\mathbb A^1 - \{ 0 \}$ and so we can't glue to a trivial surface over $\infty$. The fact that the monodromies exactly pair off to allow this is remarkable!
Jan 19, 2022 at 20:55	history	edited	Enclitic Sarcool	CC BY-SA 4.0	added 1 character in body
Jan 19, 2022 at 20:42	comment	added	Dori Bejleri		Such a surface you tried to construct over $\mathbb{P}^1$ doesn't exist. When you extend your surface over $\mathbb{A}^1$ to one over $\mathbb{P}^1$, it will necessarily have a singular fiber over $\infty$ and that singular fiber will exactly "balance" the Euler characteristic to get a multiple of $12$.
Jan 19, 2022 at 19:33	history	asked	Enclitic Sarcool	CC BY-SA 4.0