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Sep 12, 2015 at 13:03 answer added David E Speyer timeline score: 2
Sep 12, 2015 at 12:45 comment added David E Speyer I'm sorry, I just redid the RH computation, and now I get $20$. So we have to work harder, to exclude the ramification profile $(2,2,2)^6$, $(2,2,1,1)$. This sounds like a question for Mike Zieve ... However, as I explain below, it is easy to find solutions by letting $R$ acquire nodes.
Sep 11, 2015 at 19:01 comment added David E Speyer At most three of those points of $R$ can map to any given point in $\mathbb{P}^1$. So we get ramification over at least $22/3 > 7$ points if we follow this route. The question is about whether we can find a way, by less obvious choices of the parameters, to get down to $7$.
Sep 11, 2015 at 19:00 comment added David E Speyer I figured I'd explain where the $7$ is presumably coming from. As Yusuf says, $X$ should be a double cover of a $\mathbb{P}^1$ bundle over $\mathbb{P}^1$. The simplest case is $\mathbb{P}^1 \times \mathbb{P}^1$. And $X$ should be ramified over some $R$ of degree $(6,n)$ in $\mathbb{P}^1 \times \mathbb{P}^1$. We must have $n$ even in order for the double cover to exist, so the simplest choice is $(6,2)$. Then, if $R$ is smooth, it is genus $5$. So Riemann-Hurwitz shows that $R \to \mathbb{P}^1$ is ramified at $22$ points of $R$. (continued)
Sep 11, 2015 at 7:36 answer added Remke Kloosterman timeline score: 4
Sep 10, 2015 at 18:36 answer added Yusuf Mustopa timeline score: 0
Sep 10, 2015 at 1:10 comment added guest Yes, I want 7 nodal fibers.
Sep 10, 2015 at 0:58 comment added Daniel Litt Just checking--do you want 7 nodal fibers? Or actually seven nodes on the surface?
S Sep 10, 2015 at 0:31 history suggested Henry.L CC BY-SA 3.0
It should be a complex manifold and I edited the title. I added a sentence in the beginning and reformat it.
Sep 10, 2015 at 0:28 review Suggested edits
S Sep 10, 2015 at 0:31
Sep 9, 2015 at 23:23 review First posts
Sep 9, 2015 at 23:42
Sep 9, 2015 at 23:19 history asked guest CC BY-SA 3.0