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Let $f:X\to \mathbf{P}^1$ be a simple cover of the Riemann sphere. This means that $f$ is a branched cover, and that each fibre has at least $\deg f-1$ points in it.

Is it true that the number of ramification points is $(\deg f -1) \cdot \# B$, where $\#B$ is the number of branch points in $\mathbf{P}^1$?

If you label the branch points $b_1,\ldots,b_r$, is there a natural way to label the set of ramification points?

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[Edit: removed some remarks inquiring about background...]... since there are at most deg $f$ points in each fiber, the constrain of "simple cover" apparently requires that the worst ramification that can occur is where (only) two sheets out of deg $f$ are joined. Thus, if the number of branch points on $\mathbb P^1$ is $B$, yes, there are $(deg\,f-1)\cdot B$ ramified points on the cover.

... There is a unique ramified point over each branch point, so they can be labelled by the branch point.

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    @'paul garrett', you are welcome !2011-11-21