1
$\begingroup$

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?

1 Answers 1

1

[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.

  • 0
    The *simple covers* are the simplest covers after étale covers. The terminology goes back at least to W. Fulton in his paper "Hurwitz schemes and irreducibility of moduli of algebraic curves", Ann. of Math. (1969).2011-11-21
  • 0
    @QiL: thanks for the info/clarification!2011-11-21
  • 0
    @'paul garrett', you are welcome !2011-11-21