Let $S$ be a (compact, connected) Riemann surface of genus $g$ and $f: S\to \mathbb CP^1$ be a degree $d$ meromorphic function. Then the Riemann-Hurwitz formula tells us that the number of ramifications of $f$ counted with multiplicity is $2(d-1)+2g-2$.

Suppose we consider instead a map $\varphi: S\to \mathbb CP^1$ of degree $d$ that is smooth, but not necessarily holomorphic. It will have singularities, like folds, etc.

Question. Can one express the number $2(d-1)+2g-2$ as a sum of contributions, involving various types of singularities of the map $\varphi$ (say, if $\varphi$ is analytic)?

  • $\begingroup$ Surely yes if the map has finitely many singularities, so we can compute the Euler class of the of the tangent bundle of $S$ in terms of the pullback of the Euler class of the tangent bundle of $\mathbb P^1$ and local contributions at singularities. $\endgroup$ – Will Sawin Dec 18 '19 at 21:57
  • $\begingroup$ Dear Will, thanks for this comment. Let's even assume for simplicity that the map is analytic or has only simple singularities (in the sense of Arnold). Do you have an idea of what will be the formula? $\endgroup$ – aglearner Dec 18 '19 at 22:54
  • $\begingroup$ If $x$ is a singularity of $f$, then a loop around $x$ will be sent to some kind of curve in a small neighborhood of $f(x)$, which for a sufficiently small loop and an analytic map won't touch $f(x)$. The local term should be the winding number minus one. $\endgroup$ – Will Sawin Dec 18 '19 at 23:12
  • $\begingroup$ But 1) folds are not isolated. 2) consider the map $z\to z^2$ and perturb it by a real linear map. The resulting map will have three cusps. Do you say that each of them contribute $\pm 1$? (I don't say this is not so, just curious). Anyway, do you think you can prove what you write in the comment? $\endgroup$ – aglearner Dec 18 '19 at 23:21
  • $\begingroup$ Will, this formula with the winding number can't work. Consider the map $(x,y)\to (x^2, y)$ then the contribution of each point with $x=0$ is equal to $-1$ according to the formula you propose... $\endgroup$ – aglearner Dec 19 '19 at 0:04

This is not possible, just using the singularities and the homological degree of the map.

Let $f_1$ be a generic degree $3$ holomorphic map $\mathbb P^1 \to \mathbb P^1$. It has four quadratic singularities. Let $E$ be an elliptic curve, and $f_2$ a degree $2$ map $E \to \mathbb P^1$, with four simple ramification points singularities.

Let $f_1'$ and $f_2'$ be $f_1$ and $f_2$ with the orientations of the source reversed.

Then we can glue $f_1$ to $f_1'$, and $f_2$ to $f_2'$, by cutting out a disc from each component, lying over the same disc of $\mathbb P^1$, and gluing the holes together, introducing orientation-reversing fold singularities.

These glued maps are from Riemann surfaces of genus $0$ and $2$ respectively, but have the same singularities, being four positive-orientation simple ramification points, four negative-orientation simple ramification points, and a loop of orientation-reversing folds, and the same degree, $0$.

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  • $\begingroup$ That's cute! Though not what I was hoping for :) ... I guess, in the end I would like to be able to find the degree of the map, provided I know the genus. $\endgroup$ – aglearner Dec 19 '19 at 8:54
  • $\begingroup$ @aglearner You can build an example with the same genus but opposite degrees using the same gluing-opposite-orientation strategy. $\endgroup$ – Will Sawin Dec 19 '19 at 14:37
  • $\begingroup$ Will, that is interesting. I wonder still if there is a positive statement. For example, singularities of maps $S\to \mathbb CP^1$, $S\to \mathbb CP^1$ of degrees $10$ and $20$ can not be the same. I would find it really surprising anything like this exists. $\endgroup$ – aglearner Dec 19 '19 at 22:33

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