[Math] Motivation behind defining the Ramification Divisor

ag.algebraic-geometrycomplex-geometrycv.complex-variablesriemann-surfaces

I would like to understand what exactly is the motivation for defining the notion of a ramification divisor of a function.

As I see the definition,

If $f$ is a meromrophic function between two Riemann surfaces – say $X$ and $X'$ then let $\nu_p(f)$ be the ramification (or order) of the function $f$ at $p$. Basically if one is working in local coordinates such that $z(p)=0$ then $f$ in a neighbourhood of $p$ looks like $f=z^{\nu_p}h(z)$ where $h(z)$ is a holomorphic function which is never $0$ in a neighbourhood of $p$.

  • In the above definition of ramification, can the function $h$ be always set to unity? By choosing coordinate in$X'$ such that $f(p)=0$? (…I am not sure..)

  • Does anything in the above definition depend on $X$ or $X'$ being compact?

Now for a similar map $f$ one defines its ramification divisor ($R_f$) as $R_f = \sum _{p \in X} (\nu_f(p) – 1)p$

  • Its not clear to me whether people define ramification divisors for meromorphic functions too since i almost seem to see the texts exclusively using it in the case of non-constant holomorphic functions. I would be glad if someone can clarify this…may be I am missing something very basic.

  • Also this definition almost exclusively seems to be used when $X$ and $X'$ are compact Riemann surfaces. Is that somehow necessary?

{I guess in all this discussion one has to keep in mind that a holomorphic function on a Riemann surface and a holomorphic function between two Riemann surfaces are defined "differently" – as i see it. I guess there is no analogue of Liouville's theorem in the later case.}

  • Why that "-1" in the definition? Is $\nu_p(f)$ always greater than $1$ ?

  • Let $q \in X'$ and let $p_1$ be a pre-image of $q$ under $f$ with multiplicity of $m_1$. Then I guess one will say that $\nu_f(p_1) = m_1$. Now is it obvious that any "small" perturbation of $q$ can only "split" $p_1$ into $m_1$ points each with $\nu_f = 1$? That nothing else can happen? For "large" enough perturbation to $q$ isn't it possible for many of its pre-images to "join up" and have larger ramifications than initially?

  • consider this set, $p \in X' \vert f^{-1} (p)$ has all points with $\nu_f(p)=1$ (called "simple points"?). Is this set open and dense in $X'$?

  • Finally a curiosity – is there a "simple" way to see the Riemann-Hurwitz formula without using the Poincare-Hopf formula?

Best Answer

A few answers:

  • As the comments mention, the "-1" is certainly needed to get a divisor in the first place, since $\nu_p$ is usually equal to 1 and exceeds 1 at the ramification points. Thus, the support of the ramification divisor as you define it is precisely the ramification locus.
  • Ramification is a local phenomenon, so compactness is totally irrelevant.
  • A meromorphic function on a Riemann surface $X$ can be interpreted as a map $X\to \mathbb{P}^1$ (the poles go to $\infty\in \mathbb{P}^1$ with ramification index equal to the degree of the pole).
  • Here's how I think of/recall the Riemann-Hurwitz formula for $f:X\to X'$: Imagine that you have triangulated $X'$ such that all ramification points (or the images thereof if you think of them on $X$) occur at vertices. Now consider the "pullback" of this triangulation to $X$ (look a the preimages of the faces, edges, and vertices). If you compute the Euler characteristic of $X$ using this pullback triangulation you will see that it differs from the degree of $f$ times the Euler characteristic of $X'$ (computed using the original triangulation) exactly by the degree of your ramification divisor, and the Riemann-Hurwitz formula drops out!
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