## Motivation behind defining the Ramification Divisor

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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?

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You have a lot of questions. I'll just address just a couple. One way to understand it is look at the  the basic example $y=x^n$, then $dy= nx^{n-1}dx$. Note the exponent is precisely the coefficient of the ramification divisor; so it measures the "differenc" between differentials upstairs and downstairs. Since the issue is local, compactness is not essential, but the divisor might be an infinite sum in general. – Donu Arapura Oct 4 2011 at 22:26
The point is not that $\nu_p(f)$ is always greater than 1 but that it is always greater than or equal to 1 and usually is just 1. Only at $p$ where there is ramification do you have $\nu_p(f) > 1$, so in the definition of $R_f$ you get a finite sum.  If the coefficient was $\nu_f(p)$ then the formal sum would have uncountably many nonzero terms! – KConrad Oct 5 2011 at 2:32
@KConrad Can you kindly explain as to why only at finitely many points is the ramification greater than 1? May be I am missing something basic! I wonder if that is related to there being only finitely many singular points on a compact Riemann surface - but I can't relate the two things. At least in local coordinates it seems that if the ramification is greater than 1 then partial derivatives there will start vanishing and that would then be a singular point by definition. Is that the argument?  – Anirbit Oct 7 2011 at 0:37

## 2 Answers

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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A few more answers:

$h$ can be made to be a unit (=nowhere zero) in a small enough neighborhood; this is just manipulating the Taylor series in local coordinates.

The set of "simple" (=non-ramified) points is open and dense (think in terms of derivatives). If you are interested in the image of the set of ramified points under $f$, Sard's theorem can help.

Regarding small perturbations, "splitting" a ramified point into simple points is not the only possible scenario. Compare two perturbations of $f(z)=z^3$ in $\mathbb{C}$: $z^3+az^2$ and $z^3+a$, $z$ and $a$ in a neighborhood of $0$. I am not sure what you mean by "large" perturbations.

An alternative intuition for Riemann-Hurwitz on compact surfaces if the critical points of the map $f$ are nondegenerate can be given using a Morse function on $X'$; see

MR2126710 (2006a:30040) Stawiska, Małgorzata: Riemann-Hurwitz formula and Morse theory. The $p$p-harmonic equation and recent advances in analysis, 209–211, Contemp. Math., 370, Amer. Math. Soc., Providence, RI, 2005

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@Margaret Thanks for your reply. Isn't the case where you propose to use the Sard's theorem a converse question of what I was asking? I was asking about the possibly open and dense nature of those points in the target Riemann surface whose preimages in the domain Riemann surface are simple (= non-ramified). May be I am confusing but I think you ate looking at the opposite case if images in the range space of the simple points in the domain space. Apologies if I am miseading!  – Anirbit Oct 7 2011 at 0:52
@Anirbit: The set of ramified points is closed, and a non-constant holomorphic map between Riemann surfaces is open. So the image of all nonramified points is open. Sard's theorem can be used to deduce density (the image of the set of ramified points, being of zero Lebesgue measure,  cannot contain an open set). For compact surfaces you don't need this argument, as a holomorphic map has then only finitely many ramified points.  – Margaret Friedland Oct 7 2011 at 17:13
@Margaret Thanks for the reply. Can you explain this in the language of arguing that the points with positive ramification are isolated? That might help understand better.   – Anirbit Oct 8 2011 at 21:50
If you think of ramification points as zeros of the derivative, then of course they are isolated in a complex domain, so you get only finitely many of them.  (This is not true in higher dimensional manifolds, and you need an argument like my previous one.)   – Margaret Friedland Oct 11 2011 at 19:16
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