Motivation behind defining the Ramification Divisor - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-18T06:01:33Z http://mathoverflow.net/feeds/question/77180 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/77180/motivation-behind-defining-the-ramification-divisor Motivation behind defining the Ramification Divisor Anirbit 2011-10-04T22:06:35Z 2011-10-06T14:26:26Z <p>I would like to understand what exactly is the motivation for defining the notion of a ramification divisor of a function. </p> <p>As I see the definition, </p> <p>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 $p$. </p> <ul> <li><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..)</p></li> <li><p>Does anything in the above definition depend on $X$ or $X'$ being compact? </p></li> </ul> <p>Now for a similar map $f$ one defines its ramification divisor ($R_f$) as $R_f = \sum _{p \in X} (\nu_f(p) - 1)p$ </p> <ul> <li><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. </p></li> <li><p>Also this definition almost exclusively seems to be used when $X$ and $X'$ are compact Riemann surfaces. Is that somehow necessary? </p></li> </ul> <p>{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.} </p> <ul> <li><p>Why that "-1" in the definition? Is $\nu_p(f)$ always greater than $1$ ? </p></li> <li><p>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? </p></li> <li><p>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 $X'$? </p></li> <li><p>Finally a curiosity - is there a "simple" way to see the Riemann-Hurwitz formula <em>without</em> using the Poincare-Hopf formula? </p></li> </ul> http://mathoverflow.net/questions/77180/motivation-behind-defining-the-ramification-divisor/77228#77228 Answer by Ramsey for Motivation behind defining the Ramification Divisor Ramsey 2011-10-05T12:22:45Z 2011-10-05T12:22:45Z <p>A few answers:</p> <ul> <li>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.</li> <li>Ramification is a local phenomenon, so compactness is totally irrelevant.</li> <li>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).</li> <li>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!</li> </ul> http://mathoverflow.net/questions/77180/motivation-behind-defining-the-ramification-divisor/77282#77282 Answer by Margaret Friedland for Motivation behind defining the Ramification Divisor Margaret Friedland 2011-10-05T19:31:21Z 2011-10-06T14:26:26Z <p>A few more answers: </p> <p>$h$ can be made to be a unit (=nowhere zero) in a small enough neighborhood; this is just manipulating the Taylor series in local coordinates.</p> <p>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.</p> <p>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. </p> <p>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 </p> <p>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 </p>