Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

This question is sort of a further extension to this question I have been asking,

Relation between n-tuple points on an algebaric curve and its pre-image in the normalizing Riemann surface

It seems that there is a method of constructing the Riemann surface of a function and its not clear to me that it is related to the idea of finding a "normalization". In the way I see the method of construction of a Riemann surface from a function is roughly like below -

First one writes the function $f$ as a polynomial of degree $n$ in (say $x$) whose coefficients are rational functions in say $y$. Then from $\mathbb{P}^1$ one removes the points where the coefficients have poles or where the discriminant of the given function is $0$. Then for any $y_0$ in this punctured sphere there exists an open disk around it such that in that disk one can solve the equation to get $n$ function elements of the form $(x_\mu(y_0),y_0)$ (for $\mu =$ 1,2,...,n) such that $f(x_\mu(y_0),y_0)=0$ These analytic elements can be continued on the sphere by monodromy such that their extension depends only on the homotopy class of the loop in the punctured sphere.

Its often not clear in the texts whether after doing this one has to - like in finding of the normalization - create symmetric functions of these local function elements which can be globally defined and unambiguous upto crossing a chosen line which has been removed from the sphere and which passes through all the punctures.

  • What is the correspondence here?

Now one is supposed to imagine that these $n$ local function elements are denoting the n-sheets of the Riemann surface over this Riemann sphere. And the removed points on the sphere lead to produce the ramification points of the projection map from the Riemann surface to this sphere.

  • Can anyone kindly make the above step precise?

One of the many things that confuses me above the above construction is that it seems to say that if I start with a degree $d$ algebaric curve then I will necessarily get the Riemann surface as a $d$ sheeted cover of $P^1$. But that is possibly not the case.

  • How do the number of sheets get reduced in the above construction?

Like I guess it is true that if there is a degree $d$ algebraic curve with a point of multiplicity of $d-2$ then it is necessarily normalized by a hyperelliptic Riemann surface which is necessarily a $2$ sheeted branched cover of the sphere.

One way I can imagine that is as follows - for every point on the Riemann sphere take its image point (under the normalization map) on the algebraic curve and try to pass a line through this point and the multiplicity point. Now this line may or may not pass through another point on the curve depending on whether this image point is a simple intersection between the curve and the line or a double intersection. In either case map the point on the Riemann surface to this line and if it happened through a simple intersection then there should exist another point on the Riemann surface which maps to this same line. The space of lines in $\mathbb{P}^2$ is $\mathbb{P}^1$ and hence I guess one has realized the Riemann surface as a ramified two sheeted cover of the sphere. (Though I don't know how to establish holomorphicity of this ramified covering map)

But doing the initial kind of argument one would have thought that one is getting a $d$ sheeted cover of the sphere!

  • Can someone kindly help reconcile these two points of view?
share|improve this question

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.