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In Miranda's "Algebraic Curves and Riemann Surfaces," there is a problem that asks the reader the show, using the compactness of a Riemann Surface $X$, that there are only a finite number of global meromorphic functions separating points and tangents on $X$? We are assuming that the field of rational functions on $X$ separates points and tangents, so there's no need to show any existence.

I've been interpreting this question to mean that there are only a finite number of global meromorphic functions on $X$ which take different values at every point on $X$ and for which, at every point $p$ in $X$, the function is either holomorphic at $p$ with the order of $f - f(p)$ at $p$ equal to one or has a simple pole at $p$, using the definitions from the text about what it means for the entire field of meromorphic functions to separate points and tangents.

Is this interpretation accurate, or am I misreading? Given that my interpretation is accurate (and I suppose even if it's not), what is the quickest solution?

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  • $\begingroup$ The first thing to show is that for $X$ compact, if its field of meromorphic functions separate the points (ie. $p\ne q\implies \exists f,f(p)\ne f(q)$) then there are finitely many meromorphic functions such that the map $p \to [f_0(p):\ldots:f_n(p)]$ is injective (can be improved to embedding which involves your simple zero/pole-local chart idea) from $X$ to $\Bbb{P}^n$. It reduces more or less to show that $\Bbb{C}(X)/\Bbb{C}(f_0)$ is a finite extension. $\endgroup$ – reuns May 1 '20 at 20:53
  • $\begingroup$ Would you care to elaborate? Admittedly, I don't see how showing those things implies what I need. The chapter shows that the space you're mentioning is finite dimensional towards the end (assuming your top space is the meromorphic functions field on $X$). How does that conclude? $\endgroup$ – Johnny Apple May 2 '20 at 7:43
  • $\begingroup$ I was never able to solve this. The question didn't make sense to me without assuming finite with respect to linear independence. $\endgroup$ – Johnny Apple May 19 '20 at 17:16
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I belive the idea is the following :

So let $X$ be the algebraic curve, first to see that it there are a finite number of functions that separate tangents we notice that the condition for separating tangents is equivalent to the condition that at every $p$ there is a local coordinate chart around $V_p$ so you can take these open sets and cover $X$ and retrieve a finite number of them since $X$ is compact and we will have the finite number of functions associated to these open sets will do what we desire.

Now to see that it separates points we know that for every $x,y\in X$ there is $f_{x,y}$ such that it separates $x$ and $y$, and so since the function is continuous there are neighborhoods $V_x$ and $V_y$ such that $f_{x,y}$ separates $V_x$ from $V_y$, now the idea is to try and cover $X\times X$ with $V_{x,y}=V_x\times V_y$ the only problem is that this open sets don't get the diagonal and for this we need the separating tangents condition again to find $f_x$ such that is bijective in a neighborhood of $x$ $V_x$ and we consider $V_x\times V_x$. Now we take a finite subcover since $X\times X$ is compact and the meromorphic functions associated to each $V_{x,y}$ will separate the points.

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  • $\begingroup$ How is separating tangents the same as having charts? We have charts automatically, since X is a Riemann Surface. $\endgroup$ – Johnny Apple Jun 17 '20 at 17:26
  • $\begingroup$ Yes but If you read the beggining of chapter 7 the author does mention it , basically when you have a function with multiplicity one it means that the associated holomorphic to the Riemann sphere will have multiplicity one so it will look like $z$ and we can see that this will gives us that locally around $p$ ,$f$ will be a coordinate chart.@JohnnyApple $\endgroup$ – Something Jun 17 '20 at 18:29
  • $\begingroup$ I suppose I'm also not sure of the premise of your answer. You start with letting X be an algebraic curve, and the show "it" separates points and tangents. What is "it"? Algebraic curves separate points and tangents by definition. $\endgroup$ – Johnny Apple Jun 17 '20 at 18:35
  • $\begingroup$ Sorry I mean that we will have a finite number of functions that separate tangents I will edit it. $\endgroup$ – Something Jun 17 '20 at 18:40

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