Why the number of points where $f$ ramifies is finite?

Question

I know roughly that there is a theorem in complex analysis saying that if $f$ has degree $e_x>1$ at point $x$, which is $f(z)=(x-z)^{e_x}g(z)$, then $f^{-1}(y)$ has $e_x$ different preimages in a neighborhood of $y$.

In a complex Riemann surface, we have the same result considering $f$ in the local coordinates.

We call that $f$ ramifies at point $x$ if $e_x>1$.

I was told that the number of points where $e_x>1$ is finite.

Can anyone tell me why? Is this result valid only in compact Riemann surface or generally?

update:

I realized that $f$ has digree $e_x$ at $x$ sould be $f(z)-f(x)=(z-x)^{e_x}g(z)$ rather than what I originally posted.

Sorry for trouble..

Doesn't $\sin^2 z$ have an infinite number of points $x$ where $e_x > 1$? — Antonio Vargas, Jan 09 '13 at 02:09
@AntonioVargas, so the result may be valid only on compact set? — hxhxhx88, Jan 09 '13 at 02:19
@AntonioVargas, just consider the Riemann Hurwitz formula, the $\sum_{x\in X}(e_x-1)$ in the formula has only finite non-zero terms. That is my question.... — hxhxhx88, Jan 09 '13 at 03:09

score 16 · Answer 1 · answered Jan 09 '13 at 03:40

If $f:X \to Y$, a non-constant holomorphic map of Riemann surfaces, has ramification degree $e$ in a n.h. of a point $x$, then we may choose local coords. in the source and target so that $x$ and $f(x)$ both lie at the origin, and $f$ is given by $z \mapsto z^e$. Note that every point other than zero in this n.h. of $x$ is not a ram. point. So every point $x$, whether or not it is a ramification point (i.e. whether or not $e > 1$) has a n.h. $U$ so that $U\setminus \{x\}$ contains no ramification points.

Conclusion: the set of ramification points is closed and discrete.

In particular, if $X$ is compact, then the set of ramification points is compact and discrete, thus finite. If $X$ is not compact, then the set of ramification points need not be finite (as the examples already given in the comments show).

Your answer is quite beautiful, thanks! – hxhxhx88 Jan 09 '13 at 05:13 — hxhxhx88, Jan 09 '13 at 05:13

score 7 · Accepted Answer · answered Jan 09 '13 at 03:23

7

Suppose there are infinitely many $x$ for which $e_x>1$. Any infinite subset of a compact set has an accumulation point in the set, so $f(z) = 0$ on a sequence which converges in your domain and must be zero by the identity theorem.

If $f$ is nonzero, it must have only finitely-many such zeros.

Now, if the domain isn't compact, then $f$ may indeed have infinitely many such $x$, for example if $f(z)=\sin^2 z$.

answered Jan 09 '13 at 03:23

Antonio Vargas

24,993

1

I'm sorry I think I mistook the meaning of degree in my post, it should be $f(z)-f(x)=(z-x)^{e_k}g(z)$. But your argument is quite fine because if $e_x>1$ for infinitely number of $x$, we can consider $f'(z)$ to say it must be $0$, then $f$ must be constant. anyway, thank you! – hxhxhx88 Jan 09 '13 at 03:38

Why the number of points where $f$ ramifies is finite?

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