2
$\begingroup$

I refer to Chapter II.4 of Rick Miranda - Algebraic curves and Riemann surfaces, which I understand says that the branch points of a nonconstant holomorphic map $F: X \to Y$ between Riemann surfaces $X$ and $Y$, both of which are not necessarily compact, form a discrete subset of range $Y$ of the map $F$. (Not sure if 'connected' is part of the definition of Riemann surfaces in other textbooks, but it is here.)

Question 1: Do I understand correctly, and is this indeed true?

What else I understand:

  1. $A$ is a discrete subspace/subset (I just assume everything here is the usual subspace topology) of $B$ if and only if every singleton subset of $A$ is open in $A$ if and only if every point of $A$ is isolated.

  2. This overflow question (Basic question about branch points on Riemann surfaces) seems to suggest this is not true but is true if $F$ is proper. However, I'm not sure the definitions of Riemann surface are the same (maybe some answers use definitions where Riemann surfaces are not necessarily connected or something).

    • 2.1. I know $F$ is 'discrete' (meaning that its fibres are discrete; here, i refer to a different text): see definition before Lemma 3.1, on p. 7), by Proposition II.3.12, and open, by open mapping theorem, which is Proposition II.3.8, but I'm not sure these help show $F$ is proper or are otherwise helpful. Ostensibly 'discrete' maps don't map discrete subsets to discrete subsets.
  3. This stackexchange question (Are branch points always isolated?) seems to suggest this is true.

  4. I think this is true if $F$ is injective or at least something like locally injective. (Please don't make me type up all the stuff I did for this part.) However based on the question in (3) and based on the '3.2. Branch points' on p. 7 in text by Armin Rainer in (2.1), which I think uses the term 'branch points' to mean the same thing as what Miranda means by 'ramification points' (and thus is different from Miranda's 'branch points'), it seems that no neighbourhood $U$ of a ramification point $p \in U \subseteq X$ of $F$ is such that the restriction $F|_U$ is injective.

  5. This page (rigtriv: Hurwitz’s Theorem) says

Now, the ramification and branch points must form a discrete set

I don't think the page assumes anything like $F$ is proper.

  1. The Armin Rainer text in (2.1) actually seems to assume proper based on the '3.7. Proper holomorphic maps' part, but said part refers to 'Lemma 3.17', so I think proper can be replaced with closed. Maybe $F$ is closed or something.

  2. S. K. Donaldson - Riemann Surfaces says

If $F$ is proper then the image $\Delta = F(R)$ is discrete in $Y$.

  • This seems to have all the same assumptions as with Miranda. Ostensibly, either Miranda is wrong or there's some higher level machinery that allows us to omit $F$ proper.

Question 2: To possibly generalise this, what are the minimum requirements on $F, X$ and $Y$ to make $F$ map discrete subsets to discrete subsets? I mean, I'm not sure we really need 'holomorphic' here. I have a feeling this applies perhaps to maps that are just open/closed, continuous and discrete and spaces that are just Hausdorff and locally compact or something.

$\endgroup$
10
  • 2
    $\begingroup$ Keep in mind that the ramification points are in $X$, while branch-points are in $Y$. The set of branch-points need not be discrete (if $F$ is not proper). If you are quoting Miranda correctly, he made a mistake here. Properness is exactly the right condition to ensure that discrete subsets map to discrete subsets. $\endgroup$ Sep 4 '20 at 22:50
  • $\begingroup$ @MoisheKohan Thanks! 1. Do you disagree with Federico Fallucca? 2. Also is it correct to say that that $F$ is discrete does NOT necessarily imply that $F$ maps discrete to discrete? (kind of weird name for a map but eh) $\endgroup$ Sep 5 '20 at 6:00
  • 1
    $\begingroup$ Discrete maps in general do not send discrete sets to discrete sets. The written answer is fine, just does not address the issue of branch points (it only deals with ramification points). $\endgroup$ Sep 5 '20 at 17:30
  • 2
    $\begingroup$ I do not know what you think, but what is written about branch points is plain wrong. $\endgroup$ Oct 31 '20 at 4:01
  • 3
    $\begingroup$ I did not notice the incorrect claim about branch-points in his answer. However, as I said, what he wrote about ramification points is correct. $\endgroup$ Oct 31 '20 at 14:13
4
$\begingroup$

If you want an explicit counter-example to Miranda's claim about discreteness of the set of branch-points, consider the entire function $f(z)=e^z (1+\sin(z))$ as a map ${\mathbb C}\to {\mathbb C}$. The set of branch-points will accumulate to zero and the latter is a branch-point. One can even construct examples where the set of branch-points is dense in the complex plane. I suspect, Miranda had in mind algebraic maps between complex algebraic curves but forgot to mention this assumption.

$\endgroup$
0
$\begingroup$

In general for any map $F: X \to Y$ of any topological spaces $X$ and $Y$ with $X$ compact and $Y$ Fréchet/T1 and for any closed discrete subspace $A$ of $X$, we have $F(A)$ discrete.

Proof: Closed discrete subspaces $A$ of compact is finite $\implies$ $A$ is finite $\implies$ $F(A)$ is finite $\implies$ $F(A)$ is discrete because finite subspaces of Fréchet/T1 are discrete. QED

Apply this to the case of $A=Ram(F)$ when $F$ is a non-constant holomorphic map between connected Riemann surfaces with $X$ compact (and thus $F$ is surjective, open, closed and proper and $Y$ is compact) to get $F(A)=Branch(F)$ is discrete.

In particular, this means we do not use that $F$ is proper, closed, open, surjective, non-constant or holomorphic or that $X$ is connected or that $Y$ is connected. We can relax this to $X$ compact (and not necessarily Riemann surface) and $Y$ Fréchet/T1 (and not necessarily Riemann surface, Hausdorff/T2 or compact).

I think finding conditions to make $F$ map discrete to discrete is another story. Actually, what we want is for $F$ to map closed discrete to closed discrete or at least closed discrete to discrete.

$\endgroup$
2
  • 1
    $\begingroup$ Yes, this is correct. $\endgroup$ Oct 31 '20 at 14:07
  • $\begingroup$ Thanks a lot, @Moishe Kohan ! $\endgroup$ Nov 3 '20 at 14:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.