If $X$ and $Y$ are Riemann surfaces (not necessarily compact), and $f:X\to Y$ is a holomorphic function, then it is obvious that the ramification points of $f$ in $X$ form a discrete subset of $X$. Is the same true of the branch points of $f$ (the set made up of the images of the ramification points)?
6 Answers
Dear Robert, there exists a holomorphic function $X\to Y $ having non discrete and even dense set of branch points, with $X=\mathbb C^\ast \setminus \{0\}$ and $Y=\mathbb C$.
Consider an enumeration $(q_n)$ of $\mathbb Q$ and the polynomials $P_n(z)=q_n + (z1/n)^2$.
A theorem due to MittagLeffler says that there exists a holomorphic function
$f:\mathbb C^\ast \setminus \{0\} \to \mathbb C$ whose Taylor development at $1/n$ is $P_n(z)$. The $q_n=f(1/n)$ , that is all of $\mathbb Q$, are then branch points of $f$.
Bibliography and comments The version of MittagLeffler used above is not so easy to find in the literature (I just checked). It is proved in AshNovinger's Complex Variables ( theorem 6.3.3 ) where they deduce from it some algebraic properties ( due to Helmer) of the ring $\mathcal O(D)$ of holomorphic functions on an open connected $D\subset \mathbb C$. It is a nonnoetherian domain, not a UFD but any collection of elements of $\mathcal O(D)$ has a GCD and all its finitely generated ideals are principal.

$\begingroup$ Dear Georges, there is something I do not understand. Locally around a branch point, in suitable coordinates a holomorphic map between Riemann surfaces can be written as $z \to z^n$, where $n$ is the ramification number (see for instance FarkasKra, p. 12) so it seems to me that the branch locus should be actually discrete. Am I missing something? $\endgroup$ Apr 5, 2011 at 13:13

$\begingroup$ Dear Francesco, I think it is "the suitable coordinates" which make it a little difficult to follow. Put $w=zq_n$ at the target and $\zeta=z1/n$ at the source . Then locally my map is described by $\zeta\mapsto w=\zeta^2$, as it should. The confusion arises because we constantly replace the values of $f$ near $f(b)$ by values near $0$ by surreptitiously changing $w$ to $wf(b)$ and calling that "suitable coordinates"! $\endgroup$ Apr 5, 2011 at 13:35

$\begingroup$ Dear Georges, I'm still not completely convinced. If the map $f$ can be locally described as $w=\zeta^2$ around $w=0$, then there is a neighborhood of $w=0$, i.e. a neighborhood of $z=q_n$, where $f$ is not branched (since outside $w=0$ we have two distinct preimages). But this seems to contradict your assertion that $f$ has dense branch locus... $\endgroup$ Apr 5, 2011 at 13:49

$\begingroup$ I find it weird that Miranda "assumes" that branch points are discrete because ramification points are discrete. I guess it's just an error in the text maybe? (page 46 at the top) $\endgroup$– rfauffarApr 5, 2011 at 14:10

1$\begingroup$ Dear Francesco, here is the subtle point which (I hope!) will take care of your objection. What you say is true but some points near $q_n$ are branch points of $f$ not because of the behaviour of $f$ near $1/n$, which as you correctly noted is étale at points near to but different from $1/n$, but because of some nasty far away $1/N$ such that $q_N$ happens to be very close to $q_n$. Once again this is hidden by those local coordinates which give us the feeling that $0$ is some determined point whereas all sorts of points get called $0$ the minute you focus your attention on them! $\endgroup$ Apr 5, 2011 at 14:34
This is true if $f$ is proper (the preimage of a compact is compact). Indeed, any compact neighborhood of $y\in Y$ contains only finitely many branch points because its preimage in $X$ contains only finitely many ramification points.
If $f$ is not proper, ramification points in $Y$ may not be discrete. Take for $X$ the union of copies of $\mathbb C$ indexed by $n=1,2,3,...$ and let $f:X\to \mathbb C$ be the holomorphic map which on the $n$th copy of $\mathbb C$ in $X$ is given by $f(z) = (z1/n)^2$.

1$\begingroup$ But in your example, $X$ isn't connected. I understand if the preimage of a compact set is compact. What if $X$ is connected? Many books simply say that branch points form a discrete set, but don't give any proof (Miranda, Algebraic Curves and Riemann Surfaces, for example). $\endgroup$– rfauffarApr 5, 2011 at 12:28

$\begingroup$ Dear Xandi, I think your map has $f(1/n)=0$ as its only branch point (=critical value). $\endgroup$ Apr 5, 2011 at 13:42

Another very concrete answer but without any formula:
I suppose you know how to construct "with papers, scissors and glue" a Riemann surface with a single branch point of order 2 (for example the surface of the function $\sqrt z$). Now take the Riemann surface of the logarithm. It has a countable infinite number of sheets. On each sheet you can add "with papers, scissors and glue" a branch point of order two, and this at any place you wish except above the origin. In this way you construct, for any given countable set $A\subset\mathbb C$ a Riemann surface $f : X \to \mathbb C$ which has a branch point above every point of A.
Moreover you see in the same way, that you can prescribe any (finite or infinite) order to each branch point (just glue more sheets, as you would do for $\sqrt[n]z$ or $\ln$); and you can also prescribe the number of branch points you want to have above each point of A (above each point of A you may want to have a countable number of distinct branch points).
No: $f: \mathbb{C}\setminus \{0\} \to \mathbb{C}$, $f(z):= \frac{1}{z} sin (z)$.

$\begingroup$ Yes, or $f:\mathbb C\to \mathbb C$, $f(z)=e^zsin z$. $\endgroup$ Apr 5, 2011 at 19:27

$\begingroup$ Dear Johannes, I had thought of that function, but it was not clear to me that the set of branch points is not discrete. How do you show that ? $\endgroup$ Apr 5, 2011 at 20:42

$\begingroup$ Dear Tom, according to my calculations, the branch points of your function do form a discrete set, with $0$ not a branch point but the unique point of accumulation of that branch locus. [ I may have made a stupid mistake, of course. Do you agree that the the set of critical points is $\pi /4+ \pi \mathbb Z$ ?] $\endgroup$ Apr 5, 2011 at 20:50

$\begingroup$ George, you are right. I was not paying attention and was thinking "closed and discrete" rather than "discrete". $\endgroup$ Apr 5, 2011 at 22:36

$\begingroup$ $e^{i(\alpha x+sin(x))}$ where $\alpha$ is irrational and between $0$ and $1$. $\endgroup$ Apr 6, 2011 at 0:23
Here is another example using only algebraic and not transcendental functions. Take three distinct complex numbers $z_1,z_2,z_3$ and let $X$ be the Riemann surface obtained by integrating the form $\sqrt{(zz_1)(zz_2)(zz_3)}dz$. The map $g: X \to\mathbb C$ which sends each germ $\varphi_z\in X$ to $z$ is a covering with infinitely many sheets above the $z$plane. (The ramification points are exactly above $z_1,z_2,z_3$, and there are infinitely many; each one is of order 2. This is so because near $z_k$ the form is like $(zz_k)^{1/2}dz$ and therefore the primitive is like $(zz_k)^{3/2}+cte$.) Now consider the holomorphic map $f: X\to\mathbb C$ which sends each germ $\varphi_z\in X$ to $\varphi_z(z)$. The branch points of $f$ are exactly the same as the branch points of $g$, but whilst they are of order 2 for $g$ they are of order 3 for $f$. (To see why this is so resolve the equation $\zeta=(zz_k)^{3/2}+cte$ in $z$.) Now comes the important point: For nearly all choices of $z_1,z_2,z_3$ the periods $a_1,a_2,a_3$ of $ydz$ on the elliptic curve $y^2=(zz_1)(zz_2)(zz_3)$ are such that the lattice $\mathbb{Z}a_1+\mathbb{Z}a_2+\mathbb{Z}a_3$ is dense in the plane. This means that the set of the images of the branch points of $f$ is dense in the plane.
In this way you can construct many other examples...

$\begingroup$ My example must be modified in order to work: instead of taking three distinct points take five (thus replacing the elliptic curve by a hyperelliptic one to assure that there are really three periods for the form ydx). $\endgroup$– MathOManApr 22, 2011 at 14:57
(I'll just say $F$ instead of $f$.)
You forgot to say $F$ is nonconstant. Then again, I guess $Ram(F)$ is not defined for $F$ nonconstant. Or you just say $Ram(F)$ is empty or something I guess.
First let's see $X$ compact:
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 nonconstant 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. ($F$ is surjective if $Y$ is connected)
In particular, this means we do not use that $F$ is proper, closed, open, surjective, nonconstant 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).
Now let's see $X$ noncompact:
The key is not really that $Ram(F)$ is discrete but that $Ram(F)$ is closed discrete. (Closed for the same reason it is discrete!) That $Branch(F)$ was discrete for the case of compact $X$ is more of a corollary of that $Branch(F)$ was finite.
With $X$ noncompact, $Ram(F)$ is no longer necessarily finite and thus $Branch(F)$ is no longer necessarily finite. You might have to now use those other conditions on $F$, $X$ or $Y$, and those might not be enough. You might have to assume further that $F$ is perfect or something. However, instead of finding out conditions for mapping discrete to discrete, we might find out conditions for mapping closed discrete to discrete because maybe that $A$ is closed could still be relevant even if $X$ no longer compact.

$\begingroup$ It is not clear how this adds anything to the answers that were already given $\endgroup$ Nov 29, 2020 at 18:21

$\begingroup$ @YemonChoi I emphasise the closedness. The closest thing this answer is to what I've seen to answers and comments already given is 'George, you are right. I was not paying attention and was thinking "closed and discrete" rather than "discrete". – Tom Goodwillie' $\endgroup$– BCLCDec 3, 2020 at 23:22

$\begingroup$ @YemonChoi Is there anything incorrect in my answer? $\endgroup$– BCLCDec 3, 2020 at 23:22

$\begingroup$ Have you read the answer by Georges Elencwajg to the original question? Your answer starts with a fairly trivial observation that closed subsets of compact spaces which are discrete in the subspace topology must be finite, and then observes that this reasoning will not suffice to answer the original question if X is not compact. But you do not actually answer the original question, while Georges has. If you are seeking to correct or clarify something said in one of the other answers then you should leave a comment on that answer. New answers should only be given if they contribute something $\endgroup$ Dec 5, 2020 at 0:46