Basic question about branch points on Riemann surfaces

Question

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)?

Answer 1

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 $P_n(z)=q_n + (z-1/n)^2$.
A theorem due to Mittag-Leffler 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 Mittag-Leffler used above is not so easy to find in the literature (I just checked). It is proved in Ash-Novinger'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 non-noetherian domain, not a UFD but any collection of elements of $\mathcal O(D)$ has a GCD and all its finitely generated ideals are principal.

Answer 2

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 $f(z) = (z-1/n)^2$.

Answer 3

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).

Answer 4

No: $f: \mathbb{C}\setminus \{0\} \to \mathbb{C}$, $f(z):= \frac{1}{z} sin (z)$.

Answer 5

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 $\sqrt{(z-z_1)(z-z_2)(z-z_3)}dz$. 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 $(z-z_k)^{1/2}dz$ and therefore the primitive is like $(z-z_k)^{3/2}+cte$.) 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 $\zeta=(z-z_k)^{3/2}+cte$ 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 $y^2=(z-z_1)(z-z_2)(z-z_3)$ 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...

Answer 6

(I'll just say $F$ instead of $f$.)

You forgot to say $F$ is non-constant. Then again, I guess $Ram(F)$ is not defined for $F$ non-constant. 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 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. ($F$ is surjective if $Y$ is connected)

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).

Now let's see $X$ non-compact:

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 $X$ non-compact, $Ram(F)$ 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.