# Riemann surface with infinite sheets but algebraic branch points

I recently read that all infinitely sheeted Riemann surfaces can be related to exponentials, which I took to mean that they occur when you have logarithmic branch points.

Is it possible to have an infinitely sheeted Riemann surface constructed with only algebraic branch points?

For example a sheet with a square-root branch point connecting it to a sheet above, which in turn is connected via a compact branch cut to a sheet above, which in turn is connected by a square-root branch cut to a sheet above that, and so on infinitely many times.

• Note that there is no such thing as a logarithmic branch point. In the exponential map, the origin is not covered, so there is no branch point there. – Rene Schipperus Dec 20 '17 at 12:40
• So I quess the question becomes is there an infinite sheeted covering (connected) of $\mathbb{C}$ ? – Rene Schipperus Dec 20 '17 at 12:41
• I see. Yes, the question is whether you can have an infinite sheeted covering with only algebraic branch points. – Aran Dec 20 '17 at 12:53

## 1 Answer

The answer is positive but you need to know some elementary algebraic topology (fundamental groups and covering spaces) in order to understand it. Let $Z\subset {\mathbb C}$ be any infinite discrete subset (i.e. a subset with no accumulation points in the complex plane). Then for $X= {\mathbb C}- Z$, we have that $\pi_1(X)$ is a free group on countably many generators $s_k$, one for each point $z_k\in Z$. Define the homomorphism $$\phi: G=\pi_1(X)\to H:=\oplus_{Z} {\mathbb Z}_2,$$ where ${\mathbb Z}_2={\mathbb Z}/2{\mathbb Z}$. There are many such homomorphisms, I will send each generator $s_k$ of $G$ to the generator of the $k$th factor of $H$ (the one indexed by $z_k$). Let $p:\tilde{X}\to X$ denote the (connected) cover associated with the kernel $K$ of $\phi$. I will equip $X$ with the complex structure obtained by pull-back of the complex structure on $X$, making $p$ a holomorphic covering map. Then $p$ has infinitely many sheets (since the group $H$ is infinite and, hence, $K$ has infinite index in $G$). For each $z_k\in Z$, take a small disk $D_k$ centered at $z_k$ and disjoint from the rest of $Z$. Set $D^*_k:= D_k-\{z_k\}$. Then each $p^{-1}(D_k^*)$ is conformal to a (countably infinite) disjoint union of once punctured disks $\tilde{D}_k^*$ such that the restriction of $p$ to each $\tilde{D}_k^*$ is a 2-fold cover. (This cover is conformally isomorphic to the map $z\mapsto z^2$ on the unit disk punctured at the origin). Hence, we can conformally embed $\tilde{X}$ in a Riemann surface $\tilde{C}$ by "filling in" the punctures of the disks $\tilde{D}_k^*$. (This is very standard can be made rigorous if you like.) The map $p$ will extend to a surjective holomorphic map $q: \tilde{C}\to {\mathbb C}$ which is an infinitely-sheeted branch cover of the complex plane, ramified (with local degree 2) only at the points of $Z$.

Lastly, you can make the point of $Z$ to be, say, integers to make them algebraic numbers (if this is what you have meant by "algebraic branch points").

One can modify this construction making the number of branch-points in the (extended) complex plane finite. Namely, start with the infinite group $H$ which has the presentation $$<a,b,c| a^3, b^3, c^3, abc>.$$ Next, let $Z\subset {\mathbb C}\subset S^2={\mathbb C}\cup\{\infty\}$ be a 3-point subset, whose elements are $A, B, C$. As in the above construction, set $X:= S^2 -Z$; then $G=\pi_1(X)$ has presentation $$<\alpha, \beta, \gamma| \alpha \beta \gamma>,$$ each generator corresponds to a small circle around the respective point $A, B, C$. You then have an epimorphism $$\phi: G\to H, \phi(\alpha)=a, \phi(\beta)=b, \phi(\gamma)=c.$$ Let $K$ be the kernel of $\phi$; it is an infinite index subgroup of $G$ since $|H|=\infty$. Now, repeat the construction I described earlier, constructing an infinite cover $p:\tilde{X}\to X$, etc. The only difference will be that you have is that the local ramification numbers have degree 3 rather than 2.