Is the Riemann surface of the square root function compact?

Question

I would like to understand the theory of Riemann surfaces, and I have a small question. Is the Riemann surface determined by the square root function compact? And if it is compact, what is its genus?

Answer 1

If you mean the Riemann surface of the square root over the Riemann sphere, yes, it is compact (see below). If just over the complex plane (thus, not including $\infty$), then it isn’t. Genus is generally defined only for compact surfaces, or else the genus of a noncompact surface is defined to be the genus of its compactification. The genus of the square root surface is 0 (see below).

Let $S$ be the Riemann sphere. The Riemann surface of the square root, call it $R$, has two points over every point of $S$, except for the so-called branch points 0 and $\infty$, where there is only one. Also, the points in $R$ lying over $z\in S$ are labeled $\pm\sqrt{z}$.

Topologically, $R$ and $S$ are equivalent. It is easiest to construct a homeomorphism $f:S\to R$. Just send a point $z\in S$ to one of the points lying over $z^2$. The map $z\mapsto z^2$ from $S$ to $S$ is 2-1, which is why we have two choices for a square root (except for 0 and $\infty$). But the surface $R$ is provided with two points above $z^2$, so $f$ is 1-1.

This is easiest to visualize if we restrict attention to the unit circle of $S$. Here, $z=e^{i\theta}\mapsto z^2=e^{2i\theta}$. Imagine two dots, $z_+$ and $z_-$, leaving 1 and traversing the semicircles: $z_+$ takes the upper semicircle and $z_-$ takes the lower semicircle. Meanwhile on $R$, we have dots $p_+$ and $p_-$. These are not above $z_+$ and $z_-$; rather, $p_+$ stays above $z_+^2$ and $p_-$ stays above $z_-^2$. When $z_+$ reaches $+i$ and $z_-$ reaches $-i$, $p_+$ is above $-1$ on one sheet of $R$ and $p_-$ is above $-1$ on the other sheet. Continuing around the circle, when both $z$'s reach $-1$, the two $p$'s reunite above 1. We have mapped the unit circle of $S$ homeomorphically onto the "doubly wrapped" circle lying above it on $R$.

You can do the same thing for any circle of radius $r$ in $S$ (with center at the origin), but then the $p$'s will lie above the circle of radius $r^2$. Here $0<r<\infty$. For $r=0$, i.e. for $z=0$, we have only one point lying above it, and $f$ maps 0 to that point. Likewise for $z=\infty$.

Now label each point $p$ with the $z$ it came from. Since $p$ lies over $z^2$, that means $p$ is labeled with the square root of the point under it. But the signs have been chosen continuously.

Since $z\mapsto z^2$ is continuous (even at 0 and $\infty$), and the map is a bijection, we have a homeomorphism provided the inverse map is also continuous. Which it is. This can be seen in a couple of ways, but intuitively, the inverse is obtained by dropping down from $p$ to $z^2$ under it and then taking the square root.

Since $R$ and $S$ are topologically identical, $R$ like $S$ is compact and has genus 0. If you delete $\infty$ from both $S$ and $R$, you are left with noncompact surfaces.

(There ought to be a good youtube video illustrating this, but I did a quick search and found nothing satisfying.)

Finally, suppose we attach two labels to each point of $R$. Since $p$ lies over $z^2$ but comes from $z$, label the point $p$ with the pair $(z,z^2)$. Or setting $w=z^2$, $(z,w)$. So we have two functions from $R$ to $S$: $p\mapsto z$ and $p\mapsto w$.

It is a general fact that any compact connected Riemann surface $R$ is a multi-sheeted covering of the Riemann sphere $S$, with two complex-valued functions (i.e., maps from $R$ to $S$), traditionally denoted $z$ and $w$. (For $S$ itself, "multi"="one".) Furthermore the pair $(z,w)$ satisfy an algebraic equation. In general, though, $R$ will not be homeomorphic to $S$.