Consider Lemma 1.7 from page 60-61 of Miranda's Algebraic Curves and Riemann Surfaces. For a link to the book, see here.

Lemma 1.7. With the above construction, $Z$ is a compact surface of genus $g$. The meromorphic function $x$ on $X$ extends to a holomorphic map $\pi: Z \to \mathbb{C}_\infty$, which has degree $2$. The branch points of $\pi$ are the roots of $h$ (and the point $\infty$ if $h$ as odd degree).

Proof. One checks readily that $Z$ is Hausdorff, and hence is a Riemann surface. $Z$ is compact, since it is the union of two compact sets$$\{(x, y) \in X : \|x\| \le 1\} \text{ and }\{(z, w) \in Y : \|z\| \le 1\}.$$

To me, I don't see why $Z$ is the union of the above two compact sets. Can anyone clarify? Thanks.


1 Answer 1


The two given sets are indeed compact, since $h$ and $k$ are bounded in modulus on the indicated unit disks. Now the point is just that in the first coordinate, the gluing map is $z = \tfrac{1}{x}$, so the exterior of the first set (i.e. $\{ (x,y)\in \mathbb C^2 \, | \, ||x|| > 1 \}$ ) is mapped to $\{ (z,w) \in \mathbb C^2 \, | \, 0 < ||z|| < 1 \}$. Thus every point of $X$ with $||x|| > 1$ is mapped to a point of $Y$ with $||z|| < 1$, and these two compacta cover $Z$.


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