The Riemann surface of $w^{2} \ =\text{sin} \ z$ in $ \mathbb{C}^{2}$ is not the interior of a compact surface-with-boundary

Question

Let $X$ be the Riemann surface of $w^{2} \ =\text{sin} \ z$ in $ \mathbb{C}^{2}$, i.e. let $X = \{(z,w): w^2 = \text{sin} \ z\}$.

The Riemann surface structure on $X$ is obtained by paramertizing by $z$ at all places where $w \ne 0$ and parametrizing by $w$ at places where $w=0$.

The question is to show is not the interior of a compact surface-with-boundary.

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My attempt:

Consider the holomorphic function $\displaystyle F:X\rightarrow \mathbb{C}$ given by $\displaystyle F:( z,w)\rightarrow w$. Now assume there is a compact surface-with-boundary $\displaystyle \tilde{X}$ such that interior of $\displaystyle \tilde{X}$ is $\displaystyle X$. It is natural to expect that $\displaystyle F$ extends to an holomorphic function $\displaystyle \tilde{F} :\tilde{X} \ \rightarrow \mathbb{C} P^{1}$.

Now observe $\displaystyle \tilde{F}^{-1}( 0)$ has a limit point as it contains an infinite sequence $\displaystyle \{( n\pi ,0)\}_{n=1}^{\infty }$ in $\displaystyle \tilde{X}$. Holomorphicity of $\displaystyle \tilde{F}$ forces $\displaystyle \tilde{F} $ to be a constant.
This is a contradiction as $\displaystyle F$ is not constant.

The issue with this proof is that there is no reason to believe that $\displaystyle F$ should have an extension $\displaystyle \tilde{F}$.

Answer 1

The thing to prove is that the surface $X$ has infinite topological type. Instead of your function $F$, I will consider the function $g(z,w)=z$ on $X$, which is (generically) 2-to-1. Critical values of this function are $z_n=\pi n$, $n\in {\mathbb Z}$ (the zeroes of the function $\sin(z)$).

Consider a sequence of closed disks $$ D_n=\bar{D}(0, \pi n + \frac{\pi}{2})\subset {\mathbb C}, $$ each containing $2n+1$ critical values of the function $g$ and no critical values on the boundary. Let $X_n:= g^{-1}(D_n)$. Then each $X_n$ has one boundary component (see below). Each critical value of $g$ will have exactly one preimage in $X$. By the Riemann-Hurwitz formula, $$ \chi(X_n)= 2\chi(D_n) - 2n-1= 1-2n. $$ Since $\partial X_n$ is connected, it follows that the genus of $X_n$ is $n$. (This calculation also shows that $\partial X_n$ cannot consist of 2 components and, since $g$ is 2-to-1, the higher number of boundary components is impossible.) Hence, $X$ has infinite genus and, therefore, is not homeomorphic to the interior of a compact surface.

Answer 2

I am trying to use some other way to prove that the Riemann surface $X=\{(z,w):w^2=\sin z\}$ has infinite genus, as shown in the answer of Kohan's. The main result I wanna to show is that:

For any positive integer $n$, there is closed, compactly supported, smooth 1-forms $\theta_1, \theta_2, \cdots, \theta_n$ (not required to be holomorphic) that is linear independent over $\Bbb{R}$, i.e. if $a_1 \theta_1 + \cdots + a_n \theta_n = \mathrm{d}u$ for some smooth function $u$, then $a_i=0$ for all $i$.

And the main proposition to use is that(It's Proposition 15 in Chapter 5 in Donaldson's Riemann Surfaces, Section II.3.3 in Farkas&Kra's Riemann Surfaces, and Formula 6.71 in Page 289 in Kodaira's Complex analysis):

Proposition 15 For any loop $\gamma$ in $X$, there is a compactly supported 1-form $\theta$ such that for any closed smooth compactly supported 1-form $\phi$,

$$\int_\gamma \phi = \int_X \theta \wedge\phi $$

All references explain this result in details. Mainly it's using partition of unity to construct a smooth function $\rho_\gamma^+$ on $X-\gamma$ and taking $\mathrm{d} \rho_\gamma^+$ as $\theta$.

Furthermore, for this 1-form $\theta$ associated to the curve $\gamma$, if $\delta$ is some other loop does not intersect the support of $\theta$ (the support of $\theta$ is near $\gamma$, it can be shown from the construction of $\theta$), then $\int_\delta\theta = 0$. And if $\delta$ intersects $\gamma$ once, then $\int_\delta\theta = 1$ (from one direction, from the other direction it's $-1$).

With this proposition, using the loops as follow: (sorry for the poor drawing, the black loops are in the $u-x$ plane, but the red loops are in the $v-y$ plane, where $w=u+iv$ and $z=x+iy$. For the loop $\gamma_0$, imagine that $z$ goes from $0$ to $\pi$, then $w$ goes from $0$ to $1$ and back to $0$, and the opposite direction is similar. I attached the picture that projecting $z=\mathrm{arcsin}(w^2)$ to the $u$ coordinate to help the illustration.)

Then for each $\gamma_n$, there is an associated closed smooth compactly supported 1-form $\theta_n$. The set $\{\theta_n\}$ is linear independent over $\Bbb{R}$:

Suppose $a_1 \theta_1 + \cdots a_n \theta_n = 0$, applying the loop $\delta_0$ to it (integrating it on $\delta_0$), we get $a_1 = 0$ since $\delta_0$ only intersects $\gamma_1$. Similarlly $a_i = 0$ for all other $i$.

Hence $X$ cannot be the interior of a compact surface-with-boundary.(It's not required to be a Riemman surface. The argument doesn't use any condition of being holomorphic)