Riemann Surface: Intro

Now we are in a position to derive some basic understanding on Riemann surfaces.

Play with Riemann Surface

Keeping in mind about the definition of “biholomorphic”, “complex charts”, “compatible”.

[Def] (Riemann Surface) A Riemann surface is a pair $(X,\Sigma)$, where $X$ is a connected manifold of complex dimension $1$ and $\Sigma$ is a complex structure on $X$.

With the knowledge on manifold, by only changing “homeomorphism” to “biholomorphic”, the definition of holomorphic function, holomorphic map and biholomorphism between Riemann surfaces are easy to write.

[Lemma] Let $X$, $Y$ be Riemann surfaces, $U\subset X$ be an open set of $X$. Then

1. the inclusion map $i:U\rightarrow X$ is holomorphic;
2. $f:Y\rightarrow U$ is holomorphic $\Leftrightarrow$ $i\circ f:Y\rightarrow X$ is holomorphic;
3. Let $\{U_i\}_{i\in I}$ be an open cover of $X$. Then $f:X\rightarrow Y$ is holomorphic $\Leftrightarrow$ $f_i:U_i\rightarrow Y$ is holomorphic $\forall i\in I$.

The most important and somehow original object that we are playing with must be the Riemann sphere.

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Riemann Sphere

[e.g.] (Riemann Sphere) Denote $\bar{\mathbb{C}}=\mathbb{C}\cup\{\infty\}$, equipped with a topological structure where open sets in $\bar{\mathbb{C}}$ is defined as

1. open sets of $\mathbb{C}$;
2. $V\cup\{\infty\}$, where $V=\mathbb{C}-K$ is the complement of a compact set $K$.

Since compact sets in $\mathbb{C}$ are closed and bounded, then $\infty$ and any $z\in\mathbb{C}$ can be separated by two open sets. Thus $\bar{\mathbb{C}}$ is Hausdorff. In addition, given any open cover $\{U_i\}_{i\in I}$ of $\bar{\mathbb{C}}$, $\exists j\in I$, s.t.

$$\infty\in U_j.$$

By the definition of open sets in $\bar{\mathbb{C}}$, the complement of $U_j$ is a compact set. Hence there exists a finite subcover $\mathcal{U}$ of $\{U_i\}_{i\in I}$ covers $U_j^c$​, which means that

$$\mathcal{U}\cup\{U_j\}$$

is a finite subcover of $\bar{\mathbb{C}}$. Therefore, $\bar{\mathbb{C}}$ is compact. To prove that $\bar{\mathbb{C}}$ is a Riemann surface, let

$$\begin{aligned} U_1&=\bar{\mathbb{C}}-\{\infty\}=\bar{\mathbb{C}},\\ U_2&=\bar{\mathbb{C}}-\{0\}=\mathbb{C}^*\cup\{\infty\}. \end{aligned}$$

and

$$\begin{aligned} \varphi_1:U_1&\rightarrow \mathbb{C}\\ z&\mapsto z\\ \varphi_2:U_2&\rightarrow \mathbb{C}\\ z\mapsto\frac{1}{z},&~\infty \mapsto 0\\ \end{aligned}$$

Then it can be verified that $\varphi_i$ are both homeomorphisms. Thus $\bar{\mathbb{C}}$ is a complex manifold of complex dimension $1$. Since $U_1$, $U_2$, and $U_1\cap U_2$ are all connected, then $\bar{\mathbb{C}}$ is also connected.

To summarize, $\bar{\mathbb{C}}$​ is a Riemann surface.

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Complex Torus

Another interesting example is complex torus. In this part, we shall take the equality of objects carefully.

[e.g.] (Complex Torus) Let $w_1,w_2\in\mathbb{C}$ be two linearly independent elements over $\mathbb{R}$. Denote the lattice

$$\begin{aligned} L: &=\{nw_1+mw_2:n,m\in\mathbb{Z}\}\\ &=\mathbb{Z}w_1\oplus\mathbb{Z}w_2 \end{aligned}$$

and the fundamental parallelogram

$$\Pi:=\{aw_1+bw_2:a,b\in[0,1]\}.$$

Define $\mathbb{C}/L:=\{z+L:z\in\mathbb{C}\}$ and there is a natural projection $\pi:\mathbb{C}\rightarrow \mathbb{C}/L$

$$\pi(z)=[z]\in\mathbb{C}/L.$$

Hence $\mathbb{C}/L$ can be equipped with quotient topology. To show that $\mathbb{C}/L$ is a, so called, “complex torus”, we shall verify that $\mathbb{C}/L$ is

1. compact;
2. connected;
3. endowed with complex structure.

Compactness: Since the fundamental parallelogram $\Pi$ is compact, and the projection map $\pi$ is continuous, then $\mathbb{C}/L$ is also compact.

Open mapping: Given an open set $V\subset\mathbb{C}$, it suffices to show that $\pi(V)$ is open. By the definition of openness in $\mathbb{C}/L$,

$$\pi(V)\text{~is~open}\Leftrightarrow\tilde{V}=\pi^{-1}(\pi(V))~\text{is~open}.$$

Since the preimage can be written as

$$\tilde{V}=\bigcup_{w\in L}(V+w)$$

the union of open sets in $\mathbb{C}$, then $\pi(V)$ is open and $\pi$ is an open map.

Complex Structure: