How to understand the Riemann surface of $\log z$ using the formal definition of Riemann surface?

Question

I am a beginner studying Riemann surface. First things first, I learned about the definition of Riemann surface as follows.

Definition: A Riemann surface $X$ is defined by the following data:

1. a connected Hausdorff topological space $X$;
2. an open cover $\{U_\alpha\}_{\alpha\in A}$ of $X$;
3. for each $\alpha\in A$, a homeomorphism $\phi_\alpha:U_\alpha\rightarrow V_\alpha$ to an open subset $V_\alpha\subset\mathbb{C}$ in such a way that for each $\alpha,\beta\in A$, if $U_\alpha \cap U_\beta\neq \emptyset$, the transition functions $$\phi_\beta\circ\phi^{-1}_\alpha:\phi_\alpha(U_\alpha\cap U_\beta)\rightarrow\phi_\beta(U_\alpha\cap U_\beta),$$ is bi-holomorphic, namely, holomorphic with inverse holomorphic.

We know the famous Riemann surface for $\log z$, which has the following shape

How can I describe such surface with the formal definition above? In other words, I need to construct a precise $U_\alpha$, $\phi_\alpha$ and transition map to describe the Riemann surface for $\log z$, not using intuitive words such as "gluing", "rotate", "visualized like parking lot" etc...

Thanks very much for your help.

Answer 1

We want to show the following defines a Riemann surface: $$\mathscr{M}:= \bigg\{(z,f(z))\text{ }\bigg|\text{ }z\in \mathbb{C}\backslash\{0\}\text{ }\text{ and }\text{ }f(z):= Log(z)\bigg\}.$$ (Graph of principle branch of log(z))

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A.) We may define a global chart into $\mathbb{C}$ by projecting onto the first coordinate: $$\varphi:\mathscr{M}\to\mathbb{C}\backslash\{0\};$$ $$(z,f(z))\mapsto z.$$ This is a homeomorphism since it has inverse: $$\varphi^{-1}:\mathbb{C}\backslash\{0\}\to \mathscr{M};$$ $$z\mapsto (z,f(z))$$ and both are arguably continuous. In particular, take a convergent domain sequence: $(z_n,f(z_n))\to(z,f(z))$ and observe that: $$\lim\limits_{n\to\infty}\varphi\big((z_n,f(z_n))\big) = \lim\limits_{n\to\infty}z_n = \varphi\big((\lim\limits_{n\to\infty}z_n,f(\lim\limits_{n\to\infty}z_n))\big) = \varphi\big(\lim\limits_{n\to\infty}(z_n,f(z_n))\big) = \varphi\big((z,f(z))\big),$$ where in the second to last equality, we apply continuity of $f$. So $\varphi$ preserves limits of sequences and is thus continuous as well. Similarly, for a convergent image sequence: $z_n\to z$, we have: $$\lim\limits_{n\to\infty}\varphi^{-1}(z_n) = \lim\limits_{n\to\infty}(z_n,f(z_n)) = (\lim\limits_{n\to\infty}z_n,f(\lim\limits_{n\to\infty}z_n)) = \varphi^{-1}(\lim\limits_{n\to\infty}z_n) = \varphi^{-1}(z),$$ where in the third to last equality we used continuity of $f(z)$.

Moreover, being the only chart, the transition maps are: $$\varphi\circ \varphi^{-1} = Id_{\mathbb{C}\backslash\{0\}}\text{ }\text{ }\text{ and }\text{ }\text{ }\varphi^{-1}\circ\varphi = Id_{\mathscr{M}},$$ which are trivially holomorphic. $\square$

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B.) Connectedness. Regardless of topologies chosen, $\varphi$ is a homeomorphism, which means $\varphi^{-1}$ is continuous. So towards contradiction, if $\mathscr{M}$ was a disjoint union of clopen, proper subsets, the $\varphi^{-1}$-pre-images (i.e. images through $\varphi$) of the components would remain clopen. Since $\varphi$ is also injective, disjointness is preserved. This shows that $\varphi(\mathscr{M}) = \mathbb{C}\backslash\{0\}$ is disconnected. But it is well known to the contrary that $\mathbb{C}\backslash\{0\}$ is connected. $\square$

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C.) Lastly, we want to prove $\mathscr{M}$ is Hausdorff. Accordingly, let $(z,f(z))\neq (w,f(w))\in\mathscr{M}$. Then $z\neq w\in \mathbb{C}\backslash\{0\}$ (using contrapositive of injectivity of $\varphi$). We know $\mathbb{C}\backslash\{0\}$ is Hausdorff, so we have the following is satisfied: $$\exists U,V\in \tau_{\mathbb{C}\backslash\{0\}}: (z\in U)\wedge (w\in V)\wedge (U\cap V = \emptyset).$$ This gives us: $$\exists \widetilde{U},\widetilde{V}\in \tau_{\mathscr{M}}: \big((z,f(z))\in \widetilde{U}\big)\wedge \big((w,f(w))\in\widetilde{V}\big)\wedge\big(\widetilde{U}\cap\widetilde{V}=\emptyset\big),$$ where $\widetilde{U} := \varphi^{-1}(U)$ and $\widetilde{V}:= \varphi^{-1}(V)$; And since $\varphi^{-1}$ is injective, it preserves disjointness. Therefore $\mathscr{M}$ is Hausdorff. $\square$

Note: The topology on $\mathscr{M}$ is implied to be the pullback topology for which ever topology is chosen on $\mathbb{C}\backslash\{0\}$.

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From $(A)$, $(B)$, and $(C)$, we may conclude that $\mathscr{M}$, given the pullback topology from its global chart, is a Riemann surface. $\blacksquare$

My text reference is Rick Miranda's "Algebraic Curves and Riemann Surfaces" (p.10).

Answer 2

I am trying to describe the Riemann surface of $\log z$ using the fundemental definition of Riemann surface.

First, constructing a surface $X$ with coordinate $(r\cos\theta,r\sin\theta,\theta)$ where $-\infty<\theta<\infty$ and $0<r<1$. Clearly,$X$ is a Hausdorff topogical space, which satisfies condition 1.

Second, constructing a bunch of open set covering $X$. For arbitrary $x_\alpha=(r_\alpha \cos\theta_\alpha,r_\alpha\sin\theta_\alpha,\theta_\alpha)\in X$, there is a isomorphism $f:\{|z-z_\alpha|<\delta\}\rightarrow U_\alpha$ where $z_\alpha = r_\alpha e^{i\theta_\alpha\pi}$ and $\delta$ is suffiently small. Therefore $U_\alpha$ is an open set, and $\{U_\alpha\}$ covers $X$.

Third, constructing $\phi_\alpha:(r \cos\theta,r\sin\theta,\theta)\in U_\alpha\rightarrow \log r+i\theta\in V_\alpha$. Clearly, $V_\alpha\subset\mathbb{C}$, and $\phi_\alpha$ is homeomorphism. And the transition function $\phi_\beta\circ\phi_\alpha^{-1}$ is just a identity function, obviously holomorphic.

Now we have proved the Riemann surface for $\log z$ satisfies all three conditions of the fundemental definition.

Is my proof correct? Thanks for your criticize.