# Abstract Nonsense

## Holomorphic Maps and Functions (Pt. II)

Point of Post: This is a continuation of this post.

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Let us now give some examples of holomorphic maps:

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Let $D$ and $D'$ be domains in $\mathbb{C}$. Then, a holomorphic map $f:D\to D'$ is just a holomorphic map using the standard definitions from complex analysis.

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Consider the Riemann sphere $\mathbb{C}_\infty$ and the map $f:\mathbb{C}_\infty\to\mathbb{C}_\infty$ given by $\displaystyle z\mapsto \frac{1}{z}$ (where we understand $\displaystyle \frac{1}{\infty}=0$ and $\displaystyle \frac{1}{0}=\infty$) is an automorphism of $\mathbb{C}_\infty$. Indeed, to see that $f$ is actually holomorphic we consider the standard atlas for $\mathbb{C}_\infty$ consisting of the charts $(\mathbb{C},\text{id})$ and $(\mathbb{C}_\infty-\{0\},\varphi)$ where $\displaystyle \varphi(z)=\frac{1}{z}$. Now, it’s easy to see that $f(\mathbb{C})=\mathbb{C}_\infty-\{0\}$ and we quickly check that $\varphi\circ f\circ\text{id}$ is just the identity map on $\mathbb{C}$, and thus evidently holomorphic. Now, it’s equally evident that $f(\mathbb{C}_\infty-\{0\})=\mathbb{C}$ and since $\text{id}^{-1}\circ f\circ\varphi$ is just the identity map on $\mathbb{C}$ again, we see that $f$ is legitimately holomorphic. Since $f$ is bijective and $f^{-1}=f$ we see that $f$ is evidently an automorphism.

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More generally, we claim that if $\displaystyle h(z)=\frac{f(z)}{g(z)}\in\mathbb{C}(z)$ is a rational function (where we assume for convenience that $f(z)$ and $g(z)$ are coprime) then $h(z)$ is a holomorphic map $\mathbb{C}_\infty$ where we define $\displaystyle h(\infty)=\lim_{z\to\infty} h(z)$ and $h(z)=\infty$ whenever $g(z)=0$. Indeed, let $a_1,\cdots,a_n$ be the roots of $g(z)$ (obviously meaning roots in $\mathbb{C}$ and counted with multiplicity). Now, we can clearly find an open subset $O\subseteq\mathbb{C}_\infty$ for which $O$ does not contain any  $a_i$ or $\infty$ (if one wants to explicit take a closed disc around each mentioned points [not intersecting each other for visual simplicity] and then take the Riemann sphere minus these discs). So, it’s definitely clear that $O\subseteq\mathbb{C}$ and that $h(O)\subseteq\mathbb{C}$ so that we need only check $\text{id}\circ h\circ \text{id}^{-1}=h$ is holomorphic, but this is evident. Thus, it remains to prove $h$ is holomorphic at each $a_i$, and at $\infty$. Now,  for each $a_i$, choose a neighborhood $U_i$ which contains no root of $f$, nor any other $a_j$, nor containing $\infty$. Clearly then we have that $h(U_i)\subseteq \mathbb{C}_\infty-\{0\}$ and so we can check holomorphicity at $a_i$ by checking that $\varphi\circ h\circ\text{id}^{-1}$ is holomorphic at $0$. But, $\varphi\circ h$ is given by

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$\displaystyle \varphi(h(z))=\begin{cases}\displaystyle \frac{1}{h(z)} & \mbox{ if }\quad z\ne a_i\\ 0 & \mbox{if}\quad z=a_i\end{cases}$

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Now, evidently $\varphi\circ h$ is holomorphic on $U_i-\{a_i\}$ and is holomorphic on $U$ because $\displaystyle \lim_{z\to a_i}\varphi(h(z))$ exists, and is equal to $0$ (this is Riemann’s Removable singularity theorem). Thus, it remains to show that $h$ is holomorphic at $\infty$. This requires two cases. If $h(\infty)\in\mathbb{C}$, choose a neighborhood $U$ of $\infty$ for which does $\varphi(U)$ not contain any $a_i$, and for which $U$ doesn’t $0$ (why can we do this?). Then, $h(U)\subseteq\mathbb{C}$, and thus we need to check that $\text{id}\circ h\circ\varphi^{-1}$ is holomorphic. Now, $\varphi(U)$ is some neighborhood of $0$, and so we see that on $\varphi(U)$ one has that

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$\displaystyle h(\varphi^{-1}(z))=\begin{cases}\displaystyle h\left(\frac{1}{z}\right) & \mbox{if}\quad z\ne 0\\ h(\infty) & \mbox{if}\quad z=0\end{cases}$

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Now, since $\varphi(U)$ doesn’t contain any roots, we know that this is holomorphic on $\varphi(U)-\{0\}$ and since it’s continuous by construction, $h$ is holomorphic by Riemann’s Removable singularity theorem. If $h(\infty)=\infty$, then we choose a neighborhood $U$ of $\infty$ for which $\varphi(U)$ doesn’t contain any roots of $f$, nor of $g$, and for which $U$ doesn’t contain $0$. Then, we have that we need only check the holomorphicity of $\varphi\circ h\circ\varphi^{-1}$ which is just

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$\displaystyle (\varphi\circ h\circ\varphi^{-1})(z)=\begin{cases}\displaystyle \frac{1}{h\left(\frac{1}{z}\right)} & \mbox{if}\quad z\ne 0\\ 0 & \mbox{if}\quad z=0\end{cases}$

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By similar reasoning, this is holomorphic on $\varphi(U)-\{0\}$ and continuous (by construction) on $\varphi(U)$, and thus holomorphic. Thus, we have shown that $h$ is holomorphic at every point of $\mathbb{C}_\infty$ and thus is holomorphic.

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It’s obvious that if $X$ is any Riemann surface and $(U,\varphi)$ is a chart on $X$, with $U$ connected. Then, $\varphi$ is a biholomorphism $U\to\varphi(U)$.

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New Concept, Old Theorems

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We now prove some of the classic theorems of complex analysis still holds true for holomorphic maps between Riemann surfaces. The basic idea in most of the proofs is that local properties transfer from subsets of $\mathbb{C}$ to Riemann surfaces, because they are locally the same. Global concepts then shall be pieced together via these local results, connectivity being an integral part of these theorems.

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Probably one of the most shocking theorems concerning the mapping properties of analytic functions is the Identity Theorem, which says that if two holomorphic functions agree on a non-discrete subset of a domain, then they are necessarily equal! This tells us, for example, that any holomorphic function $f$ on the unit disc $\mathbb{D}$ is determined by its values $\displaystyle \left\{f\left(\frac{1}{n}\right)\right\}_{n\in\mathbb{N}}\cup\{f(0)\}$, or more generally that if $\{z_n\}$ is any convergent sequence of points, then we need only specify a function on $\displaystyle \{f(a_n)\}\cup\{f\left(\lim a_n\right)\}$. As one can probably guess, this is true for general Riemann surface.

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References:

[1] Varolin, Dror. Riemann Surfaces by Way of Complex Analytic Geometry. Providence, RI: American Mathematical Society, 2011. Print.

[2] Miranda, Rick. Algebraic Curves and Riemann Surfaces. Providence, RI: American Mathematical Society, 1995. Print.

[3] Forster, Otto. Lectures on Riemann Surfaces. New York: Springer-Verlag, 1981. Print.

[4] Conway, John B. Functions of One Complex Variable. New York: Springer-Verlag, 1978. Print.

[5] Gong, Sheng. Concise Complex Analysis. Singapore: World Scientific, 2001. Print

[6] Rudin, Walter. Real and Complex Analysis. New York: McGraw-Hill, 1966. Print.