# Abstract Nonsense

## Meromorphic Functions on Riemann Surfaces (Pt. II)

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

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Let $X$ be a Riemann surface and suppose that $f:X\to\mathbb{C}$ is meromorphic at $p$. We know then that for each chart $(U,\varphi)$ at $p$ one has that $f\circ\varphi^{-1}:\varphi(U)\to\mathbb{C}$ is meromorphic at $p$. Thus, from basic complex analysis we can form the Laurent series at $p$ relative to $\varphi$ $\displaystyle \sum_{n\in\mathbb{Z}}a_n(z-\varphi(p))^n$, which is defined on a punctured disc $D-\{p\}\subseteq\varphi(U)$. We then define the order of $f$ relative $\varphi$ to be $\inf\{n:a_n\ne 0\}$. As one would hope, despite the fact that the Laurent series of $f$ varies greatly depending on which chart one takes it relative to, the order of $f$ does not!

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Theorem: Let $X$ be a Riemann surface and $f:X\to\mathbb{C}$ be meromorphic at $p$. Then, the order of $f$ is independent of chart.

Proof: Let $(U,\varphi)$ and $(V,\psi)$ be any two charts at $p$. We first consider the Laurent series of $f$ relative $\varphi$, so that on some disc $D_{r_1}(\varphi(p))-\{\varphi(p)\}\subseteq U$ we have that

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$\displaystyle (f\circ\varphi^{-1})(z)=\sum_{n=m}^{\infty}a_n (z-\varphi(p))^n$

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where $m\in\mathbb{Z}$ is the order of $f$ relative $\varphi$ (note that this order is finite by the assumption that $f\circ\varphi^{-1}$ is meromorphic at $\varphi(p)$). Now, we consider a disc $D_r(\varphi(p))-\{\varphi(p)\}$ contained in $\varphi(U)\cap\psi(U)$. We see then that on this disc

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$f\circ\psi^{-1}=(f\circ\varphi^{-1})\circ\underbrace{(\varphi\circ\psi^{-1})}_{T}$

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Now, since $T$ is biholomorphic on a neighborhood of $\psi(p)$ we have that we can extend $T$ into it’s Taylor series

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$\displaystyle T(z)=\sum_{k=0}^{\infty}b_n(z-\psi(p))^k$

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Now, note that $b_0=T(\psi(p))=\varphi(p)$, and $b_1=T'(\psi(p))\ne 0$. This second fact is because if $S=T^{-1}$ [remember that $T$ is biholomorphic!) then $S(T(w))=w$ for all $w$ in some neighborhood of $\psi(p)$, and so $S'(T(w))T'(w)=1$, so $T'(w)\ne 0$. Thus, we see that

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\displaystyle \begin{aligned}(f\circ\psi^{-1})(z) &= (f\circ \varphi^{-1})\circ T\\ &= \sum_{n=m}^{\infty}a_n(T(z)-\varphi(p))^n\\ &= \sum_{n=m}^{\infty}\left(\sum_{k=0}^{\infty}(z-\psi(p))^k-\varphi(p)\right)\\ &= \sum_{n=m}^{\infty}\left(\sum_{k=1}^{\infty}(z-\psi(p))^k\right)^n\\ &= a_mb_1(z-\psi(p))^m+\sum_{n=m+1}^{\infty}c_n(z-\psi(p))^n\end{aligned}

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where $c_n$ are some coefficients in $\mathbb{C}$. Now, by the uniqueness of the Laurent series on a punctured disc, this must be the Laurent series of $f$ at $p$ relative $\psi$. But, note that $a_m$ and $b_1$ are both nonzero so that $a_mb_1$ is non-zero. It follows easily that the order of $f$ at $p$ relative $\psi$ is also $m$. Since $(U,\varphi)$ and $(V,\psi)$ were arbitrary the conclusion follows. $\blacksquare$

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Thus, if $f:X\to\mathbb{C}$ is a meromorphic function at $p$, we may unabashedly define the order of $f$ at $p$ to be its order at $p$ relatively any chart at $p$. We denote the order of $f$ at $p$ by $\text{ord}_p(f)$ or sometimes (although rarely, if at all) by $v_p(f)$ (this is supposed to be suggestive to anyone familiar with some parts of algebra!).

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Let’s give an example of a meromorphic function (finally, right?). Consider the extended complex plane $\mathbb{C}_\infty$ and a rational function $\displaystyle h(z)=\frac{f(z)}{g(z)}\in\mathbb{C}(z)$. It’s trivial that $h$ is meromorphic at each point of $\mathbb{C}$, and checking the $h$ is meromorphic at $\infty$ is equivalent to checking that $\displaystyle h\left(\frac{1}{z}\right)$ is meromorphic at $0$–a classical exercise in basic analysis. This should be expected considering our intuition that meromorphic functions are just holomorphic functions to the Riemann sphere, since we have already shown that $h$ is a holomorphic function $\mathbb{C}_\infty\to\mathbb{C}_\infty$.

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Now, having the notion of order allows us to classify singularities. Namely, let $f:X\to\mathbb{C}$ be meromorphic at $p$, then we say that $p$ is a removable singularity if $\text{ord}_p(f)\geqslant 0$ and that $p$ is a pole if $\text{ord}_p(f)<0$. Note that the name ‘removable singularity’ makes sense since evidently if $\text{ord}_p(f)\geqslant 0$ then $|f|$ is bounded in a neighborhood of $p$ and so Riemann’s Removable Singularity Theorem allows us to holomorphically extend $f$ to $p$. In particular, we see that $f$ is holomorphic at $p$ if and only if $\text{ord}_p(f)\geqslant 0$. Along with this, it’s easy to see that $f$ has a pole at $p$ if and only if $\displaystyle \lim_{z\to p}|f(z)|=\infty$.

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The other thing to notice about this definition is that the poles of $f$ are necessarily discrete in $X$. Indeed, if $f$ is meromorphic at $p$ then we know that $f\circ\varphi^{-1}$ is meromorphic at $\varphi(p)$ for any chart $(U,\varphi)$ at $p$. Thus, we know there exists a neighborhood $p\in O\subseteq U$ such that $f\circ\varphi^{-1}$ is holomorphic on $O-\{p\}$ and thus we see that $\varphi^{-1}(O)-\{p\}$ is a neighborhood of $p$ which intersects the poles of $f$ only at $p$–and thus discreteness follows. This is comforting since we always had the discreteness of poles for meromorphic functions on subsets of $\mathbb{C}$. In fact, not only is this comforting, but it should be expected. For, if meromorphic $f:X\to\mathbb{C}$ are really just holomorphic $F:X\to\mathbb{C}_\infty$ then the poles of $f$ should be $F^{-1}(\infty)$ and fibers are discrete!

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This above analysis gives us an equivalent way of defining holomorphic functions on a Riemann surface $X$–by specifying poles. Namely, suppose that $X$ is a Riemann surface and $W\subseteq X$ is open. Then, we say that a function $f:W\to\mathbb{C}$ is a meromorphic function on $X$ if it is a holomorphic function on $W$, if $X-W$ is discrete, and if $\displaystyle \lim_{z\to p}|f(z)|=\infty$ for each $p\in X-W$. In essence, this says that a meromorphic function on $X$ is a holomorphic function with a discrete set of poles.

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Now, we leave it to the reader to verify the following simple properties of the order:

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Theorem: Let $f,g:X\to\mathbb{C}$ be meromorphic at $p$, then:

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\begin{aligned}&\mathbf{(1)}\quad \text{ord}_p(fg)=\text{ord}_p(f)+\text{ord}_p(g)\\ &\mathbf{(2)}\quad \text{ord}_p(f\pm g)\geqslant \inf\{\text{ord}_p(f),\text{ord}_p(g)\}\end{aligned}

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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