# Abstract Nonsense

## Meromorphic Functions on the Riemann Sphere (Pt. I)

Point of Post: In this post we classify the meromorphic functions on the Riemann sphere $\mathbb{C}_\infty$.

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Motivation

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If a random kid off the street asked you “what are the continuous functions $(0,1)\to(0,1)$?” or “what are all the smooth  maps $S^4\to\mathbb{R}$?” you would probably replay with a definitive “Ehrm…well…they’re just…” This is because such a description (besides “they’re just the continuous functions!”)  is beyond comprehension in those cases! For example, it would take quite a bit of ingenuity to come up with something like the Blancmange function–in fact, such crazy continuous everywhere, differentiable nowhere functions are, in a sense dense in the space of continuous functions.

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Thus, it should come as somewhat of a surprise that after reading this post you will be able to answer a kid asking “what are all the meromorphic functions on the Riemann sphere?” with a “Ha, that’s simple. They’re just…”. To be precise, we have already proven that $\mathbb{C}(z)\subseteq\mathcal{M}(\mathbb{C}_\infty)$, and we shall now show that the reverse inclusion is true!  Now, more generally, we shall be able to give a satisfactory (algebraic!) of the meromorphic functions of any compact Riemann surface. While this won’t be quite as impressive as the explicit, simple characterization of $\mathcal{M}(\mathbb{C}_\infty)$ but still a far cry from our situation with trying to characterize $C([0,1],[0,1])$, since we don’t really even have an algebraic (ring theoretic) description of this (in terms of more familiar objects). This should be, once again, another indication that the function theory of compact Riemann surfaces is very rigid–they admit meromorphic functions, but not so many that the computation (algebraically) of their meromorphic function field is untenable.

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Ok, now that we have had some discussion about the philosophical implications of actually being able to describe $\mathcal{M}(\mathbb{C}_\infty)$ let’s discuss how we are actually going to prove $\mathcal{M}(\mathbb{C}_\infty)=\mathbb{C}(z)$. The basic idea comes from the fact that we can actually find a function $r$ which has prescribed zeros $\lambda_1,\cdots,\lambda_n$ and poles $p_1,\cdots,p_m$ such that $\text{ord}_{\lambda_i}(r)=e_i$ and $\text{ord}_{p_i}(r)=-g_i$ for any $e_i,g_i\in\mathbb{N}$–namely, the function $\displaystyle r(z)$ given by $(z-\lambda_1)^{e_1}\cdots(z-\lambda_n)^{e_n}(z-p_1)^{-g_1}\cdots(z-p_m)^{-g_m}$. Thus, if $f$ is a meromorphic function on $\mathbb{C}_\infty$ with zeroes and poles described as in the last sentence we see that $\displaystyle \frac{f}{r}$ is a meromorphic function on $\mathbb{C}_\infty$ and which has no zeros or poles on $\mathbb{C}$. In particular, $\displaystyle h=\frac{f}{r}$ is a function meromorphic on $\mathbb{C}_\infty$ but holomorphic on $\mathbb{C}$, and with no zeros. Now, it’s a common fact from complex analysis that the only entire function with a pole at infinity (recall that the somewhat confusing definition of pole at infinity now makes a lot more sense!) is a polynomial. Thus, $h$ is a polynomial, but since $h$ has no zeros on $\mathbb{C}$ we know from the fundamental theorem of algebra that $h$ is constant. Thus, $f$ is really just a constant multiple of $r(z)$! Note that the key to this proof is that ability specify poles and zeros of a given multiplicity, except perhaps specifying a pole at infinity, and that the point infinity is well-behaved (in the sense that things that have poles there are pretty tame).

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Meromorphic Functions on $\mathbb{C}_\infty$

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Ok, let’s now try to make the above argument rigorous:

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Theorem: $\mathcal{M}(\mathbb{C}_\infty)=\mathbb{C}(z)$.

Proof: We have already proven that $\mathbb{C}(z)\subseteq\mathcal{M}(\infty)$ and so it suffices to prove the reverse inclusion. To do this, let $f\in\mathcal{M}(\mathbb{C}_\infty)$. Since $\mathbb{C}_\infty$ is compact we know that the zeros $f^{-1}(0)$ and poles of $f^{-1}(\infty)$ are finite. To be explicit, let the zeros of $f$ in $\mathbb{C}$ be $\lambda_1,\cdots,\lambda_n$ with $\text{ord}_{\lambda_i}(f)=e_i$, and let the poles of $f$ in $\mathbb{C}$ be $p_1,\cdots,p_m$ with $\text{ord}_{p_j}=-g_j$. Now, consider the function $r(z)\in\mathbb{C}(z)$ given by

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$\displaystyle r(z)=\frac{\displaystyle \prod_{i=1}^{n}(z-\lambda_i)^{e_i}}{\displaystyle \prod_{j=1}^{m}(z-p_j)^{g_j}}$

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Now, we know that $\displaystyle \frac{1}{r}$ is a meromorphic function on $\mathbb{C}_\infty$ (it’s still in $\mathbb{C}(z)$!) and since $\mathcal{M}(\mathbb{C}_\infty)$ is evidently a ring we see that $\displaystyle \frac{f}{r}$, which we shall call $h$, is in $\mathcal{M}(\mathbb{C}_\infty)$. Now, by construction $\text{ord}_p(h)=0$ for all $p\in\mathbb{C}$, and thus we see that $h$ is entire, and furthermore has no zeros. Now, since $h$ is entire it has a globally valid representation as a Maclaurin series, say

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$\displaystyle h(z)=\sum_{n=0}^{\infty}a_nz^n$

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Now, since $h$ is mermorphic at $\infty$, we know that $\displaystyle h\left(\frac{1}{z}\right)$ is meromorphic at $0$. Just to be clear, this is because $(\mathbb{C}_\infty-\{0\},\varphi$ with $\displaystyle \varphi(z)=\frac{1}{z}$ is a chart at $\infty$, $h(\varphi^{-1}(z))=h\left(\frac{1}{z}\right)$, and $\varphi(\infty)=0$–thus this is the definition of $h$ being meromorphic at $\infty$. That said, we know that locally at $0$ one has that

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$\displaystyle h\left(\frac{1}{z}\right)=\sum_{n=0}^{\infty}a_n z^{-n}$

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Now, since $h$ must have finite order at $\infty$, we see that, in particular, $a_n=0$ for sufficiently large $n$–in other words, $h$ is a polynomial. That said, we see then that $h(z)$ is a polynomial on $\mathbb{C}$ which, by prior comment, has no zeros on $\mathbb{C}$. By the Fundamental Theorem of Algebra, this implies that $h$ is constant, say $h(z)=C$. Thus,

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$\displaystyle f(z)=C\; \frac{\displaystyle \prod_{i=1}^{n}(z-\lambda_i)^{e_i}}{\displaystyle \prod_{j=1}^{m}(z-p_j)^{g_j}}$

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in particular, $f\in\mathbb{C}(z)$ as desired. $\blacksquare$

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More specifically, the above really proves that a meromorphic function (not presented to us as a rational function in $z$) is really just the rational function in $z$ which has the same poles and zeros as $f$ (obviously!).

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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