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Notation: If $X$ is a Riemann surface, $\mathscr O(X)$ is the ring of holomorphic functions on $X$ and $\mathscr M(X)$ is the field of meromorphic functions on $X$.


If $X$ and $Y$ are two isomorphic Riemann surfaces, then is not true that $\mathscr M(X)\cong \mathscr M(Y)$ in fact for example $\mathscr M(\,\overline{\mathbb C}\,)=\mathbb C(z)$ while $$\mathscr M(\mathbb P^1)=\left\{\frac{p}{q}\in\mathbb C(z,w)\,:\,\textrm{$p$ and $q$ are homogeneous with the same degree}\right\}$$

For holomorphic functions, is it true the following proposition?

$X$ and $Y$ are isomorphic as Riemann surfaces if and only if $\mathscr O(X)\cong \mathscr O(Y)$

Where can I find the proof?


Edit: The above statement about meromorphic functions is wrong. I apologize.

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  • $\begingroup$ Sure you have $\mathscr{M}(X) \cong \mathscr{M}(Y)$ if $X$ and $Y$ are isomorphic Riemann surfaces, $f \mapsto f\circ \varphi$ is an isomorphism for every biholomorphic $\varphi \colon Y\to X$. $\endgroup$ – Daniel Fischer Dec 14 '13 at 12:31
  • $\begingroup$ Probably I don't see the isomorphism between $\mathscr M(\mathbb P^1)$ and $\mathscr M(\overline{\mathbb C})$. Can you help me? $\endgroup$ – Dubious Dec 14 '13 at 12:35
  • $\begingroup$ Maybe the map $\frac{p(z,w)}{q(z,w)}\mapsto\frac{p(z,1)}{q(z,1)}$? $\endgroup$ – Dubious Dec 14 '13 at 12:39
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    $\begingroup$ Yes, that map gives you an isomorphism. $\endgroup$ – Daniel Fischer Dec 14 '13 at 12:41
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    $\begingroup$ If you have $\frac{p(z,w)}{q(z,w)} = \frac{\tilde{p}(z,w)}{\tilde{q}(w,z)}$, that means $p(z,w)\tilde{q}(z,w) = \tilde{p}(z,w) q(z,w)$. Then you clearly have $p(z,1)\tilde{q}(z,1) = \tilde{p}(z,1) q(z,1)$, and thus $\frac{p(z,1)}{q(z,1)} = \frac{\tilde{p}(z,1)}{\tilde{q}(z,1)}$. $\endgroup$ – Daniel Fischer Dec 14 '13 at 13:41
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For isomorphic Riemann surfaces $X$ and $Y$, every biholomorphic $\varphi \colon Y \to X$ induces isomorphisms $\varphi^\ast \colon \mathscr{O}(X) \to \mathscr{O}(Y)$ and $\overline{\varphi}\colon \mathscr{M}(X) \to \mathscr{M}(Y)$ by composition with $\varphi$, we have $\overline{\varphi}(f) = f \circ \varphi$, and $\varphi^\ast$ is the restriction of $\overline{\varphi}$ to $\mathscr{O}(X)$.

However, for every compact Riemann surface $C$, we have $\mathscr{O}(C) \cong \mathbb{C}$, and since there are non-isomorphic compact Riemann surfaces, the isomorphism of the rings of holomorphic functions doesn't imply an isomorphism of the surfaces.

A $\mathbb{C}$-algebra isomorphism between the fields of meromorphic functions induces a biholomorphism between the surfaces, and for open surfaces also a $\mathbb{C}$-algebra isomorphism between the rings of holomorphic functions induces a biholomorphism.

Farkas/Kra, Riemann Surfaces, $2^{\text{nd}}$ edition, GTM 71, Springer-Verlag, prove that for the fields of meromorphic functions on compact Riemann surfaces in $\mathrm{IV}.11.16$ and following, and remark the analogue(s) for the open surfaces thereafter.

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  • $\begingroup$ It is a general fact that isomorphism of fields of merimorphic functions w/o essential singularities implies birational isomorphism, hence, isomorphism in the case of complex curves. $\endgroup$ – Moishe Kohan Dec 14 '13 at 12:53
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The result that you highlight is not true! Take $X = \Bbb{P}^1_{\Bbb{C}}$ and $Y =\Bbb{P}^2_{\Bbb{C}}$. The only holomorphic functions on $X = \Bbb{P}^1_{\Bbb{C}}$ and $Y = \Bbb{P}^2_{\Bbb{C}}$ are the constants and so $\mathcal{O}(X) \cong \mathcal{O}(Y)$. However $X \not\cong Y$ because they have different dimensions as complex manifolds!

Those familiar with algebraic geometry will notice this is a special case of the following fact: The ring of regular functions is not enough to tell if two projective varieties are isomorphic.

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As noted above the claim is false for compact Riemannian surfaces.

Nevertheless, the claim is true for open Riemannian surfaces. If you consider also the topology on $\mathscr O(X)$ and $\mathscr O(Y)$ then the claim generalizes to a statement about Stein spaces and Stein algebras: Every open Riemannian surface is Stein. And two Stein spaces are isomorphic iff their algebras of holomorphic functions are homeomorphic as topological algebras, see 'Anhang zu Kap. VI, Satz 8' in Behnke, H.; Thullen, P.: Theorie der Funktionen mehrerer komplexer Veraenderlichen, 2. erweiterte Auflage, 1970".

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