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I am writing a thesis on Uniformisation theorem and I'm quite closely following Donaldson's Riemann Surfaces. I (and Donaldson) have covered topics of covering spaces, discrete möbius groups of $S^1,\mathbb{C},D(0,1)$, differential forms, de Rham cohomology, decomposition of one forms, Laplace operator, Dirichlet (semi) inner product and (semi) norm. These are the topics that I've found necessary to prove Uniformisation so far.

You can find first half of his book as online pdf version hosted on Imperial College London website here: http://wwwf.imperial.ac.uk/~skdona/RSPREF.PDF. I assume it is there legally and that I don't violate any rules by sharing this link. If I do violate some regulation then feel free to notify me.

I'm finding trouble proving Corollary 4 on page 101. It states:

Any compact Riemann surface with $g = 1$ is equivalent to a torus $\mathbb{C}/\Lambda$.

After proving Theorem 4 (p. 99) we get Theorem 5 (p. 100) by doing some diagram chasing. We also have Proposition 25 (p. 99). The diagram is given by these spaces:

  • $H^{0,0}=\ker \overline{\partial}\colon \Omega^0\to\Omega^{0,1}$
  • $H^{1,0}=\ker \overline{\partial}\colon \Omega^{1,0}\to\Omega^{2}$
  • $H^{0,1}=\text{coker} \overline{\partial}\colon \Omega^{0}\to\Omega^{0,1}$
  • $H^{1,1}=\text{coker} \overline{\partial}\colon \Omega^{1,0}\to\Omega^{2}$

So given compact genus 1 Riemann surface $X$. I have following facts:

  • For distinct points $p_1,p_2\in X$, there exists meromorphic function $F\colon X \to S^1$ with $F^{-1}(\infty) = \left\lbrace p_1,p_2\right\rbrace$. These poles are simple.
  • $H^{0,0}$ a.k.a space of holomorphic functions is equal to space of constants and also equals to $H^{0}$ of de Rham cohomology.
  • $H^{1,0}$ a.k.a space of holomorphic one forms has one real dimension.

As my thesis is already packed with stuff I am seeking solution with minimal effort (extra pages written). These are solutions that I can think of:

  • Find nowhere vanishing holomorphic one form and prove theorem that states that: If there exists nowhere vanishing holomorphic one form on compact $X$ then it is a torus. (p. 80, Thm 3)
  • Have some fundamental understanding of critical values (branch points, ramification points) of the existing meromorfic $F$ or just know the number of critical points. Then use Riemann existence theorem (p.45, Thm 2) to show that pair $(X,F)$ is equivalent to Weierstrass function (p. 82) or theta function on torus.

I have a real copy of the book and theta functions are introduced in it right after Weierstrass function. Basically we get double periodic meromorphic function (like Weierstrass) which is double cover of Riemann sphere with four branch points.

The problem with these two approaches is that I can't find nowhere vanishing one form, can't prove it's existence and that I only know that there are at least three critical values for meromorphic $F$. This is obtained by investigating the construction before Riemann existence theorem and deducing that fundamental groups of once or twice punctured sphere are too small or simple.

Unfortunately this quote before Corollary 4 doesn't help me: "We can give some simple consequences of the above."

Could someone help me with those two proposed paths? Also a proof that avoids those steps mentioned above will be equally appreciated.

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Here's a sketch of your first strategy that uses only what you have:

There exists a non-trivial holomorphic one-form $\eta$ on $X$. (Incidentally, the space of one-forms is one-dimensional as a complex vector space, not as a real vector space. ;)

If $\eta$ vanished at some point $p_{1}$, then choosing $p_{2} \neq p_{1}$ and a meromorphic function $f$ with simple poles at $p_{1}$ and $p_{2}$ (and holomorphic elsewhere), the meromorphic one-form $f\eta$ would have a single pole at $p_{2}$ and be holomorphic elsewhere, contradicting the residue theorem. (The sum of the residues of a meromorphic form on a compact Riemann surface vanishes.)

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  • $\begingroup$ So we conclude that every holomorphic one form is nowhere vanishing? Because if we have one with a zero then we use that meromorphic $F$ like you did. What does the non-trivial property stand for here? It is not a constant? $\endgroup$ – VadaVad Sep 18 '16 at 13:48
  • $\begingroup$ By "non-trivial", I meant "not identically zero". Yes, the conclusion is that $\eta$ is nowhere-vanishing. :) $\endgroup$ – Andrew D. Hwang Sep 18 '16 at 14:06
  • $\begingroup$ Yes, this is it. Thanks. :) $\endgroup$ – VadaVad Sep 18 '16 at 15:22

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