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Nevanlinna in his book Analytic functions seems to state the following (at the very end of Ch. X): For every compact Riemann surface $X$ of genus $g\geq 2$ there is a non-constant holomorphic map $f : X \to Y$ to some hyperelliptic Riemann surface $Y$. How is this proved?

I explain in more detail. Nevanlinna proves the theorem of Picard that there is no non-constant map of the complex plane to a compact Riemann surface of genus $g \geq 2$. The proof begins with the sentence:

As Picard noticed, it is enough to prove this for a hyperelliptic curve.

And proceeds to prove the theorem for hyperelliptic case.

The corresponding paper of Picard (Acta 11, 1887, p. 11-12) refers in the relevant place to a private letter of Hurwitz. It says:

Soit $f(x, y)=0$, la relation que l'on ne suppose pas hyperelliptique, et pour laquelle on a par consequent $g>2$. A l'equation prcedente, le savant geomere associe une relation $$f_1(x,y_1)=0\quad\mbox{de genre}\quad g=2$$ jouissant des proprietes suivantes: les points de ramification de la fonction algebrique $y_1$ de $x$ sont tous compris parmi les points de ramification de la fonction algebrique $y$ de $x$ (on suppose, pour plus de simplicite, et comme il est permis, que tous les points de ramification de la fonction donnent seulement des cycles de deux racines) et dans le voisinage de tout point analytique $(x,y)$ de fonction $y$, la fonction $y_1$ peut etre consideree comme une fonction uniforme du point $(x,y)$.

Can anyone explain what Picard says here, and why is this correct?

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    $\begingroup$ This is not true: a general curve $X$ of genus $\geq 2$ has no nontrivial map to a curve of genus $\geq 1$. I couldn't find anything in Nevanlinna's book that resembles this statement. $\endgroup$
    – abx
    Commented Aug 2, 2023 at 17:46
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    $\begingroup$ @abx: And what about the identity map $X\to X$ ? $\endgroup$ Commented Aug 2, 2023 at 20:21
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    $\begingroup$ @abx: It is the last paragraph of Chap. 10, the proof of a theorem of Picard. He refers to Picard and Bloch on this, but none of them gives a clear statement, not speaking of a proof. $\endgroup$ Commented Aug 2, 2023 at 20:49
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    $\begingroup$ Hmmm. Is every Riemann surface $X$ branched covered by a hyperelliptic surface? This would also give a proof (because the plane is simply connected). $\endgroup$
    – Sam Nead
    Commented Aug 3, 2023 at 7:42
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    $\begingroup$ @SamNead if i'm not mistaken, wouldn't that mean that the corresponding finite extension $K/\mathbb{C}(t)$ of the field of rational functions embeds into a degree 2 extension $𝐿\supset 𝐾\supset \mathbb{C}(t)$ and then the Galois group $K/\mathbb{C}(t)$ should be a quotient of $\mathbb{Z}_2$ but the Galois group of a generic finite extension $K/\mathbb{C}(t)$ is $S_n$ or something. $\endgroup$ Commented Aug 3, 2023 at 16:01

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I think this is what Picard says, translated in modern algebraic geometry language:

You have a curve $X$ with a map $\pi :X\rightarrow \mathbb{P}^1$. Assume for simplicity that for each $z\in \mathbb{P}^1$, $\pi ^{-1}(z)$ contains at most one ramification point, with ramification index 2. Choose 6 branch points of $\pi $, and let $\rho :C\rightarrow \mathbb{P}^1$ be the double covering branched at these 6 points (so $g(C)=2$). Now consider the fiber product $X\times _{\mathbb{P}^1}C$. It has a node over each of the 6 branch points of $\rho$; take its normalization, say $\tilde{X} $. Now $\tilde{\rho }:\tilde{X} \rightarrow X $ is étale (that is, analytically, a local isomorphism), and we have a (nontrivial) map $\tilde{X}\rightarrow C$.

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    $\begingroup$ Could you please extend your explanation a bit? Suppose $X$ receives a non-constant holomorphic map from the complex pane. Why $\tilde{X}$ does? $\endgroup$ Commented Aug 3, 2023 at 12:27
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    $\begingroup$ I am sorry I don't understand your question. $X$ has genus $\geq 2$, there is no non-constant holomorphic map from the complex plane to $X$. $\endgroup$
    – abx
    Commented Aug 3, 2023 at 12:42
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    $\begingroup$ Correct. But Picard used this construction to PROVE this fact, that there is no map from C to a curve of genus >2. How exactly did he reduce the proof of this theorem to hyperelliptic case? $\endgroup$ Commented Aug 3, 2023 at 16:06
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    $\begingroup$ Oh I see, sorry I didn't pay attention to this. Well, suppose you have a nontrivial map $\mathbb{C}\rightarrow X$; using the notation in my previous comment, since $\tilde{\rho} $ is étale, this map lifts to $\tilde{X} $, then by projection to $C$ gives a nontrivial map $\mathbb{C}\rightarrow C$. $\endgroup$
    – abx
    Commented Aug 3, 2023 at 17:12
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    $\begingroup$ @Christian Remling: yes, of course. But the question is to understand how Picard (or Nevanlinna?) wanted to proceed. $\endgroup$
    – abx
    Commented Aug 3, 2023 at 19:08

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