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I have some troubles in understanding the proof of the well-known assertion

The only quadratic differential on $S^2$ is identically zero.

Here $S^2$ is the standard 2D sphere. This can be proved without using the Riemann-Roch theorem, taking advantage of the Liouville theorem for holomorphic functions on $\mathbb{C}$. More precisely, the standard strategy (see, e.g., last page here) consists in noting that $S^2$ can be covered with the two charts given by the stereographic projections from the "north pole" and from "south pole". If $z \in \mathbb{C}$ is the complex in the first chart and $\Psi$ is a quadratic differential, one writes in this chart $\Psi = \psi \mbox{d}z^2$, with $\psi$ holomorphic in $\mathbb{C}$. On the other hand, the complex variable in the other chart is $w = 1 / z$ and, by change of variables, one obtains $\psi(z) \mbox{d}z^2 = \frac{\psi(z)}{w^4} \mbox{d}w^2$ where the two charts are compatible. That's fine. The passage I find rather vague is the following key observation:

by considering what happens as $w \to 0$, we see that $\frac{\psi(z)}{w^4}$ must remain bounded [...]

More or less, these words are the same in any proof I know of this fact. So, what does happen, exactly, when $w \to 0$ (i.e., when $z \to \infty$)? I'm not sure to understand if and how holomorphicity of $\psi$ implies the assertion. Said another way, it is not clear to me why $\Psi$ is continuous as $z \to \infty$.

Remark. As far as I know, a holomorphic quadratic differential is a holomorphic section of the symmetrized holomorphic cotangent bundle $\odot^2 T_{1,0}^* S^2 \to S^2$. As far as I can understand, this notion of holomorphicity is equivalent to the fact $\psi$ as above is holomorphic for any complex coordinate.

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What your highlighted passage should say is that $\dfrac{\psi(1/w)}{w^4}$ must have a removable singularity at $w=0$.

But here's a different approach for you. Since $\Psi$ is a global quadratic differential, $\sqrt{\Psi} = \sqrt\psi\,dz$ is a (not necessarily well-defined) holomorphic differential. Be that as it may, the quantity $\sqrt\Psi\wedge\overline{\sqrt\Psi} = |\psi|dz\wedge d\bar z$ is well-defined and must have finite integral over $\Bbb C$. The only $L^1$ entire function is identically zero.

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  • $\begingroup$ I agree with you, my point is that is not really clear to me why the singularity must be removable. $\endgroup$
    – user91126
    Jun 15, 2019 at 20:18
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    $\begingroup$ Because the original section must be holomorphic at infinity! $\endgroup$
    – Ted Shifrin
    Jun 15, 2019 at 20:57
  • $\begingroup$ Ok. Conversely, if I am able to show that $\psi$ is holomorphic for any complex coordinate, then I can deduce $\Psi$ is globally holomorphic (i.e., even at infinity), right? (I'm thinking to harmonic maps from $S^2$ into Riemannian manifold and their Hopf differentials) $\endgroup$
    – user91126
    Jun 15, 2019 at 21:04
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    $\begingroup$ Be careful: You're reasoning backwards. It's $\Psi$ that must be a holomorphic section in any coordinate chart, and from that you deduce that the corresponding $\psi$ must be holomorphic in the respective chart. (Think about how to compute the residue of a meromorphic $1$-form at infinity in $\Bbb CP^1$. This is the same sort of computation.) $\endgroup$
    – Ted Shifrin
    Jun 15, 2019 at 21:23
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    $\begingroup$ OK, we're quibbling over language here. We are starting with the fact that $\Psi$ is a global holomorphic quadratic differential. This in turn implies the result in coordinate charts. Of course, they are equivalent statements. $\endgroup$
    – Ted Shifrin
    Jun 16, 2019 at 20:02

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