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

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.

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.
• 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) Jun 15, 2019 at 21:04
• 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.) Jun 15, 2019 at 21:23
• 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. Jun 16, 2019 at 20:02