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Let $f(z)dz^2$ be a holomorphic quadratic differential on the punctured disk $\{0<|z|<1\}$, which gives rise to a Riemannian metric $g=|f(z)|\,|dz|^2$ and hence a volume form $\nu=|f(z)| dx\,dy$.

Problem: prove that if $f$ has essential singularity at $0$, then

(1) the total area of $\nu$ is infinite;

(2) the metric $g$ is incomplete at $0$, i.e. there is a sequence of points $(z_n)_{n\geq 0}$ which converges to $0$ while the distances $d(z_0, z_n)$ (defined with respect to $g$) are bounded.

I don't know much about essential singularities except Picard's theorem, which is not enough for either of the questions. Standard techniques such as Maximum Modulus Principle don't seem enough either.

In Strebel's book Quadratic Differentials, it is claimed (in the paragraph following Definition 5.3) that a neighborhood of $0$ has finite area with respect to $\nu$ if and only if $0$ is a first order pole of $f$. Proposition (1) above is the nontrivial portion of this claim. But I don't know whether Strebel had considered essential singularities seriously since this is discussed nowhere in his book.

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For statement 2, you have to specify whether $f$ is allowed to have zeros. (If yes, this is a metric with isolated singularities, but can be complete. If $f$ is free of zeros, it is always incomlete by a deep result of A. Huber MR0094452)

To prove statement 1, it is better to make the change of the variable $z=1/w$, then we have to prove that $$\int_1^\infty\int_{-\pi}^\pi |f(re^{it})|dt \; r^{-3}dr<\infty$$ implies that $f$ has a simple pole at $\infty$. This can be done with the help of the inequality $$\max_{t}u(re^{it})\leq \frac{R+r}{R-r}\int u(Re^{it})dt +O(1), \quad 2< r< R,$$ which is true for every subharmonic function $u$, and we apply it to $u=|f|$. The inequality follows from the Poisson formula applied to the least harmonic majorant of $u$ in the ring $2< |z|\leq R$.

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Another proof of (the contrapositive of) 1 is the following: writing $f(z)=\sum a_nz^n$, if $f$ has a singularity worse than a pole there exists $k\ge2$ such that $a_{-k}\ne0$. Then $$\eqalign{\int_{\Delta^*}|f(z)|dxdxy&=\int_0^1\Big(\int_0^{2\pi}\big|\sum a_nr^ne^{in\theta}\big|d\theta\Big)rdr\cr &\ge\int_0^12\pi|a_{-k}|r^{-k}rdr=+\infty.\cr}$$

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  • $\begingroup$ How do you conclude your inequality? After all, there may well be some places where $f$ is much smaller than the corresponding term in the Laurent series, so it seems that you need to give some justification. $\endgroup$ Oct 2 '14 at 11:56
  • $\begingroup$ Multiply by $e^{-ik\theta}$ and take out the modulus... $\endgroup$ Dec 1 '16 at 19:26
  • $\begingroup$ Ah, how silly of me! That's rather nice. $\endgroup$ Dec 2 '16 at 0:08

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