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In his pioneering paper An extension of Schwarz's lemma, Ahlfors proves the lower bound on the Bloch constant $B \geq \frac{\sqrt{3}}{4}$. The proof of this lower bound proceeds as follows:

Let $W$ be a Riemann surface. For an arbitrary point $\mathfrak{m}$ of $W$, let $\rho(\mathfrak{m})$ denote the radius of the largest simple circle of centre $\mathfrak{m}$ contained in $W$. It is clear that $\rho(\mathfrak{m})$ is continuous, and is zero only at the branch points. We introduce the metric $ds = \lambda | dw |$, with $$\lambda = \frac{A}{2 \sqrt{\rho} (A^2-\rho)},$$ where $\rho = \rho(\mathfrak{m})$ and $w$ denotes the variable on the function plane (not the uniformizing variable), and $A^2>B(f)$ (the Bloch constant of $f$).

If $\mathfrak{a}$ is a branch point, then $\rho = | w - \mathfrak{a} |$. Let $n$ be the multiplicity of $\mathfrak{a}$. Then $w_1 = (w-\mathfrak{a})^{\frac{1}{n}}$ is a uniformizing variable the corresponding $\lambda_1$ is determined from $\lambda_1 | dw_1 | = \lambda | dw |$. Explicitly, $$\lambda_1 = n \rho^{\frac{1}{2} - \frac{1}{n}} / 2(A^2 - \rho).$$ For $n=2$ the metric is regular, and for $n>2$, the metric is zero.

Ahlfors gives no explanation for why the $\lambda$ above is chosen. Is it simply constructed so that, with respect to the uniformizing variable, we obtain a continuous pseudo metric? Can one choose another metric to start with?

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1 Answer 1

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He explains his choice in lines 8-9 on p. 364 of the paper:

"This metric has the curvature $-4$ for it is obtained from the hyperbolic metric by the transformation $w'=w^{1/2}$ ."

Remarks. By more sophisticated choices of metrics later authors, M. Bonk (MR0979048) and Chen and Gauthier (MR1428103) were able to improve Ahlfors's estimate of the Bloch constant.

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    $\begingroup$ Thank you for your response (I was hoping you'd see this, considering you're one of the world leaders on the subject). Your remark is exactly what I was looking for. I should've been more precise in my question: Are there better choices of the metric? Thanks again :) $\endgroup$
    – AshyK
    Commented Sep 9, 2021 at 22:23

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