Let $\mathcal{L}$ denote extremal length. The following theorem appears in http://arxiv.org/abs/math/0505191.

Theorem Let $U$ and $V$ be Riemann surfaces, and let $f:U\rightarrow V$ be a holomorphic map (possibly of infinite degree). If $\Gamma$ is a curve family on $U$, then $\mathcal{L}(f\circ\Gamma)\geq\mathcal{L}(\Gamma)$.

The proof there states that one should take an admissible (measurable) conformal metric $\mu$ on $U$, think of it as a measure $\mu^2$, push it forward to a obtain a measure $f_*\mu^2$ on $V$, and convert this measure to a measurable conformal metric (defined up to measure 0) $\nu$ on $V$. Then it is claimed that $f^*\nu\geq\mu$, but I don't understand this. We want to see that $\int_\gamma f^*\nu\geq\int_\gamma\mu$. Intuitively, I suppose we should have something like $\int_\gamma f^*\nu\geq\int_{f^{-1}(f\circ\gamma)}\mu\geq\int_\gamma\mu$, but I don't see how to show this rigorously.

An alternative approach to prove this theorem might be to take a conformal metric $\mu$ on $U$, and define $\nu(w)$, $w\in V$, to be the supremum of the pushforwards of $\mu(z)$, where $f(z)=w$. When I try to do this, I'm unable to obtain the required area inequality.

Can anyone help me to see why this theorem is true?

Edit 1: The usual thing to try, taking a metric on $V$ and pulling it back to a metric on $U$, cannot work since the inequalities will go in the wrong direction.

Edit 2: This is an edit in response to Professor Eremenko's answer. If I use $\nu(w)=\sum_{z\in f^{-1}(w)}\frac{\mu(z)}{|f'(z)|}$, I see that $f^*\nu\geq\mu$. However, I should check that the area of $\nu^2$ is less than or equal to the area of $\mu^2$, but I think it does not work.

I think we should have pushed forward $\mu^2$ to obtain $\rho^2(w)=\sum_{z\in f^{-1}(w)}\frac{\mu(z)^2}{|f'(z)|^2}.$ The length element $\rho$ also gives the desired inequality $f^*\rho\geq\mu$, and I think it should give the right inequality for area as well, but I'm still having trouble seeing this. (Perhaps my trouble is in failing to understand Tonelli or change of variables properly, but I have tried to no avail)


1 Answer 1


EDIT. Push forward is $$\nu(w)|dw|=\sum_{z:f(z)=w}\mu(z)|dz|,$$ or $$\nu(w)=\sum_{z:f(z)=w}\frac{\mu(z)}{|f'(z)|}.$$ The density of the pull back $f^*\nu$ is $$\nu(f(z))|f'(z)|=\sum_{\zeta:f(\zeta)=f(z)}\frac{\mu(\zeta)}{|f'(\zeta)|}|f'(z)|.$$ Dropping all positive summands except one we obtain the required inequality $f^*\nu\geq\mu$.


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