Series Law for Extremal Length

Theorem

Let $X$ be a Riemann surface.

Let $\Gamma_1$, $\Gamma_2$ and $\Gamma$ be families of rectifiable curves (or, more generally, families of unions of rectifiable curves) on $X$.

Let every $\gamma \in \Gamma$ contain a $\gamma_1 \in \Gamma_1$ and a $\gamma_2 \in \Gamma_2$ such that $\gamma_1 \cap \gamma_2 = \O$.

Then the extremal lengths of $\Gamma_1$, $\Gamma_2$ and $\Gamma$ satisfy:

$\map \lambda \Gamma \ge \map \lambda {\Gamma_1} + \map \lambda {\Gamma_2}$

Proof

Let $\rho_1 = \map {\rho_1} z \size {\d z}$ and $\rho_2 = \map {\rho_2} z \size {\d z}$ be conformal metrics as in the definition of extremal length.

It can be assumed that these are normalized:

$\map A {\rho_j} = \map L {\Gamma_j, \rho_j}$ for $j \in \set {1, 2}$.

We define another metric $\rho = \map \rho z \size {\d z}$ by:

$\map \rho z := \map \max {\map {\rho_1} z, \map {\rho_2} z}$

Note that this is a well-defined metric.

By definition, the area form $\map {\rho^2} z \size {\d z}$ satisfies:

$\map {\rho^2} z \size {\d z}^2 = \map \max {\map {\rho_1} z^2, \map {\rho_2} z^2} \size {\d z}^2 \le \paren {\map {\rho_1} z^2 + \map {\rho_2} z^2} \size {\d z}^2$

Hence:

$\map A \rho \le \map A {\rho_1} + \map A {\rho_2} = \map L {\Gamma_1, \rho_1} + \map L {\Gamma_2, \rho_2}$

On the other hand, let $\gamma \in \Gamma$.

Let $\gamma_1$, $\gamma_2$ be as in the assumption.

Then:

 $\ds \map L {\gamma, \rho}$ $\ge$ $\ds \map L {\gamma_1, \rho} + \map L {\gamma_2, \rho}$ as $\gamma_1$ and $\gamma_2$ are disjoint $\ds$ $\ge$ $\ds \map L {\gamma_1, \rho_1} + \map L {\gamma_2, \rho_2}$ Definition of $\rho$ $\ds$ $\ge$ $\ds \map L {\Gamma_1, \rho_1} + \map L {\Gamma_2, \rho_2}$ Definition of $\map L {\Gamma_j, \rho_j}$

Thus

$\map L {\Gamma, \rho} \ge \map L {\Gamma_1, \rho_1} + \map L {\Gamma_2, \rho_2}$

Combining this with the inequality for the area:

$\dfrac {\map L {\Gamma, \rho}^2} {\map A \rho} \ge \dfrac {\paren {\map L {\Gamma_1, \rho_1} + \map L {\Gamma_2, \rho_2} }^2} {\map L {\Gamma_1, \rho_1} + \map L {\Gamma_2, \rho_2} } = \map L {\Gamma_1, \rho_1} + \map L {\Gamma_2, \rho_2}$

Taking the supremum over all metrics $\rho_1$ and $\rho_2$ as above:

$\map L \Gamma \ge \map L {\Gamma_1} + \map L {\Gamma_2}$

as claimed.

$\blacksquare$

Also known as

The series law and the parallel law are also referred to collectively as the composition laws of extremal length.