Extremal Length of Composition

Theorem

Let $\Gamma_1$ and $\Gamma_2$ be families of (unions of) rectifiable curves on a Riemann surface $X$.

Let $\Gamma_1$ and $\Gamma_2$ be disjoint in the sense that there exist disjoint Borel sets $E_1 \subseteq X$ and $E_2 \subseteq X$ with $\bigcup \Gamma_1 \subset E_1$ and $\bigcup \Gamma_2 \subset E_2$.

The extremal length of the family:

$\Gamma := \set {\gamma_1 \cup \gamma_2:\ \gamma_1 \in \Gamma_1 \text{ and }\gamma_2 \in \Gamma_2}$

satisfies:

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

Proof

By the Series Law for Extremal Length:

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

Hence it remains only to prove the opposite inequality.

Let $\rho$ be a metric as in the definition of extremal length, normalized such that $\map A \rho = 1$.

The claim to be proved is that:

$\paren {\map L {\Gamma, \rho} }^2 \le \map \lambda {\Gamma_1} + \map \lambda {\Gamma_2}$

Define $\alpha_j := \sqrt {\map A {\rho {\restriction_{E_j} } } }$ for $j = 1, 2$.

Suppose either $\map \lambda {\Gamma_1}$ or $\map \lambda {\Gamma_2}$ is infinite.

It trivially follows that both $\alpha_1$ and $\alpha_2$ are positive.

Assume both extremal lengths are finite.

Let $\alpha_j = 0$.

Then:

$\map L {\Gamma_j, \rho} = 0$

Indeed, we can choose a metric $\tilde \rho$ that agrees with $\rho$ on $A_j$ and has total area $\epsilon > 0$.

Then:

$\paren {\map L {\Gamma_j, \rho} }^2 = \paren {\map L {\Gamma_j, \tilde \rho} }^2 \le \epsilon \cdot \map \lambda {\Gamma_j}$

Because $\epsilon$ was arbitrary, the claim follows.

Without loss of generality, suppose $\alpha_2 = 0$.

Then:

$\paren {\map L {\Gamma_j, \rho} }^2 = \paren {\map L {\Gamma_1, \rho} + \map L {\Gamma_2, \rho} }^2 = \paren {\map L {\Gamma_j, \rho} }^2 \le \map \lambda {\Gamma_1} \le \map \lambda {\Gamma_1} + \map \lambda {\Gamma_2}$

Thus both $\alpha_1$ and $\alpha_2$ are positive.

Let:

$\rho_j := \dfrac {\rho {\restriction_{E_j} } } {\alpha_j}$

Then $\map A \rho = 1$.

So:

\(\ds \paren {\map L {\Gamma, \rho} }^2\)

\(=\)

\(\ds \paren {\map L {\Gamma_1, \rho} + \map L {\Gamma_2, \rho} }^2\)

Definition of $\Gamma$

\(\ds \)

\(=\)

\(\ds \paren {\map L {\Gamma_1, \alpha_1 \rho_1} + \map L {\Gamma_2, \alpha_2 \rho_2} }^2\)

Definition of $\rho_j$ and disjointness of $E_1$ and $E_2$

\(\ds \)

\(=\)

\(\ds \paren {\alpha_1 \map L {\Gamma_1, \rho_1} + \alpha_2 \map L {\Gamma_2, \rho_2} }^2\)

Definition of $\map L {\Gamma_j, \rho_j}$)

\(\ds \)

\(\le\)

\(\ds \paren {\alpha_1^2 + \alpha_2^2} \paren {\map L {\Gamma_1, \rho_1}^2 + \map L {\Gamma_2, \rho_2}^2}\)

Cauchy-Bunyakovsky-Schwarz Inequality

\(\ds \)

\(\le\)

\(\ds \map L {\Gamma_1, \rho_1}^2 + \map L {\Gamma_2, \rho_2}^2\)

as $\alpha_1^2 + \alpha_2^2 \le \map A \rho = 1$

\(\ds \)

\(\le\)

\(\ds \map \lambda {\Gamma_1} + \map \lambda {\Gamma_2}\)

Definition of Extremal Length

In conclusion:

$\map \lambda \Gamma = \sup_\rho \map L {\Gamma, \rho} \le \map \lambda {\Gamma_1} + \map \lambda {\Gamma_2}$

as claimed.

$\blacksquare$