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:
 * $\Gamma := \left\{{\gamma_1 \cup \gamma_2:\ \gamma_1 \in \Gamma_1 \text{ and }\gamma_2 \in \Gamma_2}\right\}$

satisfies:
 * $\lambda \left({\Gamma}\right) = \lambda \left({\Gamma_1}\right) + \lambda \left({\Gamma_2}\right)$

Proof
By the Series Law for Extremal Length, we have
 * $\lambda \left({\Gamma}\right) \ge \lambda \left({\Gamma_1}\right) + \lambda \left({\Gamma_2}\right)$

Hence it remains only to prove the opposite inequality.

Let $\rho$ be a metric as in the definition of extremal length, normalized such that $A \left({\rho}\right) = 1$.

The claim to be proved is that:
 * $\left({L \left({\Gamma, \rho}\right)}\right)^2 \le \lambda \left({\Gamma_1}\right) + \lambda \left({\Gamma_2}\right)$

Define $\alpha_j := \sqrt {A \left({\rho {\restriction_{E_j} } }\right)}$ for $j = 1, 2$.

Suppose either $\lambda \left({\Gamma_1}\right)$ or $\lambda \left({\Gamma_2}\right)$ 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:


 * $L \left({\Gamma_j, \rho}\right) = 0$

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

Then:
 * $\left({L \left({\Gamma_j, \rho}\right)}\right)^2 = \left({L \left({\Gamma_j, \tilde \rho}\right)}\right)^2 \le \varepsilon \cdot \lambda \left({\Gamma_j}\right)$

Because $\varepsilon$ was arbitary, the claim follows.

, suppose that $\alpha_2 = 0$.

Then:
 * $\left({L \left({\Gamma_j, \rho}\right)}\right)^2 = \left({L \left({\Gamma_1, \rho}\right) + L \left({\Gamma_2, \rho}\right)}\right)^2 = \left({L \left({\Gamma_j, \rho}\right)}\right)^2 \le \lambda \left({\Gamma_1}\right) \le \lambda \left({\Gamma_1}\right) + \lambda \left({\Gamma_2}\right)$

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

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

Then $A \left({\rho}\right) = 1$.

So:

In conclusion:
 * $\lambda \left({\Gamma}\right) = \sup_\rho L \left({\Gamma, \rho}\right) \le \lambda \left({\Gamma_1}\right) + \lambda \left({\Gamma_2}\right)$

as claimed.