Extremal Length of Composition

Proposition

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

Also suppose that $\Gamma_1$ and $\Gamma_2$ are disjoint in the sense that there exist disjoint Borel sets $E_1, E_2 \subseteq X$ with $\bigcup\Gamma_1\subset E_1$ and $\bigcup \Gamma_2\subset E_2$.

Then the extremal length of the family:

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

satisfies

$\lambda(\Gamma) = \lambda(\Gamma_1)+\lambda(\Gamma_2)$

Proof

By the Series Law for Extremal Length, we have

$\lambda(\Gamma) \geq \lambda(\Gamma_1)+\lambda(\Gamma_2)$

Hence it remains only to prove the opposite inequality.

Let $\rho$ be a metric as in the definition of extremal length, normalized such that $A(\rho)=1$. We claim that:

$(L(\Gamma,\rho))^2 \leq \lambda(\Gamma_1) + \lambda(\Gamma_2)$

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

We may assume that both values are positive.

If we define:

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

then $A(\rho_j)=1$.

So we have:

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle (L(\Gamma,\rho))^2\)	\(=\)	\(\displaystyle (L(\Gamma_1,\rho)+L(\Gamma_2,\rho))^2\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	(by Definition of $\Gamma$)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle (L(\Gamma_1,\alpha_1\rho_1)+L(\Gamma_2,\alpha_2\rho_2))^2\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	(by definition of $\rho_j$ and disjointness of $E_1$ and $E_2$)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle (\alpha_1L(\Gamma_1,\rho_1)+\alpha_2L(\Gamma_2,\rho_2))^2\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	(by definition of $L(\Gamma_j,\rho_j)$)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\leq\)	\(\displaystyle (\alpha_1^2+\alpha_2^2)(L(\Gamma_1,\rho_1)^2+L(\Gamma_2,\rho_2)^2)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	(by the Cauchy-Schwarz Inequality)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\leq\)	\(\displaystyle L(\Gamma_1,\rho_1)^2+L(\Gamma_2,\rho_2)^2\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	(since $\alpha_1^2+\alpha_2^2\leq A(\rho)=1$)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\leq\)	\(\displaystyle \lambda(\Gamma_1)+\lambda(\Gamma_2)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	(by definition of extremal length)

In conclusion:

$ \lambda(\Gamma) = \sup_{\rho} L(\Gamma,\rho) \leq \lambda(\Gamma_1)+\lambda(\Gamma_2)$

as claimed.

$\blacksquare$