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 $X$.

Assume that every $\gamma\in\Gamma$ contains a $\gamma_1 \in \Gamma_1$ and a $\gamma_2 \in \Gamma_2$ such that $\gamma_1 \cap \gamma_2 = \emptyset$.

Then the extremal lengths of $\Gamma_1$, $\Gamma_2$ and $\Gamma$ satisfy
 * $ \lambda(\Gamma) \geq \lambda(\Gamma_1) + \lambda(\Gamma_2)$

Proof
Let $\rho_1 = \rho_1(z)|dz|$ and $\rho_2 = \rho_2(z)|dz|$ be conformal metrics as in the definition of extremal length.

We may assume that these are normalized so that $ A(\rho_j) = L (\Gamma_j, \rho_j)$ for $j \in \{1, 2\}$.

We define another metric $\rho = \rho(z)|dz|$ by:
 * $ \rho(z) := \max(\rho_1(z), \rho_2(z))$

Note that this is a well-defined metric.

By definition, the area form $\rho^2(z)|dz|$ satisfies
 * $\rho^2(z)|dz|^2 = \max(\rho_1(z)^2,\rho_2(z)^2)|dz|^2 \leq (\rho_1(z)^2 + \rho_2(z)^2)|dz|^2$

Hence
 * $A(\rho) \leq A(\rho_1) + A(\rho_2) = L(\Gamma_1,\rho_1) + L(\Gamma_2,\rho_2)$

On the other hand, let $\gamma\in \Gamma$, and let $\gamma_1$, $\gamma_2$ be as in the assumption. Then

Thus
 * $ L (\Gamma, \rho) \geq L (\Gamma_1, \rho_1) + L (\Gamma_2, \rho_2)$

Combining this with the inequality for the area, we see that:
 * $\dfrac {L (\Gamma, \rho)^2} {A(\rho)} \geq \dfrac {(L (\Gamma_1, \rho_1) + L (\Gamma_2, \rho_2))^2} {L (\Gamma_1, \rho_1) + L (\Gamma_2,\rho_2)} = L (\Gamma_1, \rho_1) + L (\Gamma_2, \rho_2)$

Taking the supremum over all metrics $\rho_1$ and $\rho_2$ as above, we see that
 * $ L (\Gamma) \geq L (\Gamma_1) + L (\Gamma_2)$

as claimed.

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