# Parallel Law for Extremal Length

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## Theorem

Let $X$ be a Riemann surface.

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

Let $\Gamma_1$ and $\Gamma_2$ be disjoint, in the sense that:

- there exist disjoint Borel subsets $A_1, A_2 \subseteq X$ such that:
- for any $\gamma_1 \in \Gamma_1$ and $\gamma_2 \in \Gamma_2$, we have $\gamma_1 \subseteq A_1$ and $\gamma_2 \subseteq A_2$.

Let $\Gamma$ be a third curve family, with the property that every element $\Gamma_1$ and every element of $\Gamma_2$ contains some element of $\Gamma$.

Then the extremal length of $\Gamma$ satisfies:

- $\dfrac 1 {\lambda \left({\Gamma}\right)} \ge \dfrac 1 {\lambda \left({\Gamma_1}\right)} + \dfrac 1 {\lambda \left({\Gamma_2}\right)}$

## Proof

The assumption means that every element of $\Gamma_1 \cup \Gamma_2$ contains some element of $\Gamma$.

Hence:

\(\ds \frac 1 {\lambda \left({\Gamma}\right)}\) | \(\ge\) | \(\ds \frac 1 {\lambda \left({\Gamma_1 \cup \Gamma_2}\right)}\) | Comparison Principle for Extremal Length | |||||||||||

\(\ds \) | \(=\) | \(\ds \frac 1 {\lambda \left({\Gamma_1}\right)} + \frac 1 {\lambda \left({\Gamma_2}\right)}\) | Extremal Length of Union |

$\blacksquare$

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