Properties of Grötzsch and Teichmüller Moduli

Lemma
For any number $R > 1$, let the modulus of the Grötzsch annulus be denoted $M \left({R}\right)$.

Then:
 * $M \left({R}\right) = 2 M \left({\dfrac {1 + R} {2 \sqrt R} }\right)$

Similarly, for any $R > 0$, let the modulus of the Teichmüller annulus be denoted $\Lambda \left({R}\right)$.

Then:
 * $\Lambda \left({R}\right) \cdot \Lambda \left({\dfrac 1 R}\right) = \dfrac 1 4$

In particular:
 * $\Lambda \left({1}\right) = \dfrac 1 2$

Furthermore, these two quantities are related by:
 * $M \left({R}\right) = \Lambda \left({\dfrac{\left({1 - R}\right)^2} {4 R} }\right)$

and:
 * $2 M \left({R}\right) = \Lambda \left({R^2 - 1}\right)$

Proof
We begin by proving the equation relating $\Lambda \left({R}\right)$ and $\Lambda \left({\dfrac 1 R}\right)$.

To do so, we consider the quadrilateral $Q \left({R}\right)$ given by the upper half plane with the boundary arcs $\left[{-1 \,.\,.\, 0}\right]$ and $\left[{R \,.\,.\, \infty}\right)$.

Then
 * $\mod \left({Q \left({R}\right)}\right) = 2\Lambda \left({R}\right)$

Now consider the quadrilateral $Q'$ that again consists of the upper half plane, but now with the boundary arcs $\left({-\infty \,.\,.\, 0}\right]$ and $\left[{0 \,.\,.\, R}\right]$.

Then:
 * $\mod \left({Q'}\right) = \dfrac 1 {\mod \left({Q \left({R}\right)}\right)} = \dfrac 1 {2 \Lambda \left({R}\right)}$

On the other hand, the function $z \mapsto \dfrac {-z} R$ takes $Q'$ conformally to $Q \left({\dfrac 1 R}\right)$.

Hence, by Invariance of Extremal Length under Conformal Mappings:
 * $2 \Lambda \left({\dfrac 1 R}\right) = \mod \left({Q'}\right) = \dfrac 1 2 \Lambda \left({R}\right)$

Rearranging yields the desired identity.

To prove the first equality (regarding $M \left({R}\right)$), let $G \left({R}\right)$ denote the Grötzsch annulus.

Consider the set
 * $U := \left\{ {z \in \C: \left|{z}\right| > 1, z \notin \left[{\sqrt R \,.\,.\, \infty}\right) \text{ and } z \notin \left({-\infty \,.\,.\, -\sqrt R}\right]}\right\}$

Then $z \mapsto z^2$ maps $U$ to $G \left({R}\right)$ as a covering map of degree $2$.

Hence
 * $M \left({R}\right) = \mod \left({G \left({R}\right)}\right) = 2 \mod \left({U}\right)$

On the other hand, the Möbius transformation:
 * $z \mapsto \dfrac {1 + \sqrt R z} {z + \sqrt R}$

is a conformal isomorphism between $U$ and $G \left({\dfrac {1 + R} {2 \sqrt R} }\right)$.

The claim now follows again from Invariance of Extremal Length under Conformal Mappings.

To prove the first relation between the Teichmüller and Grötzsch moduli, observe that the Koebe Function:
 * $z \mapsto \dfrac z {\left({1 + z}\right)^2}$

maps $G \left({R}\right)$ conformally onto the set:
 * $V := \C \setminus \left({\left[{\dfrac 1 4 \,.\,.\, \infty}\right) \cup \left[{0 \,.\,.\, \dfrac R {\left({1 + R}\right)^2} }\right]}\right)$

On the other hand, the Möbius transformation:
 * $z \mapsto z \cdot \dfrac{\left({1 + R}\right)^2} R - 1$

takes $V$ conformally onto the Teichmüller domain for:
 * $\dfrac 1 4 \cdot \dfrac{\left({1 + R}\right)^2} R - 1 = \dfrac{\left({1 - R}\right)^2} {4 R}$

So:
 * $M \left({R}\right) = \Lambda \left({\dfrac {\left({1 - R}\right)^2} {4 R} }\right)$

as claimed.

The second relation can be proved from the first, together with the property of $M \left({R}\right)$ that we proved above.

Indeed, choose $Q$ such that:


 * $R = \dfrac {1 + Q} {2 \sqrt Q}$

This is possible because the right-hand side is a strictly-increasing function from the interval $\left[{1 \,.\,.\, \infty}\right)$ to itself.

Then:

Alternatively, we can also prove the last equality directly: reflection of the Grötzsch annulus in the unit circle yields the set:
 * $W := \C \setminus \left({\left[{0 \,.\,.\, \dfrac 1 R}\right] \cup \left[{R \,.\,.\, \infty}\right)}\right)$

It follows that:
 * $\mod \left({W}\right) = 2 M \left({R}\right)$

On the other hand, the map:
 * $z \mapsto R z - 1$

takes $W$ to the Teichmüller domain for $R^2 - 1$.