Properties of Grötzsch and Teichmüller Moduli

From ProofWiki
Jump to navigation Jump to search





Lemma

For any number $R > 1$, let the modulus of the Grötzsch annulus be denoted $\map M R$.



Then:

$\map M R = 2 \map M {\dfrac {1 + R} {2 \sqrt R} }$

Similarly, for any $R > 0$, let the modulus of the Teichmüller annulus be denoted $\map \Lambda R$.

Then:

$\map \Lambda R \cdot \map \Lambda {\dfrac 1 R} = \dfrac 1 4$

In particular:

$\map \Lambda 1 = \dfrac 1 2$

Furthermore, these two quantities are related by:

$\map M R = \map \Lambda {\dfrac {\paren {1 - R}^2} {4 R} }$

and:

$2 \map M R = \map \Lambda {R^2 - 1}$


Proof

We begin by proving the equation relating $\map \Lambda R$ and $\map \Lambda {\dfrac 1 R}$.

To do so, we consider the quadrilateral $\map Q R$ given by the upper half plane with the boundary arcs $\closedint {-1} 0$ and $\hointr R \infty$.

Then

$\map \mod {\map Q R} = 2 \map \Lambda R$

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

Then:

$\map \mod {Q'} = \dfrac 1 {\map \mod {\map Q R} } = \dfrac 1 {2 \map \Lambda R}$

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



Hence, by Invariance of Extremal Length under Conformal Mappings:

$2 \map \Lambda {\dfrac 1 R} = \map \mod {Q'} = \dfrac 1 2 \map \Lambda R$

Rearranging yields the desired identity.


To prove the first equality (regarding $\map M R$), let $\map G R$ denote the Grötzsch annulus.

Consider the set

$U := \set {z \in \C: \cmod z > 1, z \notin \hointr {\sqrt R} \infty \text{ and } z \notin \hointl {-\infty} {\sqrt R} }$

Then $z \mapsto z^2$ maps $U$ to $\map G R$ as a covering map of degree $2$.

Hence

$\map M R = \map \mod {\map G R} = 2 \map \mod U$

On the other hand, the Möbius transformation:

$z \mapsto \dfrac {1 + \sqrt R z} {z + \sqrt R}$

is a conformal isomorphism between $U$ and $\map G {\dfrac {1 + R} {2 \sqrt R} }$.

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:

$z \mapsto \dfrac z {\paren {1 + z}^2}$

maps $\map G R$ conformally onto the set:

$V := \C \setminus \paren {\hointr {\dfrac 1 4} \infty \cup \closedint 0 {\dfrac R {\paren {1 + R}^2} } }$

On the other hand, the Möbius transformation:

$z \mapsto z \cdot \dfrac {\paren {1 + R}^2} R - 1$

takes $V$ conformally onto the Teichmüller domain for:

$\dfrac 1 4 \cdot \dfrac {\paren {1 + R}^2} R - 1 = \dfrac {\paren {1 - R}^2} {4 R}$

So:

$\map M R = \map \Lambda {\dfrac {\paren {1 - R}^2} {4 R} }$

as claimed.


The second relation can be proved from the first, together with the property of $\map M R$ 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 $\hointr 1 \infty$ to itself.

Then:

\(\ds \map \Lambda {R^2 - 1}\) \(=\) \(\ds \map \Lambda {\dfrac {\paren {1 + Q}^2} {4 Q} - 1}\)
\(\ds \) \(=\) \(\ds \map \Lambda {\dfrac {\paren {1 - Q}^2} {4 Q} }\)
\(\ds \) \(=\) \(\ds \map M Q\)
\(\ds \) \(=\) \(\ds 2 \map M R\)




Alternatively, we can also prove the last equality directly: reflection of the Grötzsch annulus in the unit circle yields the set:

$W := \C \setminus \paren {\closedint 0 {\dfrac 1 R} \cup \hointr R \infty}$

It follows that:

$\map \mod W = 2 \map M R$

On the other hand, the map:

$z \mapsto R z - 1$

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

$\blacksquare$


Sources