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For any element m of $Mob^+(\mathbb H)$ other than the identity, define its displacement $disp(m)$ to be

$disp(m) = inf\{d_\mathbb H(z, m(z)) | z ∈ \mathbb H\}$

Show that if $m$ is parabolic then $disp(m)=0$

(m is parabolic if it has one fixed point in $\mathbb R$ in which case it is conjugate in $Mob( \mathbb H)$ to $q(z)=z+1$)

I have shown that if $m$ is loxodromic then $disp(m)>0$ and that if $m$ is elliptic then $disp(m)=0$ but I am struggling with the parabolic case.

user2973447
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1 Answers1

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As the conjugation in $Mob(\mathbb{H})$ are the isometries, without loss of generality we can assume that the parabolic element $m(z)=q(z)=z+1$. Therefore the fixed point of $m$ is $\infty$. consider two sequence of points $z_n=in$ and $m(z_n)=1+in$ where $n\in \mathbb{N}$. Now remember the hyperbolic distance $d_{\mathbb{H}}=\frac{d_{E}}{y}$ where $d_E$ is the Euclidean distance. Therefore $d_{\mathbb{H}}(x_n,m(z_n))=\frac{1}{n}\rightarrow 0$ as $n\rightarrow \infty$. Hence the imfimum is zero.

tessellation
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