Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 133 / 4

For $x,y\in X$, the triangle inequality and the fact that $\varphi$ is an isometry giveThus the displacement function is $2$-Lipschitz, hence continuous.

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Choose $y_n\in X$ with $d_{\varphi }(y_n)\to 0$. By the cocompact group action, there is a compact set $K\subseteq X$ whose translates cover $X$. Choose ${\gamma }_n\in \Gamma$ such thatIsometric invariance givesFor a fixed basepoint $x_0$, compactness of $K$ gives $C={\max }_{x\in K}d(x_0,x)<\infty$, so $d(x_0,x_n)\le C$.

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The bounded sequence $(x_n)$ lies in a compact set because $X$ is a proper metric space. After taking a subsequence, $x_n\to x$. Since $d(x_n,{\varphi }_n{x}_n)\to 0$, both $x_n$ and ${\varphi }_n{x}_n$ eventually lie in one compact neighborhood $L$ of $x$. A properly discontinuous group action has only finitely many $g\in \Gamma$ with $gL\cap L\ne \varnothing$, so some conjugate $\psi$ occurs as ${\varphi }_n$ along an infinite subsequence. Continuity then givesThus $\psi$ fixes $x$. Since $\psi =\gamma \varphi {\gamma }^{-1}$ for some $\gamma$, the original element $\varphi$ fixes ${\gamma }^{-1}x$.

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Suppose a cocompact Fuchsian group contained a nonidentity parabolic isometry of the hyperbolic plane $\varphi$. After conjugating in the upper half-plane model, write $\varphi (z)=z+c$ with $c\ne 0$. The supplied estimate givesso the infimum of the displacement function is zero.

A Fuchsian action is properly discontinuous, and the action is cocompact by hypothesis. Parts b and c therefore imply that $\varphi$ fixes a point of the hyperbolic plane. A nonidentity parabolic isometry has no fixed point inside the plane, only one on its ideal boundary. This contradiction excludes nontrivial parabolic elements.

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