Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 156 / 1

Send each orientation-preserving isometry to its isotopy class. This is a group homomorphismFor $g\ge 2$, a hyperbolic isometry isotopic to the identity is the identity. One proof lifts it to the hyperbolic plane: after composing with a deck transformation, its lift commutes with the surface fundamental group. It consequently fixes the endpoints at infinity of every hyperbolic deck transformation. Those endpoints are dense in the circle at infinity, so the lift fixes that circle pointwise and is the identity. The homomorphism is therefore injective, proving the injection of a finite hyperbolic isometry group into a mapping class group.

Both low-genus analogues fail. Every orientation-preserving homeomorphism of the unit sphere is isotopic to the identity, but a round sphere has nontrivial finite rotation groups. On a flat torus, translation by a nonzero torsion point is a finite-order orientation-preserving isometry isotopic to the identity.

Let $G$ be generated by $r$ elements and choose $h\ge \max \{2,r\}$. The fundamental group of a closed orientable surface has the presentationMapping $a_1,\dots ,a_r$ to generators of $G$ and every $b_i$ and remaining $a_i$ to the identity defines a surjective group homomorphism ${\pi }_1(S_h)\twoheadrightarrow G$.

Its kernel determines a connected regular covering $S\to S_h$ with deck transformation group $G$. The action is free and orientation preserving. The covering surface has genus at least two, and the lifted hyperbolic metric makes every deck transformation an isometry. Part a now gives $G\hookrightarrow \mathrm{Mod}(S)$. This is the realization of a finite group as a surface deck group.

Because the action is free, the quotient $S_g/G$ is a closed orientable surface $S_h$, and $S_g\to S_h$ is a covering of degree $|G|$. The Euler characteristic under a finite covering givesor equivalentlyThe quotient inherits a hyperbolic metric, so $h\ge 2$. Hence $h-1\ge 1$ and $|G|\le g-1$.

Take the genus-two surface obtained from a regular hyperbolic octagon fundamental polygon by identifying opposite sides. Rotation of the octagon through $\pi /4$ respects the side pairing and descends to an orientation-preserving isometry of order eight. Thus $G=C_8$ acts on $S_2$ andThe fixed image of the octagon centre explains why this does not contradict part c.