Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 156 (Mapping Class Groups)

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.

The arcs $\alpha$ and $\beta$ bound a bigon if there are subarcs with common endpoints whose union is the boundary of an embedded disc and whose interiors are disjoint. They are in minimal position if $|\alpha \cap \beta |$ is the least possible intersection count among proper arcs isotopic to $\alpha$ and $\beta$ relative to their ends.

Let $\tilde{\alpha }$ and $\tilde{\beta }$ be lifts to the compactified hyperbolic plane $\overline{{\mathbb{H}}^2}$. Suppose they have two common points. Choose two consecutive common points along one lift. If both lie in ${\mathbb{H}}^2$, the intervening subarcs contain an innermost embedded disc. The covering projection is injective on its interior; otherwise a nontrivial deck translate would produce a still smaller such disc. Its projection is a bigon between $\alpha$ and $\beta$, contrary to the hypothesis.

The same innermost-disc argument works when one corner is on the circle at infinity, after deleting a sufficiently small horoball about that corner. The only possible obstruction would identify the ideal corner by both a hyperbolic deck transformation associated with an essential return and a parabolic deck transformation stabilizing the relevant puncture. A hyperbolic and a parabolic element of the surface group cannot have that fixed point in common, so the truncated disc again projects to a bigon. If both common points are ideal, truncate at both ends and apply the same argument. Therefore any pair of lifts intersects in at most one point of $\overline{{\mathbb{H}}^2}$.

If two arcs bound a bigon, pushing one side across its disc is an isotopy relative to the ends that removes the two corners. Such representatives cannot be in minimal position.

Conversely, compare $\alpha$ with an isotopic representative having the fewest intersections with $\beta$, and lift the isotopy to $\overline{{\mathbb{H}}^2}$. At the first stage where the original excess intersections disappear, two lifted arcs enclose an innermost disc. Part b rules out escape through a puncture or repeated intersections at the ideal boundary, so this disc projects injectively to a bigon on $S$. Thus absence of a bigon implies minimal position. This proves the bigon criterion for essential simple proper arcs.

Whenever $\alpha$ and $\beta$ bound a bigon, isotope one side across the bigon. The two removed crossings have opposite local signs: the induced directions around the two corners of an oriented disc are opposite. Thus the move reduces the geometric intersection number by two while leaving the algebraic intersection number of curves on an oriented surface unchanged.

Repeatedly remove bigons. The process terminates because the intersection count is a nonnegative integer, and the bigon criterion says that the resulting curves ${\alpha }_0$ and ${\beta }_0$ are in minimal position. Since every removal cancelled one positive and one negative crossing,

Choose a generic isotopy from ${\alpha }_0$ to ${\alpha }_1$ while keeping ${\beta }_0$ fixed. Except at finitely many times, all intersections are transverse. At each exceptional time a tangency creates or removes a pair of crossings. In oriented local coordinates the two crossings have opposite signs, so their contributions cancel. The signed sum is constant throughout the isotopy, and thereforeThis is the isotopy invariance of algebraic intersection number.

At a crossing $x$, the sign is the orientation of the ordered pair of tangent vectors $(\dot{\alpha },\dot{\beta })$. Exchanging the two vectors reverses orientation, soSumming over the same finite set of crossings gives

On a genus-two surface, let $\alpha$ be a separating simple closed curve that cuts the surface into two once-punctured tori. Choose a simple closed curve $\beta$ that passes from one side to the other and back, with the two crossings arranged in minimal position. The crossings have opposite signs, sobut the bigon criterion shows that they cannot be removed and henceEquivalently, $\alpha$ is separating and therefore represents zero in first homology, forcing its algebraic intersection with every curve to vanish even though its geometric intersection need not vanish.

Define the arc complex $\mathcal{A}(S)$ as follows. Its vertices are isotopy classes, relative to the punctures, of unoriented essential simple proper arcs. A set of $k+1$ vertices spans a $k$-dimensional simplex when the classes have representatives with pairwise disjoint interiors. Every orientation-preserving homeomorphism sends such representatives to such representatives, and isotopic homeomorphisms induce the same permutation of classes. Henceby simplicial automorphisms.

Fix punctures $p,q,r$. There is one isotopy class ${\ell }_p$ of essential arc from $p$ back to $p$: by the Jordan curve theorem, it separates $q$ from $r$, and any two such arcs are isotopic relative to the punctures.

Altogether $\mathcal{A}(S_{0,3})$ has six vertices:The joining arcs $a_{pq},a_{pr},a_{qr}$ span a triangle. For each puncture $p$, the three verticesspan another triangle. These are all the maximal simplices: ${\ell }_p$ must cross $a_{qr}$, and returning arcs based at distinct punctures cannot be made disjoint.

Every orientation-preserving permutation of the three punctures is realizable, while a homeomorphism acting trivially on the punctures is isotopic to the identity. ThusIt permutes the labels in the displayed description. There are two vertex orbits, the three joining arcs and the three returning arcs, as summarized by the arc complex of the three-punctured sphere.

Choose an essential returning arc $\alpha$ based at $p$ and an essential simple closed curve $\gamma$ with $i(\alpha ,\gamma )>0$. The iteratesare again simple proper arcs based at $p$. Their geometric intersection number with a fixed transverse arc grows linearly with $|n|$, so they represent infinitely many isotopy classes. This is the standard Dehn twist construction.

There are exactly two $\mathrm{Mod}(S_{0,4})$-orbits of vertices. Equality or inequality of the two endpoints is preserved by every homeomorphism. Conversely, a homeomorphism can send any ordered configuration of punctures and complementary discs of an arc to any other of the same endpoint type. Thus all arcs joining distinct punctures lie in one orbit, and all arcs returning to one puncture lie in the other. These are the arc-complex vertex orbits of the four-punctured sphere.