Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 156 (Mapping Class Groups)

A sufficient condition is that both collections are Alexander systems: within each collection the essential simple closed curves are pairwise nonisotopic, are in pairwise minimal position of curves or arcs, have no triple intersection points, and no three curves intersect pairwise. If ${\alpha }_i$ is isotopic to ${\beta }_i$ for every $i$, the simultaneous-isotopy lemma for Alexander systems gives an ambient isotopy $H_t$ with $H_1({\alpha }_i)={\beta }_i$ for all $i$. Successive applications of the bigon criterion prove the lemma while preserving the curves already matched.

The structure graph of an Alexander system is the embedded graphwith vertices at the intersection points, punctures, and any proper-arc endpoints. Its edges are the curve, arc, and boundary segments between consecutive vertices.

The Alexander method says that if an Alexander system $\Gamma$ fills $S$, then a homeomorphism preserving the isotopy class of every member is determined up to isotopy by its induced structure-graph automorphism. In particular, a homeomorphism inducing the identity is isotopic to the identity, and the stabilizer of all curve classes is finite.

To prove this, use part a to isotope the homeomorphism so that it carries the entire embedded union $|\Gamma |$ to itself. Its remaining action on that union is exactly the structure-graph automorphism. If this action is trivial, another isotopy fixes $|\Gamma |$ pointwise. Since $\Gamma$ fills, every component of $S\setminus |\Gamma |$ is a disc. The restriction to each closed complementary disc fixes its boundary, so the Alexander trick isotopes it to the identity there. These isotopies agree on their fixed boundaries and combine into an isotopy of the whole surface. The same argument shows that two homeomorphisms with the same graph action are isotopic.

Choose essential curves $\alpha ,\beta$ on $T^2$ with $i(\alpha ,\beta )=1$. They fill the torus. Let $[f]$ be central in $\mathrm{Mod}(T^2)$. For every curve $\gamma$,We also use that equality of Dehn twists about essential curves implies equality of their unoriented isotopy classes. Since $[f]$ commutes with $T_{\alpha }$ and $T_{\beta }$, it preserves both $\alpha$ and $\beta$.

The structure graph of this filling pair has one intersection vertex in the embedded union. Its orientation-preserving symmetries induced by a torus homeomorphism are the identity and simultaneous reversal of both curves. By the Alexander method, the corresponding mapping classes are the identity and the elliptic involutionThe involution commutes with every torus mapping class, as is also clear from the identification $\mathrm{Mod}(T^2)\cong S{L}_2(\mathbb{Z})$ where it is $-I$. Therefore the center of the mapping class group of the torus is

Fix a basepoint $x_0$. A homeomorphism $f:S\to S$ induces an isomorphism from ${\pi }_1(S,x_0)$ to ${\pi }_1(S,f(x_0))$. Choosing a path from $x_0$ to $f(x_0)$ identifies the latter group with $G={\pi }_1(S,x_0)$. Changing the path conjugates the resulting automorphism by an element of $G$, while an isotopy changes nothing in the outer automorphism group. Hence there is a natural homomorphismthe mapping-class action on the outer automorphism group of the fundamental group.

The commutator subgroup $[G,G]$ is characteristic: every automorphism sends commutators to commutators and therefore preserves the subgroup they generate. Thus an automorphism $A$ inducesThis is independent of the coset representative $g$. If $A^{\prime }=c_h\circ A$ differs from $A$ by an inner automorphism, thenbecause the abelianization is abelian. Hence $q([A])=\overline{A}$ is independent of the representative of the outer class. Finally $\overline{AB}=\overline{A}\overline{B}$, sois a well-defined group homomorphism.

For the punctured torus, $G=⟨\alpha ,\beta ⟩\cong F_2$ and $G^{\mathrm{ab}}\cong {\mathbb{Z}}^2$. Dehn twists about the two standard curves act on homology, in suitable oriented bases, byThe supplied result says that these matrices generate $S{L}_2(\mathbb{Z})$. Thus the image ofcontains the infinite group $S{L}_2(\mathbb{Z})$. Its intermediate group $\mathrm{Out}(F_2)$ must therefore be infinite.

Take the three-punctured sphere $S_{0,3}$. Its full mapping class group is finite, isomorphic to the permutation group $S_3$, whereasand $\mathrm{Out}(F_2)$ is infinite by part c. Hence the finite subgroup $p(\mathrm{Mod}(S_{0,3}))$ has infinite index in $\mathrm{Out}({\pi }_1(S_{0,3}))$.

The pure mapping class group $\mathrm{PMod}(S)$ consists of orientation-preserving mapping classes fixing every puncture individually. If $S^{\ast }$ adds a distinguished puncture and $\gamma$ is a loop in $S$, the point-pushing map moves that puncture once around $\gamma$ while leaving the old punctures fixed. For a simple loop, a thin annular neighborhood of $\gamma$ has boundary curves ${\gamma }_{+}$ and ${\gamma }_{-}$, and with consistent twist conventionsThe Birman exact sequence isfor the finite-type negative-Euler-characteristic surfaces under consideration.

On $S_{0,3}$, a simple proper arc joining two distinct specified punctures is unique up to isotopy relative to its ends. Indeed, a small regular neighborhood of the arc and its two ends has one boundary curve separating those two punctures from the third; the Jordan curve theorem gives the unique such separation, and the disc it bounds gives the isotopy between any two choices.

Choose the three joining arcs, one for each puncture pair, with disjoint interiors. They form an ideal triangle graph whose complement consists of two discs. A pure homeomorphism fixes all three endpoints and carries each arc to an isotopic arc. The simultaneous-isotopy lemma makes it fix the three arcs, and the Alexander trick on each complementary disc makes it isotopic to the identity. Therefore

First forget one puncture from the four-punctured sphere. Parts a and b giveNow forget the fifth puncture of $\Sigma =S_{0,5}$. The Birman exact sequence gives

Choose a minimal free generating pair $x,y$ of $F_2$ and lifts $\tilde{x},\tilde{y}$ in $\mathrm{PMod}(\Sigma )$. Let $F=⟨\tilde{x},\tilde{y}⟩$. The restriction $\pi {|}_F:F\to F_2$ is surjective. Since $F$ is generated by two elements and $F_2$ has minimal generator number two, the stated Hopf-type fact makes $\pi {|}_F$ an isomorphism. Thus $F\cong F_2$ and $F\cap K=1$.

For any $g\in \mathrm{PMod}(\Sigma )$, choose $f\in F$ with $\pi (f)=\pi (g)$. Then $f^{-1}g\in K$, so $g\in FK$. The kernel is normal by exactness. We have proved the semidirect-product decomposition of the pure mapping class group of the five-punctured sphereThe minimal generating-set sizes are therefore

The curve complex $\mathcal{C}(S)$ has one vertex for each isotopy class of essential simple closed curves, and a set of vertices spans a simplex when it has pairwise disjoint representatives.

To prove connectedness, put two curves $\alpha ,\beta$ in minimal position and induct on $i(\alpha ,\beta )$. If it is zero, their vertices are equal or joined by an edge. If it is positive, surgery of $\alpha$ along an outermost segment of $\beta$ produces a boundary component $\gamma$ of a regular neighborhood of the surgery. At least one choice is essential; it is disjoint from $\beta$ and satisfiesThe induction hypothesis gives a path from $\alpha$ to $\gamma$, and the edge from $\gamma$ to $\beta$ completes it. Thus $\mathcal{C}(S)$ is connected. This is the connectedness of the curve complex.

Curves ${\alpha }_1,\dots ,{\alpha }_n$ in minimal position fill a surface $S$ when every essential simple closed curve has positive geometric intersection with at least one ${\alpha }_i$. On a closed surface this is equivalent to every component ofbeing a disc: a non-disc complementary component contains an essential curve, while any curve disjoint from the collection lies in such a component.

Write the genus-two surface aswhere $\alpha$ is separating and each $T_i$ is a one-holed torus. In each $T_i$, choose two disjoint essential proper arcs $p_i,q_i$ from the boundary to itself that form a cut system, so $T_i\setminus (p_i\cup q_i)$ is a disc. Arrange their four endpoints on each copy of $\alpha$ and glue the boundaries so thatjoin cyclically into one simple closed curve $\beta$. After smoothing at the four gluing points, $i(\alpha ,\beta )=4$. Cutting along $\alpha$ and these four arcs leaves one disc from each $T_i$, so $S\setminus (\alpha \cup \beta )$ consists of two discs. Hence $(\alpha ,\beta )$ is the filling pair on a closed genus-two surface.

If their curve-complex distance were at most two, there would be an essential curve $\gamma$ disjoint from both: it would be the intermediate vertex of a length-two path. This contradicts filling. Therefore