топология / Farb, Margalit, A primer on mapping class groups

components of N that are isotopic to αi in S.

Let S − N denote the surface obtained from S −N by capping each boundary component with a punctured disk. The surface S − N is naturally homeomorphic to S − αi and thus there is a canonical isomorphism τ : Mod(S − N ) → Mod(S − αi).

By Theorem 3.12, the kernel of the homomorphism η2 : Mod(S − N ) → Mod(S − N ) is the group K2 generated by the Tα+i and Tα−i .

We consider the following diagram.

K1

composition is nothing other than the map ζ defined above, and so we are

COMPUTATIONS OF MAPPING CLASS GROUPS VIA CAPPING

We can use Proposition 3.13 to determine the mapping class groups of some surfaces with boundary.

Let P denote a pair of pants, that is, a compact surface of genus 0 with 3 boundary components (and no marked points). Recall from Proposition 2.3 that PMod(S0,3) = 1. Starting from this fact and applying Proposition 3.13 three times, we obtain the isomorphism

Mod(P ) ≈ Z3.

Let S11 denote a torus minus an open disk. We will show that

1 ^

Mod(S1 ) ≈ SL(2, Z)

where ^ denotes the universal central extension of . We will

SL(2, Z) SL(2, Z) need the following group presentations (see [154, §1.5]):

SL(2, Z) ≈ ha, b | aba = bab, (ab)6 = 1i

^

SL(2, Z) ≈ ha, b | aba = babi

From these presentations one sees that there is a surjective homomorphism

^ 6

SL(2, Z) → SL(2, Z) sending a to a and b to b with kernel h(ab) i = Z.

Mod(S1,1), where in each case the generators a and b map to the Dehn twists about the latitude and longitude curves. These maps fit into t he following diagram of exact sequences, where each square commutes:

The desired isomorphism follows from the five lemma.

We mention that the group ^ is also isomorphic to the braid group

SL(2, Z)

on 3 strands (see Chapter 9), the fundamental group of the complement of the trefoil knot in S3, as well as the local fundamental group of the ordinary cusp singularity, that is, the fundamental group of the complement in C2 of the affine curve x2 = y3.

3.4 RELATIONS BETWEEN TWO DEHN TWISTS

The goal of this section is to answer the question: what algebraic relations can occur between two Dehn twists? In fact we answer the more general question where powers of Dehn twists are allowed. We have already seen that Dehn twists about disjoint curves commute in the mapping class group. The next most basic relation between twists is the braid relation. Except in a few cases, we will see that there are no other relations between Dehn twists.

THE BRAID RELATION

The following proposition gives a basic relation between Dehn twists in

Mod(S), called the braid relation.

Proposition 3.15 (Braid relation) If a and b are isotopy classes of simple closed curves with i(a, b) = 1 then

β

Ta

TaTb(α)

isotopy

β

Figure 3.9 The proof of Proposition 3.16.

Proof. The relation

TaTbTa = TbTaTb

is equivalent to the relation

(TaTb)Ta(TaTb)−1 = Tb.

By Fact 3.6, this is equivalent to the relation

TTa Tb(a) = Tb.

Applying Fact 3.5, this is equivalent to the equality

TaTb(a) = b.

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By the change of coordinates principle, it suffices to check t he last statement for any two isotopy classes a and b with i(a, b) = 1. The computation is shown in Figure 3.9, where α is some representative of a and β is some

isomorphism of Theorem 2.5 the braid relation corresponds to the familiar

The next proposition records our rephrasing of the braid relation for use in our proof of Theorem 5.1 below.

Proposition 3.16 If a and b are isotopy classes of simple closed curves that satisfy i(a, b) = 1, then TaTb(a) = b.

The braid relation gets its name from the fact that it resembles one of the defining relations for the braid group. This is not a coincide nce; see Section 9.4.

One can ask for a converse to the braid relation: if two Dehn twists satisfy the braid relation algebraically, then do the corresponding curves necessarily have intersection number one? McCarthy gave the following proof that the answer is yes [115]. Theorem 3.18 below is a much more general fact; we consider Proposition 3.17 as a warmup.

Proposition 3.17 If a and b are distinct isotopy classes of simple closed curves and the Dehn twists Ta and Tb satisfy TaTbTa = TbTaTb, then i(a, b) = 1.

Proof. As in the proof of Proposition 3.15, the relation TaTbTa = TbTaTb is equivalent to the statement that TaTb(a) = b, which implies

i(a, TaTb(a)) = i(a, b).

Applying Ta−1 to both curves on the left hand side of the equation, we see that

i(a, Tb(a)) = i(a, b).

Now, by Proposition 3.2, we have that

i(a, b)2 = i(a, b).

And so i(a, b) is either equal to 0 or 1. If i(a, b) were 0, an application of Fact 3.8 reduces the relation to Ta = Tb, which, by Fact 3.5, contradicts the

THEOREM 3.18 Let a and b be two simple closed curves in a surface S. If i(a, b) ≥ 2, then the group generated by Ta and Tb is isomorphic to the free group F2 of rank 2.

We can also say what happens in the other cases. If a = b, then hTa, Tbi ≈ Z since Taj = Tbk if and only if a = b and j = k. If a 6= b and i(a, b) = 0, then hTa, Tbi is isomorphic to Z2 by Fact 3.8 plus the fact that Taj = Tbk if and only if a = b and j = k. When i(a, b) = 1 we have that

hTa, Tbi ≈ Mod(S11) ≈ hx, y | xyx = yxyi,

where S11 is a torus with an open disk removed (see above).

We remark that the question of which groups can be generated by three Dehn twists is completely open. See Section 6.1 for one relation between three Dehn twists.

Below we give the proof of Theorem 3.18 published by Ishida and HamidiTehrani [84, 65]. The theorem, though, was apparently known to Ivanov (and perhaps others) in the early 1980's. We first introduce t he so-called “ping pong lemma,” which is a basic and fundamental tool from geometric group theory. It is a method to prove that a group is free by understanding how it acts on a set. Poincar´e used this method to prove that if two hyperbolic translations have different axes, then sufficiently h igh powers of these elements generate a free group of rank 2.

Lemma 3.19 (Ping pong lemma) Let G be a group acting on a set X. Let g1, . . . , gn be elements of G. Suppose that there are nonempty, disjoint subsets X1, . . . , Xn of X with the property that, for each i and each j 6= i, we have gik (Xj ) Xi for every nonzero integer k. Then the group generated by the gi is a free group of rank n.

86 CHAPTER 3

Proof. We need to show that any nontrivial freely reduced word in the gi represents a nontrivial element of G. First suppose that w is a freely reduced word that starts and ends with a nontrivial power of g1. Then for any x X2, we have w(x) X1, and so w(x) 6= x since X1 ∩ X2 = . Thus w represents a nontrivial element of g. Since any other freely reduced word in the gi is conjugate to a word that starts and ends with g1, every freely reduced word in the gi represents an element of G that is conjugate to a

Proof of Theorem 3.18. Suppose that i(a, b) ≥ 2. Let G be the group generated by g1 = Ta and g2 = Tb, and let X be the set of isotopy classes of simple closed curves in S. The group G acts on X. With the ping pong lemma in mind, we define sets Xa and Xb as follows:

Xa = {c X : i(c, b) > i(c, a)}

Xb = {c X : i(c, a) > i(c, b)}

These sets are obviously disjoint, and they are nonempty since a Xa and b Xb.

By the ping pong lemma, the proof is reduced to checking that Tak (Xb) Xa and Tbk(Xa) Xb for k 6= 0. By symmetry, we only need to check the former inclusion.

−i(b, c) ≤ i(Tak (c), b) − |k|i(a, b)i(a, c) ≤ i(b, c).

If c Xb, then i(a, c) > i(b, c). Since k 6= 0, the left-hand inequality implies

i(Tak (c), b) ≥ |k|i(a, b)i(a, c) − i(b, c)

≥2|k|i(a, c) − i(b, c) > 2|k|i(a, c) − i(a, c) = (2|k| − 1)i(a, c)

≥i(a, c)

=i(Tak (a), Tak (c))

=i(a, Tak (c)).

A free group in SL(2, Z). The proof of Theorem 3.18 given above is inspired by a proof that the matrices

The classification of groups generated by two Dehn twists. With a little more care, the method of proof of Theorem 3.18 can be applied to give the stronger statement that hTaj , Tbk i ≈ F2 except if i(a, b) = 0 or if i(a, b) = 1 and the set {j, k} is equal to {1}, {1, 2}, or {1, 3}. When j = k = 1 we already know that we have the braid relation. And in the other exceptional cases, there exist nontrivial relations as well. For instance, if i(a, b) = 1, then Ta2 and Tb satisfy the relation

Ta2TbTa2Tb = TbTa2TbTa2

and Ta3 and Tb satisfy

Ta3TbTa3TbTa3Tb = TbTa3TbTa3TbTa3.

What is more, it turns out that these are the defining relation s for the groups hTa2, Tbi and hTa3, Tbi. The group hTa2, Tbi corresponds to a well-known index 3 subgroup of B3 (the subgroup “fixing” the first strand). The group hTa3, Tbi does not seem to be a well-known subgroup of B3. Luis Paris has explained to us that this is an index 8 subgroup of B3 and he has used the Reidemeister–Schreier algorithm to give an elementary pro of that the stated relation is the unique defining relation; see [140].

Combining the results from this section, we can completely list all possibilities for groups generated by powers of two Dehn twists. In the table we assume that a and b are nontrivial and that j ≥ k > 0, and that the underlying surface is not T 2 or S1,1.

88 CHAPTER 3

Chapter Four

The Dehn–Nielsen–Baer Theorem

The Dehn–Nielsen–Baer Theorem states that Mod(Sg ) is isomorphic to an index two subgroup of the group Out(π1(Sg )) of outer automorphisms of π1(Sg ). This is a beautiful example of the interplay between topology and algebra in the mapping class group. It relates a purely topological object, Mod(Sg ), to a purely algebraic one, Out(π1(Sg )). Further, these are related via hyperbolic geometry!

We begin by defining the objects in the statement of the theore m.

Extended mapping class group. The extended mapping class group, denoted Mod±(S), is the group of isotopy classes of all homeomorphisms of S, including the orientation-reversing ones.1 The group Mod(S) is an index two subgroup of Mod±(S). There is a homomorphism Mod±(S) → Z/2Z which records whether or not an element is orientation preserving or not, and we have the short exact sequence:

For any S, there is an order 2 element of Mod±(S) that reverses orientation, and so the sequence (4.1) is split.

As a first example, we have Mod±(S2) Z/2Z. Also, it follows from the

=

fact that Mod(T 2) ≈ SL(2, Z) (Theorem 2.5) that

Mod±(T 2) ≈ GL(2, Z).

1For the surfaces S0,1 and S0,2, we must be careful to define Mod±(S) as the group of isotopy classes of homeomorphisms; for these surfaces, every homeomorphism is homotopic to an orientation-preserving homeomorphism.

Similarly, we have:

Mod±(S0,3) ≈ Σ3 × Z/2Z

Mod±(S0,4) ≈ PGL(2, Z) (Z/2Z × Z/2Z)

Mod±(S1,1) ≈ GL(2, Z)

We remark that, the way we have defined things, we do not automa tically have a definition of the extended mapping class group for a sur face S with boundary, since a homeomorphism that is the identity on ∂S is necessarily orientation preserving.

Outer automorphism groups. For a group G, let Aut(G) denote the group of automorphisms of G. For any h G, there is an associated inner automorphism Ih : G → G given by

g 7→hgh−1

for all g G. For Φ Aut(G) and h G, we have

Φ ◦ Ih ◦ Φ−1 = IΦ(h).

Thus the inner automorphisms form a normal subgroup of Aut(G), called the inner automorphism group of G, denoted Inn(G).

The outer automorphism group of G is defined as the quotient

Out(G) = Aut(G)/ Inn(G).

In other words, Out(G) is the group of automorphisms of G, considered up to conjugation. Note that while an element of Out(G) does not act on the set of elements in G, it does act on the set of conjugacy classes of elements in G.

A natural homomorphism. Let S be a surface with χ(S) ≤ 0. The universal cover of S is contractible, and so S is a K(π1(S), 1) space. As mentioned in the proof of Theorem 2.5, we thus have a correspondence:

Let p S. Given a map φ : S → S, and a path γ from p to φ(p), we obtain a homomorphism φ : π1(S, p) → π1(S, p) as follows. For a loop α based at p, we set

φ ([α]) = [γ φ(α) γ−1].