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LECTURES ON THE MAPPING CLASS GROUP OF A
SURFACE
THOMAS KWOK-KEUNG AU, FENG LUO, AND TIAN YANG
Abstract. In these lectures, we give the proofs of two basic theorems onsurface topology, namely, the work of Dehn and Lickorish on generating themapping class group of a surface by Dehn-twists; and the work of Dehn andNielsen on relating self-homeomorphisms of a surface and automorphisms ofthe fundamental group of the surface. Some of the basic materials on hyper-bolic geometry and large scale geometry are introduced.
Contents
Introduction 1
1. Mapping Class Group 2
2. Dehn-Lickorish Theorem 13
3. Hyperbolic Plane and Hyperbolic Surfaces 22
3.1. A Crash Introduction to the Hyperbolic Plane 22
3.2. Hyperbolic Geometry on Surfaces 29
4. Quasi-Isometry and Large Scale Geometry 36
5. Dehn-Nielsen Theorem 44
5.1. Injectivity of Ψ 45
5.2. Surjectivity of Ψ 46
References 52
2010 Mathematics Subject Classification. Primary: 57N05; Secondary: 57M60, 57S05.Key words and phrases. Mapping class group, Dehn-Lickorish, Dehn-Nielsen.
i
ii
LECTURES ON MAPPING CLASS GROUPS 1
Introduction
The purpose of this paper is to give a quick introduction to the mapping class
group of a surface. We will prove two main theorems in the theory, namely, the
theorem of Dehn-Lickorish that the mapping class group is generated by Dehn
twists and the theorem of Dehn-Nielsen that the mapping class group is equal
to the outer-automorphism group of the fundamental group. We will present a
proof of Dehn-Nielsen realization theorem following the argument of B. Farb and
D. Margalit, [De, FM]. Along the way of the proof, we will introduce some of the
basic notions and tools used in the surface theory.
This paper is the result of a series of lectures given by the second author in the
Center of Mathematical Sciences, Zhejiang University, China during the summer
of 2008. It aims at providing a concise understanding of the surface theory,
especially suitable for graduate students. The prerequisites for these lectures have
been kept to a minimum. Basic knowledge of algebraic topology and Riemannian
geometry is all one needs to follow the lectures.
Let Σ be a compact oriented surface. A homeomorphism φ : Σ → Σ is called a
Dehn twist if its support A, the closure of x ∈ Σ | φ(x) 6= x, is homeomorphic
to an annulus under a homeomorphism I : S1×[0, 2] → A and I−1φI is a 2π-twist
on the annulus. A more precise definition will be given later in Definition 1.16.
The goal of these lectures is to prove the following two theorems.
Dehn-Lickorish Theorem. Suppose h : Σ → Σ is an orientation preserving
self-homeomorphism of a surface Σ so that h|∂Σ = id. Then there exists a finite
set of Dehn twists φ1, . . . , φn of Σ such that h is homotopic to the composition
φ1 · · · φn.
In fact, Dehn and Lickorish proved a stronger theorem that φi’s can be chosen
from a fixed finite set.
2 THOMAS AU, FENG LUO, AND TIAN YANG
Dehn-Nielsen Theorem . Let Σ be a closed surface of nonzero genus. Then
the group of self-homeomorphisms on Σ modulo homotopy is isomorphic to the
outer automorphism group of the fundamental group of Σ.
The theorem may be stated in a simplier form in terms of the notion of mapping
class groups, which we will discuss later.
In the first section, we will introduce the notions of mapping class groups and
Dehn twists. Simple examples of mapping class groups and basic properties of
Dehn twists will also be given. In the second section, how the mapping class
group of a surface is generated by Dehn twists will be discussed. Indeed, we will
prove the Dehn-Lickorish Theorem. Section three is a discussion of the hyperbolic
geometry of surfaces. It consists of a quick introduction of the hyperbolic plane
and the geometric interpretation of the fundamental group of a surface. In section
four, quasi-isometries and large scale geometry are introduced. Our attention is
on facts pertinent to our study of mapping class groups. In the last section, we
will prove the Dehn-Nielsen Theorem.
The authors are very grateful to the Center of Mathematical Sciences and the
hosts for the wonderful and quiet environment where people could think and work
well. The workshop also provided enlightenment to a number of mathematicians
and graduate students. Particular thanks should be addressed to the organiz-
ers especially Professor Lizhen Ji. We also thank the referee for many helpful
suggestions on improving the exposition in this paper.
1. Mapping Class Group
In this section, we will introduce basic notions about mapping class groups and
elementary techniques in handling Dehn twists.
Let Σ = Σg,r be the compact oriented surface of genus g with r boundary com-
ponents, r ≥ 0.
LECTURES ON MAPPING CLASS GROUPS 3
Definition 1.1. The mapping class group of Σ is given by
Γ(Σ)def:== Homeo(Σ)/ ' ,
where Homeo(Σ) is the set of homeomorphisms from Σ to Σ and the relation 'is homotopy of maps.
Remark . The relationship between homotopic and isotopic homeomorphisms on
surfaces was established by Baer. Baer’s theorem says that they are the same in
the following sense.
Theorem 1.2. (Baer) Suppose f and g are two homotopic homeomorphisms of
a compact oriented surface X so that f = g on the boundary of X. Then f is
isotopic to g by an isotopy leaving each point in the boundary of X fixed.
One may consult [St2, ch. 6] or [Ep] for a proof, which includes a discussion
on isotopic curve systems on surfaces. For this reason, we will not address the
isotopy issues in this paper.
The theory of mapping class groups plays an important role in the study of
low-dimensional topology.
Example 1.3. Let Hg be the genus g handlebody, that is, the regular neigh-
borhood of a wedge of g circles in the 3-space. The boundary of Hg is Σg,0. It
is well-known that for each closed orientable 3-manifold M , there is a positive
number g and an h ∈ Γ(Σg,0) such that
M = Hg ∪h Hg .
Such a decomposition of a 3-manifold is called a Heegaard Splitting. In other
words, M is obtained by gluing up two copies of Hg by a homeomorphism h of
their boundary surfaces Σg,0. The resulting manifold M depends only on the
homotopy class of h.
4 THOMAS AU, FENG LUO, AND TIAN YANG
Example 1.4. Thurston’s Virtual Fibre Conjecture states that any closed hy-
perbolic 3-manifold M has a finite cover N such that
N = Σg,0 × [0, 1]/ (x, 0) ∼ (h(x), 1) : x ∈ Σg,0 ,
where h is a homeomorphism on Σg,0. In short, N is a surface bundle over the
circle. This is one of the main conjectures in 3-manifold theory after the resolution
of the Geometrization Conjecture.
Example 1.5. Mapping class groups are needed in several other important areas.
For example, in the theory of Lefschetz fibrations of 4-manifolds, [Au, Don, Gom];
in the Teichmuller theory of surfaces in which the mapping class group acts on the
Teichmmuller space, [BH, Pa]; and in the theory of the Moduli space of algebraic
curves and Riemann surfaces, [HL].
We will assume basic facts from topology. For example, the Euler Characteristic
χ(Σg,r) of a surface satisfies
χ(Σg,r) = 2− 2g − r .
Surfaces Σ = Σg,r with χ(Σ) > 0 are the sphere S2 and the disk D2; for χ(Σ) = 0,
we have the torus T2 and the annulus S1 × [0, 1]; all others are of χ(Σ) < 0.
As a beginning, the reader is encouraged to work out the following.
Example 1.6. Γ(S2) ∼= Z/(2Z) and Γ(D2) = id .
The following was proved by J. Nielsen in his Ph.D. thesis in 1913.
Theorem 1.7. (Nielsen) Γ(T2) ∼= GL(2,Z).
Proof. We will represent T2 as the quotient space R2/Z2. There is a natural
homomorphism
ρ : Γ(T2) → Aut(H1(T2,Z)) = GL(2,Z)
that takes a homeomorphism class [h] to the induced map h∗ on the first homology
group.
LECTURES ON MAPPING CLASS GROUPS 5
First, we will show that ρ is surjective, i.e., Image(ρ) = GL(2,Z).
Let A ∈ GL(2,Z). The integral matrix A can be seen as the linear map (by abuse
of language)
A : R2 → R2, x 7→ Ax for x ∈ R2 ,
which preserves the integer lattice, i.e., A(x+ n) = A(x) +A(n) with A(n) ∈ Z2
for every x ∈ R2 and n ∈ Z2. This clearly induces a homeomorphism on the
quotient
A : T2 → T2, [x] 7→ [Ax] .
It is easy to verify that ρ(A) = (A)∗ = A with a suitable identification of n and
A(n) ∈ Z2 with the standard basis of R2.
Second, to prove that ρ is injective, we will show ker(ρ) = id .
Let [h] ∈ Γ(T2) where h ∈ Homeo(T2) and h∗ = id. Represent T2 as the quotient
space of [0, 1] × [0, 1] by gluing [0, 1] × 0 to [0, 1] × 1 and 0 × [0, 1] to
1 × [0, 1]. Let a, b denote the curves in T2 corresponding to quotient classes of
[0, 1]× 0, 1 and 0, 1 × [0, 1], respectively. Let P be the point of intersection
of a and b.
a
P
bb
a
Figure 1.1
Then H1(T2,Z) ∼= π1(T2, P ) = 〈a, b〉 ∼= Z2. Also, T2 \ a, b is a disk.
Since h∗ = id, we have h|a ' ida and h|b ' idb. Using such facts, we will construct
below a continuous function H from T2 × [0, 1] to T2 satisfying
• H|T2× 0 = id and H|T2× 1 = h, and
• H|a×[0,1] is a homotopy of h|a to ida, and
• H|b×[0,1] is a homotopy of h|b to idb.
6 THOMAS AU, FENG LUO, AND TIAN YANG
Let X = (T2 × 0, 1 )∪ ((a ∪ b)× [0, 1]). First of all, we define H from X to T2
as specified by the above three conditions. Let ℘ : [0, 1]2 → T be the quotient
map, which induces the map ℘ × id : [0, 1]3 → T × [0, 1]. Consider the map
H (℘× id) : ∂([0, 1]3) → T2. Since π2(T2) = 0, the map H (℘× id) extends to a
map F : [0, 1]3 → T2. According to this construction, the map F clearly induces
a continuous map T2 × [0, 1] → T2. This map is the required homotopy between
idT2 and h. Hence [h] = 1 ∈ Γ(T2). ¤
Exercise 1.1. A similar proof also shows that if χ(Σg,r) ≤ 0 and h ∈ Homeo(Σg,r)
satisfies h∗ = id in π1(Σg,r), then h ' id rel ∂Σg,r. Here, we need to use
π2(Σg,r) = 0 for surfaces with χ(Σg,r) ≤ 0.
This theorem about T2 is the key to the understanding of Γ(Σg,r) for all g, r.
Moreover, it can be seen that the essential conditions are that h∗ = id on π1 and
πk = 0 for k ≥ 2.
Definition 1.8. Let Homeo+(Σ, ∂Σ) be the group of all orientation preserving
self-homeomorphisms
h : (Σ, ∂Σ) → (Σ, ∂Σ), h|∂Σ = id .
The relative mapping class group is given by
Γ∗(Σ)def:== Homeo+(Σ, ∂Σ)
/('
∂Σ) ,
where'∂Σ
is the homotopy relative to ∂Σ, i.e., homotopy that leaves ∂Σ pointwise
fixed.
Example 1.9. Standard Dehn twist on the annulus
Consider Σ0,2 = S1 × [0, 2], we will demonstrate a generator for Γ∗(Σ0,2). Let
c = S1 × 1 be the central curve of the annulus. Define
Dc : S1 × [0, 2] → S1 × [0, 2], Dc(eiθ, t) = (ei(θ+πt), t) .
LECTURES ON MAPPING CLASS GROUPS 7
Clearly, Dc ∈ Homeo∗(Σ0,2, ∂Σ0,2). It is called the standard Dehn twist on the
annulus along the curve c. Note that it is a twist of one full turn on the annulus
leaving the boundary circles pointwise fixed.
α
( )αDc
Figure 1.2: The orientation of S1 × [0, 2] gives a normal into the paper.
Remark . The above picture is drawn according to traditional convention that a
positive turn is counterclockwise. That is equivalent to taking a normal into the
paper. In the future, pictures of Dehn twists on surfaces are drawn with respect
to the right hand orientation of the front face.
If Σ0,2 is drawn as a cylinder with outward normal orientation, the Dehn twist
along the central meridian curve c is drawn as follows. The rule is that if one
traces from either end of α, the image curveDc(α) turns right into c and follows c,
then turns left into the other end of α.
D ( )αα
c
c
Figure 1.3: Drawn with outward normal orientation.
Exercise 1.2. Show that Dc ' id but Dc 6' id rel ∂Σ0,2.
Proposition 1.10. Γ∗(S1 × [0, 2]) ' Z.
Sketch of proof. An outline of the the proof is given here. Details are left as
exercises to the reader. For simplicity, let A = S1 × [0, 2].
8 THOMAS AU, FENG LUO, AND TIAN YANG
Consider the homomorphism ϕ : Γ∗(A) → H1(A, ∂A) defined by ϕ([h]) = [h(α)]
where α is the arc e2πi × [0, 2]. The homomorphism ϕ is surjective as shown in
Example 1.9. To show that ϕ is injective, suppose that [h(α)] = 0 ∈ H1(A, ∂A).
Then the curves α and h(α) are isotopic by an isotopy leaving ∂A pointwise
fixed. Using the fact that A \ (α ∩ ∂A) is an open disk, one obtains a homotopy
between h and the identity. ¤
We will assume certain basic facts from surface theory without proof. Readers
may refer to [St, CB, Ro].
Example 1.11. For the torus, T2 = R2/Z2, let a, b be the curves defined by
a = ℘(0 × R) and b = ℘(R × 0), where ℘ : R2 → T2 is the covering projection.
Two homeomorphisms Da and Db on T2 are induced by the following linear
transformations (using the same notations) on R2
Da =
(1 −10 1
), Db =
(1 01 1
).
As a side remark, the homeomorphisms Da and Db are in fact Dehn twists on T2.
Since SL(2,Z) is clearly generated by these two matrices, it follows that Da
and Db generate the index two subgroup of Γ(T2) corresponding to orientation
preserving homeomorphisms. The curve Da(b) is shown in the picture.
b
a
Da ( )ba
Figure 1.4
Exercise 1.3. Show that Da(b) = a− b where the curves are also interpreted as
homology classes.
LECTURES ON MAPPING CLASS GROUPS 9
Example 1.12. The group Γ(Σg,r) acts transitively on the r circle boundary
components. Thus, if r ≥ 2, then the mapping class group and the relative map-
ping class group are different. Note that Dehn twists do not permute boundary
components.
πif is oddr
Figure 1.5
The proof of Dehn-Lickorish Theorem depends on the study of simple loops on
surfaces. In our context, unless it is specified otherwise, a simple loop in a
surface Σ is a simple closed curve in the interior Int(Σ). Let us begin with the
following.
Definition 1.13. Let Σ be a connected surface. A simple loop s ⊂ Int(Σ) is
nonseparating if Σ \ s is connected. Otherwise, it is separating.
In the case that a simple loop s is separating in a surface Σg,r with boundary,
we say that s bounds the boundary components b1, . . . , bk of Σg,r if b1, . . . , bk are
those boundary components other than s of a connected component of Σg,r \ s.If s bounds a single boundary component b, we say that s is parallel to the
boundary b.
Remark . A simple loop s bounds the boundary components b1, . . . , bk of Σg,r if
and only if it also bounds bk+1, . . . , br. Any simple loop in Σ0,r is separating.
Lemma 1.14. If s, s′ are two nonseparating simple loops in Int(Σg,r), then there
is an h ∈ Homeo+(Σg,r, ∂Σg,r) such that h(s) = s′.
Proof. Since both s and s′ are nonseparating, the surfaces Σs and Σs′ obtained by
removing small open neighborhoods of s and s′ have genus g−1 and r+2 boundary
10 THOMAS AU, FENG LUO, AND TIAN YANG
circles. Let us denote the new boundary circles s± and s′± respectively in the
surfaces Σs and Σs′ . Since Γ(Σg−1,r+2) induces transitive actions on the boundary
circles, one may choose a homeomorphism h0 from Σs to Σs′ such that h0(s±) =
s′± and h0|∂Σ = id. Now, if one reglues the surfaces Σs and Σs′ along s and s′
respectively, one obtains a homeomorphism h on Σg,r, which is induced by h0
and it satisfies h(s) = s′. ¤
Lemma 1.15. If s, s′ are two simple loops in Σ0,r bounding the same boundary
components, then there exists an h ∈ Homeo+(Σ0,r, ∂Σ0,r) such that h(s) = s′.
s
s’
Figure 1.6
Exercise 1.4. Prove this lemma, using the technique similar to that in Lemma 1.14.
Definition 1.16. (Dehn Twist) Let s be a simple loop in Int(Σg,r). A positive
Dehn twist or simply Dehn twist along s, Ds : Σg,r → Σg,r, is a homeomorphism
defined as follows.
Figure 1.7
Let c = S1 × 1 ∈ S1 × [0, 2] = Σ0,2 and
I : (Σ0,2, c) → (N (s), s) ⊂ Σg,r
be an orientation preserving embedding onto a neighborhood N (s) of s such that
the central curve c ⊂ Σ0,2 is mapped to I(c) = s ⊂ Σg,r. Denote Dc the Standard
LECTURES ON MAPPING CLASS GROUPS 11
Dehn twist on the annulus along c (see Example 1.9). Define
Dsdef:== I Dc I−1
and extend it to Σg,r by the identity map outside N (s).
It will be important in our study to know how to do computations with Dehn
twists. We will finish the section with a discussion on it.
Let a, b be simple loops in a surface that intersect transversely in k points. One
way to find Da(b) is by using the resolution of the intersection points.
Definition 1.17. Let x, y be two smooth arcs in Σ intersecting transversely at
a point p ∈ Σ. The resolution of x ∪ y at p from x to y is defined as follows (the
picture is drawn with right-hand orientation on the plane).
y
x resolution
to
tox y
y x
p
Figure 1.8
Take any orientation of x and determine the orientation of y at the intersection
point p such that the orientations on x and y from x to y are consistent with
the orientation of Σ. Then resolve the intersection point p according to the
orientations on x and y as in the picture above.
Note that if the orientation of x is reversed, so is the one of y; and so the resolution
does not change. Thus, the resolution depends only on the order of (x, y) and
the orientation of the surface Σ.
12 THOMAS AU, FENG LUO, AND TIAN YANG
Lemma 1.18. Computation of Da(b). Assume that |a ∩ b| = k ∈ Z trans-
versely, then Da(b) is obtained by taking k parallel copies ka of a and resolving
all intersections of (ka) ∩ b from ka to b.
Remark. In this case, one has to perform resolution at a total of k2 intersections.
At each intersection, it is sufficient to choose the compatible orientations of the
curves locally.
D
a
b
a ( )b
Figure 1.9: An example of Da(b) with k = 2.
Lemma 1.19. If a, b are simple loops in Σg,r such that they intersect trans-
versely at a single point (i.e., |a t b| = 1), then DaDb(a) ' b . In addition,
(DbDaDaDb)(a) ' a with orientation reversed.
Proof. Let I : S1 × [0, 2] → N (a) ⊂ Σg,r be an embedding of the annulus onto a
regular neighborhood N (a) of a such that b∩N (a) = I( 1 × [0, 2]). Moreover,
let a ∩ b ∈ N (a) be shown as in the first picture of Figure 1.10 below. Then,
after resolving the intersection a ∩ b, Db(a) intersects a at one point in N (a) as
in the second picture. A further resolution at this point gives DaDb(a) ⊂ N (a)
as in the third picture. On the other hand, outside N (a), the curves Db(a) and
DaDb(a) coincide with b. From the third picture, DaDb(a) is homotopic to b with
a homotopy supported in N (a).
LECTURES ON MAPPING CLASS GROUPS 13
a Dba
bDb( )a
( )aD
Figure 1.10
The second result is proved similarly and is illustrated by the following pictures
in Figure 1.11. In the first picture, p, p′ = b ∩ ∂N (a). After resolving the
intersection, we have the second picture where p, p′ = Da(b) ∩ ∂N (a) and
q, q′ = b ∩ ∂N (a). Then, a further resolution leads to the third picture.
a
bp p
q
pDa ( )b DbDa( )b
q
p p p
q q
’ ’
’ ’
’
Figure 1.11
It is indicated by the arrows in the last picture that DbDa(b) ' a with orientation
reversed. ¤
2. Dehn-Lickorish Theorem
The main objective of this section is to prove the following fundamental theorem.
Dehn-Lickorish Theorem. For any orientable compact surface Σ, the mapping
class group relative boundary
Γ∗(Σ) = Homeo+(Σ, ∂Σ)/('
∂Σ)
is generated by Dehn twists.
Remark . As it is mentioned before, a Dehn twist does not permute boundary
components.
14 THOMAS AU, FENG LUO, AND TIAN YANG
The proof of the Dehn-Lickorish Theorem is essentially following the idea of the
torus case in Theorem 1.7. One first establishes by induction a homotopy relative
boundary between a homeomorphism and a composition of Dehn twists on a set
of simple closed curves. Then this homotopy is extended to the whole surface.
For this purpose, we will begin with the study of the action of Dehn twists on
curves.
Definition 2.1. Let a, b ⊂ Int(Σ) be two simple closed curves. We say that
a ∼ b if there exists a finite sequence of simple closed curves c1, . . . , cn such that
Dε1c1 Dε2
c2 · · · Dε2cn(a) = b ,
where εj ∈ Z, j = 1, . . . , n.
Example 2.2. If |a t b| = 1, then a ∼ b. This is shown in Lemma 1.19 in the
previous section. Moreover, if a−1 denotes the same curve a with orientation
reversed, by choosing b with |a t b| = 1, one has a ∼ a−1.
Here is a simple way to characterize nonseparating curves (recall Definition 1.13)
and see their equivalences. The proof of the lemma is left as an exercise.
Lemma 2.3. A simple closed curve a ⊂ Σ is nonseparating if and only if there
is a simple closed curve b ⊂ Σ such that b t a (transversely) at a point.
Proposition 2.4. (Dehn-Lickorish)
(a) If a, b are nonseparating, then a ∼ b.
(b) If a, b are separating in Σ0,r and they bound the same boundary compo-
nents of Σ, then a ∼ b.
Proof. For (a), let a, b ⊂ Int(Σ) be nonseparating simple closed curves. We will
prove by induction on
I(a, b) = min ∣∣a′ ∩ b′
∣∣ : a′ ' a, b′ ' b, a′, b′ ⊂ Σ,
where ' denotes isotopy between curves in Σ.
LECTURES ON MAPPING CLASS GROUPS 15
We will handle the proof in four cases:
Case 1. I(a, b) = 0;
Case 2. |a ∩ b| = 2 with opposite algebraic intersection signs at the two intersec-
tion points;
Case 3. I(a, b) ≥ 2 with the same algebraic intersection sign at two adjacent
intersection points along the curve a;
Case 4. I(a, b) ≥ 3 with alternating algebraic intersection signs at three consecu-
tive intersection points along the curve a.
Case 1: Observe that if I(a, b) = 0, no matter whether Σ \ (a ∪ b) is connected,
there is a simple closed curve c such that |a ∩ c| = |b ∩ c| = 1. Indeed, assume
that Σ\ (a∪ b) is disconnected, then it has two connected components Σ1 and Σ2
because both a, b are nonseparating. For each j = 1, 2, Σj has boundary curves
aj , bj corresponding to a, b respectively. Choose a simple arc cj to connect aj to
bj in Σj . Then c is formed by c1, c2. The construction for connected Σ \ (a ∪ b)
is similar (see Figure 2.1 below). In any case, there is a simple loop c ⊂ Int(Σ)
such that c ∼ a and c ∼ b, as mentioned in Example 2.2 above. Thus, a ∼ b.
b
ac
b
ca
Figure 2.1
Case 2: Suppose a ∩ b = p, q with the signs at the intersections are opposite.
b
b
q
a
pb bL R
Figure 2.2
16 THOMAS AU, FENG LUO, AND TIAN YANG
The points p, q divide the curve b into segments bL and bR as in Figure 2.2. Let
b1 and b2 be the two boundary components of a regular neighborhood of a ∪ bR
such that b1 and b2 are not homotopic to a as in Figure 2.3 below.
a
p
q
b
b
1
2b
Figure 2.3
The three curves a, b1, b2 bound a pair of pants, Σ0,3 in Σ. Since the curve
a is nonseparating, at least one of b1 or b2 is also nonseparating. Say, b1 is
nonseparating. Note, |a ∩ b1| = |b ∩ b1| = 0. By the first case above, we have
a ∼ b1 and b1 ∼ b and hence a ∼ b.
Now, we deal with the remaining two cases which are illustrated in Figure 2.4, in
which p, q, r are consecutive intersection points along the curve a.
ba
pq
rb
a
b
b
b
r
q
p
Figure 2.4
Each of the two cases can be handled by replacing the curve b with another
nonseparating curve b′ that satisfies b′ ∼ b and 0 < |b′ ∩ a| < |b ∩ a|. Then,
eventually we have |b′ ∩ a| = 1 by induction and so b′ ∼ a.
Case 3: There are two points p, q ∈ a∩ b that are adjacent along the curve a such
that the signs of intersection are the same. In this case, we perform a surgery on
the curve b to obtain a curve b′ as in Figure 2.5 below.
LECTURES ON MAPPING CLASS GROUPS 17
b
b
a
b ’
Figure 2.5
Then, |b′ t b| = 1 so b′ is nonseparating by Lemma 2.3 and thus b ∼ b′ by
Lemma 1.19. Furthermore, |b′ ∩ a| < |b ∩ a|.
Case 4: There are three points p, q, r ∈ a∩b that are consecutive along the curve a
such that the signs of intersection alternate. We will work on a neighborhood U of
the subarc of the curve a containing p, q, r , as illustrated in Figure 2.6 below.
Note that there are two possible ways of connecting the three subarcs of the
curve b as shown. Nevertheless, we will only provide detailed illustrations for the
first one. Then, it can be seen that the same proof applies to the other situation.
b
r b
ap
b
q b
r b
ap
b
q
Figure 2.6
For the picture on the left hand side above, we perform a surgery on the curve b to
obtain a curve c with |c ∩ b| = 2 as in Figure 2.7(a) below. Then, to obtain Dc(b),
we need to consider the double of c as in Figure 2.7(b).
18 THOMAS AU, FENG LUO, AND TIAN YANG
b
r b
ap
b
q
c
Figure 2.7(a)
b
b
a b
2c
Figure 2.7(b)
Then after the resolution of b ∩ (2c), we obtain b′ = Dc(b). By definition of the
relation ∼, we have b′ ∼ b. The curves a, b′ within the neighborhood U are shown
as in Figure 2.8(a) below.
a
b ’
Figure 2.8(a)
a
b ’
Figure 2.8(b)
Then, up to isotopy supported within a neighborhood of b, the curve b′ can be
simplified as in Figure 2.8(b). Note that the dotted parts of b in Figure 2.7(a,b)
may intersect a, say k times, outside the neighborhood U of p, q, r . This
gives rise to 3k points of intersection between a and the dotted parts of b′ in
Figure 2.8(a). Nevertheless, after the isotopy as in Figure 2.8(b), the dotted part
only have k points of intersection with a. Hence, |b′ ∩ a| < |b ∩ a|. In addition,
the curve b′ is the homeomorphic image of a nonseparating loop b, thus b′ is also
nonseparating.
Note that the above proof is also valid for the second connection of subarcs of the
curve b. It is because intersection resolutions are done within the neighborhood
U of p, q, r and isotopies are supported in a neighborhood of b.
For the proof of statement (b), note that on Σ0,r, if a ∩ b = ∅ and if they bound
the same boundary components, then a is isotopic to b.
LECTURES ON MAPPING CLASS GROUPS 19
a b
Figure 2.9
Thus, we only need to consider the case that a∩b 6= ∅ and |a ∩ b| is even. Observe
that in this situation, all adjacent intersection points have opposite signs.
Now, we claim that |a ∩ b| ≥ 4. Otherwise, |a ∩ b| = 2 and then they must bound
different boundary components.
a
b
Figure 2.10
Therefore, we have |a ∩ b| ≥ 4 where all adjacent intersection points have opposite
signs. This has been dealt with above. Hence a ∼ b. ¤
We will need the following basic property of Dehn twists in relation to a homeo-
morphism on the surface.
Lemma 2.5. Let a, b ⊂ Σ be simple loops and ϕ : Σ → Σ be an orientation
preserving homeomorphism such that ϕ(a) = b. Then Db ' ϕ Da ϕ−1.
Proof. The proof is left as an exercise. ¤
Now, we are ready to prove the Dehn-Lickorish Theorem. It will be proved by
induction on the number |Σg,r| = 3g + r. We will only consider the case that
Euler Characteristic χ(Σ) < 0. The cases that χ(Σ) = 0, i.e., the annulus and
the torus, have been dealt with before in Theorem 1.7 and Proposition 1.10.
20 THOMAS AU, FENG LUO, AND TIAN YANG
It should be noted that the starting case is the 3-hole sphere, which is the unique
surface with |Σ0,3| = 3.
Lemma 2.6. Γ∗(Σ0,3) is generated by Dehn twists along the three boundaries.
Proof. The proof makes use of the following two facts, of which the proofs will
be left as exercises.
Fact 1. If α, β are two embedded arcs in Int(D2) with the same end points,
α(0) = β(0) and α(1) = β(1), then α is isotopic to β by an isotopy leaving ∂D2
and the end points fixed.
Fact 2. Let α, β be two arcs in the annulus Σ0,2 having the same starting point
in a boundary component and the same terminal point in another boundary
component. Then α is isotopic to Dnc (β) for some n ∈ Z where c is the central
curve in Σ0,2.
Let f : Σ0,3 → Σ0,3 be a homeomorphism with f |∂Σ0,3 = id and let D1, D2, and
D3 are the Dehn twists along the three boundary components b1, b2, and b3 of
Σ0,3 respectively. We are going to show that f is isotopic to Dm1 Dn
2D`3 for some
m,n, ` ∈ Z. Take an embedded arc s ∈ Σ0,3 with s(0) ∈ b1 and s(1) ∈ b2. Then
f(s) is also an embedded arc joining s(0) to s(1). Let E = D2 be the quotient
space obtained by identifying b1 to a point and also b2 to a point. Then the
quotient arcs [s] and [f(s)] are both embedded arcs in Int(E) = Int(D2) joining
the quotient points [b1] and [b2]. By Fact 1, [s] and [f(s)] are isotopic in E.
Since the isotopy leaves ∂E fixed, it may be lifted to an isotopy in Σ0,3. Without
loss of generality, by applying isotopic changes, we may assume that f(s) = s on
Σ0,3 \ (N(b1) ∪N(b2)) where N(bj) is a regular neighborhood of bj for j = 1, 2.
Applying Fact 2 on N(b1) and N(b2), which are both homeomorphic to Σ0,2,
we may assume that f |s = (Dm1 Dn
2 )|s for some m,n ∈ Z.
By a further isotopy of f , we may assume that
f |N(s∪b1∪b2) = (Dm1 Dn
2 )|N(s∪b1∪b2)
LECTURES ON MAPPING CLASS GROUPS 21
where N(s ∪ b1 ∪ b2) is a regular neighborhood of s ∪ b1 ∪ b2. Now, consider the
set A = Σ0,3 \ Int(N(s ∪ b1 ∪ b2)), which is an annulus Σ0,2. The restriction map
(Dm1 Dn
2 f)|A : Σ0,3 \ Int(N(s ∪ b1 ∪ b2)) → Σ0,3 \ Int(N(s ∪ b1 ∪ b2))
is a self-homeomorphism of the annulus A and it equals the identity on ∂A. Thus,
by the known fact (Proposition 1.10) that Γ∗(Σ0,2) is generated by the Dehn twist
along a boundary component, say b3, we conclude that the map (Dm1 Dn
2 f)|A is
isotopic to D`3, for some ` ∈ Z, by an isotopy leaving boundary pointwise fixed.
Then we simply extend this isotopy to an isotopy on Σ0,3 by the identity on
N(s ∪ b1 ∪ b2). As a result, f is isotopic to Dm1 Dn
2D`3 for m,n, ` ∈ Z. ¤
Lemma 2.7. Let s ∈ Σ be a simple loop that is neither null-homotopic nor
boundary parallel. If Σ′ is a connected component obtained from Σ by cutting
along s, then |Σ′| < |Σ|.
The proof is left as an exercise.
Proof of Dehn-Lickorish Theorem. Suppose that the statement of the theorem
holds for all Σ with |Σ| < k where k ≥ 3. Let Σ be a surface Σg,r with |Σ| = k
and h ∈ Homeo+(Σ, ∂Σ) be a homeomorphism.
Assume first that genus(Σ) > 0. Choose a nonseparating loop s ⊂ Σ. Consider s
and its image h(s). Since both are nonseparating, Proposition 2.4 gives that
h(s) ∼ s. In other words, there is an orientation preserving homeomorphism ϕ
of Σ, which is a composition of Dehn twists, such that ϕ(s) = h(s). By Ex-
ample 2.2, if s is oriented, we may assume that h(s) and ϕ(s) have the same
orientation. Hence, one may further assume that h|s = ϕ|s, i.e., (ϕ−1 h)|s = id.
Now, cut Σ open along the loop s to obtain a surface Σ′ with |Σ′| < |Σ|. In addi-
tion, we can regard the homeomorphism ϕ−1h as an element of Homeo+(Σ′, ∂Σ′).
By the induction hypothesis, ϕ−1 h is a product of Dehn twists on Σ′, which
corresponds to a product of Dehn twists on Σ. Thus, h is a product of Dehn
twists on Σ.
22 THOMAS AU, FENG LUO, AND TIAN YANG
For a surface Σ with genus(Σ) = 0, we may take a simple loop s ⊂ Σ such that
it bounds two boundary components. Then we may apply the same argument as
above on s and h(s) and conclude by induction. ¤
3. Hyperbolic Plane and Hyperbolic Surfaces
In this section, we will first recall briefly the geometry of the hyperbolic plane
and hyperbolic structures on surfaces. Then we will discuss the relationship
between the fundamental group of a surface, the deck transformation group, and
the geometric action of groups on surfaces. These notions will be frequently used
in the proof of the Dehn-Nielsen Theorem.
3.1. A Crash Introduction to the Hyperbolic Plane. We will mainly use
the upper half plane model of the Hyperbolic Plane. The upper half plane is
given by
H2 = z ∈ C | Im(z) > 0 , Im(z) = y where z = x+ iy, x, y ∈ R .
Its boundary in the Riemann sphere C = C ∪ ∞ is denoted R = R ∪ ∞ and
is called the circle of infinity.
The hyperbolic metric on H2 is the Riemannian metric defined by the symmetric
2-tensor
ds2 =dx2 + dy2
y2=
−4dzdz
(z − z)2.
The area form of the metric isdx ∧ dy
y2. If the length of a smooth curve γ ⊂ H2
is denoted by L(γ), then
L(γ) =
∫ b
a
|γ′(t)|Im(γ(t))
dt .
There is a way to identify the group of orientation preserving isometries of the hy-
perbolic plane, Isom+(H2), with the matrix group PSL(2,R) = SL(2,R)/±I .Let us start with three specific isometries. Let fλ(z) = λz where 0 < λ ∈ R and
gµ(z) = z + µ where µ ∈ R. Then by a simple calculation, both fλ and gµ are
orientation preserving hyperbolic isometries. In addition, the map h(z) =−1
z
LECTURES ON MAPPING CLASS GROUPS 23
preserves the hyperbolic metric. Indeed, let w = h(z). Then, the pull-back metric
h∗(ds2) is given by
h∗(ds2) =−4 dw dw
(w − w)2=
−4d(1z )d(1z )
(1z − 1z )
2
= −41/(z2z2)dzdz
(z − z)2/(z2z2)=
−4 dz dz
(z − z)2.
Thus h : H2 → H2 is an orientation preserving isometry.
A Mobius transformation
F (z) =az + b
cz + d, where ad− bc = 1,
preserves the upper half plane if and only if it has real coefficients a, b, c, d.
Now it is well known that each Mobius transformation F above is a composition
of fλ, gλ and h for λ ∈ R>0, µ ∈ R. Thus, F ∈ Isom+(H2). It follows that
PSL(2,R) acts on H2 as a group of isometries by the correspondence
(a bc d
)7→ az + b
cz + d.
We leave it as an exercise for the readers to verify that the group of all orien-
tation preserving isometries of H2 is PSL(2,R). Hint: show that SL(2,R) acts
transitively on the unit tangent vectors in H2.
We will give a classification of the orientation preserving isometries of H2 by
looking at the number of fixed points of the isometry in H2, where H2 is the
closure of H2 in the Riemann sphere C.
Let γ ∈ Isom+(H2) be γ(z) = az+bcz+d , a, b, c, d ∈ R, ad − bc = 1. Let z ∈ H2 be a
fixed point of γ, by definition, it means that
az + b
cz + d= z, i.e.,
(1) cz2 + (d− a)z − b = 0.
Note that Equation (1) is quadratic if c 6= 0 and is linear if c = 0. This equation
always has a solution z in the Riemann sphere. Note that its conjugate z is also
24 THOMAS AU, FENG LUO, AND TIAN YANG
a solution to (1) because of the fact that a, b, c, d ∈ R. We may thus assume that
z is a solution to (1) in H2.
If c 6= 0 and the discriminant ∆ = (d− a)2 +4bc < 0, then the point z lies in H2.
In this case, γ has a unique fixed point in H2. If c 6= 0 and ∆ = 0, then the
quadratic equation has a unique real solution a−d2c , which is again the only fixed
point of γ lying in R.
If c 6= 0 and ∆ > 0, then the quadratic equation has two distinct real solutions,
which are the two fixed points of γ lying in R.
Finally, if c = 0, d−a 6= 0, then γ(z) = adz+
bd . Thus, γ has two fixed points b
d−a
and ∞. If c = 0, d− a = 0, then γ(z) = z + bd , and ∞ is the unique fixed point
for γ 6= id.
To summarize the discussion above, we have the following classification.
Definition 3.1. We say that γ ∈ Isom+(H2) is of
• elliptic type if γ 6= id and it has a fixed point in H2;
• parabolic type if γ has no fixed point in H2 and only one fixed point
in H2;
• hyperbolic type if γ has no fixed point in H2 and two distinct fixed points
in H2.
It is clear from the definition that conjugate isometries, γ and σγσ−1, where
γ, σ ∈ Isom+(H2), are of the same type.
Below is an algebraic characterization of the types of isometries.
Proposition 3.2. Let A =
(a bc d
)∈ SL(2,R) be a matrix representative of
γ ∈ Isom+(H2), γ 6= id, then
(1) γ is of elliptic type if and only if |tr(A)| < 2;
(2) γ is of parabolic type if and only if |tr(A)| = 2;
LECTURES ON MAPPING CLASS GROUPS 25
(3) γ is of hyperbolic type if and only if |tr(A)| > 2.
Proof. We use the same notations introduced above. That is,
γ(z) =az + b
cz + d, a, b, c, d ∈ R , ad− bc = 1 ;
and z ∈ H2 is a fixed point of γ in H2 such that Equation (1) holds.
Let us begin with the case that c 6= 0. In this case, | tr(A)|2 = 4 + ∆ where
∆ = (d− a)2+4bc is the discriminant of the quadratic equation (1). Since c 6= 0,
any fixed point of γ in H2 is a root of the quadratic equation (1). Now ∆ < 0
if and only if the equation has no real root, i.e., γ has a unique fixed point in
H2. This shows statement (1). Next, ∆ = 0 if and only if the equation has a a
unique real root, i.e., γ has a unique fixed point in R. This verifies statement (2).
Finally, ∆ > 0 if and only if the equation has two distinct real roots, i.e., γ has
two distinct fixed points in R. This shows statement (3).
In the case that c = 0, we have ∆ = tr(A)−4 = (d−a)2 > 0 and ad = ad−bc = 1.
Now |tr(A)| = 2 if and only if a = d 6= 0, in which case γ(z) = z+ ba is parabolic.
This shows statement (2). Finally, |tr(A)| > 2 if and only if a 6= d, in which case
γ(z) = adz +
bd is hyperbolic. This verifies statement (3). ¤
From this proposition, it is easy to verify that fλ(z) = λz with 0 < λ ∈ R \ 1is hyperbolic; gµ(z) = z + µ with µ ∈ R is parabolic; and h(z) = −1/z is elliptic.
Our next task is to find all the geodesics in the hyperbolic plane. A geodesic is a
curve with the local distance minimizing property. We will adopt the terminology
that a hyperbolic geodesic line is a geodesic in H2 that is isometric to R. Also,
a geodesic ray is a subset of a geodesic line isometric to [0,∞). Since isometries
preserve geodesics, we will find one geodesic and obtain others as images of it
under isometries. Here is a typical geodesic in the upper half plane model found
by simple calculations.
26 THOMAS AU, FENG LUO, AND TIAN YANG
Example 3.3. The positive y-axis Y = iy | y > 0 is a geodesic in H2.
Let z(t), t ∈ [0, 1], parametrize a path from ia to ib in H2 where b > a > 0. Write
z(t) = x(t) + iy(t), where y(t) > 0. Then z′(t) = x′(t) + iy′(t) and the length of
the path is given by the integral∫ 1
0
√x′(t)2 + y′(t)2
y(t)dt ≥
∫ 1
0
√y′(t)2
y(t)dt ≥
∣∣∣∣∫ 1
0
y′(t)y(t)
dt
∣∣∣∣ = lnb
a.
Note that the above equalities hold if and only if x(t) = 0 and y′(t) ≥ 0. That
is the same as that z(t) ∈ Y and y(t) is monotonic increasing. Thus the positive
y-axis is a geodesic. Moreover, one obtains explicitly the distance between ia and
ib, that is,
dH2(ia, ib) = ln
(b
a
).
Y
a
b
i
i
x
z ( )t
Figure 3.1
Definition 3.4. The cross ratio of four complex numbers z1, z2, z3, z4 ∈ C is
defined to be (z1, z2, z3, z4)def:==
z1 − z3z1 − z4
· z2 − z4z2 − z3
.
The distance between ia and ib can be expressed in terms of cross ratio by
dH2(ia, ib) = ln(ia, ib,∞, 0) .
Let z 6= w ∈ H2 and α be the geodesic line passing through z, w. Then there
is a unique Mobius transformation ϕ ∈ Isom+(H2) taking ia, ib, Y to z, w, α
respectively. Let z = ϕ(0) and w = ϕ(∞) in R (see Figure 3.2 below). Then
ϕ(Y ) is the geodesic in H2 ending at z and w. More precisely, its closure in H2
intersects R at z and w. Furthermore, since ϕ preserves the cross ratio,
(?) dH2(z, w) = ln(z, w, w, z), z, w ∈ H2 .
LECTURES ON MAPPING CLASS GROUPS 27
A direct calculation shows that PSL(2,R) acts transitively on UTH2, the unit
tangent bundle of H2. Indeed, for all z = x + iy ∈ H2,
(√y x√
y
0 1√y
)maps i
to z, which shows that PSL(2,R) acts transitively on H2, and the subgroup(cos θ sin θ− sin θ cos θ
)| θ ∈ (0, 2π]
of PSL(2,R) acts transitively on the unit tan-
gent vectors at i as rotations. As a consequence, we have the following corollary
whose proof is left as an exercise.
Corollary 3.5. Each geodesic line in H2 is either a vertical line or a semicircle
perpendicular to the x-axis.
z w
z
w
~ ~
upper half plane model disk model
Figure 3.2
Remark . There is an analogous discussion using the disk model. One simply
replaces H2 by D2 and PSL(2,R) by the isometry groupz 7→ eiθ
(z − a
az − 1
)| a ∈ C, |a| < 1
.
In our pictures, we often use disk model whenever it is more illustrative.
A geodesic line α in H2 is determined by its end points x, y in ∂H2. For instance,
the end points of the positive y-axis Y are 0,∞.
Definition 3.6. Let γ ∈ Isom+(H2) be of hyperbolic type. The axis of γ, denoted
by Axis(γ), is the geodesic line with the two ends equal to the fixed points of γ.
By definition, we have Axis(σγσ−1) = σ(Axis(γ)) for σ ∈ Isom+(H2). Since γ
sends Axis(γ) to a geodesic line with the same end points, γ(Axis(γ)) = Axis(γ).
On the other hand, if Axis(γ1) = Axis(γ2), then γ1 and γ2 have the same fixed
28 THOMAS AU, FENG LUO, AND TIAN YANG
points. After a conjugation, one may assume their fixed points to be 0 and ∞.
It is then easy to show that there exists γ ∈ Isom+(H2) of hyperbolic type and
m,n ∈ Z such that γ1 = γm and γ2 = γn.
Next, we will describe the δ-neighborhood of a geodesic line s, namely, the set
Nδ(s) =z ∈ H2 : dH2(z, s) < δ
, δ > 0.
Let us begin with the following example.
Example 3.7. Let Y be the positive y-axis, then the set Nδ(Y ) is bounded by
two straight lines from 0 to ∞ making the same angle θ with Y at 0, namely
Nδ(Y ) = x+ iy : y > |x| cot(θ) , where tanh
(δ
2
)= tan
(θ
2
).
Indeed, let z ∈ Nδ(Y ), then dH2(z, Y ) < δ. Since for each 0 < λ ∈ R, the
hyperbolic isometry fλ leaves Y invariant, we have
dH2(λz, Y ) = dH2(λz, fλ(Y )) = dH2(z, Y ) < δ .
Therefore, if z ∈ Nδ(Y ), the whole straight line Lz = λz : 0 < λ ∈ R ⊂Nδ(Y ). To see that Nδ(Y ) is symmetric about Y , it suffices to observe that
dH2(x+ iy, Y ) = dH2(−x+ iy, Y ) , x ∈ R, y > 0 .
This shows that Nδ(Y ) is a sector at the origin.
To prove the formula for θ, let z0 = ei(π2−θ) be a point on the boundary of Nδ(Y ).
By the formula (?) of hyperbolic distance in terms of cross ratio above, we have
δ = dH2(i, z0) = ln(i, z0, 1,−1) = lni− 1
i+ 1· z0 + 1
z0 − 1= ln
(−1 + ieiθ
i− eiθ
).
In other words, eδ =−1 + ieiθ
i− eiθ. Consequently,
tanh
(δ
2
)=
eδ − 1
eδ − 1=
−1 + ieiθ
i− eiθ− 1
−1 + ieiθ
i− eiθ+ 1
= tan
(θ
2
).
Using an isometry in Isom+(H2), we obtain the δ-neighborhood of a general
geodesic line in H2 (Figure 3.3 below).
LECTURES ON MAPPING CLASS GROUPS 29
Corollary 3.8. Let s be a geodesic line in H2 ending at z0, z∞ in R. Then
for each δ > 0, ∂Nδ(s) consists of two circular arcs passing through z0 and z∞
so that the angle between each of the arcs and s is equal to θ where
tanh
(δ
2
)= tan
(θ
2
).
In particular,
Nδ(s) ∩ ∂H2 = s ∩ ∂H2 .
L L z z '
δ
θ
x
y
0
s
δ
N (s) δ
Figure 3.3
3.2. Hyperbolic Geometry on Surfaces. A hyperbolic surface Σ is a smooth
surface together with a complete Riemannian metric of constant curvature −1.
If the surface Σ is closed, we may equivalently define a hyperbolic structure as
a collection of smooth charts(Ui, φi) | φi(Ui) ⊂ H2
so that Σ =
⋃i Ui and
each transition function φi φ−1j on φj(Ui ∩ Uj) is a restriction of an isometry
in Isom+(H2). See [BP] for more details.
Proposition 3.9. Every closed orientable surface Σg,0 with g > 2 has a hyper-
bolic structure.
Proof. The surface Σg,0 is the quotient space of a regular Euclidean 4g-gon by
identifying pairs of opposite sides by Euclidean translations. If the sides are
labeled cyclically as a1, . . . , a2g, the algebraic relation on the boundary is given
by
a1a2 · · · a2g a−11 a−1
2 · · · a−12g .
Note that all the 4g vertices of the polygon are identified.
30 THOMAS AU, FENG LUO, AND TIAN YANG
2
a2
a1a 1
a3
a4
a4
a3
a
= 2g
Figure 3.4
To construct a hyperbolic structure on Σg,0, consider the set of all regular hy-
perbolic 4g-gons in H2. Such a polygon is determined up to isometry by its edge
length, denoted by t. Let α(t) be its inner angle. Then we have
limt→+∞α(t) = 0 and lim
t→0α(t) =
(4g − 2)π
4g
t goes to 0 t goes to ∞
Figure 3.5
Indeed, the first equation comes from the fact that when t → +∞, the adjacent
edges of the polygon are tangent to each other at the circle of infinity. The
second equation comes from the fact that when t → 0, the polygon becomes
asymptotically Euclidean, and (4g−2)π4g is the inner angle of the regular Euclidean
4g-gon. Here we have used the fact that the notion of angles in the disk model
of the hyperbolic plane coincides with that of the Euclidean space.
Since α(t) is a continuous function of t according to the cosine law, there exists
a t0 such that α(t0) =2π4g . Take this regular hyperbolic 4g-gon with inner angle
2π4g and identify opposite sides by hyperbolic isometries. Then the quotient space
Σg,0 has the induced hyperbolic structure. This is due to the Poincare Polyhedron
LECTURES ON MAPPING CLASS GROUPS 31
Theorem, see Maskit, [Ma], or Beardon, [Be]. Note that the key condition that
the sum of angles at all the vertices in the polygon is 2π is satisfied by the choice
of 2π/4g. ¤
By Proposition 3.9, we can identify the universal cover of the closed surface Σg,0
with the hyperbolic plane H2. Let
Θ : H2 → Σg = Σg,0 .
be the covering projection. Let p0 ∈ Σg and choose z0 ∈ H2 such that Θ(z0) = p0.
Then the fundamental group π1(Σg, p0) can be identified with the deck transfor-
mation group G of the universal cover.
For each γ ∈ G, let α be the geodesic segment from z0 to γ(z0). Then aγ = Θ(α)
is a loop based at p0 in Σg. The map Φ : G → π1(Σg, p0) having Φ(γ) = [aγ ]
is an isomorphism between the groups. For any other point z in H2, let s be a
path from z to γ(z) in H. Then the quotient of s is a loop Θ(s) in the surface Σg
freely homotopic to aγ . It is known that the free homotopy class of a loop in Σg
corresponds to a conjugacy class of an element in π1(Σg, p0).
p
aγ
0 Θ ( )s
Θ
α
s
z
z
γ( )z
0
( )γ z0
Figure 3.6
For this reason, we will take G = π1(Σg, p0) and consider an element of it as
either an isometry on H2 or a homotopy class of a loop at p0 in Σg. For z ∈ H2,
the point Θ(z) ∈ Σg is also denoted by [z]. In the rest of the paper, we will
always equip Σg with a fixed hyperbolic structure. Since the action of G on H2
is free, there is no elliptic element in G.
32 THOMAS AU, FENG LUO, AND TIAN YANG
Recall that the injectivity radius ε(q) at q ∈ Σg is the supremum over all positive
real numbers δ > 0 such that the δ-disk at q is isometric to a δ-disk in H2. The
injectivity radius ε(Σg) of Σg is defined by
ε(Σg)def:== inf
q∈Σg
ε(q) .
Lemma 3.10. Under the above identification Σg = H2/G where G = π1(Σg, p0),
we have the following:
(1) Each γ ∈ G− id is an isometry of hyperbolic type.
(2) For any γ ∈ G − id , the projection of Axis(γ) onto Σg is the unique
closed geodesic in the conjugacy class of γ ∈ π1(Σg, p0).
Proof. Let ε be the injective radius of Σg. Since Σg is compact, ε(Σg) > 0. Then,
by definition, any loop in Σg of length less than ε/2 is null homotopic.
For statement (1), suppose that there is a parabolic element γ ∈ G\ id . By the
classification of isometries, one may assume that γ(z) = z + a for z ∈ R, wherea ∈ R, after taking conjugation. By a direct calculation, dH2(z, z + a) = |a|
Im(z) .
So, there exists a point z ∈ H2, with sufficiently large Im(z), such that
dH2(z, γ(z)) <ε
2.
This shows that the projection of the geodesic segment from z to γ(z) is a loop ρ
in Σg of length at most ε/2 based at the point [z]. By the construction, ρ
is freely homotopic to a representative of γ, and so γ must be trivial, which
contradicts that γ 6= id. Since γ cannot be an elliptic element, the first statement
is established.
For statement (2), let a = Θ(Axis(γ)), then it is a geodesic in the conjugacy class
of γ. Let b be another closed geodesic freely homotopic to a and F : S1 × [0, 1] → Σg
be a smooth homotopy between a and b. Define
F : R× [0, 1] → Σg by F (s, t) = F (eis, t) .
LECTURES ON MAPPING CLASS GROUPS 33
Then F can be lifted to F ∗ : R × [0, 1] → H so that F ∗(R, 0) and F ∗(R, 1) are
two geodesic lines satisfying Θ F ∗(R, 0) = a and Θ F ∗(R, 1) = b respectively.
Since F is smooth, there is c > 0 such that the hyperbolic length of any curve
F (eis, t), t ∈ [0, 1], is bounded by c for each choice of s. This shows that
dH(F∗(s, 0), F ∗(s, 1)) 6 c for s ∈ R. Thus, these two geodesic lines, F ∗(R, 0)
and F ∗(R, 1), are within bounded distance. Thus, they must coincide. Hence
a=b and uniqueness is established. ¤
Let G = π1(Σg, p0) and γ ∈ G− id , we will call the geodesic Θ(Axis(γ)) ⊂ Σg
the geodesic representative of γ. Note that if τ ∈ G, then τγτ−1 and γ have the
same geodesic representative since Axis(τγτ−1) = τ(Axis(γ)) and Θ = Θ τ .
Recall that a loop ρ : S1 → Σg is called simple if ρ is an embedding.
Definition 3.11. An element γ ∈ G− id is call simple if its geodesic represen-
tative is an embedded circle in Σg. Two elements γ1, γ2 ∈ G− id are called dis-
joint if their geodesic representatives are disjoint. Two elements γ1, γ2 ∈ G− id are called linked if Axis(γ1) ∩Axis(γ2) 6= ∅ and Axis(γ1) 6= Axis(γ2)).
γ is simple,
Axis( )γ1
Axis( )γ2
Axis( )σ
γτ
is non−simpleτ
Figure 3.7
For γ1, γ2 ∈ G−id such that γ1 is not conjugate to γ2, the geodesic intersection
number I(γ1, γ2) is the number of intersection points between the closed geodesic
representatives of γ1 and γ2.
Lemma 3.12. Let the deck transformation group G be identified with π1(Σg, p0)
as before. Then
34 THOMAS AU, FENG LUO, AND TIAN YANG
(1) An element γ ∈ G− id is simple if and only if
g(Axis(γ)) ∩Axis(γ) = ∅ for all g ∈ G− γn | n ∈ Z .
(2) If there exists a simple loop ρ freely homotopic to a representative in
γ ∈ G− id , then γ is simple.
(3) Two elements γ1, γ2 ∈ G− id are disjoint if and only if
G(Axis(γ1)) ∩G(Axis(γ2)) = ∅ .
(4) Suppose γ1, γ2 ∈ G− id are simple such that Axis(γ1) ∩ Axis(γ2) 6= ∅.Then I(γ1, γ2) = 1 if and only if for all σ ∈ G− γn2 | n ∈ Z,
σ(Axis(γ1)) ∩Axis(γ2) = ∅ .
Proof. To see (1), suppose that there is g ∈ G − γn | n ∈ Z and there exists
z ∈ g(Axis(γ)) ∩Axis(γ) 6= ∅ . Since g 6= γn,
Axis(gγg−1) = g(Axis(γ)) 6= Axis(γ) .
Therefore, the fixed point sets of gγg−1 and γ separate each other in R and so the
curves g(Axis(γ)) and Axis(γ) intersect transversally in H2. These two curves
are projected to the same geodesic, Θ(Axis(γ)) = Θ(gAxis(γ)) ⊂ Σg. Then,
Θ(z) ∈ Σg is a transversal self intersection of it; equivalently, γ is not simple.
Conversely, suppose that the quotient Θ(Axis(γ)) intersects itself at q ∈ Σg. Let
w ∈ H2 such that Θ(w) = q. Consider the lifting of Θ(Axis(γ)) in a neighborhood
of w. Since Θ is a local isometry, there are two distinct geodesic lines si, i = 1, 2
intersecting at w, which are projected to Θ(Axis(γ)). Thus, there are σi ∈ G
such that si = σi(Axis(γ)). Assume that s1 6= Axis(γ). Then σ1 6∈ γn | n ∈ Z .It follows that the orbit of Axis(γ) under the action of G is not pairwise disjoint.
This completes the proof of (1).
For statement (2), suppose otherwise that γ is not simple. By (1), there are
γ1, γ2 ∈ G such that γ1γ−12 6∈ γn | n ∈ Z and
γ1(Axis(γ)) ∩ γ2(Axis(γ)) 6= ∅ .
LECTURES ON MAPPING CLASS GROUPS 35
Let ρ be a simple loop freely homotopic to Θ(Axis(γ)) and ρ be a lifting of ρ.
Since ρ is compact, it is of bounded distance to Axis(γ). Further let ρi = γi (ρ)
for i = 1, 2. We have ρi ⊂ Nc(γi(Axis(γ))), i = 1, 2, for some constant c > 0. It
follows that ρ1 ∩ ρ2 6= ∅. In addition, Θ(ρi) = ρ for i = 1, 2; so ρ is not simple,
which contradicts the assumption.
ρ ρ
~
~
1
2
γ
γ
γ Axis( )
Axis( )
1
γ 2
Figure 3.8
To see (3), note that γ1, γ2 are disjoint if and only if Θ(Axis(γ1))∩Θ(Axis(γ2) is
empty. The latter condition is the same as that G(Axis(γ1)) ∩G(Axis(γ2)) = ∅.
Finally, for (4), let z1 ∈ Axis(γ1) ∩ Axis(γ2) 6= ∅. Then [z1] = Θ(z1) ∈ Σg is
an intersection point of the geodesic representatives of γ1 and γ2. For i = 1, 2,
Θ−1 [z1] ∩ Axis(γi) = G(z1) ∩ Axis(γi) where G(z1) is the G-orbit of z1. Since
both γi are simple, by (1), G(z1) ∩Axis(γi) = γni (z1) | n ∈ Z for i = 1, 2.
Suppose that I(γ1, γ2) = 1, that is, the geodesic representatives of γ1 and γ2
intersect only at the point [z1]. If w ∈ G(Axis(γ1)) ∩ Axis(γ2), then Θ(w) = [z1]
and so w ∈ G(z1). It follows that w ∈ G(z1) ∩ Axis(γ2). From the discussion in
the previous paragraph, w = γn2 (z1) for some n ∈ Z.
Axis( )γ1
Axis( )γ2
γ2 Axis( )γ1n
γ2 Axis( )γ1−m
Θcovering
γ2
γ1
[ ]z1
Figure 3.9
36 THOMAS AU, FENG LUO, AND TIAN YANG
Hence, if I(γ1, γ2) = 1, then G(Axis(γ1)) ∩ Axis(γ2) ⊂ γn2 (z1) | n ∈ Z . This
shows that for σ ∈ G− γn2 | n ∈ Z
σ(Axis(γ1)) ∩Axis(γ2) = ∅.
Conversely, suppose that there exists σ ∈ G − γn2 | n ∈ Z and z2 ∈ H2 such
that
z2 ∈ σ(Axis(γ1)) ∩Axis(γ2) .
We will show that I(γ1, γ2) > 2 by proving that [z1] 6= [z2]. Assume otherwise
that [z1] = [z2], then z2 = g(z1) for some g ∈ G.
Now, both z1 and z2 are in Axis(γ2). Thus, z2 = g(z1) ∈ G(z1) ∩ Axis(γ). Since
γ2 is simple, by (1), we have z2 = γm2 (z1) for some m ∈ Z.
Axis( )γ2
Θ
γ2Axis( )γ1σ
γ1
[ ]z
[ ]z1
2z1
Axis( )γ1
z2
Axis( )γ1mγ2
Figure 3.10
On the other hand, it has been assumed that z2 ∈ σ(Axis(γ1)) where σ 6∈ γn2 | n ∈ Z . Thus, σ(Axis(γ1)) 6= γm2 (Axis(γ1)). However, they are projected
to the same geodesic representative Θ(Axis(γ1)) with a self intersection point [z2].
This contradicts that γ1 is simple. ¤
4. Quasi-Isometry and Large Scale Geometry
In this section, we study large scale geometry, in particular, the properties of a
quasi-isometry on the hyperbolic plane. The main result of this section is the
result of Milnor that relates the isometric actions of the fundamental group with
the quasi-isometric geometry of the universal cover. These notions constitute the
key idea of Farb-Margalit’s proof of Dehn-Nielsen theorem, [FM].
LECTURES ON MAPPING CLASS GROUPS 37
Definition 4.1. Suppose (X, dX), (Y, dY ) are two metric spaces and F : X → Y
is a map which may not be continuous. We say F is a quasi-isometry (or more
precisely K-quasi-isometry) if there exists a constant K > 0 such that
(1) for all x1, x2 ∈ X,
1
KdX(x1, x2)−K 6 dY (f(x1), f(x2)) 6 KdX(x1, x2) +K ; and
(2) Y = NK(f(X)).= y ∈ Y | dY (y, f(x)) 6 K for some x ∈ X .
It follows directly from the definition that compositions of quasi-isometries are
quasi-isometries. The followings are some examples of quasi-isometries.
Example 4.2. Let X ⊂ Y be a subset such that
NK(X) = y ∈ Y | dY (x, y) 6 K for some x ∈ X
and dX(x1, x2) = dY (i(x1), i(x2)) where i : X → Y is the inclusion map. Then
the inclusion map i is a quasi-isometry. For specific examples, we may takeX = Z
and Y = R, or more generally X = Zn and Y = Rn in the standard Euclidean
metrics.
Example 4.3. Suppose (Mn, g) is a compact Riemannian manifold with p0 ∈ M .
Moreover, suppose Y = (Mn, g) is the universal cover ofM with y0 ∈ Y projected
to p0 so that the covering map is a local isometry. Let X ⊂ Y be the orbit
of a point y ∈ Y under the isometric action of the deck transformation group
π1(M,p0). Then, by construction, ND(X) = Y where D is the diameter of
M . Thus the inclusion map i : X → Y is a quasi-isometry. Note that X is
a bijective image of the fundamental group π1(M,p0) via the action map γ 7→γ(y) : π1(M,p0) → X.
Our main interests are in the quasi-isometry classes of a finitely generated groupG.
Recall that a symmetric generating set S for G is a set which generates G and it
satisfies that if x ∈ S, then x−1 ∈ S.
38 THOMAS AU, FENG LUO, AND TIAN YANG
Definition 4.4. (Cayley Graph) Suppose G is a finitely generated group with
a symmetric finite generating set S. The Cayley graph Γ(G,S) is the graph of
which the vertex set is G and two vertices g1, g2 ∈ G are joined by an edge if
g−11 g2 ∈ S, i.e., g2 = g1s for some s ∈ S. A metric is given to Γ(G,S) such that
each edge has length one. Consequently, the distance between two vertices g1, g2
of Γ(G,S) is the minimum length of all possible edge paths from g1 to g2.
Remark. For any edge between g1, g2 and any g ∈ G, a unique edge is determined
by gg1, gg2. Thus, the left action of G on Γ(G,S) is an isometric action.
Example 4.5. Let G = Z, S = ±1 , then Γ(G,S) is isometric to R1 with the
standard metric.
Example 4.6. Let G = Z2, S = ±(1, 0),±(0, 1) , then Γ(G,S) is isometric to
the standard lattice grid in the plane.
. . g
g -(0,1)
g +(0,1)
g -(1,0) g +(1,0)
Figure 4.1
For a finitely generated group G with a symmetric generating set S, the word
length, |w|S of an element w ∈ G is the distance from w to the identity element id
in the Cayley graph Γ(G,S), namely,
|w|S = min k ∈ N | w = s1s2 · · · sk, for some s1, . . . , sk ∈ S .
For instance, let F2 denote the free group on two generators x, y and let S =x, x−1, y, y−1
. Then we have
∣∣x3y−1xy−2x−5∣∣S= 12 in Γ(F2, S).
LECTURES ON MAPPING CLASS GROUPS 39
Example 4.7. If T is another symmetric generating set for the group G, then
for any w ∈ G
(2) |w|S 6 K |w|T ,
where K = max |t|S | t ∈ T .
Indeed, suppose that |w|T = N and w = t1 · · · tN where t` ∈ T . Now each t` can
be in turns written in terms of S, namely,
t` = si`,1 · · · si`,n`for si`,j ∈ S ,
where n` = |t`|S 6 K for ` = 1, . . . , N . It follows that
w = si1,1 · · · si1,n1 · si2,1 · · · si2,n2 · · · · · · · · siN ,1 · · · siN ,nN , where si`,j ∈ S.
Thus |w|S 6N∑
i=1
ni 6N∑
i=1
K 6 K |w|T .
Lemma 4.8. Given any two symmetric generating sets S and T of a group G,
their Cayley graphs Γ(G,S) and Γ(G,T ) are quasi-isometric.
Proof. Denote dS and dT the metrics on Γ(G,S) and Γ(G,T ) respectively. Note
that from Example 4.7 above, the identity map from (G, dS) to (G, dT ) is a
quasi-isometry. Moreover, from Example 4.7, the inclusion (G, dT ) → Γ(G,T ) is
a quasi-isometry. Since a composition of quasi-isometries is still a quasi-isometry,
it suffices to establish a quasi-isometry p : Γ(G,S) → (G, dS) such that p|G is the
identity map on G. Indeed, for x ∈ Γ(G,S), simply take p(x) = x if x ∈ G and
p(x) to be any one of the two vertices of the edge containing x. Then p is clearly
a quasi-isometry. ¤
Due to Lemma 4.8, we see that the quasi-isometry class of the Cayley graph
Γ(G,S) is independent of the choice of the finite symmetric generating set. From
now on we may discuss if a map φ : G → G is a quasi-isometric map in the sense
that G is given the metric with respect to any finite symmetric generating set.
Corollary 4.9. If G is a finitely generated group and φ : G → G is an isomor-
phism, then φ is a quasi-isometric map.
40 THOMAS AU, FENG LUO, AND TIAN YANG
Proof. Indeed, the map φ is a composition of f : Γ(G,S) → Γ(G,φ(S)) and
g : Γ(G,φ(S))id→ Γ(G,S). Here f is a graph isomorphism and g|G = idG is the
identity map on G. By definition f is an isometry and by Lemma 4.8, g is a
quasi-isometry. Thus the corollary follows. ¤
Theorem 4.10. (Milnor-Svarc) Suppose (M, g) is a closed Riemannian man-
ifold with the universal cover (M, g) such that G = π1(M,p0) acts isometrically
on M as the group of deck transformations. Then G is quasi-isometric to M via
the map sending γ to γ(z) where z ∈ M is a chosen point.
Proof. Let Ω be a connected compact fundamental domain of the G action on M
and K1 = diam(Ω). Define S = γ ∈ G | γ(Ω) ∩ Ω 6= ∅ . Note that S is a sym-
metric finite set since γ(Ω)∩Ω 6= ∅ is equivalent to Ω∩γ−1(Ω) 6= ∅. Furthermore,
it is known that G is generated by S. We choose the base point z to be in the
interior of Ω and consider the map P : G → M given by
P (γ) = γ(z).
First of all, by definition of the diameter K1, one has NK1(P (G)) = M .
Since the left action of G on both (Γ(G,S), dS) and (M, d) are isometries, it
suffices to verify the quasi-isometry condition that there exists a constant K such
that for each γ ∈ G,
(3)1
KdS(1, γ)−K 6 d(P (1), P (γ)) 6 K dS(1, γ) +K ,
Indeed, we have dS(γ1, γ2) = dS(1, γ−11 γ2) and
d(P (γ1), P (γ2)) = d(γ1(z), γ2(z)) = d(z, γ−11 γ2(z)) = d(P (1), P (γ−1
1 γ2)) .
Now, we claim that the second inequality in (3) holds, namely,
(3a) d(P (1), P (γ)) 6 K2 dS(1, γ) +K2 where K2 = max(2K1, 1).
LECTURES ON MAPPING CLASS GROUPS 41
Indeed, let dS(1, γ) = n and write γ = γ1γ2 · · · γn, where γi ∈ S. Then
d(P (1), P (γ)) = d(z, γ(z)) = d(z, γ1γ2 · · · γn(z))
6n−1∑
i=1
d(γ1 · · · γi−1(z), γ1 · · · γi(z))
6n−1∑
i=1
d(z, γi(z)) 6n−1∑
i=1
2K1
6 2K1n = 2K1dS(1, γ) 6 K2dS(1, γ).
Here we have used the fact that z and γi(z) lie in Ω ∪ γi(Ω) with Ω ∩ γi(Ω) 6= ∅,and thus
d(z, γi(z)) 6 diam (Ω ∪ γi(Ω)) 6 2K1.
To see the first inequality in (3), that
(3b) dS(1, γ) 6 K d(P (1), P (γ)) +K2 ,
for some K, we let
δ =1
2infξ∈G
d(Ω, ξ(Ω)) | Ω ∩ ξ(Ω) = ∅ > 0 .
By the definition of δ, if ξ ∈ G satisfies that d(ξ(z), z) 6 δ, then ξ ∈ S. Choose
the shortest geodesic segment L in M joining z to γ(z). Write
d(γ(z), z) = nδ + δ′ where 0 6 δ′ < δ, n ∈ Z>0 .
Let z1 = z, z2, . . . , zn, zn+1 = γ(z) be points along the geodesic L so that the
distance from zi to z along L is (i− 1) · δ, i = 1, . . . , n. In particular,
d(zi, zi+1) = δ, i 6 n− 1, and
d(zn, zn+1) = d(zn, γ(z)) = δ′ < δ .
For each such zi along L, zi ∈ hi(Ω) for some hi ∈ G. Now d(zi, zi+1) 6 δ implies
that
d(Ω, h−1i hi+1(Ω)) = d(hi(Ω), hi+1(Ω)) 6 δ .
Thus, we have γi = h−1i hi+1 ∈ S. As a consequence, one obtains
hi+1 = hi · γi = γ1γ2 · · · γi for each i = 1, . . . , n.
42 THOMAS AU, FENG LUO, AND TIAN YANG
Now, by the construction, both zn+1 = γ(z) ∈ γ(Ω) and zn+1 ∈ hn+1(Ω). Thus
γ(Ω) = hn+1(Ω), that is,
γ = hn+1 = γ1γ2 · · · γn where γi ∈ S.
In particular, |γ|S 6 n according to definition. In other words,
dS(1, γ) = |γ|S 6 n =
(1
δ
)δn 6 1
δ
n∑
i=1
d(zi, zi+1) 61
δd(z, γ(z)) .
This shows that δ dS(1, γ) 6 d(P (1), P (γ) and Inequality (3b) follows.
In summary, the condition (3) for P being a quasi-isometry is satisfied by taking
K = max(1δ , 1,K2). ¤
We now come to the main technical result for proving Dehn-Nielsen Theorem.
Proposition 4.11. Suppose φ : π1(Σg, p0) → π1(Σg, p0) is an isomorphism. If
α, β ∈ π1(Σg, p0)− id satisfy Axis(α) ∩ Axis(β) = ∅, then
Axis(φ(α)) ∩ Axis(φ(β)) = ∅ .
We will proceed by assuming that Axis(φ(α)) ∩ Axis(φ(β)) 6= ∅ and derive a
contradiction. Before doing so, we need to consider a quasi-isometry on H2 and
establish a lemma.
Let G = π1(Σg, p0). By Corollary 4.9, and Theorem 4.10, we may replace the
isomorphism φ : G → G by a K-quasi-isometric map F : H2 → H2 such that the
following diagram commutes up to quasi-isometry where P is the evaluation map
defined by P (γ) = γ(z0) for a fixed chosen point z0 ∈ H2.
G G
H2 H2-
-
? ?
φ
F
P P
Lemma 4.12. If F : H2 → H2 is a K-quasi-isometry, then there exists a
constant c = c(K) such that for any geodesic line L, F (L) ⊂ Nc(L′) for some
geodesic line L′.
LECTURES ON MAPPING CLASS GROUPS 43
We omit the details of the proof of the lemma. See Lemma 3.43, page 51, in
Kapovich’s book [Kap] for a proof.
Proof of Proposition 4.11. By Lemma 4.12, we see that the quasi-isometry F
sends the two geodesic lines A = Axis(α) and B = Axis(β) into Nc(A′) and
Nc(B′), where A′, B′ are two geodesic lines. Moreover, if pn, qn ∈ ∂H2 are
sequences converging to the end points of A, then A′ has end points limF (pn)
and limF (qn). Since F P = P φ, by taking pn = αn(z0) and qn = α−n(z0),
one can see that the end points of A′ are the same as those of Axis(φ(α)). Thus,
A′ = Axis(φ(α)) and similarly, B′ = Axis(φ(β)).
Recall that we start with the assumption that Axis(φ(α)) ∩ Axis(φ(β)) 6= ∅.Thus, as in Figure 4.2, we have A′ ∩B′ 6= ∅ and the end points of A′ and B′ are
respectively denoted a1, a2 and b1, b2 .
FA
B
a
a
b
b
1
2
1
2
AB’
’
Figure 4.2
Choose two curves A ⊂ NR(A) and B ⊂ NR(B) of constant distance R to A and
B respectively so that the distance from A to B is at least 2R.
44 THOMAS AU, FENG LUO, AND TIAN YANG
FA
B
A~
~
A’
B’B( )F ~
B
( )F A~
Figure 4.3
Since F is a quasi-isometry, by taking R large, we can make dist(F (A), F (B))
arbitrarily large.
In the special case that F |A and F |B are continuous, we choose R such that
dist(F (A), F (B)) ≥ 1. It is easy to conclude from continuity of F that both
H2 \ F (A) and H2 \ F (B) are disconnected. According to the assumption that
Axis(A′) ∩ Axis(B′) 6= ∅, we have F (A) ∩ F (B) 6= ∅ for any choice of R. This
contradicts that dist(F (A), F (B)) ≥ 1.
In the general case, since F (A) ⊂ Nc(A′) and F (B) ⊂ Nc(B
′) and F is a quasi-
isometry, for any curve A (respectively B) of constant distance to A (respectively
B), there exists 0 < c′ ∈ R such that F (A) ⊂ Nc′(A′) and F (B) ⊂ Nc′(B
′). Now
by the assumption that A′ ∩ B′ 6= ∅, it follows that for any choices of A and B
there is a point z which is within a constant distance c∗ to both F (A) and F (B)
where c∗ depends only on K. However, as we see from the above case, by taking
R large, we can make dist(F (A), F (B)) > 2c∗. This is a contradiction. ¤
5. Dehn-Nielsen Theorem
We follow the proof appeared in Farb-Margalit’s book to prove Dehn-Nielsen
Theorem in this section. The basic idea of the proof is the same as that of Dehn.
LECTURES ON MAPPING CLASS GROUPS 45
Given a group G, we use Aut(G) to denote the group of all self-isomorphisms
of G. Furthermore, let
Inn(G) =φ ∈ Aut(G) | φ(x) = g−1xg for all x ∈ G , for some g ∈ G
be the group of inner automorphisms of G. It is easy to check that Inn(G) is a
normal subgroup of Aut(G). We define the outer automorphism group of G to
be the quotient group
Out(G) = Aut(G)/ Inn(G).
For instance, Out (Zn) = GL(n,Z).
Given a surface Σ with a base point p0 ∈ Σ, a self-homeomorphism h on Σ induces
an isomorphism h∗ : π1(Σ, p0) → π1(Σ, h(p0)). Let
θξ : π1(Σ, h(p0)) → π1(Σ, p0)
be an isomorphism induced by a path ξ from h(p0) to p0. Then θξ h∗ ∈Aut(π1(Σ, p0)). Moreover, different choices of paths ξ, η from h(p0) to p0 will
result in two automorphisms, θξ h∗ and θη h∗. They are obviously related by an
inner automorphism. Thus h∗ represents a well defined element in Out(π1(Σ, p0)),
which we still denote h∗ for simplicity. Furthermore, (h1 h2)∗ = (h1)∗(h2)∗ for
self-homeomorphisms h1, h2 of Σ. In particular, by sending the homotopy class [h]
to h∗, we produce a group homomorphism
Ψ : Γ(Σ) = Homeo(Σ)/ ' −−→ Out(π1(Σ, p0)).
Dehn-Nielsen Theorem. Suppose Σ = Σg,0 is a closed surface of genus g > 1,
then Ψ is an isomorphism.
5.1. Injectivity of Ψ. The proof is essentially the same as the proof for the
torus given in Theorem 1.7. We sketch the argument as follows.
First take a collection of simple closed curves a1, . . . , a2g in Σ so that
(1)⋂2g
i=1 ai = p0 = aj ∩ ak for all pair j, k of indices; and
(2) Σ−⋃2gi=1 ai is a topological disk.
46 THOMAS AU, FENG LUO, AND TIAN YANG
p0
a j ak
Figure 5.1
It is known from surface topology that π1(Σ, p0) is generated by the homotopy
classes [a1], . . . , [a2g].
Let [h] ∈ kerΨ ⊂ Γ(Σ). That is, h ∈ Homeo(Σ) such that h∗ ∈ Inn(π1(Σ, p0)).
Then, there is a loop b based at p0 such that for each i = 1, 2, . . . , 2g, the loop
h(ai) is homotopic to baib−1 relative p0. We want to show that h is homotopic to
the identity map. To construct the homotopy H from h to idΣ, let us define H
on Σ× 0, 1 ∪⋃2gi=1 ai × I by,
H(x, 0) = h(x) , x ∈ Σ
H(x, 1) = x , x ∈ Σ
H(p0, t) = b(t), t ∈ [0, 1]
H|ai×[0,1] = the homotopy from ai to h(ai), i = 1, . . . , 2g.
Now using the facts that π2(Σ, p0) = 0 and (Σ × [0, 1]) \ (⋃2g
i=1 ai × [0, 1]) is a
3-cell with H defined on the boundary of the 3-cell, we conclude that H extends
to Σ× [0, 1] and becomes a homotopy between h and idΣ. ¤
5.2. Surjectivity of Ψ. Suppose φ : π1(Σ, p0) → π1(Σ, p0) is an isomorphism.
We will show that φ = h∗ for some h ∈ Homeo(Σ).
First of all, we may assume that the genus g > 2 since the result for g = 1 was
proved in Theorem 1.7. In this case, we will identify the universal cover of Σ
with the hyperbolic plane H2 and π1(Σ, p0) with the deck transformation group
G acting isometrically on H2.
LECTURES ON MAPPING CLASS GROUPS 47
In the rest of the proof, we use [s] to denote the free homotopy class of a loop s
in Σ. We will also identify [s] with the conjugacy class of an element in π1(Σ, p0).
Lemma 5.1. For the automorphism φ : π1(Σ, p0) → π1(Σ, p0) and α, β ∈π1(Σ, p0),
(1) if α is simple, then so is φ(α)
(2) if α, β are disjoint, then φ(α) and φ(β) are disjoint
(3) if α, β are simple, and I(α, β) = 1, then I(φ(α), φ(β)) = 1.
Proof. By Proposition 4.11, we identify G with the orbit G(z) ⊂ H2 and extend
φ : G → G to a quasi-isometry on H2, which is still denoted by φ : H2 → H2.
By choosing a hyperbolic metric on Σ, we can identify each conjugacy class in
π1(Σ, p0)− id with its geodesic representative.
For (1), let α be simple, i.e., its geodesic representative s is simple. By definition,
s is simple if and only if Θ−1(s) is a disjoint union of geodesics, where Θ : H2 → Σ
is the universal covering projection.
Let us pick a y ∈ G such that α =xyx−1 | x ∈ G
. Then
Θ−1(s) =⋃
x∈Gx(Axis(y))
Then s is simple if and only if Axis(x1yx−11 ) ∩ Axis(x2yx
−12 ) = ∅ for all pairs of
x1, x2 ∈ G with x1x−12 6∈ yn | n ∈ Z. According to Proposition 4.11 that every
quasi-isometry preserves the disjointness of axes, together with the fact that for
each z ∈ G, Axis(φ(z)) and φ(Axis(z)) have the same end points in ∂H2, we have
Axis(φ(x1)φ(y)φ(x1)−1) ∩Axis(φ(x2)φ(y)φ(x2)
−1) = ∅ .
On the other hand, it follows from the surjectivity of φ : G → G that
φ(α) =xφ(y)x−1 | x ∈ G
and therefore
Axis(x1φ(y)x−11 ) ∩Axis(x2φ(y)x
−12 ) = ∅ ,
for each pair of x1, x2 ∈ G with x1x−12 6∈ φ(y)n | n ∈ Z. Thus φ(α) is simple.
48 THOMAS AU, FENG LUO, AND TIAN YANG
The proof of (2) is similar. Namely, α, β are disjoint if and only if for any x ∈ α
and y ∈ β, Axis(x) ∩ Axis(y) = ∅. Again, the quasi-isometry φ preserves the
disjointness of axes. Thus the result follows.
For statement (3), we can use Proposition 4.11 together with Lemma 3.12. ¤
Finally, we come to the proof of Dehn-Nielsen’s theorem.
Let φ : π1(Σ, p0) → π1(Σ, p0) be an automorphism. Choose oriented simple loops
c1, . . . , c2g as shown in Figure 5.2. Note that Σ \ ∪2gi=1ci is connected.
2gcc4c 2
c1
c 3
Figure 5.2
More precisely, if we denote [ci] the free homotopy class of the unoriented loop ci
for i = 1, . . . , 2g, the sequence ci : i = 1, . . . , 2g satisfies
(1) I([ci], [ci+1]) = 1 for i = 1, . . . , 2g − 1;
(2) I([ci], [cj ]) = 0 if |i− j| > 2 for i, j = 1, . . . , 2g;
(3) the algebraic intersection sign at ci ∩ ci+1 from ci to ci+1 is 1 for each
i = 1, . . . , 2g − 1.
By Lemma 5.1, φ([ci]) is simple for each i. Then, there is a new sequence
d1, . . . , d2g of simple loops such that [di] = φ([ci]). Moreover, by Lemma 5.1
again, I(φ([ci]), φ([ci+1])) = 1 and I(φ([ci]), φ([cj ])) = 0 for |i− j| > 2. By taking
di to be the geodesic representative of [di], we may further assume that
|di ∩ di+1| = 1 for i = 1, . . . , 2g − 1 and |di ∩ dj | = 0 for |i− j| > 2 .
LECTURES ON MAPPING CLASS GROUPS 49
By the classification of surfaces, there exists a homeomorphism h ∈ Homeo(Σ)
such that
h(ci) = di for i = 1, 2, . . . , 2g ,
and hence, h∗([ci]) = φ([ci]) for all i. Indeed, since both c1, d1 are nonseparating,
we may find h1 ∈ Homeo(Σ) so that h1(c1) = d1. Thus, we may assume c1 = d1
after composition with h1. Next, since both c2, d2 are nonseparating in with
|c1 ∩ c2| = |c1 ∩ d2| = 1, we are able to find h2 ∈ Homeo(Σ) so that
h2(c1) = c1 and h2(c2) = d2.
Inductively, after taking composition, we find h ∈ Homeo(Σ) with the required
property, h(ci) = di, i = 1, ..., 2g. Note that in the process, each ck is nonsepa-
rating in the surface obtained by cutting along c1, . . . , ck−1.
We further claim that there is an involution τ of Σ such that
h∗ = φ or h∗ = τ∗ φ in the group Out(π1(Σ, p0)).
To see this, let us consider the automorphism φ h−1∗ ∈ Aut(π1(Σ, p0)), which
takes [ci] to [ci]. Thus, we may simply assume that φ fixes each free homotopy
class [ci] of the unoriented curve ci. With this assumption, the goal is to show
φ ∈ Inn(G) or τ∗ φ ∈ Inn(G).
To this end, choose a set of generators z1, . . . , z2g for π1(Σ, p0) as shown so that
zi and ci are freely homotopic as oriented loops.
z2
z1
p0
z4
z3
Figure 5.3
There are two involutions τ1 and τ2 of Σ leaving each ci invariant. The first
involution τ1 is the hyper-elliptic involution which reverses orientation of each ci.
50 THOMAS AU, FENG LUO, AND TIAN YANG
The second involution τ2 is the reflection of Σ about a “plane” which leaves the
orientation of c2i+1 invariant and reverses the orientation of each c2i.
τ τ 1 2
Figure 5.4
Since φ([c1]) = [c1], we have φ(z1) = az1a−1 or φ(z1) = az−1
1 a−1 for some a ∈ G.
By composing φ with the inner automorphism x 7→ a−1xa and the involution
(τ1)∗ if necessary, we may assume that φ(z1) = z1.
Now, due to the fact that Axis(z1) ∩Axis(z2) 6= ∅, we conclude that
Axis(z1) ∩Axis(φ(z2)) 6= ∅ .
Similarly, φ([c2]) = [c2] implies that φ(z2) = bz±12 b−1 for some b ∈ G. By
Lemma 3.12, statement (4), and I([c1], [c2]) = 1, we conclude
φ(z2) = zn1 z±12 z−n
1 for some n ∈ Z.
By composing φ with the inner automorphism x 7→ zn1 xz−n1 and the involutions
(τ1)∗ and (τ2)∗ if necessary, we may assume that
φ(z2) = z2.
We claim that φ(zi) = zi for i > 3.
Indeed, to see φ(z3) = z3, we only need the above assumptions of φ(z1) = z1 and
φ(z2) = z2 together with the fact that φ([c3]) = [c3].
Due to I(φ[c3], φ[c2]) = I([c3], [c2]) = 1, and Axis(z2)∩Axis(z3) 6= ∅, we concludethat Axis(φ(z2)) ∩Axis(φ(z3)) 6= ∅ and so
φ(z3) = zn2 z±13 z−n
2 for some n ∈ Z.
LECTURES ON MAPPING CLASS GROUPS 51
Suppose that n 6= 0 and then we will derive a contradiction. In such case, the
self-composition φk of φ satisfies
φk(z3) = znk2 z±13 z−nk
2 .
As a result, Axis(φk(z3)) = znk2 (Axis(z3)) converges to the end points of Axis(z2)
in H2 as k → ∞ in the Hausdorff metric.
On the other hand, φk(z1z2) = z1z2 is represented by a simple loop of which the
geodesic representative intersects z3 at one point. In particular
Axis(z3) ∩Axis(z1z2) 6= ∅ .
p0
z1 z 2
z3
Figure 5.5
Thus, for all k, φk(Axis(z3)) ∩ φk(Axis(z1z2)) 6= ∅. On the other hand,
φk(Axis(z3)) = Axis(φk(z3)) = znk2 (Axis(z3)) .
Therefore, znk2 (Axis(z3))∩Axis(z1z2) 6= ∅ for all k large. This is impossible since
Axis(z1z2) is disjoint from the end-points of Axis(z2) in H2.
Axis( z ) 2
Axis( z ) 3
φ (Axis( z )) k 3
Axis( z ) 2
Axis( z z ) 1 2
Figure 5.6
52 THOMAS AU, FENG LUO, AND TIAN YANG
It follows that φ(z3) = z±13 . We claim φ(z3) = z−1
3 is also impossible. If otherwise,
φ(z3) = z−13 , we derive a contradiction as follows. By Figure 5.7, we can see that
I(z2z3, z−11 z2) = 0 and I(z2z
−13 , z−1
1 z2) 6= 0.
−1zz 21
zz2 3−1
2 zz 3
Figure 5.7
Now all of the classes z2z3, z−11 z2, z2z
−13 are simple as shown. Thus using
I(z2z3, z−11 z2) = 0 and statement (2) of Lemma 5.1, we conclude that the conju-
gacy classes of φ(z2z3) = z2z−13 and φ(z−1
1 z2) = z−11 z2 are represented by disjoint
simple loops. This contradicts that I(z2z−13 , z−1
1 z2) 6= 0.
2 zz 3
−1zz 21 zz2 3−1
−1zz 21
Figure 5.8
Thus we conclude that φ(z3) = z3. Next, the facts that φ(z2) = z2, φ(z3) = z3
and φ([c4]) = [c4] lead to I([z2], [c4]) = 0 and I([z3], [c4]) = 1. Following exactly
the same argument above and using these facts, we conclude that φ(z4) = z4.
Inductively, we conclude that φ(zi) = zi for each i = 1, . . . , 2g. Hence φ = idΣ.
The surjectivity of Ψ is thus established. ¤
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Mathematics Department, Princeton University, New Jersey, 1979.
Department of Mathematics, The Chinese University of Hong Kong, Shatin, NT,Hong Kong SAR, China
E-mail address: thomasau@cuhk.edu.hk
Department of Mathematics, Rutgers University, Piscataway, New Jersey, USA
E-mail address: fluo@math.rutgers.edu
Department of Mathematics, Rutgers University, Piscataway, New Jersey, USA
E-mail address: tianyang@math.rutgers.edu
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