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dissertation entitled

COMPLEXES 0F CURVES
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MAPPING CLASS GROUPS

presented by

Mustafa Korkmaz

has been accepted towards fulﬁllment
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COMPLEXES OF CURVES
AND
MAPPING CLASS GROUPS

By

Mustafa Korkmaz

A DISSERTATION

Submitted to
Michigan State University
in partial fulﬁllment of the requirements
for the degree of

DOCTOR OF PHILOSOPHY

Department of Mathematics

1996

ABSTRACT

COMPLEXES OF CURVES
AND
MAPPING CLASS GROUPS

By

Mustafa Korkmaz

This thesis is composed of two parts.

In the first part of this thesis, we study the complexes of curves on orientable
surfaces of small genus in order to better understand the mapping class groups of
such surfaces. Our main result is that the group of automorphisms of the complex of
curves of a surface is isomorphic to the extended mapping class group of the surface,
if the surface is a sphere with at least ﬁve punctures or is a tori with at least three
punctures. As an application we prove that any isomorphism between two ﬁnite index
subgroups of the extended mapping class group is induced by an inner automorphism
of the extended mapping class group. We conclude that the outer automorphism

group of a ﬁnite index subgroup of the extended mapping class group is ﬁnite.

The second part concerns closed nonorientable surfaces. Namely, we compute the
ﬁrst homology of the mapping class group of such a surface. It turns out that this
group is cyclic of order two if the genus of the surface is at least seven. We also
show that in this case the the subgroup of the mapping class group generated by
Dehn twists is perfect. As an algebraic application, we conclude that the group of
isometries of a vector space of dimension n 2 7 over the ﬁnite ﬁeld of order two
equipped with the symmetric bilinear form ( ,) deﬁned by (vhvj) = 6,-1- on a basis

{221, 122,. . . , vn} is perfect.

To my parents, brothers, sisters and to my wife

iii

ACKNOWLEDGMENTS

First, I must thank my advisor, Dr. Nikolai V. Ivanov for his excellent guidance,
help and encouragement throughout all of my study at Michigan State University, and

bringing to my attention various mathematical research problems, including those

addressed in this thesis.

I am very grateful for Dr. John D. McCarthy, who always made for me, answering

all of my questions with great patience.

Many thanks to the members of my advisory committee Dr. Selman Akbulut, Dr.
Ronald Fintushel, and Dr. Jon G. Wolfson for their many helpful conversations with

me and their review of my work.

I also would like to thank Belgin Korkmaz and Bill Vautaw for reading my work

carefully and pointing out some grammatical errors.

Finally, I would like to thank The Turkish Ministry of Education for their support
during my study at Middle East Technical University and at Michigan State Univer-ﬁ
sity, and Dr. Hatice Kandamar of Adnan Menderes University, Dr. Nursevin Oztop
and Dr. Muharrem Soytiirk of Cumhuriyet University, Turkey, for their help to go to

study abroad.

iv

Contents

0 Introduction 1
1 Complexes of Curves and Extended Mapping Class Groups 3
1.1 Introduction and Preliminaries ...................... 3
1.1.1 Circles on surfaces ........................ 3

1.1.2 The complex C(S) ........................ 4

1.2 The complex B (S ) and ideal triangulations ............... 6
1.3 The extended mapping class group and its actions ........... 9

2 Automorphisms of 0(3) and Mg 14
2.1 Introduction ................................ 14
2.2 Punctured Spheres ............................ 15
2.2.1 Simple pairs in C(S) ....................... 16

2.2.2 The map Aut C(S) —> Aut 8(5) ................ 22

2.2.3 Automorphisms of C (S) and Mg ................ 29

2.3 Punctured Tori .............................. 32
2.3.1 Automorphisms of C (S) and M3 ................ 32

2.3.2 Proof of Theorem 2.16 for punctured tori ............ 40

2.4 Subgroups of Mg ............................. 48

First Homology Groups of Mapping Class Groups of Nonorientable

Surfaces 52
3.1 Introduction ................................ 52
3.2 Mapping Class Groups of Nonorientable Surfaces ........... 53
3.3 Groups of isometries of vector spaces over the ﬁeld of order 2 ..... 59
3.4 The ﬁrst homology groups ........................ 64

vi

List of Figures

1.1 A simple pair and a chain. ........................ 8
1.2 Triangles in an ideal triangulation .................... 9
1.3 Various conﬁgurations. .......................... 12
2.1 A conﬁguration of circles on a punctured sphere ............. 18
2.2 The disc D with four punctures ...................... 20
2.3 Construction of d’ if i = 0. ........................ 24
2.4 Construction of d’ if 2' > 0. ........................ 24
2.5 Seven possible cases for a’ and b’. .................... 26
2.6 Case (vi). ................................. 28
2.7 Case (vii) .................................. 29
2.8 A pentagon in C(S'). ................ f ........... 33
2.9 The surface S... .............................. 34
2.10 A nonseparating and an n-separating circle. .............. 35
2.11 Some conﬁgurations of circles and arcs. . ................. 41
2.12 Existence of various conﬁgurations. ................... 43
2.13 Two possible cases for (v) ......................... 46

vii

2.14 The arcs a’ and b’. ............................ 47

3.1 A Y-homeomorphism. .......................... 55
3.2 Models for nonorientable surfaces ..................... 57
3.3 Another model if g is even ......................... 57
3.4 The circles el,f1 and f2. ......................... 57
3.5 Sphere with four holes 50 and torus with two holes. .......... 58
3.6 Surface S(g). ............................... 59
3.7 Circles ql , Q2, . . . , q, ............................. 66
3.8 Embedding of So .............................. 67
3.9 Embedding of So .............................. 71
3.10 Orientable surfaces of genus one and two ................. 73

viii

Chapter 0

Introduction

The complex of curves was introduced by Harvey [11] to deﬁne a certain boundary
to the Teichmiiller space. Later, this complex is used for many authors in the study
of the algebraic structures of the (extended) mapping class groups of surfaces. In
particular, Ivanov [16] proved that the canonical map from the extended mapping
class group M; of a surface S to the group of automorphisms of the complex of
curves C (S ) is onto, provided the genus of the surface S is at least two. In fact, this
natural map is an isomorphism if in addition, the surface S is not a closed surface
of genus two. Using this result, he proved that any isomorphism between two ﬁnite
index subgroups of the (extended) mapping class group is induced by some inner
automorphism of Mg, concluding that the outer automorphism group of a subgroup
of M; of ﬁnite index is ﬁnite. This is a generalization of the fact that the outer
automorphism group of the (extended) mapping class group is ﬁnite (cf. [15], [22]).
He also gave a new proof of Royden’s Theorem saying that the isometries of the

Teichmiiller space are all induced by diifeomorphisms of the surface.

In this work, we prove that for a surface S the natural map from M; to the group
of automorphisms of the complex of curves C (S ) is an isomorphism if S is either a

sphere with at least ﬁve punctures or a torus with at least three punctures. After

proving this result, the result of Ivanov about the ﬁnite index subgroups of mapping
class groups follows for the surfaces we consider. Since this proof has not been written
yet, we give it here. We also prove that (with one exception) the topological type of

the surfaces of genus at most one are determined by their complexes of curves.

In Chapter 1, we give the relevant deﬁnitions and preliminary information on
orientable surfaces. It is well-known to the experts that the map Mfg —) Aut C (S ) is
injective if S is not a torus with two punctures or a closed surface of genus two. We

give a proof of this fact in this section.

We prove our main results about the mapping class group and the automorphisms
of the complex of curves in Chapter 2. The proofs for the case of the punctured sphere
and for torus are given separately. Although the main ideas of our proofs are the same
as that of Ivanov’s, the proof in the punctured sphere case differs substantially from

the other cases.

We prove our results on nonorientable surfaces in Chapter 3. In this chapter,
we ﬁrst give the necessary background on the mapping class groups of nonorientable
surfaces and the vector spaces over the ﬁnite ﬁeld of order two. Relationship between
these two concepts comes from a theorem of McCarthy and Pinkall [23], Theorem 2.
Then, we compute the ﬁrst homology group of the mapping class group of a nonori-
entable surface. It turns out that if the genus of the surface is at least seven, then
the ﬁrst homology group of the mapping class group is cyclic of order two. Another
result in this chapter is that the subgroup of the mapping class group generated by
the Dehn twists about 2-sided simple closed curves is perfect. We then conclude that
the group of isometries of a vector space V over the ﬁnite ﬁeld F2 with a symmetric
bilinear form (,) deﬁned on a basis {121,122, . . . ,vn} by (vhvj) = 6,5 is perfect if the

dimension of V is at least seven.

Chapter 1

Complexes of Curves and
Extended Mapping Class Groups

In this chapter, we study the complex of curves of an orientable surface and the action
of the extended mapping class group on this complex. The necessary deﬁnitions and

preliminary informations are given.

1.1 Introduction and Preliminaries

1.1.1 Circles on surfaces

Let S be a connected orientable surface without boundary. Although a simple closed
curve on S is an embedding S1 —> S', by a ‘simple closed curve’ on S (or ‘circle’ on

S) we will mean the image of the map under consideration.

A circle on S is said to be nontrivial if it neither bounds a disc nor bounds a disc
with one puncture. We denote by S the set of isotopy classes of nontrivial circles.
Circles will be denoted by lower-case letters a, b, c, etc., and their isotopy classes by

the Greek letters a, 6,7, etc.

Let C be a collection of pairwise disjoint circles on S. The surface obtained from

S by cutting along C is denoted by So.

A nontrivial circle a on S is called lc-separating if the surface Sa is disconnected
and one of its components is a disc with k punctures. If S, is connected we call a
a nonseparating circle. The isotopy class of a circle is called nonseparating (or k-
separating) if the circle is nonseparating (or k-separating). Note that on a sphere
with n punctures, a k-separating circle is also (n — k)-separating. In this case, we
may take k S ’21.

Two circles a,b on a surface S are called t0pologically equivalent if there exists a
homeomorphism F: S —> S such that F (a) = b. It is clear from the classiﬁcation of

surfaces that two circles on a surface are topologically equivalent if and only if they

are either both nonseparating or both Ira-separating for some k.

Let a, ,6 E S. The geometric intersection number i(a, 6) of a and ﬂ is the inﬁmum
of the cardinality of a (1 b, where a 6 a and b E ,8. By the geometric intersection of

two circles we will mean the geometric intersection number of their classes.

1.1.2 The complex C(S)

Let V be a nonempty set. An abstract simplicial complex K with vertices V is a

collection of nonempty ﬁnite subsets of V satisfying the following two conditions:

1) If a: E V then {3:} E K.

2) If a E K and if 0’ C a is a nonempty subset of V, then 0’ E K.

In the second case, 0’ is called a face of a. The dimension dim a of a simplex a is
card a - 1, where card a is the cardinality of a. A simplex a is called a q-simplex
if dim a = q. The supremum of the dimensions of the simplices of K is called the

dimension of K , denoted by dim K.

Let K be an abstract simplicial complex and L a subcomplex of K, i.e., L is itself
a simplicial complex and L C K. L is said to be a full subcomplex if whenever a set

of vertices of L is a simplex in K, it is also a simplex in L.

The complex of curves C (S) on an orientable surface S is the abstract simplicial
complex with vertex set S such that a set of vertices {00, al, . . . , 09} forms a q-simplex

if and only if a0, (11,. . . , 01., have representatives which are pairwise disjoint.

The complex of curves can also be deﬁned for a surface with boundary in a similar
manner. In this case a circle is nontrivial if it does not bound a disc with at most one
puncture or it is not parallel to a boundary component of S. Clearly, the complex
of curves of a surface of genus g with b boundary components and with n punctures,
and that of a surface of genus g with n + b punctures are isomorphic. Therefore, we
will consider only punctured surfaces. We will think of punctures on the surface S as

distinguished points.

An alternative deﬁnition of the complex C (S) can be given as follows. The set of
vertices is the same as before. Now {010, 011, . . . , aq} forms a q—simplex if and only if

i(a,~,aj) = 0 for all i,j.

If the Euler characteristic x(S) of S is negative, then S can be endowed with
a hyperbolic metric, and there exists a unique geodesic in the isotopy class of each
nontrivial circle (cf. [1], [4]). A simple closed geodesic is automatically nontrivial.
The geometric intersection number of two classes are realized by the unique geodesics
they contain. In this case, a third deﬁnition of C (S) can be given: the vertices are
the simple closed geodesics on S and a set of vertices forms a simplex if and only if
the geodesics in the set are pairwise disjoint. This deﬁnition of the complex of curves
is independent of the choice of the metric chosen in the sense that different metrics

give rise to isomorphic complexes.

If S is a sphere with at most three punctures, then there are no nontrivial circles
on S. Hence C (S) = 0. If S is a closed torus and a is a nontrivial circle on S, then
the surface 5., is an annulus. Therefore, any nontrivial circle on S other than a either
is isotopic to a or intersect a. If S is not a sphere with at most three punctures
or a closed torus, then the Euler characteristic of S is negative and the number of
nontrivial simple closed curves on S is 39 — 3 + n, where g is the genus of S and n is

the number of punctures on S. Hence the dimension of C (S) is 39 — 4 + 12.

Suppose that dim C (S) __>_' 1. A pentagon in the complex C (S) is an ordered 5-
tuple (a1, 02, 03,04,a5), deﬁned up to cyclic permutations, of vertices of C (S) such
that i(aj,aj+1) = 0 for j = 1,2,3,4,5 and i(a,-,ak) 7é 0 otherwise (as = (11).

Let a be a vertex of C (S) The link L(a) of a is the full subcomplex of C (S )

whose vertex set is the set of vertices of C (S) which form an edge together with a,

i.e., the vertex set of L(a) is
{[3 = ﬂ e a and new) = 0}.

Given a subcomplex L of C (S), we deﬁne the dual link Ld of L to be the graph
whose vertices are the vertices of L, and whose edges are those pairs of vertices which

do not form an edge in L. We denote the dual link of L(oz) by Ld(a).

1.2 The complex B(S) and ideal triangulations

Let S be a surface with n 2 1 punctures. The simplicial complex B(S) is deﬁned as
follows. The vertices of B ( S) are the isotopy classes of nontrivial embedded arcs on S
joining punctures. (By deﬁnition, an arc is called nontrivial if it is not deformable to
a puncture.) A set of vertices forms a simplex if and only if the vertices in the set have

representatives which are pairwise disjoint. A simple Euler characteristic argument

shows that dim B(S) is ﬁnite and all maximal simplices have the same dimension.
As in the deﬁnition of a circle, by an are we mean the image of (0,1) under a map
9: [0,1] -) S such that g(0) and g(1) are punctures of S (thinking the punctures on
S as distinguished points). Therefore we do not include endpoints into arcs. We will

denote arcs by the letters a’, b’, c’ etc, and their isotopy class by a’, 6’, 7’ etc.

The geometric intersection number of isotopy classes of two arcs is deﬁned similarly
to that of circles: if a’, ,B’ are two vertices of B(S), then i(a’, ,B’) is the inﬁmum of the
cardinality of a’ (I b’, where a' E a’,b’ E ,8’. Then an alternative deﬁnition of B(S)
can be given. The vertex set is the same and a set of vertices form a simplex if and

only if any two classes of arcs in the set have geometric intersection number zero.

Similarly, for a vertex 0: of C (S) and a vertex ,6’ of B(S), we can deﬁne the
geometric intersection number i(a,ﬂ’) as the inﬁmum of the cardinality of a n b’,
where a 6 a and b’ E ,B’. We deﬁne the geometric intersection number of two arcs or

an arc and a circle in the obvious way, similar to that of two circles.

The following lemma is proved in [10], Expose 2, III, and it will be used throughout

this work.

Lemma 1.1 Let S be sphere with three punctures. Then

(i) up to isotopy there exists a unique nontrivial embedded are joining a puncture

P to itself, or P to another puncture Q.

(ii) any circle on S can be deformed to a puncture.

Now suppose that S is a sphere with at least ﬁve punctures or a torus with at
least three punctures for the next three paragraphs. If a is 2-separating circle on
S, by Lemma 1.1 there exists up to isotopy a unique nontrivial embedded are a’

on the twice-punctured disc component of Sn joining two punctures. Conversely, an

arc a’ joining two different punctures of S determines uniquely a 2—separating circle
up to isotopy, namely the boundary of a regular neighborhood of the arc and its
endpoints (as punctures). This gives a one-to-one correspondence between the set
of 2-separating isotopy classes in S and the set of isotopy classes of embedded arcs

joining different punctures.

If a and b are two 2-separating circles on S, and a and S are their isotopy classes,
such that the corresponding arcs a’ and b’ can be chosen disjoint with exactly one
common endpoint P (see Figure 1.1 (a)), then we say that a and b constitute a simple
pair of circles and denote it by (a; b). Similarly, (a’; b’) is called a simple pair of arcs.

We call P the center of the simple pair. We also call (0:; ﬂ) and (a’; 6’) simple pairs.

Let a’1,a’2, . . . ,a]c be embedded pairwise disjoint arcs, P-._1 and H the endpoints
of a2, with P,- 76 P,- for i 75 j, 0 S i,j S k. Thus (af;a$+l) is a simple pair of arcs
with center P,- for each 1 S i S k — 1. Let a1,a2, . . . ,ak be the corresponding circles.
These circles are well—deﬁned up to isotopy by Lemma 1.1. We call a’1,a’2,...,a[c
(respectively, a1, a2, . . . , ak) a chain of arcs (respectively, a chain of circles) and denote

it by (a’1;a’2;.. .;ak) )(respectively, (a1;a2;.. .;ak)) (see Figure 1.1 (b)).

A

Figure 1.1: A simple pair and a. chain.

A

   

     

(b)

If the Euler characteristic x(S) of S is at most —2, then it follows from Theorem

3.3 of [14] that the geometric realization [B(S)] of B(S) is connected.

An ideal triangulation of S is a triangulation whose vertex set is the set of punc—
tures of S, in the sense that vertices of a triangle can coincide, as can a pair of

edges. The importance of B(S) comes from its close connection with ideal triangu-

9

lations of the surface S. Note that the set of isotopy classes of the edges of an ideal
triangulation form a codimension-zero simplex in B(S). Conversely, a set of nonin-
tersecting (embedded) representatives of a codimension—zero simplex determines an

ideal triangulation of S.

A ‘good’ ideal triangle is by deﬁnition a set {a’,b’,c’} of nontrivial embedded
disjoint arcs such that
(i) a' , b’ and c’ join, say, P1 to P2, P2 to P3 and P3 to P1, respectively, for three

different punctures P1, P2 and P3,

(ii) a’ U U U c’ bound a disc in S.

A

A good triangle Not a good triangle

Figure 1.2: Triangles in an ideal triangulation

The following theorem is proved by Hatcher [11].

Theorem 1.2 Let S be a connected orientable surface with at least one puncture.
Then for any two codimension-zero simplices 0’ and 0’ of B(S) there exists a se-
quence of codimension-zero simplices a = 00,01, . . . ,0]. = 0' such that 0,-_1 ﬂ 0.- is a

codimension-one simpler for each i, 1 S i S k.
1.3 The extended mapping class group and its ac-

tions

Let S be an orientable surface. Let Diff S be the group of self-diffeomorphisms of

S and Diﬁ'o S the normal subgroup of Diff S consisting of diffeomorphisms isotopic

10

to the identity. The extended mapping class group of S is the group Diff S/Diffo S
and we denote it by M3. Alternatively, one can deﬁne the extended mapping class
group as the quotient Homeo S by the normal subgroup Homeoo S, where Homeo S
is the group of homeomorphisms of S and Homeoo S is the subgroup consisting of

homeomorphisms isotopic to the identity. It is well-known that these two deﬁnitions

agree.

In the above deﬁnition, if we take only orientation preserving diffeomorphisms
or homeomorphisms, the group obtained is called the mapping class group of S and

usually denoted by M 5.

We deﬁne an action of M; on C (S) as follows. For a mapping class f and an
0: E 8, choose representatives F 6 f and a E 0. Deﬁne f (a) to be the class of the
circle F (a). If F and G are two isotopic diffeomorphisms of S, then F(a) is isotopic
to 0(a). If a and b are two isotopic circles, F (a) and F (b) are isotopic. It follows that
M; has a well-deﬁned action on 5. Let 01,3 6 S be such that i(a,ﬂ) = 0. Choose
a E a, b E ,3 with aﬂb = (0.1ff 6 Mg and ifF E f, then F(a) and F(b) are
disjoint. Therefore i( f (a), f (ﬂ)) = 0. This implies that f is, in fact, a simplicial map
C (S) —-) C (S) Clearly, it is injective, and surjective. Hence it is an automorphism

of C (S) Therefore, there exists a natural map M; —> Aut C (S)

Now suppose that S has at least one puncture. For f 6 M g and a' 6 B (S), choose
F E f and a’ E 0’. Deﬁne f(a’) to be the class of the arc F(a’). Again, it is easy to

see that this gives a well-deﬁned action and MI“; acts on B (S) by automorphisms.

Theorem 1.3 Let S be a connected oriented surface. If S is not a sphere with at
most four punctures or a torus with at most two punctures or a closed surface of

genus two, then the natural map M; —> Aut C (S) is injective.

Proof: Suppose that a is a 2-separating circle on S with the isotopy class a. Denote

11

by D the twice-punctured disc component of Sa. By interchanging two punctures on
D leaving the boundary component pointwise ﬁxed, we get a homeomorphism of D.
Extension of this homeomorphism to S \D by the identity gives a homeomorphism
of S. Let té be the isotopy class of this homeomorphism. Note that we can choose
the twisting of the punctures on D so that (ti)2 = to, where to is the right Dehn
twist about a. It seems right to call t: a ‘half twist’ about o. It is clear from the
deﬁnition that f t3!- f‘1 = tic) if f is orientation preserving and f ti”- f ’1 = (tied-l

if f is orientation reversing.

We claim that an orientation reversing mapping class cannot act as the identity
on the complex 0 (S) Suppose on the contrary that f 6 Mg is orientation reversing
and f (a) = a for all vertices a of C (S) First, suppose that S is a punctured sphere.
By hypothesis, S has at least ﬁve punctures. Let 7 be a 2-separating vertex of C (S ),
c E 7 and F 6 f such that F (c) = c. Orient c arbitrarily. Since the components of S;
are not homeomorphic to each other, F preserves each component of SC and reverses
the orientation of c. Thus F permutes the punctures on the twice-punctured disc
component of Sc. Considering another circle d such that (c; d) is a simple pair, we
see that F, in fact, ﬁxes each puncture. Now let a and b be two 2-separating circles
intersecting as in Figure 1.3 (a). Clearly, we can choose an F E f such that F (a) = a
and F (b) = b. Let us orient a and b arbitrarily. Then F (a) = a'1 and F (b) = b"1 as
oriented circles. Since F ﬁxes each puncture, it follows that F (“I X,y]) = “DCXI’ where
“[X,Y] represent the segment of the oriented circle a from X to Y by following the
orientation of a. In particular, F(X) = Y. On the other hand, F (b[X,Z]) = b[z,x], and

hence F (X ) = Z. By this contradiction f cannot be orientation reversing.

Suppose now that the genus of S is at least one. We can consider S in R3 in such
a way that it is invariant under the reﬂection p across X Y plane (see Figure 1.3 (b),

(c)). Let us denote the isotopy class of p by 9. Then h = p f is orientation preserving.

12

 

 

 

 

 

 

 

C 29 \C’)

Figure 1.3: Various conﬁgurations.

For any vertex 0: of C (S ), we have

Ma) = e(f(a)) = 9(0)-

Consider the circles a, b, d1, d2 (and c if the genus of S is greater than one) illustrated
in the ﬁgure. Let P denote the once-punctured annulus bounded by a and (11 if S is
a punctured torus or the sphere with three holes bounded by a, c and d1 otherwise.
Since S is not a torus with at most two punctures or a closed surface of genus two, d1 '
and at; are not isotopic to each other. Let Q denote p( P). It is clear that there exists
a homeomorphism H E h such that H(a) = p(a) = a, H(dl) 2 d2 (and H(c) = c
if the genus of S is greater than one) as unoriented circles. Then H (P) = Q. Since
H is orientation reversing, H reverses the orientation of a, i.e., H (a) = a'l. Again
since H is orientation reversing, we must have H (b) = b‘". Then H..([a]) = —[a] and
H..([b]) = —[b], where H... is the automorphism of H1(S,Z) induced by H and [a] is

the homology class of a. It follows that
[H (ta(b))] = H.([ta(b)l) = H-(lal + M) = —[al — [b]-
On the other hand
[p(ta(b))l = p-([ta(b)]) = p.([a] + [b]) = [a] - [5]-

Note that ta(b) is a circle. If two oriented circles c1 and C; are isotopic as unoriented

circles, then [c1] = :t[c2]. Therefore, H (ta(b)) is not isotopic to p(ta(b)) or p((ta(b))'1).

13

As a consequence, if a and ,8 denote the isotopy classes of a and b, respectively, then
tow) is, of course, a vertex of C(S) and h(ta(ﬂ)) 75 g(ta(ﬂ)). This contradiction

proves the claim.

Suppose now that the mapping class f acts trivially on C (S), i.e., f (a) = a for
all 0. Then f 6 M5. Hence fté f“’ = tic) = t: if a is a 2-separating vertex
of C(S) and ftgf-l = tf(ﬁ) = t3 for any vertex B of C(S). That is, ftci = téf
and f t3 = t3 f. If S is a punctured sphere, since the mapping class group M5 is
generated by the ‘half twists’ for a number of 2-separating vertices [2], Theorem 4.5,
we conclude that f E C (Ms), where C(Ms) is the center of M5. On the other
hand, C(Ms) is trivial [17]. Hence f = 1. It is well-known that if if the genus of
S is positive, the subgroup of Ms ﬁxing each puncture is generated by the Dehn
twists about (nonseparating) vertices of C (S) (a proof of this may be found in [17]). '
From this it is not difﬁcult to conclude that M5 is generated by the Dehn twists
about vertices of C (S) and the ‘half twists’ about 2-separating vertices. Therefore

f 6 C(Ms). But C(Ms) is trivial [17]. Hence f = 1. The proof of the theorem is

now, complete. C1

Chapter 2

Automorphisms of C(S) and Mfg

2.1 Introduction

The main purpose of this chapter is to prove that if S is a sphere with at least ﬁve
punctures or a torus with at least three punctures, then the natural map M; —>
Aut C (S), described in Section 1.3, is an isomorphism. They are stated as Theorem
2.12 and Theorem 2.18. We have already proved in Chapter 1 that this map is
injective.

Section 2.2 discusses the proof of Theorem 2.12. Clearly, every mapping class takes
a simple pair of circles to a simple pair of circles. So, if automorphisms of C (S) are
induced by mapping classes, then the image of any simple pair under an automorphism
must be a simple pair. This is the starting point. We ﬁrst prove Theorem 2.2, which
enables us to recognize simple pairs of circles in the complex C (S) After proving
that Aut C (S) preserves the topological type of vertices of C (S), we conclude that
simple pairs are preserved under the action of Aut C (S) Next, we deﬁne actions of
Aut C (S) on the punctures of S and on the set of vertices of B(S). This gives rise
to a natural injective homomorphism Aut C (S) -> Aut B(S). Then, we show that
any element of Aut B (S) induced by an element of Aut C (S) agrees with a mapping

class on a codimension-zero simplex, and then on all of B(S). Finally, this mapping

14

15

class agrees with the original element of Aut C (S) on C (S)

The main results of Section 2.3 are Theorem 2.18 and Theorem 2.19. Instead
of simple pairs, we work with the pairs of circles intersecting transversally at one
point. The ﬁrst step is to recognize these conﬁgurations in the complex C (S) The
difﬁculty here is to show that the group Aut C (S) preserves the topological type of
vertices of C (S) We are able to overcome this difﬁculty when S has three punctures.
Then induction on the number of punctures proves Theorem 2.18. In the proof we
use Theorem 2.16, which Ivanov proved, but unfortunately, has not been published
yet. We give our own proof of it in the case of a torus with at least three punctures.
This proof is basically the same as that given in Section 2.2; we only point out
the differences. Theorem 2.19 asserts that the surfaces of genus at most one are

determined by their complexes of curves.

In Section 2.4, we give an application of the main results of Section 2.2 and Section
2.3. Namely, we prove that if S is a sphere with at least ﬁve punctures or a torus
with at least three punctures, then any isomorphism between two subgroups of M;
of ﬁnite index is the restriction of an inner automorphism of Mg. This implies that
two subgroups of ﬁnite index are isomorphic if and only if they are conjugate, and

that the outer automorphism group of a subgroup of ﬁnite index is ﬁnite.

2.2 Punctured Spheres

In this section S will, unless otherwise stated, denote a sphere with n punctures, and

n will be greater than or equal to ﬁve.

16
2.2.1 Simple pairs in C(S)

Lemma 2.1 Let S be a sphere with n _>_ 5 punctures and a be a codimension-zero

simpler of C(S). Then at least two vertices of a are 2-separating.

Proof: Recall that if R is a sphere with four punctures, then dim C (R) = 0. Also

note that any nontrivial circle on R is 2—separating.

We now prove the lemma by induction on 12. If S is a sphere with ﬁve punctures,
then since dim C (S) = 1, (hence card a = 2) and since every nontrivial circle on S is

2-separating, we are done.

Let n 2 6. If all vertices in a are 2-separating, then we are done since card a Z 3.
So suppose 0 contains a k-separating vertex (1 for some 3 S k S g. Choose a circle
a E (1. Then the surface 3,. is a disjoint union of S,’,, a disc with k punctures, and
S;’, a disc with n - k punctures. Note that C(53) and C (S: ) are isomorphic to the
complexes of curves of spheres with k + 1 and n — k + 1 punctures, respectively. Also
a — {a} is the union of two simplices o’ and a”, where o’ and a” are codimension-zero

simplices of the complexes C (5;) and C (S: ), respectively.

If k = 3 and n = 6 then C (5;) and C (S: ) are isomorphic to the complex of curves
of a sphere with four punctures. Hence dim 0’ = dim a” = 0. Then the vertex of a’

and that of o” are 2-separating on S.

If k = 3 and n 2 7 then the only vertex of a’ is 2-separating on S. Also, since
dim C (Sf: ) = n—k+1 Z 5, by the induction hypothesis there exist at least two vertices

in a” which are 2-separating on 55;. Then one of these vertices is 2—separating on S.

Iflc Z 4 then dimC(S,’,) = k+12 5 and dimC(S;') = n—k+1 Z 5. By
induction hypothesis, 0" contains at least two vertices which are 2-separating on S,’,.

One of these vertices must be 2-separating on S. Similarly, one of the vertices of a”

17

must be 2—separating on S. Therefore, we are done. El

Since any diffeomorphism of S takes a simple pair of circles to a simple pair of
circles, any automorphism of C(S) must take a simple pair of vertices to a simple
pair in order to have a hope that it is induced by some diffeomorphism of S. The

following theorem enables us to recognize simple pairs in the complex C ( S)

Theorem 2.2 Let a and S be two 2-separating vertices of C(S). Then (a;,8) is a
simple pair if and only if there exist vertices 71,72,73, . . . ,7n_2 of C(S) satisfying the

following conditions.

(i) (7137290173,:B) is a pentagon in C(S))

(ii) ’71 and 7n_2 are 2-separating, 72 is 3-separating, and 7;, and 7,,4, are k-

separating for 3 S k S 121,

(iii) {0373:74375w-‘a7n-2}: {0,72,74,75an'37n-2}: {(6,73a74375w'w7n-2} and

{71, 72, 74, 75, . . . ,7n_2} are codimension-zero simplices.

Proof: The ‘only if’ part of the proof is very easy. Let a E a and b 6 S such that
(a; b) is a simple pair. It is clear that any two simple pairs of circles are topologically
equivalent, i.e., if (a1; b1) is any other simple pair, then there exists a homeomorphism
F : S ——) S such that (F(al); F(b1)) = (a;b). Hence we can assume that a and b are
the circles illustrated in Figure 2.1. The ﬁgure represents the case n = 8. In the
ﬁgure, we think of the sphere as the one point compactiﬁcation of the plane. Then

the isotopy classes 7,- of the circles c,- satisfy (i)-(iii).

Now we prove the converse. Assume that conditions (i)-(iii) above hold. For a

k—separating circle c on S with 2 S k S %, let S", and S2’ denote the connected

71

components of Sc having k and n — lc punctures, respectively. In the case of k = E

18

 

 

 

 

 

f
(fr \63 c4 c5 c6

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

k \
\K JJJ

Figure 2.1: A conﬁguration of circles on a punctured sphere.

k

either of them may represent either component. Let a, b and c, be the representatives

of a, ﬂ and 7;, respectively, intersecting each other minimally.

We claim that a and b lie on a thrice-punctured disc bounded by c3. By (i),
a and b intersect transversally at least once, because the number of points in the
intersection a n b is the geometric intersection number i(oz, B) of a and )8, which is
nonzero. Also a U b does not intersect c3. Hence they lie on the same component of
Scs. For n = 5 or n = 6 since c3 is 3-separating, each component of Sc3 is either a
disc with two punctures or a disc with three punctures. Since there is no nontrivial
circle on a disc with two punctures, the claim is obvious. So suppose that n 2 7. Let
C = c3 U c; U. . . U cn_3 and consider the surface Sc. We ﬁrst prove that So is a union
of two discs with three punctures, whose boundaries are c3 and cn_3, and a number
of annuli with one puncture. Since every circle on S is separating, the number of
components of Sc is n — 4. Also, for any circle d on S, x(S) = x(S,’,) + x(SQ’). From

this it follows that

2-n =X(5) =ZX(R)

where R runs over the components of Sc. Since all c,- are non-isotopic, x(R) is

negative for all R. Hence either there is only one R with x(R) = —3, or there are

19

precisely two components with x(R) = —2, and the rest of the components have Euler
characteristic —1. If there exists a component R with x(R) = —3, then R is a disc
with a number of holes and punctures, the total number of holes and punctures on
the disc R being four. We then can ﬁnd two circles d1 and (12 on R such that at most
one of d1 and d2 is 2-separating on S. Then {61,62,73,'y4, . . . ,7n-3} is a maximal
simplex of C (S) containing at most one 2-separating vertex, where 6,; is the class of
d.- for i = 1,2. This is a contradiction to Lemma 2.1. Thus So has two components

with Euler characteristic —2, namely S; and Saks.

Since a 0 b is nonempty and since {0,73, . . . , 7n..3} and {ﬂ,73, . . . , 7,,_3} are sim-
plices of C (S), a U b lies either on S23 or on Sén—E}, both of which are discs with three
punctures. Now let 6 be the boundary of the thrice-punctured disc on which a and b
lie. Since c2 is a 3-separating circle intersecting b and is not isotopic to 0, it follows
that C; and 8 intersect nontrivially. As i(72,7n_3) = 0, we must have 0 = c3. This

proves the claim.

The condition (iii) implies that the circles a,b, c1, c2 and c3 all lie on SCI, where
C" = c., U c5 U .. . U c,,_2. (If n = 5 then c2 and c3 are 2-separating circles disjoint
from a. Hence the corresponding arcs c’2 and c3 have a common endpoint. In this
case, we take 0’ to be a trivial simple closed curve deformable to this puncture.) By
arguing as above, one can see that the surface So: is the disjoint union of a number of
surfaces of Euler characteristic -1 and a disc D with four punctures, with boundary

c4, after changing the roles of c4 and cn_4 if necessary (see Figure 2.2).

Keeping the correspondence between 2-separating vertices and the arcs in mind,
suppose that the endpoints of a’ are P and Q. Then c3 separates P, Q and another
puncture, say, R from the forth puncture T on D. Up to a homeomorphism of S, the

picture of a’, c3 and c4 are as illustrated Figure 2.2 (a).

20

 

(a)

Figure 2.2: The disc D with four punctures.

Since i(a, '72) = 0 and i(72, 73) 75 0, each component of car‘lSé3 is an arc connecting
two points on c3, and isotopic to each other by an isotopy of SL3 leaving the endpoints
of the are on c3 by Lemma 1.1. (Here 523 denotes the thrice-punctured disc component
of S which does not contain the circle c4). Let A be one of these arcs and let A 0 c3 =
{X , Y}. Then a component u of c3—{X, Y} and A bound a disc D’ with two punctures

P and Q. Note that P and Q are on $22, which is a thrice—punctured disc.

Since b’ is on S23 , its endpoints are among P, Q and R. So one of them must be P,
by changing the roles of P and Q if necessary. If the other endpoint of b’ is Q, then
the endpoints of c’l must be T and R since c; is disjoint from b. This and c1 0 c2 = 0
imply that R and T are on S22, too, since at least one of R and T is on 522. This is
a contradiction because the four punctures P, Q, R, T cannot be on S22 all together.

By this contradiction the endpoints of b’ are P and R, and that of c’l are Q and T.

Since c1 0 62 = 0, c’1 does not intersect A and hence it intersects p. Let Z be the
ﬁrst point where c’l meets u starting from Q. Denote the segment of c’l between Q
and Z by (f). Up to an isotopy of D’ leaving the endpoint at Q ﬁxed and keeping the

other endpoint on A U u, such an arc is unique by Lemma 1.1. So 43 can be chosen so

21

that it does not intersect a’.

Finally, by cutting Séa along a regular neighborhood of ¢U {Q} we get a disc with
two punctures P and R. The arc b’ must lie on this disc since it meets neither c’l nor
c3. Again, up to an isotopy of this disc there is only one arc joining the punctures P

and R, which can be chosen disjoint from a’. This ﬁnishes the proof. Cl

Lemma 2.3 The group Aut C(S) preserves the topological type of the vertices of
C (S)

Proof: Note that all vertices of C (S) are separating. For a vertex a of C (S), the
dual link L"(oz) is connected if and only if a is 2-separating. This implies that each

automorphism of C (S) permutes the set of 2-separating vertices.

If a is a k-separating vertex for some 2 < k S %, then the dual link L"(a) has
exactly two connected components. Let us denote these components by Lg(a) and
L‘l’(a). The vertices of these components are the isotopy classes of the circles on the
two components of S“, where a 6 01. We then deﬁne two full subcomplexes Lo(oz) and
L1(a) of the link L(a) of a as follows: The vertices of Lj(oz) are those of L‘J’(oz) for
j = 0,1. We can choose Lg(a) and Li’(a) so that Lo(a) and L1(a) are isomorphic to
the complexes of curves of, respectively, the k-punctured and the (n — k)-punctured

disc components of Sa. Hence dim Lo(a) = k — 3 and dim L1(a) = n — k — 3.

Let a be a k-separating vertex for some 2 < k S 52’- and f be an automorphism of
C (S) It is clear that f induces an isomorphism from the disjoint union Lo(a) U L1(a)
to the disjoint union Lo(f(a)) U L1(f(a)). Since dim Lo(a) S dim L1(a), it follows
that dim f(Lo(a)) S dim f(L1(a)) and hence f(Lo(a)) = Lo(f(a)). (In the case
k = g we may change the role of Lo(f(oz)) and L1(f(a)) if necessary.) Since a is

k-separating, dim Lo(a) = dim Lo(f(o)) = k — 3. Since Lo(f(a)) is isomorphic to

22

the complexes of curves of one of the components of S f(a) for f (a) E f (a), the circle

f (a) must be k-separating. D

Corollary 2.4 Let f be an automorphism of C(S). If (mﬂ) (and hence (a';ﬂ')) is
a simple pair, then so is (f(a);f(ﬁ)) (and hence (f(a’); (ﬂ’))). Similarly, the image

of a chain in C(S) under f is also a chain.

Proof: It is clear that the conditions (i) and (iii) of Theorem 2.2 are invariant under the
automorphisms of C (S) The fact that the condition (ii) is invariant under Aut C (S)

is proved in Lemma 2.3 above.

The second part of the corollary follows easily from the ﬁrst part. B

2.2.2 The map Aut C(S) —> Aut B(S)

In this subsection we show that every automorphism of C (S ) gives rise to an auto-
morphism of B (S) in a natural way. In fact, we will have an injective homomorphism

Aut C(S) —-> Aut B(S).

The ﬁrst step is to deﬁne an action of Aut C (S) on the punctures of S. We
deﬁne this action as follows. For f E Aut C (S) and for a puncture P of S, take any
simple pair (a’;ﬂ’) with center P, and deﬁne f(P) to be the center of the simple
pair ( f (a’ ); f (3’ )) Note that by the one-to-one correspondence between the set of
2-separating vertices of C (S ) and the set of those vertices of B (S) which join different

punctures, Aut C (S) has a well-deﬁned action on the latter set.

Lemma 2.5 The deﬁnition of the action of Aut C(S) on the punctures ofS is in-

dependent of the choice of the simple pair.

23

Proof: Let a’ 6 a’ and b’ E ,6’ be such that a’ and b’ are disjoint, i.e., (a’;b’) is a
simple pair with center P. Let f (a’ ) and f (b’ ) be the disjoint representatives of f (a’ )

and f (,B’ ), respectively, and P the center of ( f (a’ ); f (b’ )) .

We show ﬁrst that if c’ is an arc joining P to some other puncture, with the class
7’, and if f(c’) is a representative of f(7’) intersecting f(a’) and f(b’) minimally,
then one of the endpoints of f(c’) is P. The proof of this is by induction on i =

i(a’,7’) + i(ﬂ’,7’). Let us denote by P1,P2 and P3 the other endpoints of a’, b’ and

c’, respectively.

As the ﬁrst step of the induction, suppose that i = 0. There are two cases to

consider.

Case 1: If (a’;7’) and (ﬂ'n’) are simple pairs also (i.e., if P3 is different from P1
and P2), then there is a fourth arc (1’ such that any two arcs in the set {a’, b’, c’, d’}
constitute a simple pair with center P (see Figure 2.3 (a)). This is because there are
at least ﬁve punctures on S. Then any two arcs in { f (a’ ), f (b’ ), f (c’ ), f (d’ )} constitute
a simple pair, where f (d’ ) is a representative of the image of the class of d’ under f.
An easy argument shows that all four arcs f (a’), f (b’ ), f(c’ ) and f (d’ ) must have a

common endpoint, which must be P.

Case 2: If the endpoints of a’ and c’ are the same, (i.e., P1 = P3), then there
exist a puncture Q different from P, P1 and P2, and an arc d’ joining Q and P not
intersecting any of the three arcs a’ ,b’ , c’ (see Figure 2.3 (b)). By an application of
Case 1 to {a’,b’,d’} and then to {b’,d’,c’}, we see that one of the endpoints of f(c’)

is P. Notice that the center of ( f (ﬂ’ ); f (7’ )) must be P.

For the case i > 0, let us orient all three arcs from P to P], for each j = 1,2,3.
Let X be the ﬁrst point where c’ meets a’ U b’. Without loss of generality we can

assume that X is on a’. Let d’ be the arc consisting of the segment of c’ from P to X

24

 

Figure 2.3: Construction of d’ if i = O.

and that of a’ from X to P1 (see Figure 2.4). Then i(oz’,6’) + i(ﬂ’,5’) = 0, (6’;ﬂ’) is a
simple pair, and i(6’,7’) + i(ﬂ’,7') < i(a’,7’) + i(ﬂ’,7’), where 6’ is the class of (1’. By
induction, one of the endpoints of f (c’ ) is the center of the simple pair ( f (6’ ); f (ﬂ’ )),

which is P by Case 2.

 

Figure 2.4: Construction of d’ if i > 0.

If (‘7’;6’) is another simple pair with center P, and c’ 6 7’ and d’ 6 6’ intersect
each other as well as a’ and b’ minimally, then by applying the argument above ﬁrst
to {a’, b’, c’ } and then to {a’, b’, d’} we see that P is an endpoint of both f (c’ ) and
f (d’ ), which must be the center of the simple pair ( f (7’ ); f (6’ )) Hence the action of

Aut C (S) on the set of punctures of S is well-deﬁned. D
For a punctured surface R, let us denote by ’P(R) the set of punctures of R.

Lemma 2.6 Let f 6 Aut C(S), a a k-separating vertex of C(S), and a E a. If S",
and Sf,’ denote the lc-punctured and (n — k)-punctured disc components of SO, then
f(P(S,’,)) = ’P( }(a)), and hence f(’P(S¢’,’)) = ’P( 320)). In the case ofk = %, we may

change the roles of S; and Sf,’ if necessary.

25

Proof: Two punctures P and Q are on the same connected component of 5,, if and
only if P and Q can be joined by an arc disjoint from a, and P and Q can be joined ’
by an arc disjoint from a if and only if f (P) and f (Q) can be joined by an arc disjoint

from f (a). Now the proof of the lemma follows. D

We can now deﬁne an action of Aut C(S) on the vertices of B(S). Let f E
Aut C (S), a’ a vertex of B(S) and let a’ 6 a’. If a’ is joining two different punctures,
then f (a’ ) is already deﬁned by the correspondence with the 2-separating vertices of
C(S) and the action of Aut C(S) on C(S). That is, f(a’) is the isotopy class of the
arc, which is unique up to isotopy, joining two punctures on the twice-punctured disc
component of S f(a) for f (a) E f ((1). Suppose now that the arc a’ is joining a puncture
P to itself. Let a1 and a2 be the boundary components of a regular neighborhood of
a' U {P} and al and (12 be their classes. Since a’ is not deformable to P, at most
one of a1 and a2 is trivial. If a1 is trivial then a2 is 2—separating, and by Lemma 2.3
f (02) is 2-separating. Hence for a (circle) representative f (a2) of f (02), one of the
components, say Sh”), of S f(a2) is a twice-punctured disc, and one of the punctures
on Shag) is f (P) by Lemma 2.6. Deﬁne f(a’) to be the class of a nontrivial simple
arc on 551%) joining f (P) to itself. Such an arc is unique up to isotopy by Lemma

1.1.

In the case that neither a1 nor a2 is trivial, we claim that f (a1) and f (a2) bound
a once-punctured annulus with only one puncture f (P) Here f (a,-) is a represen-
tative of ﬂu.) For the proof of this, suppose that the set of punctures on Sf,l
and S"; are P(S;l) = {P1,...,Pk} and P(S;’2) = {Q1,...,Q,,_k_1}, respectively.
Then P,- ;e Q,- for all i, j. By Lemma 3.5, r( he”) = {f(Pl),...,f(Pk)} and
’P(S}’(az)) = {f(Ql),...,f(Q,,_k_1)}. It follows that, since f(al) and f(ag) are dis-

joint and nonisotopic, they must bound an annulus with only one puncture f(P).

26

Then f (a’ ) is deﬁned to be the isotopy class of the unique arc (up to isotopy) on this

annulus joining f (P) to itself.

Lemma 2.7 Let f be an automorphism of C (S) and a’ and ,B’ be two distinct vertices
of B(S) such that i(a’,ﬂ’) = 0. Then i(f(a’),f(ﬂ’)) = 0. Hence every automorphism
of C(S) induces an automorphism of B(S).

Proof: Let a’ and b’ be two disjoint representatives of a’ and B’, respectively. There
are seven cases to consider (see Figure 2.5). In the ﬁgure, we assume that the arc on

the left is a’ and the one on the right is b’.

If a’ (resp. b’) is joining two different punctures, let us denote by a (resp. S)
the 2-separating vertex of C (S) corresponding to a’ (resp. ﬂ’), and by a (resp. b) a
representative of a (resp. S).

If a’ (resp. b’) is connecting a puncture P to itself, let us denote by a1 and a2
(resp. b1 and b2) the boundary components of a regular neighborhood of a’ U {P}
(resp. b’ U {P}). Note that we use the classes of a,- (resp. b.) to deﬁne f (a’ ) (resp.

f(ﬁ’)). We also denote representatives of f(a), f(a’) by f(a), f(a’) etc.

If a1 or b; is trivial, we work with them by considering the trivial simple closed

curve deformable to a puncture to be that puncture.

// /<) O O /\ C3. ——0 (>0
0) (ii) (iii) (W) (V) (vi) (vii)

Figure 2.5: Seven possible cases for a’ and b’.

We examine each of the seven cases illustrated above.

(i) Here a and b are disjoint, so i(a, ﬂ) = 0. Since f is an automorphism of C (S),
this implies that i( f (a), f (ﬂ)) = 0. Another way of saying this is that f (a) and f (,6)

have distinct disjoint representatives f (a) and f(b). Hence a’ and b’ are disjoint.

27

(ii) On the annulus determined by f (b1) and f (b2) there is only one puncture.
Since f is an automorphism of C (S), the 2—separating circles f(a), f (b1) and f (b2)
are all distinct and no two are isotopic. Therefore f (a) cannot lie on this annulus.

Since f (b’ ) is on the annulus, we are done.

(iii) The once-punctured annuli determined by a1 and a2, and b1 and b; are disjoint.
Since a1, a2,b1 and by are pairwise disjoint, so are f(al), f(ag),f(b1) and f(bg). So the

annuli determined by f (a1) and f (a2), and f (b1) and f(bg) are disjoint.
(iv) Corollary 2.4.

(v) Suppose the endpoints of a’ and b’ are P and Q. Let R be any puncture other
than P and Q. Then there exist arcs c’ and d’ and a 3—separating circle e satisfying

the following conditions:
(1) c’ and d’ are disjoint from a’ U b’,
(2) c’ joins P and R and d’ joins Q and R,
(3) b’ U c’ U (1’ bounds a disc on S, and
(4) b’, c’ and b’ lie on a thrice-punctured disc component of 5,.

Note that any are joining P and Q which is disjoint from c’ and d’ is isotopic to
an arc disjoint from b’. (Note that any are isotopic to b’ can be isotoped to an arc
disjoint from b’) Then f (b’ ) U f (c’ ) U f (d’ ) lies on a thrice-punctured disc component
of 51(5) and bounds a disc on S. The latter follows from a similar argument given in
the ﬁrst step of the induction in the proof of Lemma 2.5. Since (a’; c’) and (b’; c’) are
simple pairs, so are (f(a’); f(c’)) and (f(b’); f(c’)). In particular, f(a’) can be chosen

disjoint from f (c’ ) and f (d’ ) Hence it can also be chosen disjoint from f (b’).

(vi) Suppose that a’ is connecting P to Q and that b’ is connecting P to itself.

We can assume that b; is disjoint from a’. Clearly, there exists an are c’ not meeting

28
a’, b’ and b1, joining P to some other puncture, say, R (see Figure 2.6).

f( b1 )
f(b'

 

f(Q) f(P)

 

Figure 2.6: Case (vi).

Then f(c’) connects f(P) to f(R) and meets f(bz), but not f(bl) or f(a’). Let
A be the segment of f (c’ ) lying on the once-punctured annulus determined by f (b1)
and f (b2), and connecting f (P) to f(bl). Note that the intersection of f (a’ ) with the
this annulus is a collection of arcs joining a point on f (b1) either with another point
on f (b1) or with f (P) Since f(a’) is disjoint from A, the intersection of f(a’) and
this annulus consists of only one arc connecting f (P) to some point on f (b1). Then
f (b’ ), which is an arc on this annulus joining f (P) to itself, can be chosen so that it

does not intersect f (a’ )

(vii) Let P be the common endpoints of a’ and b’. We can assume that al does
not meet b1 and b2. Let P1, . . . , Pk be the punctures on the component of So, which
does not contain a’ Ub’. Choose k arcs do, c’1,. . . , CL, such that c: joins P,- to Pg.“ for
i = 0,1,. . . ,k -1, where P0 = P, and {a’,B’,7(’,,7{,. . .,7,"_1} is a simplex of B(S).
Then the are f(e) joins f(a) and f(Pa+1)a and {f(a’),f(76),f('vi)e---,f(75.-1)} and
{f(ﬂ’). ms), f(vi), . . . , f(7£-1)} ere simplices of B(S) by (i), (ii). (iv) and (vi). Since
any arc joining P to itself which is disjoint from cﬁ,c’l, . . . , cf“, is isotopic to an arc
disjoint from a’, any arc joining f(P) to itself disjoint from f(cg),f(c’l), . . . ,f(cj,_1)
is isotopic to an arc disjoint from f (a’ ) Therefore, f (b’ ) can be chosen disjoint from

f (a’ ) This completes the proof. 0

29

 

Figure 2.7: Case (vii).

Proposition 2.8 The group Aut C(S) is naturally isomorphic to a subgroup of the
group Aut B(S).

Proof: By Lemma 2.7 every element of Aut C (S) induces an element of Aut B (S)
Clearly, the map taking an element of Aut C (S) to the induced element of Aut B (S)
is a homomorphism. It is not very difficult to show that if an automorphism of C (S )

induces the identity automorphism of B(S), then it is, in fact, the identity. Cl

2.2.3 Automorphisms of C(S) and Mg

In this subsection, S will denote a sphere with at least ﬁve punctures.

Lemma 2.9 Let f and g be two automorphisms of B(S ) If they agree on a codimension-

zero simplex, then they agree on all of B(S).

Proof: Let a be a codimension-zero simplex of B(S). Suppose that f is equal to
g on a. If 0’ is another codimension—zero simplex, then by Theorem 1.2 there exist
codimension-zero simplices a = 0’0, 01, . . . , 0‘;‘ = 0’ such that 0,400,- is a codimension-
one simplex for each i. Since any codimension-one simplex is a face of either one or

two codimension—zero simplices, if two automorphisms B (S) agree on 0,-_1 then they

30

agree on 0,. Clearly, this implies that f must be equal to g on 0’. Since every simplex

of B(S) is a face of a codimension—zero simplex, we are done. D

Recall the correspondence between the codimension-zero simplices of B (S) and
the isotopy classes of ideal triangulations of S. Since all maximal simplices in B(S)
have the same dimension, we have a well-deﬁned action of the group Aut B(S) on

the codimension-zero simplices.

Lemma 2.10 Let f 6 Aut C(S), A = {a’, b’, c’} be a ‘good’ ideal triangle on S and
let a’, ,B’ and 7’ be the isotopy classes of a’, b’ and c’, respectively. Then {a’, ﬁ’, 7’}, and
hence 0130{f(a'),f(ﬂ'),f(7')}, is a 242'"!le in B(S)~ 1ff(A) = {f(a’), f(b'), f(C')}

is a realization of the latter simplex, then it is a ‘good’ ideal triangle on S.

Proof: Recall that a ‘good’ ideal triangle has three different vertices and (a’; ,B’),
(ﬂ’n’) and (01’ ;7’) are simple pairs. It follows that f(A) is an ideal triangle all of
whose vertices are different. Let P and Q be any two punctures different from the
vertices of f (A) Since f "(P) and f ”1(Q) can be joined by an are d’ not intersecting
any of the edges of A, P and Q can be joined by an arc f(d’). Since f(d’) has the
geometric intersection number zero with each arc in f (A), it can be chosen so that it

does not intersect any of them. Thus the proof of the lemma is complete. D

Lemma 2.11 Let f E Aut C(S). Suppose that there exists a homeomorphism F :
S -+ S such that [F] and f are equal as automorphisms of B(S), where [F] denotes

the isotopy class of F. Then [F] = f as automorphisms of C(S).

Proof: Let a be a k-separating vertex of C(S), a E a and let ’P(S,’,) = {P1, . . . , Pk},
’P(S,’,’) = {Q1, . . . , Qn_k}. Choose two arbitrary chains (b’l; . . .; [_1) and (Ci; . . . ; 4,4,4)
such that b:- and c’ are disjoint from a for i and 3'. Then (f(b’l); . ..;f(b’_1)) and

J

31

(f(c’l); . . . ; f(cﬁ,_,,_1)) are two chains such that f(bﬁ) and f(cg) are disjoint from f(a)
for all i, j. Since [F ](,B,’) = f (3f), [F](7;-) = f (7;) by assumption, and since there is
only one (kt-separating) circle up to isotopy disjoint from the chains ( f (b’1 ); . . . ; f ( [_1 ))
and (f(Ci);---;f(ct-k-1))e namely f(a) E f(a), we must have WHO!) = f(a) This

proves the lemma. D

We are now ready to prove the main result of Section 2.2.

Theorem 2.12 Let S be a sphere with at least ﬁve punctures. Then the natural map

Mg —> Aut C(S) is an isomorphism.

Proof: Injectiveness of the map Mg —-> Aut C (S) was proved in Theorem 1.3. Let
us now show that it is onto. Let f E Aut C(S) and let 0’ be an arbitrary ideal
triangulation of S such that each triangle has three different vertices, i.e., a ‘good’
triangulation. Existence of such a triangulation is clear. Let a be the isotopy class
of 0'. Then a is a codimension—zero simplex of B(S). By Lemma 2.10, ‘good’ ideal
triangles are mapped to ‘good’ ideal triangles by f, and it is a well-known fact that f
can be realized by a homeomorphism on each such triangle. Since each edge of C’ is
an edge of exactly two ‘good’ ideal triangles, the homeomorphisms of these triangles
give rise to a homeomorphism F of S. If [F] denotes the isotopy class of F, then f
agrees with [F] on a. By Lemma 2.9, they agree on B(S). Finally, by Lemma 2.11,
[F] is equal to f on C(S). Cl

32

2.3 Punctured Tori

In this section, unless otherwise stated, S denotes a torus with n punctures.

2.3.1 Automorphisms of C(S) and Mg

The following theorem enables us to recognize whether or not two vertices of C (S)

have geometric intersection number one, by looking at the complex C (S)

Theorem 2.13 Let S be a torus with at least two punctures, and let a and ,8 be two
vertices of C(S). Then the geometric intersection number i(a,,8) = 1 if and only if

there exist three vertices 71,72 and 73 of C(S) such that

(i) (7190,72iﬂi73) is a pentagon in C(S)) and

(ii) 0,,6 and 73 are nonseparating, and 71 and 72 are n-separating.

Proof: Let us ﬁrst prove the ‘only if’ clause of the theorem. Clearly, i(a,,8) = 1
implies that a and ,8 are both nonseparating. Thus there exist a E a and b E H
such that a and b intersect transversally at only one point. It is well known that if c
and d are any other pair of circles intersecting transversally at only one point, then
there exists a homeomorphism F :S —+ S such that F (a) = F (c) and F (b) = F (d)
Hence we can assume that a and b are the standard circles in Figure 2.8 (a). The
existence of the other circles whose isotopy classes satisfy (i) and (ii) is now obvious

from Figure 2.8 (b).

For the converse, let a 6 a,b E B and c,- 6 '7,- intersect each other minimally
pairwise. Since a is nonseparating, the surface 5,, is an annulus with n punctures
and two boundary components q] and q2. Then S is a quotient space of So. Let

p : S,, —> S be the quotient map, so p(q1) = p(q2) = a. Up to a homeomorphism of SO

33

© sets!

(a) (b)

Figure 2.8: A pentagon in C(S).

preserving ql and q;, we can assume that the picture of p‘1(c2) in So is as in Figure

2.9 (a). In the ﬁgures, c, represents p‘1(c,).

Let i(a, ,6) = m, which is the cardinality of aﬂb. Since a and 6 are not connected
by an edge in the pentagon (and hence in C (S )), this geometric intersection number

m must be positive.

We now consider the components of the preimage p‘1(b) of b, which is a collection
of arcs. Since i(ﬂ,72) = 0, the components of p”(b) lie on a disc with two holes
whose boundary components are q1, q; and c2, and they do not intersect c2. Thus
it follows from Lemma 1.1 that each arc in p”(b) joins either a point on ql to a
point on q;, or, two points on (11, or, two points on Q2. Let mm, mm mm be the
number of these components joining ql to q1, q; to Q2 and q1 to Q2, respectively. Then
m = m12 + 2m“ = 77112 + 2m22 and hence mu = m22. On the other hand, if A is
an embedded arc on 5,, connecting two points on ql such that A is not isotopic to
a segment of ql and does not intersect c2, then every embedded arc connecting two
points on q; which is disjoint from A must be trivial, i.e., isotopic to a segment of Q2.
Therefore mm and m2; must be zero, so each component of p"(b) connects a point

on ql to a point on qg. Therefore the picture of these arcs on S. is as in Figure 2.9
(b).
Let us now orient ql and q; so that the induced orientations of p(q1) and p(q2)

agree in S, and let X1, X2, . . . , Xm and Y1, Y2, . . . , Ym be the consecutive intersection

34

 

 

 

 

(c) ((9

Figure 2.9: The surface Sa.

points of p"(b) with q; and Q2, respectively, such that X,- is joined with Y,- by some
arc in p"(b) for each i = 1,2,...,m. It is clear that there exists a k, 0 S k < m,

such that p(X,) = p(Y,+k) for each i. By convention we set Ym+,- = Y,.

Since i(ﬂ, 71) aé 0 and i(a, 71) = 0, the preimage of Cl, also denoted by CI in the ﬁg-
ure, intersects every component of p‘1(b) (see Figure 2.9 (c)). As i (73, 71) = i(73, ﬂ) =
0, each component of p‘1(c3), intersects ql only in the open interval [Xm X1] and q;
in ]Ym,Y1[. Therefore p(]Xm,X1[) = p([Ymal/ll) in S, i.e., p(X1) = p(Yl) and hence
k = 0. Finally, p(b) is a connected curve only if m is equal to 1. This ﬁnishes the

proof of the theorem. Cl

Lemma 2.14 Let n 2 3. Let (1,3 and ’7 be distinct vertices of C(S). [fa is non-
separating, ,8 is n-separating and '7 is separating, and i(a,,8) = i(ﬂ,7) = 0, then

i(a,7) = 0.

35

Proof: Let a,b and c be representatives of a,ﬂ and 7 in minimal position. The
nonseparating circle a and the separating circle c are, respectively, nonseparating and
separating on the surface Sb, the surface obtained from S by cutting along b. But
nonseparating and separating circles on 5;, lie on different components (see Figure

2.10). [:1

 

 

Figure 2.10: A nonseparating and an n-separating circle.

Lemma 2.15 Let n 2 2 and let S and 5’ denote a torus with n punctures and
a sphere with n + 3 punctures, respectively. If every automorphism of the complex

C (S) is induced by some self-homeomorphism of S, then C(S) and C(S’) are not

isomorphic.

Proof: Suppose that p is an isomorphism from C (S) to C (S’ ) Then cp induces a
group isomorphism go. : Aut C(S) —-> Aut C(S’), deﬁned by cp...( f) = cpfcp'l for each
f E Aut C (S) This implies that

Aut C(S’) = {cpfcp'l : f 6 Aut C(S)}.

We now show that this is impossible. Note that for a vertex oz of C (S), the dual
link L"(a) of a is connected if and only if a is either nonseparating or 2-separating,
and for a vertex [3 of C (S’ ), the dual link L"(ﬂ) of S is connected if and only if
B is 2—separating. From this it follows that the image of the union of the set of
nonseparating vertices and the set of 2-separating vertices of C(S) is precisely the set

of 2—separating vertices of C (S’ )

36

Let a be a nonseparating vertex of C (S), and choose a 2-separating vertex 3 of
C(S) such that i(a,ﬂ) = 0, i.e., a and S are joined by an edge in C(S). Then 90(0)
and (p(ﬁ) are two 2-separating vertices of C (S’ ) and are joined by an edge in C (S’ ).
Let c and d be representatives of 90(0) and (p(ﬂ), respectively. Then c and d are
two disjoint 2-separating circles on S’. By the classiﬁcation of surfaces there exists
a homeomorphism C of S’ such that C(c) = C(d). Hence g(cp(a)) = (p(ﬁ), where g
is the isotopy class of G'. Then the automorphism cp:1(g) = go'lgcp of C(S) takes
the nonseparating vertex 0: to the separating vertex ,6. This is impossible since every
automorphism of C (S) is induced by a homeomorphism of S, by hypothesis. Hence

we have the lemma. C]

We need the following theorem of Ivanov [16]. The proof has not been published
yet, so for completeness we give a proof of it in the next section for tori with at
least three punctures. The proof we give is basically the same as the one we give for

punctured spheres.

Theorem 2.16 Let S be an orientable surface of genus at least one. Suppose that
f E Aut C(S) and a and ,8 two vertices of C(S) with i(a,,6) = 1 imply that
i(f(a),f(ﬁ)) = 1. Then every element of Aut C(S) is induced by some homeo-

morphism of S.

Lemma 2.17 Let n 2 3. If S is a torus with n punctures and S’ is a sphere with

n + 3 punctures, then C(S) and C(S’) are not isomorphic.
We prove this lemma together with the main theorem of this section.

Theorem 2.18 Let S be a torus with n 2 3 punctures. Then the natural map Mg —)

Aut C(S) is an isomorphism.

37

Proofs of Lemma 2.17 and Theorem 2.18: As we have mentioned in the proof of
Lemma 2.15, for a vertex a of C(S), L"(oz) is connected if and only if 01 is either
nonseparating or 2-separating. Hence every element of the group Aut C (S) maps a
nonseparating vertex either to a nonseparating vertex or to a 2-separating one, and

a k-separating vertex to a k’-separating one for k, k’ 2 3.

Let a be a k-separating vertex of C (S) with k 2 3 and let a E a. Let us denote
by S50) and S5,” the components of S“ of genus zero and of genus one, respectively.
The graph L"(a) has exactly two connected components, say, Lg(oz) and L‘i‘(a). The
vertices of these components correspond to the isotopy classes of circles on the con-
nected components of So. We can choose Lg(a) and Li’(oz) so that the vertices of
Lﬁ’(oz) are the isotopy classes of circles on 35:). We then deﬁne two full subcomplexes
Lo(a) and L1(a) of C(S) as follows. The set of vertices of L,-(a) are those of L§’(a).
Then L,(a) is isomorphic to C (5,9)).

If f is an automorphism of 0(5) and a is a k-separating vertex with k 2 3,
then f restricts to an isomorphism from L"(a) = Lg(a) u L‘,’(a) to L"(f(a)) =
Lg(f(a)) u L‘1’(f(a)). Since Lj(a) and Lj(f(a)) are connected components, we get
f(L3(a)) = L3(f(a)) and f(Li‘(a)) = Li-.(f(a)) for some 1‘ = 0 or r = 1. Then
f(Lo(a)) = L..(f(a)) and f(L1(a)) = L1_.-(f(a))- Since dim Lo(a) = dim C(59)) =
k — 3 and dim L1(a) = dim C(59)) = n — k, dim L,.(a) = lc — 3 and dim L,_,(o)

n — k. From this it is easy to conclude that if r = 0 (resp. r=1), then f(a) is k-
separating (resp. (n — k + 3)-separating). The proofs now proceed simultaneously by

induction on n.

Suppose that n = 3. Let f 6 Aut C (S) and let a and ,3 be two vertices of
C(S) with i(a,ﬂ) = 1. By Theorem 2.13, there exist vertices 71,72 and 73 of C(S)

such that (71,01,73,ﬂ,73) is a pentagon, 71 and 73 are 3—separating, and 73 is 2-

38.

separating. Then (f(71),f(a),f(72),f(8),f(73)) is a pentagon in C(S). From the
discussion given in the preceeding paragraph, it follows that f (71) and f (72) are 3-
separating. Note that any two distinct nonisotopic 3-separating circles on S, a torus
with three punctures, must intersect. Therefore none of the vertices f (a), f (8) and
f (73) can be 3-separating. By the same argument, any two distinct nonisotopic 2-
separating circles on S must intersect. We conclude that one of the vertices f (,8 ) and
f (73), say f (8), is nonseparating. By applying Lemma 2.14 twice to the pentagon
(f(71)tf(a),f(72)tf(ﬂ),f(73))t we see ﬁrst that f(a), and then f(73), is nonsep-
arating, i.e., f(a),f(8),f(71),f(72) and f(73) satisfy the conditions (i) and (ii) of
Theorem 2.13. Hence i(f(a),f(,8)) = 1. This is true for any automorphism of C(S).
Therefore, by Theorem 2.16 every automorphism of C (S) is induced by some home—
omorphism of S. Using this and Lemma 2.15, we see that C(S) and C(S’) are not

isomorphic if S’ is a sphere with six punctures.

Now suppose that n 2 4. Let a and 8 be nonseparating and 2-separating vertices
of C (S) and let a E a and b E ,8. Then L(oz) is isomorphic to the complex of curves on
Sa, a sphere with n + 2 punctures, and L(8) is isomorphic to the complex of curves on
Sf”, a torus with n — 1 punctures. By the induction hypothesis, every automorphism
of C (5,) is induced by a self-homeomorphism of 50. Then by Lemma 2.15, C (SQ) is
not isomorphic to C(Sgl’) = C(Sb), i.e., L(a) is not isomorphic to L(,8). It follows
that if f E Aut C (S) then f(a) cannot be 2-separating, i.e., nonseparating vertices

are preserved under the action of Aut C (S)

To show that n-separating vertices are also preserved under the action of Aut C ( S),
we assume the converse. Suppose that there exist f 6 Aut C (S) and an n-separating
vertex 0: of C (S) such that f (a) = 8 is not n-separating. Then 8 is 3-separating by
the discussion given above. The automorphism f restricts to an isomorphism from

the disjoint union Lo(a) U L1(a) to the disjoint union Lo(,8) U L1(8). Since L1(a)

39

and Lo(8) are discrete, and since Lo(oz) and L1(8) are, for instance, connected, we
must have f (L1(a)) = Lo(,8). But this means that f takes the nonseparating vertices
in the link of a to separating vertices, a contradiction. Thus the automorphisms of

C (S) preserve the set of n-separating vertices as well.

Now it follows from Theorem 2.13 that if f 6 Aut C(S) and if i(a,8) = 1, then
i(f(a), f(8)) = 1. Then Theorem 2.16 implies that automorphisms of C(S) are
induced by self-homeomorphisms of S. This completes the proof of Theorem 2.18.
Again, that C (S) and C (S’ ) are not isomorphic if S’ is a sphere with n + 3 punctures

follows from Lemma 2.15. So the proof of Lemma 2.17 is complete, too. D
As a corollary to Lemma 2.17 we can state the following theorem.

Theorem 2.19 Let S be a sphere with at least ﬁve punctures, or a torus with at least
three punctures. Let S’ be a connected orientable surface of genus at most one. In
the case that S is a sphere with ﬁve punctures, suppose, in addition, that S’ is not
a torus with two punctures. If C(S) and C(S’) are isomorphic, then S and S’ are

homeomorphic.

Proof: If S and S’ are either both punctured spheres or both punctured tori, and if
C (S) and C (S ’ ) are isomorphic, then their dimensions, hence the number of punctures
on S and S’, are equal. By the classiﬁcation of surfaces, these two surfaces are

diffeomorphic.

Let S be a sphere with at least ﬁve punctures and let S’ be torus with at least
three punctures. Certainly, if the dimensions of C (S) and C (S’ ) are not equal, then
S and S’ are not homeomorphic. But if their dimensions are the same, then S has
three more punctures than S’. In this case C(S) and C(S’) are not isomorphic by

Lemma 2.17. D

40

2.3.2 Proof of Theorem 2.16 for punctured tori

The purpose of this section is to give a proof of Theorem 2.16 for tori with n 2 3
punctures. So let S be a torus with at least three punctures, and assume that f is

an automorphism of C(S) and 0 and 8 two vertices of C (S) with i(0,8) = 1 imply
that i( f (0), f (8)) = 1. The idea of the proof is that of Section 2.2. Recall that for a

2-separating vertex 0 of C (S), the corresponding vertex of B (S ) is denoted by 0’.

Lemma 2.20 (i) Let f be an automorphism of C(S) and let 01 and 03 be two non-
separating vertices of C (S) such that al and a3 bound an annulus with one puncture
for a,- E 0,. Then f(al) 6 f(01) and f(ag) 6 f(02) bound an annulus with one

puncture.

(ii) Let f be an automorphism of C(S) and let 0 and 8 be nonseparating and
2-separating vertices of C(S). Ifi(0,8’) = 1, then i(f(0),f(8’)) = 1.

(iii) The group Aut C(S) preserves the topological type of the vertices of C(S).

Proof: (i) Note that for a nonseparating vertex 0 of C (S), there exists a vertex 8
such that i(0,8) = 1. By assumption, i(g(0),g(8)) = 1 for g E Aut C(S). Hence

9(a) is nonseparating. Therefore Aut C (S) preserves nonseparating vertices.

Now let 0 and 8 be two nonseparating vertices of C(S) such that {0,8} is a
l-simplex of C (S) We deﬁne the link L(0, 8) of {0, 8} to be the full subcomplex of

C (S) with the vertex set

{7 E 0(5) = 7 # an 9‘ 8,217,631) = i(mﬂ) = 0}-

In fact, L(0,8) = L(0) ﬂ L(8) and L"(0,8) = L"(0) ﬂ L"(8).

Let a E 0 and b E 8 be disjoint representatives. Then the surface Saul, has two

connected components. The vertices of L(0,8) is the isotopy classes of nontrivial

41

circles on these two components. That is, L(0,8) is isomorphic to the complex of

curves C (Snub).

Since the circles on different components do not intersect, if two vertices of L"(0, 8)
form an edge then their representatives can be isotoped to circles on the same con-
nected components of Snub. It follows that L"(0, 8) is connected if and only if the
complex of curves of one of the connected components of Saul, is empty. That is,
L“(0, 8) is connected if and only if one of the components of Snub is a once-punctured

annulus.

Since g(L"(0, 8)) = L"(g(0), g(8)) for any automorphism g of C (S), the proof of
(i) follows.

(ii) We can find nonseparating vertices 00,01,...,0,, of C(S) such that 01 =
0, i(01,8’) = i(0o,0,-) = 1, 1 S i S n, and all of the unmentioned intersection
numbers are zero (see Figure 2.11 (a)). Then by using the assumption of Theorem
2.16, and part (i), we see that the conﬁguration formed by minimally intersecting
representatives f (a,) of f (01-) is homeomorphic to the one formed by a,. (Such a
homeomorphism is constructed in [15] and [22].) Hence we can assume that f (0,) =
0,- for all j. Then up to isotopy there exists a unique 2-separating circle, which must

be b, disjoint from every a,- for j 79 1. Hence the conclusion follows.

 

(a) (b)

Figure 2.11: Some conﬁgurations of circles and arcs.

42

(iii) The fact that Aut C (S) preserves the nonseparating vertices is proved in (i).
Let 0 be a k-separating vertex of C (S) with 2 S k S n. We complete the proof by

induction on k.

Let lc = 2. The dual link of a vertex of C (S) is connected if and only if the vertex
is either nonseparating or 2-separating. Since Aut C (S) preserves the nonseparating

vertices, it must preserve the 2-separating ones as well.

Suppose now that k 2 3 and that the assertion is true for all l—separating ver-
tices with l < k. This implies that f(0) is at least k-separating. Note that there
exist nonseparating vertices 01, 03, . . . , 0,,_k+1 such that {0, 01, 03, . . . , 0,,_k+1} is a
simplex of C (S), and that a,- and aj+1 bound an annulus with one puncture for each
1 S j S n — k, where a, E 0,- (see Figure 2.11 (b)). By part (i), f(aj) and f(aJ-H)
bound an annulus with one puncture. It follows that the surface obtained by cutting S C
along the circles f (a1), f (a3), . . . , f (an_k+1) is a disjoint union of n—k once—punctured
annuli and an annulus A with k punctures. Since the circle f (a) is on the annulus A,

it can be at most k-separating. D

By the correspondence between the set of 2-separating vertices of C (S ) and the
set of isotopy classes of arcs joining different punctures on S, there is a well-deﬁned

action of Aut C (S) on the latter set.

The next step is to prove that Aut C (S) takes simple pair of circles to simple pair

of circles.

Lemma 2.21 [f(0; 8) is a simple pair, then so isi(f(0); f(8)) for any automorphism
f 0f 0(3)-

Proof: Note that any simple pair (0; 8) is determined uniquely by the existence of

vertices 70,71, . . . ,7" such that

43

(a) each 7,- is nonseparating, 0 S i S n,

(b) i(7o,7,-)=1 and i(7,-,7,-) = 0, l S i,j S n,

(C) i(a’a’h) = i(ﬂ'ﬂz) = 1, 311d

((1) all the other unmentioned intersection numbers of 0, and 0, or 8 are zero (see
Figure 2.12 (a)).

Since Aut C (S) preserves the conditions (a)-(d) and 2-separating circles, the sim-

ple pairs are preserved by Aut C (S) U I

92$ v

«Q .
a?»

(a) (b)

Figure 2.12: Existence of various conﬁgurations.

By the correspondence with 2-separating circles and the arcs connecting different

punctures, the group Aut C (S) preserves the simple pairs of arcs, too.

The action of Aut C (S) on the punctures of S is deﬁned in the same way as before.
For f E Aut C (S) and a puncture P on S, choose a simple pair (0’; 8’) with center
P and deﬁne f (P) to be the center of the simple pair ( f (0’ ); f (8’ ))

Lemma 2.22 The deﬁnition of the action of Aut C(S) on the set of punctures of S

is independent of the choice of the simple pair.

Proof: We use the technique and the notation of Lemma 2.5. We induct on i =

1i(a’ﬂ') + i(ﬁ’, 7’)-

44

If S has at least ﬁve punctures, the proof of Lemma 2.5 works here, too. Hence

we may assume that S is a torus with three or four punctures.
Suppose that i = 0. There are two cases to consider.

Case 1: If P1,P3 and P3 are all different, then S has four punctures and any
two circles in the set {a’ , b’, c’ } constitute a simple pair. Let d be the boundary of
a regular neighborhood of the union a’ U b’ U c’ U {P,P1,P3,P3}. Clearly, d is 4-
separating. There exists a nonseparating circle, say, 6 such that i(e, a’) = i(e, b’) = 0
and i(e,c’) = 1. Then f(d) is 4-separating and the arcs f(a’),f(b’) and f(c’) lie on
the four-punctured disc component of S f(d). Let us denote the endpoints of f (a’ ) and
f(b’) by P, P1, P3. Since any two circles in the set {f(a’), f(b’),f(c’)} form a simple
pair, either one of the endpoints of f(c’) is P, or else the endpoints of f (c’ ) are P1
and P2. Since there exists a nonseparating circle, namely f (e), intersecting f (c’ ) and

not intersecting f (a’ ) U f (b’ ), the latter case cannot hold.

Case 2: Suppose that P3 is one of P1 and P3. Without loss of generality we may

assume that P3 = P1. Now the proof splits into two subcases.

Subcase 1: First suppose that the boundaries of a regular neighborhood of a’ U
c’ U {P, P1} are nonseparating. If S has three (resp. four) punctures, then there exist
two (resp. three) nonseparating circles a1,a3 (resp. a1,a2,a3) disjoint from a’ and c’
such that Sawa, (resp. Smuuma) is a disjoint union of an annulus with two punctures
P, P1, and one (resp. two) once-punctured annulus (resp. annuli) (see Figure 2.12
(b)). Then only one of the components of Sf(a,)uf(a,) (resp. Sj(al)uf(a2)uf(aa)) is a
twice-punctured annulus, on which the arcs f (a’), f (c’ ) lie. Hence the endpoints of

f(c’) are those of f(a’).

Subcase 2: Now suppose that the boundaries of a regular neighborhood of a’ U c’ U

{P, P1} are separating. It is clear that there exists an arc d’ joining P to P1 disjoint

45

from a’ Ub’ U c’ such that the components of the boundaries of a regular neighborhood
of a’ U (1’ U {P, P1} are nonseparating. Now we apply Subcase 1 to {a’, d’, b’ } to see
that one of the endpoints of f (d’ ) is the center of the simple pair ( f (a’ ); f (b’ )) Note
that ( f (d’ ); f (b’ )) is a simple pair with center that of ( f (a’ ); f (b’ )) We apply Subcase

1 to {d’, c’, b’ } again to conclude that one of the endpoints of f (c’ ) is the center of the
simple pair (f (d’); f (5’))-

The proof now proceeds as the proof of Lemma 2.5. D

We now deﬁne an action of Aut C (S) on the vertices of the complex B(S), as
in the punctured sphere case in Section 2. This action of Aut C (S) on the isotopy
classes of arcs joining different punctures is deﬁned by the correspondence with the

isotopy classes of 2-separating circles.

Let us deﬁne the action on the classes of arcs joining a puncture to itself. For
this purpose let f 6 Aut C(S) and let 0’ be a vertex of B(S) such that a’ E 0’
is connecting a puncture P to itself. Consider the circles a1 and a2 which are the
components of the boundary of a regular neighborhood of a’ U {P}. Then either both
of a,- are nonseparating, or al is k-separating and a2 is (k + 1)-separating for some
k 2 1. (By convention, a l-separating circle is the puncture to which the trivial simple
closed curve is deformable.) If a1 and a2 are both nonseparating, then they bound an
annulus with one puncture. By Lemma 2.20, f (a1) and f (a2) are nonseparating and
bound an annulus with one puncture. Clearly, this puncture must be f (P) We deﬁne
f (0’ ) to be the isotopy class of the unique (up to isotopy) nontrivial embedded arc on
this annulus joining f (P) to itself. If a1 is k-separating and a2 is (k + 1)-separating,
then f(al) is k-separating and f(ag) is (k + 1)-separating. Using the technique of
Lemma 2.20 (iii), it is not hard to see that f (a1) and f (a3) bound an annulus with one

puncture, f (P) Then f (0’ ) is deﬁned as the isotopy class of any nontrivial embedded

46

arc on this annulus joining f (P) to itself.

Lemma 2.23 Every automorphism of the complex C(S) induces an automorphism

of B(S).

Proof: Let f E Aut C(S) and let 0’ and 8’ be two distinct vertices of B(S) with
i(0’,8’) = 0. It sufﬁces to show that i(f(0’),f(8’)) = 0. As in the proof of Lemma
2.7, there are seven cases to consider (see Figure 2.5). If a’ (resp. b’) joins a puncture

to itself, let a1,a2 (resp. b1, b3) be the circles used to deﬁne f(0’) (resp. f(8’)).
(i),(ii),(iii) The proof is the same as that of Lemma 2.7 (i),(ii),(iii).
(iv) Lemma 2.21.

(v) Let P and Q be the endpoints of a’ and b’. There are two possible cases. The
boundary components a, b of a regular neighborhood of a’ Ub’ U {P, Q} are either both
nonseparating or both separating (see Figure 2.13). In either case, a and b bound an
annulus with two punctures on which a’ and b’ lie. Moreover, there exists a circle c
such that i(7,0’) at 0 and i(7,8’) = 0, where 7 is the isotopy class of c. Let 0 and
8 be the isotopy classes a and b respectively. By looking at the link and the dual
link of {0,8} as in the proof of Lemma 2.20 (i), it is easy to see that f(a) and f(b)
bound an annulus with two punctures, where f (a) 6 f (0) and f (b) E f (8) Clearly,
the arcs f(a’) and f(b’) lie on this annulus. Since i(f(7),f(0’)) 75 O, a segment of
f (c) connects f (a) to f (b) Then any arc disjoint from this segment can be isotoped

to an arc disjoint from f (a’ )

(vi) The argument in the proof of Lemma 2.7 works when b1 and b2 are nonsepa-

rating circles as well.

(vii) Suppose that a’ and b’ join the puncture P to itself. If a1 and a3, or b1

and b2 are separating, then the arguments given in the proof of part (vii) of Lemma

 

 

47

 

(a) (b)

Figure 2.13: Two possible cases for (v).

2.7 ﬁnish the proof. So suppose that a1,a3,b1 and b3 are all nonseparating. Then
i(8’, 01) = i(8’,03) = 1, or i(8’, 01) = 0 and i(8’,03) = 2 (by changing the roles of
a1 and a; if necessary). By composing f with a mapping class we can assume that

f(01) —— 01,f(03) = 03 and hence f(0’)— — 0’.

Figure 2.14: The arcs a’ and b’.

If i(8’, 01) = i(8’,03) = 1 (see Figure 2.14 (a)), then by Lemma 2.20 (ii) we have
i(f(ﬂ'LGI) = z'(f(ﬂ’)tf(or1)) = 1 and i(f(ﬁ’LCIz) = z'(f(ﬂ’),f(012)) = 1- Choose
f (b’ ) E f (8’ ) so that it intersects a1 and a3 only once. The intersection of f (b’ ) with
the once-punctured annulus determined by a; and a2 has two components, one joining
P to al and the other joining P to a2. Now it is obvious that these components can

be isotoped so that they do not intersect a’. Therefore, i( f (0’ ), f (8’ )) = 0.

Suppose now that i(8',01) = 0 and i(8’,02) = 2 (see Figure 2.14 (b)). Clearly,
there exists a nontrivial embedded are c’ joining P with itself such that it is disjoint

from a’ and b’, and i(7’,01) = i(7’,03) = 1 and i(8’,71) = i(ﬂ’ﬂz) = 1, where 7’

48

is the isotopy class of d. Then i(f(i’),f(a’)) = i(f(v’La’) = 0. i(f(7’),f(ﬁ’)) = 0
and i( f (7’ ), 01) = i( f (7’ ), 02) = 1. Choose a representative f (c’ ) of f (7’ ) intersecting
a’, a1 and a2 minimally. A representative f (b’ ) of f (8’) can be chosen so that it is
disjoint from a1 (as i(f(8’),01) = 0) and f(c’). But then f(b’) can be isotoped to a

curve disjoint from a’. This completes the proof of the lemma. Cl
Corollary 2.24 Aut C (S) is naturally isomorphic to a subgroup of Aut B(S).

The proof of Theorem 2.16 for tori with n 2 3 punctures proceeds now as in
Section 2.2.3. In fact, Lemmas 2.9, 2.10 and 2.11 hold for punctured tori, too. Then

the proof of Theorem 2.12 ﬁnishes the proof of Theorem 2.16.

2.4 Subgroups of Mg

In this section, S denotes an oriented surface. The purpose of this section is to prove

Theorem 2.28 as an application of our main results in Sections 2.2 and 2.3.

If 0 is a vertex of C (S), we denote by to, the right Dehn twist about 0. It is well-
known that for f 6 Mg, f to, f’1 = t f(a) if f is orientation preserving and f to, f '1 =
tile) if f is orientation reversing. An immediate consequence of the deﬁnition of Dehn

twists is

Theorem 2.25 Let 0 and 8 be two vertices of C(S) and let N, M be two nonzero

integers. Then t5 = tg’ if and only if0 = 8 and'N = M.

The following relations between Dehn twists are well known. We will use this
theorem to prove our results in this section and in the next chapter. A proof of the

theorem may be found in [15] or [22].

49

Theorem 2.26 Let 0 and 8 be two vertices of C(S) and let N,M be two nonzero

integers. Then
(i) i(0,8) = 0 if and only if t? ti,” = tg’ t5.

(ii) (braid relations) i(0,8) = 1 if and only if ta, ta to = t3 ta, ta.

Let G be a group and let f E C. We denote by Cc;( f) the centralizer of f in G,
i.e.,
CCU) = {96 G=9f=f9l

We denote the center of G by C (G)

Let m 2 3 be an integer. Let F be a subgroup of ﬁnite index of the kernel of the
natural homomorphism Mg —> Aut H1(S,Zm). Then clearly I‘ is of ﬁnite index in
Mg.

Theorem 2.27 Let S be a sphere with at least ﬁve punctures or a torus with at least

three punctures. An element f E I‘ is a power ofa Dehn twist if and only if
(i) C(Cp(f)) is isomorphic to Z, and

(ii) C(Cr(f)) is not isomorphic to Cr(f).

This theorem is proved in the same manner as Theorem (2.3) of [15]. We now

ready to prove

Theorem 2.28 Let S be a sphere with at least ﬁve punctures or a torus with at least
three punctures. Let GI and 02 be two subgroups of Mg of ﬁnite index. Then any
isomorphism G1 —) G2 is induced by some inner automorphism of Mg. In particular,
two subgroups of Mg of ﬁnite index are isomorphic if and only if they are conjugate.

Also, if G is a subgroup of Mg of ﬁnite index, then the outer automorphism group
Out G ofG is ﬁnite.

50

Proof: Let (I) : C1 ——> G'3 be an isomorphism. Consider the subgroup (1)-1(Cgﬂl‘)ﬂf‘ of
Mg of ﬁnite index. Let [‘1 be a ﬁnite index subgroup of <I>"1(Cg 0 F) D F and let F3 =
<I>(F1). Then, F1 and F3 are of ﬁnite index in Mg and (I) restricts to an isomorphism
from 1‘1 to 1‘2. Clearly. ‘1’(C(Cr,(f))) = C(Ca(<l>(f))) and <1>(Cr.(f)) = chum».
It follows from Theorem 2.27 that (I) takes sufﬁciently high powers of Dehn twists to
powers of Dehn twists. More precisely, for a vertex 0 of C (S), since the index of F1 is
ﬁnite, there exists a nonzero integer N such that t3,” 6 F1. If t3,” 6 F; then (Mtg) = tg’

for some vertex 8. If t”1 t”’2 6 F1 and <I>(t(’,”‘) = tgf‘, then

0’0

tg’lez : Q(t2rl)N2 = @0552)!“ =tg2N1°

It follows from Theorem 2.25 that 81 = 83. That is, we have a well-deﬁned map 90 from
the vertex set of C (S) to itself, deﬁned by the equation @(tév) = ting), independent
of the choice of the powers involved.

Next, we show that 90 is an automorphism C (S) —> C (S) Obviously, we also have
a map 90'1 from the vertex set of C (S) to itself induced by Q”. Now for any 0 of
C(S)

t” = <I>“(<I>(tf,v)) = (1)-10M ) = iii-Mal)

0 90(0)
and

t” = <Ii(<1>"(tf)) = Mtg-1(a)) = tine-1(a))

0

for some appropriate integers N, M, T, K, L. Then again by Theorem 2.25, we have

90"(90(0)) = 0 and 90(90”(0)) = 0, so 90 is a bijection.

Let 0, 8 be two vertices of C(S) with i(0, 8) = 0, then t2" commutes with tgv’, and

M
hence t d;

) commutes with tib’l’ﬁ)’ It follows from Theorem 2.26 that i(cp(0), <p(8)) = O,

i.e., 90 is an automorphism of the complex of curves C (S)

Now, the automorphism 9p : C (S) -> C (S) is induced by a mapping class f of S,

(i.e., 90(0) = f(a)), by Theorem 2.12 if S is a punctured sphere, or by Theorem 2.18

_.__-ve_..h~ ____‘_, _ _ 7

51

if S is a punctured torus. Then
is?) = iii.) 741,70, = its“ r1.

Let g E G1. If 0 is a vertex of C (S), then for appropriate integers N, M and K,
9(th g“) = <I>(g)<1>(t§’) My") = <1>(g)t.,’5’(a)<I>(g)‘1 = motif.) (F(a)-1 = tilg‘man
and

that? g“) = 44651)) = tiles» = lilac»-

Then @(g)(f(a)) = f (9(0)), or, equivalently, (@(g)f)(a) = (fg)(a), and thus (fg)"<1>(g)f
is in the kernel of the map Mg —> Aut C(S), which is trivial by Theorem 1.3. So

<I>(g)f = f9, i-es, M9) = fgf".
The second conclusion follows easily from the ﬁrst.
Let G be a subgroup of Mg of ﬁnite index. Let us denote by NM§(C) the
normalizer of C in Mg. That is,
NM°5(G) = {f e M; : fo-l c G}.

For each f E N MoS(G), the inner automorphism I f of Mg maps G to itself. This gives
a homomorphism w : N M's(G) —> Aut G deﬁned by 9b( f) = I f. This homomorphism
is surjective by the ﬁrst part of the theorem. Hence we have a surjective homomor-
phism 1,7; : N M‘s(G) —> Out G. Clearly, G is in the kernel of J. Consequently, the
order of Out G is

[Out G : 1] = [NM;(G') :kerE] S [NM33(G) : C] S [Mg : G]

which is ﬁnite by assumption. U

Remark Except, possibly, for closed surfaces of genus two, the conclusion of Theo—
rem 2.28 is, in fact, true for all surfaces of genus 2 2. A sketch of this is given in [16],

and the proof of Theorem 2.28 above is almost identical to that in [16].

Chapter 3

First Homology Groups of
Mapping Class Groups of
N onorientable Surfaces

3.1 Introduction

In this chapter, we calculate the ﬁrst homology groups of the mapping class groups of
closed nonorientable surfaces. It turns out that they are isomorphic to Z2 if the genus
of the surface is at least seven. Another result is that the subgroup of the mapping
class group generated by Dehn twists about 2-sided circles has trivial ﬁrst homology
group, again, if the genus of the surface is at least seven. As an algebraic application,

we deduce Corollary 3.15.

The ﬁrst homology groups of the mapping class groups of closed orientable surfaces
are well-known. Let F be a closed orientable surface of genus g. If F is a sphere
(g = 0) then Mp is trivial, and hence H1(Mp) = O. Ifg Z 3 then H1(Mp) is again
trivial. This result is due to Powell [26]. The group H1(Mp) is Z10 if g = 2, proved

by Mumford [24], and Z12 if g = 1.

If g = 0 or g 2 3, then it follows from H1(Mp) = 0 and the fact that Mp is

of index two in M} that the ﬁrst homology group H1(Mg) is Z3. Also, H1(M;.~)

52

53

is Z2 EB Z3 if g = 1,2. Since this is not so well-known as the other results stated
above, we give a proof at the end of the present chapter. The case g = 1 is, probably,

well-known to algebraists. The case -g = 2 is proved at the end of (5.5) in [15].

In Section 3.2, we give the preliminary information on nonorientable surfaces. We
state various results regarding the mapping class groups of nonorientable surfaces,

and prove some lemmas which will be used in the proofs of the main results.

In Section 3.3, we discuss vector spaces over the ﬁnite ﬁeld F2 of order 2. We are
mainly interested in the group of isometries of such a vector space with a symmetric
bilinear form. This subject is related to the mapping class group of nonorientable

surfaces via a theorem of McCarthy and Pinkall.

We prove our main results of this chapter in the last section, Section 3.4. In
the proofs, we use known relations between Dehn twists to get an upper bound
on the order of the ﬁrst homology groups. A lower bound is obtained by ﬁnding

homomorphisms onto abelian groups.

3.2 Mapping Class Groups of Nonorientable Sur-
faces

Let S be a closed connected nonorientable surface of genus g. For nonorientable
surfaces, the genus is defined to be the number of real projective planes in a connected
sum decomposition. The mapping class group Mg is deﬁned as the group of isotopy
classes of self-homeomorphisms of S. Let M s be the subgroup of Mg generated by
twists about 2-sided circles. It turns out that this subgroup is the analogue of M p

in the orientable case.

It is known that the mapping class group of a projective plane is trivial and that

of a Klein bottle is Z3 EB Z2 (cf. [19]). In addition, the subgroup of the mapping

54

class group of a Klein bottle generated by the Dehn twists about 2-sided circles is Z3.
Therefore, in the rest of the chapter we restrict ourself to surfaces of genus at least

three.

Recall that for a simple closed curve (a circle) a on a surface S we denote by 5,, the
surface obtained from S by cutting along a. Two circles a and b on S are topologically

equivalent if there exists a homeomorphism H : S -> S such that H (a) = b.

Lickorish [19] proved that the group Mg is generated by the isotopy classes of
Dehn twists about two-sided nontrivial circles and that of so—called Y-homeomorphisms
(crosscap slides). He also showed that the isotopy classes of Dehn twists about two-
sided nontrivial circles do not generate Mg. In fact, the subgroup M 8 generated
by the isotopy classes of Dehn twists about 2-sided circles is of index 2 in Mg (cf.
[21]). Later, Chillingworth [8] found a ﬁnite set of generators. Using the results of

Humphries [13], one can eliminate some of Chillingworth’s generators of Mg.

Let us describe the Y-homeomorphisms of nonorientable surfaces. Consider a
Mbbius band M with one hole. Let us attach a M6bius band M’ to M along the
boundary of the hole. The resulting surface K is a Klein bottle with one hole. By
moving M’ once along the core of M we get a homeomorphism of K which is the
identity on the boundary of K (see Figure 3.1). If K is embedded in a surface S,
we can extend this homeomorphism by the identity to a homeomorphism of S. This
homeomorphism is called a Y-homeomorphism or a crosscap slide. It is clear from the
deﬁnition that the square of a Y-homeomorphism is a Dehn twist about the boundary
of the Klein bottle K with hole. We note that Y-homeomorphisms act as the identity

on the Zg-homology.

Let S be a closed connected nonorientable surface of genus g. In the rest of this

chapter, we will mainly use the models given in Figure 3.2. If g is even, we will also

 

“ Nﬂvcnn '. . -

55

 

Figure 3.1: A Y-homeomorphism.

use the model for S given in Figure 3.3. In the ﬁgures, the interiors of the shaded discs

are to be removed and the antipodal points on each boundary component identiﬁed.

For a 2-sided circle a on the nonorientable surface S, let us denote by ta one of the
two Dehn twists about a. Consider the models in Figure 3.2. Let d be the boundary
of the shaded part if g is odd and d = b,“ if g is even. Clearly, 5.1 is orientable. After

orienting Sd arbitrarily, we can assume that the twists to are actually the right twists

on S1 for a = a,,b,-,c,-.

Theorem 3.1 (Chillingworth, [8]) Let S be a closed connected nonorientable sur-

face of genus g. Then Mg is generated by
{tantbntcﬂz : 1 S i S r,l Sj S r— l} ifg: 2r+1,
{to.«,tb,,to.~,tb,+l,2 : 1 S i S r} ifg = 2r +2,

where z is the isotopy class of a Y-homeomorphism such that 22 is a twist about 6 if

g is odd and about f ifg is even.

Theorem 3.2 (Chillingworth, [8]) Let S be a closed connected nonorientable sur-
face of genus g. Then M5 is generated by

{ta,.,tg,,.,t¢j,tc,tcl :1 SiSr,1 Sj S r—l} ifg: 2r+1,

and

{tagatbi3tCiatbr+1atfytfl$tf2 : l S i _<_ T} if g = 2T+2

56

where el,f1, f2 are the circles in Figure 3.4.

The generators of M 5 can also be described by using the oriented double cover

of S (cf. [4]).

Theorem 3.3 (McCarthy-Pinkall, [23]) Let S be a closed nonorientable surface.
If L is an automorphism of H1(S, Z2) which preserves the Zz-valued intersection pair-

ing, then L is induced by a diﬂeomorphism which is a product of Dehn twists.

In other words the natural map
M3 —i ISO H1(S,Zz)

is onto, where Iso H1(S, Z3) is the group of isomorphisms of H1(S, 22) preserving the

Zg-valued intersection pairing.

Let us recall the lantern relation discovered by Dehn [9] and rediscovered by
Johnson [18]. Let So be a sphere with four holes with boundary components do, d1, d2
and d3. For 1 S i < j S 3, let dij denote a circle encircling d,- and dj as in Figure
3.5 (a). If we consider the diffeomorphisms of So ﬁxing BSD, and if we embed So
in a surface R, then we can consider the right twists t3, and t3”. as twists in R by

extending them by the identity to the complement of So. Then we have

Lemma 3.4 In this situation tdotd1td2td3 = td12td13td23 .

Lemma 3.5 Consider the torus with two holes in Figure 3.5(b) as embedded in a

surface. Then

tqltq, = (talctb)2 (tbtatc)2 = (tatctb)".

57

 

 

 

 

 

 

g=2r+2

 

 

 

 

g=2r+2

 

 

 

 

   

 

 

 

 

 

 

   

g=2r+2

Figure 3.4: The circles e1, f1 and f2.

 

58

 

(a) (b)

Figure 3.5: Sphere with four holes So and torus with two holes.

Proof: By Lemma3 of [20], tmtq, = (tatct(,)2 (tbtatc)2. The equality (tbtatc)2 = (tatctbl2

follows from Theorem 2.26:
(tbtatc)2 = tbtatctbtatc = tbtatctbtcta = tbtatbtctbta = tatbtatctbta

= tatbtctatbta = tatbtctbtatb = tatctbtctatb = tatctbtatctb = (tatctb)2. Cl

Lemma 3.6 Let S be a closed connected nonorientable surface. Then there exists a

mapping class 9 such that gtahg"l = tgl’.

Proof: Consider the models given in Figure 3.3. Let d be the boundary of the shaded
part if g is odd and d = br+1 if g is even. Then S is a quotient of the surface 53.
Let us embed S3 in R3 in such a way that it is invariant under the reﬂection )5 across
the XY-plane and [f(d) = d. Then ,5 induces a homeomorphism p of S. Let p be the
isotopy class of p. Since p"(a1) = a1 and since 8 is orientation reversing, it follows that

ptmp‘1 =t;11. Cl

Lemma 3.7 Let S be a closed connected nonorientable surface of even genus g 2 4.
Suppose that a is a separating circle on S such that one of the components of 5., is
a Klein bottle with one hole, and the other component is also nonorientable. Then

there exists a mapping class 9 such that ptag‘l = t;’.

59

Proof: Let g = 2r. Consider the sphere 82 as the one point compactiﬁcation of R2.
Remove from S2 the interiors of 9 discs of radius % with centers (:l:1, 0), (i2, 0),... ,
(:tr, 0). Let S(g) be the resulting surface. Let it be the circle of radius % with center
(0,0) and p be the reﬂection across Y-axis. Then p(&) = ii. Let us realize S as
the quotient of S(g) obtained by identifying the antipodal points on each boundary
component. We may assume that a is the image of ii. The homeomorphism p descends
to a homeomorphism of S. Denote the isotopy class of this homeomorphism of S by

9. Since p is orientation reversing, we have ptap‘l = ta‘l. D

“Y

 

Figure 3.6: Surface S(g).

Lemma 3.8 Let C be a group and H a normal subgroup of G'. Then [H, H] is a

normal subgroup of G

Proof: The proof of the lemma follows immediately from the following fact:

g"[h1,h3]g = lg—lhlg, 94/129]. 0

3.3 Groups of isometries of vector spaces over the
ﬁeld of order 2

Let V be a vector space space of dimension m over F3, the ﬁnite ﬁeld of order 2,
let {v1, v3, . . . , vm} be a basis of V, and let (,) be the symmetric bilinear form on V

deﬁned by (v,, v,) = 6.3. Let us denote by 130 V the group of isometries of V.

60

If m is odd, say m = 2n + 1, then the restriction of (,) to the 2n-dimensional

subspace

W={vEV:(v,v)=0}

is nondegenerate, and hence it is a symplectic form. Recall that a symmetric bilinear
form (,) on a vector space W over a ﬁeld of characteristic 2 is called symplectic if
it is nondegenerate and (w, w) = 0 for all w E W. Any isometry of V induces an
isometry of the symplectic space (W, ( , )), and any isometry h of W can be extended

uniquely to an isometry of V by requiring that

2n+l 2n+l
h( 2 vi) = 2 v,.
i=1 i=1
From this observation it follows easily that there is an isomorphism between Iso V and

Iso W. The group 180 W is called the symplectic group and is denoted by Sp(2n, 2).

It is known that Sp(2n,2) is perfect for n 2 3 (see, for instance, [27]).

A proof of the next lemma may be found in [25], Chapter 3, or in [27], Chapter 8.
Lemma 3.9 The order of Sp(2n,2) is 2"2 ?=1(22’ — 1).

Lemma 3.10 The group Sp(4,2) is isomorphic to E6, and Sp(2,2) is isomorphic to

23, where E, is the symmetric group on r letters.

Proof: A proof of the ﬁrst assertion may be found in [25], Chapter 3, or in [27],
Chapter 8, so let us prove the second. Let W be a vector space over F; of dimension
2 with a basis {101,102} and let (,) be deﬁned by.(w,-,w,-) = 1+ 5.5. Then (W, ( , )) is
a symplectic space and hence 130 W is isomorphic to Sp(2, 2). Since any permutation
of {w1,w3, wl + 102} induces an isometry, E3 is isomorphic to a subgroup of I30 W.

Also, since the orders of 23 and 130 W are both 6, they are isomorphic. Cl

 

61

Lemma 3.11 Suppose that m = 2n. Let us consider V as a subspace of a vector
space V’ over F3 of dimension 2n + 1 with a basis {v1,. . . ,vgn,v3,,+1}. Extend the
form (,) to V’ by (v;,v3,,+1) = 6.93"“. Then Iso V is isomorphic to the stabilizer of
v3n+1 in Iso V’ and the order [Iso V : 1] of I80 V is

n-l
[Iso V: 1] = 2”2 H(22’ — 1).
1

i:

Proof: We deﬁne a map
9b: Iso V —> Stabl,o v:(v2,,+1)

by
h(v) if v E V

¢(h)(v) = { 1,2,,“ if v = vzn+1

for h E 180 V, where Stabl,o Vl(v3,,+1) is the stabilizer of v2n+1 under the canonical

action of Iso V’. If v E V then

(i/J(h)(v),w(h)(v2n+1)> = (h(v),v2,,+1) = 0 = (vsvzn+l)-

As a consequence of this, l/JUI), the extension of h to V’, is an isometry ﬁxing Ugn+1.

Also, for any h E Stabl,o Vt(v3,,+1)
W) = h({vaniil = {seam = v.

Therefore the map if) is onto. The injectiveness of 9b is clear. Hence, the ﬁrst assertion

follows.

Now we calculate card Orb(v3,,+1), the cardinality of the orbit of v3n+1 under

the canonical action of Iso V’. We claim that Orb(v3n+1) is the set of sums of odd

2n+l

numbers of basis elements, except for 2-

:21 v,-. It is obvious that for any k < n

there exists an h E 130 V’ such that h(:§:'[lv{1) = 23:,” v,- if i,- 5£ ijl for j at j’.

 

62

Therefore we only need to show that for any k < n there exists an h E Iso V’ such
that h(v2n+1)— — 23"“ v,-. This 13 true if and only if 22"? v,- can be completed to an

orthonormal basis of V’. For this purpose let I: < n, and let

2k+l

vza+1=Z v,-

and
v,_ v.-+vok+o+v;,,+1 ifl Sig2k+1
“ v,+1 if2k+2§iS2n.
It is easy to check that {v[,v§, . . . ,vgn +1} is an orthonormal basis of V’. Since

(v,v) = 1 if and only if v is an odd number of sums of basis elements and since

h(Z?"1"lv.-)= 2,2"? v,- for all h E Iso V’, the claim follows.
Now it follows that card Orb(van) = 22" —- 1.

Recall that for any ﬁnite group C acting on a ﬁnite set X,
[G2 1] = [Staba(x) : 1] card Orb(x)
for x E X, where [G' : 1] is the order of G. Hence we have
[180 V’: 1]: [I30 V. 11(22" — 1).

Now, since Iso V’ is isomorphic to Sp(2n, 2), the conclusion follows from Lemma 3.9.

E]

The next lemma is stated in [27] as Exercise 1 on page 174.

Lemma 3.12 Suppose that m = 2n. Then there is a surjective homomorphism from

Iso V to Sp(2n — 2,2).

Proof: Let W = {v E V: (v,v) = 0}. If w,- = v1+ 1).-+1 then {w1,w3, . . . ,w2n_1} is a
basis of W. Clearly, W’" = {0,v1 + v; + . . . + 02"} and W’- C W. Then W/WJ' with

63

the induced form is a symplectic space of dimension 2n — 2. Clearly, Iso W/Wi is
isomorphic to Sp(2n — 2, 2).

Let h E Iso V. Then h(W) = W. That is, h induces an isometry of W, and
h(W'L) = W". This implies that h induces an isometry of W/Wi. It follows that
there is a homomorphism from 150 V to I80 W/Wl. We show that this homomor-

phism is surjective.

Let h be an isometry of W/Wi. Then
h(w, + W") = w; + W”

for some w: E W. If w’1 + w; + + w’2n_1 = 0 then by replacing w2n—1 with
w§n_1+ v1 + v2 + ...+ v3“ we may assume that w’1 + w; + . . . + w§n_1 # 0. The set

{w’1,w§, . . . ,w§n_1} is a basis of W and

<w2,w3-> = <w£+W*,w;-+W*>
= (F(w,+Wi),E(w,-+W*))
= (w,+WL,w,-+W‘L)

= (w,,wj).

Then h(w,) = w: deﬁnes an isometry of W which induces h on W/Wi.

It remains to prove that isometries of W can be extended to V. So let h be an
isometry of W. Consider the subspace U of W generated by {w1,w2,...,w2,,-3}.
Then

U'L ={0,v2mv1+v2+...+v2,,_1,,v1+v2+...+v3n}.

Hence the restriction of (,) to U is nondegenerate. Since h is an isometry between
U and h(U), the restriction of (,) to h(U) is also nondegenerate. This implies that

V = h(U) EB h(U)". In particular, h(U)J‘ is not a subspace of W. Let us choose

64

v E h(U)i\W. Then (v,v) = 1 and (v,h(v1+ v,)) = 0 for 1 S i S 2n — 1. If
(v,h(v,~ + v3")) = 0 for some 1 S i S 2n — 1, we would conclude, since W is also

generated by
{h(v1+ 02), h(v1+ Us), - - - 7 h(v1'l' v2n—l)a h(vi + vzn)},

that v E W* C W, a contradiction. Therefore, (v,h(v,- + Uzn» = 1 for all 1 S i S

2n — 1. Hence (v,h(v,- + vzn)) = 1 + 62",;. Now deﬁne k : V —> V by the formula

k(v,) = v + h(v; + v2"). Then

(Hwhﬁmh==<v+Mw+eahe+Mw+eah
= (v.12) + (h(vs + van), v) + (v, h(va + Uzn» + (h(vt + van), h(va- + 222..»
= 1+1 + 62...,- +1 + 6m.- + (vs + vomv, + van)
= claw-+1 + 62a,.-+ (vavj) + (vi1v2n>+ (vzn,vj) + (v2nav2n>
= 6a,,- + 1 + 62....- + 6:,- + 62....- + 62..., + 1

= 5,,

Clearly, k deﬁnes an isometry of V which is an extension of h. D

3.4 The ﬁrst homology groups

We are now ready to prove the main results of this chapter. Recall that for a group G,

H1(G) = G/[G, G]. We will use multiplicative notation for the operation in H1(G).

Theorem 3.13 Let S be a closed connected nonorientable surface of genus g. Then

H1(Mg) is equal to
(0089=L

(ii) 22 ea z2 ifg = 2,3,5,6,

65

(iii) z, e Z: e Z: ifg = 4,

Proof: If g 2 2, then the subgroup M5 is of index 2 in Mg (cf. [21]). Hence, it is
a normal subgroup. Then the quotient Mg / M s is of order 2, and hence is abelian.
Clearly, this implies that the index of the commutator subgroup [M g, Mg] is at least
2 in Mg, i.e., the order of H1(Mg) is at least 2. It is well—known that if a group G is
generated by X, then G/[G, C] is generated by X = {x[G', C] : x E X}. For h E Mg
let us denote by h the class of h in Mg/[Mg, Mg].

If g = 1, then Mg is trivial. Hence so is H1(Mg).
Ifg = 2, then ME = Z2 EB Zg (Cf. [19]). SO H1(Mg~) = Z2 EB Z2.

If g = 3, Birman and Chillingworth [4] proved that Mg admits the presentation:
(a,b,y : aba = bab,yay‘l = a‘1,yby—l = b'l,y2 = 1, (aba)4 = 1).
It follows that Mg/[Mg, Mg] has the presentation:

M‘s/[Mg,Ms] = (as : 62 = t2 1.6? = W)-

Therefore H1(Mg) = Z3 69 Zz. This can also be deduced from the arguments below.
We now assume that g 2 4. The rest of the proof splits into two cases.

Case 1: g is odd. Let g = 2r + 1. From the braid relations it follows that

ta. f5, = to, in H1(Mg). Hence, it follows from Theorem 3.1 that H1(Mg) is

I

generated by {fall ,3}.
Lemma 3.6 implies that 2:, = 1. We now show that 22 = 1.

Notice that e is a separating circle and one of the components of S3 is a Klein

bottle with one hole. There exist circles q1,q2, . . . , q, = e on S such that ql bounds a

66

M6bius band, and that q,- and q,“ bound a torus with two holes for i = 1, 2, . . . , r — 1

(see Figure 3.7). Then by Lemma 3.5
tQit91+l = (taIthtbf)4

for some nonseparating 2-sided circles af, b1- and c:- each of whose complement is nonori—

entable. Hence, a£,b:-,c’- are topologically equivalent to al. Therefore, each of the

1

Dehn twists tahtb: and to; is conjugate to either ta, or tgll. Since the conjugate ele-

-1
we have

ments induces the same element in the abelianization and since L, = fa, ,

to: = {3: = to: 2 fan. This implies that

Since the Dehn twist about a 2-sided circle which bounds a Mbbius band is isotopic

to the identity, tql = 1. Now it follows that in = 1 for all i. Therefore 32 = tq, 1.

 

qu=e

   

§bo

 

Figure 3.7: Circles ql, Q2, . . . , qr.

From these discussions it follows that the generators of H 1(M g) satisfy the rela-
tions f2, = 1,32 = 1 and the obvious commutation relation. Hence H1(Mg) is either

Z3 or Z3 69 Z2, as its order is at least 2.

Let g = 5. In view of Section 3.3, Iso H1(S, Z3) is isomorphic to Sp(4,2), where

H1(S, Z2) is considered with the Zg-valued intersection pairing. Let
90: Mg —> 150 H1(S,Z3) E Sp(4,2)

be the canonical map. By Theorem 3.3 the restriction of cp to M5 is onto. By

Lemma 3.10, Sp(4,2) is isomorphic to 26, the symmetric group on six letters. Since

67

the commutator subgroup of 26 is of index 2 in 23, the commutator subgroup of
Sp(4, 2) is of index 2. In particular, there is a homomorphism from Sp(4, 2) onto Z2.
Composition of this homomorphism with cp gives a homomorphism a from Mg onto
Z3. Since the restriction of ¢ to M5 is also onto, ker? 0 M5 is normal of index 4
in Mg. Recall that for any prime number p, any group of order p2 is abelian. Hence
the group Mg/ker9-0' n M s is abelian. It follows that the index of the commutator
subgroup of Mg is at least 4. Therefore H1(Mg) = Z3 EB Z2.

If g 2 7 then the sphere with four holes of Figure 3.5 (a) can be embedded in
S so that do = b1,d1 = c1,d2 = c2,d3 = b3. Then each d;,- is 2-sided nonseparating
with a nonorientable complement (see Figure 3.8). Since a1 intersects d1; and d13
only once and since a2 intersects d33 once, it follows from the braid relations that

id, = id” = in. From the lantern relation we have

{do {(11 {d2 Eds = £112 Zdlszdzs

and hence t2, = t3]. Thus ta, = 1. Therefore H1(Mg) = 22.

 

 

 

 

 

 

Figure 3.8: Embedding of So.

Case 2: g is even. Let g = 2r + 2. It follows from the braid relations that
ta. = tb. = fa, for 1 S i,j,lc S r. Hence, H1(Mg) is generated by {tanfbrﬂfi} by

Theorem 3.1.
By Lemma 3.6, ptmp‘1 = t: for some mapping class 9. It follows that t; = 1.

We now claim that if,“ = 1. Let us consider the model for S given in Figure 3.3.

Let p : 33 -—> 3,; be the reﬂection across the XY-plane, where d is a circle isotopic

 

68

to br+1. The homeomorphism p induces a homeomorphism p. Since ,5 is orientation

reversing, since ﬁ(b,.+1) = bf.+1 on S; and since b,.+1 is isotopic to b’, +1 on S, we have
—1 _ —1 __ -1
9tbr+19 — tb’,+, - tbr+l’
where p is the isotopy class of p. The claim now follows.

The next step is to prove that t; = if: = fl = 1. Consider the generators of M 3
given by Theorem 3.2. Note that f is a separating circle such that the surface S I is a
disjoint union of a Klein bottle with one hole, and a nonorientable surface. It follows
from Lemma 3.7 that t; = 1. Since f1 intersects b, at only one point, from the braid
relations we have ff, 2 fbr. That is, t}! = 1. Also, since f3 is 2-sided nonseparating
and since S f2 is orientable, f2 is topologically equivalent to br+1. In another words,

t [2 is conjugate to ti: 1' This implies that {is = {:11 = 1.
It is clear now from Theorem 3.2 that if h E M 5 then R2 = 1.

Now let q be a separating circle on S such that one of the components of 5,, is a
Klein bottle K with one hole and the other is an orientable surface. Let 20 be the
isotopy class of a Y-homeomorphism supported in K such that 23 = tq. We now show
that Z, = 1. On the orientable component of Sq, there are circles qo, q1, q3, . . . , q, = q
such that go bounds a disc, and q,-_1 and q,- bound a torus with two holes for each
i = 1, 2,. . . ,r. Then by Lemma 3.5, tq,_,tq, is the fourth power of a product of isotopy
classes of twists about circles which are topologically equivalent to al. This implies

that fq,_,fq, = 1. Since tqo 2' 1, it follows that f, = 33 = 1.

Finally, we show that E2 = 1. As indicated in [19], the automorphism of H1(S, R)
induced by zo has determinant -—1. However, Dehn twists about 2-sided circles induce
automorphisms with determinant +1. Therefore, 20 E M5. Since M 5 is of index

two in Mg and since 2 9! M3, 20 = hz for some h E M5. Hence,

_. "'2_ _
1=z§=h22=22.

 

69

By putting the results of the last six paragraphs together, we conclude that the
generators of H1(Mg) satisfy f3, = fir +1 = 'z'2 = 1 and the commutation relations.

Therefore, H1(Mg) is Z2, Z2 69 Z2, or Z; G) Z2 EB 22.

Let us consider the case g = 4 ﬁrst. Consider H1(S,Z2) with the Zg-valued

intersection pairing and let
WM; —> ISO H1(S, Z2)

be the natural homomorphism. The restriction of 9p to M 5 is onto by Theorem 3.3.

It follows from Lemma 3.11 that the order of Iso H1(S, Z2) is 48. Since any per-
mutation of an orthonormal basis of H1(S,Z2) induces an isometry of H1(S,Z3),
Iso H1(S, Z3) has a subgroup isomorphic to the symmetric group 24. Let us identify
this subgroup with 24. Since 24 is of index 2 in Iso H1(S,Z3), it is normal. Re-
call that the commutator subgroup [24, E4] of E4 is the alternating group, which is
of index 2 in 24, and hence of index 4 in Iso H1(S,Z3). By Lemma 3.8, [24,534]
is normal in Iso H1(S,Z2). Note that any group of order 4 is abelian. There—
fore, Iso H1(S,Zg)/[E4,E4] is abelian. It follows that the commutator subgroup
of Iso H1(S,Z2) is equal to [24,24]. In particular, there is a homomorphism from

Iso H1(S, Z2) onto an abelian group C of order 4.

Let 90’ be the composition of tp with the quotient map from Iso H1(S, Z3) to G.
Then [Mg,Mg] C kera. Since [Mg,Mg] C M5, we have [Mg,Mg] C kerﬁﬂMs,
and since the restriction of (p? to M5 is onto, ker ¢ﬂMs is of index 8 in Mg. Therefore,

the order of H1(Mg) is at least 8. Consequently, H1(Mg) = Z2 69 Z2 EB Z2.

Suppose now that g _>_ 6. The sphere with four holes of Figure 3.5 (a) can be
embedded in S in such a way that do = b,_1,d1 = c,_1, d3 = c,,d3 = br+l~ Then each
dij is 2-sided nonseparating with a nonorientable complement. Since d1, d3, d3, dlg, c113

and d33 are all topologically equivalent to al, and L, = f: , it follows that id” = id, =

 

70

L, for 1 S i < j S 3 and 1 S k S 3. Hence, as a consequence of the lantern relation
we have

dozdjdzfds = d122d13-{d23i

which implies that to = 1. Hence H1(Mg) is Z3 or Z2 69 Z2.

r+1

Let g = 6. Consider H1(S, Z2) with the Zg-valued intersection pairing again. Since
H1(S, Z2) is isomorphic to Z3, by Lemma 3.12 we have a surjective homomorphism
from Iso H1(S, Z3) to Sp(4,2). Since Sp(4,2) is isomorphic to 26, there is a natural
map from Sp(4, 2) onto Z3. Therefore, the composition of the following natural maps
is onto:

M;- —> ISO H1(S,Z2) —'> Sp(4,2) —) Z2.

In fact, even the restriction of ¢ to M s is onto by Theorem 3.3. Therefore keraﬂMs
is normal and of index 4 in Mg. Clearly, this implies that the order of H1(Mg) is at

least 4. It follows that H1(Mg) = Z2 EB Z2.

If g 2 8 then, as in the case of g odd, the lantern relation implies that ta, = 1,
because the sphere with four holes of Figure 3.5 (a) can be embedded in S such that
do = b1,d1 = c1,d2 = c2 and d3 = b3. Hence, the order of H1(Mg) is at most 2, and

thus exactly 2. The proof of Theorem 3.13 is now complete. Cl

Theorem 3.14 Let S be a closed connected nonorientable surface of genus g. Then

H1(M3) is equal to

(ii)0ifg=10rg27.

Proof: If g = 1 then M5 is trivial. If g = 2 then it is easy to conclude from
Mg = Z2 63 22 that M s = Z3. Hence, the proof of the theorem follows in these two

cases .

71

Suppose now that g 2 7. Let us denote by to the class of to in the quotient

M s/ [M s, M 5]. Consider the generators of M 5 given by Theorem 3.2.

If g is odd, then for 1 S i S r the relations ta, = to, = ta, = 1 follow from the
braid relations and the lantern relations. In the proof of Theorem 3.13, to prove the
triviality of to = 32 in H 1(M g), we used only Dehn twists. The same argument shows
that to = 1 in H1(Ms). Since e1 intersects a, once, the braid relations imply that

t3, = fa, = 1. This implies the theorem for g odd.

Suppose g is even. Again, for 1 S i S r the equalities to, = to, = t3, = 1 follow
from the braid relations and lantern relation as in the proof of Theorem 3.13. The
sphere with four holes of Figure 3.5 (a) can be embedded in S so that do = b,_1, d1 =
c,-1, d3 = c,, d3 = br+1. By the lantern relation, to,“ = 1. Since S f is a disjoint union
of a Klein bottle with one hole and a nonorientable surface of genus Z 6, the sphere
with four holes of Figure 3.5 (a) can be embedded in S in such a way that do = f
and the circles d1, d3, d3,d12, d13, d23 are all 2—sided nonseparating with nonorientable

complements (see Figure 3.9). Now by the lantern relation

tftd1td2td3 = td12 td13 tdss °

It follows that if = 1. Since f1 intersects b, once, we have if, = 1 by the braid
relations. The circle f; is nonseparating and its complement is orientable. Hence it is
topologically equivalent to b,.+1. By the same reason that to, +1 = 1, we have if, = 1.

This completes the proof of the theorem. Cl

 

1

 

 

 

 

,
l:

J

 

Figure 3.9: Embedding of So.

 

72

Remark: If g = 3, 4, 5, 6, the problem of the computation of H1(M5) remains open.

Our theorems support a little surprising similarity between the subgroups M3 and
Mp of Mg and Mg respectively: they are both of index 2 and perfect for sufﬁciently

high genera.

Corollary 3.15 Let V be a vector space over the ﬁnite ﬁeld F; of order 2 with a basis
{v1,v3, . . . ,vn} and let (,) be the symmetric bilinear form deﬁned by (v,,v,-) = 6,5. If

n 2 7 then the group 180 V is perfect.

Proof: This is a consequence of Theorem 3.3 and Theorem 3.14. D

Remark: If the dimension n of V is odd, then 130 V is the symplectic group
Sp(n — 1, 2). This group is known to be perfect if n 2 7 (cf. [25], [27]).

Theorem 3.16 Let F be a closed connected orientable surface of genus g. Then
H1(Mg) is equal to
(02289-10 0r923,

(ii) Z2€BZ2 ifg=1 org=2.

Proof: Ifg = 0 then M; is equal to Z2 and hence H1(Mg) = Z3. For g 2 3, since
H1(Mp) = 0, [Mp,Mp] = Mp. Since Mp is of index 2 in M}, it follows that
[Mg,Mg] = Mp. Hence H1(M';.s) = 22. I

We now prove (ii). Let us realize F in R3 invariant under the reﬂections across

coordinate planes, as in Figure 3.10. Recall that Mp is generated by {t A, , t A,} if g = 1

and by {L4, , tA2, tA3, tA4 , tAs} ifg = 2 (cf. [20]). Then these generators, together with

 

 

 

 

 

 

 

Figure 3.10: Orientable surfaces of genus one and two.

the isotopy class of an orientation reversing homeomorphism, generate M 3:. Let p be
the isotopy class of the reﬂection p across the XY-plane. Then, M}; is generated by
{tA,,tA,,p} ifg =1 and by {tA,,tA,,tA3,tA4,tA5,p} ifg = 2. Ifg = 1 then it follows
from Theorem 2.26 that it, = {42. Similarly, it, = {4, = {43 = i4, = {.45 if g = 2.
This implies that H1(M}‘.~) is generated by {imp}. Since p is orientation reversing
and p(Al) = A1, we have ptA,p'l =t211. Hence {:1 = 1. Also, since p2 = 1, '9'" = 1.

Since Mg has a normal subgroup of index 2, namely Mp, the ﬁrst homology group
H1(M}‘.~) cannot be 0. Therefore, it is either Z2 or Z3 63 Z3.

We now prove that the order of H1(Mg) is at least 4. Consider H1(F, Z2) with
the Zg-valued intersection pairing. Note that the group 130 H1(F, Z3) is isomorphic
to Sp(2,2) ’-_‘-’ 23 ifg = 1 and Sp(4,2) ’5 23 ifg = 2, i.e., Iso H1(F,Z2) ’5 239 for

g = 1, 2. Consider the natural homomorphism
(p 2 M; —) ISO H1(F,Z2) g 239.

It is well-known that the restriction of 90 to Mp is onto. Let ¢ be the composition
of 90 with the quotient map from 23, onto 2398239, 239] ’5 Z3. Then ker'95 ﬂ Mp is
a normal subgroup of Mg of index 4. Therefore, the order of H1(Mg) is at least 4.

This implies the theorem. CJ

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74

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