<< ñòð. 11(âñåãî 12)ÑÎÄÅÐÆÀÍÈÅ >>
2 distinct points of C , then f is constant.
For our immediate purposes, the most important fact is that the universal
covering of every Riemann surface of genus g  2 is isomorphic to H . More
generally, we have the following interpretation of the stability condition 2g
2 + n > 0.
Exercise 4.7. Show that if U is a Riemann surface of the form X F
where X is a compact Riemann surface genus g and F a nite subset of X
of cardinality n, then the universal covering of U is isomorphic to H if and
Moduli of Riemann surfaces, transcendental aspects 321

only if 2g 2 + n > 0. Observe that 2 2g n is the topological Euler
characteristic of U.
5. Teichmuller Space

Suppose that g  2 and that S is a compact oriented surface of genus
g  2. Fix a base point xo 2 S and set  = 1(S; xo ). Let
conjugacy classes of injective representations  :)
(
Xg =  ,! PSL2(R) such that im  acts freely on H and
im nH is a compact Riemann surface of genus g
It is standard that we can choose generators a1 ; : : : ; ag , b1 ; : : : ; bg of  that
are subject to the relation
g
Y
[aj ; bj ] = 1:
j=1
where [x; y] denotes the commutator xyx 1 y 1 .
Exercise 5.1. Show that Xg can also be identi ed naturally with the set
 
compact Riemann surfaces C plus a conjugacy class of isomor-
phisms   1 (C) modulo the obvious isomorphisms
=
The group PSL2 (R) is a real ane algebraic group. It can be realized as
a closed subgroup of GL3 (R) de ned by real polynomial equations.
Exercise 5.2. Denote the standard 2-dimensional representation of SL2 (R)
by V . Denote the second symmetric power of V by S 2 V . Show that this
is a 3-dimensional representation of SL2 (R). Show that I 2 SL2 (R) acts
trivially on S 2 V , so that S 2 V is a 3-dimensional representation of PSL2 (R).
Show that this representation is faithful and that its image is de ned by
polynomial equations. Deduce that PSL2 (R) is a real ane algebraic group.
Because of this, we will think of elements of PSL2 (R) as 3  3 matrices
whose entries satisfy certain polynomial equations. A representation  :  !
PSL2(R) thus corresponds to a collection of matrices
A1 ; : : : ; Ag ; B1 ; : : : ; Bg
in PSL2 (R)  GL3 (R) that satisfy the polynomial equation
g
Y
I [Aj ; Bj ] = 0;
(4)
j=1
the correspondence is given by setting Aj = (aj ) and Bj = (bj ).
322

The set of all representations  :  ! PSL2 (R) is the real algebraic
subvariety R of PSL2 (R)2g consisting of all 2g-tuples
(A1 ; : : : ; Ag ; B1 ; : : : ; Bg )
that satisfy (4). This is a closed subvariety of PSL2 (R)2g and is therefore
an ane variety. Note that5
dim R  2g  dim PSL2 (R) dim PSL2 (R) = 6g 3:
Observe that PSL2 (R) acts on R on the right by conjugation:
A :  7! fg 7! A 1 (g)Ag:
Exercise 5.3. Show that
 :  ,! PSL2 (R) such that im  acts)
(
U := freely on H and im nH is a compact Rie-
mann surface of genus g
is an open subset of R and that it is closed under the PSL2 (R)-action.
Hint: One can construct a fundamental domain F for the action of  on H
given by  as follows. First choose a point x 2 H . Then take F to be all
points in H that are closer to x than to any of the points (g)x (with respect
to hyperbolic distance) where g 2  is non-trivial. Then F is a compact
subset of H whose boundary is a union of geodesic segments. As  varies
continuously, the orbit of x varies continuously. Now choose x to be generic
enough and study the change in F as  varies.
Note that Xg = U=PSL2 (R).
Exercise 5.4. Show that the center of PSL2(R) is trivial.
Proposition 5.5. If  2 U, then the stabilizer in PSL2 (R) of  is trivial.
Proof. The rst step is to show that if  2 U, then im  is Zariski dense
in PSL2 (R). To prove this, it suces to show that im  is Zariski dense in
the set of complex points PSL2 (C ). From Lie theory (or algebraic group
theory), we know that all proper subgroups of PSL2 (C ) are extensions of a
nite group by a solvable group. Since im  is isomorphic to , and since  is
not solvable (as it contains a free group of rank g), im  cannot be contained
in any proper algebraic subgroup of PSL2 (R), and is therefore Zariski dense.
One has to be more careful here as, in real algebraic geometry, ane varieties of higher
5

codimension can all be cut out by just one equation. Probably the best way to proceed is
to compute the derivative of the product of commutators map PSL2 (R)2g ! PSL2 (R) at
points in the ber over the identity.
Moduli of Riemann surfaces, transcendental aspects 323

An element A of PSL2 (R) stabilizes  if and only if A 1 (g)A = (g) for
all g 2  | that is, if and only if A centralizes im . But A centralizes im 
if and only if it centralizes the Zariski closure of im . Since im  is Zariski
dense in PSL2 (R), and since the center of PSL2 (R) is trivial, it follows that
A is trivial.
Theorem 5.6. The set U is a smooth manifold of real dimension 6g 3. The
group PGL2(R) acts principally on U, so that quotient Xg = U=PSL2 (R) is
a manifold of real dimension 6g 6.
Sketch of Proof. It can be shown, using deformation theory, that the Zariski
tangent space of R at the representation  of  is given by the relative
cohomology group:6
T R = H 1 (S; xo ; A )
where A is the local system (i.e., locally constant sheaf) over S whose ber
over xo is 1 (R) and whose monodromy representation is the homomorphism
sl

! PSL2(R)
 2
Here sl2 (R) denotes the Lie algebra of PSL2 (R), which is the set of 2  2
A 7! fX 7! AXA 1 g:
A result equivalent to this was rst proved by Andr Weil [26]. The long
e
exact sequence of the pair (S; xo ) with coecients in A gives a short exact
sequence
0 ! 2 (R) ! H 1 (S; xo ; A ) ! H 1 (S; A ) ! 0
sl

and an isomorphism H 2 (S; xo ; A )  H 2 (S; A ).
=
This is a non-standard way of giving this group | Weil used cocycles. Here A (S; A )
6

is the complex of di erential forms on S with coecients in the local system A . It is a
di erential graded Lie algebra. Restricting to the base point x one obtains an augmenta-
tion A(S; A ) ! 2 (R). The complex A (S; x ; A ) is the kernel of this, and H  (S; x ; A )
o

sl o o

is de ned to be the cohomology of this complex. Brie
y, this is related to deformations
of  as follows: when the representation  is deformed, the vector bundle underlying the
local system corresponding to  does not change, only the connection on it. But a new
connection on the bundle corresponding to  will di er from the original connection by
an element of A1 (S; A ). So a deformation of  corresponds to a family of sections w(t) of
A1 (S; A ) where w(0) = 0. For each t the new connection is
at. This will imply that the
t derivative of w(t) at t = 0 is closed in A (S; A ). Changes of gauge that do not alter the
marking of the ber over x are, to rst order, elements of A0 (S; x ; A ). The t derivative
o o

of such rst order changes of gauge gives the trivial rst order deformations of .
324

Deformation theory also tells us that if H 2 (S; xo ; A ) vanishes, then R is
smooth at [] 2 R.7 The Killing form is the symmetric bilinear form on
2 (R) given by
sl

b(X; Y ) = tr(XY ):
It is non-degenerate and invariant under Ad . It follows that A is isomor-
phic to its dual A  as a local system. Thus it follows from Poinar duality
e
(with twisted coecients) that
H 2 (S; A )  H 0 (S; A ) :
=
But if  is an element of U, it follows from Proposition 5.5 that H 0 (S; A )
vanishes. Consequently, if  2 U, then
dim H 1 (S; A ) = (S; A ) = (rank A )  (S) = 3(2g 2) = 6g 6
and
dim H 1 (S; xo ; A ) = 6g 3:
We saw above that U has dimension  6g 3, but dim U is not bigger
than the dimension of any of its Zariski tangent spaces, which we have just
shown is 6g 3. It follows that dim U = 6g 3 and that it is smooth as
its dimension equals the rank of its Zariski tangent space. Finally, since the
PSL2 (R)-action on U is principal, Xg = U=PSL2 (R) is smooth of dimension
(6g 3) dim PSL2 (R) = 6g 6.
It is not very dicult to show that the tangent space of Xg at [] is
naturally isomorphic to H 1 (S; A ).

6. Mapping Class Groups
The mapping class group g plays a role for higher genus (g  2) Riemann
surfaces analogous to that played by SL2 (Z) in the theory of genus 1 curves.
As in the previous section, S is a xed compact oriented surface of genus g,
xo 2 S, and  = 1 (S; xo ). But unlike the last section, we allow any g  1.
The mapping class group S of S is de ned to be the group of connected
components of the group of orientation preserving di eomorphisms of S:
S = 0 Di + S:

A proof of this and also a proof of Weil's result can be found in the beautiful paper of
7

Goldman and Millson [10].
Moduli of Riemann surfaces, transcendental aspects 325

Here Di S is given the compact open topology. It is the nest topology on
Di S such that a function f : K ! Di S from a compact space into Di S
is continuous if and only if the map f : K  S ! S de ned by
f (k; x) = f(k)(x)
is continuous.
Exercise 6.1. Suppose that M is a smooth manifold. Show that a path in
Di M from  to is a homotopy : [0; 1]M ! M from  to such that
for each t 2 [0; 1], the function (t; ) : M ! M that takes x to (t; x) is a
di eomorphism. Such a homotopy is called an isotopy between  and .
Recall that the outer automorphism group of a group G is de ned to be
the quotient
Out G = Aut G= Inn G
of the automorphism group Aut G of G by the subgroup Inn G of inner
automorphisms.
Exercise 6.2. Suppose that M is a smooth manifold. Show that each dif-
feomorphism of M induces an outer automorphism of its fundamental group
and that the corresponding function
Di M ! Out (M)
is a homomorphism. (Hint: Fix a base point of M and show that every dif-
feomorphism of M is isotopic to one that xes the base point.) Show that the
kernel of this homomorphism contains the subgroup of all di eomorphisms
isotopic to the identity. Deduce that there is a homomorphism
0 (Di M) ! Out 1 (M):
Taking M to be S, our reference surface, we see that there is a natural
homomorphism
 : S ! Out :
It is a remarkable fact, due to Dehn [6] and Nielsen [22], that this map is
almost an isomorphism. (I believe there is a proof of this in the translation
of some papers of Dehn on topology by Stillwell.)
Theorem 6.3 (Dehn-Nielsen). For all g  1, the homomorphism  is in-
jective and the image of  is a subgroup of index 2.
326

One can describe the index 2 subgroup of Out  as follows. Since S has
genus  1, the universal covering of S is contractible, and there is a natural
isomorphism Hi(S; Z)  Hi (; Z). Each element of Out  induces an auto-
=
morphism of Hi (; Z) as inner automorphisms of a group act trivially on its
cohomology. The image of  is the kernel of the natural homomorphism
Out  ! Aut H2 (; Z)  f1g:
=
When g = 1,   Z2, and
=
Out  = Aut   GL2 (Z):
=
Exercise 6.4. Show that when g = 1, the homomorphism Out  ! f1g
corresponds to the determinant GL2(Z) ! f1g.
The genus 1 case of the Dehn-Nielsen Theorem can be proved by elemen-
tary means:
Exercise 6.5. Show that if g = 1, then the homomorphism S ! SL2(Z) is
an isomorphism. Hint: Construct a homomorphism from SL2 (Z) to Di S.
Since two compact orientable surfaces are di eomorphic if and only if they
have the same genus, it follows that the group S depends only on the genus
of S. For this reason, we de ne g to be S where S is any compact genus
g surface.
7. The Moduli Space
In this section, S is once again a compact oriented surface of genus g  2
and  denotes its fundamental group. The mapping class group g = S
acts smoothly on Teichmller space Xg on the left via the homomorphism
u
g ! Out .
Exercise 7.1. Describe this action explicitly. Show that the isotropy group
of [] 2 Xg is naturally isomorphic to the automorphism group of the Rie-
mann surface im nH . (Compare this with the genus 1 case.) Deduce that
all isotropy groups are nite.
Exercise 7.2. Show that two points in Xg are in the same g orbit if and
only if they determine the same Riemann surface. Hint: Use the Uniformiza-
tion Theorem.
Rephrasing this, we get:
Theorem 7.3. If g  2, then Mg is naturally isomorphic to the quotient
of Xg by g .
Moduli of Riemann surfaces, transcendental aspects 327

This result allows us to put a topology on Mg . We give it the unique
topology so that Xg ! Mg is a quotient mapping. To try to understand the
topology of Xg we shall need the following fundamental result.
Theorem 7.4 (Teichmller). For all g  2, the Teichmller space Xg is
u u
contractible and the action of g on it is properly discontinuous.
A natural way to approach the proof of this theorem is via hyperbolic
geometry. We do this in the next two sections.
8. Hyperbolic Geometry
Perhaps the most direct way to approach the study of Teichmller space is
u
via hyperbolic geometry. The link between complex analysis and hyperbolic
geometry comes from the fact that the upper half plane has a complete
metric
ds2 = y12 (dx2 + dy2 )
of constant curvature 1 whose group of orientation preserving isometries is
PSL2(R). The Riemannian manifold (H ; ds2 ) is called the Poincar upper
e
half plane. A good reference for hyperbolic geometry is the book by Beardon
[2].
Exercise 8.1. Show that
(i) every element of PSL2 (R) is an orientation preserving isometry of
(H ; ds2 );
(ii) PSL2 (R) acts transitively on H and that the stabilizer of one point
(say i), and therefore all points, of H is isomorphic to SO(2) and acts
transitively on the unit tangent space of H at this point.
Deduce that the Poincar metric ds2 has constant curvature and that PSL2 (R)
e
is the group of all orientation preserving isometries of the Poincar upper
e
half plane.
The fact that the group of biholomorphisms of H coincides with the group
of orientation preserving isometries of H is fundamental and is used as fol-
lows. Suppose that X is a Riemann surface whose universal covering is H .
Then, by Exercise 4.4, X is biholomorphic to the quotient of H by a discrete
subgroup of PSL2 (R) isomorphic to 1 (X). Now, since PSL2 (R) is also
the group of orientation preserving isometries of H , the Poincar metric ds2
e
is invariant under and therefore descends to X = nH . This metric is
complete as the Poincar metric is.
e
328

Exercise 8.2. Prove the converse of this: if X is an oriented surface with a
complete hyperbolic metric, then X has a natural complex structure whose
canonical orientation agrees with the original orientation.
These two constructions are mutually inverse. Thus we conclude that
when g  2, there is a natural one-to-one correspondence
n
isomorphism classes of compacto \$
(5) Riemann surfaces of genus g
n
isometry classes of compact oriented surfaceso
of genus g with a hyperbolic metric
It follows that if g  2, then
 
isometry classes of compact oriented surfaces :
Mg = of genus g with a hyperbolic metric
Remark 8.3. Likewise, a Riemann surface of genus 1 has a
at metric (unique
up to rescaling) which determines the complex structure. So M1 can be
regarded as the moduli space of (conformal classes of)
at tori.
Exercise 8.4. Show that if X is a Riemann surface whose universal covering
is H , then the group of orientation preserving isometries of X equals the
group of biholomorphisms of X.
9. Fenchel-Nielsen Coordinates
Once again, we assume that g  2 and that S is a reference surface of
genus g. The interpretation of Teichmller space in terms of hyperbolic
u
geometry allows us to de ne coordinates on Xg . To do this we decompose
the surface into \pants."
A pair of pants is a compact oriented surface of genus 0 with 3 boundary
components. Alternatively, a pair of pants is a disk with 2 holes. A simple

Figure 2. pairs of pants

closed curve (SCC) on S is an imbedded circle. A pants decomposition of
Moduli of Riemann surfaces, transcendental aspects 329

Figure 3. a pants decomposition

S is a set of disjoint simple closed curves in S that divides S into pairs of
pants.
Proposition 9.1. If the simple closed curves C1; : : : ; Cn divide S into m
pairs of pants, then n = 3g 3 and m = 2g 2.
Proof. Since a pair of pants has the homotopy type of a bouquet of 2 circles,
it has Euler characteristic 1. Since the Euler characteristic of a disjoint
union of circles is 0, we have
2 2g = (S) = m  (a pair of pants) = m:
Thus m = 2g 2. Since each pair of pants has 3 boundary components, and
since each circle lies on the boundary of two pairs of pants, we see that
n = 3m=2 = 3g 3:

If T is a compact oriented surface with boundary and (T) < 0, then
T has a hyperbolic metric such that each boundary component is totally
geodesic. As in the case where the boundary of T is empty, the set of all
hyperbolic structures on T (modulo di eomorphisms isotopic to the identity)
can be described as a representation variety. I will omit the details. This
space is called the Teichmller space of T and will be denoted by XT .
u
Proposition 9.2. If P is a pair of pants, then XP is a manifold di eomor-
phic to R3 . The di eomorphism is given by taking a hyperbolic structure on
P to the lengths of its 3 boundary components.
Discussion of the Proof. A proof can be found in the Expos 3, partie II by
e
Ponaru in the book [7]. The basic idea is that a pair of pants can be cut
e
into two isomorphic hyperbolic right hexagons, and that two right hyperbolic
hexagons are equivalent if the lengths of every other side of one equal the
lengths of the corresponding sides of the other. Also, one can construct
330

hyperbolic right hexagons where these lengths are arbitrary positive real
numbers.
A basic fact in hyperbolic geometry is that if one has a compact hyperbolic
surface T (not assumed to be connected) with totally geodesic boundary,
then one can identify pairs of boundary components of the same lengths to
obtain a hyperbolic surface whose new hyperbolic structure agrees with that
on T @T. Another elementary fact we shall need is that if C is a SCC in
a compact hyperbolic surface T with totally geodesic boundary, then there
is a unique closed geodesic
: S 1 ! T that is freely homotopic to C.
Now suppose that P = fP1 ; : : : ; P2g 2 g is a pants decomposition of S
determined by the SCCs C1 ; : : : ; C3g 3 . For each hyperbolic structure on S,
we can nd closed geodesics
1 ; : : : ;
3g 3 that are isotopic to the Cj . Taking
the lengths of these, we obtain a function
` : XS ! R3g 3 :
+
It is not dicult to show that this mapping is continuous. (See [7], for
example.) On the other hand, we can de ne an action of R3g 3 on XS and
whose orbits lie in the bers of `. To de ne the action of (1 ; : : : ; 3g 3 ) on a
hyperbolic surface S, write S as the union of 2g 2 pairs of hyperbolic pants
whose boundaries are represented by geodesics
1 ; : : : ;
3g 3 isotopic to the
Cj . Then twist the gluing map at
j by an angle j . This is also continuous.
Theorem 9.3 (Douady,[7, Expos 7]). If g  2, then the map ` : XS !
e
3g 3 is a principal R3g 3 bundle with the action described above.
R+

Corollary 9.4. For all g  2, Teichmller space Xg is di eomorphic to
u
6g 6 .
R

Exercise 9.5. At rst it may appear that ` : Xg ! R3g 3 should be a
+
1 )3g 3 bundle. Show that rotation by 2 about one of the SCCs
principal (S
Cj alters the representation  by an automorphism of  that is not inner.
Hint: it may help to rst read the part of Section 15 on Dehn twists.
10. The Complex Structure
Suppose that C is a compact Riemann surface. Recall that a deformation
of C is the germ about to 2 T of a proper analytic mapping f : C ! T,
where T is an analytic variety, and an isomorphism j : C ! f 1 (to ). One
de nes maps between deformations to be cartesian squares. A deformation
of C is called universal if every other deformation is pulled back from it.
Moduli of Riemann surfaces, transcendental aspects 331

A standard result in deformation theory is that every Riemann surface of
genus g  2 has a universal deformation. This has the property that T is
smooth at to and that T is of complex dimension 3g 3 with tangent space
at to canonically isomorphic to H 1 (C; C ), where C denotes the sheaf of
holomorphic sections of the tangent bundle of C.
As explained in Looijenga's lectures, Mg can be obtained by patching
such local deformation spaces together. If C is a smooth projective curve of
genus g and  :  ! PSL2 (R) is a representation such that C  im nH ,
=
then [] 2 Xg goes to [C] 2 Mg under the projection Xg ! Mg . The
following fact can be proved using partial di erential equations.
Theorem 10.1. If (T; to ) is a universal deformation space for C, then there
is a smooth mapping : T ! Xg such that (to ) = [] which is a di eo-
morphism in a neighbourhood of to . Moreover, the composition of with the
projection Xg ! Mg is the canonical mapping that classi es the universal
deformation of C.
Corollary 10.2. There is a canonical g invariant complex structure on Xg
such that the projection Xg ! Mg is a complex analytic mapping.
It should be noted that, although Xg is a complex manifold di eomorphic
to R6g 6 , it is not biholomorphic to either a complex 3g 3 ball or C 3g 3
when g > 1.
Combining these results, we obtain the following basic result:
Theorem 10.3. The moduli space Mg is a complex analytic variety whose
singularities are all nite quotient singularities. Furthermore, Mg can be
regarded as an orbifold whose universal covering is the complex manifold Xg
and whose orbifold fundamental groups can be identi ed, up to conjugacy,
with g . More precisely, if C is a compact genus g curve with no non-trivial
automorphisms, then there is a canonical isomorphism
1 (Mg ; [C])  0 Di + C:
orb =
11. The Teichmuller Space Xg;n

It is natural to expect that there is a complex manifold Xg;n and a discrete
group g;n that acts holomorphically on Xg;n in such a way that Mg;n is
isomorphic to the quotient g;nnXg;n . We give a brief sketch of how this
may be deduced from the results when n = 0 and g  2.
First, we x an n-pointed reference surface of genus g where g  2. That
is, we x a compact oriented surface S and a subset P = fx1 ; : : : ; xn g of n
332

distinct points of S. De ne
g;n := 0 Di + (S; P)
By de nition, elements of Di + (S; P) are orientation preserving and act
trivially on P.
Here is a sketch of a construction of Xg;1 . One can construct the Xg;n
when n > 1 in a similar fashion.
There is a universal curve C ! Xg . This can be constructed using defor-
mation theory. This is a ber bundle in the topological sense and it is not
dicult to see that the action of the mapping class group g on Xg can be
lifted to an action on C such that the projection is equivariant. Since the
ber of C ! Xg is a compact surface of genus g, and since Xg is contractible,
C has the homotopy type of a surface of genus g and therefore has funda-
mental group isomorphic to 1 (S). De ne Xg;1 to be the universal covering
of C. This is a complex manifold as C is. The bers of the projection Xg;1
are all isomorphic to H . Since Xg is contractible, so is Xg;1 .
Exercise 11.1. Show that there is a natural bijection
8 9
<conjugacy classes of representations  :  !=
Xg;1 = :PSL2 (R) such that C := im nH is of genus g,; :
plus a point x 2 C
Use this (or otherwise) to show that g;1 acts on Xg;1 and that the quotient
is Mg;1 . Show that the isotropy group of any point of Xg;n lying above
[C; x] 2 Mg;n is naturally isomorphic to Aut(C; x).
We shall regard Mg;n as the orbifold g;nnXg;n . There is a natural iso-
morphism
1 (Mg;n ; [C; x1 ; : : : ; xn ])  0 Di + (C; fx1 ; : : : ; xn g)
orb =
provided Aut(C; x1 ; : : : ; xn ) is trivial.
12. Level Structures
Level structures are useful technical devices for rigidifying curves. Suppose
that ` is a positive integer. A level ` structure on a compact Riemann surface
is the choice of a symplectic basis of H1 (C; Z=`Z).
Exercise 12.1. Show that if C is a compact Riemann surface of genus g,
then there is a canonical isomorphism between H1 (C; Z=`Z) and the `-torsion
points in Pic0 C. (Remark: the intersection pairing on H1 (C; Z=`Z) corre-
sponds to the Weil pairing on the `-torsion points.)
Moduli of Riemann surfaces, transcendental aspects 333

Denote the moduli space of n-pointed, smooth, genus g curves with a level
` structure by Mg;n[l]. This can be described as a quotient of Teichmller
u
space by a subgroup of g;n that we now describe.
Fix an n-pointed compact oriented reference surface S of genus g. The
mapping class group g;n acts naturally on H1 (S; Z). Since it preserves the
intersection pairing, this leads to a homomorphism
 : g;n ! Aut(H1(S; Z); intersection form):
De ne the level ` subgroup of g;n to be the kernel of the homomorphism
` : g;n ! Aut(H1 (S; Z=`Z); intersection form):
This homomorphism is surjective.
Exercise 12.2. Show that g;n[`] is a normal subgroup of nite index in
g;n and that the quotient is isomorphic to Spg (Z=`Z).
Exercise 12.3. Show that there is a natural bijection between Mg;n[`] and
the quotient of Xg;n by g;n [`]. Show that the quotient mapping Mg;n [`] !
Mg;n that forgets the level structure has nite degree and is Galois with
Galois group Spg (Z=`Z).
Choosing a symplectic basis of H1 (S; Z) gives an isomorphism
Aut(H1 (S; Z); intersection form)  Spg (Z):
=
We therefore have a homomorphism
 : g;n ! Spg (Z):
De ne the level ` subgroup Spg (Z)[`] of Spg (Z) to be the kernel of the
reduction mod ` homomorphism
Spg (Z) ! Spg (Z=`Z):
This homomorphism is surjective.
Exercise 12.4. (i) Show that g;n[`] is the inverse image of Spg (Z)[`]
under .
(ii) Show that Sp1 (Z) is isomorphic to SL2 (Z) and that  is the standard
representation when g = 1.
Theorem 12.5 (Minkowski). The group Spg (Z)[`] is torsion free when ` 
3.
Proposition 12.6. If 2g 2 + n > 0, g  1 and `  3, then the mapping
class group g;n[`] is torsion free and acts xed point freely on Xg;n.
334

Sketch of Proof. The case g = 1 is left as an exercise. Suppose that g  2.
We rst show that g;n[`] acts xed point freely on Xg;n. The isotropy group
of a point in Xg;n lying over [C; x1 ; : : : ; xn ] is isomorphic to Aut(C; x1 ; : : : ; xn ).
This is a nite group, and is a subgroup of Aut C. It is standard that the
natural representation Aut C ! Aut H 0 (C;
1 ) is injective (Exercise: prove
C
this. Hint: use Riemann-Roch). It follows that the natural representation
Aut C ! Aut(H1 (C; Z); intersection pairing)  Spg (Z):=
is injective and that Aut(C; x1 ; : : : ; xn ) \ g;n[`] is trivial. (Here we are
realizing Aut(C; x1 ; : : : ; xn ) as a subgroup of g;n as an isotropy group.) It
follows from Minkowski's Theorem that if `  3, then g;n[`] acts xed point
freely on Xg;n.
The rest of the proof is standard topology. If g;n[`] has a torsion element,
then it contains a subgroup G of prime order, p say. Since this acts xed
point freely on the contractible space Xg;n, it follows that the quotient GnXg;n
is a model of the classifying space B(Z=pZ) of the cyclic group of order p.
Since the model is a manifold of real dimension 6g 6+2n, this implies that
H k (Z=pZ; Z=pZ) vanishes when k > 6g 6 + 2n. But this contradicts the
known computation that H k (Z=pZ; Z=pZ) is non-trivial for all k  0. The
result follows.
Putting together the results of this section, we have:
Corollary 12.7. If g  1, n  0 and `  3, then Mg;n[`] is smooth and the
mapping Xg;n ! Mg;n [`] is unrami ed with Galois group g;n; the covering
Mg;n[`] ! Mg;n is a nite (rami ed) Galois covering with Galois group
Spg (Z=`Z).
With this information, we are able to prove a non-trivial result about the
mapping class group.
Corollary 12.8 (McCool, Hatcher-Thurston). For all g and n, the map-
ping class group is nitely presented.
Proof. We shall use the fact that each Mg;n [`] is a quasi-projective variety.
As we have seen, this is smooth when `  3 and has fundamental group
isomorphic to g;n [`]. But a well known result Lojasiewicz [19] (see also
[18]) implies that every smooth quasi-projective variety has the homotopy
type of a nite complex. It follows that g;n [`] is nitely presented when
`  3. But since g;n [`] has nite index in g;n , this implies that g;n is also
nitely presented.
Moduli of Riemann surfaces, transcendental aspects 335

13. Cohomology
One way to de ne the homology and cohomology of a group G is to nd
a topological space X such that
(
j (X; ) = G j = 1;
0 j 6= 1:
This occurs, for example, when the universal covering of X is contractible.
Such a space is called a classifying space of G. Under some mild hypotheses,
it is unique up to homotopy. One then de nes the cohomology of G with
coecients in the G-module V by
H j (G; V ) = H j (X; V)
where V is the locally constant sheaf over X whose ber is V and whose
monodromy is given by the action of G on V . It is well de ned. Homology
is de ned similarly in terms of the homology of X with coecients in V.
Exercise 13.1. Suppose that V is a g;n-module and V is the corresponding
locally constant sheaf over Mg;n [`], where `  3. Show that there is a natural
isomorphism
H  (Mg;n[`]; V)  H  ( g;n[`]; V ):
=
Show that the group Spg (Z=`Z) acts on both sides and that the isomorphism
is Spg (Z=`Z)-equivariant.
Since g;n does not act xed point freely on Xg;n , Mg;n is not a classifying
space for g;n. Nonetheless, standard topological arguments give
H  (Mg;n ; Q )  H (Mg;n; Q )Sp Z=`Z
= g

and
H  ( g;n; Q )  H  ( g;n ; Q )Sp Z=`Z:
= g

It follows that:
Theorem 13.2. There is a natural isomorphism
H  ( g;n; Q )  H  (Mg;n ; Q ):
=
If V is a g;n-module and g;n has xed points in Xg;n, we cannot always
de ne a local system V over Mg;n corresponding to V . However, we can
formally de ne
H  (Mg ; V
Q) := H  (Mg;n [`]; V
Q )Sp Z=`Z:
g
336

where the superscript Spg (Z=`Z) means that we take the Spg (Z=`Z) invari-
ant part. This should be regarded as the cohomology of the orbifold Mg;n
with coecients in the orbifold local system corresponding to V .
Exercise 13.3. Show that this de nition is independent of the choice of the
level `  3.
Let A be the locus of curves in Mg with non-trivial automorphisms. The
goal of the following exercise is to show that A is an analytic subvariety of
Mg , each of whose components has codimension  g 2. Set
M0g = Mg A:
Exercise 13.4. Suppose that X is a Riemann surface of genus g  2 and
that G is a nite subgroup of Aut X.
(i) Set Y = GnX. Show that if X is compact, then
X
g(X) 1 = d(g(Y ) 1) + (d jOj)=2
O
where O ranges over the orbits of G acting on X and where d is the
order of G.
(ii) Show that if g  3, then each component of the locus of curves in Mg
that have automorphisms is a proper subvariety of Mg .
(iii) Show that the hyperelliptic locus in Mg has codimension g 2.
(iv) Show that codimension of each component of the locus in Mg with
curves with a non-trivial automorphism is  g 2, with equality if and
only if the component is the hyperelliptic locus. Hint: reduce to the
case where d is prime.
(v) Give an argument that there are only nitely many components of the
locus in Mg of curves with a non-trivial automorphism.
Deduce that when g  3, the set of points of Mg corresponding to curves
without automorphisms is Zariski dense in Mg .
Since each component of A has real codimension  2g 4, it follows from
standard topological arguments (transversality) that the inverse image Xg0
of Mg in Xg has the property
j (Xg0 ) = 0 if j < 2g 5:
From this, one can use standard topology to show that if g  3, and if V is
any g -module, then there is a natural mapping
H k ( g ; V ) ! H k (M0g ; V)
Moduli of Riemann surfaces, transcendental aspects 337

which is an isomorphism when k < 2g 5 and injective when k = 2g 5.8
In particular, there is a natural isomorphism
H k ( g ; Z)  H k (M0g ; Z)
=
when k  2g 5.
There are similar results for Mg;n, but the codimension of the locus of
curves with automorphisms rises with n. For example, if n > 2g + 2, then
Aut(C; x1 ; : : : ; xn ) is always trivial by the Lefschetz xed point formula.
(Exercise: prove this.)
Lecture 3: The Picard Group
In this lecture, we compute the orbifold Picard group of Mg for all g  1.
Recall that an orbifold line bundle over Mg is a holomorphic line bundle
L over Teichmller space Xg together with an action of the mapping class
u
group g on it such that the projection L ! Xg is g -equivariant. An
orbifold section of this line bundle is a holomorphic g -equivariant section
Xg ! L of L. This is easily seen to be equivalent to xing a level `  3 and
considering holomorphic line bundles over Mg [`] with an Spg (Z=`Z)-action
such that the projection is Spg (Z=`Z)-equivariant. Working on Mg [`] has
the advantage that we can talk about algebraic line bundles more easily.
An algebraic orbifold line bundle over Mg is an algebraic line bundle
over Mg [`] for some ` equipped with an action of Spg (Z=`Z) such that the
projection to Mg [`] is Spg (Z=`Z)equivariant. A section of such a line bundle
is simply an Spg (Z=`Z)-equivariant section de ned over Mg [`]. Isomorphism
of such orbifold line bundles is de ned in the obvious way. Let
Picorb Mg
denote the group of isomorphisms classes of algebraic orbifold line bundles
over Mg . Our goal in this lecture is to compute this group. It is rst useful
to review some facts about the Picard group of a smooth projective variety.
14. General Facts
Recall that if X is a compact Khler manifold (such as a smooth projective
a
variety), then the exponential sequence gives an exact sequence
0 ! H 1 (X; Z) ! H 1 (X; OX ) ! Pic X ! H 2 (X; Z)
There is a similar result for homology with the arrows reversed and injectivity replaced
8

by surjectivity.
338

where the last map is the rst Chern class c1 . The quotient
Pic0 X := H 1 (X; OX )=H 1 (X; Z)
is the group of topologically trivial complex line bundles and is a compact
complex torus (in fact, an abelian variety). Note that if H 1 (X; Z) vanishes,
then Pic0 X vanishes and the rst Chern class c1 : Pic X ! H 2 (X; Z) is
injective.
If H 1 (X; Z) vanishes, it follows from the Universal Coecient Theorem
that the torsion subgroup of H 2 (X; Z) is Hom(H1 (X; Z); C  ). Every torsion
element of H 2 (X; Z) is the Chern class of a holomorphic line bundle as
a homomorphism  : H1 (X; Z) ! C  gives rise to a
at (and therefore
holomorphic) line bundle over X.
Exercise 14.1. Suppose that X is a compact Khler manifold with H 1 (X; Z) =
a
0. Show that if L is the
at line bundle over whose monodromy is given by
the homomorphism  : H1 (X; Z) ! C  , then
c1 (L) =  2 Hom(H1 (X; Z); C  )  H 2(X; Z):
These basic facts generalize to non-compact varieties. Suppose that X is
a smooth quasi-projective variety. De ne Pic X to be the group of isomor-
phism classes of algebraic line bundles over X, and Pic0 X to be the kernel
of the Chern class mapping
c1 : Pic X ! H 2 (X; Z):
Theorem 14.2. Suppose that X is a smooth quasi-projective variety. If
H 1(X; Q ) vanishes, then Pic0 X = 0 and the torsion subgroup of Pic X is
naturally isomorphic to Hom(H1 (X; Z); C  ).
Sketch of Proof. There are several ways to prove this. One is to use Deligne
cohomology which gives a Hodge theoretic computation of Pic X. Details can
be found in [13], for example. A more elementary approach goes as follows.
First pick a smooth compacti cation X of X. Each line bundle L over X
can be extended to a line bundle over X. Any two extensions di er by twists
by the divisors in X that lie in X X. After twisting by suitable boundary
components, we may assume that the extended line bundle also has trivial
c1 in H 2(X; Z). (Prove this using the Gysin sequence.) It therefore gives an
element of Pic0 X, which, by the discussion at the beginning of the section,
is
at. This implies that the original line bundle L over X is also
at. But
if H 1 (X; Q ) vanishes, then H1 (X; Z) is torsion and c1 (L) 2 H 2 (X; Z) is the
Moduli of Riemann surfaces, transcendental aspects 339

corresponding character  : H1 (X; Z) ! C  . But since c1 (L) is trivial, this
implies that  is the trivial character and that L is trivial.
This result extends to the orbifold case. By a smooth quasi-projective
orbifold, we mean an orbifold nX where has a subgroup 0 of nite index
that acts xed point freely on X and where 0 nX is a smooth quasi-projective
variety. There is a Chern class mapping
c1 : Picorb ( nX) ! H 2 ( ; Z):
(The Picard group is constructed using equivariant algebraic line bundles on
nite covers of nX.) The Chern class c1 can be constructed using the Borel
construction, for example. De ne Pic0 ( nX) to be the kernel of c1 .
orb
Theorem 14.3. Suppose that nX is a smooth quasi-projective orbifold. If
H 1(X; Q ) vanishes, then Pic0 X = 0 and the torsion subgroup of Pic X is
orb
naturally isomorphic to Hom(H1 ( ; Z); C  ).
Sketch of Proof. Fix a nite orbifold covering of 0 nX of nX where 0 is
normal in and acts xed point freely on X. The Galois group of the
covering is G = = 0 . Suppose that L is an orbifold line bundle over nX
with c1 (L) = 0 in H 2 ( ; Z). This implies that the rst Chern class of the
pullback of L to 0 nX is also trivial. Since this pullback has a natural G-
action, this means that the corresponding point in Pic0 ( 0 nX) is G-invariant.
Since
H 1 ( 0 nX; Q )G  H 1 ( nX; Q ) = 0
=
it follows that the G-invariant part of Pic( 0 nX) is nite. It follows that
some power L
N of the pullback of L to 0 nX is trivial. This also has a
G-action. If s is a trivializing section of L
N , then the product
O
gs
g2G
section of L
NjGj. It follows that L has a
at structure
is a G-invariant
invariant under the G-action. But since L has trivial c1 , it must have trivial
monodromy. It is therefore trivial. It follows that Picorb ( nX) is trivial.
Assembling the pieces, we have:
Corollary 14.4. If H 1 ( g ; Q ) vanishes, then the Chern class mapping
Picorb Mg ! H 2 ( g ; Z)
is injective.
340

15. Relations in g
In this section we write down some well known relations that hold in
various mapping class groups. These will be enough to compute H1 ( g ),
which we shall do in the next section.
First some notation. We shall let S denote any compact oriented surface
with (possibly empty) boundary. The corresponding mapping class group is
de ned to be
S := 0 Di + (S; @S):
That is, S consists of isotopy classes of orientation preserving di eomor-
phisms that act trivially on the boundary @S of S.
Exercise 15.1. Show that elements of S can be represented by di eomor-
phisms that equal the identity in a neighbourhood of @S.
Exercise 15.2. Show that if S is a compact oriented surface and T is a com-
pact subsurface, then there is a natural homomorphism T ! S obtained
by extending elements of T to be the identity outside T.
An important special case is where we take T to be the cylinder [0; 1]S 1
(with the product orientation). One element of T is the di eomorphism
: (t; ) 7! (t;  + 2t):

Figure 4. a positive Dehn twist

Theorem 15.3. If T is the cylinder, then T is in nite cyclic and is gen-
erated by .
Moduli of Riemann surfaces, transcendental aspects 341

Now, if S is any surface, and A is a (smoothly imbedded) simple closed
curve (SCC), then A has a neighbourhood that is di eomorphic to the cylin-
der T = [0; 1]  S 1 as in Figure 5. Denote the image of  under the homo-
morphism T ! S by tA . It is called the Dehn twist about A.
A

T

Figure 5

Theorem 15.4. The class of tA in S is independent of the choice of the
tubular neighbourhood T of A and depends only on the isotopy class of A.
Every mapping class group is generated by Dehn twists.
Exercise 15.5. Show that if  2 S and A is a SCC in S, then
tA = tA  1 :
Observe that if two elements of S can be represented by di eomorphisms
with disjoint support, then they commute. In particular:
Proposition 15.6. If A and B are disjoint SCCs on S, then tA and tB
commute in S .
Exercise 15.7. Show that if S is a surface with boundary and A and B
are each non-separating SCCs in S, then there is an orientation preserving
di eomorphism  of S that takes one onto the other. Deduce that tA and
tB are conjugate in S . Give an example to show that not all Dehn twists
about bounding SCCs are conjugate in S when g  3.
Next we consider what happens when A and B are two simple closed
curves in S that meet transversally in one point as in Figure 6.
Exercise 15.8. Show that if A and B are two SCCs in S that meet transver-
sally in one point, then there is a neighbourhood of their union that is a
compact genus 1 surface with one boundary component. Hint: Compute the
homology of a small regular neighbourhood of the union and then apply the
classi cation of compact oriented surfaces.
342

A
B

Figure 6
So any relation that holds between Dehn twists such curves in a genus
one surface with one boundary component will hold in all surfaces. Let S
be a compact genus 1 surface with one boundary component and let A and
B be the two SCCs in the diagram:

A

B

Figure 7

Theorem 15.9. With notation as above, we have
tAtB tA = tB tA tB
in S and therefore in all mapping class groups.
This relation is called the braid relation as it comes from the braid group
on 3 strings9 using the following technique.
Denote the unit disk by D. We view this as a manifold with boundary.
Suppose that P is a set of n distinct points of D, none of which lies on @D.
The braid group Bn is de ned to be
Bn := 0 Di +(D; (P ));
where Di + (D; (P )) denotes the group of orientation preserving di eomor-
phisms of D that x P as a set, but may permute its elements. There is a
surjective homomorphism
Bn ! Aut P  n
=
A good reference for braid groups is Joan Birman's book [4].
9
Moduli of Riemann surfaces, transcendental aspects 343

onto the symmetric group on n letters.
Suppose that U is a disk imbedded in D such that
@U \ P = ;
and U \ P consists of two distinct points x and y of P. Then there is an
element U of Bn whose square is the Dehn twist t@U about the boundary
of U and which swaps x and y.
It can be represented schematically as in Figure 8: The braid group Bn
x y

x y

Figure 8. a basic braid
is generated by the braids 1 ; : : : ; n 1 illustrated in Figure 9. Note that i
1 i i+1 n-1 n

... ...

1 i i+1 n-1 n

Figure 9. the generator i
and j commute when ji jj > 1, and that
ii+1 i = i+1i i+1 :
(6)
Now suppose that (S; @S) ! (D; @D) is a branched covering, unrami ed
over the boundary. Suppose that the image of the branch points is P. Then
there is a natural homomorphism
Bn ! S :
344

So relations in Bn will give relations in S . The relations we are interested
in come from certain double branched coverings of the disk.
Suppose that (S; @S) ! (D; @D) is a 2-fold branched covering. The in-
verse image of a smooth arc joining two critical values p1 ; p2 2 P and
avoiding P otherwise, is an SCC, say A, in S. There is a small neighbour-
hood U of that is di eomorphic to a disk and whose intersection with P
is fp1 ; p2 g as in Figure 10. There is a braid  supported in U that swaps p1
and p2 by rotating in the positive direction about .

P1

U
Î±

P2

Figure 10

Proposition 15.10. The image of  under the homomorphism Bn ! S
is the Dehn twist tA about A.
Exercise 15.11. Show that a genus 1 surface S with one boundary compo-
nent can be realized as a 2:1 covering of D, branched over 3 points.
We therefore have a homomorphism B3 ! S . Note that the inverse
image of the two arcs and in Figure 11 under the covering of the disk
branched over fp1 ; p2 ; p3 g is a pair of SCCs in S that intersect transversally
in one point. The braid relation in S now follows as we have the braid
relation (6)
1 21 = 2 1 2
in B3 .
More relations can be obtained this way. Suppose that S is a compact
genus 1 surface with 2 boundary components.
Let A, B, C, T1 and T2 be the SCCs in Figure 12. Denote the Dehn twists
about A, B and C by a, b and c, and those about T1 and T2 by t1 and t2 .
Moduli of Riemann surfaces, transcendental aspects 345

Î± Î²

p p p
1 2 3

Figure 11
A
B
T1 T2

C

Figure 12
Theorem 15.12. The relation
(abc)4 = t1 t2
holds in the mapping class group S .
Corollary 15.13. If S is a compact genus 1 surface with one boundary
component, then the relation
(ab)6 = t
holds in S , where a and b denote Dehn twists about a pair of SCCs that
intersect transversally in one point and t denotes a Dehn twist about the
boundary.
Exercise 15.14. Deduce Corollary 15.13 from Theorem 15.12 by capping
o one boundary component and using the braid relation, Theorem 15.9.
Theorem 15.12 is proved in the following exercise.
Exercise 15.15. Suppose that S is a genus 1 surface with two boundary
components. Show that S is a double covering of the disk, branched over 4
points. Use this to construct a homomorphism from the braid group B4 on
4 strings into S . Use this to prove the relation
(abc)4 = t1 t2
346

where a, b, c, t1 and t2 denote Dehn twists on the SCCs A, B, C, T1 and T2
in the diagram above. Note that in the braid group, we have the relation
t = 1 2 3
where t is Dehn twist about the boundary of the disk and i is the ith
standard generator of B4 .
There is one nal relation. It is called the lantern relation and is due
to Johnson and Harer independently. Let S be a disk with 3 holes. (That
is, a genus 0 surface with 4 boundary components.) Consider the SCCs in
Figure 13.

A
0
A1

A12 A
31

A2 A3

A23

Figure 13. the lantern con guration

Theorem 15.16. The relation
a0 a1 a2 a3 = a12 a23 a31
holds in S where ai denotes the Dehn twist about Ai and aij denotes the
16. Computation of H1 ( g ; Z)
The fact g is generated by Dehn twists and the relations given in the
previous section allow the computation of H1 ( g ).
Theorem 16.1 (Harer). If g  1, then 8
>Z=12Z g = 1;
<
 Z=10Z g = 2;
H1 ( g ; Z) = >
g  3:
:
0
Moduli of Riemann surfaces, transcendental aspects 347

Proof. We begin with the observation that if S is a compact oriented genus
g surface, then all Dehn twists on non-separating SCCs lie in the same
homology class as they are conjugate by Exercise 15.5. Denote their common
homology class by L. Next, using the relations coming from an imbedded
genus 1 surface with one boundary component, we see that the homology
class of any separating SCC that divides S into a genus 1 and genus g 1
surfaces has homology class 12L. Using the relation that comes from an
imbedded genus 1 surface with 2 boundary components, we see that the
homology class of every separating SCC is an integer multiple of L. It
follows that H1 ( g ) is cyclic and generated by L.
Now suppose that g  3. Then we can nd an imbedded lantern as
in Figure 14 Since each of the curves in this lantern is non-separating, the

rest of surface

lantern

Figure 14
lantern relation tells us that 3L = 4L, which implies that L = 0. This proves
the vanishing of H1 ( g ) when g  3.
When g = 1, the relation for a genus 1 surface with one boundary com-
ponent implies that 12L = 0 as the twist about the boundary is trivial in
1 . Thus H1 ( 1 ) is a quotient of Z=12Z. But, as we shall explain a lit-
tle later, the fact that Picorb M1 is at least as big as Z=12 implies that
H1 ( 1 ) = Z=12Z.10
The abelianization of SL2 (Z) can also be computed using, for example, the presen-
10

tation of PSL2 (Z) given in [24].
348

A genus 2 surface can be obtained from a genus 1 surface with two bound-
ary components by identifying the boundary components. The relation ob-
tained for a genus 1 surface with 2 boundary components gives 12L = 2L,
so that 10L = 0. This shows that H1 ( 2 ) is a quotient of Z=10Z. But, as
in the genus 1 case, the theory of Siegel modular forms shows that it cannot
be any smaller. So we have H1 ( 2 ) = Z=10Z.
Corollary 16.2. For all g  1, H 1 ( g ; Z) vanishes.

17. Computation of Picorb Mg
Since H 1 ( g ; Q ) is torsion, it follows from Corollary 14.4 that
c1 : Picorb Mg ! H 2 ( g ; Z)
is injective. Since H1 ( g ) is torsion, the Universal Coecient Theorem im-
plies that the sequence
0 ! Hom(H1 ( g ; Z); C  ) ! H 2 ( g ; Z) ! Hom(H2 ( g ); Z) ! 1
is exact. To determine the rank of H 2 ( g ), we need the following fundamen-
tal and dicult result of Harer [15].
Theorem 17.1. The rank of H2( g ; Q ) is 0 if g  2 and 1 if g  3.
Combining this with our previous discussion, we have:
Theorem 17.2. If g  1, then
8
g = 1;
>Z=12Z
<

H 2( g ; Z) = >Z=10Z g = 2;
g  3:
:
Z

There is one obvious orbifold line bundle over Mg . Namely, if we take
the universal curve  : C ! Mg over the orbifold Mg , then we can form the
line bundle
L := det !C=M g

over Mg , where !C=M denotes the relative dualizing sheaf. The ber over
[C] 2 Mg is
g

g H 0 (C;
1 ):
C
Moduli of Riemann surfaces, transcendental aspects 349

Theorem 17.3. If g  1, the orbifold Picard group of Mg is cyclic, gener-
ated by L and given by
8
g = 1;
>Z=12Z
<
Picorb Mg  g = 2;
= >Z=10Z
g  3:
:
Z

Proof. All but the generation by L follows from preceding results. In genus
1 and 2, the generation by L follows from the theory of modular forms.
Suppose that g  3. Denote the rst Chern class of L by . To prove that
L generates Picorb Mg , it suces to show that  generates H 2 ( g ; Z). The
following proof of this I learned from Shigeyuki Morita. It assumes some
knowledge of characteristic classes. A good reference for this topic is the
book [21] by Milnor and Stashe .
We begin by recalling the de nition of the signature of a compact ori-
ented 4-manifold. Every symmetric bilinear form on a real vector space can
be represented by a symmetric matrix. The signature of a non-degenerate
symmetric bilinear form is the number of positive eigenvalues of a repre-
senting matrix minus the number of negative eigenvalues. To each compact
oriented 4-manifold X, we associate the symmetric bilinear form
H 2 (X; R)
H 2 (X; R) ! R
de ned by Z
1
2 7! 1 ^ 2 :
X
Poincar duality implies that it is non-degenerate. The signature (X) of
e
X is de ned to be the signature of this bilinear form. It is a cobordism
invariant.
The Hirzebruch Signature Theorem (see [21, Theorem 19.4], for example)
asserts that
1 Z p (X)
(X) = 3 1
X
where p1 (X) 2 H 4 (X; Z) is the rst Pontrjagin class of X. When X is a
complex manifold,
p1 (X) = c1 (X)2 2c2 (X):
(7)
Now suppose that X is a smooth algebraic surface and that T is a smooth
algebraic curve. Suppose that  : X ! T is a family whose bers are smooth
curves of genus g  3.11 Denote the relative cotangent bundle !X=T of  by
Such families exist | see [17, p. 45, p. 55].
11
350

!. Then it follows from the exact sequence
0 ! 
1 !
2 ! !X=T ! 0
S X
that
c1 (X) =  c1 (T ) c1 (!) and c2 (X) = c1 (!) ^ c1 (T ):
Plugging these into (7) we see that p1 (X) = c1 (!)2 . Using integration over
the ber, we have
1 Z c (!)2 = 1 Z  (c (!)2 ):
(X) = 3 1 3T1
X
It is standard to denote  (c1 (!)2 ) by 1 . Thus we have
1Z  :
(X) = 3 1
T
An easy consequence of the Grothendieck-Riemann-Roch Theorem is that
1 = 12. This is proved in detail in the book of Harris and Morrison [17,
pp. 155{156]. It follows that for a family  : X ! T of smooth curves
Z
(X) = 4 :
T
The last step is topological. Suppose that F is a compact oriented surface
and that p : W ! F is an oriented surface bundle over F where the bers of
p are compact oriented surfaces of genus g  3. Denote the local system of
the rst integral homology groups of the bers by H . There is a symmetric
bilinear form
H 1 (F; H R )
H 1 (F; H R ) ! R
(8)
obtained from the cup product and the intersection form. Poincar duality
e
implies that it is non-degenerate. It follows from the Leray-Serre spectral
sequence of p that
(W) = the signature of the pairing (8):
The local system H over F corresponds to a mapping  : F ! BSpg (Z)
into the classifying space of the symplectic group. Meyer [20] shows that
there is a cohomology class m 2 H 2 (Spg (Z); Z) whose value on  is the
signature of the pairing (8). It follows from this and the discussion above,
that under the mapping
 : H 2 (Spg (Z); Z) ! H 2 ( g ; Z)
Moduli of Riemann surfaces, transcendental aspects 351

induced by the canonical homomorphism  : g ! Spg (Z), m goes to 4.
Mayer also shows that the image of
m : H 2 ( g ; Z) ! Z
is exactly 4Z, which implies that  generates H 2 ( g ; Z).
As a corollary of the proof, we have:
Corollary 17.4. For all g  3, both H 2 (Spg (Z); Z) and H 2( g ; Z) are gen-
erated by  and the natural mapping
 : H 2 (Spg (Z); Z) ! H 2 ( g ; Z)
is an isomorphism.
Note that H 2 (Sp2 (Z); Z) is in nite cyclic, while H 2 ( 2 ; Z) is cyclic of
order 10. So  is not an isomorphism in genus 2.
Acknowledgements. It is a great pleasure to thank all those with whom I have
to thank Robert Bryant, John Harer, Eduard Looijenga, Makoto Matsumoto
and Mark Stern.
352

References
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Arithmetic fundamental groups
and moduli of curves
Makoto Matsumoto

Department of Mathematics, Kyushu University,
6-10-1 Hakozaki, Higashi-ku, Fukuoka 812-8581 Japan

Lecture given at the
School on Algebraic Geometry
Trieste, 26 July { 13 August 1999

LNS001008

 matumoto@math.kyushu-u.ac.jp
Abstract
This is a short note on the algebraic (or sometimes called arithmetic)
fundamental groups of an algebraic variety, which connects classical funda-
mental groups with Galois groups of elds. A large part of this note describes
the algebraic fundamental groups in a concrete manner.
This note gives only a sketch of the fundamental groups of the algebraic
stack of moduli of curves. Some application to a purely topological state-
ment, i.e., an obstruction to the surjectivity of Johnson homomorphisms in
the mapping class groups, which comes from Galois group of Q , is explained.
Contents
1. Galois groups of eld extension 359
2. A short way to arithmetic fundamental groups 360
2.1. Pro nite groups and algebraic fundamental groups 361
2.2. Arithmetic fundamental groups 364
2.3. Arithmetic monodromy 365
2.4. The projective line minus two points 366
2.5. The projective line minus three points 367
3. Arithmetic fundamental groups by etale topology 368
3.1. Unrami ed coverings of a topological space and ber functors 368
3.2. Finite coverings 371
3.3. Etale fundamental groups 372
4. Arithmetic mapping class groups 375
4.1. The algebraic stack Mg;n over SpecZ 375
4.2. The arithmetic fundamental group of the moduli stack 375
5. A conjecture of Takayuki Oda 377
5.1. Weight ltration on the fundamental group 377
5.2. Obstruction to the surjectivity of Johnson morphisms 379
5.3. The projective line minus three points again 380
References 382
Arithmetic fundamental groups 359

1. Galois groups of field extension
We recall some basic notions of classical Galois theory of elds. Let K

be a eld, and x its algebraic closure K. Let L be a nite extension of K.

An extension L=K is said to be separable if the cardinality of HomK (L; K)
is the same as the extension degree [L : K]. Let Aut(L=K) be the group of
automorphisms of L as a eld, xing each element in K. Then, Aut(L=K)

acts on HomK (L; K). If the action is transitive, then the extension L=K is
said to be a normal extension. A nite separable normal extension L=K is
called a nite Galois extension, and Aut(L=K) is called the Galois group of
L=K and denoted by G(L=K).
Suppose that L=K is a nite Galois extension. Galois theory asserts that
there exists a one-to-one correspondence between
fM : intermediate eld between L and Kg
and
fN : subgroup of G(L=K)g
given by M 7! G(L=M) and N 7! LN , where LN denotes the sub eld of ele-
ments xed by the action of all elements of N. The correspondence reverses
the inclusion relation, i.e., if M  M 0 then G(L=M)  G(L=M 0 ). If we use
the terminology of categories, then the correspondence is a contravariant
equivalence.
There is an in nite version. Let L=K be an algebraic extension, which is
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