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a nite connected graph that may have multiple bonds and for which we
allow loose ends (that is, edges of which only one end is attached to a vertex,
some call them legs). We need the additional data consisting of
 for each vertex a nonnegative integer gv and
 a numbering of the loose ends by f1; 2; : : : ; ng.
We say that these data de ne a stable graph if
 for every vertex v we haveP v 2 + deg(v) > 0.
2g
We de ne the genus by g( ) := v gv + b1 ( ), and we call the pair (g( ); n)
the type of the stable graph. The recipe for assigning a stable graph of
type (g; n) to (Sg ; p1 ; : : : ; pn ; (e )e2E ) is as follows: the vertex set is the set
of connected components of Sg fp1 ; : : : ; pn g [ee , the set of bonds is
indexed by E: the two sides of e de ne one or two connected components
and we insert a bond between the corresponding vertices (so this might be
a loop), and we attach the ith loose end to the vertex v if the corresponding
connected component contains pi . You should check that the topological
284

type is faithfully represented in this way. Note that the stable graph of a
smooth curve of type (g; n) is like a star: n loose ends attached to a single
vertex of weight g.
Let Mg;n denote the set of isomorphism classes of stable curves of type
(g; n). We have the expected theorem:
Theorem 6.1 (Knudsen-Mumford). The universal deformations of stable
curves of type (g; n) put a complex-analytic orbifold structure on Mg;n of
dimension 3g 3 + n. The space Mg;n is compact and the locus @Mg;n
parametrizing singular curves is a normal crossing divisor in the orbifold
sense.
The construction of Mg;n can also be obtained by means of geometric
invariant theory, which implies that it is a projective orbifold. (The role of
the dualizing sheaf is taken by !C (x1 +    + xn ); details can be found in
a forthcoming sequel to .) The irreducible components of the boundary
@Mg;n are in bijective correspondence with the embedded circles in S
fp1 ; : : : ; png given up to orientation preserving di eomorphism. These are:
0 : C is irreducible or
f(g0 ;I 0 );(g00 ;I 00 )g : C is a one point union of smooth curves of genera g0 and
g00 with the former containing the points xi with i 2 I 0 and the latter
the points indexed by I 00 (so fI 0 ; I 00 g is a partition of f1; : : : ; ng). We
allow g0 to be zero, provided that jI 0 j  2 and similarly for g00 .
6.1. The universal stable curve. Let (C; x1 ; : : : ; xn ) be a stable curve
of type (g; n). Let us show that any x 2 C determines a stable curve
˜˜
(C; x1 ; : : : ; xn+1 ) of type (g; n + 1).
˜
˜˜
 If x 2 Creg fx1; : : : ; xng, then take (C; x1 ; : : : ; xn+1 ) = (C; x1; : : : ; xn; x).
˜
˜
 If x = xi for some i, we let C be the disjoint union of C and P1 with the
points xi and 1 indenti ed. We let xi = 1 2 P1 and xn+1 = 0 2 P1 ;
˜ ˜
˜
whereas for j 6= i; n + 1, xj = xj , viewed as a point of C. We denote
˜
this (n + 1)-pointed curve by i (C; x1 ; : : : ; xn ).
˜
 If x 2 Csing , then C is obtained by separating the branches of C in x
(i.e., we normalize C in this point only) and by putting back a copy of
P1 with f0; 1g identi ed with the preimage of x. Then xn+1 = 1 2 P1 ˜
˜
and for i  n, xi = xi , viewed as a point of C.
˜
We thus have de ned a map C ! Mg;n+1 that maps xi to i ((C; x1 ; : : : ; xn ).
˜˜
There is also a converse construction: given stable curve (C; x1 ; : : : ; xn+1 ) ˜
of type (g; n + 1), then we can associate to it a stable curve (C; x1 ; : : : ; xn )
A minicourse of moduli of curves 285

of type (g; n) basically by forgetting xn+1 ; this yields a stable pointed curve
˜
unless xn+1 lies on a smooth rational component which has only two other
˜
˜
special points. Let C be obtained by contracting this component and let xi
be the image of xi (i  n). This de nes a map  : Mg;n+1 ! Mg;n. Notice
˜
that the map C ! Mg;n+1 de ned above parametrizes the ber of  over
the point de ned by (C; x1 ; : : : ; xn ).
Proposition 6.2. The map  : Mg;n+1 ! Mg;n is morphism and so are its
sections 1 ; : : : ; n . The ber of  over the point de ned by (C; x1 ; : : : ; xn )
can be identi ed with the quotient of C by Aut(C; x1 ; : : : ; xn ).
This Proposition says that in a sense the projection  with its n sections
de nes the universal stable curve of type (g; n). For this reason we often
refer to Mg;1 as the universal smooth genus g curve (g  2) and denote it
by Cg . Likewise C g := Mg;1 is the universal stable genus g curve.
6.2. Strati cation of Mg;n. The normal crossing boundary @Mg;n deter-
mines a strati cation of Mg;n in an obvious way: a stratum is by de nition
a connected component of the locus of points of Mg;n where the number of
local branches of @Mg;n at that point is equal to xed number. That number
may be zero, so that Mg;n is a stratum. It is clear that the strata decom-
pose Mg;n into subvarieties. A stratum of codimension k parametrizes stable
curves of type (g; n) with xed topological type (and k singular points). You
may check that distinct strata correspond to distinct topological types.
We next show that the closure of every stratum is naturally covered
by a product of moduli spaces of stable curves. Let Y be a stratum.
If (C; x1 ; : : : ; xn ) represents a point of Y , then consider the normaliza-
^ ^
tion n : C ! C and the preimage of the set of special points X :=
^^
n 1 (Csing [ fx1 ; : : : ; xn g). The connected components of the pair (C; X)
^
are stable curves, at least after suitably (re)numbering the points of X,
^^
for every component of C X maps homeomorphically to a component of
Creg fx1 ; : : : ; xng, hence has negative euler characteristic. So if the types of
Q
the stable curves are (gi ; ni )i2I , then we nd an element of i2I Mgi ;ni . Con-
versely if we are given a nite collection of smooth curves of type (gi ; ni )i2I ,
then by identifying some pairs of points we nd a stable curve of type (g; n).
In terms of stable graphs: the rst procedure amounts to cutting all the
bonds in the middle of the stable graph associated to Y so that we end
up with a nite set of stars, whereas the second builds out of stars a sta-
ble graph by identifying certain pairs of edges. This recipe de nes a map
286

Q
i Mgi ;ni ! Y of which it is not dicult to see that it is a nite surjec-
tive morphism. The same recipe can be used to glue stable (not necessarily
smooth) curves of type (gi ; ni ), so that the map extends to
Y
f : Mgi;ni ! Mg;n :
i
It is easy to verify that this is a nite morphism with image the closure of
Y . (It is in fact an orbifold cover of that closure.) The section i of the
universal curve is a special case of this construction.
7. Tautological classes
Throughout the discussion that follows we take rational coecients for
our cohomology groups. One reason is that the space underlying an orbifold
is a rational homology manifold, so that it satis es Poincar duality when
e
oriented (which is always the case in the complex-analytic setting). We shall
be dealing with complex quasi-projective orbifolds and the coholomology
classes that we consider happen to have Poincar duals that are Q -linear
e
combinations of closed subvarieties. Such classes are called algebraic classes.
7.1. The Witten classes. Given a stable curve (C; x1 ; : : : ; xn ), then for
a xed i 2 f1; : : : ; ng, we may associate to it the one dimensional complex

vector space Txi C. This generalizes to families: if (C ! B : (i : B !
C)n ) is a family of stable curves, then for a xed i, the conormal bundle
i=1
of i de nes a line bundle over B. In the universal example (Mg;n+1 !
Mg;n; 1; : : : ; n) this produces an orbifold line bundle over Mg;n. Its rst
Chern class, denoted i 2 H 2 (Mg;n), is called the ith Witten class. Since
the notation wants to travel lightly, it is a little ambiguous. For example,
it is not true that the image of i in H 2 (Mg;n+1 is the ith Witten class
of Mg;n+1. The euler class of the normal bundle of a parametrization f :
Q
i Mgi ;ni ! Mg;n of a closed stratum is a product of Witten classes of
the factors. This illustrates a point we are going to make, namely, that all
classes of interest appear to be obtainable from the Witten classes.
7.2. The Mumford classes. Mumford de ned these classes for Mg before
the Witten classes were considered; Arbarello-Cornalba used the Witten
classes to extend Mumford's de nition to the pointed case. It goes as follows:
if ˜n+1 denotes the last Witten class on H 2 (Mg;n+1), and r is a nonnegative
integer, then the rth Mumford class is
r :=  ( ˜n+1 ) 2 H 2r (Mg;n):
r+1
A minicourse of moduli of curves 287

(It is an interesting exercise to check that 0 = 2g 2 + n.) Arbarello and
Cornalba showed that 1 is the rst Chern class of an ample line bundle.
They also proved that H 2 (Mg;n) is generated by 1 ; : : : ; n ; 1 and the
Poincar duals of the irreducible components of the boundary @Mg;n and
e
that these classes form a basis when g  3.
7.3. The tautological algebra. It is convenient to make an auxiliary
de nition rst: de ne the basic algebra B(Mg;n) of Mg;n as the subal-
gebra of H  (Mg;n ) generated by the Witten classes ( j )j and the Mum-
ford classes (r )r0 . Then the tautological algebra of Mg;n, R(Mg;n ), is
by de nition the subalgebra of H  (Mg;n) generated by the direct images
Q
f (
iB(Mgi;ni )  H even (Mg;n), where the maps f : i Mgi;ni ! Mg;n
run over the parametrizations of the strata. Since the tautological algebra
is made of algebraic classes we grade it by half the cohomological degree:
Rk (Mg;n)  H 2k (Mg;n). It is clear that the tautological algebra and the
basic algebra have the same restriction to Mg;n ; we denote that restriction
by R(Mg;n ) and call it the tautological algebra of Mg;n .
It is remarkable that all the known algebraic classes on Mg;n are in the
tautological subalgebra. We illustrate this with two examples.
Example 7.1 (The Hodge bundle). If C is a stable curve of genus g  2,
then H 0 (C; !C ) is a g-dimensional vector space. On the universal example
this gives a rank g vector bundle E over Mg called the Hodge bundle. (Since
H 0(C; !C ) only depends on the (generalized) Jacobian of C, E is the pull-
back of a bundle that is naturally de ned on a certain compacti cation of
the moduli space of principally polarized abelian varieties of dimension g.)
Mumford  expressed the Chern class i := ci (E) 2 H 2i (Mg ) as an element
of Ri (Mg ).
Example 7.2 (The Weierstra loci). Suppose C is a smooth, connected pro-
jective curve C of genus g  2, p 2 C, and an l a positive integer. It is easy
to see that the following conditions are equivalent:
 The linear system jl(p)j is of dimension  1.
 There exists a nonconstant regular function on C fpg that has in p
with a pole of order  l.
 There exists a nite morphism C ! P1 of degree  l that is totally
rami ed in p.
By Riemann-Roch these conditions are always ful lled if l  g + 1. If l = 2,
then the morphism f appearing in the last item must have degree 2 so that
C is hyperelliptic and p is a Weierstra point.
288

These equivalent conditions de ne a closed subvariety Wl of the universal
smooth curve of genus g, Cg . Arbarello, who introduced these varieties, noted
that Wl is irreducible of codimension g + 1 l in Cg . Mumford expressed
the corresponding class in H 2(g+1 l) (Cg ) as an element of Rg+1 l (Cg ).
The validity of Grothendieck's standard conjectures implies that the al-
gebraic classes on Mg;n make up a nondegenerate subspace of H even (Mg;n )
with respect to the intersection product. Since we do not know any algebraic
class that is not tautological, we ask:
Question 7.3. Does R(Mg;n) satisfy Poincar duality?
e
Since all tautological classes originate from Witten classes, this question
could, in principle, be answered for a given pair (g0 ; n0 ) if we would know
all the intersection numbers
Z
k1    kn 2 Q;
n
1
Mg;n
(where it is of course understood that this number is zero if the degree
k1 +    + kn of the integrand fails to equal 3g 3 + n) for all (g; n) with
2g + n  2g0 + n0 . A marvelous conjecture of Witten (which is a conjecture
no longer) predicts the values of these numbers. We shall state it in a form
that exhibits the algebro-geometric content best (this formulation is due
to Dijkgraaf-E. Verlinde-H. Verlinde). For this purpose it is convenient to
renormalize the intersection numbers as follows:
Z
k1    kn ;
[k1 k2    kn ]g := (2k1 + 1)!!(2k2 + 1)!!    (2kn + 1)!! n
1
Mg;n
where (2k + 1)!! = 1:3:5:    :(2k + 1). We use these to form the series in the
variables t0 ; t1 ; t2 ; : : : :
1 1
X X
n! k1 0;;k2 0;:::;kn0[k1 k2    kn ]g tk1 tk2    tkn :
Fg :=
n=1
It is invariant under permutation of variables. The Witten conjectures assert
that these polynomials satisfy a series of di erential equations indexed by
the integers  1. For index 1 this is the string equation:
@Fg = X (2m + 1)t @Fg + 1  t2 ;
m @t
@t0 m1 0;g 0
m1 2
A minicourse of moduli of curves 289

for index 0 the dilaton equation:
@Fg = X (2m + 1)t @Fg + 1  t2
m @t
@t1 m0 1;g 0
m8
and for k  1 we get:
@Fg = X (2m + 1)t @Fg
m @t
@tk+1 m1 m+k
@Fg 1
X
>+ 2
1

m0 +m00 =k 1 @tm @tm
0 00

X @Fg0 @Fg00
X
@tm0 @tm00 :
+2
1

m0 +m00 =k 1 g0 +g00 =g

You may verify that these equations determine the functions Fg completely
(note that Fg has no constant term). The string equation and the dilaton
equation involve a single genus only and were veri ed by Witten using stan-
dard arguments from algebriac geometry. But the equations for k  1 were
proved in an entirely di erent manner: Kontsevich gave an amazing proof
based on a triangulation of Teichmller space on which the intersection num-
u
bers appear as integrals of explicitly given di erential forms. Yet there are
reasons to wish for a proof within the realm of algebraic geometry. The form
of the equations is suggestive in this respect: the rst line involves the Mg;n ,
but the second and third seem to be about intersection numbers formed on
irreducible components of @Mg; n (0 and the f(g0 ;I 0 );(g00 ;I 00 )g respectively).
Consider this a challenge.
7.4. Faber's conjectures. These concern the structure of R(Mg ). From
the de nition it is clear that these are generated by the restrictions of the
Mumford classes. Let us for convenience denote these classes r instead of
r jMg . Faber made the essential part of his conjectures around 1993. We
shall not state them in their most precise form (we refer to  for that).
1. Rg is zero in degree  g 1 and is of dimension one in degree g 2
and the cup product
Ri (M)g)  Rg 2 i(Mg ) ! Rg 2 (Mg )  Q =
is nondegenerate.
2. The classes 1 ; : : : ; [g=3] generate R(Mg ) and satisfy no polynomial
relation in degree  [g=3].
290

Let us review the status of these conjectures. First, there is the evidence
provided Faber himself: he checked his conjectures up to genus 15.
As to conjecture 1: the vanishing assertion was proved in a paper of
mine where it was also shown that dim Rg 2 (Mg )  1. Subsequently Faber
proved that the dimension is in fact equal to one. The rest of conjecture
1 remains open. Conjecture 2 now seems settled: Harer had shown around
the time that Faber made his conjectures that the kappa classes have no
polynomial relations in degree  [g=3] and Morita has recently announced
that 1 ; : : : ; [g=3] generate R(Mg ).
Let me close with saying a bit more about Faber's nonvanishing proof.
He observes that the class g g 1 2 R2g 1 (Mg ) restricts to zero on the
boundary @Mg . This implies that for every u 2 Rg 2 (Mg ), the intersection
R
product Mg g g 1 u only depends on ujMg . So this de nes a `trace' t :
R(Mg ) ! Q . Faber proves that this trace is nonzero on g 2 . The unproven
part of the conjecture can be phrased as saying that the associated form
(u; v) 2 R(Mg )  R(Mg ) 7! t(uv) is nondegenerate. Faber has also an
explicit proposal for the value of the trace on any monomial in the Mumford
classes.
A minicourse of moduli of curves 291

References
 E. Arbarello, M. Cornalba, P.A. Griths, J. Harris: Geometry of Algebraic Curves,
Vol. I, Grundl. der Math. Wiss. 267, Springer 1985.
 J. Harris and I. Morrison: Moduli of Curves, Graduate texts in Math. 187, Springer
1998.
 D. Mumford Towards an enumerative geometry of the moduli space of curves, In
Arithmic and Geometry, II, Prgress in Math. 36, 271{326. Bikhaser 1983.
u
 Moduli of Curves and Abelian varieties, C. Faber and E. Looijenga eds., Aspects of
Mathematics E33, Vieweg 1999.
Moduli of Riemann surfaces, transcendental aspects
Richard Hain

Department of Mathematics, Duke University, Durham, NC 27708, USA

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

LNS001007

hain@math.duke.edu

Contents
LECTURE 1: Low Genus Examples 297
1. Genus 0 298
2. Genus 1 301
2.1. Understanding SL2 (Z)nH 306
2.2. Automorphisms 308
2.3. Families of Genus 1 and Elliptic Curves 309
3. Orbifolds 311
3.1. The Universal Elliptic Curve 316
3.2. Modular Forms 317
LECTURE 2: Teichmller Theory
u 319
4. The Uniformization Theorem 319
5. Teichmller Space
u 321
6. Mapping Class Groups 324
7. The Moduli Space 326
8. Hyperbolic Geometry 327
9. Fenchel-Nielsen Coordinates 328
10. The Complex Structure 330
11. The Teichmller Space Xg;n
u 331
12. Level Structures 332
13. Cohomology 335
LECTURE 3: The Picard Group 337
14. General Facts 337
15. Relations in g 340
16. Computation of H1 ( g ; Z) 346
17. Computation of Picorb Mg 348
References 352
Moduli of Riemann surfaces, transcendental aspects 297

These notes are an informal introduction to moduli spaces of compact
Riemann surfaces via complex analysis, topology and Hodge Theory. The
prerequisites for the rst lecture are just basic complex variables, basic Rie-
mann surface theory up to at least the Riemann-Roch formula, and some
algebraic topology, especially covering space theory. Some good references
for this material include  for complex analysis,  and  for the basic the-
ory of Riemann surfaces, and  for algebraic topology. For later lectures
I will assume more. The book by Clemens  and Chapter 2 of Griths
and Harris  are excellent and are highly recommended. Other useful
references include the surveys  and  and the book .
The rst lecture covers moduli in genus 0 and genus 1 as these can be
understood using relatively elementary methods, but illustrate many of the
points which arise in higher genus. The notes cover more material than was
covered in the lectures, and sometimes the order of topics in the notes di ers
from that in the lectures. I hope to add the material from the last lecture
on the Torelli group and Morita's approach to the tautological classes in a
future version.
Lecture 1: Low Genus Examples
Suppose that g and n are non-negative integers. An n-pointed Riemann
surface (C; x1 ; : : : ; xn ) of genus g is a compact Riemann surface C of genus g
together with an ordered n-tuple of distinct points (x1 ; : : : ; xn ) of C. Two n-
pointed Riemann surfaces (C; x1 ; : : : ; xn ) and (C 0 ; x01 ; : : : ; x0n ) are isomorphic
if there is a biholomorphism f : C ! C 0 such that f(xj ) = x0j when 1  j 
n. The principal objects of study in these lectures are the spaces
 
isomorphism classes of n-pointed compact
Mg;n = Riemann surfaces C of genus g
At the moment all we can say is that these are sets. One of the main
objectives of these lectures is to show that each Mg;n is a complex analytic
variety with very mild singularities.
Later we will only consider Mg;n when the stability condition
2g 2 + n > 0
(1)
is satis ed. But for the time being we will consider all possible values of g
and n. When n = 0, we will simply write Mg instead of Mg;0 .
The space Mg;n is called the moduli space of n-pointed curves (or Riemann
surfaces) of genus g. The isomorphism class of (C; x1 ; : : : ; xn ) is called the
moduli point of (C; x1 ; : : : ; xn ) and will be denoted by [C; x1 ; : : : ; xn ].
298

There are (at least) two notions of the genus of a compact Riemann surface
C. First there is the (analytic) genus
g(C) := dim H 0 (C;
1 );
C
the dimension of the space of global holomorphic 1-forms on C. Second there
is the topological genus
gtop (C) := 1 rank H1 (C; Z):
2
Intuitively, this is the `number of holes' in C. A basic fact is that these are
equal. There are various ways to prove this, but perhaps the most standard
is to use the Hodge Theorem (reference) which implies that
H 1(C; C )  fholomorphic 1-formsg  fanti-holomorphic 1-formsg:
=
The equality of gtop (C) and g(C) follows immediately as complex conjugation
interchanges the holomorphic and antiholomorphic di erentials.
Finally, we shall use the terms \complex curve" and \Riemann surface"
interchangeably.
1. Genus 0
It follows from Riemann-Roch formula that if X is a compact Riemann
surface of genus 0, then X is biholomorphic to the Riemann sphere P1 . So
M0 consists of a single point.
An automorphism of a Riemann surface X is simply a biholomorphism
f : X ! X. The set of all automorphisms of X forms a group Aut X. The
group GL2 (C ) acts in P1 via fractional linear transformations:
 
a b : z 7! az + b
cd cz + d
The scalar matrices S act trivially, and so we have a homomorphism
PGL2 (C ) ! Aut P1
where for any eld F
PGLn(F ) = GLn(F )=fscalar matricesg
and
PSLn(F ) = SLn (F )=fscalar matrices of determinant 1g:
Exercise 1.1. Prove that PGL2(C )  PSL2(C ) and that these are isomor-
=
1.
phic to Aut P
Moduli of Riemann surfaces, transcendental aspects 299

Exercise 1.2. Prove that Aut P1 acts 3-transitively on P1 . That is, given
any two ordered 3-tuples (a1 ; a2 ; a3 ) and (b1 ; b2 ; b3 ) of distinct points of P1 ,
there is an element f of Aut P1 such that f(aj ) = bj for j = 1; 2; 3. Show
that f is unique.
Exercise 1.3. Prove that if X is a compact Riemann surface of genus 0,
then X is biholomorphic to the Riemann sphere.
Exercise 1.4. Show that the automorphism group of an n-pointed curve
of genus g is nite if and only if the stability condition (1) is satis ed.
(Depending on what you know, you may nd this a little dicult at present.
More techniques will become available soon.)
Since Aut P1 acts 3-transitively on P1 , we have:
Proposition 1.5. Every n-pointed Riemann surface of genus 0 is isomor-
phic to
(P1 ; 1) if n = 1;
1 ; 0; 1) if n = 2;
(P
1 ; 0; 1; 1) if n = 3:
(P

Corollary 1.6. If 0  n  3, then M0;n consists of a single point.
The rst interesting case is when n = 4. If (X; x1 ; x2 ; x3 ; x4 ) is a 4-pointed
Riemann surface of genus 0, then there is a unique biholomorphism f : X !
P1 with f(x2 ) = 1, f(x3 ) = 0 and f(x4 ) = 1. The value of f(x1 ) is forced
by these conditions. Since the xj are distinct and f is a biholomorphism,
f(x1) 2 C f0; 1g. It is therefore an invariant of (X; x1 ; x2 ; x3; x4 ).
Exercise 1.7. Show that if g : X ! P1 is any biholomorphism, then f(x1)
is the cross ratio
(g(x1 ) : g(x2 ) : g(x3 ) : g(x4 ))
of g(x1 ), g(x2 ), g(x3 ), g(x4 ). Recall that the cross ratio of four distinct
points x1 , x2 , x3 , x4 in P1 is de ned by
(x x )=(x x )
(x1 : x2 : x3 : x4 ) = (x1 x3 )=(x2 x3 )
1 4 2 4
The result of the previous exercise can be rephrased as a statement about
moduli spaces:
Proposition 1.8. The moduli space M0;4 can be identi ed naturally with
f0; 1g. The moduli point [P1 ; x1 ; x2; x3 ; x4 ] is identi ed with the cross
C
ratio (x1 : x2 : x3 : x4 ) 2 C f0; 1g.
300

It is now easy to generalize this to general n  4. Since every genus 0
Riemann surface is biholomorphic to P1 , we need only consider n-pointed
curves of the form (P1 ; x1 ; : : : ; xn ). There is a unique automorphism f of P1
such that f(x1 ) = 0; f(x2 ) = 1 and f(x3 ) = 1. So every n-pointed Riemann
surface of genus 0 is isomorphic to exactly one of the form
(P1 ; 0; 1; 1; y1 ; : : : ; yn 3 ):
To say that this is an n-pointed curve is to say that the points 0; 1; 1,
y1; : : : ; yn 3 are distinct. That is,
(y1 ; : : : ; yn 3 ) 2 (C f0; 1g)n 3 
where  = [j<k jk is the union of the diagonals
jk = f(y1 ; : : : ; yn 3 ) : yj = yk g:
This is an ane algebraic variety as it is the complement of a divisor in an
ane space. This shows that:
Theorem 1.9. If n  3, then M0;n is a smooth ane algebraic variety of
dimension n 3 isomorphic to
f0; 1g)n 3 :
(C
The symmetric group n acts on M0;n by
 : (P1 ; x1 ; : : : ; xn) 7! (P1 ; x1 ; : : : ; xn ):
Exercise 1.10. Show that each  2 n acts on M0;n as a regular mapping.
(Hint: it suces to consider the case of a transposition.)
Exercise 1.11. Suppose that n  3. Construct a universal n-pointed
genus 0 curve M0;n  P1 ! M0;n that is equipped with n disjoint sections
1 ; : : : ; n such that
j ([P1 ; x1 ; : : : ; xn ]) = ([P1 ; x1 ; : : : ; xn]; xj ):
Show that it is universal in the sense that if f : X ! T is a family of
smooth genus 0 curves over a smooth variety T and if the family has n
sections s1 ; : : : ; sn that are disjoint, then there is a holomorphic mapping
f : T ! M0;n such that the pullback of the universal family is f and the
pullback of j is sj .
Moduli of Riemann surfaces, transcendental aspects 301

2. Genus 1
The study of the moduli space of genus 1 compact Riemann surfaces is
very rich and has a long history because of its fundamental connections to
number theory and the theory of plane cubic curves. We will take a tran-
scendental approach to understanding M1 which will reveal the connection
with modular forms. Our rst task is show that genus 1 Riemann surfaces
can always be represented as the quotient of C by a lattice.
One way to construct a Riemann surface of genus 1 is to take the quotient
of C by a lattice. Recall that a lattice in a nite dimensional real vector
space V is a nitely generated (and therefore free abelian) subgroup  of
V with the property that a basis of  as an abelian group is also a basis
of V as a real vector space. A lattice in C is thus a subgroup  of C that
is isomorphic to Z2 and is generated by two complex numbers that are not
real multiples of each other.
Exercise 2.1. Show that if  is a lattice in V and if dimR V = d, then V=
is a compact manifold of real dimension d, which is di eomorphic to the
d-torus (R=Z)d .
If  is a lattice in C , the quotient group C = is a compact Riemann
surface which is di eomorphic to the product of two circles, and so of genus
1.
Theorem 2.2. If C is a compact Riemann surface of genus 1, then there is
a lattice  in C and an isomorphism  : C ! C =. If xo 2 C, then we may
choose  such that (xo ) = 0.
The proof follows from the sequence of exercises below. Let C be a com-
pact Riemann surface of genus 1.
Exercise 2.3. Show that a non-zero holomorphic di erential on C has no
zeros. Hint: Use Riemann-Roch.
Since gtop (C) = g(C) = 1, we know that H1 (C; Z) is free of rank 2.
Fix a non-zero holomorphic di erential w on C. Every other holomorphic
di erential is a multiple of w. The period lattice of C is de ned to be
Z

w : c 2 H1 (C; Z) :
=
c
This is easily seen to be a subgroup of C .
302

Exercise 2.4. Show that  is a Rlattice in C . Hint: Choose a basis a; b of
R
H1(C; Z). Show that if a w and b w are linearly independent over R, then
this would contradict the Hodge decomposition of H 1 (C; C ).
Let E = C =. Our next task is to construct a holomorphic mapping from
C to E.
Exercise 2.5. Fix a base point xo of C. De ne a mapping  : C ! E by
Z
(x) = w

where
is any smooth path in C that goes from xo to x. Show that
(i)  is well de ned;
(ii)  is holomorphic;
(iii)  has nowhere vanishing di erential, and is therefore a covering map;
(iv) the homomorphism  : 1 (C; xo ) ! 1 (E; 0)   is surjective, and
=
therefore an isomorphism.
Deduce that  is a biholomorphism.
This completes the proof of Theorem 2.2. It has the following important
consequence:
Corollary 2.6. If C is a compact Riemann surface of genus 1, then the
automorphism group of C acts transitively on C. Consequently, the natural
mapping M1;1 ! M1 that takes [C; x] to [C] is a bijection.
Proof. This follows as every genus 1 Riemann surface is isomorphic to one
of the form C =. For such Riemann surfaces, we have the homomorphism
C = ! Aut(C =)

that takes the coset a +  to the translation z +  7! z + a + .
De nition 2.7. An elliptic curve is a a genus 1 curve C together with a
point xo 2 C.
The previous result says that if C is a genus 1 curve and xo and yo are
points of C, then the elliptic curves (C; xo ) and (C; yo ) are isomorphic.
The moduli space of elliptic curves is M1;1 .
Exercise 2.8. Suppose that f : C ! C = is a holomorphic mapping from
an arbitrary Riemann surface to C =. Let xo be a base point of C. The 1-
form dz on C descends to a holomorphic di erential w on C =. Its pullback
Moduli of Riemann surfaces, transcendental aspects 303

f w is a holomorphic di erential on C. Z that for all x 2 C,
Show
f(x) = f(xo) + f w + 

where
is a path in C from xo to x.
Exercise 2.9. Use the results of the previous exercise to prove the following
result.
Corollary 2.10. If 1 and 2 are lattices in C , then C =1 is isomorphic
to C =2 if and only if there exists  2 C  such that 1 = 2 .
Exercise 2.11. Show that if (C; xo ) is an elliptic curve, then C has a natural
group structure with identity xo .
We are nally ready to give a construction of M1 . Recall that the complex
structure on a Riemann surface C gives it a canonical orientation. This can
be thought of as giving a direction of \positive rotation" about each point
in the surface | the positive direction being that given by turning counter-
clockwise about the point in any local holomorphic coordinate system. If C
is compact, this orientation allows us to de ne the intersection number of
two transversally intersecting closed curves on C. It depends only on the
homology classes of the two curves and therefore de nes the intersection
pairing
h ; i : H1(C; Z)
H1(C; Z) ! Z:
If , is a basis of H1 (C; Z), then h ; i = 1. We shall call the basis
positive if h ; i = 1.
A framing of Riemann surface C of genus 1 is a positive basis ; of
its rst homology group. We will refer to (C : ; ) as a framed genus 1
Riemann surface. Two framed genus 1 Riemann surfaces (C : ; ) and
(C 0 : 0 ; 0 ) are isomorphic if there is a biholomorphism f : C ! C 0 such
that 0 = f and 0 = f .
Let  
isomorphism classes of framed :
X1 = Riemann surfaces of genus 1
At the moment, this is just a set. But soon we will see that it is itself a
Riemann surface. Note that forgetting the framing de nes a a function
: X1 ! M1 :
Denote the isomorphism class of (C : ; ) by [C : ; ]. If C is a genus 1
Riemann surface, then
1 ([C]) = f[C : ; ] : ( ; ) is a positive basis of H1(C; Z)g:
304

Exercise 2.12. Show that if ( ; ) and ( 0 ; 0 ) are two positive bases of
H1(C; Z), then there is a unique element
 
ab
cd
of SL2 (Z) such that  0   
=ab
0 cd
(The reason for writing the basis vectors in the reverse order will become
apparent shortly.)
De ne an action of SL2 (Z) on X1 by
 
a b [C : ; ] = [C : 0 ; b0 ]
cd
where  0   
=ab
0 cd
Exercise 2.13. Show that there is a natural bijection
M1  SL2(Z)nX1:
=
At present, X1 is just a set, but we now show that it is naturally a Riemann
surface. We know from Theorem 2.2 that every element of X1 is of the
form [C = : ; ]. But, by standard algebraic topology, there is a natural
isomorphism
  H1 (C; Z):
=
Thus a basis of H1 (C; Z) corresponds to a basis of .
Exercise 2.14. Suppose that ; is a basis of H1(C =; Z) and that !1; !2
is the corresponding basis of . Show that ; is positive if and only if
!2=!1 has positive imaginary part.
It follows from this and Corollary 2.10 that
X1 = f[C = : !1; !2] : Im(!2=!1) > 0g=C 
where the C  -action is de ned by
  [C = : !1 ; !2 ] = [C = : !1 ; !2 ]:
We can go even further: since the basis !1 , !2 determines the lattice,
 = Z!1  Z!2;
Moduli of Riemann surfaces, transcendental aspects 305

we can dispense with the lattice altogether. We have:
X1 = f(!1 ; !2) : !1; !2 2 C and Im(!2 =!1) > 0g=C 
(2)
where C  acts on (!1 ; !2 ) by scalar multiplication.
Denote the upper half plane fz 2 C : Im z > 0g by H . Each  2 H
determines the element
[C =(Z  Z ) : 1; ]
of X1 . This de nes a function : H ! X1 . Under the identi cation (2),
() = the C  -orbit of (1; )
Since (!1 ; !2 ) and (1; !2 =!1 ) are in the same orbit, we have proved:
Theorem 2.15. The function : H ! X1 is a bijection.
The group PSL2 (C ) acts on P1 by fractional linear transformations.
Exercise 2.16. Show that T 2 PSL2(C ) satis es T(H )  H if and only if
T 2 PSL2(R).
Thus the group SL2 (R) acts on H by fractional linear transformations:
 
a b  = a + b
cd c + d
Exercise 2.17. Show that : H ! X1 is SL2(Z)-equivariant. That is, if
T 2 SL2(Z), then T () = (T ).
Theorem 2.18. There are natural bijections
M1  M1;1  SL2(Z)nH :
= =
Proof. The rst bijection was established in Corollary 2.6. The second fol-
lows from Exercise 2.13, Theorem 2.15 and Exercise 2.17.
Exercise 2.19. Suppose that C is a genus 1 Riemann surface. Show that
the point of SL2 (Z)nH that corresponds to [C] 2 M1 is the SL2 (Z) orbit of
Z Z 
w 2H
w

is any non-zero element of H 0 (C;
1 ) and
where w , is a positive basis
C
of H1 (C; Z).
306

2.1. Understanding SL2 (Z)nH . The Riemann surface structure on H de-
scends to a Riemann surface structure on SL2 (Z)nH . Good references for
this are Chapter VII of Serre's book , and Chapter 3 of Clemens' book
. We'll sketch part of the proof of the following fundamental theorem.
Theorem 2.20. The quotient of H by SL2(Z) has a unique structure of a
Riemann surface such that the projection H ! SL2 (Z)nH is holomorphic.
Moreover, there is a biholomorphism between SL2 (Z)nH and C which can be
given by the modular function j : H ! C , where
j() = 1 + 744 + 196 884 q + 21 493 760 q2 +   
q
and q = e2i .
The following exercises will allow you to construct most of the proof. The
rest can be found in  and .
Let P1 (R) = R [f1g. This is a circle on the Riemann sphere which forms
the boundary of H . Let H be the closure of H in the Riemann sphere P1 ; it is
the union of H and P1 (R). Recall that every non-trivial element of PSL2 (C )
has at most two xed points in P1 . Note that the xed points of elements of
PSL2(R) are real or occur in complex conjugate pairs.
Exercise 2.21. Suppose that T 2 SL2(Z) is not a scalar matrix. Show that
T has exactly
(i) one xed point in H if and only if j tr Tj < 2;
(ii) one xed point in P1 (R) if and only if j tr Tj = 2;
(iii) two xed points in P1 (R) if and only if j tr Tj > 2.
Show that T 2 SL2 (R) has nite order if and only if T has a xed point in
H.

Fix an integer l  0. The level l subgroup of SL2 (Z) is the subgroup of
SL2(Z) consisting of those matrices congruent to the identity mod l. We
shall denote it by SL2 (Z)[l]. Since it is the kernel of the homomorphism
SL2 (Z) ! SL2(Z=l), it is normal and of nite index in SL2 (Z).
Exercise 2.22. Show that SL2(Z)[l] is torsion free for all l  3. Hint: use
the previous exercise. Deduce that SL2 (Z)[l]nH is a Riemann surface with
fundamental group SL2 (Z)[l] and universal covering H whenever l  3.
The quotient of SL2 (Z) by its level l subgroup is SL2 (Z=l), which is a
nite group. It follows that the projection
SL2 (Z)[l]nH ! SL2 (Z)nH
Moduli of Riemann surfaces, transcendental aspects 307

is nite-to-one of degree equal to the half order of SL2 (Z=l) when l > 2.
Exercise 2.23. Show that the quotient GnC of a Riemann surface C by
a nite subgroup G of Aut C has the structure of a Riemann surface such
that the projection C ! GnC is holomorphic. Hint: rst show that for each
point x of C, the isotropy group
Gx := fg 2 G : gx = xg
is cyclic. This can be done by considering the actions of Gx on Tx X and
OX;x.
This result, combined with the previous exercises, establishes that M1 =
SL2(Z)nH has a natural structure of a Riemann surface such that the pro-
jection H ! M1 is holomorphic.
Serre [24, p. 78] proves that a fundamental domain for the action of SL2 (Z)
on H is the region
F = f 2 H : j Re j  1=2; jj  1g:
Points of F can be thought of as giving a canonical framing of a lattice 

i

ПЃ2 ПЃ

1
-1 -1/2 0 1/2

Figure 1. A a fundamental domain of SL2 (Z) in H

in C . Such a framing is given as follows: the rst basis element is a non-zero
vector of shortest length in , the second basis element is a shortest vector
in  that is not a multiple of the rst.
308

Exercise 2.24. Show that  2 F if and only if (1; ) is such a canonical
basis of the lattice Z  Z .
Serre [24, p. 78] also proves that PSL2 (Z) is generated by  7! 1= and
7!  + 1.
Exercise 2.25. Use this to prove that SL2(Z)nH is the quotient of the
fundamental domain F obtained by identifying the opposite vertical sides,
and by identifying the arc of the circle jj = 1 from  to i with the arc from
2 to i. Deduce that M1 = SL2 (Z)nH is a Riemann surface homeomorphic
to a disk.
It is conceivable that M1 is biholomorphic to a disk, for example. But
this is not the case as M1 can be compacti ed by adding one point.
Exercise 2.26. Show that M1 can be compacti ed by adding a single point
1. A coordinate neighbourhood of 1 is the unit disk . Denote the
holomorphic coordinate in it by q. The point  of M1 = SL2 (Z)nH is
identi ed with the point e2i of . Show that
M1 [ f1g
is a compact Riemann surface of genus 0 where q is a local parameter about
1 and where M1 holomorphically embedded. Deduce that M1 is biholo-
morphic to C .
Remark 2.27. This compacti cation is the moduli space M1;1 .
Since every compact Riemann surface is canonically a complex algebraic
curve, this shows that M1 is an algebraic variety.
2.2. Automorphisms. The automorphisms of an elliptic curve are inti-
mately related with the set of elements of SL2 (Z) that stabilize the points
corresponding to it in H .
Exercise 2.28. Suppose that C is a genus 1 Riemann surface curve. Sup-
pose that  : C ! C is an automorphism of C. Show that
(i)  is a translation if and only if  : H1 (C; Z) ! H1 (C; Z) is the identity;
(ii) if ; is a positive basis of H1 (C; Z), then
[C : ; ] = [C :  ;  ] 2 X1 ;
Deduce that there is a natural isomorphism
Aut(C; xo )  stabilizer in SL2 (Z) of [C : ; ] 2 X1 :
=
Moduli of Riemann surfaces, transcendental aspects 309

Note that any two points of H that lie in the same orbit of SL2 (Z) have
isomorphic stabilizers. Consequently, to nd all elliptic curves with auto-
morphism groups larger than Z=2Z, one only has to look for points in the
fundamental domain with stabilizers larger than Z=2Z.
Exercise 2.29. Show that the stabilizer in SL2(Z) of  2 H is
8
>Z=2Z  2 orbit of i and ;
=
<
Z=4Z  2 orbit of i;
>
Z=6Z  2 orbit of :
:

An immediate consequence of this computation is the following:
Theorem 2.30. If (C; xo) is an elliptic curve, then
8
>Z=4Z if (C; xo )  (C =Z[i]; 0)
=
<
Aut(C; xo )  >Z=6Z if (C; xo )  (C =Z[]; 0)
= =
:
Z=2Z otherwise.

It is easy to see from this, for example, that the Fermat cubic
x3 + y3 + z3 = 0
in P2 has automorphism group Z=6Z and therefore is isomorphic to Z=Z[].
2.3. Families of Genus 1 and Elliptic Curves. Suppose that X is a
complex analytic manifold and that f : X ! T is a holomorphic mapping
to another complex manifold each of whose bers is a genus 1 curve. We say
that f is a family of genus 1 curves. If there is a section s : T ! X of f,
then the ber of f over t 2 T is an elliptic curve with identity s(t). We say
that f is a family of elliptic curves.
Each such family gives rise to a function
f : T ! M1
that is de ned by taking t 2 T to the moduli point [Xt ] of the ber Xt of f
over t. We shall call it the period mapping of the family.
Theorem 2.31. The period map f is holomorphic.
Sketch of Proof. For simplicity, we suppose that T has complex dimension 1.
The relative holomorphic tangent bundle of f is the holomorphic line bundle
Tf0 on X consisting of holomorphic tangent vectors to X that are tangent to
the bers of f. That is,
Tf0 = kerfT 0 X ! T 0Tg:
310

The sheaf of holomorphic sections of its dual is called the relative dualizing
sheaf and is often denoted by !X=T . The push forward f !X=T of this sheaf
to T is a holomorphic line bundle over T and has ber
H 0 (Xt ;
1 )
X t

over t 2 T. Fix a reference point to 2 T. Let w(t) be a local holomorphic
section of !X=T de ned in a contractible neighbourhood U of to . Since bundle
is locally topologically trivial, f 1 (U) is homeomorphic to U  Xt . We can
thus identify H1 (Xt ; Z) with H1 (Xt ; Z) for each t 2 U. Fix a positive basis
o

o

; of H1(Xt ; Z). This can be viewed as a positive basis of H1(Xt ; Z) for
all t 2 U.
o

It is not dicult to show Z that Z
w(t) and w(t)

vary holomorphically with t 2 U. It follows that
Z Z 
(t) := w(t) w(t)

varies holomorphically with t 2 U. This shows that map f is holomorphic
in the neighbourhood U of to . It follows that f is holomorphic.
If you examine the proof, you will see that we really proved two extra
facts. First, the period mapping f : T ! M1 associated to every family
f : X ! T of genus 1 curves is locally liftable to a holomorphic mapping to
H . If, in addition, f is a family of elliptic curves, then the period mapping
f determines the family f and the section s:
Proposition 2.32. The period mapping f associated to a family f : X !
T of genus 1 curves is locally liftable to a holomorphic mapping to H . If f
is a family of elliptic curves, then f can be globally lifted to a holomorphic
˜˜
mapping f : T ! H and there is a homomorphism f : 1 (T; to ) ! SL2 (Z)
(unique up to conjugacy) such that the diagram
˜˜
T ! H f

? ?
? ?
y y

! M1
T
f

commutes and such that
˜ ˜
f (
 x) = f (
)  f (x)
Moduli of Riemann surfaces, transcendental aspects 311

˜
for all x 2 T and all
2 1 (T; to ).
The proof is left as an exercise. We shall see in a moment that the converse
of this result is also true.
This result has an important consequence | and that is that not every
holomorphic mapping T ! M1 is the period mapping of a holomorphic
family of elliptic (or even genus 1) curves. The reason for this is that not
every mapping T ! M1 is locally liftable.
Exercise 2.33. Show that the identity mapping M1 ! M1 is not locally
liftable. Deduce that there is no family of genus 1 curves over M1 whose
period mapping is the identity mapping. In particular, show that there is
no universal elliptic curve over M1 .
To each genus 1 curve C, we can canonically associate the elliptic curve
Jac C := Pic0 C which we shall call the jacobian of C. Abel's Theorem tells
us that C and Jac C are isomorphic as genus 1 curves, but the isomorphism
depends on the choice of a base point of C.
Exercise 2.34. Show that for each family f : X ! T of genus 1 curves the
corresponding family of jacobians is a family of elliptic curves. Show that
these two families have the same period mapping. Show that if f is a family
of elliptic curves, then the family of jacobians is canonically isomorphic to
the original family f.
3. Orbifolds
The discussion in the previous section suggests that M1 should be viewed
as SL2 (Z)nH rather than as C | or that M1 is not an algebraic variety
or a manifold, but rather something whose local structure includes the in-
formation of how it is locally the quotient of a disk by a nite group. In
topology such objects are called orbifolds and in algebraic geometry stacks.
Very roughly speaking, orbifolds are to manifolds as stacks are to varieties.
The moduli spaces Mg;n are often conveniently viewed as orbifolds or as
stacks.
For us, an orbifold is a topological space that is the quotient of a simply
connected topological space X by a group . The group is required to act
properly discontinuously on X and all isotropy groups are required to be
nite.1 For example, M1 can be viewed as an orbifold as the quotient of H
by SL2 (Z).
We could have added some other natural conditions, such as requiring that there be
1

a nite index subgroup of that acts on X xed point freely. We could also require that
312

This de nition is not as general as it could be, but since all of our moduli
spaces Mg;n are of this form, it is good enough for our purposes. The general
de nition is obtained by \shea fying" this one | i.e., general orbifolds are
locally the quotient of a simply connected space by a nite group. A general
de nition along these lines can be found in Chapter 13 of . Mumford's
de nition of stacks can be found in .
A morphism f : 1 nX1 ! 2 nX2 of orbifolds is a continuous map that
arises as follows: there is a homomorphism f : 1 ! 2 and a continuous
mapping f˜ : X1 ! X2 that is equivariant with respect to f ; that is, the
diagram

! X2
X1
? ?
g? ?
yf g
y

! X2
X1
commutes for all g 2 1 .
At this stage, I should point out that every reasonable topological space
˜
X can be viewed as the quotient 1 (X; x)nX of its universal covering by its
fundamental group. In this way, every topological space can be regarded as
an orbifold. It thus it makes sense to talk about orbifold mappings between
orbifolds and ordinary topological spaces.
An orbifold nX can have an enriched structure | such as a smooth, Rie-
mannian, Khler, or algebraic structure. One just insists that its \orbifold
a
universal covering" X has such a structure and that act on X as automor-
phisms of this structure. Maps between two orbifolds with the same kind
of enriched structure are de ned in the obvious way. For example, a map
between two orbifolds with complex structures is given by an equivariant
holomorphic map between their orbifold universal coverings.
Many orbifolds are given as quotients of non-simply connected spaces by
a group that acts discontinuously, but not xed point freely. Such quotients
have canonical orbifold structures: If M is a topological space and G a group
that acts discontinuously on M, then
GnM  nM
=f

the set of elements of G that act trivially on X is central in . Both of these conditions
are natural and are satis ed by all of our primary examples of orbifolds, the M .
g;n
Moduli of Riemann surfaces, transcendental aspects 313

where p : M ! M is the universal covering2 of M and
f

= f(; g) : where  : M ! M covers g 2 Gg:
f f

Here  covers g 2 G means that the diagram

!M
M
f f
? ?
p? ?p
y y

!M
M g
commutes.
There is a notion of the orbifold fundamental group 1 ( nX; xo ) of a con-
orb
nected pointed orbifold. It is a variant of the de nition of the fundamental
group of a pointed topological space.
First, we'll x the convention that the composition of two paths
and in a space is de ned when (1) = (0). The product path is the one
obtained by traversing rst, then . This is the convention used in ,
for example, and is the opposite of the convention used by many algebraic
geometers such as Deligne.
We need to impose several conditions on X and on xo in order for the
de nition to make sense. Let p : X ! nX be the projection. Note that
if x; y 2 p 1 (xo ), then the isotropy groups y and z are conjugate in .
Our rst assumption is that y is contained in the center of for one (and
hence all) y 2 p 1 (xo ). For such xo , we shall denote the common isotropy
group of the points lying over xo by x . Our second assumption is that X
o

is connected, locally path connected, and locally simply connected. These
conditions are satis ed by all connected complex algebraic varieties. Finally,
we shall assume that if g 2 acts trivially in the neighbourhood of some
point of X, it acts trivially on all of X. All three of these conditions are
natural and will be satis ed in cases of interest to us.
Let
P(xo ) = fhomotopy classes of paths ([0; 1]; f0; 1g) ! (X; p 1 (xo ))g:
Since acts on the left of X, this is a left -set. Let
Q(xo ) = f(g;
) 2  P(xo) : g 1 
(0) =
(1)g:
Of course, I am assuming that M is nice enough as a topological space to have a uni-
2

versal covering. We assume, for example, that M is locally simply connected, a condition
satis ed by all complex algebraic varieties and all manifolds.
314

This has a natural left -action given by
g : (h;
) 7! (ghg 1 ; g 
):
De ne 1 ( nX; xo ) to be the quotient nQ(xo ). This has a natural group
orb
structure which can be understood by noting that Q(xo ) is a groupoid. The
composition of two elements (g;
) and (h; ) is de ned when
(1) = (0).
It is then given by
(g;
)  (h; ) = (gh;
):
orb
To multiply two elements of 1 (GnX; xo ), translate one of them until they
are composable. For example, if we x a point xo of p 1 (xo ), then each
˜
orb (GnX; xo ) has a representative of the form (g;
) where
(0) =
element of 1
xo . This representation is unique up to translation by an element of x .
˜ o

To multiply two elements (g;
) and (h; ) starting at xo , multiply (h; ) by
˜
g so that it can be composed with (g;
). Then the product of these two
orb
elements in 1 (GnX; xo ) is represented by the path
(g;
)  (g 1 hg; g 1  ) = (hg;
(g 1  )):
Exercise 3.1. Show that if k 2 x , then k acts trivially on P(xo). Deduce
o

that the multiplication above is well de ned. (This is where we need x to o

be central in .)
The following exercises should help give some understanding of orbifold
fundamental groups.
Exercise 3.2. Show that if acts xed point freely on X, then there is
a natural isomorphism between 1 ( nX; xo ) and the usual fundamental
orb
group 1 ( nX; xo ). In particular, if we view a topological space X as an
orbifold, then the two notions of fundamental group agree.
Exercise 3.3. Show that if is an abelian group that acts trivially on a one
point space X = fg, then 1 ( nX; xo ) is de ned and there is a natural
orb
isomorphism
1 ( nX; xo )  :
orb =
Exercise 3.4. Let X = C and = Z=nZ. De ne an action on X by
letting the generator 1 of Z=nZ act by multiplication by e2i=n . Show that
1 ( nX; xo ) is de ned for all xo 2 nC , and for each xo there is a natural
orb
isomorphism
1 ( nX; xo )  Z=nZ:
orb =
Moduli of Riemann surfaces, transcendental aspects 315

Exercise 3.5. Show that for each choice of a point x 2 p 1(xo ), there is a
natural isomorphism
orb
x : 1 ( nX; xo ) !
de ned by x (g;
) = g 1 when
(0) = x. Show that if y = hx, then
y = hx h 1 .
Remark 3.6. One can also de ne the orbifold fundamental group of nX
using the Borel construction. First, nd a contractible space E on which
acts discontinuously and xed point freely (any one will do). There is a
canonical construction of such spaces. (See for example [3, p. 19].) Fix a
base point eo 2 E . The diagonal action of on E X, a simply connected
space, is xed point free and the map
q : E  X ! n(E  X)
is a covering mapping with Galois group . For x 2 X, de ne
orb
1 ( nX; q(eo ; x)) = 1 ( (E  X); q(eo ; x)):
If x is central in , then this depends only on xo = p(x). It is not dicult
to show that, in this case, this de nition agrees with the more elementary
one given above.
The moduli space M1 is naturally an orbifold, being the quotient of H
by SL2 (Z). The condition on xo is satis ed for all points not in the orbit
orb
of i or . Consequently, 1 (M1 ; xo ) is de ned for all xo other than those
corresponding to the orbits of i and , and for each such xo , there is an
isomorphism
1 (M1 ; xo )  SL2 (Z)
orb =
which is well de ned up to conjugacy.
This gives the following restatement of Proposition 2.32.
Proposition 3.7. If f : X ! T is a holomorphic family of genus 1 curves,
then the period mapping f : T ! M1 is a morphism of orbifolds.
Remark 3.8. Since M1 can also be written as PSL2 (Z)nH , it is natural
to ask why we are not giving M1 this orbifold structure. This question
will be fully answered in the section on the universal elliptic curve and in
subsequent sections on curves of higher genus. For the time being, just note
that if we give M1 the orbifold structure PSL2 (Z)nH , then orb (M1 ) would
be PSL2 (Z) instead of SL2 (Z).
316

3.1. The Universal Elliptic Curve. Let's attempt to construct a \uni-
versal elliptic curve" over M1 . We begin by constructing one over H . The
group Z2 acts on C  H by
(n; m) : (z; ) 7! (z + n + m; ):
This action is xed point free, so the quotient Z2n(C  H ) is a complex
manifold. The ber of the projection
Z2n(C  H ) ! H

over  is simply the elliptic curve C =ZZ. The family has the section that
takes  2 H to the identity element of the ber lying over it. So this really
is a family of elliptic curves.
Let's see what happens if we try to quotient out by Z2 and SL2 (Z) at
the same time. First note that SL2 (Z) acts on Z2 on the right by matrix
multiplication. We can thus form the semi-direct product SL2 (Z)nZ2. This
is the group whose underlying set is SL2 (Z)  Z2 and whose multiplication
is given by
(A1 ; (n1 ; m1 ))(A1 ; (n2 ; m2 )) = (A1 A2 ; (n1 ; m1 )A2 + (n2 ; m2 )):
Exercise 3.9. Show that the action of SL2(Z) n Z2 on C  H given by
    
a b ; (n; m) : (z; ) 7! z + n + m ; a + b
cd c + d c + b
is well de ned. Show that there is a well de ned projection
(SL2 (Z) n Z2)n(C  H ) ! SL2 (Z)nH :
This is a reasonable candidate for the universal curve. But we should be
careful.
Exercise 3.10. Show that the ber of the natural projection
(SL2 (Z) n Z2)n(C  H ) ! SL2 (Z)nH
(3)
over the point corresponding to the elliptic curve (C; 0) is the quotient of C
by the nite group Aut(C; 0). In particular, the ber of the projection over
[(C; 0)] is always a quotient of C=  P1 , and is never C.
=
However, it is more natural to regard E := (SL2 (Z) n Z2) n H as an
orbifold. We shall do this. First note that the projection (3) has an orbifold
section that is induced by the mapping
!C H
H
Moduli of Riemann surfaces, transcendental aspects 317

that takes  to (0; ). Note that if we consider M1 as the orbifold PSL2 (Z)nH ,
then the section would not exist. It is for this reason that we give M1
the orbifold structure with orbifold fundamental group SL2 (Z) instead of
PSL2(Z).
The following should illustrate why it is more natural to view M1 as an
orbifold than as a variety.
Theorem 3.11. There is a natural one-to-one correspondence between holo-
morphic orbifold mappings from a smooth complex curve (or variety) T to
M1 and families of elliptic curves over T. The correspondence is given by
pullback.
We could de ne a `geometric points' of an orbifold X to be an orbifold map
from a point with trivial fundamental group to X. For example, suppose
is any point of H . Denote its isotropy group in SL2 (Z) by  . Then  nfg
is a point of SL2 (Z)nH , but is not a geometric point as it has fundamental
group  , which is always non-trivial as it contains Z=2Z. The corresponding
geometric point corresponds to the `universal covering' fg !  nfg of this
point. One can de ne pullbacks of orbifolds in such a way that the ber of
the pullback of the universal curve to the geometric point  of M1 is the
corresponding elliptic curve C =(Z  Z ).
Denote the n-fold bered product of E ! M1 with itself by E n ! M1 .
It is naturally an orbifold (exercise). This has divisors jk that consist of
the points of E n where the jth and kth points agree. It also has divisors
j where the jth point is zero.
Exercise 3.12. Show that the moduli space M1;n can be identi ed with
[ [ 
E n 1 jk [ j :
j
j<k
3.2. Modular Forms. There is a natural orbifold line bundle L over M1 .
It is the quotient of H  C by SL2 (Z) where SL2 (Z) acts via the formula
   
a b : (; z) ! a + b ; (c + d)z :
cd c + d
Exercise 3.13. Show that this is indeed an action. Show that the kth
power L
k of this line bundle is the quotient of H  C by the action
   
a b : (; z) ! a + b ; (c + d)k z :
cd c + d
318

Orbifold sections of L
k correspond to holomorphic functions f : H ! C
for which the mapping
H !H C  7! (; f())
is SL2 (Z)-equivariant.
Exercise 3.14. Show that f : H ! C corresponds to a section of L
k if
and only if  
a b  = (c + d)k f():
f cd
Such a function is called a modular form of weight k for SL2 (Z).
Exercise 3.15. Show that f : H ! C is a modular form of weight 2k if
and only if the k-di erential (i.e., section of the kth power of the canonical
bundle)
f()d
k
is invariant under SL2 (Z). Deduce that the (orbifold) canonical bundle of
M1 is isomorphic to L
2.
Some basic properties and applications of modular forms are given in
Chapter VII of .
Just as one can de ne the Picard group of a complex analytic variety to
be the group of isomorphism classes of holomorphic line bundles over it, one
can de ne the Picard group Picorb X of a holomorphic orbifold X as the
group of isomorphism classes of orbifold line bundles over X. The details
are left as a straightforward exercise. The following result will be proved
later in the lectures.
Theorem 3.16. The Picard group of the orbifold M1 is cyclic of order 12
and generated by the class of L.
It is easy to see that the class of L in Picorb M1 as L
12 is trivialized by
the cusp form
1
12 q Y (1 qn )24
(q) = (2)
n=1
of weight 12 which has no zeros or poles in H . Also, the non-existence of
meromorphic modular forms of weight k with 0 < k < 12 and no zeros or
poles in H implies that L is an element of order 12 in Picorb M1 . We will
show later that the order of Picorb M1 is of order 12, from which it will
follow that Picorb M1 is isomorphic to Z=12 and generated by the class of
L.
Moduli of Riemann surfaces, transcendental aspects 319


Lecture 2: Teichmuller Theory
We have seen in genus 1 case that M1 is the quotient
1 nX1
of a contractible complex manifold X1 = H by a discrete group 1 = SL2 (Z).
The action of 1 on X1 is said to be virtually free | that is, 1 has a nite
index subgroup which acts ( xed point) freely on X1 .3 In this section we
will generalize this to all g  1 | we will sketch a proof that there is a
contractible complex manifold Xg , called Teichmller space, and a group
u
g , called the mapping class group, which acts virtually freely on Xg . The
moduli space of genus g compact Riemann surfaces is the quotient:
Mg = g nXg :
This will imply that Mg has the structure of a complex analytic variety with
nite quotient singularities.
Teichmller theory is a dicult and technical subject. Because of this, it
u
is only possible to give an overview.
4. The Uniformization Theorem
Our basic tool is the the generalization of the Riemann Mapping Theorem
known as the Uniformization Theorem. You can nd a proof, for example,
in  and .
Theorem 4.1. Every simply connected Riemann surface4 is biholomorphic
to either P1 , C or H .
Exercise 4.2. Show that no two of P1 , H and C are isomorphic as Riemann
surfaces.
We have already seen that Aut P1  PSL2 (C ) and that PSL2 (R) 
=
Aut H .
Exercise 4.3. (i) Show that
Aut C = fz 7! az + b : a 2 C  and b 2 C g:
(ii) Show that there is an element T of Aut P1 that restricts to an isomor-
phism T :  ! H between the unit disk and the upper half plane.
More generally, we say that a group G has a property P virtually if P holds for some
3

nite index subgroup of G.
We follow the convention that every Riemann surface is, by de nition, connected.
4
320

(iii) Show that every element of Aut  is a fractional linear transformation.
Hint: Use the Schwartz Lemma.
(iv) Deduce that every element of Aut H is a fractional linear transformation
and that Aut H  PSL2 (R).
=
If X is a Riemann surface and x 2 X, then 1 (X; x) acts on the universal
covering X as a group of biholomorphisms. This action is xed point free.
e

Exercise 4.4. Show that if X is a Riemann surface whose universal covering
is isomorphic to H and x 2 X, then there is a natural injective homomorph-
ism  : 1 (X; x) ! PSL2 (R) which is injective and has discrete image.
Show that this homomorphism is unique up to conjugation by an element of
PSL2(R), and that the conjugacy class of  is independent of the choice of
x. Show that X is isomorphic to im nH and that the conjugacy class of 
determines X up to isomorphism.
This will give a direct method of putting a topology on Mg when g  2.
But rst some preliminaries.
Exercise 4.5. Show that if X is a Riemann surface whose universal covering
is
(i) P1 , then X is isomorphic to P1 ;
(ii) C , then X is isomorphic to C , C  or is a genus 1 curve.
Hint: Classify the subgroups of Aut P1 and Aut C that act properly discon-
tinuously and freely.
So we come to the striking conclusion that the universal covering of a
Riemann surface not isomorphic to P1 , C , C  or a genus 1 curves must be
isomorphic to H . In particular, the universal covering of C f0; 1g is H .
Picard's Little Theorem is a consequence.
Exercise 4.6. Show that if f : C ! C is a holomorphic mapping that omits
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