Notes on Math 600D: Modular Forms

Ed Belk

Fall, 2018

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1 Lecture One

Instructor: Julia Gordon; gor@math.ubc.ca

Text: Modular Functions and Modular Forms, J. Milne

References to algebraic geometry will be only occasional, and can often be safelyignored. We closely follow Milne’s introduction to the subject.

We begin with a broad goal: to study all holomorphic (or meromorphic) functions onall Riemann surfaces. Recall that a Riemann surface is a one-dimensional complexmanifold; that is, it is a connected, Hausdorff topological space X equipped with afamily of charts .U; �/, with

� W U ! C

a homeomorphism onto an open subset of C, and for which, whenever .U1; �1/ and.U2; �2/ are charts with U1 \ U2 ¤ ;, the map

�2 ı ��11 W �1.U /! C

is holomorphic. Such a covering by charts is called a complex structure on X .

Examples:

1. C, or any connected open subset of C.

2. The Riemann sphere P1.C/. We briefly review its construction: this is the setof dimension one subspaces in C2, described in homogeneous coordinates:

aX C bY D 0 corresponds to .a W b/ corresponds to .1 W ba/:

3. The standard torus. This (as we will soon see) admits infinitely many non-isomorphic complex structures.

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2 Lecture Two

There will be extra classes Fridays at 16.00-17.30 on days when colloquium is off, orirrelevant. The first homework assignment has been posted.

The goal for this class is to carefully describe the affine modular curve �nH, where� is some group of interest, as:

1. a topological space;

2. a Riemann surface; and, later,

3. an algebraic variety.

We begin with some careful review of group actions. Let G be a topological group;that is, a group which is a Hausdorff topological space, and for which the multipli-cation homomorphism G � G ! G and the inversion homomorphism G ! G arecontinuous with respect to this topology. If X is a topological space, we say that Gacts continuously on X if the map

G �X �! X

.g; x/ 7�! g � x

is continuous. From now on, we will assume that all actions of topological groups ontopological spaces are continuous.

Definition: The stabilizer of x 2 X in G is the subgroup

StabG.x/ D fg 2 G W g � x D xg:

If H � G is a subgroup of our topological group, then a topology (known as the quo-tient topology) is placed on the coset space G=H by choosing the finest topologyon G=H for which the canonical projection G ! G=H is continuous. More precisely:a subset U � G=H is open in the quotient topology if and only if the union of cosetsin U is an open subset of G. In the same way, we can define a topology on the orbitspace GnX of a topological space X under a continuous action; this will be the finesttopology on the space of orbits such that the map X ! GnX is continuous.

Facts: (to be solved as homework)

1. If X is Hausdorff, then for all x 2 G, StabG.x/ is a closed subgroup of G.

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2. The quotient G=H is Hausdorff if and only if H is closed in G (as such, G=His a topological group if and only if H is normal and closed).

3. Suppose the action of G on X is continuous and transitive (i.e., X consists ofa single orbit), and that G is locally compact and second-countable; then thebijection G= StabG.x/! X is, in fact, a homeomorphism.

4. Assume further that StabG.x/ is compact in G for all x 2 X ; then � � G isdiscrete if and only if � acts properly discontinuously on X .

5. With the same assumptions: if � is discrete, then �nX is Hausdorff.

Aside: The dedicated reader should take the time to read pp. 16-18 in Milne’s notes,on the topic of “Riemann surfaces as ringed spaces.”

Thus, with � D SL2.Z/ acting on H, we see that �nH is a Hausdorff topologicalspace. If �, a discrete subgroup of SL2.R/, or �=f˙I g acts freely on H (meaningthat Stab�.x/ D feg for all x 2 H), then there is a unique complex structure on �nHsuch that, for any open set U � H, a function f is holomorphic on U if and only iff ı p is holomorphic on �nH, where p W H! �nH.

In our case, the group �.1/ D SL2.Z/ does not act freely on H; however, we will stillget a complex structure on �nH, and so we are motivated to study stabilizers. Todo this, we want to:

1. Investigate Aut.H/;

2. Classify fractional linear transformations and study their fixed points; and

3. Find a fundamental domain for �.1/, and give it a complex structure.

Later, we will perform the same task for �.N/.

Aside: We take a moment to investigate the unusual origins of the fractional lineartransformations. Consider the action of GL2.C/ on P1.C/; this is given by�

a b

c d

��z0z1

�D

�az0 C bz1cz0 C dz1

�:

Suppose now that � is one of the two standard charts on P1.C/, so that

C��1

�! P1.C/ n fpt.g��! CI

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then, writing z 2 C as z0=z1, we have�a b

c d

�� z D

az0 C bz1

cz0 C dz1Daz C b

cz C d;

and we complete the picture by identifying the point z1 D 0 with 1, the “point atinfinity” on the Riemann sphere.

Thus we are led to classify the fractional linear transformations�a b

c d

�� z D

az C b

cz C d; ad � bc ¤ 0:

Some are obvious: for � 2 C, � � Id acts trivially. Furthermore, we notice that, ifg 2 GL2.C/ has fixed point x, then hg0h

�1 has fixed point h �x. As such, the Jordancanonical form guarantees that every linear fractional transformation is similar toone represented by one of the matrices�

� 1

�

�; or

��

�

�; with � ¤ �:

If D

�a b

c d

�gives rise to a linear fractional transformation similar to the first

type, then we call parabolic. If gives rise to a linear fractional transformationsimilar to the second type, then we call elliptic if �=� has absolute value one,hyperbolic if � D � is positive and real, and loxodromic otherwise. Typically, wewill only be interested in parabolic, elliptic, and hyperbolic matrices.

Remark: Observe that the characteristic polynomial of is X2�Tr. /XC1, whichhas discriminant � D Tr. /2 � 4.

Exercise: Suppose G D SL2.C/. Show that two non-scalar matrices are conjugateif and only if they have the same trace. Show that ˛ 2 G is:

1. parabolic if and only if Tr.˛/ D ˙2;

2. elliptic if and only if Tr.˛/ 2 R and jTr.˛/j < 2;

3. hyperbolic if and only if Tr.˛/ 2 R and jTr.˛/j > 2;

4. loxodromic if and only if Tr.˛/ … R.

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If instead G D SL2.R/, show that ˛ 2 G is parabolic if and only if it has a singlefixed point on R [ f1g.

We remark on the usefulness of giving many perspectives on the objects that wework with. For instance, a fixed point on P1.C/ corresponds to a fixed line in C2,

i.e., and eigenvector for the action of GL2.C/ described above. The matrix

�� 1

�

�has only one such eigenvector, up to scaling. In the case of a hyperbolic matrix��

�

�with �;� 2 R, we have two fixed points in R. In the elliptic case, we have

two eigenvectors which are complex conjugates; that is, there is a single fixed pointin H.

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3 Lecture Three

Last time, we saw that if � � SL2.R/ is discrete (perhaps among other assumptions),then �nH is a Hausdorff topological space, and we classified the action of an elementof SL2.R/ as elliptic, parabolic, or hyperbolic, according to their traces. Our goaltoday is to equip �nH with a complex structure.

First, we will realize H as a quotient of SL2.R/, and establish rigorously the fact that

SL2.R/=f˙I g Š Aut.H/;

where Aut.H/ denotes the set of (biholomorphic) automorphisms of H. We will thenfind a fundamental domain for the action of SL2.Z/ on H.

Proposition 3.1. One has:

1. The action of SL2.R/ on H is transitive.

2. There is an isomorphism SL2.R/=f˙I g Š Aut.H/.

3. There is a bijection StabSL2.R/.i/ D SO2.R/, where i Dp�1.

4. The map SL2.R/= SO2.R/! H given by Œ˛� 7! ˛.i/ is a homeo-morphism.

We recall that SO2.R/ consists of all linear transformations of R2 which preserveorientation and inner products, and so in particular can be written

SO2.R/ D f

�cos � sin �� sin � cos �

�W � 2 Rg:

Proof. Let ˛ D

�a b

c d

�2 SL2.R/.

1. It suffices to show, for all z D x C iy 2 H, that there exists ˛ 2 SL2.R/ such

that ˛ � i D z. Because y is assumed positive, we can take

�y x

1

�2 SL2.R/,

and we are done.

2. A quick calculation shows that Im.˛ � z/ D Im.z/jczCd j2

, so there exists a homomor-

phism SL2.R/! Aut.H/ which descends to a homomorphism SL2.R/=f˙I g !

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Aut.H/. Another easy calculation shows that this latter map is injective; thus,it suffices to show that it is surjective.

Suppose 2 Aut.H/; we from from above that there exists ˛ 2 SL2.R/ suchthat ˛ �i D �i . Thus, replacing with ı˛�1, we can assume that stabilizesi . The homomorphism

�.z/ Dz � i

z C i

maps H to the unit disc, i to 0, and the real line to the unit circle; as such,�ı ı��1 is an automorphism of the unit disc, which fixes zero. By the Schwarzlemma, it follows that

j� ı ı ��1.z/j � jzj

for all z; jzj < 1. The same argument applies mutatis mutandis to the auto-morphism ��1 ı ı �, from which it follows that

j� ı ı ��1.z/j D jzj whenever jzj � 1I

from this, the Schwarz lemma gives the existence of some � 2 C; j�j D 1 suchthat

� ı ı ��1.z/ D e2�i�z for some � 2 R:

It follows that 2 SO2.R/.

3. This is immediate from the previous part; it can also be proven using theorbit-stabilizer theorem.

4. Because SO2.R/ is the stabilizer of an element of a space with a continuousgroup action, we know it is a closed subgroup of SO2.R/; the result now followsimmediately from the associated homework problem.

Definition: Let � be a discrete subgroup of SL2.R/. A fundamental domain for� in H is a connected, open set D such that:

1. For any z ¤ z0 2 D, there is no 2 � such that z D z0, and

2. One has [ 2�

D D H;

where D is the topological closure of D.

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It is a fact that a fundamental domain always exists for any discrete subgroup ofSL2.R/; we will prove this fact only for the congruence subgroups. Today, we willprove that there exists a fundamental domain for the action of �.1/.

We note the important fact that the group SL2.Z/=f˙I g is generated by the elements

S D

��1

1

�and T D

�1 1

1

�:

Observe that, for z 2 H, we have S � z D �1z

and T � z D z C 1.

Theorem 3.2. A fundamental domain for the action of �.1/ on H isgiven by the set

D D fz 2 H W jzj > 1g \ fz 2 H W �12

Re.z/ < 12g;

whose boundary consists of the two vertical rays

f12C iy W y >

p32g and f�1

2C iy W y >

p32g;

as well as the arcfei� W �

3< � < 2�

3g:

Furthermore, if z D z0 for some 2 SL2.Z/; z ¤ z0 2 D, thenRe.z/ D ˙1

2and z0 D �1, or jzj D 1 and z0 D S � z; in particular, if

the stabilizer of z is not f˙I g, then z D i; �; or �2, where �2 D ei�D3,and the respective stabilizers are hSi; hTSi; or hST i. Finally, �.1/ DhT; Si, and SL2.Z/ has presentation

SL2.Z/ D hS; T jS2D .ST /3 D I iI

thus, SL2.Z/ is the free product of two cyclic groups of prime order.

Proof. Let � 0 be the subgroup of �.1/ generated by S and T ; we will show first thatD is a fundamental domain for � 0.

Observe that, if z D xC iy 2 H and N is some fixed positive integer, then there areonly finitely many pairs of integers .c; d/ such that jcz C d j < N . Indeed, if this

is so, then jcj � N 2

yis satisfied by only finitely many integers, and the claim is now

obvious. It follows from this that, in the orbit of z under � 0, there exists an element˛ � z with minimal imaginary part.

We now claim that there exists some 2 � 0 and some z0 2 D such that z0 D z.Clearly there exists some integer n 2 Z such that �1

2� Re.T n/.˛ � z/ � 1

2; write

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z0 for T n.˛ � z/. We must now have jz0j � 1; otherwise, we apply S and we havejSz0j � 1, and repeat (clean this up later).

Finally, let z ¤ z0 2 D and suppose that 2 �.1/ has z0 D � z. Without loss ofgenerality, we will assume Im.z/ � Im.z0/; then

jcz C d j D .cx C d/2 C c2y2 � 1I

in D we must have y2 � 34, and so we can only have c D 0 or ˙1. If c D 0 we

have d D ˙1 and so D T ˙1, and z; z0 lie on the parallel rays which make up theboundary of D; if c D 1 we must have jz C d j � 1 and so ˙1 and z D � or �2, ord D 0.

It remains only to show that � 0 D �.1/, and the rest of the theorem will be proven.Let 2 �.1/ and choose z0 2 D; we know from our above work that � 0 �D D H,and so there exist 0 2 � 0; z0 2 D such that

0z0 D z0:

Consider the point . 0/�1 z0 D z0 2 D: the only way both z0 and z0 lie in D is if z0

lies on the boundary, which is not the case (clean this up later).

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4 Lecture Four

We briefly return to correct an error in the last lecture: we recall that the minimumvalue of jcz C d j corresponds to the maximum value of Im. z/. The element 0should have been chosen to give this maximum value Im. z0/.

Today we will talk about the complex structure on modular curves, and compactifythe upper half-plane; we begin with two guiding examples.

Let D denote the unit disc, and consider the group Aut.D; 0/ of automorphisms ofD fixing zero. As we saw last time, the Schwarz lemma implies that every elementof Aut.D; 0/ is a rotation of the form z 7! �z, where � 2 C; j�j D 1. Let � be afinite group acting by automorphisms on D, assuming that no nontrivial subgroupof � acts trivially on D. Because there is an inclusion � ,! S1 Š R=Z, we knowthat � has to be cyclic; its image will be the group of roots of unity of some degreem (say).

Let �0 be a generator of �; we want to place a complex structure on �nD. Equiv-alently, we want to determine which functions on �nD are holomorphic. We notethat the maps z 7! zm; m 2 Z, are holomorphic on D which are invariant under �,and so by the universal property of quotients they descend to a “holomorphic” mapon �nD.

Note that the map �nD ! D; z 7! zm is a homeomorphism. First of all, it iswell-defined: Œz1� D Œz2� in �nD if and only if z1 D �k0 z2 for some k 2 Z, and sozm1 D zm2 . It is also injective: if zm1 D zm2 in D, then jz1j D jz2j, and so there existssome � 2 S1; �m D 1 such that z1 D �z2, hence

zm1 D �mzm2 :

If z1 D z2 D 0 then we are done; otherwise we must have � 2 �, and the claim isproven. Surjectivity is clear, and is left as an exercise.

Thus we can define a complex structure on �nD via the atlas consisting of the singlechart

�nD ! D; z 7! zm;

where holomorphic functions on �nD are holomorphic functions of zm.

For our second example, we put

X D fz 2 C W Im.z/ > c � 0g;

and we fix h 2 Z. Let Z act on X by translations:

n:z D z C nh:

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We add to X a point 1, called the “point at infinity,” to get a new space X� DX [ f1g; the action of Z on X is extended to X� via n:1D1. We can topologizeX� as follows: for x 2 X , a fundamental system of neighbourhoods is given baseof neighbourhoods giving its topology in X , and a base of neighbourhoods of 1 isgiven by the open sets UN D fIm.z/ > N g for N 2 Z�0. With this new topology,the action of Z on X� is continuous.

We obtain a map q from X� to the open disc of radius e�2�c=h which is Z-periodicby defining

q.z/ D

(e2�iz=h if z ¤1;

0 if z D1:

Then q is in fact a homeomorphism, which we can use as the coordinate function onZnX�, so that again holomorphic functions on ZnX� are holomorphic functions ofq.z/.

With these two examples in mind, we can now define a complex structure on �nHand �nH�, where � D SL2.Z/. First, denote by p the projection map H ! �nH,let P 2 �nH, and let Q 2 p�1.P /. If Q is not an elliptic point, then p is a localhomeomorphism around Q. If Q � i (that is, Q is in the same orbit as i), then takeQ D i for simplicity. The map

�.z/ WDz � i

z C i

sends H to the unit disc and maps i to 0. If D0 is some neighbourhood of 0 in theunit disc, then D00 D ��1.D0/ is some neighbourhood of i in H which is invariantunder the action of S 2 �; S.z/ D �1

z; this is to be checked as an exercise, with the

hint that ��1 ı S ı � is the map z 7! �z on the unit disc. We can therefore equiphSinD00 with a complex structure, with chart

z 7!

�z � i

z C i

�2:

We will play the same game near the points � and �2 in H: for �2 we choose coordinatechart

z 7!

�z C �2

z C N�2

�3which is a conformal map sending H to the unit disc and �2 to 0. The exponent of3 takes care of the action of ST , the stabilizer of �2.

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We claim next that �nH is, in fact, the Riemann sphere minus one point. Moreprecisely, we will show that �nH admits a one-point compactification which has thestructure of a Riemann sphere.

As in example 2: let H� D H [ f1g and topologize H� in the same way. As in thatcase, we have T � 1 D 1, and also S � 1 D 0 (which we will explain in a moment).The complex structure on �nH� is constructed in the same way as example 2 for theco-ordinate chart near 1.

More precisely: a point s 2 R [ f1g is called a cusp for some discrete subgroup �of SL2.R/ if it is the fixed point of some (parabolic) element of �.

Exercise: The cusps of �.1/ are Q [ f1g; moreover, they all lie in the same orbitunder �.1/.

Thus motivated, we choose instead to define H� D H [ Q [ f1g; then �nH� iscompact, and has the complex structure of the Riemann sphere.

Next time, we will discuss differences that arise when we replace �.1/ with �.N/.

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5 Lecture Five

Last time, we closed by putting a complex structure on �.1/nH�. Today we will dothe same for �.N/nH�, where N is arbitrary.

Proposition 5.1. Let � 0 be a finite-index subgroup of a discrete sub-group � of SL2.R/, let D be a fundamental domain for �, and let 1; : : : ; m 2 � be a complete set of coset representatives for cosets of� 0=f˙I g in �=f˙I g; then

D0 WD

m[iD1

iD

is a fundamental domain for � 0.

Proof. Exercise.

It does not follow that the set D0 is connected, but it is possible to show that theinterior of the closure of D0 is connected.

For the moment we will leave aside the cusps, and consider the complex structure on�.N/nH: it is that which arises from H. Indeed, if N > 1, the points i; �; and �2,which were elliptic in the case N D 1, are not fixed by any elements of �.N/ exceptf˙I g, and so there are no elliptic points in the N > 1 case, and �.N/ is a normalsubgroup of �.1/.

However, there is more than one orbit P1.Q/ under �.N/ when N > 1, giving finitelymany inequivalent cusps; the exact number of such cusps we will compute next time.There is a natural covering map (with possible branch points)

�.N/nH� ! �.1/nH�

of the Riemann sphere by �.N/nH�; this map is m-to-one, where m D Œ�.1/ W �.N/�,away from branch points.

We now take an aside, per Milne, to discuss Riemann surfaces as ringed spaces. Letk be a field.

Definition: By a ringed space is meant a pair .X;OX/ consisting of a topologicalspace X and a sheaf OX of k-algebras on X . We recall that this means that OX is afunctor from the category of open subsets of X , whose morphisms are inclusions, to

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the category of k-algebras, satisfying the sheaf axiom: If U D [iUi is an open coverof some open subset U of X , and fi 2 OX.Ui/ is a collection of sections such that

fi jUi\Uj D fj jUi\Uj

for every i , then there exists some unique f 2 OX.U / such that f jUi D fi for everyi .

For this course, we will use the following convention: If V � C is open and U � Vany open subset, the symbol OV .U / will denote the set of holomorphic functionf W U ! C. The pair .V;OV / will be called the standard ringed space.

It should now be clear that to give a complex structure on a topological space X isto give a sheaf of C-algebras on X satisfying the condition that there exists an opencover fUig of X such that each ringed space .Ui ;OX jUi / is isomorphic to a standardringed space. Here, OX jUi denotes the sheaf U 7! OX.U /, when U � Ui .

The next terminology we introduce will be that of differential forms. Let U � Cbe an open subset; a differential (one)-form on U is an expression of the form! D f .z/ dz. To a (meromorphic, say) function f W U ! C we associate the one-form !f WD f

0.z/ dz. If � W U ! W is a holomorphic map between open subsets ofC, then the pullback of the form (on W ) ! D f .w/ dw is the one-form (on U )

��.!/ WD f .�.z//� 0.z/ dz:

Note that, if U has z-co-ordinates and W has w-co-ordinates, then we can write

dw D � 0.z/ dz:

More generally: LetX be a Riemann surface and let .Ui ; zi/ be a co-ordinate covering.A differential one-form on X is a collection .!i/i of differential one-forms on thesubsets Ui such that, if

!i D fi.zi/ dzi and zi D �ij .zj /;

then ��ij .!i/ D !j ; that is,

fj .zj / dzj D fi.�ij .zj //�0ij .zj / dzj :

Exercise: Phrase this observation as a statement of sheaves of differential formsand OX -modules.

We emphasize the contrast between the transformation of functions with that ofdifferential forms. We mention here also the fact that a compact, complex manifoldcan be realized as the C-points of a projective algebraic variety.

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Example: What are the meromorphic functions on S D P1.C/? By Liouville’stheorem, we know that the only holomorphic functions on S are constants. Workingwith co-ordinates makes it clear that, to be meromorphic on S, a function f .z/ onC must be meromorphic in z, and also meromorphic in 1

z. From this it is clear that

the only rational functions on S are the rational functions.

We might ask the same question about differentials on S? The same analysis tells usthat a meromorphic one-form on S must have both f .z/ meromorphic and �f .1=z/

z2

meromorphic, whenever z ¤ 0.

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6 Lecture Six

It is a fact that a meromorphic function f on a compact Riemann surface X has thesame number of poles as it has zeroes, counting multiplicity. If ! is a one-form onX , then the sum of the residues of ! at its poles will be zero.

The sketch of the proof is as follows: one can triangulate X in such a way that nopole lies on a vertex or an edge. On each face F of the triangulation, we will haveby Cauchy the equation Z

@F

D 2�iX

poles p in F

Res.!Ip/:

From this, it follows that Xall poles p

Res.!Ip/ D1

2�i

XF

Z@F

!:

The first statement follows from the second by setting ! D dff

.

Proposition 6.1. Let f be a non-constant, meromorphic function ona compact Riemann surface X . Then f attains every value in its rangeexactly n times (counting multiplicity), where n is the sum of the ordersof the poles of f . This value is known as the valence of f .

Proof. (sketch) Let c 2 C be a constant, and apply the previous statement to thefunction f � c. The argument will then show, as a corollary, that the range of fmust be all of C.

Thus, the statement that f has valence n is the same as the statement that theinduced map f W X ! S onto the Riemann sphere is an n-to-one algebraic cover.

Definition: A point p 2 S is called a ramification point for the function f W

X ! S if f �1.p/ contains fewer than n distinct points.

Now, recall that the meromorphic functions on S are precisely the rational functionsC.z/.

Proposition 6.2. If X is some compact Riemann surface and f W

X ! S is some given meromorphic function of valence n, then everymeromorphic function on X is the root of a polynomial of degree atmost n, with coefficients in C.z/.

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Proof. (sketch) Let c 2 S be a point for which f �1.c/ has exactly n points (that is,c is not a ramification point), and label

f �1.c/ D fP1.c/; : : : ; Pn.c/g:

Let z 2 X be some value such that f .z/ D c, and let g W X ! C be meromorphic.Then

nYiD1

.g.z/ � g.Pi.c/// D 0

by construction; expanding the product gives

g.z/n C r1.c/g.z/n�1C � � � C rn.c/ D 0;

where ri.c/ are (symmetric) polynomials in g.Pi.c//. Considered as functions of c,therefore, we see that g.Pi.c// and ri.c/ are both meromorphic functions on S , andso in particular the ri.c/ are rational functions on S . The proof now follows whenwe take c D f .z/.

Another definition: a divisor on X is a formal finite sum of the form

D DXP2X

nP � P;

where nP 2 Z and nP D 0 for all but finitely many P . Denote by Div.X/ the freeabelian group on the points of X , and define a homomorphism of groups

deg W Div.X/! Z

via

deg

XP2X

nP � P

!D

XP2X

nP :

A divisor can be attached to any meromorphic function f on X via

D.f / DXP2X

ordP .f / � P I

a divisor D is called principal if it equals D.f / for some meromorphic function f .Our work above implies that degD.f / D 0 whenever X is compact. Two divisorsD1;D2 are called linearly equivalent if D1�D2 is a principal divisor, and a divisoris called effective (denoted D � 0) if all its coefficients nP are nonnegative.

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Now, let ! be a differential on X . We can associate to ! a differential in anotherway, namely: for each P 2 X we can take a chart .Ui ; zi/ containing P in which !has co-ordinates f .zi/ dzi ; we then define

ordP .!/ D ordP .fi/:

This map is well-defined: if �ij .z/ is some transition function on the appropriatechart, then

d! D � 0ij .z/fj .z/ dz;

with � 0ij .z/ ¤ 0 anywhere.

Fact: If ! is a nonzero (meromorphic) differential on the compact Riemann surfaceX , then every other differential onX has the form f �!, where f is some meromorphicfunction on X . It follows from this that all divisors on X are linearly equivalent.The equivalence class of the divisor associated to ! is denoted K, and is called thecanonical divisor of X .

For an example: when X D P1.C/, the differential dz on the usual chart correspondsto the differential dw D �dz

z2on the chart which identifies w D 1

z. This differential

has a pole of order 2 at1, and so in this case we will have K D �2 � f1g. Obviously,the canonical divisor in this case has degree 2.

Definition: If D is a divisor on X , put

L.D/ D fmeromorphic f W X ! C W div.f /CD � 0g [ f0gI

then L.D/ is a vector space, of finite dimension `.D/, and its dimension dependsonly on the linear equivalence class of D. We can now state the important theoremof our lecture.

Theorem 6.3 (Riemann-Roch theorem). Let X be a compact Rie-mann surface. There exists an integer g 2 Z�0 such that, for allD 2 Div.X/, one has

`.D/ D deg.D/C 1 � g C `.K �D/;

where K is the canonical divisor of X .

The quantity g occurring in the theorem is called the genus of X .

This theorem is very useful in certain special cases. For instance, if deg.D/ < 0,then `.D/ D 0; indeed, f 2 L.D/ if and only if div.f /CD � 0. But deg.div.f //Cdeg.D/ D deg.D/ < 0, so the divisor cannot possibly be effective.

19

Example: Let D D m � f1g on S D P1.C/, with m > 0. Then div.f / C D � 0implies that the only possible pole of f is at 1, of order at most m, and so f mustbe a polynomial of degree at most m; this condition is also sufficient. Thus

`.D/ D mC 1 D deg.D/C 1:

We also have K �D D �.mC 2/ � f1g, so `.K �D/ D 0 by the same argument; itfollows that g D 0 works for this divisor D. Because all divisors on S are linearlyequivalent, this shows that g D 0 generally.

Corollaries:

1. Taking D D 0 gives `.D/ D 1 and L.D/ equals the space of constant functions.In this case, Riemann-Roch yields the equation

1 D `.0/ D 0C 1 � g C `.K/ H) g D `.K/:

From this it follows that the dimension of the space of holomorphic one-formson a compact Riemann surface X of genus g, is g itself.

2. Taking D D K gives the equation

`.K/ D g D deg.K/C 1 � g C `.0/ H) deg.K/ D 2g � 2:

An important point in all of this discussion is this: if X is a compact Riemannsurface, then there exists a nonconstant meromorphic function f W X ! S ; thecorresponding statement is not true in higher dimensions.

We close with a more general topological statement: let X be a topological spaceand let �.X/ be the Euler-Poincare characteristic (or topological genus), whichis the (well-defined) quantity

�.E/ D V C F �E;

where V; F; and E denote respectively the number of vertices, faces, and edges in agiven triangulation of X . It is a fact that the topological genus equals the genus gfrom the Riemann-Roch theorem, as we will see next time.

20

7 Lecture Seven

We finish our discussion on genus, from Friday. We saw in that lecture the state-ment of the Riemann-Roch theorem, and saw some examples of its computationalusefulness. Today, we we begin with the Riemann-Hurwitz formula.

Theorem 7.1. Let f W Y ! X be a holomorphic surjection of com-pact Riemann surfaces, that is m-to-one, counting multiplicity (that is,ramification points are permitted); assume without loss of generalitythat there are only finitely many ramification points. For P 2 Y , leteP be the multiplicity of P in the fibre over F.P /: then

2g.Y / � 2 D m.2g.X/ � 2/CXP2Y

.eP � 1/;

the sum taken over all points of Y .

An example of the situation that arises in the theorem comes from considering themap from the unit disc D to itself, z 7! ze. If w D ze, then dw D eze�1 dz, hence

f �.dw/ D eze�1dz

has a zero of order e � 1 at zero.

Proof. Choose a differential ! on X such that ! has no zero or pole at any ramifi-cation point of f ; then f �! has, at every pre-image of a zero or pole of !, a zeroor pole respectively of the same order. From this discussion, if we put D! for thedivisor associated to !, we see that

deg.Df �!/ D m degD! C

XP over a

ramificationpoint

.eP � 1/:

The result is now immediate from the Riemann-Roch theorem.

Now: let � � �.1/ be a discrete subgroup. Recall that X.�/ D �n�, and the mapp W H� ! X.�.1//; let � W X.�/ ! X.�.1// be the induced map. Ignoring cusps,the map p ramifies only over p.i/ and p.�/ D p.�2/, with respective ramificationindices 2 and 3 (this is the cardinality of the stabilizer subgroup). As such, � canonly possibly ramify at p.i/; p.�/, or at the cusps.

21

Theorem 7.2. Suppose Œ�.1/ W �� D m. Let �2 be the number of�-inequivalent points of order 2 in H, which are elliptic for � (i.e.,j Stab�=f˙I g.z0/j D 2), and let �3 be the corresponding number of pointsof order 3. Similarly, let �1 be the number of �-inequivalent cusps,which is exactly the number of orbits of � in Q[f1g. Then the genusof X.�/ is

g D 1Cm

12��2

4��3

3��1

2:

Proof. See Milne, pp. 37-38.

If we lighten notation by writing X.N/ for �.N/nH�, then the above theorem allowsus to calculate the genus of X.N/.

Facts:

1. One hasŒ�.1/ W �.N/� D N 3

YpjN

.1 � p�2/;

the product taken over all primes p which divide N .

2. If N ¤ 2, then

Œ N�.1/ W N�.N/� D1

2Œ�.1/ W �.N/�;

where N�.N/ D �.N/=f˙I g. Moreover, Œ N�.1/ W N�.2/� D 6.

3. Put �N D Œ N�.1/ W N�.N/�; then

g.X.N// D 1C �N.N � 6/

12N:

The proofs are left as homework exercises.

Now: to every pair of complex numbers !1; !2 which are linearly independent overR, we can associate the lattice ƒ.!1; !2/ D Z!1 C Z!2. Assuming without loss ofgenerality that the quotient � D !1

!2has positive imaginary part, we will normalize

our lattices so that we have the association

ƒ.�/ D ZC Z�:

22

We might ask when two different bases will yield the same lattice? Clearly, .!1; !2/generates the same lattice as .!01; !

02/ if and only if there exist integers a; b; c; d with

ad � bc D ˙1, such that �a b

c d

��!1!2

�D

�!01!02

�:

As such, we deduce that ƒ.�/ D ƒ.� 0/ (with � 2 H/) if and only if � and � 0 are �.1/-equivalent. It follows that the points of �.1/nH are in one-to-one correspondencewith the set of all lattices ƒ.�/.

Given a lattice ƒ, we can consider the complex torus C=ƒ. This quotient inherits thestructure of a Riemann surface; as we will see, different lattices will inherit differentcomplex structures. In fact, it is possible to equip each Riemann surface C=ƒ.�/with the structure of an elliptic curve.

As we will see, a meromorphic function on the quotient C=ƒ corresponds to a mero-morphic function on C which is doubly periodic, under both generators of ƒ. Thefield of such functions is C.}.z/; } 0.z//, where } is the Weierstrass }-function at-tached to the lattice, and the functions satisfy the equation

} 0.z/2 D 4}.z/3 � g2}.z/ � g3;

where g2 and g3 are (known) constants.

23

8 Lecture Eight

Let � D �.1/. Recall from before our identification of real manifolds

�n SL2.R/= SO2.R/$ �nHI

as a Riemann surface, �nH has the structure of a Riemann sphere minus a point,and so it has the structure of an algebraic variety. We also saw a one-to-one corre-spondence between �nH and the set of full-rank lattices in C; if ƒ� is the lattice inC generated by 1 and � , then the correspondence is given by Œ� �$ ƒ� . The quotientspace C=ƒ� is in every case a Riemann surface of genus one. Today, we will giveC=ƒ� the structure of an elliptic curve.

By a doubly periodic function on C is meant a meromorphic function f W C! Cfor which there exists a full-rank lattice ƒ in C such that

f .z C �/ D f .z/ for all � 2 ƒ:

The function f is said to be doubly periodic with respect to ƒ. By Liouville’stheorem, the only holomorphic doubly periodic functions are constants.

Proposition 8.1. Let f .z/ be a nonzero, doubly periodic function, andlet D be a fundamental parallelogram such that f has neither poles norzeroes on the boundary; thenX

P2D

ordP .f / DX

poles p

Res.f Ip/ D 0;

andPP2D ordP .f / � P Š 0 mod ƒ.

Proof. We proved all but the last statement before; for this, we consider the differ-ential ! D f 0.z/

f .z/, and apply our result [ref] from above.

Corollary: A non-constant, doubly periodic function has at least two poles, countingmultiplicity.

Given these restrictions, we will attempt to create a doubly periodic function withrespect to a given lattice ƒ D ƒ� . First of all: let G be a finite group acting on theset X . A G-invariant function on X arises from any function f on X , via

h.x/ WDXg2G

f .g � x/:

24

Indeed,

h.g0 � x/ DXg2G

f .g � .g0 � x// DXg2G

f ..gg0/ � x/ DXg 02G

f .g0 � x/ D h.x/;

where the last sum has been re-indexed via g0 D gg0; notice that the terms in thesummation have been reordered. An analogous construction will therefore extend toinfinite groups as long as the sum over g 2 G converges absolutely. Moreover, thesum will be meromorphic if it converges absolutely and uniformly on compact sets.

We will attempt this with the function f .z/ D 1z2

; define the function

h.z/ D1

z2C

X.m;n/2Z2

.m;n/¤.0;0/

1

.m� C nC z/2;

which does not converge absolutely. However, replacing the exponent in the denom-inator by a 3 does give an absolutely convergent series; thus we define

} 0ƒ.z/ WD �2X�2ƒ

1

.z � !/3;

which is meromorphic, and therefore admits an antiderivative, viz.

}ƒ.z/ D1

z2C

X�2ƒ�¤0

�1

.z � �/2�1

�2

�:

This is known as the Weierstraß }-function.

Fact: Consider the map C=ƒ! P2.C/ given by

z 7!

(.}.z/ W } 0.z/ W 1/ if z ¤ 0;

.0 W 1 W 0/ if z D 0:

The image of this map is the curve with equation

Y 2Z D 4X � g2XZ2� g3Z

3;

where g2 and g3 are constants defined in the homework; that is, } satisfies thedifferential equation

.} 0/2 D 4}3 � g2} � g3:

25

(Hint: the difference does not have enough poles).

Now, let k 2 N, and write

Gk.ƒ/ DX�2ƒ�¤0

��2kI

then Gk.z/ gives us a function on H by defining Gk.�/ D Gk.ƒ�/. This function isknown as the Eisenstein series associated to ƒ.

Proposition 8.2. For k > 1, the function Gk.z/ converges to a holo-morphic function on H, and takes the value 2�.2k/ at 1.

Proof. We compute

lim�!i1

X�2ƒ�

��2k DX�2ƒ�

lim�!i1

��2k

D

X.m;n/2Z2

.m;n/¤.0;0/

lim�!i1

1

.m� C n/2kD

Xn2Zn¤0

1

n2kD 2�.2k/:

We will write g2 D 60G2.ƒ/ and g3 D 140G3.ƒ/.

Theorem 8.3. One has .} 0/2 D 4}3 � g2} � g3, and the field ofƒ-doubly periodic functions is C.}ƒ; }

0ƒ/.

Proof. We prove only the last statement, leaving the first as an exercise.

Suppose f is an even, meromorphic function with a zero or pole of order m at z0;then f also has a zero or pole of the same order at �z0. We consider the two casesz0 Š �z0 mod ƒ and z0 6Š �z0 mod ƒ. In the first case, because the odd derivativesof f are odd functions, we get the equation

f .2kC1/.z/ D �f .2kC1/.�z/;

from which it follows that m is even. Then consider the function

g.z/ DYzi

.}.z/ � }.zi//mi

YziŠ�zi mod ƒ

.}.z/ � }.zi//mi=2 ;

where the first product is taken over all zeroes and poles zi of f .z/ for which zi 6Š�zi mod ƒ; then f=g is a constant, and the result is now immediate.

26

As we consider elliptic curves, we will assume that the characteristic of our basefield is not 2 or 3. For the moment, we will also assume our field is algebraicallyclosed. From the Riemann-Roch theorem, we know that projective curves over thatfield have an associated value g (the genus). Let E be such a curve, and considerthe divisor D D 2 � f0g; as usual, K denotes the canonical divisor. We know fromRiemann-Roch that

deg.K/ D 2g � 2;

which vanishes when g D 1. We have

`.D/ D dimff W div.f /CD � 0g D deg.D/C 1 � g C `.K �D/:

As g D 1 and degD > 0, this equation reduces to `.D/ D deg.D/ D 2. It followsthat the space of functions which have a pole of order at most 2 at zero, is two. Onefunction, call it X , will have a pole of order 2 at zero; another function Y has a poleof order 3. Consider the functions

1;X; Y;XY;X2; Y 2; X3I

because there are seven functions, and the dimension of L.6f0g/ is six, we know theremust be a linear relation between the functions. We will explore this further nexttime.

27

9 Lecture Nine

We continue from last time. If ƒ;ƒ0 are two full-rank matrices in C, then we saythat ƒ is equivalent to ƒ0 and write ƒ � ƒ0 if there exists some c 2 C� such thatƒ D cƒ0.

Fact: there is a one-to-one correspondence between the set of equivalence classes offull-rank lattices in C and �.1/nH.

We saw last time how, given a lattice ƒ� , we can equip the quotient space C=ƒ withthe structure of a Riemann surface. It is a fact that

C=ƒ Š C=ƒ0 ” ƒ � ƒ0:

We begin by discussing endomorphisms of the quotient C=ƒ, by which we mean agroup homomorphism which is also holomorphic. Suppose ƒ0 is a lattice satisfying

˛ƒ � ƒ0; some ˛ 2 C�I

then there is a homomorphism

�˛ W C=ƒ! C=ƒ0

sending the class of z to the class of ˛z. Clearly, �˛.Œ0�ƒ/ D Œ0�ƒ0 .

Proposition 9.1. Every holomorphic map � W C=ƒ! C=ƒ0 such that�.Œ0�ƒ/ D Œ0�ƒ0 equals �˛ for some ˛.

Proof. From topology, we know there exists a continuous function Q� W C! C satis-fying Q�.0/ D 0. Because the projection maps

C! C=ƒ and C! C=ƒ0

are local isomorphisms, it follows that Q� is holomorphic.

Take � 2 ƒ; then Q�.z C �/ � Q�.z/ takes values in ƒ0, and so by Liouville’s theoremmust be constant. Thus Q� 0 is doubly-periodic with respect to ƒ; by our work fromthe previous lecture, it must be a constant itself. That is, Q�.z/ D ˛z for some˛; ˇ 2 C, and the result is now clear.

Corollary: There is an isomorphism of Riemann surfaces C=ƒ Š C=ƒ0 if and onlyif there exists some ˛ 2 C� such that ˛ƒ D ƒ0.

28

Proposition 9.2. One has:

1. End.C=ƒ/ � Z;

2. if End.C=ƒ/ ¤ Z, then there exists a quadratic extension K ofQ with ring of integers OK, and End.C=ƒ/ is a subring of OK,of rank 2 over Z.

Proof. Write ƒ D Z!1CZ!2, and put � D !1=!2. If ˛ 2 C has ˛ƒ � ƒ, then thereexist integers a; b; c; d such that

˛!1 D a!1 C b!2 and ˛!2 D c!1 C d!2I

that is,˛� D a� C b and ˛ D c� C d:

Thusc�2 C .d � a/� � b D 0 D ˛2 � .aC d/˛ C .ad � bc/ D 0;

and the result now follows.

Theorem 9.3. There are natural equivalences between any pair of thefollowing three categories:

� The category of full-rank lattices ƒ � C whose morphisms are

Hom.ƒ;ƒ0/ D f˛ 2 C W ˛ƒ � ƒ0g:

� The category of compact Riemann surfaces of genus 1 and dis-tinguished basepoint 0, whose morphisms are holomorphic mapssatisfying 0 7! 0.

� The category of elliptic curves over C with basepoint, i.e. pro-jective, smooth algebraic curves of (algebraic) genus 1 with dis-tinguished point 0, whose morphisms are regular maps satisfying0 7! 0.

Proof. Exercise.

29

We have already done most of the work involved in this proof; there is a single missingingredient, namely how elliptic curves arise from lattices, which we will discuss now.

Suppose E is the elliptic curve defined by the equation y2 D 4x3�g2x�g3, and let� D g32 � 27g

23. Then, up to some multiplicative constant, we have

� DYi¤j

.xi � xj /2D Res.f; f 0/;

where the product is taken over all roots of the polynomial f .x/ D y2, and Resdenotes the resultant. It is a fact that E is nonsingular if and only if � ¤ 0.

Definition: If E is defined by the equation y2 D 4x3�ax�b, define the associatedj -invariant to be

j.E/ D1728a3

�:

It is a fact that, if E and E 0 are two elliptic curves defined over an algebraicallyclosed field, then j.E/ D j.E 0/ if and only if E and E 0 are isomorphic. We will shownow that the induced function j W H! C is surjective.

Recall that if � is a subgroup of finite index in �.1/, then a modular function for� is a function on H which is �-invariant, meromorphic, and meromorphic at thecusps, where the last condition means that the function

z 7! f .e2�iz=h/

is holomorphic as z ! i1. Here, h is the width of the cusp at infinity, whichwe define to be the unique positive integer such that the stabilizer of i1 in � is

generated by the matrix

�1 h

1

�.

Definition: A modular form of weight 2k for � is a function f .z/ on H which isholomorphic on H, holomorphic at all cusps of �, and which satisfies

f . � z/ D .cz C d/2kf .z/

(the so-called automorphy condition) for every

�a b

c d

�2 � and z 2 H. The notion

of “holomorphy at the cusps” is similarly contrived to our above construction of“meromorphy at [the cusp] infinity.”

We note that modular functions coincide exactly with the meromorphic functionson �nH�, whereas modular forms coincide with holomorphic differentials on �nH.

30

Observe also that the quotient of two modular forms of a given weight is a modularfunction.

Example: Consider the Eisenstein series we introduced above; recall we defined

G2k.z/ DX�2ƒz�¤0

1

�2k:

An easy calculation shows that

G2k.˛z/ D ˛�2kG2k.z/ for any ˛ 2 C�I

then G2k.z/ is a weakly modular function, meaning it satisfies the automorphy con-dition, for the full modular group. This is an immediate consequence of the following

Lemma: Let F be a function on lattices of weight 2k, so that

F.˛ƒ/ D ˛�2kF.ƒ/ for any ˛ 2 C�I

then f .z/ WD F.ƒz/ is a weakly modular function of weight 2k, and all weaklymodular functions of weight 2k are of this form.

Proof. We have, for any D

�a b

c d

�2 �.1/, that

F.ƒ.!1; !2// D F.ƒ.a!1 C b!2; c!1 C d!2//:

Put � D !1=!2, and put f .�/ D F.�; 1/ D !2k2 F.!1; !2/. Then

f

�a� C b

c� C d

�D F

�ƒ

�a!1 C b!2

c!1 C d!2; 1

��D .c!1 C d!2/

2kF.ƒ.a!1 C b!2; c!1 C d!2// D .c!1 C d!2/2kF.ƒ.!1; !2//

D .c!1 C d!2/2k!�2k2 F.ƒ.�; 1// D .c� C d/2kf .�/;

and the result is clear.

Definition: By a cusp form is meant a modular form which vanishes at every cusp.We denote by Mk.�/ the space of weight k modular formms for �, and by Sk.�/

the subspace of cusp forms. Note that

Mk.�/Mn.�/ DMkCn.�/:

31

The ring of modular forms M .�/ for � is therefore graded by weight:

M .�/ DMk

Mk.�/:

Example: Recall from above the j -invariant, defined

j.z/ D1728g32�

; where � D g32 � g23:

We showed g2 D 60G2 and g3 D 140G3 to be modular forms for �.1/ already, ofrespective weight 4 and 6, from which it follows that � is a modular form for �.1/of weight 12. Because the numerator is as well, we must have that j.z/ is a modularfunction.

As we will see soon, � has no zeroes on H and has only a simple zero at the cusp(exercise: show that this fact implies that 8�.4/3 � 108�.6/2 D 0) Thus, the j -invariant has a single pole on �.1/nH�, and so has valence 1. It follows that it takesevery complex value exactly once.

32

10 Lecture Ten

Last time, we saw the j -invariant, and its isomorphism from �.1/nH� to the Riemannsphere. We proved almost everything, except for the fact that �.z/ has a simple zeroat i1; we will prove this today.

We begin with a warm-up calculation. Consider the meromorphic differential ! Df .z/ dz on H, and suppose � � �.1/ is a finite-index subgroup. When is ! invariantunder the action of �? that is: When does ! descend to a meromorphic differentialon the modular curve �nH? Write

D

�a b

c d

�2 �;

so that

.z/ Daz C b

cz C dI

then �! D f . .z// d. .z// D f . .z//.cz C d/�2 dz:

It follows that ! gives a differential on �nH if and only if f .z/ is weakly modular ofweight 2 for �. Therefore we have one-to-one correspondences between each of thefollowing sets:

� the set of meromorphic modular forms on �nH�;

� the set of meromorphic differential forms on H� which are invariant under �;

� the set of meromorphic modular forms of weight 2 for �.

Now: by a k-fold differential on a Riemann surface is meant a formal expressionin local co-ordinates of the form ! D f .z/ .dz/k, where k � 1 is an integer. If wetake new co-ordinates u D u.z/, then ! transforms under the co-ordinate change via

u�! D f .u.z// .du/k D f .u.z//.u0.z//k .dz/k:

Note that if � is the sheaf of differentials on our Riemann surface, then k-folddifferentials are sections of �˝k. The same analysis establishes a correspondencebetween the set of meromorphic modular forms of weight 2k for �, and the set ofmeromorphic k-fold differentials on �nH�.

We note that the order of a zero or pole at P 2 �nH� is well defined for a k-folddifferential, because u0.z/ ¤ 0 anywhere. We might reasonably ask therefore: how

33

do the zeroes and poles of a modular form relate to the zeroes and poles of thecorresponding differential?

Example: Consider the map w from the unit disc D to itself, defined w.z/ D ze;we will investigate what becomes of the differential around zero. If f W D ! C isany function, then w�.f / D f .ze/; so, if f has Laurent series f .w/ D anw

n C � � � ,it follows that

w�f .z/ D anzenC � � �

Thus, the order at zero of w�f is e ord0.f /. If ! is a k-fold differential on thecodomain D, then ! D f .w/ .dw/k for some function f .w/, and so

!� WD w�! D f .ze/ .d.ze//k D f .ze/.eze�1/k .dz/kI

thusord0.!

�/ D e ord0.!/C k.e � 1/:

Proposition 10.1. Let f be a meromorphic modular form of weight2k for some subgroup finite-index subgroup � � �.1/, and let ! bethe corresponding k-fold differential on �nH�. If Q 2 H� lies aboveP 2 �nH� with multiplicity e, then:

1. If Q is elliptic, then ordQ.f / D e ordP .!/C k.e � 1/;

2. If Q is a cusp, then ordQ.f / D ordP .!/C k; and

3. In any other case, ordQ.f / D ordP .!/.

Recall that the multiplicity of P above Q is 1 if Q is a cusp, and otherwise is theorder of the stabilizer of Q in �.

Proof. The first and third claim are clear in light of our calculations above; as such,it remains only to prove the second point. Consider the image of the cusp Q in �nH,and let U be a neighbourhood containing this cusp. A map �nH to the puncturedunit disc D� is given by q D q.z/ D exp.2�iz=h/, as in the previous lecture. If ! isa k-fold differential on D�, we can write in local co-ordinates

! D g.q/ .dq/k; hence !� D g.q.z//

�2�i

hq.z/

�k.dz/k:

The result is now obvious.

34

We now investigate higher-weight modular forms. We know that f 2M2k.�/ if andonly if:

� eP ordP .!f /C k.eP � 1/ � 0 for all point P ;

� ordP .!/C k � 0 at every cusp P ; and

� !f is holomorphic elsewhere.

If we fix some k-fold differential !0 on �nH�; then any ! will be of the form h.z/ �!0,where h.z/ is some meromorphic function. With this notation, our conditions aboveare equivalent to the conditions:

� ordP .h/C ordP .!0/C k.1 � 1=eP / � 0 at all elliptic points P ;

� ordP .h/C ordP .!0/C k � 0 at every cusp P ; and

� ordP .h/C ordP .!0/ � 0 at every other point P .

We can reformulate these conditions once again to the equivalent statement: f 2M2k.�/ if and only if div.h/CD � 0, where

D D div.!0/CXcusps

k � Pi CX

elliptic points

�k.1 � 1=ePi /

�Pi ;

and Œx� is the greatest integer part of x.

Theorem 10.2. Suppose � � �.1/ has finite index; then

dim.M2k.�// D

8<:0 if k � �1;

1 if k D 0;

1 � g C k.2g � 2/C k�1 CPP Œk.1 � 1=eP /� if k � 1;

where in the last case g is the genus of X.�/ D �nH�, and �1 is thenumber of �-inequivalent cusps.

Proof. (sketch) We use the Riemann-Roch theorem, using the fact that `.K�D/ D 0because deg.K �D/ < 0.

35

In the special case � D �.1/, this formula simplifies considerably: we have

dim.M2k/ D 1 � k C

�k

2

�C

�2k

3

�D

(�k6

�if k � 1 mod 6;

1C�k6

�if k 6� 1 mod 6;

whenever k � 0.

This formula allows us easily to compute the dimensions of the various spaces M2k,and our previous work will give us generators.

k 1 2 3 4 5 6 7weight 2 4 6 8 10 12 14

dimension 0 1 1 1 1 2 1generator(s) 0 G2 G3 G22 G2G3 G32 ; G

23 G22G3

Indeed, it may be shown that, as a ring, M2k is generated by G2 and G3.

We close with a few words on the zeroes of modular forms. With deg.D/ D k.2g�2/,we have X

Q

�ordQ.f /

1eQ� k.1 � 1

eQ/�D k.2g � 2/C k�1

where the sum is taken over a set of representatives Q of zeroes, poles, and cusps off in �nH. For �.1/, this becomes

ordi1.f /C1

2ordi.f /C

1

3ord�.f /C

XQ

ordQ.f / Dk

6;

where the sum is taken over the set of all points Q on the interior of a fundamentaldomain for �.1/. For G2, we have a simple pole at �, and no others; for G3, we havea simple zero at i , and no others; for �, which has no zero in H, has a simple zeroat i1.

36

11 Lecture Eleven

Recall that, at the end of the last lecture, we calculated the dimensions of M2k D

M2k.�.1//. Today, we will show that

M WD

Mk�1

M2k D CŒG2; G3�;

where G2 and G3 are the Eisenstein series from before.

Lemma 1. The map f 7! f� is an isomorphism of M2k onto S2kC12.

Proof. That this map is a homomorphism is clear, and injectivity is similarly obvious;thus it suffices to show surjectivity. Let g 2 S2kC12, and consider the quotient g

�;

because � has no zeroes in H and has only a simple zero at i1, it is easy to see thatg

�2M2k. This completes the proof.

This lemma will make it easy to prove that G2 and G3 generate M . We will useinduction on k, the cases k � 6 having already been established. If k > 6, we willwrite k D 2n C 3m (with m; n � 0) and consider the modular form h D Gn2G

m3 ,

which has weight 2k. It is nonzero at i1; in fact, it can be shown that its valuethere is

.2�.4//n.2�.6//m ¤ 0:

Now, for any f 2M2k, we will put

Qf D f �f .i1/

h.i1/h 2 CŒG2; G3�:

And this... completes the proof? (Clean this up)

We now want to prove that G2 and G3 are algebraically independent; we will do thisby proving directly that G32 and G23 are algebraically independent, from which ourdesired result will follow. We begin with another

Lemma 2. Let f; g 2M2k; then either f and g are proportional, orthey are algebraically independent.

We know that G2 vanishes at � while G3 does not.

37

Proof. Suppose P.f; g/ D 0 for some polynomial P.X; Y / 2 CŒX; Y �, and write Pdfor the degree d homogeneous component of P ; then Pd .f; g/ is a modular form ofweight dk. Writing

P DXd�0

Pd ;

we see that Pd .f; g/ D 0 for every d . We define a function of a single variable via

p.f=g/ WDPd .f; g/

gd;

which must then be identically zero as well. Because p is a nonconstant polynomial,it follows that we must have that f=g is a constant, and so f and g are proportional,as claimed.

We remark here that our j -invariant from before, defined

j.z/ D1728g32.z/

�.z/;

gives an isomorphism �.1/nH� ! S , where S is the unit sphere. Indeed: j is theunique isomorphism �.1/nH� ! S satisfying

j.i/ D 1728; j.�/ D 0; j.i1/ D1:

We now discuss Fourier expansions. Suppose � � �.1/ has finite index, and recallthe generator T 2 �.1/. For some positive integer h (the width of the cusp, definedabove), we will have T h 2 �, and so we have for any modular form f 2M2k.�/ theidentity

f .z C h/ D f .z/;

and so f is periodic with period h. It follows that we can take its Fourier seriesexpansion. For the moment, we restrict our attention to � D �.1/. Writing q Dexp.2�iz/ as before, we will obtain the series expansion

f .z/ DXn�0

anqn;

for some an 2 C; that the series has nonnegative coefficients follows from the holo-morphy of f at the cusp. We will compute the Fourier series expansion for G2k.

38

We have the identity

� cot.�z/ D limN!1

NXnD�N

1

z C nI

we call the expression on the right the principal value of the (not absolutely conver-gent) series

Pn2Z.z C n/

�1; thus

� cot.�z/ D1

zC

1XnD1

�1

z C nC

1

z � n

�D1

zC 2

1XnD1

z

z2 � n2:

We have also the Weierstrass factorization

sin.�x/ D �x

1YnD1

�1 �

x2

n2

�I

in fact, the first identity is a consequence of the second, by taking logarithmic deriva-tives.

We will prove the first identity: � cot.�z/ has simple poles (with residue one) at allintegers, and nowhere else. The series

1

zC

1XnD1

�1

z C nC

1

z � n

�also has simple poles with residue one at each integer. Moreover, both functions areclearly periodic, with period one, and are holomorphic away from the poles; it mayalso be shown that both functions are bounded in vertical strips. It follows thattheir difference is holomorphic on all of C, which is bounded everywhere, and musttherefore be a constant. The value of this constant is shown to be zero by evaluatingboth functions at 1

2.

Now: an easy calculation (left as an exercise) shows that

cot.z/ D ie2iz C 1

e2iz � 1D i C

2i

e2iz � 1:

Proposition 11.1. For k � 2, one has

Gk.z/ D 2�.2k/C 2.2�/2k

.2k � 1/Š

Xn�1

�2k�1.n/qn;

where �`.n/ is the `-th power sum-of-divisors function

�`.n/ DXd jn

d `:

39

Proof. We have

� cot.�z/ D �i � 2�i

1XnD1

qn D1

zC

1XnD1

�1

z C nC

1

z � n

�I

differentiating both sides k � 1 times (with respect to z) gives

.�2�i/k

.k � 1/Š

1XnD1

nk�1qn DXn2Z

1

.nC z/k:

Thus

Gk.z/ DX

.m;n/¤.0;0/

1

.mz C n/2kD 2�.2k/C 2

1XnD1

Xm2Z

1

.nz Cm/2k

D.�2�i/k

.k � 1/Š

1XmD1

m2k�1qmn D 2�.2k/C2.2�i/2k

.2k � 1/Š

1XnD1

1XmD1

m2k�1qmnI

careful rearrangement of the last series then yield the result.

We close with an application. Define constants Bk by the equation

x

ex � 1D 1 �

x

2C

1XkD1

.�1/kC1Bkx2k

.2k/ŠI

then Euler was able to show

�.2k/ D22k�1

.2k/ŠBk�

2k:

40

12 Lecture Twelve

Last time, we closed by introducing the Bernoulli numbers and evaluating the Rie-mann zeta function at even positive integers. It is an easy consequence of the propo-sition that closed the last lecture, that 1

�2kGk has rational Fourier coefficients; with

more work, it can be shown that

�.z/ D .2�/12.q � 24q2 C 252q3 � 1472q4 C � � � /

In fact, it may be shown that � admits an infinite product factorization: as we willprove later,

�.z/ D .2�/12q

1YnD1

.1 � qn/24;

which itself proves that �.2�/12

has integer Fourier coefficients.

Definition: the Ramanujan �-function is the arithmetic function � W N ! Cdefined by

�.z/ D .2�/12q

1YnD1

�.n/qn:

Ramanujan conjectured several properties of the � -function:

� it is multiplicative; that is, �.mn/ D �.m/�.n/ if .m; n/ D 1;

� if p is a prime and k � 1, then �.pkC1/ D �.p/�.pk/ � p11�.pk�1/; and

� if p is prime, then j�.p/j � 2p11=2.

None of these facts are obviously true from the definition of the function. The firsttwo conjectures were proven by Mordell in 1917; the last was proven by Deligne in1976, as part of his proof of the Weil conjectures.

To � , and other functions more generally, we can associate a series. Indeed, if f isa modular form with Fourier expansion

f .z/ D

1XnD0

anqn;

then we can define a Dirichlet series

Lf .s/ D

1XnD0

an

nsI

41

bounds on the coefficients an can then be found by investigating the analytic prop-erties of Lf . The Shimura-Taniyama-Weil conjecture (now proven) provides a cor-respondence between such L-functions and the zeta function �E attached to ellipticcurves over finite fields; the conjectured properties of such �E were ultimately provenby Wiles in his proof of Fermat’s last theorem. We hope to focus on the details ofthis work during the remainder of the course.

For the moment, consider our modular function j D1728g32�

as a formal power seriesin q; we know

�.z/ D .2�/12q.1C � � � /;

and similarly thatg32.z/ D .2�/

12.1C � � � /;

from which it follows that

1728g32�

D q�1 C c C � � �

for some constant c. In fact, the first few coefficients are known:

j.z/ D q�1 C 744C 196884q C � � �

Of course, we observe that 196884 is one greater than the dimension of the smallestnontrivial irreducible representation of the Monster group.

Recall the correspondence we established between Y.1/, i.e. the affine modular curve�.1/nH, and the set of equivalence classes of elliptic curves over C; namely, j�1. IfL=Q is an algebraic extension of Q in C, and an elliptic curve E is defined by theequation

y2 D 4x3 � ax � b;

with a; b 2 L, then the restriction of j�1 to the set of (equivalence classes of) ellipticcurves defined over L defines a bijection onto the set Y.1/.L/ of L-points of Y.1/. Ifwe take � 2 H with ŒQ.�/ W Q� D 2, then it will follow from these observations thatj.�/ 2 Q.�/.

Theorem 12.1. Let � 2 H be a quadratic irrational number; then j.�/is algebraic, and � together with j.�/ generates the Hilbert class fieldof Q.�/. That is, ŒQ.�; j.�// W Q� D 2h, where h is the class number ofQ.�/.

Proof. Omitted.

42

In fact, more can be said: suppose we know that, for some particular � 2 H, thevalue j.�/ is algebraic; then � is itself imaginary quadratic.

We pause now to see how the j -invariant helps us to approximate � : if L D Q.�/is some quadratic extension with class number 1, then j.�/ 2 Q; for instance, if

� D 1Cip163

2, then

q D exp.2�i�/ D exp.i� � �p163/ D �e��

p163;

and so j.�/ D q�1C744CqC� � � � 744Cq�1. Apparently, this... lets us approximate�?

What we want to prove now is the following: the function

f .z/ D q

1YnD1

.1 � qn/24

is a cusp form of weight 12, and so the product is absolutely convergent everywhere onH. Observe that z 2 H implies jqj < 1, and therefore that f .z/ converges absolutelyon H, and it suffices to check the transformation property. Periodicity is clear, andit remains only to check that f .�1=z/ D z12f .z/.

At the same time, we will consider the “quasi-Eisenstein series”

G1.z/ DXn�1

1

n2kC

Xm¤0

Xn2Z

1

.mz C n/2I

this series converges, but not absolutely, for z 2 H. We avoid this problem byredefining G1 according to the general Fourier expansion of Gk we found before:

G1.z/ D .2�i/2

�B1

4C

1XnD1

�.n/qn

!;

which has constant term �.2/ D �2

6. We simplify notation by defining generally:

Ek D1

Gk.0/GkI

then Ek.z/ has constant term 1.

Proposition 12.2. For any a; b; c; d 2 Z with ad � bc D 1, we have

G1

�az C b

cz C d

�D .cz C d/2G1.z/ � �ic.cz C d/:

43

We will consider the proof next time. With this knowledge in hand, however, we cancompute

1

2�ilog f .z/ D 1 � 24

1XnD1

ne2�inz

1 � e2�inzD E1.z/:

44

13 Lecture Thirteen

Today we will introduce the Petersson inner product. We begin by placing a metricon H.

Recall that H is isomorphic to the Poincare disc D; this is the open unit disc, whosegeometry is given by the Euclidean axioms, apart from the parallel postulate, whichis replaced by the axiom:

� Given a line L and a point p not on L, there exist more than one line passingthrough p which do not meet L.

The resulting geometry has, for its straight lines, all diameters of the disc, as well asall arcs of Euclidean circles contained in D, which are orthogonal to the boundary ofD. We then transform this under the correct Mobius transformation to obtain themetric on H.

Alternatively, we can define the geometry on H by defining its metric. For z1; z2 2 H,let 11;12 be the points on R which intersect the (hyperbolic) line joining z1 andz2; assume without loss of generality that the points can be traversed in the order11; z1; z2;12. Define the cross-ratio of z1; z2; z3; z4 to be

D.z1; z2; z3; z4/ D.z1 � z3/.z2 � z4/

.z2 � z3/.z1 � z4/I

we can then define the metric on H via

d.z1; z2/ D logD.z1; z2;11;12/

(a simple calculation shows that D.z1; z2;11;12/ is real and positive). Locally, themeasure on H is given by

ds2 Ddx2 C dy2

y2;

considered as a real manifold. The group PSL2.R/ acts naturally on this mani-fold, and the cross-ratio is invariant under this transformation. The correspondingPSL2.R/-invariant volume form on H is dx dy

y2; this follows (as an exercise) from the

isomorphismSL2.R/ D NAK;

where K D SO2.R/, A is the subgroup of diagonal matrices, and N is the subgroupof upper- (or lower-) triangular matrices.

45

With this volume form, we can compute the (finite) volume of �nH: for � D �.1/,we have

vol.�.1/nH/ D

ZD

dx dy

y2D

Z 1=2

�1=2

Z 1p1�x2

dy

y2dx D Œarcsin.x/�1=2

xD�1=2D�

3:

For more general �, we have by Stokes’ theorem (exercise)Z�nH

dx dy

y2D 2�

2g � 2C �1 C

XP

.1 � 1=eP /

!:

Moreover, if f is a modular form of weight 2k for �, then the number of zeroes off counted with multiplicity equals k vol.�nH/

4�. We remark that the ubiquitous factor

112

appears here as the quotient of vol.�.1/nH/ by 4� , i.e. the normalized volume ofthe 2-sphere.

We will now place an inner product on the space M2k.�/, for some arbitrary finite-index subgroup � of �.1/.

Proposition 13.1. Let z D x C iy, and suppose f; g 2M2k.�/; then

f .z/g.z/y2k�2dx dy D f .z/g.z/ Im.z/2k�2dx dy

is invariant under the transformation z 7! z; 2 �.

Proof. For 2 �, write D

�a b

c d

�; then

Im. z/=1

jcz C d j2Im.z/;

and sinced

dz. z/ D

1

.cz C d/2;

we know that

f . z/ D .cz C d/2kf .z/; g. z/ D .cz C d/2kg.z/:

An easy calculation (exercise; see p. 64 in Milne) shows that

�.dx dy/ Ddx dy

jcz C d j4:

The result is now immediate.

46

We remark that the case k D 0 and f D g D 1 gives a proof of the invariance of themetric dx dy

y2.

Proposition 13.2. If f is a cusp form for �.1/, then

jf .x C iy/j D O.e�2�y/

as y !1.

Proof. (sketch) We use the Fourier expansion

f .z/ D

1XnD1

anqn;

with q D exp.2�iz/ D exp.�2�y/ exp.2�ix/. The absolute value of the dominantterm is thus clearly exp.�2�y/.

A more general result exists for cusp forms of arbitrary level, which we will see later.Retaining the above notation for the Fourier expansion, we can also obtain growthestimates for the coefficients an.

Proposition 13.3. If f is a cusp form of weight 2k for �.1/, thenjanj � cn

k for some (absolute) constant c.

Proof. It is clear from our above remarks that the map z 7! jf .z/j.Im z/k is �.1/-invariant, and therefore bounded on H. It follows that there exists a constant c suchthat

jf .z/j �c

ykfor all z D x C iy 2 H:

We compute the Fourier coefficients by the definition: we have

e�2�nyan D

Z 1

0

f .x C iy/e�2�inxdx;

and taking y D 1n

gives

e�2� janj D

ˇZ 1

0

f .x C i=n/e2�inxˇ:

The integrand is dominated by cy2

, as remarked above, and so the left-hand side is

dominated by cnk.

47

Now: we define the Petersson inner product of f; g 2M2k.�/ by the formula

hf; gi D

ZD

f .z/g.z/y2k�2dx dy;

where D is a fundamental domain for the action of � on H, and one of f; g is a cuspform.

Proposition 13.4. The Petersson inner-product is a positive-definiteHermitian product; thus, S2k.�/ is a finite-dimensional Hilbert space.

Tomorrow, we will discuss Hecke operators.

48

14 Lecture Fourteen

Today, we define the Hecke operators. Recall that we have already seen the relation-ship between modular forms and functions on lattices; we will define operators onlattices, and thereby obtain an action on modular forms. Let D be the free abeliangroup on the set of full-rank lattices

ƒ.!1; !2/ D Z!1 C Z!2I

an element of D therefore can be written

mXiD1

ni Œƒi �;

with ni 2 Z. We remark that Œƒ� is not an equivalence class, and that this notationis used only to emphasize the lattice ƒ as an element of D .

Definition: for n 2 N, define the operator

T .n/ W D ! D

by the formula

T .n/.Œƒ�/ DXƒ0�ƒ

ŒƒWƒ0�Dn

Œƒ0�:

We also define the operator R.n/ W D ! D via R.n/.Œƒ�/ D Œnƒ�.

Proposition 14.1. Let T .n/; R.n/ be as above; then

� one has T .m/ ı T .n/ D T .mn/ if .m; n/ D 1; and

� if p is a prime and n � 1, then

T .pn/ ı T .p/ D T .pnC1/C pR.p/ ı T .pn�1/:

We remark that an immediate corollary of this proposition is the fact that the T .p/(with p prime) generate the same algebra as hT .n/i.

Proof. By the correspondence theorem, every index mn sublattice is an index sub-lattice of a unique index m sublattice, when m and n are coprime. From this, wededuce that

ƒ=ƒ0 Š Z=.mn/Z Š .Z=mZ/˚ .Z=nZ/;

49

and the result now follows. For the second point: we know that both left- and right-hand sides are formal sums of sublattices of index pnC1. If ƒ0 is an index pnC1

sublattice of ƒ, we can compare its multiplicities on both sides of the equation.

In the first case, we have ƒ0 � pƒ D R.p/.Œƒ�/; we know Œƒ W pƒ� D p2. So on theright-hand side, ƒ0 occurs with multiplicity p C 1. On the left-hand side, becauseƒ0 is contained in pƒ, it must lie in every sublattice of index p. Therefore, themultiplicity of ƒ0 on the left-hand side is the number of sublattices of index p in ƒ;we have

ƒ=pƒ Š Z=pZ˚ Z=pZ Š F2p:

The number of sublattices of index p in ƒ is precisely the number of lines in F2p,which is exactly

p2 � 1

p � 1D p C 1;

and the fact is proven.

In the second case, when ƒ0 6� pƒ, then ƒ0 appears not at all in the second termon the right-hand side, and with multiplicity one in the first term. On the left-handside, we claim that there is only a single index p sublattice of ƒ which contains ƒ0;indeed, suppose ƒ1; ƒ2 both contain ƒ0; then

ƒ0 � ƒ1 \ƒ2 D pƒ;

which is a contradiction, and the result is proven.

We remark that R.p/ and T .p/ commute for every prime p, and also that T .pn/and T .p/ commute for every p. It follows that the algebra generated by the T .n/ iscommutative. Today, we will build an algebra of Hecke operators acting on Mk.�.1//

for every k; in general, this can be done for arbitrary subgroups.

For the rest of this lecture, we fix � D �.1/. Recall that we established a correspon-dence between weakly modular functions f .z/ of weight 2k, and weighted functionson lattices F.ƒ/; F.�ƒ/ D ��2kF.ƒ/. If T is an operator on lattices, define theaction of T on the set of functions of lattices via

.T � F /.ƒ/ D F.T .ƒ//I

thenT .n/ � F D

Xƒ0�ƒ

F.ƒ0/ and R.n/ � F D n�2kF:

50

We will continue to normalize our lattices so that one generator is 1, and the otherhas positive imaginary part:

ƒ.!1; !2/ D ƒ.1;!1

!2/ D ƒ.1; z/ DW ƒz:

To the weakly modular function f we define the function F on lattices by the formulaF.ƒz/ D f .z/; it is an exercise to show that this is well-defined.

Definition: Define the action of Hecke operators T .n/ on the set of weakly modularfunctions f by the equation

T .n/ � f .z/ WD n2k�1T .n/ � F.ƒz/:

It is easy to check that T .n/ � f is also weakly modular of weight 2k.

Now, we claim that

T .n/ � f .z/ D n2k�1Xa;b;d

d�2kf

�az C b

d

�;

where the sum is taken over all triples .a; b; d/ such that

ad D n; a � 1; and 0 � b < d:

We give a rough justification: the formula aries from an explicit description of sub-lattices of ƒ of index n. Such a lattice is the image of ƒ under the action of someelement 2 M2.Z/ of determinant ˙n. It may be shown that the set M2.n/ of all2 � 2 matrices with integer entries and determinant n has decomposition

M2.n/ Daa;b;d

SL2.Z/

�a b

d

�;

the coproduct taken over the same index as above. Thus, taking the quotient ofM2.n/ by the action (by left-multiplication) of SL2.Z/ gives, as a set of coset repre-sentatives, our set of triples .a; b; d/ from before.

Having introduced the Hecke operators on the space of weakly modular functions, wemight ask: how do Hecke operators affect Fourier expansions? Recall our operatorsT .N / acting on M2k.�.1//.

Theorem 14.2. Fix a weight k and let f 2M2k.�.1// have Fourierexpansion

f .z/ D

1XmD0

cmqm:

51

Then

T .n/ � f .z/ D

1XmD0

mqm;

where m D

Xd j.m;n/d�1

d 2k�1cmn=d2 :

Proof. Omitted. (see Milne, proposition 5.16(b))

Corollary 1: The constant term of the Fourier expansion of T .n/ �f is c0 ��2k�1.n/,and the coefficient of the linear term is cn.

Corollary 2: If n D p is prime, then

m D

(cpm if p - mIcpm C p

2k�1cm=p if pjm:

Corollary 3: Every T .n/ maps cusp forms to cusp forms.

Fact: For � D �.1/, the Hecke operators T .n/ are Hermitian with respect to thePetersson inner product; hence, the space fT .n/g have a common basis of eigenformson Sk.

We close now with the motivation for our next lecture: if f 2M2k.�.1// is a nonzeroeigenform for T .n/, and has Fourier expansion

f .z/ D

1XnD0

cnqn;

then we have cn D �.n/c1, where �.n/ is the eigenvalue of T .n/ on f . In particular,the Ramanujan � -function from Lecture Twelve is multiplicative (because S12 isone-dimensional). We will pick up here next time.

52

15 Lecture Fifteen

Last time, we introduced the Hecke operators T .n/ for M2k.�.1//, and saw thatit satisfied various properties. In particular, we saw how the Fourier coefficients ofHecke eigenforms are determined by its eigenvalues and constant term.

Proposition 15.1. Let Gk be the Eisenstein series of weight 2k for�.1/; then Gk is an eigenform of T .p/ with eigenvalue �2k�1.p/ Dp2k�1 C 1.

Proof. Omitted.

Proposition 15.2. Let Sk be the space of cusp forms of weight 2k for�.1/; then

hf;Gki D 0 for all f 2 Sk;

where h; i is the Petersson inner product.

These results mean that we have a direct sum decomposition

M2k.�.1// D S2k ˚ hGkiI

we also recall our isomorphismMk

M2k.�.1// D CŒE2; E3�;

where Ek D .2�.2k//�1Gk are the normalized Eisenstein series.

Theorem 15.3. Let M2k.Z/ be the Z-submodule of M2k.�.1// con-sisting of modular forms with integer Fourier coefficients; thenM

k

M2k.Z/ D ZŒE2; E3�:

Proof. We have shown that the Ek are Hecke eigenforms, and we know that everyT .p/ has eigenvalues in Z on Ek; hence, every T .n/ has eigenvalues in Z, and so itfollows from our work in the previous lecture that the Fourier coefficients of Ek mustbe integers.

We also know that � has eigenvalues in Z, and that f 7! f� gives an isomorphismM2k ! S2kC12; the result is now immediate.

53

We remark that, as the action of each T .n/ on S2k.�.1// is Hermitian, then T .n/must have real eigenvalues. Corollary: The eigenvalues of T .n/ are algebraic inte-gers, and so lie in a totally real finite extension of Q.

Proof. The action of T .n/ on M2k.�.1// preserves the lattice M2k.Z/; thus, it actsby a matrix with Z coefficients once we have chosen a basis arising from this lattice.In particular, if the Fourier expansion is normalized so that the coefficient of q is 1,then all coefficients must be algebraic integers.

We will now change our perspective slightly, and consider modular forms for gen-eral congruence subgroups. Recall that we have defined the principal congruencesubgroups

�.N/ D

� D

�a b

c d

�W � I mod N

�I

we showed in homework that �.N/ is normal in SL2.Z/, and that

SL2.Z/=�.N / Š SL2.Z=NZ/

has cardinality N 3YpjN

�1 �

1

p2

�. We called a congruence subgroup any subgroup

� of �.1/ of finite index which contains some �.N/. We have also establishedthat the cusps of �nH (for an arbitrary congruence subgroup �) are in one-to-onecorrespondence with the �-orbits of P1.Q/.

We used the Riemann-Roch theorem to prove that

dim M2k.�/ D .2k � 1/.g � 1/C

�k

2

�"2 C

�2k

3

�C k�1;

where g is the genus of �nH, "` is the number of `-elliptic points (for ` D 2; 3),and �1 is the number of �-inequivalent cusps of H. If N� denotes the image of � inPSL2.Z/, then

�1 D

The stabilizer P of i1 in �.1/ is generated by

�1 1

1

�, while the stabilizer of i1

in � is generated by

�1 h

1

�for some positive integer h. These observations give a

bijection�n SL2.Z/=P ! fcusps of �

54

via

�

��a b

c d

��P 7! � �

�ac

�:

Fact: When k � 2, one has

dim S2k.�/ D dim M2k.�/ � �1;

whiledim S2.�/ D dim M2.�/ � .�1 � 1/:

In the remainder of today’s lecture, we will compute the Eisenstein series associatedto a given cusp for � (for k � 2); this will be nonzero at our distinguished cusp, andzero at every other cusp. In every case, the Eisenstein series we construct will beorthogonal (with respect to the Petersson inner product) to the space of cusp forms.

In the �.1/ case, we have from the definition

Gk.z/ D

0X.c;d/2Z2

.c;d/¤.0;0/

1

.cz C d/2kD

Xm2Z

X.c;d/2Z2

gcd.c;d/Dm

1

.cz C d/2k

D

Xm�1

1

m2k

Xgcd.c;d/D1

1

.cz C d/2kD �.2k/

Xgcd.c;d/D1

1

.cz C d/2k:

That is,

E2k.z/ D1

2�.2k/Gk.z/ D

1

2

Xgcd.c;d/D1

1

.cz C d/2k;

where E2k from now on denotes the normalized Eisenstein series of weight2k. Forgeneral �.N/, let Nv 2 .Z=NZ/2 be an element of order N ; we will consider Nv as arow vector Œ Ncv; Ndv�, where Na is the image of a modulo N . Let

ı D

�a b

cv dv

�2 SL2.Z/I

we know that some such ı must exist because Nv has order N . We now introduce twoimportant pieces of notation.

Definition: The automorphy factor j. ; z/ is the function

j W SL2.Z/ �H! C

55

defined

j. ; z/ D cz C d for D

�a b

c d

�(our notation is from Diamond-Shurman; Milne has j. ; z/ D .cz C d/2 instead). Asimple calculation shows that the automorphy factor satisfies the cocycle relation

j. 0; z/ D j. ; 0 � z/j. 0; z/ for all z 2 H; ; 0 2 SL2.Z/:

We pause here for a moment to remark that the modularity condition (of weight 2k)can be written as the statement that

f . � z/ D j. ; z/2kf .z/ for all 2 SL2.Z/; z 2 HI

in fact, any function j. ; z/ satisfying the cocycle condition leads to a reasonabletheory.

Definition: Suppose f is a complex-valued function on H, and let 2 SL2.Z/. Aright action of SL2.Z/ on f is given by

.f Œ �k/.z/ D j. ; z/�kf . .z//I

clearly, f is a modular form of weight k for �.1/ if and only if it is holomorphic onH, and Œ �k acts trivially on f for all 2 SL2.Z/.

Having established these notations, we will begin the next lecture by constructingthe previously-promised Eisenstein series for �.N/. Our method will be to takean arbitrary function on H, sum over its values on �-translations (weighted by theautomorphy factor).

56

16 Lecture Sixteen

We continue with the construction of Eisenstein series associated to the principalcongruence subgroups �.N/. Let Nv 2 .Z=NZ/2 have order N , and write Nv D . Ncv; Ndv/with cv; dv 2 Z; then there exists a matrix

ı D

�a b

cv dv

�2 SL2.Z/:

We define a function (per Milne’s notation): for k � 2, let

E Nvk .z/ D "NX

gcd.c;d/D1.c;d/�v mod N

.cz C d/�2k;

where "N D12

if N D 1; 2, and equals 1 otherwise (Diamond-Shurman would callthis function E Nv

2k). We can also write

E Nvk .z/ DX

2PC\�.N/n�.N/ı

j. ; z/�2k;

where

PC D hT i D

��1 n

1

�W n 2 Z

�and j. ; z/ D .cz C d/ is the automorphy factor we introduced last time. A proofsimilar to the last time will show that this series converges absolutely and uniformlyon compact subsets.

Proposition 16.1. For 2 SL2.Z/, one has

.E Nvk Œ �2k/.z/ D E. z/

k.z/;

where .f Œ �2k/ D j. ; z/�kf . z/ as before.[

Proof. Omitted.

Corollary: One has E Nvk2M2k.�.N //.

Proposition 16.2. One has

limIm.z/!1

E Nvk .z/ D

(1 if Nv � ˙.0; 1/;

0 otherwise.

57

Corollary: The value of E Nvk.z/ at the cusp corresponding to Nv is 1, and is zero at

every other cusp.

We outline the proof of the corollary: let ı D

�a b

cv dv

�as before, and let s D a0

c0be

a cusp for �.N/. Then there exists a Mobius transformation

˛ D

�a0 b0

c0 d 0

�that sends i1 to s. The behaviour of E Nv

k.z/ at s is therefore determined by

E NvkŒ˛�2k.z/ at i1; but

E Nvk Œ˛�2k.z/ D E.˛z/

k.z/ D E

.0;1/ı˛

k.z/;

which is nonzero at i1 if and only if

.0; 1/ı˛ D ˙.0; 1/;

if and only if �a0

c0

�� ˙

��dvcv

�mod N;

if and only if �.N/s D �.N/.�dv=cv/. It follows that the cusps of �.N/ are inbijection with the orbits of �.N/ on Q[fi1g. We have a decomposition, orthogonalwith respect to the Petersson inner product, for k � 2:

M2k.�.N // DM

Nv2.Z=NZ/2

ord. Nv/DN

C �E Nvk ˚S2k.�.N //:

Parts of this statement are obvious (e.g. that E Nvk

is a modular form), other parts areknown, and the rest will just be believed.

We will now generalize our earlier construction of the Hecke operators. Let �1; �2be two congruence subgroups of �.1/; they can also be considered as subgroups ofGLC2 .Q/ (i.e. rational matrices with positive determinant). We want to consider thedouble coset

�1˛�2 D f 1˛ 2 W i 2 �ig;

where ˛ 2 GLC2 .Q/. Consider the space �1n�1˛�2 of left �1-orbits; we have thedecomposition

�1n�1˛�2 Daˇi2�2

˛ˇi :

58

Each double coset �1˛�2 will give us an operator acting on Mk.�1.N //, where �1.N /is a particular congruence subgroup which we will define below. As we saw in thecase of the full modular group, the double-coset decomposition will give relationsbetween the Hecke operators we will construct.

Now: there is a one-to-one correspondence between orbit spaces

�1n�1˛�2 $ ..˛�1�1˛/ \ �2/n�2I

the proof of this fact is an easy exercise. The action of SL2.Z/ on functions f W H! Cis extended to an action of GLC2 .Q/ via

.f Œ˛�k/.z/ D .det˛/k�1j.˛; z/�kf .˛ � z/I

notice that the condition that det˛ > 0 means that f Œ˛�k is indeed a function onH, and not on NH.

Definition: The weight k double coset operator of �1˛�2 on function f W H ! Cis defined

f Œ�1˛�2�k DXj

f Œˇj �k;

where the sum runs over a complete set of coset representatives

�1˛�2 Daj

�1ˇj :

We first claim that there are only finitely many coset representatives ˇj ; our corre-spondence above shows that it is enough to show that there are only finitely manycosets ˛�1�1˛ \ �2n�1, which is clear because ˛�1�1˛ is a congruence subgroup of�.1/ which contains �2.

Fact: Any two congruence subgroups are commensurable, meaning that theirintersection has finite index in both.

Moreover, the function does not depend on choice of coset representatives: this isclear in the case f 2M2k.�.1//. The general case is left as an exercise.

Proposition 16.3. The action Œ�1˛�2� is a map M2k.�1/!M2k.�2/.

Proof. (sketch) We check the condition of weak modularity. Take 2 2 �2; thenif fˇj g are a set of representatives for the �1 orbits, then fˇj 2g is also a set ofrepresentatives, and ˇ1 2; ˇ2 2 must lie in different �1-cosets. This is the case if andonly if ˇ1; ˇ2 lie in different �1-cosets. It follows that the action of 2 permutes theterms in our sum.

59

In the special case �1 � �2, we obtain an inclusion

Œ�1�2� WMk.�1/ ,!Mk.�2/

by taking ˛ D id. In the special case ˛�1�1˛ D �2, we obtain an isomorphism

Œ�1˛�2� WMk.�1/��!Mk.�2/:

Finally, if �1 � �2, we again take ˛ D id, and our sums are indexed by the cosets�1n�2. Then Œ�1�2� is a projection

Œ�1�2� WMk.�1/� Mk.�2/:

In general, any operator Œ�1˛�2� can be written as a composition of these three maps:inside �1 is the subgroup �1\˛�2˛

�1, whose modular forms are isomorphic to thoseof ˛�1�1˛ \ �2, which lives inside �2. Thus:

Œ.˛�1�1˛\�2/�2�Œ.�1\˛�2˛�1/.˛�1�1˛\�2/�Œ�1.�1\˛�2˛

�1/� WMk.�1/!Mk.�2/:

Now: define the subgroups

�0.N / D

��a b

c d

�2 SL2.Z/ W c � 0 mod N

�and

�1.N / D

��a b

c d

�2 SL2.Z/ W a � b � 1 mod N and c � 0 mod N

�I

clearly we have inclusions

�.N/ � �1.N / � �0.N /:

Consider the homomorphism �0.N /! .Z=NZ/� sending

�a b

d

�to d mod N ; this

is a surjective homomorphism with kernel �1.N /, and so �1.N / is normal in �0.N /,and has quotient

�0.N /=�1.N / Š .Z=NZ/�:

The double-coset operators we have just constructed give us an action of �0.N / onMk.�1.N //: take ˛ 2 �0.N /, and consider the operator Œ�1.N /˛�1.N /�k. Because�1 is normal in �0 (in fact, �0.N /=�1.N / Š .Z=NZ/�), we know that

Œ�1.N /˛�1.N /�k WMk.�1.N //!Mk.�1.N //:

60

Moreover, this depends only on the image of ˛ 2 �0=�1; that is, only on the imageof d modulo N . We therefore have an action of .Z=NZ/� on the finite-dimensionalcomplex vector space Mk.�1.N //, and so there is an orthogonal decomposition

Mk.�1.N // DM

�22.Z=NZ/�

Mk.�1.N /; �/;

where Mk.�1.N /; �/ is the subspace of Mk.�1.N // on which �0 acts by the char-acter �. Note that we have slightly abused notation by considering � as a Dirichletcharacter

� W .Z=NZ/� ! C�:

We will pick up here next time.

61

17 Lecture Seventeen

We can define an operator hd i WMk.�1.N //!Mk.�1.N // via

hd if D f Œ˛�k;

where

˛ D

�a b

c ı

�2 �0.N /; ı � d mod N:

This is known as the diamond operator; in this case, the operator Œ˛�k coincideswith the operator Œ�1.N /˛�1.N /�k. This allows us to rewrite our subspaces fromlast lecture via

Mk.N; �/ D ff 2Mk.�1.N // W hd if D �.d/f g:

We will now construct another sort of Hecke operator: take ˛ D

�1

p

�2 GLC2 .Q/,

with p prime, and put �1 D �2 D �1.N /. We define an operator

Tp WMk.�1.N //!Mk.�1.N //

by putting Tp D Œ�1.N /�1p

��1.N /�k.

Proposition 17.1. If pjN , then

Tpf D

p�1XjD0

f��1 jp

��k;

and otherwise

Tpf D

p�1XjD0

f��1 jp

��kC f

��m nN p

� �p1

��k;

where mp �Nn D 1.

The proof follows from the fact that

�1.N /

�1

p

��1.N / �

�1 x

p

�mod N

62

for some x modulo N , and hence

�1.N /

�1

p

��1.N / D

aj

�1.N /ˇj

for an appropriate set fˇj g of coset representatives (we will skip the details).

We compare the operators Tp with the operators T .p/ on Mk.�.1// from before:the sum over j for Tp is the same as what we had for T.p/, and of course if N D 1

then p - N for all p, and so the extra term

�p

1

�does indeed appear. We state

without proof the following properties of the operators Tp:

� Suppose f 2Mk.�0.N // DMk.N; 1/ (where 1 denotes the trivial character ).If an.f / denotes the nth Fourier coefficient of the function f , then an.Tpf / Danp.f /C1N .p/p

k�1an=p.hpif /, where 1N .p/ is 1 if pjN and is zero otherwise.

� The operators hd i and Tp commute, and for p ¤ q both prime, we haveTp � Tq D Tpq.

Definition: Suppose n 2 N; then define an operator on Mk.�1.N // via

Tnf D

(hn mod N i if .n;N / D 1;

0 if .n;N / > 1:

It may be easily shown from our properties above that

Tm � Tn D Tmn if .m; n/ D 1

andTpr D TpTpr�1 � p

k�1hpiTpr�2

whenever r � 2. Moreover, if f 2Mk.�1.N //, then

am.Tnf / DX

d j.m;n/

dk�1amn=d2.hd if /I

in particular, if f 2Mk.N; �/, then

am.Tnf / DX

d j.m;n/

�.d/dk�1amn=d2.f /:

63

We remark that Mk.N; �/ can be thought of as the �-eigenspace of hd i in Mk.�1.N //.By construction, the algebra of operators generated by all Tn and hmi is commutative.Finally, the algebra of Hecke operators preserves Sk.�1.N //.

We now return briefly to the Petersson inner product, viz.

hf; gi� D1

vol.�/

Z�nH

f .z/ Ng.z/y.z/kdxdy

y2;

wherevol.�/ D Œ N�.1/ W N�� � vol.�.1/nH/;

and N� denotes the image of � in PSL2. We observe that, if �1 � �2 and f; g 2

Sk.�2/, thenhf; gi�1 D hf; gi�2 :

If � � �.1/ is a congruence subgroup and ˛ 2 GLC2 .Q/, put ˛0 D det.˛/˛�1. Then:if ˛�1�˛ � �.1/, then for f 2 Sk.�/; g 2 Sk.˛

�1�˛/, one has

hf Œ˛�k; gi˛�1�˛ D hf; gŒ˛0�ki� :

Moreover, if f; g 2 Sk.�/, then

hf Œ�˛��k; gi D hf; gŒ�˛0��ki:

Now: recall that the adjoint operator T � of a given operator T satisfies

hv; T wi D hT �v;wi;

and that T is called normal if it commutes with T �.

We will use the fact that hpik D hpi�1; this follows from our above calculationsapplied to ˛ D

�1p

�, for which ˛0 D ˛. Similarly, we can show

T �p hpi�1Tp for p - N:

64

18 Lecture Eighteen

We begin by correcting a statement from the previous lecture (regarding the coeffi-cients am.f /). Today, we will discuss old forms and new forms.

There are two ways to fit Sk.�1.N // into Sk.�1.M// when M jN : one is via theidentity inclusion, as every modular form for �1.M/ is necessarily a modular formfor �1.N /. There is another way: if d jN

M, put

˛d D

�d

1

�2 GLC2 .Q/;

so thatf Œ˛d �k.z/ D d

k�1f .dz/

for any function f W H! C. An easy calculation shows

˛d�1.N /˛�1d � �1.M/;

and so Œ˛�d takes modular forms for �1.M/ to modular forms for �1.N /. Thus, ford jN , we have a map

�d W Sk.�1.N=d// �Sk.�1.N=d//! Sk.�1.N //

defined by .f; g/ 7! f C gŒ˛d �k. We define the space of old forms of level N to bethe space

Sk.�1.N //oldD

XpjN

�p.Sk.�1.N=p///:

We define the space Sk.�1.N //new to be the orthogonal complement of Sk.�1.N //

old

in the space Sk.�1.N //.

Fact: The operators Tn and hni preserve Sk.�1.N //old and Sk.�1.N //

new, for all n.

To prove this fact, one treats separately the old and the new cases. The calculationfor the old formms is straightforward, but is not so easy for the new space: for these,we show instead that the space is invariant under the adjoints T �n and hni�.

We want to investigate the Fourier expansions of the forms in our spaces. Using ournotation from above, we have

�d D d1�kŒ˛d �k and �d .f /.z/ D f .dz/;

65

whenever f 2 Sk.�1.N //. Suppose f has Fourier expansion

f .z/ D

1XnD1

anqnI

then �d has Fourier expansion

.�df /.z/ D

1XnD1

anqnd :

In particular, if f is an old form, write

f DXpjN

�pfp

as before, with each fp 2 Sk.�1.N=p//. Then an.f / D 0 for all n such that.n;N / D 1; in fact, the converse is true.

Lemma 3. If f 2 Sk.�1.N //, then f is an old form if and only ifan.f / D 0 for all n with .n;N / D 1.

Proof. There are three main steps in the proof; we outline them first.

In the first step, we will change � so that �p becomes the identity inclusion. Then,we consider the �.1/ action on Sk.�1/, for some subgroup �1 � �.1/; if �2 � �1has finite index, then the elements of Sk.�1/ are the vectors of Sk.�2/ which arefixed under the action of �1. Finally, we will use representation theory, applied tothe finite group �1=�2.

First: put

�1.N / D

��a b

c d

�2 �.1/ W b � 0 mod N and a � d � 1 mod N

�I

then˛M�1.M/˛�1M D �

1.M/;

and so M k�1Œ˛�1M �k W Sk.�1.M// ! Sk.�1.M// is an isomorphism. We have the

66

following commutative diagram:

Sk.�1.M// Sk.�1.N //

Sk.�1.M// Sk.�

1.N //

id

�d

Writing qM D q1=M ; qN D q1=N , we use the relation N D Md to obtain qdnN D qnMfor any n. The image of f D

Pn anq

n in Sk.�1.M// in Sk.�1.N // is

Pn anq

dnN .

Next, we reformulate things in terms of representation theory. Define the projection

�d W Sk.�.N //! Sk.�d /;

where�d D �1.N / \ �

0.N=d/;

and

�0.N / D

��a b

c d

�2 �.1/ W b � 0 mod N

�:

The inclusions �.N/ � �d � �1.N / allow us to write in co-ordinates

�d

1XnD1

anqnN

!D1

d

d�1XbD0

f��1 bN=d

1

��k:

If we put

� DYpjN

.1 � �p/;

then the statement that f is a linear combination of the terms �pfp is the statementthat f 2 ker� . In fact, it can be shown that

ker� DXpjN

Im.�p/:

67

Thus we have obtained a projection � W Sk.�.N // ! Sk.�.N //. Let G D �.1/,and let H D N�1.N / D �1.N / D �.N/: we will find the H -fixed vectors in ker� .We have by a previous observation that

ker� DXpjN

Sk.�p/I

by our representation theory reformulation, we that

Sk.�.N // \

0@XpjN

Sk.�.N //

1A�p=�.N/ DXpjN

Sk.�1.N=p//; (*)

where the superscript �p=�.N / denotes the subset of vectors fixed under the actionof that subgroup. Note that we can write

Sk.�1.N=p// D Sk.�.N //

�1.N=p/I

thusSk.�p/ D Sk.�.N //

�1.N/\�0.N=d/:

We know abstractly that if a group G has a (finite) direct sum decomposition G DQi Gi , and if V is an irreducible representation of G, then for any subgroups H DQi Hi ; K D

Qi Ki , one has

V H \Xi

V Ki DXi

V hHi ;Ki i; (�)

where hHi ; Kii is the subgroup of G generated by Hi and Ki .

In our situation, we will take G D �.1/;Gi D SL2.Z=peii Z/ (where N D

Qi p

eii ),

Hi D �1.p

eii /=�.p

eii /, and

Ki D �1.peii / \ �

0.pei�1i /=�.p

eii /:

To complete the proof, it remains to identify equation * with the result �; this is leftas an exercise.

We meet here again in a week and a half.

68

19 Lecture Nineteen

We closed last time with a discussion of the decomposition of the space of cuspforms into the new and old. These complementary subspaces are preserved underthe Petersson inner product. Recall that we saw that f 2 Sk.�1.N //

old implies thatan.f / D 0 whenever .n;N / D 1, where an.f / is the nth Fourier coefficient of f .

Definition: Suppose f 2 Mk.�1.N //; we say that f is a Hecke eigenform (orsimply an eigenform) if it is an eigenvector for all Tn and all hni.

If we consider only common eigenvectors for those Tn and hni such that .n;N / D 1,then we can show that Sk.�1.N // has a basis of such eigenvectors. The key pointin this proof is the fact that, if f 2 Sk.�1.N //

new is an eigenvector for all Tn andhni with .n;N / D 1, then it is an eigenform. This follows from the formula

an.f / D �na1.f / if .n;N / D 1;

where �n is the eigenvalue of Tn acting on f . In particular, we must have a1.f / ¤ 0;otherwise, f would not be a new form. Normalizing f so that a1.f / D 1, let m 2 Nand let

gm D Tmf � am.f /f:

ThenTngm D Tn.Tmf / � am.f /Tnf D �n.Tmf � am.f /f / D �ngm;

and so gm is an eigenvector for all Tn with .n;N / D 1. We have

a1.gm/ D a1.Tmf / � a1.am.f /f / D am.f / � am.f /a1.f / D 0;

because of our normalization of f ; it follows that gm is old. But gm D Tmf �am.f /fis also a new form, and so must equal zero, and we have proven our claim.

Definition: A new form is a normalized eigenform in Sk.�1.N //new.

Theorem 19.1. The eigenspaces of Tn on Sk.�1.N //new are one-

dimensional (this is known as the multiplicity one result). Further-more, the new forms in Sk.�1.N //

new form an orthonormal basis forthat space; the nth Fourier coefficient of each new form is its eigen-value on Tn; each new form lies in Sk.N; �/ for some �; and, iff 2 Sk.�1.N // is an eigenform, then it is either old or new (it isnot the sum of a nonzero new form and a nonzero old form).

The proof is left as an exercise.

69

We now begin our discussion on L-functions, which will probably take us to the endof the course. Suppose I have a modular form f with Fourier expansion

f .z/ D

1XnD0

anqn;

where q D exp.2�iz/. Define the Dirichlet series associated to f to be the formalseries

L.s; f / D

1XnD1

ann�s;

where s 2 C. Recall that, several weeks ago, we calculated an estimate for themagnitude of the Fourier coefficients of a cusp form; namely, we showed that if fis a cusp form of weight k, then janj � Cn1=2 for some absolute constant C . Aneasy exercise is to show that this implies that L.s; f / will converge absolutely forRe.s/ > 1C k

2when f is a cusp form.

It can be shown (with more work) that, if f is an Eisenstein series, our explicitformula for its Fourier coefficients gives the bound janj � Cn

k�1, which will implythat L.s; f / converges absolutely when Re.s/ > k.

Theorem 19.2. Let f 2 Mk.N; �/, and suppose f has Fourier ex-pansion

f .�/ D

1XnD1

anqn:

Then f is a normalized eigenform if and only if

L.s; f / DYp

.1 � app�sC �.p/pk�1�2s/�1;

where the product is taken over all primes.

Proof. (sketch) We have

1XnD0

aprp�rsD

1

1 � aprp�rsD

1

1 � app�s C �.p/pk�1�2sI

the normalization a1 D 1 corresponds to the case r D 0. Thus1XnD1

ann�sD

Yp

� 1XrD0

aprp�rs

�;

valid when Re.s/� 0.

70

Now, a brief aside on the Mellin transform; this is the Fourier transform on thelocally compact abelian group R>0. First, we discuss the general Fourier tarnsform.

Recall that if G is a locally compact abelian group, we denote by OG its Pontryagindual, consisting of all multiplicative homomorphisms

� W G ! C�:

The Fourier transform of a function f W G ! C is the function Of W OG ! C definedby

Of .�/ D

ZG

f .g/ N�.g/ dg:

For instance, if G D S1 D C�, then OG consists of maps of the form z 7! zn, withn 2 Z. Hence OG Š Z, and the Fourier coefficient at n isZ 1

0

f .x/ exp.�2�inx/ dx:

If G D R, then OG consists of maps of the form x 7! exp.2�ix/, so OG Š G Š R, and

Of .y/ D

ZR

f .x/ exp.�2�ixy/ dx:

Finally, if G D R>0, then characters on G have the form x 7! xs for some s 2 C.Hence OG Š C, and the Fourier transform is in this case called the Mellin transformand takes the form

g.s/ D

Z 10

f .it/t sdt

t:

Theorem 19.3. Let f 2 Sk.�1.N // and let g.s/ be the Mellin trans-form of f ; then

g.s/ D .2�/�s�.s/L.s; f /;

where

�.s/ D

Z 10

xs�1e�x dx

is the usual �-function.

Proof. Omitted.

71

Theorem 19.4. Let ƒN .s/ be the “completed L-function of f ”, de-fined

ƒN .s/ D Ns=2.2�/�s�.s/L.s; f /;

and let wN W Sk.�1.N //! Sk.�1.N // be the function

wN .f / D ikN 1�k=2f

��0 �1N 0

��k:

Then wN is idempotent (meaning w2N D id) and self-adjoint; there isthe decomposition

Sk.�1.N // D S C

k˚S �

k ;

where S "k

denotes the "-eigenspace of wN in Sk.�1.N / (for " 2 f˙1g);and, moreover, if f 2 S "

k, then ƒN .s/ extends to a function which is

entire on C, and which satisfies the functional equation

ƒN .s/ D "ƒN .k � s/:

72