Topics In Algebraic Geometry- Hilbert Modular Surfaces

Matthew Barrass Blowen Bates

0826227

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Abstract

Develop some basic algebraic geometry theory, including the de�nition

of dimension, singularity and the notion of blow-ups. Examine cyclic quo-

tients of C2 and their relation to the so-called Hirzebruch-Jung continued

fractions. Show that such quotients are a�ne varieties and give a method

of constructing the coordinate ring and demonstrate how to resolve the

quotient singularity. Brie�y consider the modular curve H/PSL2Z as mo-

tivation for the study of Hilbert modular surfaces. A Hilbert modular

surface is a two-dimensional analogue of a modular curve. Conclude by

demonstrating how to resolve the cusp singularities.

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Contents

1 Results from algebraic geometry 4

1.1 A�ne space . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41.2 Projective space . . . . . . . . . . . . . . . . . . . . . . . . . . . 51.3 Weighted projective space . . . . . . . . . . . . . . . . . . . . . . 61.4 Nomenclature . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71.5 Abstract Varieties . . . . . . . . . . . . . . . . . . . . . . . . . . 71.6 Local theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 81.7 Resolution of singularities . . . . . . . . . . . . . . . . . . . . . . 10

2 Surface cyclic quotient singularities and Herzebruch-Jung res-

olution 11

2.1 Hirzebruch-Jung continued fractions . . . . . . . . . . . . . . . . 132.2 Newton polygon . . . . . . . . . . . . . . . . . . . . . . . . . . . 132.3 Quotients of the form 1

r (1, b) . . . . . . . . . . . . . . . . . . . . 142.4 Resolution of singularities . . . . . . . . . . . . . . . . . . . . . . 16

3 Hilbert Modular Surfaces 16

3.1 Motivation from one dimensional modular curves . . . . . . . . . 163.2 Hilbert modular groups . . . . . . . . . . . . . . . . . . . . . . . 203.3 Isotropy groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . 223.4 The quotient space . . . . . . . . . . . . . . . . . . . . . . . . . . 243.5 Resolution of cusps . . . . . . . . . . . . . . . . . . . . . . . . . . 26

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1 Results from algebraic geometry

1.1 A�ne space

Much of what follows can be generalised to arbitrary �elds, we will consideronly the �eld C. This is in a sense the easiest �eld, being algebraically closedand of characteristic zero.

Recall that C[X1, ..., Xn] is a �nitely generated noetherian C-algebra, byHilbert Basis Theorem all ideals are �nitely generated. Let J = (f1, ..., fk) ≤C[X1, .., Xn] be an ideal, it is meaningful to consider the vanishing set of theideal i.e.

V (J) = {x ∈ Cn : f(x) = 0, ∀f ∈ J}.

We de�ne a topology on Cn by de�ning a set, X ⊆ Cn, to be closed i�X = V (J),for some J ∈ C[X1, ..., Xn]. This topology is called the Zariski topology and wecall V (J) Zariski closed. Note, the Zariski topology is strictly coarser than theclassical topology on C, by this I mean that every Zariski closed (open) set isclosed (open) in the classical sense but the converse does not hold. Given anyclosed set V be de�ne the ideal

I(V ) = {f ∈ C[X1, ..., Xn] : f(a) = 0, ∀a ∈ V }.

We always have that X = V (I(X)), but the reverse in not quite as nice, i.e.I(V (J)) =

√J . This is the so called Nullstellensatz, the truth of which relies

on the fact that C is algebraically closed. From the Nullstellensatz we see acorrespondence between points of Cn and maximal ideals (X1−a1, ..., Xn−an)of C[X1, ..., Xn]. Moreover, we get the following inclusion reversing one-to-onecorrespondence,

Closed sets U ⊆ Vl l

Radical ideals J ⊇ I

De�nition 1.1. Given an algebraically closed set V , de�ne the coordinate ring

C[V ], to be C[X1,...,Xn]I[V ] .

In a sense, the coordinate ring is the ring of polynomials de�ned on V . Notethere is a correspondence between points of V and maximal ideals of C[V ]. Thus,talking about a closed set V is equivalent to talking about C[V ]. A morphismof closed sets is a polynomial map, f : V → U , this induces a homomorphismf∗ : C[U ] → C[V ] of C-algebras (by the pull-back). Two closed sets are calledisomorphic if there exists a bijective morphism between them with a inversemorphism. Note, two a�ne closed sets are isomorphic i� there coordinate ringsare isomorphic as C-algebras.[11]

We can classify which algebras occur as a�ne coordinate rings.

Proposition 1.1. [11]A C-algebra A occurs as an a�ne coordinate ring i� Ais �nitely generated and contains no non-zero nilpotents.

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The nilpotent condition can be seen as a consequence of Hilbert Nullstel-lensatz. Given a a�ne coordinate ring A we construct the corresponding a�neclosed set, SpecA, as follows: A is �nitely generated, thus A ∼= C[X1, .., Xn]/Ifor some ideal I, further I is radical since A contains no nilpotents, de�neSpecA = V (I).

We emphasize the relation between the closed set V and the coordinatering C[V ], by saying V = SpecC[V ]. The identi�cation is that points of V areidenti�ed to maximal ideals of SpecC[V ]. Thus it may be more intuitive to sayV = SpecmC[V ]; this is purely a notational matter.

An a�ne closed set V is said to be irreducible if it cannot be written asV1 ∪ V2, where Vi 6= V and Vi closed. There is a notion of unique factorisationfor closed sets. i.e. every closed set V can be wrote as a �nite union, V =

⋃Vi,

where Vi irreducible and Vi *⋃j 6=i Vj . Such Vi as above are called irreducible

components of V . I could have equivalently said that C[V ] has a minimalprimary decomposition, which follows from the fact that C[V ] is noetherian.

Note the following equivalences, V is irreducible i� I(V ) is prime i� C[V ]is an integral domain. Thus, when V is irreducible we can form the �eld offractions FracC[V ], this is denoted C(V ). A general element f

g ∈ C(V ) will

be such that V (g) ∩ V 6= ∅. Thus fg need not well de�ned on the whole of V ,

however it is de�ned on a non-empty open subset V − V (g). To highlight thisissue we use the augmented notation f

g : V 99K C. We call functions in C(V )rational functions.

De�nition 1.2. [6]An irreducible closed a�ne set V is called normal if C[V ] isintegrally closed.

All unique factorisation domains are integrally closed. Thus, Cn with itscorresponding coordinate ring C[X1, ..., Xn] is normal.

Given an irreducible a�ne closed set, V , we can construct a unique normalclosed set, V ′, by taking the closed set associated to the integral closure ofthe coordinate ring C[V ]. i.e. if C[V ]′ is the integral closure of C[V ], thenthe normalisation of V is given by V ′ = Spec(C[V ]′). Note the process ofnormailising V gives the following diagram,

C[V ] ↪→ C[V ]′ = C[V ′]l lV ←↩ V ′

we call the induced inclusion V ′ ↪→ V the normalisation map[11].

1.2 Projective space

De�nition 1.3. The nth dimensional complex projective space Pn, is the setof all 1-dimensional linear subspaces of Cn+1. A point of Pn is represented by(x0; ...;xn), which is unique up to C×.

We will now try to reformulate much of the machinery developed above fora�ne spaces to projective spaces. First note that the notion of a polynomial

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function on Pn in general does not make sense. This is because for any functionto be well de�ned on Pn it must by invariant under C×. With this in mind we sayf ∈ C[X0, ..., Xn] is homogeneous of degree d if f(λx0; ...;λxn) = λdf(x0; ...;xn)for all λ ∈ C×. Thus, for a homogenous polynomial f , it is meaningful to askwhen f(x) = 0. An ideal I ≤ C[X0, ..., Xn] is called homogenous if it has agenerating set of homogeneous polynomials. Given such an ideal I we constructV (I) = {x ∈ Cn+1 : f(x) = 0, ∀f ∈ I} ⊆ Pn.

De�nition 1.4. A set V ⊆ Pn is (Zariski) closed if it is equal to V (I), for somehomogeneous ideal I.

The notions of irreducible closed sets, subsets and (Zariski) closure are de-�ned as they are for the a�ne case. Further, given any projective closed set V ,we can de�ne the ideal I(V ) = {f ∈ C[X0, ..., Xn] : f(a) = 0, ∀a ∈ V }. Itis true (though not obvious) that if V is a projective closed set, then I(V ) is aprojective ideal[11].

We now consider functions on Pn. Note, f ∈ C[X0, ..., Xn] does not ingeneral de�ne a function on Pn, since any function on Pn must be invariantunder C×. How every, the ratio of two homogeneous polynomials of the samedegree does de�ne a function on Pn, in some sense. Similarly to the function�eld for a�ne closed sets, we have f/g : Pn − V (g) → C. We use the samenotation to denote this as for the a�ne case, i.e. f/g : Pn 99K C. We call f/g,f, g ∈ C[X0, ..., Xn] with g 6= 0, a rational function, an denote the �eld of allrational functions on the closed set V by C(V ).

We now consider maps between projective closed sets. Let V be a closedset, if f0,, ..., fm ∈ C[X0, ..., Xn] are homogeneous of the same degree, such thatV ∩V (f0, ..., fm) = ∅, then we get an induced well de�ned function F : V → Pm,de�ned by

(x0; ...;xn) 7−→ (f0(x0; ...;xn); ...; fm(x0; ...;xn)).

Maps between projective closed sets have a special property that the imageof a projective closed set is again closed. This is in contrast to a�ne closed sets.For example, V = V (XY − 1) ⊂ C2 and F : V → C de�ned by F (X,Y ) = X.One sees that F (V ) = C×, but C× is not an a�ne closed set[6].

Pn is covered by n+ 1 copies of C via the maps,

ϕi : (x0; ...;xi−1;xi;xi+1; ...;xn)→ (x0, ..., xi−1, 1, xi+1, ..., xn).

We denote the image of ϕi by Ui, and call the the ith a�ne piece of Pn. Noticethat Pn = ∪Ui. Further, if V is a closed projective set, then the ith a�ne pieceof V is de�ned as Vi = V ∩Ui. Hence we see that V = ∪Vi, i.e. every projectiveclosed set is a union of a�ne closed sets.

1.3 Weighted projective space

Projective space, Pn, can be thought of as the quotient space of Cn+1 − {0}under an action of C× , i.e. λ ∈ C×, (x0, ..., xn) 7→ (λx0, ..., λxn). This is notthe only way C× can act.

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De�nition 1.5. [6]The weighted projective space P(a0, ..., an) is the quotientspace of Cn+1−{0} under the action λ ∈ C×, (x0, ..., xn) 7→ (λa0x0, ..., λ

anxn).

The ai are called weights. Under this construction we see that Pn ∼= P(1, ..., 1).A much more interesting example is P(4, 6). Recall that two elliptic curvesy2 = x3 + ax + b and y2 = x3 + a′x + b′ are isomorphic i� there exists t ∈ C×such that a = t4a′ and b = t6b′. Thus we see that P(4, 6) is the moduli spacefor elliptic curves. In general, weighted projective space is very di�erent to pro-jective space. For example, weighted projective spaces can have singularities.

De�nition 1.6. f ∈ C[X1, ..., Xn] is weighted homogeneous of degree d iff(λa0x0; ...;λanxn) = λdf(x0; ...;xn), λ ∈ C×.

This allows us to de�ne weighted homogeneous polynomial ideals and thecorresponding weighted projective closed sets in the obvious way. In a similarway to Pn we can cover P(a0, ..., an) by a�ne open subsets.

1.4 Nomenclature

De�nition 1.7. An irreducible projective closed set is called a quasi-projectivevariety.

De�nition 1.8. A quasi-projective variety which is also an a�ne closed set iscalled an a�ne variety.

Not every a�ne closed set is an a�ne variety.

De�nition 1.9. A quasi-projective variety which is not an a�ne variety iscalled a projective variety.

The base of a variety V is the �eld over which it is de�ned. In this paper wecan assume all varieties have base C.

1.5 Abstract Varieties

It is a common theme in geometry to try and work independent of a particularcoordinate choice. This is the motivation behind de�ning an abstract variety.Before we give a formal de�nition, �rst recall that any projective variety canbe wrote as a union of a�ne varieties. In a very informal sense, a projectivevariety is a bunch of a�ne varieties 'glued together' in a nice way. The same istrue for weighted projective varieties.

De�nition 1.10. [6]Given a collection ({Vi}I , {Vi,j}I,J , {ϕi,j : Vi,j → Vj,i}I,J),such that

• Vi is an a�ne variety,

• Vi,j ⊂ Vi is an open set,

• ϕi,j : Vi,j → Vj,i is a isomorphism,

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• ϕii = Id|Vi ,

• ϕj,k|Vj,k∩Vj,i◦ ϕi,j |Vi,j∩Vi,k

= ϕi,k|Vi,k∩Vi,j.

We de�ne the abstract variety X as

X :=⋃Vi/ ∼,

Where the equivalence relation is given by a ∈ Vi is equivalent to b ∈ Vj i�ϕi,j(a) = b.

Informally, we say that X is obtained from the Vi by gluing them togetheralong the Vi,j using ϕi,j .

Example 1.1. Let X be the abstract variety de�ned by identifying two copiesof C along C× using the gluing map ϕ(x) = x−1. Then X ∼= P.

Given two a�ne varieties, V, W , we can de�ne there Cartesian product. IfV = V (f1, ..., fn), W = V (g1, ..., gm) then V ×W := V (f1, ..., fn, g1, ..., gm) ⊂C[X1, ..., Xn, Y1, ..., Ym] , where the fi are functions of Xi and the gi functionsof Yi. One can show C[V ×W ] ∼= C[V ]⊗C C[W ][6].

It is possible to generalise the notion of Cartesian products to projective andeven abstract varieties. This is much harder though, for example there is nonatural way to de�ne Pn × Pm.

1.6 Local theory

In this section we will examine some local properties of irreducible varieties.

De�nition 1.11. A local property is a property of a point x ∈ X which remainsunchanged when we replace X with an neighborhood around x.

We have shown above that every variety can be seen as a bunch of a�ne vari-eties �glued together�. Hence any point of a variety has an a�ne neighborhood,thus it su�ces to consider a�ne varieties.

De�nition 1.12. Let x ∈ X be a point on an irreducible a�ne variety, wede�ne the maximal ideal at x to be

mx = {f ∈ C[X] : f(x) = 0}.

De�nition 1.13. [11]Let x ∈ X be a point in an irreducible a�ne variety, wede�ne the local ring at x to be,

Ox = C[X]mx.

Although the above de�nition is nice in its brevity, it may be more illumi-nating to think of Ox as the subring of C(X) consisting of functions which areregular at x ∈ X. Note that Ox is intrinsic to X, i.e. does not notice where youembed X geometrically.

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Remark 1.1. Notice that the point x ∈ X is an irreducible subvariety, con-structing the local ring Ox kind of corresponds to disregarding all informationnot local to the subvariety x. A similar construction can be done with anyirreducible subvariety Y ⊆ X. i.e. Let aY = {f ∈ C[X] : f(y) = 0, ∀y ∈ Y },then de�ne

OY,X = C[X]aY.

This is the ring of functions which are regular at some point in Y .

We will now consider a very important local property, the tangent space.Geometrically, the tangent space can be de�ned as the union of all lines �tan-gent� to a given point. Fixing a coordinate system gives a presentation of thea�ne variety X ⊆ Cn as V (f1(x1, ..., xn), ..., fr(x1, ..., xn)). Assuming we havea hypersurface (i.e. r = 1) the tangent space at the point a = (a1, ..., an) has anice form, i.e.

TaX = V (∑i

∂f

∂xi(a) · (xi − ai)).

This is exactly the de�nition of a tangent space one sees when studying vectorcalculus. When r 6= 1 we de�ne the tangent space as follows.

De�nition 1.14. [5]Let X = V (f1(x1, ..., xn), ..., fr(x1, ..., xn)) be an a�nevariety and a = (a1, ..., an) a point on X, then the tangent space of X at a is,

TaX =⋂k

V (∑i

∂fk∂xi

(a) · (xi − ai))

= V (∑i

∂f1∂xi

(a) · (xi − ai), ...,∑i

∂fr∂xi

(a) · (xi − ai)).

In words, when X is the intersection of a bunch of hypersurfaces, the tangentspace is the intersection of the tangent spaces of the corresponding hypersur-faces. Notice that TaX has a natural vector space structure. This de�nition isgood to use for calculations, but it has the draw back that it is not obviously�intrinsic�, i.e. it requires one to choose a coordinate system. For example, itis not clear whether under an isomorphism ϕ : X → Y , the tangent space atx ∈ X andϕ(x) ∈ Y are isomorphic.

Proposition 1.2. [5]Let x ∈ X be a point on an irreducible variety then,

mx/m2x∼= (TxX)∨.

This show that indeed the tangent space at a point is intrinsic to a variety.One calls the vector space mx/m

2x the cotangent space.

We now consider the what it means to de�ne dimension of a irreduciblevariety.

De�nition 1.15. [11]Let X be an irreducible variety, de�ne the dimension ofX to be,

dimX = minxdim(TxX).

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We are now able to precisely generalise the notion of a �non-smooth� curve(i.e. y2 = x3) to arbitrary varieties.

De�nition 1.16. Let x ∈ X be a point on an irreducible variety. We say x isa singularity of X if dimTxX > dimX, else we call x non-singular.

De�nition 1.17. A irreducible variety X is called non-singular if it is on-singular at every point. Else it is called singular.

Next we will give a proposition which makes calculating whether a point isa singularity relatively straight forward.

Proposition 1.3. Let X = V (f1, ..., fm) ∈ Cn be an irreducible variety. Thepoint P ∈ V is a singularity if the Jacobian of the functions fi at P is singular.i.e. ∣∣∣∣∣∣∣

∂f1∂X1

. . . ∂f1∂Xn

.... . .

...∂fm∂X1

. . . ∂fm∂Xn

∣∣∣∣∣∣∣P

= 0.

1.7 Resolution of singularities

Given a singular a�ne variety V , one could ask if we can �transform� V into asmooth variety. i.e. if there exists a non-singular variety (not necessarily a�ne)W , which is isomorphic to V in some possibly weaker category. We formalisethis as follows:

De�nition 1.18. [7]Let V ⊆ An be a singular a�ne variety. A resolution of Vis a non-singular variety W ⊆ An×Pn−1 together with a proper birational mapV 99KW .

The map is required to be proper to prevent trivial solutions, i.e. lettingW = V −{Singular points}. Note, being birationally equivalent is much weakerthan being isomorphic. For example, one can show that for any characteristic

zero �eld k, and binary quadratic form q, k[X,Y ](q) is birationally equivalent to k.

Thus, a circle is birationally equivalent to a line!One can think of a birational equivalence as an isomorphism almost every-

where.One of the most standard ways of resolving a singularity is using a blowup.

For now let us consider blowing-up An at a point P . First, we can preform achange of coordinates, shifting P to the origin O. We now de�ne a new variety,the blowup surface.

De�nition 1.19. [7]The blowup surface B of An is the ordered set consistingof points of An and a line in An through p and O. Formally,

B = {(p,−→p ) : p ∈ An} ⊆ An × Pn−1.

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Note, the line through p and O is uniquely determined whenever p 6= O.Thus we see the natural projection π : B → An is actually an isomorphismwhen restricted to the dense open subset p 6= O. O has �bre π−1(O) ∼= Pn−1which is called the exceptional divisor of B. We can give B the structure of aquasi-projective variety.

Proposition 1.4. [7]If (x1, ..., xn, y1; ...; yn) are coordinates of An×Pn−1 then

B = V ((xiyj − xjyi)i<j) ⊆ An × Pn−1.

Proof. Note, the point (x1, ..., xn, y1; ...; yn) is in B i� (x1, ..., xn) = λ(y1, ..., yn)for some λ ∈ C×. This is equivalent to saying the following matrix has rank≤ 1, [

x1 . . . xny1 · · · yn

]this happens i� all the minors vanish.

The blowup surface B is sometimes denoted BO(An), it is the simplest ex-ample of a blowup, i.e. the blowup of An at O. We de�ne Bp(An) to be theblowup surface of An such that the exceptional divisor lies over p; it can beobtained from BO(An) via a change of coordinates.

We can now give the de�nition of a blowup for a general a�ne variety V .

De�nition 1.20. Let V ⊆ An be an a�ne variety, π : Bp(An)→ An the blowupsurface at point p ∈ V ⊆ An. The blowup of V at point p ∈ V is the Zariskiclosure of π−1(V − {p}) in Bp(An), together with the restricted projection π.It is denoted Bp(V ).

Note, the de�ning equations for Bp(V ) are the de�ning equations of Bp(An)and V .

Can a blowup of V ∈ An at p ∈ V be de�ned by Bp(V ) = {(x,−→x ) : x ∈V } ⊆ An × Pn−1, where −→x is a line from p to x.

2 Surface cyclic quotient singularities and Herzebruch-

Jung resolution

Throughout this section we will assume V is an a�ne variety. De�ne Aut(V ) ={f ∈ C[X1, ..., Xn] : f(V ) = V }. In this section we will study a very simplegroup action on a very simple variety.

De�nition 2.1. Let G ⊆ Aut(V ) be a subgroup, V an a�ne variety. We letC[V ]G denote the elements of C[V ] invariant under G. i.e. {f ∈ C[V ] : fG =f}.

C[V ]Gis an a�ne ring, i.e. �nitely generated and contains no nillpotents.C[V ]G contains no nilpotents since it is a subring of C[V ] .

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De�nition 2.2. Let G ⊆ Aut(V ) be a subgroup, V an a�ne variety. Thequotient variety,V/G, is the a�ne variety de�ned by Spec(C[V ]G).

There is a natural projection map V → V/G.

Proposition 2.1. [4]Points of V/G are in one-to-one correspondence with G-orbits of V .

Example 2.1. V = {x-axis ∪y-axis} and let G ⊆ Aut(V ) be the group of order

two generated by re�ection in the line y = x. Now, V/G = Spec(

C[X,Y ](XY )

)G. We

see that every polynomial of C[X,Y ](XY ) is of the form

∑aiX

i +∑biY

i, thus we

see Spec(

C[X,Y ](XY )

)G ∼= C[X] . Thus, V/G is isomorphic to a C.

From now one we will only concern ourselves with a very simple class ofquotient varieties, those where V = C2 and G ∼= Z/nZ. Where G has thenatural action induced by the identi�cation

G =

⟨(ζa 00 ζb

)⟩,

where ζ is a primitive nth root of unity. Such examples are called �surface cyclicsingularities of type 1

n (a, b)�.[4]

Example 2.2. Consider the case 12 (1, 1), i.e. the group G = Z/2Z is acting

on C2 via (x, y) → (−x,−y). Note C[X,Y ]G clearly contains the monomialsX2, XY and Y 2, and it is not hard to see that these generate C[X,Y ]G. ThusC2/G = Spec

(C[X2, XY, Y 2]

). This is isomorphic to the ordinary quadratic

cone Spec(

C[U,V,W ](UW=V 2)

)via the isomorphism (X,Y )→ (X2, XY, Y 2).

Even this very simple example highlights a very important fact; C2/G canbe singular. In particular, 1

2 (1, 1) has a singularity at (0, 0). In this case, wecan resolve the singularity via �nding the corresponding cylinder of the coneX := 1

2 (1, 1). De�ne the cylinder of X to be the cylinder at O, i.e.

Y := {(x,−→x ) : x ∈ X, x ∈ −→x } ⊆ X × P.

Note, Y is non-singular. The projection π : Y → X is an isomorphismon Y − π−1(0) to X − 0; thus π is a birational morphism. Hence, (Y, π) is aresolution of X.

Here are the big questions we try to answer,

1. How do we give an a�ne variety structure to the quotient

2. Can we describe such a quotients coordinate ring

3. If the quotient is singular, can we �nd a resolution of singularities.

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2.1 Hirzebruch-Jung continued fractions

De�nition 2.3. [4]Let r, b be coprime integers such that r > b > 0. TheHirzebruch-Jung continued fraction of r/b is

r

b= a1 −

1

a2 − 1a3−...

=: [a1, ..., an].

Note, the fraction does indeed sequence does indeed terminate after �nitesteps; this would not be case of r, b were not assumed to be rational.

Example 2.3. 74 = 3− 1

3/2 = 3− 12− 1

2

= [3, 2, 2].

Example 2.4. nn−1 = 2− n−2

n−1 = ... = [2, ..., 2].

2.2 Newton polygon

De�nition 2.4. Let r, b be coprime integers such that r > b > 0. We de�ne alattice as follows, Lr,b := (0, 1)Z + (1, 0)Z + 1

r (1, b)Z ⊂ R.

Note, Lr,b contains the lattice Z2 as a sublattice of index r. Thus, the unitsquare properly contains exactly r − 1 lattice points, {( ir ,

ibr − [ ibr ]}i<r.

De�nition 2.5. [4]The Newton polygon of a lattice Lr,b is the convex hull ofthe non-zero lattice points in the �rst quadrant. We label the lattice points ofthe Newton polygon from left to right as,

e0 = (0, 1), e1 =1

r(1, b), e2, ..., ek, ek+1 = (1, 0).

The values of e2, ..., ek are in general not obvious without resorting to draw-ing the actual lattice. Later we will show that they are in fact closely relatedto the Hirebruch-Jung continued fraction of rb .

Proposition 2.2. [4]Any two consecutive ei, ei−1 form an orientated basis ofLr,b.

Proof. Consider the parallelogram < (0, 0), ei, ei−1, ei + ei−1 >. We will showthat this is a fundamental domain for the lattice Lr,b. By the de�nition ofthe Newton polygon, the lower triangle < (0, 0), ei, ei−1 > contains no latticepoints other than its vertices, further, for any point v in the upper triangle< ei, ei−1, ei + ei−1 > the point ei + ei−1− v is in the lower triangle. Thus theparallelogram < (0, 0), ei, ei−1, ei+ ei−1 >contains no lattice points except itsvertices, thus it is a fundamental domain for the lattice Lr,b.

This arguments fails for lattices of dimension ≥ 3. This is because the convexhull of n �consecutive vectors in an n-dimensional lattice so no longer half of thefundamental domain of those vectors.

13

Proposition 2.3. [4]Any three consecutive lattice points ei−1, ei, ei+1 satisfy arelation of the form,

ei−1 + ei+1 = aiei

for some integer ai ≥ 2.

Proof. From the previous proposition we have ei−1, ei and ei, ei+1 each formbasis, thus

ei+i = α1ei + β1ei−1

ei−1 = α2ei + β2ei+1,

we have ei−1 + ei+1 = aiei and ai > 0. Finally note, ai 6= 1 since then ei wouldbe properly contained in the Newton polygon.

Note that ai = 2 implies that ei lies on the line joining ei−1, ei+1 and ai > 2implies ei is properly under the line.

Proposition 2.4. The ai from the previous proposition satisfy,

r

b= [a1, ..., ak].

Thus we see that we can de�ne the ei recursively via,

e0 = (0, 1)

e1 =1

r(1, b)

...

ei+1 = aiei − ei−1...

ek+1 = (1, 0),

where the ai are such that rb = [a1, ..., ak].

2.3 Quotients of the form 1r(1, b)

Recall, we write 1r (1, b) for the quotient C2/G where G ∼= Z/rZ acts via the

matrix (x, y) 7→ (εx, εby), it will always be assumed that (r, b) = 1. In thissection we will give a explicit description of the ideal I(C2/G).

Proposition 2.5. [4]Let e0, ..., ek+1 be the vertices of the Newton polygon of

14

the lattice Lr,r−b, then the invariant monomials, C[x, y]G, are generated by

u0 = xr

u1 = xr−by

...

ui = xrαiyrβi

...

uk+1 = yr

where αi is the y-coordinate of ei, βi is the x-coordinate of ei. Moreover, the uisatisfy the following relation,

ui−1ui+1 = uaii ,

where the ai are such that rr−b = [a1, ..., ak].

Proof. g =

(ε 00 εb

)∈ G acts on xαyβ by g(xαyβ) = (εx)α(εby)β = εα+bβxαyβ .

Thus, our monomial is invariant i� α+ bβ ≡ 0 (mod r). α = r− b, β = 1 is onesuch solution, consider the lattice Lr,r−b. Let (β0, α0) ∈ Lr,r−b be a lattice pointin the positive quadrant, so (β0, α0) = (n,m)+λ 1

r (1, r−b) for some n,m, λ ∈ Z.Thus α0 + bβ0 ∈ Z, hence xrβ0yrα0 is an invariant monomial. Considering theNewton polygon of Lr,r−b we see Lr,r−b is generated by u0, ..., uk+1 as required.

Why is this all the invariant monomials? Because it is all solutions to thecongruence equation.

We can now de�ne a recurrence relation for the monomials ui. Consider thecase 1

r (1, b), let rr−b = [a1, ..., ak]. The generating polynomials are given by,

u0 = xr

u1 = xr−by

...

ui+1 =uaiiui−1

...

uk = yr.

Example 2.5. 17 (1, 4). Note 7

7−4 = [3, 2, 2], so the Newton polygon of L7,3 has

vertices (0, 1), 17 (1, 3), 1

7 (3, 2), 17 (5, 1) and (1, 0) thus the invariant polynomials

C[x, y]G are generated by u0 = x7, u1 = x3y, u2 = x2y3, u3 = xy5 and u4 = y7.Hence we have 1

7 (1, 4) ∼= C[u0, u1,u2, u3, u4]/(u0u2− u31, u1u3− u22, u2u4− u23).

We can also consider some more general examples,

15

Example 2.6. 1r (1, 1). Note, r

r−1 = [2, 2, ..., 2] r − 1−times, so the Newton

polygon of Lr,r−1 has vertices (0, 1), 1r (1, r− 1), ..., 1r (r− 1, 1), (1, 0). Thus the

invariant monomials, C[x, y]G, are generated by u0 = xr, u1 = xr−1y,...,ur−1 =xyr−1, ur = yr. Hence we have 1

r (1, 1) ∼= C[u0, ..., ur]/(ui−1ui+1 − u2i ).

Note, the ideal in the previous example can be wrote as

Rank

(u0 · · · ur−1u1 · · · ur

)≤ 1.

We call 1r (1, 1) a called determinantal variety.

Example 2.7. 1r (1, r − 1). Note, r

r−r+1 = [r], so the Newton polygon of Lr,1has vertices (0, 1), 1

r (1, 1), (1, 0). Thus the invariant monomials, C[x, y]G, aregenerated by u0 = xr, u1 = xy, u2 = yr. Hence we have 1

r (1, r − 1) ∼=C[u0, u1, u2]/(u0u2 − ur1).

2.4 Resolution of singularities

In this section we will �nd a resolution of singularities of 1r (1, b). We write,

N := Lr,b = (0, 1)Z + (1, 0)Z +1

r(1, b)Z ⊂ R2,

M := N� = {(α, β) ∈ Z2 : α+ bβ ≡ 0 (mod r)} = Lr,r−b.

Let Xb = C2/(Z/r) be the a�ne variety associated to the quotient withgroup action 1

r (1, b).

Proposition 2.6. [4]Let rb = [a1, ..., al], and e0, ..., ek be the usual vertices of

the Newton polygon of M . For i = 0, ..., k, let ξi, ηi be monomials forming thedual basis to M at ei, ei+1. i.e.

ei(ξi) = 1, ei(ηi) = 0

ei+1(ξi) = 0 ei+1(ηi) = 1.

Then Xb has resolution of singularity π : Y → Xb, as follows,

Y = Y0 ∪ ... ∪ Yl,

where Yi ∼= C2, with coordinates ξi, ηi. With gluing relations Yi − (ηi = 0) ∼=Yi − (ξi = 0), de�ned by ξi+1 = η−1i , ηi+1 = ξiη

bii .

3 Hilbert Modular Surfaces

3.1 Motivation from one dimensional modular curves

Consider an arbitrary Riemann surface, X. i.e. a one dimensional complexmanifold. X is a topological manifold and so has a unique universal covering

16

space, π : Y � X, which is a simply connected topological space and π is acovering map. This induces an unique complex structure on Y which makes πinto a local biholomorphism.

Proposition 3.1. If Γ is the group of deck transformations of Y/X, then X ∼=Y/Γ.

We now ask, what are the possible simply connected Riemann surfaces?Luckily, there is a very simple answer,

Proposition 3.2. Every simply connected Riemann surface is isomorphic toone of the following,

PC Riemann sphere

C Complex plane

H Upper half plane

Proof. This is the famous Uniformization Theorem, it is very hard to prove.

We will be interested in quotients ofH. We now ask what are the possibilitiesfor the group Γ.

The group GL2C acts on PC via

(a bc d

)· z := az+b

cz+d . Considered as a

Riemann surface, the biholomorphic automorphisms of PC, Aut(PC), are all ofthe form

z 7−→ az + b

cz + d,

for some a, b, c, d ∈ C, ad− bc 6= 0[15]. Thus, every automorphism of PC arisesas the action of some γ ∈ GL2C. We are interested with the action on the upperhalf plane, H.

Proposition 3.3. GL+2 R := {γ ∈ GL2R : detγ > 0} acts on the complex

upper half plane, H, with the induced action fromGL2C.

Proof. If you believe that GL2C acts on PC, then all I need show is z ∈ H, γ ∈GL+

2 R =⇒ γ · z ∈ H. This follows from the identity

Im(γ · z) = Im(az + b

cx+ d) =

(ad− bc)Im(z)

|cz + d|2> 0.

In a sense, GL+2 R is too big, i.e. there exists elements of GL+

2 R which acttrivially. Hence we consider the smaller group,

PSL2R = GL+2 R/{

(a 00 a

): a 6= 0}.

Recall that a group G is said to act e�ectively (faithfully) on a set X ifg 6= idG ⇒ g · x 6= x for some x ∈ X . i.e. no element of G acts trivially. Theword faithful comes from the fact that the induced group action map G→ AutXis injective.

17

Proposition 3.4. The action on PSL2R on H is e�ective (faithful).

Proof. Suppose az+bcz+d ≡ z, then cz2 + (d − a)z − b ≡ 0 thus b = 0, a = d and

c = 0.

Proposition 3.5. [8]AutH = PSL2R.

De�nition 3.1. A subgroup, H, of a topological group, G, is called discrete ifthe induced topology on H is the discrete topology. Equivalently, IdH , is openin H.

Example 3.1. Z is a discrete subgroup of R, with the Euclidean topology.

We can genaralise the last example,

Example 3.2. Given a number �eld K/Q, the ring OK is a discrete subgroupof K. This follows directly from Dirichlet's unit theorem.

Recall that there is a natural euclidean topology on PSL2R induced via theinclusion PSL2R ↪→ R4.

Example 3.3. PSL2Z is a discrete subgroup of PSL2R.

The group PSL2Z is usually called the modular group. It is a very importantgroup, and in some sense it is the simplest discrete subgroup of PSL2R.

With a bit of work one can �nd a simple set of generators of PSL2Z , i.e.

PSL2Z =

⟨(1 10 1

),

(0 −1−1 0

)⟩.

Hence, we see that PSL2Z acts on H via a series of translations

((1 10 1

)),

and inversions

((0 −1−1 0

)). One can further show that PSL2Z has a nice

fundamental domain, given by

{z ∈ H :−1

2≤ Re(z) <

1

2, |z|2 ≥ 1}.

Writing z = x+ iy ∈ H we have the standard Riemann metric on H, i.e.

dω =(dx)2 + (dy)2

y2.

Proposition 3.6. [1]The metric dω is invariant under the action of PSL2Z.

Thus, we have a well de�ned metric on the quotient H/PSL2Z. From thiswe get the so called Gauss-Bonnet form

ω = − 1

2π

dx ∧ dyy2

.

18

Example 3.4. We will work out the volume of the standard modular curve,ˆH/PSL2Z

ω = − 1

2π

ˆ 0.5

−0.5

ˆ ∞√1−x2

y−2dydx

= − 1

2π

ˆ 0.5

−0.5

√1− x2dx

= − 1

2π(π

3+π

3)

= −1

6.

We will now have a quick reminder of some basic theory of group actions.

De�nition 3.2. A group G is said to act freely on a set X if g · x = h · x =⇒g = h, equivalently, every x ∈ X has trivial isotropy group.

Note that a free action is much better than a faithful action.

Example 3.5. G = Z/rZ acting on C2 via 1r (1, b) is a free action. In this case

we have a nice quotient, but this is not always so! For example, Q acting on Rvia translation is free, but the quotient is horrible.

Proposition 3.7. A discrete subgroup Γ < PSL2R acts freely i� Γ has noelements of �nite order.

De�nition 3.3. A group G acts properly discontinuously on a locally compactHausdor� space X if for any x, y ∈ X there exists neighborhoods x ∈ U, y ∈ Vsuch that,

#{g ∈ G : (g · U) ∩ V 6= ∅} <∞.Proposition 3.8. PSL2Z acts properly discontinuously on H.

Being properly discontinuous is very good, since loosely speaking, it preservesa degree of separation when we move to the quotient.

Proposition 3.9. [16]If Γ acts properly discontinuously on a locally compactHausdor� space X, then the quotient space X is Hausdor�.

Proof. Let x, y ∈ X be points in distinct orbits of Γ. Let U 3 x, V 3 y bea compact neighborhoods. By assumption only �nitely many translates of Vintersect U , so we may assume, by intersecting to make U, V smaller if necessary,that the translates of U and V are disjoint. Taking V ′ = ΓV , and U ′ = ΓUgives invariant neighbourhoods of x, y which are disjoint; this shows that X/Γis Hausdor�.

Thus H/PSL2Z is a Hausdor� space.

De�nition 3.4. We denote the quotient H/PSL2Z by X0.

Note that X0 is not compact, it has a cusp at ∞. We construct the newspace, X, by performing the one-point compactifaction at ∞. Now X0 is openand dense in X. Further, there is a unique complex structure on X induced bythe covering map H → X0 → X. Thus, X is a compact connected Riemannsurface. One can show that X ∼= PC.

19

3.2 Hilbert modular groups

We now generalise the results from the previous section to �dimension two�. Weconsider quotients of H2 = H×H. Note (PSL2R)2 acts on H component wise.Moreover, we have the following short exact sequence,[2]

0→ (PSL2R)2 → AutH2 → Z/2→ 0,

where the Z/2 corresponds to switching the factors of H2.

Remark 3.1. We actually have a presentation of AutH2 as the semi-direct(PSL2R)2 n Z/2.

We shall study special discrete subgroups of (PSL2R)2.Let d > 1 be a square free integer. Then K = Q(

√d) ⊂ R is a real quadratic

�eld. It is well known that K has ring of integers

OK =

{Z + 1+

√d

2 Z if d ≡ 1 (mod 4)

Z +√dZ if d ≡ 2, 3 (mod 4).

Q(√d) has two real embeddings into R, identity (a + b

√d 7→ a + b

√d) and

conjugation (a + b√d 7→ a − b

√d); given an element α ∈ K, we implicitly

associate it with its image in R and we denote its conjugate image in R byα′. Recall, the norm and trace of an element α ∈ K are de�ned as follows,N(α) = αα′, Tr(α) = α+ α′.

Proposition 3.10. The discriminant, D, of the �eld extension Q(√d)/Q, is

D = 4d if d ≡ 2, 3 (mod 4)

D = d if d ≡ 1 (mod 4).

De�nition 3.5. [9]An element α ∈ K is called totally positive if both α > 0and α′ > 0, this is denoted α � 0.

We now take a closer look at the units. Firstly, to make notation easier wede�ne U := O×K . By Dirichlet's unit theorem we have that there exists somefundamental unit ε > 0 such that U = {±εn : n ∈ Z}, thus U ∼= Z × Z

2 . LetU+ := {u ∈ U : u � 0} be the group of totally positive units.

Proposition 3.11. U2 ⊆ U+.

Proof. Let u ∈ U , then both u2 > 0 and (u′)2 > 0. Thus, u2 � 0 for allu ∈ U .

Note, U+ = {ε2n : n ∈ Z} ∼= Z has rank 1.We can now de�ne an two dimensional analogue of GL2R.

De�nition 3.6. [2]Let K = Q(√d) be a totally real �eld extension. The full

Hilbert modular group[13] is de�ned to be,

GL+2 K = {

(a bc d

): a, b, c, d ∈ K, ad− bc � 0}.

20

Proposition 3.12. The group GL+2 K has a induced action on H2 via the in-

clusion GL+2 K ↪→ (PSL2R)2 given by,(

a bc d

)7−→

((a bc d

),

(a′ b′

c′ d′

)).

Just as for the one dimensional case we see that the group GL+2 K is too big.

i.e. not e�ective. Thus we consider the smaller group,

PSL2K = GL+2 K/{

(a 00 a

): a 6= 0}.

It follows from the one dimensional case that PSL2K is e�ective. We nowlook at some of the discrete subgroups of PSL2K,

G := PSL2OK = {(a bc d

): a, b, c, d ∈ OK , ad− bc = 1}/{

(±1 00 ±1

)},

G := {(a bc d

): a, b, c, d ∈ OK , ad− bc ∈ U+}/{

(a 00 a

): a ∈ U}.

We call the group G the Hilbert modular group and G the extended Hilbertmodular group. These are two of the simplest examples.

Proposition 3.13. [2]There is an exact sequence,

0→ G→ G→ U+/U2 → 0,

induced by the determinant map on G modulo U2.

Note that both U+ and U2 are in�nite cyclic, and U+/U2 has order 1 or 2.If we restrict our attention to �eld extensions of the form Q(

√p) for a prime p

we have the following nice classi�cation of U+/U2.

Proposition 3.14. [2]De�nitions as above,

U+/U2 =

{0 p = 2 or p ≡ 1 (mod 4)

Z/2 p ≡ 3 (mod 4).

Remark 3.2. It is much harder to �nd a set of generators for PSL2OK than itwas for PSL2Z. This makes it harder to describe the fundamental domain ofPSL2OK .

De�nition 3.7. Let Γ ≤ PSL2K be a subgroup. A cusp of Γ is de�ned to bean orbits of Γ in PK, under the normal action.

Further, we can take a representative of any cusp with integral elements.

Proposition 3.15. [1]The map,

ϕ : PK/PSL2OK → Cl(K),

(α : β) 7→ αOK + βOK .

21

This is a very nice generalisation of the one dimensional case. The proof canbe found in [1].

Corollary 3.1. The number of cusps of PSL2OK is equal to the class numberh(OK). Hence, PSL2OK has a �nite number of cusps.

This is nice, since the quotient H2/PSL2OK has �bad points� correspondingto the cusps.

Proposition 3.16. [3]The Hilbert modular group PSL2OK is a discrete sub-group of (PSL2K)2, and it acts properly discontinuously on H2.

The group PSL2OK is irreducible in the sense that if γ ∈ PSL2OK and ifeither γ or γ′ act trivially on H, then γ = Id.

3.3 Isotropy groups

In this section we will examine the isotropy groups of the Hilbert modular group(G) and the extended one (G). Denote by Gz ≤ G the isotropy of G at z ∈ H2,and similarly for Gz ≤ G.

De�nition 3.8. [3]An element γ ∈ GL2OK is called elliptic if tr(γ)2−4det(γ) ≺0.

Proposition 3.17. An elliptic element γ has exactly one �xed point in H2.

Proof. It su�ces to show either real embedding of γ has exactly one �xed point

in H. Let γ =

(a bc d

)∈ GL2R, then az+b

cz+d = z i� cz2 + (d − a)z − b = 0. We

consider the discriminant (d−a)2+4bc = a2−2ad+d2+4bc = (a+d)2−4(ad−bc).Hence, if tr(γ)2 − 4det(γ) < 0 then we have two complex conjugate solutions.Thus, exactly one in H.

An element z ∈ H is called an elliptic point under the group G if it occursas the �xed point of some elliptic element γ ∈ G.

Proposition 3.18. [2]The isotropy groups Gz, z ∈ H2 are �nite cyclic and theset {z ∈ H2 : #Gz > 1} projects down to a �nite set in H2/PSL2OK . ThusH2/PSL2OK is a complex space with �nitely many singularities.

Corollary 3.2. The same proposition holds for G.

Proposition 3.19. [9]The isotropy groups arising as non-trivial stablisers areexactly the maximal subgroups of G, they are all cyclic.

De�nition 3.9. [2]Let ar(G) be the number of points of H2/PSL2OK withisotropy group of size r, similarly for ar(G).

One can show that there exists a fundamental domain for G such that allthe points with non-trivial isotropy groups lie on the boundary[3], this is veryanalogous to the one-dimensional case. i.e.

22

Example 3.6. Consider PSL2Z acting on H. Let F = {z ∈ H : −12 ≤ Re(z) <12 , |z|

2 ≥ 1}, be a fundamental domain. Then,

a2 = 1 (z = i)

a3 = 1 (z = ρ)

ai>3 = 0.

Later we will need to know the value of ar(G) for general G. They weredetermined for a general quadratic �eld extension K = Q(

√d) by Prestel, we

will present a simpli�ed version of his result.

Proposition 3.20. [2]Let d ≥ 7 be square free integer not divisible by 2 or 3,and K = Q(

√d). Then for the Hilbert modular group G(K) we have,

if d ≡ 1 (mod 4) then

a2(G) = h(−4d)

a3(G) = h(−3d)

ar(G) = 0 for r 6= 2, 3,

if d ≡ 3 (mod 8) then

a2(G) = 10h(−d)

a3(G) = h(−12d)

ar(G) = 0 for r 6= 2, 3,

if d ≡ 7 (mod 8) then

a2(G) = 4h(−d)

a3(G) = h(−12d)

ar(G) = 0 for r 6= 2, 3.

Further, for prime d ≡ 3 (mod 4) and d 6= 3 we have the following result,if d ≡ 3 (mod 8), then

a2(G) = 3h(−d) + h(−8d)

a3(G) = h(−12d)/2

a4(G) = 4h(−d)

ar(G) = 0 for r 6= 2, 3, 4,

if d ≡ 7 (mod 8), then

a2(G) = h(−d) + h(−8d)

a3(G) = h(−12d)/2

a4(G) = 2h(−d)

ar(G) = 0 for r 6= 2, 3, 4.

23

Prestel also gave the values of ar(G) and ar(G) for d = 2, 3, 5. The proof ofthis proposition is quite long and requires a detail look at the so called ellipticpoints of G. For now, let us consider a simple example of the proposition inaction.

Example 3.7. Let G = PSL2OK where K = Q(√−11), note 11 ≡ 3 (mod 8).

We have that ar(G) = 0 for r 6= 2, 3. One can show h(Q(√−11)) = 1, thus

we have a2(G) = 10. Further, we have that h(Q(√−132)) = 4 [?](note 132 =

11× 12) so a3(G) = 4.

Example 3.8. Let G = PSL2OK were K = Q(√−17), note 17 ≡ 1 (mod 4).

We have that ar(G) = 0 for r 6= 2, 3. One can show h(Q(√−68)) = 4 [?](note

68 = 17× 4), thus we have a2(G) = 4. Further, we have that h(Q(√−51)) = 2

[?](note 51 = 17× 3) so a3(G) = 2.

We will see later that being able to calculate ai(G) is very important sincethey correspond to quotient singularities in the quotient H2/G.

3.4 The quotient space

From now on we will consider only the group G ∼= PSL2OK . In this section weexamine some basic properties of the quotient H2/PSL2OK = H2/G.

De�nition 3.10. X(K) is de�ned to be the quotient H2/PSL2OK .

Proposition 3.21. [9]X(K) is a non-compact complex surface with �nitelymany singularities which arise from points in H2 which non-trivial isotropygroups.

As we know from the previous section both G have points with non-trivialisotropy groups, thus the quotient H2/G will have some quotient singularities.One could instead considered a �nite index subgroup of G which acts freely, andthus eliminate all quotient singularities. A detailed example of this was doneby van der Geer and Zagier[?] on the group

Γ = {(a bc d

)∈ SL2OQ(

√13) : a ≡ d ≡ 1, b ≡ c ≡ 0 (mod 2)}.

We will continue with our consideration of G.We now consider the Euler number of the space X(K).

Proposition 3.22. [2]The Euler number of the space X(K) is given by

χ(X(K)) =

ˆX(K)

ω +∑r≥2

arr − 1

r,

where the di�erential form ω is comes from the standard Riemann metric on H.i.e. ω = 1

(2π)2dx1∧dy1

y21∧ dx2∧dy2

y22.

24

Example 3.9. In the case K = Q, we have ω = (−1) 12π

dx∧dyy2 and χ(X) =

− 16 + 1

2 + 13 = 1. Which makes sense since H/PSL2Z is biholomorphic to C.

We want to apply the above proposition to the quotients H2/G. To do thiswe will need to be able to calculate

´X(K)

ω. To do this we will relate´X(K)

ω

to the generalised zeta-function.

De�nition 3.11. The ζfunction of the �eld K is de�ned as

ζK(s) =∑a≤OK

1

N(a)s,

where a ≤ OK is a non-zero ideal.

I am not going to go into details concerning convergence proofs. We willaccept it as fact that ζK converges uniformally on any compact U ⊆ {s ∈ C :Re(s) > 1} and can be holomorphically extended to C − {1}. A theorem ofSiegel shows that this function is very closely to

´X(K)

ω.

Proposition 3.23. [9]´X(K)

ω = 2ζK(−1).

We now ask how to compute ζK(−1)? For this we again use a theorem ofSiegel.

Proposition 3.24. [9]Let D be the discriminant of K, then

ζK(−1) =1

30

∑i2<D

4|D−i2

σ1

(D − i2

4

),

where σ1(n) =∑d|n d.

Example 3.10. Lets consider the case K = Q(√

17). We know from theprevious section that a2(G) = 4 and a3(G) = 2. Note that the discriminantis D = 4. First we calculate

ζK(−1) =1

30

∑i2<17

4|17−i2

σ1(17− i2

4)

=1

30

(σ1(

17− 1

4) + σ1(

17− 16

4)

)=

1

30(σ1(4) + σ1(1)) =

1

30(1 + 2 + 4 + 1 + 2)

=1

3,

thus we have χ(X(K)) =´X(K)

ω +∑r≥2 ar

r−1r = 2ζK(−1) + a2(G)

2 +2a3(G)

3 = 23 + 2 + 2×2

3 = 4.

25

As a corollary of the preceding results we can obtain a purely number theo-retic result.

Proposition 3.25. Let d > 5 a square-free integer such that d ≡ 1 (mod 4),then ∑

n2<dn odd

σ1

(d− n2

4

)≡ 0 (mod 5).

Proof. We know that the Euler characteristic is always an integer, and that

χ =1

15

∑σ1

(d− n2

4

)+a22

+2a33

=1

30

(2∑

σ1

(d− n2

4

)+ 15a2 + 20a3

),

now note that 5|15a2 and 5|20a3, so 5 must also divide 2∑σ1

(d−n2

4

).

X(K) will not be compact, we compactify X(K) by adding points at thecusps. We have shown above that there are only �nitely many cusps.

3.5 Resolution of cusps

In this section we will outline a method of resolving the singularities at thecusps. Let M be a subgroup of K = Q(

√d) which is free and of rank 2; we call

M a complete Z-module of K.

De�nition 3.12. [2]Let M be a complete Z-module, then de�ne U+M := {ε ∈

K× : εM = M}.

Remark 3.3. Any ε ∈ K such that εM = M is automatically an algebraicinteger and a unit. Further, U+

M is a free group of rank 1.

De�ne an equivalence on the set of complete Z-modules via M1 v M2 i�there exists a totally positive λ ∈ K such that λM1 = M2. Note, M1 v M2 ⇒U+M1

= U+M2

. Given and cusp, c (i.e. a PSL2OK orbit of PK), there existsρ ∈ PSL2K such that ρ(x) = ∞. If we denote the isotropy group of c by Gc,then we have the following exact sequence,[2]

0→M → ρGcρ−1 → V → 1,

where M is a complete Z-module and V is a subgroup of U+M of rank 1.

Proposition 3.26. [2]Up to the above equivalence, M and V are completelydetermined by the cusp c and do not depend on the choice of ρ.

Given an exact sequence as above, we say the cusp c is of type (M,V ).

26

De�nition 3.13. [2]Given a cusp c of type (M,V ) we de�ne the group,

G(M,V ) = {(a b0 1

): a ∈M, b ∈ V }

= M o V.

Remark 3.4. In fact, ρ can be chosen so that ρGcρ−1 = G(M,V ).

The group G(M,V ) acts freely and properly discontinuously on H2. More-over, there equivalence on the complete Z-modules induces an equivalence onthe group actions. Note the quotient H2/G(M,V ) is a complex manifold. Wenow add an extra point to H2/G(M,V ) called ∞, and de�ne a system of openneighbourhoods about ∞ as follows:

(·W (d)/G(M,V )) ∪ {∞},

where, for d > 0

·W (d) = {(z1, z2) ∈ H2 : Im(z1)× Im(z2) > d}.

De�ne H2/G(M,V ) = H2/G(M,V ) ∪ {∞} with the above topology. Let Obe the local ring at in�nity, in words it is the ring of functions which are holo-morphic (regular) in some neighbourhood of ∞. We now examine the structure

of O. Let f be holomorphic on some H2/·W , the pull-back of f to

·W is invariant

under the translation z 7→ z + µ with µ ∈ M . Therefore, f has an absolutelyconvergent Fourier series,

f(z) = a0 +∑v∈M∗

avexp(2πi·tr(vz)).

Where M∗ is de�ned as follows.

De�nition 3.14. [3]Given a module M we de�ne its dual(with respect to thetrace) to be M∗ where

M∗ = {x ∈ K : tr(xa) ∈ Z, ∀a ∈M}.

Let M∗ be a complete Z-module which is dual to M .

Proposition 3.27. [3]The local ring at in�nity, O, is the ring of all Fourierseries,

f(z) = a0 +∑

v∈M∗v�0

avexp(2πi·tr(vz)),

converging on some neighbourhood of ∞, satisfying aεv = av for all ε ∈ V .

Proof. [3]

Proposition 3.28. [3]The space H2/G(M,V ) with the local ring at ∞, O, isa normal complex space.

27

Recall, for dimension greater than one, being normal does not imply beingnonsingular. In fact, it has been shown that in our current case, the point ∞is always singular. We now take what may seem like a digression to constructsome normal singularities of complex surfaces.

Let {bk ≥ 2}k=0 ∈ NN be an arbitrary sequence, for each k ∈ Z we de�neRK ∼= C2 with coordinates uk, vk. Let Rk,u = Rk − V (uk) and Rk,v = Rk −V (vk).

Proposition 3.29. [2]The equations,

uk+1 = ubkk vk

vk+1 = u−1k ,

de�ne a biholomorphic map ϕk : Rk,u → Rk+1,v.

We now glue all of these pieces together. i.e. de�ne the set

Y :=⊔Rk/ ∼,

where ∼ is the equivalence relation induced by the gluing maps ϕk. Note thatY is covered by the Rk, and each Rk maps biholomorphically onto C2. Thisde�nes an atlas on Y . By construction, a subset of Y is open i� it is openintersection with Rk for all k. Further, Y is second countable.

Proposition 3.30. [2]The space Y is Hausdor�, and thus a complex manifoldof dimension 2.

Proof. [2]

Taking stock, we see that given any sequence {bk ≥ 2}k=0 ∈ NN we canconstructed a complex manifold Y of (complex) dimension two.

[2]Given Y as above, there exists a family of compact rational curves Sk ⊆ Y ,k ∈ Z, non-singularly embedded. The curve Sk is given by uk+1 = 0 in the(k+1)th coordinate system and vk = 0 in the kth coordinate system. Moreover,Sk and Sk+1 intersect in just one point, the origin of the (k + 1)th coordinatesystem. Further, if |i− j| > 1 then Si does not intersect Sj .

Proposition 3.31. [3]The self intersection number of Sk is equal to −bk.

Notice that the construction of Y is very similar to the resolution of cyclicquotient singularities we examined earlier.

From now on we will assume that our sequence of integers, {bk} is periodic,i.e. there exists some r > 0 such that bk+r = bk for all k.

De�nition 3.15. Let {bk} be such a sequence, de�ne

wn = [bn, bn+1, ...].

28

We assume that bj > 2 for some j, else we have wn = 1 for all n. Under suchan assumption one can show that all the wn are quadratic irrationals, greaterthan one. Consider the complete Z − module M = w0Z + Z. The module Macts freely on C2 by,

a · (z1, z2) = (z1 + a, z2 + a′),

for a ∈ M (a, a′ are the two real embeddings of a). Let Y be the complexmanifold associated to our sequence {bk}. Consider the following map,

Φ : Y − ∪Sj → C2/M

Φ : (u0, v0) 7→ (z1, z2),

de�ned by,

2πiz1 = w0logu0 + logv0

2πiz2 = w′0logu0 + logv0.

Proposition 3.32. [2]Φ is a biholomorphism.

De�nition 3.16. Ak is de�ned inductively for any integer k as,

A0 = 1

Ak+1 = w′k+1Ak.

Note, 0 < Ak+1 < Ak for all k and Ak 6= 1 for k 6= 0. Further we havethat wk = bk − 1

wk+1and bkAk = Ak−1 +Ak+1. For any integer k, the numbers

Ak, Ak−1 form a basis for M . This is all very analogous to chapter two.[2]From the gluing relations on Y we get an expression for the map Φ in the

kth coordinate system,

2πiz1 = Ak−1logu0 +Aklogv0

2πiz2 = A′k−1logu0 +A′klogv0.

By the periodicity of bk we have that wk+r = wk for all k.

Proposition 3.33. [2]Let Ak be de�ned as above then

Ak+r = ArAk

(Ak)n = Ank,

moreover, ArM = M .

As a corollary of this we have that Ar is an algebraic integer and a unit(6= 1).

Proposition 3.34. [2]Both wk and Ak are totally positive.

29

Denote by V , the in�nite cyclic subgroup of U+M generated by Ar. Let us

once more take a moment to re�ect on our current situation. Given a sequence{bk} of period r we associate the complex manifoldY , the pair (M,V ) and thusthe group G(M,V ). This determines a cusp singularity. We will construct aresolution of this singularity using the complex manifold Y .

Restrict the biholomorphism Φ to the open set Φ−1(H2/M

)⊆ Y .

Proposition 3.35. [2]The set Φ−1(H2/M

)is given by,

Ak−1log|u0|+Aklog|v0| < 0

A′k−1log|u0|+A′klog|v0| < 0.

It follows thatY + := Φ−1(H2/M) ∪

⋃k

Sk,

is an open subset of Y . There is a well de�ned group action of V on Y + [3].Thus we have the following exact sequence,

0→M → G(M,V )→ V → 1.

This induces an action of V on H2/M . Note that, the map

Φ : Y + −⋃k

Sk → H2/M

is a biholomorphism. Moreover, the action of V is compatible with Φ.

Proposition 3.36. [3]The action of V on Y +is free and properly discontinuous.

Proof. Very long, can be found in[2].

The orbit space Y +/V is a complex manifold, moreover the curve Sk ismapped onto the curve Sk+r. Hence Sk and Sk+r become the same curve inY +/V .

Proposition 3.37. [2]If r > 2 then in Y +/V we have a cycle on non-singularrational curves S0, S1, ..., Sr−1, such that Sk, Sk+1 intersect transversely in ex-actly one point and the self intersection number of Sk is Sk ·Sk = −bk. Otherwisethere are no intersections.

Hirzebruch also details the cases r = 2, 1 in [2]. I omit them for brevity.Note that for a given integral sequence {bk ≥ 2} of period r we can associate

the complex manifold Y +/V . From now on we will denote this complex manifoldby Y (b0, ..., br−1). The corresponding matrix of intersection numbers is

−b0 1 0 · · · 0 11 −b1 1 0 · · · 0

0 1. . .

......

. . . 1 0

0... 1 −br−2 1

1 0 · · · 0 1 −br−1

.

30

Assuming that at least one bi > 2 implies that this matrix is negative de�nite.Thus (a theorem of Grauert) the con�guration (S0, S1, ..., Sr−1) can be blowndown to give an isolated normal point P in a complex space Y (b0, ..., br−1).[2]

Proposition 3.38. The following map is holomorphic

σ : Y (b0, ..., br−1)→ Y (b0, ..., br−1),

given by,

σ

(⋃k

Sk

)= P.

Moreover, the map,

σ : Y (b0, ..., br−1)−⋃Sk → Y (b0, ..., br−1)− {P},

is a biholomorphism.

The space Y (b0, ..., br−1) is a normal complex space with one isolated singularpoint P .

We have a natural map

Y (b0, ..., br−1)→ H2/G(M,V ),

and a commutative diagram,Y (b0, ..., br−1) → H2/G(M,V )↓ σ ↗ σ

Y (b0, ..., br−1)

.

Where σ is a biholomorphism and σ(P ) =∞.

Proposition 3.39. [3]Y (b0, ..., br−1)−{P} ∼= H2/G(M,V ), moreover the localring of Y (b0, ..., br−1) at P is isomorphic to O.

Thus H2/G(M,V ) can be identi�ed with Y (b0, ..., br−1) and hence is a nor-mal complex surface. We can now state the main theorem of the paper.

Proposition 3.40. [3]The space H2/G(M,V ) with local ring at in�nity O isa normal complex space and Y (b0, ..., br−1) is the unique minimal resolution ofthe singular point ∞ of H2/G(M,V ).

31

References

[1] Bruinier, Jan Hendrik. Hilbert modular forms and their applications.The 1-2-3 of modular forms, 105�179, Universitext, Springer, Berlin, 2008.

[2] Hirzebruch, Friedrich. Hilbert modular surfaces. Unpublished lecturenotes

[3] van der Geer, Gerard. Hilbert modular surfaces. Ergebnisse der Math-ematik und ihrer Grenzgebiete (3) [Results in Mathematics and Related Ar-eas (3)], 16. Springer-Verlag, Berlin, 1988. x+291 pp. ISBN: 3-540-17601-2

[4] Reid, Miles. Cyclic surface quotient singularities.HTTP://homepages.Warwick.ac.UK/~Masada/MA4A5/

[5] Reid, Miles. Undergraduate algebraic geometry. London mathematics so-ciety.

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[9] Hirzebruch, Friedrich; van der Geer, Gerard. Lectures on Hilbertmodular surfaces. Based on notes taken by W. Hausmann and F. J. Koll.Séminaire de Mathématiques Supérieures [Seminar on Higher Mathemat-ics], 77. Presses de l'Université de Montréal, Montreal, Que., 1981. 193 pp.ISBN: 2-7606-0562-0

[10] van der Geer, G.; Zagier, D. The Hilbert modular group for the �eldQ(√

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[11] Shafarevich, Igor R. Basic algebraic geometry. 1. Varieties in projectivespace. Second edition. Translated from the 1988 Russian edition and withnotes by Miles Reid. Springer-Verlag, Berlin, 1994. xx+303 pp. ISBN: 3-540-54812-2

[12] Hirzebruch, F. The Hilbert modular group, resolution of the singulari-ties at the cusps and related problems. Séminaire Bourbaki, 23ème année(1970/1971), Exp. No. 396, pp. 275�288. Lecture Notes in Math., Vol. 244,Springer, Berlin, 1971.

[13] Wikipedia, HTTP://en.Wikipedia.org/wiki/Hilbert_modular_group

[14] Milne, J.S. Modular Functions and Modular Forms (Elliptic ModularCurves), HTTP://WWW.jmilne.org/math/CourseNotes/MF.pdf

[15] Epstein, Adam. Riemann Surfaces course, MA475.

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[16] ,Thurston, William P. Three-dimensional geometry and topology. Vol.1. Edited by Silvio Levy. Princeton Mathematical Series, 35. PrincetonUniversity Press, Princeton, NJ, 1997. x+311 pp. ISBN: 0-691-08304-5

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