HP.masterThesis

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Radboud University Nijmegen

Master Thesis

Hyperbolic Reflection Groups and theLeech Lattice

Date: June 2011Author: Hester Pieters, 0537381

Supervisor: Prof. Dr. G.J. HeckmanSecond Reader: Dr. B.D. Souvignier


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Contents

Introduction 5

Acknowledgements 6

1 Hyperbolic Space 7

2 Hyperbolic Reflection Groups 11

3 Lattices 17

3.1 Root lattices and reflection groups . . . . . . . . . . . . . . . . . . . . . . . . . . . 173.2 Covering radius and holes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20

4 The Algorithm of Vinberg 23

5 Even Unimodular Lattices 27

5.1 Norm zero vectors in hyperbolic lattices . . . . . . . . . . . . . . . . . . . . . . . . 295.2 The lattices II9,1 and II17,1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 305.3 Niemeier lattices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33

6 The Leech Lattice 396.1 The existence of the Leech lattice . . . . . . . . . . . . . . . . . . . . . . . . . . . . 396.2 The uniqueness of the Leech lattice . . . . . . . . . . . . . . . . . . . . . . . . . . . 406.3 Theorem of Conway . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 416.4 Covering radius . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 436.5 Deep holes. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45

The holy constructions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47

Bibliography 51

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4 CONTENTS


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Introduction

The Leech lattice is the unique 24 dimensional unimodular even lattice without roots. It wasdiscovered by Leech in 1965. In his paper from 1985 on the Leech lattice ( [3]), Richard Borcherdsgave new more conceptual proofs then those known before of the existence and uniqueness of theLeech lattice and of the fact that it has covering radius

2. He also gave a uniform proof of the

correctness of the holy constructions of the Leech lattice which are described in [6], Chapter24. An important goal of this thesis is to present these proofs. They depend on the theory of

hyperbolic geometry and hyperbolic reflection groups. The first two chapters give an introductionto these and contain all that will be needed.We will furthermore elaborate on the deep holes of the Leech lattice in relation to the classificationof all 24 dimensional unimodular even lattices, the Niemeier lattices. In more detail, the thesis isorganized as follows.

In Chapter 1 three models for the hyperbolic space Hn are described together with some ele-mentary properties of this space.

In Chapter 2 we describe hyperbolic reflection groups. We show that a discrete group W gen-erated by reflections is in fact generated by the reflections in the hyperplanes that bound a convexpolyhedron D, a fundamental domain for the group W. HenceW has a Coxeter representation.We then describe criteria to determine whether the polyhedron D is bounded or has finite volume.

Chapter 3 deals with lattices. We describe the root lattices that will later play an importantrole in the classification of Niemeier lattices. We also describe how a lattice determines a reflectiongroup. Finally, we describe the theory of gluing and define the covering radius and holes of a lattice.

In Chapter 4 we describe Vinbergs algorithm. This is an algorithm to determine the fundamentaldomain D Hn of a discrete reflection group W. We emphasize the case where the reflectiongroupWis the reflection group of an even hyperbolic lattice.

In Chapter 5 we turn to even unimodular lattices. First we describe what is known about theclassification of such lattices. Then we describe the relation between primitive norm zero vectorof a lattice L of type II8n+1,1 (so L is even unimodular with signature (8n+ 1, 1)) and lattices of

type II8n. After that, we treat the examples II9,1 and II17,1 applying the theory from chapters 2and 4.We then discuss the classification of the Niemeier lattices, presenting Boris Venkovs proof of thisclassification. We end the chapter with some results on Niemeier lattices that we will need in thelast chapter.

Finally, in Chapter 6 we discuss the Leech lattice. We present Borcherds proofs of the exis-tence and uniqueness of the Leech lattice and the fact that it has covering radius

2. We end the

chapter with a section that discusses the deep holes of the Leech lattice.

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Acknowledgements

This thesis was supervised by Gert Heckman. I want to thank him for suggesting the very inter-esting subject and for the many inspiring meetings we had. His lecture notes on the Leech lattice[10] have greatly influenced the set-up of this thesis. I would also like to thank Bernd Souvignier

for spending his time to act as the second reader.

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Chapter 1

Hyperbolic Space

In this chapter we describe some standard knowledge on hyperbolic geometry. It can all be foundin [5] and[16]. We will first describe three models for the hyperbolic space. The description of

these models is taken from [16]. All these models are differentiable manifolds with a Riemannianmetric. Each model is defined on a different subset ofRn,1 (n 2), i.e. Rn+1 with symmetricbilinear form (x, y) =x1y1 + ... + xnyn xn+1yn+1. This subset is called thedomainof the model.The first is the ball model(see [16]4.5). Here the domain is

Bn = {(x1,...,xn, 0) | x21+...+x2n < 1},and the Riemannian metric is

ds2 = 4 dx21+...+dx

2n

(1 x21 ... x2n)2.

The associated volume form is

dV = 2n dx1...dxn(1 |x|2)n .

The geodesics are the circles orthogonal to the boundary sphereBn =Sn1 = {(x1,...,xn, 0) | x21+...+x2n = 1}, the points at infinity.The second model is the upper half-space model(see [16]4.6). Here the domain is

Un = {(x1,...,xn, 0) | xn > 0},and the Riemannian metric is

ds2 = dx21+...+dx

2n

x2n.


dV = dx1...dxn

xnn.

The geodesics are half circles and half lines orthogonal to the boundaryRn1 = {x Rn,1 | xn = xn+1= 0}. The set of points at infinity is Un = Rn1 {}.The last model is the hyperboloid model(see[16] Chapter 3). Here the domain is

Hn = {(x1,...,xn, xn+1) | x21+...+x2n x2n+1= 1 and xn+1> 0},and the Riemannian metric is

ds2 =dx21+...+dx2n dx2n+1.


dV = dx1...dxn

(1 +x21+...+x2n)

12

.

The geodesics are intersections of two-dimensional vector subspaces ofRn,1 with the hyperboloid.If we identify each point in Hn with the (unique) one-dimensional vector subspace ofRn,1 that

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contains that point (interpret the hyperboloid model as lying in projective space) then it is clearthat we can identify the boundary points with those lines that lie in the boundary of the coneV+= {x Rn,1 | (x, x)< 0 and xn+1> 0}.

To show that the three models are equivalent we have to describe isometries between them. The

isometryf :Hn Bn is the central projection from the point (0, ..., 0, 1) (see [16]4.5):

f : (x1,...,xn+1) ( x11 +xn+1

, ..., xn

1 +xn+1, 0).

The isometry g : Bn Un (see [16]4.6) is given by:

g: x = (x1,...,xn, 0) 1||x en||2 (2x1, ..., 2xn1, 1n

i=1

x2i , 0), (with en the standard basis vector).

Hereg = , where is the reflection ofRn in the sphere S(en,

2) with center en and radius2 and is the reflection ofRn in the boundary of Un. This isometry can be extended to the

boundary. The same formula can be used except for x = en, for this set g (en) = . These threemodels and the maps between them can also be found in [5] and [9]. The isometries do indeedinduce the different metrics as they are defined above (see for example section 7 in [ 5]).

From now on in this chapter we use Hn as a model for the hyperbolic space. The Lorentz group isdefined to be O(n, 1) = {A Gl(n+ 1,R) | (Ax,Ay) = (x, y) for all x, y Rn,1}. ForA O(n, 1)we have either AHn = Hn or AHn =Hn. Hence O+(n, 1) ={A O(n, 1)| A(Hn) = Hn}has index 2 in O(n, 1). Elements ofO+(n, 1) are sometimes called orthochronous transformations.The restriction of an element ofO+(n, 1) to H

n is called a linear isometry ofHn. A Riemannianisometry f : Hn Hn ofHn is a diffeomorphism ofHn that preserves the Riemannian metric.These two notions turn out to be equivalent (see [5]Theorem 10.2).The group O+(n, 1) acts transitively on H

n. To show this consider two points x, y Hn. Thenthe orthogonal reflection in the linear hyperplane perpendicular to x

y sends x to y and clearly

is an element ofO+(n, 1). The connected groupSO+(n, 1) ={A O+(n, 1)| det(A) = 1} actsalso transitively on Hn since ifx, y Hn and z Hn is an arbitrary fixed point then there areorthogonal reflections A, B O+(n, 1) for which Ax = z and Bz = y. Reflection matrices havedeterminant1 soB A SO+(n, 1) and this element sends x to y .

A linear subspace ofRn,1 is said to be hyperbolic (elliptic, parabolic) if the bilinear form in-duced on it is nondegenerate and indefinite (positive, degenerate). We remark that the orthogonalcomplement of a hyperbolic (elliptic, parabolic) subspace is an elliptic (hyperbolic,parabolic) sub-space of complementary dimension (see [16]3.1, note that in this book elliptic, hyperbolic andparabolic subspaces are called space-like, light-like and time-like).

Definition 1. A hyperbolicm-plane ofHn is the nonempty intersection ofHn with a hyperbolic

subspace ofRn,1

of dimensionm+ 1. A hyperbolic (n 1)-plane ofHn

is called a hyperplane ofHn. A hyperbolic1-plane ofHn is called a line or geodesic ofHn.

Lemma 1. Letp, q Hn. Then(p, q) = cosh(d(p, q)). (Where the distanced(p, q) is the lengthof the geodesic fromp to q.)

Proof. sketch (following page 83 of [5]): d(p, q) is invariant under isometries sop and qcan be firsttranslated to a standard position. In fact, they can be translated to the plane spanned by en anden+1as follows: Definean to be the unit tangent vector at p in the direction of the geodesic from ptoqand definean+1= p. By the Gram-Schmidt process the set of orthonormal vectors {an, an+1}can be extended to an orthonormal basis{a1,...,an, an+1} for Rn,1 (so:(ai, aj) = (ei, ej).) Thematrix A with columns a1,...,an+1 is a linear isometry of H

n. Its inverse A1 takes p to en+1and the plane spanned by p and qto the plane P spanned by en and en+1. The intersection of

P with Hn is (one branch of) the standard hyperbola and is the unique geodesic passing throughA1(p) and A1(q) (see below). Now A1(p) = (0, ..., 0, 1) and d(p, q) = d(A1(p), A1(q)). It

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can therefore be assumed that A1(q) = (0,..., 0, sinh(d(p, q)), cosh(d(p, q))). Hence:

(p, q) = (A1(p), A1(q))= ((0,..., 0, 1), (0,..., 0, sinh(d(p, q)), cosh(d(p, q))))

= cosh(d(p, q))

It follows from the proof above that the group of isometries acts transitively on point pairs{p, q} in Hn with the same distance. Indeed, there is an isometry that sends{p, q} to{en+1,q}with qa point in the plane spanned by en anden+1such thatd(p, q) =d(en+1,q). With a rotationthat leaves en+1 fixed qcan be mapped to any another point in this plane at the same distancefrom en+1. Reversing this argument we can now send these two points to two arbitrary pointsp,q in Hn with d(p, q) = d(p,q). ThusHn is two-point homogeneous for the group O+(n, 1) ofisometries ofHn.

Geodesics

Letp, q Hn. Then the geodesic betweenp and qis indeed the intersection of the two-dimensionalvector subspace ofRn,1 containingp and qwith Hn:Since Hn is two-point homogeneous for its group of isometries O+(n, 1) we can always assumethat p = (0, ..., 0, pn, pn+1) and q = (0,..., 0, qn, qn+1). Now let : [a, b] Hn be a path fromp to q and suppose that lies not in the plane spanned by the standard basis vectors en and

en+1. Denote by the projection ofonto the xn xn+1 plane. Then L() =ba||(t)||(t)dt=b

a

g(t)((t), (t))dt >

ba

g(t)((t), (t))dt = L(), since g =dx21+...+dx

2n dx2n+1 and

thus g(t)(t)((t) (t), (t) (t)) 0 for allt [a, b]. Hence the geodesic between p and qlies in thexn xn+1 plane and is therefore equal to the intersection of this plane with Hn. It canbe shown that under the isometries f and g as defined above these geodesics are mapped to thegeodesics as they were defined in the other two models of the hyperbolic space.

For more about hyperbolic geometry we refer to [5]and [16].

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Chapter 2

Hyperbolic Reflection Groups

Let V =Rn,1 and let Hn be the hyperboloid model of the hyperbolic space. The negative normvectors in V consist of two connected components. Denote byV+ the connected component that

contains Hn ={x = (x1,...,xn+1) V | x2 =1 and xn+1 > 0} (so V+ is the interior of theforward light cone). Ifx Vhas positive norm then the (linear) hyperplane perpendicular to xgives a hyperplane ofHn and reflection in this hyperplane is an isometry ofHn. If the norm ofx V is negative then, after multiplication by an element ofR, x corresponds to a point in Hn.If the norm ofx = 0 is zero then x corresponds to a point at infinity.Let V be such that (, ) > 0. Then the reflection r O+(n, 1) in the hyperplaneH :={x V | (x, ) = 0} orthogonal to is given by r : x x 2 (,x)(,) for all x V.LetW < O+(n, 1) be a discrete group generated by reflections. BecauseWis a discrete group theset of mirrors{H}rW is locally finite. They divide Hn into convex regions.

In general, if U is a linear hyperplane of V with nonzero intersection with V+ then we writeU = U Hn for the corresponding hyperbolic hyperplane. Denote byU+ and U the two openhalf-spaces inVthat are bounded by Uand let U

+

and U be the corresponding open half-spacesin Hn bounded by U. Let Cbe a nonempty intersection of such half-spaces, that is

C=iI

Ui .

Then the closure ofC, denoted by D, is a convex polyhedron if{Ui}iI is locally finite. It ispossible that the hyperplanes Ui accumulate towards some point in Hn. We will always assumethat no Ui contains the intersection of the remaining half-spaces. In this case the half-spacesU

i

are uniquely determined by the polyhedron D. For more background on convex sets and polyhedrasee[1] and [2].

Now we return to the discrete reflection group W < O+(n, 1). The locally finite set of mir-rors H ={x V | (x, ) = 0} Hn divides Hn into convex polyhedra. It is shown below thatWacts transitively on these polyhedra and each of them is a fundamental domain for the actionofW.Pick an arbitrary convex polyhedron D, i.e. D is the closure of a connected component CHn \ rW H. Denote by D and Cthe connected components ofV\ rWH that containD and C, respectively. So D is a convex cone and D= D Hn (see Figure2.1).

We will first show that D is a fundamental domain for the action of the group W generatedby the reflections in the hyperplanes bounding D, i.e. W =< r| H D= >. After that wewill show thatW= W, thus Wis generated by the reflections in the hyperplanes bounding D.

Lemma 2. Letp

Hn. ThenWp

D

=

.

Proof. Let p+ C and choose q Wp for which|qp+|2 |wq p+|2 for all w W. Now

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D

Hn

V+

Figure 2.1: Sketch of coneD in V+.

suppose q / D. Then there exists an i such that q / Hi and thus (i, q)> 0. So

(ri(q), p+) = (q 2(i, q)i, p+)= (q, p+) 2(i, q)(i, p+)> (q, p+),

since (i, p+)< 0. Hence

|ri(q) p+|2 = ri(q)2 2(ri(q), p+) +p2+= q2 2(ri(q), p+) +p2+< q2 2(q, p+) +p2+= |qp+|2.

This contradicts the fact that qwas chosen such that|qp+|2 |wqp+|2 for all w W. Thusq D andWp D = .

So each W-orbit meets Din at least one point. By Theorem 5.13 in[12]it now follows that itmeets each orbit in exactly one point. Hence D is a fundamental domain for the action ofW.

Lemma 3. Letr W. Then there is aw W such thatrw W.Proof. Let p H be a general point, i.e. p H \

{=|rW} H . It follows from Lemma 2

that there is a w W such that w p D. Then (wp,w) = (p, ) = 0. Sowp Hw D. ThusHw is a hyperplane bounding D and therefore rw W.

Thus any generator r W can be written as a product of reflections in W, in the notationabove r = w

1rww with w , rw W.

The results in the remainder of this chapter are all from [19]and[20], also see[18]. Leti V, i I,

be vectors such that (i, i) = 1 and D= iI Hi (where it is again assumed that no Hi con-tains the intersection of the other half-planes). The Gram matrix of the vector system{i} will

be called the Gram matrix of the polyhedron D. It contains the following geometric information(see[20]1.1.3):

1. The hyperplanes Hi and Hj intersect if and only if |(i, j)| , where mii = 1 and mij = ifHi and Hj aredisjoint (in this case there is no relation between ri and rj ).

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2. The hyperplanes Hi and Hj do not intersect if and only if|(i, j)| 1. Note that(i, j) 1 is impossible since then Hi Hj or Hj Hi which contradicts theassumption that no half space contains the intersection of the others. So (i, j) 1. If(i, j) = 1 then Hi and Hj meet at infinity and we say that the hyperplanes areparallel.Otherwise, they are ultraparalleland there is a unique geodesic from one to the other andorthogonal to both. The length of this geodesic is the distance d(Hi , Hj ) between thehyperplanes and is determined by the formula

cosh(d(Hi ,Hj )) = (i, j).

The convex polyhedron D =

i Hi is nondegenerate if the hyperplanes Hi have no common

point in Hn or Hn and there is no hyperplane orthogonal to all the Hi . Furthermore,D is

calledfinite if it is the intersection of a finite number of half-spaces H+i . Obviously every boundedpolyhedron is finite.We will now describe the conditions to determine whether a polyhedron is bounded or has finitevolume. They can be found in[19] 1. A finite convex polyhedron D Hn is bounded if and onlyif

D V+ {0}.This is clear since ifD V+ {0} then there exists a ball B Hn with finite radius such thatD B.Furthermore, a finite convex polyhedron D Hn has finite volume if and only if

D V+.This can easiest be seen in the upper half-space model. IfD V+ then the convex cone D cancontain rays in the boundary ofV+. These rays correspond to vertices at infinity ofD. Now let qbe a vertex at infinity ofD. We can assume that qis the pointin the upper-half space model.Let R > 0 be such that the horizontal hyperplane at height R only intersects the hyperplanesthat pass through. So the intersection of a horizontal hyperplane at height R with D has aconstant volume in this plane. Then the volume of the part ofD with xn

R is:

R

dxnxnn

=

n+ 1Rn+1 < .

So a finite convex polyhedron D with vertices at infinity indeed has finite volume. It is clear thatifD is not contained in V+ then the volume ofD will not be finite.

The above inclusions can be translated in conditions for the Gram matrix ofD =

iI Hi .

Before stating this result some further notation has to be fixed. Let Abe a symmetric matrix. Itwill be called the direct sum of matrices A1,...,Ak if by a permutation of its rows and the samepermutation of its columns it can be written in the form

A1 0

.. .

0 Ak

In this case we write A = A1 ...Ak. A matrix that is not a direct sum of two or morenonempty matrices is indecomposable. Every symmetric matrix A can be represented uniquely asa direct sum of indecomposable matrices. These matrices will be called the componentsofA. Nowdefine A+ to be the direct sum of all positive definite components, A0 to be the direct sum of alldegenerate nonnegative definite components and Ato be the direct sum of all components whichare not nonnegative definite.Finally, forG = (gij)i,jI= ((i, j))i,jIthe Gram matrix of a nondegenerate convex polyhedronand S I letGS := (gij)i,jSbe the principal submatrix ofGobtained by deleting the kth rowsand columns for k / Sand let

CS= iS

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be the corresponding facet ofD. Clearly C = C and CI= {0}.

Suppose now thatG is an indecomposable symmetric matrix that satisfies

gii = 1 and gij

0 for i

=j. (2.1)

Note that in the case that G is the Gram matrix of a nondegenerate finite convex polyhedronD Hn this is equivalent to the assumption that no angle ofD exceeds /2. The matrix G canbe either nonnegative definite or indeterminate. IfG is positive definite then all the elements ofG1 are nonnegative (see for example [7], Lemma 9.1). IfG is nonnegative definite and degeneratethen by the Perron-Frobenius lemma ([12], Section 2.6) its kernel is one-dimensional and spannedby a vector =

iIkii with all ki > 0. Furthermore, ifG is nonnegative definite then all its

proper principal submatrices are positive definite.A matrixG = (gij) that satisfies condition (2.1) is calledcriticalif it is not positive definite, yet allits proper principal submatrices are positive definite. A critical matrix is always indecomposable.

Proposition 1. (Theorem 1 in [19]) LetD =

iIHi Hn be a nondegenerate finite convex

polyhedron with finite volume and with indecomposable Gram matrixG that satisfies condition2.1.Then

1. GS=G0S is a principal submatrix of rankn 1ofGCS Hn is a vertex at infinity ofD.

2. GS = G+S is a principal submatrix of rankm ofG CS Hn is an ordinary face ofD of

codimensionm.

Proof. (the proof here basically follows the proofs of Lemma 3 and Lemma 5 in [18])

1.: Suppose GS = G0S, where I S ={1,...,l}. Let ki be the positive coefficients of thelinear dependence between the rows of the matrix GS. Suppose that= 0. Then forj > lwe have 0 = (, j) =

li=1kigij . Because ki > 0 and gij 0 for all i it then follows that

gij = 0 for all i and hence the Gram matrix G would be decomposable. So it follows that= 0 andis an isotropic vector that generates the one dimensional orthogonal complementof< i| i S >. Now let j > l. Since (j , i) 0 for all i S, (j, ) 0. Suppose that(j , ) = 0. Then (j , i) = 0 for all iSand thus j is in the orthogonal complement ofU =< i| i S >. The subspaceU is an n-dimensional parabolic subspace orthogonal toRand hence j = for a R. This is impossible since 2j = 1. So 2 = 0, (j , )< 0for all j I\ Sand (i, ) = 0 for all i S. Hence CS= R is a ray at the boundary ofV+that corresponds to a vertex at infinity ofD.: Suppose that the facet CSis a ray on the boundary ofV+ corresponding to a vertex qatinfinity ofD. The linear planeCS =< i| i S >orthogonal to this ray is an n-dimensionalparabolic subspace ofV. Thus the matrixGSis nonnegative and hence by Perron-Frobeniushas a 1-dimensional kernel. Now suppose that GS is decomposable, say GS=A1 A2 withA1= A

+1. Suppose furthermore that S=

{1,...,k

}and that 1,...,l (l < k) are the vectors

that participate in the formation ofA1. These vectors span an elliptic subspaceM1. Now letxM1 be a nonzero vector that satisfies (i, x)0 for i= 1, . . . , l. SincexM1 we alsohave (x, i) = 0 for i = l + 1, . . . , k. Iff CS V+ then for >0 small enough f+ x D.But also (f+x)2 =2x2 >0 sof+ x / V+ which contradicts the assumption that D hasfinite volume. So the matrix GSis indecomposable and it thus follows thatGS is a principalsubmatrix of rank n 1 and GS=G0S.

2.: Suppose GS = G+S is a principal submatrix of rank m, say S = 1,...,m. Then thevectors 1,...,m form a basis of anm-dimensional elliptic subspace ofRn,1. The orthogonalcomplement M of M =< i| i S > is a hyperbolic subspace of codimension m and

M = M Hn is a hyperbolic plane of codimension m. The orthogonal projection ofx

Rn,1 on M is

xM =x i,jS

hij(x, i)j ,

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where thehij 0 are the entries ofG1S . Then for x D andk / Swe see that

(xM , k) = (x, i) i,jS

hij(x, i)(j , k) 0.

Since furthermore (xM , k) = 0 for k Sit then follows thatxM D. Thus also ifx Dthen xM D. Hence the orthogonal projection ofD on M lies in D. Since it contains anonempty open subset ofM, M D= CS Hn is a face of codimension m ofD.: Now suppose that CS Hn is an ordinary face ofD of codimension m and let M bethe linear plane that contains the facet CS. Since CS Hn is an ordinary face this is anmdimensional hyperbolic subspace. The vectors i with i Sgenerate the orthogonalcomplement M of M that is thus an (nm)-dimensional elliptic subspace. If GS isnot positive definite then there must be a linear dependence with nonnegative coefficientsbetween the rows ofGS (GScannot b e indeterminate since thei withi Sspan an ellipticsubspace). But if this was true there would also be a linear dependence between the i withiS. As this is not the case, it follows thatGS =G+S and GS is a principal submatrix ofrankm.

The polyhedron D can be shown to be bounded or to have finite volume without completelydetermining its combinatorial structure by verifying the following conditions (Proposition 1 in[19]):

1. Dhas finite volume if and only if for any critical principal submatrix GSof the Gram matrixG

(a) if GS = G0S, then there exists a subset T I that contains S, GT = G0T and rankGT =n 1.

(b) ifGS=GS, then CS= {0}.

2. D is bounded if and only if the Gram matrix G contains no degenerate nonnegative defi-nite principal submatrices and any indeterminate critical principal submatrix ofG satisfiescondition (b).

Proof. 1.: Suppose that D has finite volume and let GSbe a critical principal submatrix ofG. First suppose thatGS = G0S. The kernel ofGS is spanned by a vector =

iSkii

with ki > 0. This is an isotropic vector that corresponds to a vertex at infinity q ofD.Now let T = {i I| (i, ) = 0} so that q= CT Hn. Then S Tand by Proposition 1,GT =G0Tand rank GT =n 1.Now suppose that GS = G

S. Since (by Perron-Frobenius) all principal submatrices of an

indecomposable semidefinite Gram matrix satisfying condition 2.1are positive definite, GScannot be contained in a semidefinite principal submatrix. It thus follows thatCScannot be

any other facet ofD then{0}.: Now assume that conditions (a) and (b) are satisfied and suppose that CS is an extremalray of the convex cone D that is not in V+. Then by Proposition 1 the matrixGS cannotbe positive definite. Hence GS contains a critical principal submatrix GT1 . SinceCS CT1 ,CT1 ={0}. Thus by condition (b) GT1 is nonnegative and therefore semidefinite. Nowit follows from condition (a) that there exists a T2 T1 such that GT2 = G0T2 and rankGT2 = n 1. By Proposition 1, T2 ={i I| (i, q) = 0} where q is a vertex at infinity.Any face ofD that contains this vertex (and is not equal to this vertex) is an ordinary faceand thus corresponds to a positive definite principal submatrix. GT1 cannot be contained ina positive definite principal submatrix so it follows that CT1 =CT2 . BecauseT1 Swe haveCS CT1 andCS corresponds to the vertexqat infinity. It thus follows that D V+ {0}and hence D has finite volume.

2.: IfDis boundedGS=G0Sis impossible since by the proof of 1. above it would then followthat D has a vertex at infinity. IfGS=G

S the proof of 1. above shows that CS= {0}.

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: Now we assume thatG contains no semidefinite critical principal submatrices and it thenfollows immediately from the discussion of case 1. that D V+ and hence D is bounded.

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Chapter 3

Lattices

In this chapter we describe the root lattices that will later play an important role in the classificationof Niemeier lattices. We also explain how a lattice determines a reflection group. Finally, we

describe the theory of gluing and define the covering radius and holes of a lattice. Most of it canbe found in[6] chapters 1,2 and 4.

Definition 2. A latticeL in a real vector spaceVof finite dimensionn is a subgroup ofV suchthat there exist anR-basis{v1,...,vn} of V which is aZ-basis of L, i.e. L = Zv1 ... Zvn.Equivalently: Lis an additive subgroup that is discrete andV\Lis compact. Furthermore, V (andthusL) is equipped with a symmetric bilinear form(, ).

We write l2 = (l, l) for the norm of l L. An n-dimensional lattice Ln =Zv1 ... Zvn iscalled integral if l2 Z for all l Ln, and even if l2 2Z for all l Ln. We will say that alattice is positive definite, hyperbolic, etc. if the bilinear form (, ) has this signature. A Grammatrix of a latticeis the Gram matrix of a basis for this lattice. Thedeterminant of a lattice isthe determinant of such a Gram matrix, it is denoted by det L. A lattice is called unimodular ifthe determinant is

1.

Thedual latticeof a lattice Ln is given by Ln = {x Rn | (x, l) Zfor all l Ln}. A lattice L isintegral if and only ifL L. IfG is the Gram matrix ofL thenL has Gram matrix G1 for thedual basis and det L = (det L)1. For an integral lattice the groupL/L of order det Lis calledits dual quotient group. So an integral lattice L is unimodular if and only ifL= L.Take a basis{v1,...,vn} ofLn and let A be then nmatrix which has as rows the vectors vi. Ais called a generator matrixfor the lattice Ln andLn consists of all vectors

A,

with Zn. The Gram matrix ofLn is G = AAt and det Ln = (det A)2. Furthermore, (A1)t isa generator matrix for Ln. For any integral lattice we have:

L L 1

det L L.

(If l L then l = (A1)t = (A1)tA1A = (det L)1adj(G)A = (det L)1M, where, Zn).From now on all lattices will be integral.

3.1 Root lattices and reflection groups

LetL be a lattice. Aroot L is an element such that reflection by its orthogonal hyperplane inVis in Aut(L), i.e. r : 2 (,)(,) is an element of Aut(L). Thusis a root if2 (,)(,) Lfor all L. The root system R(L) ofL is the set of all roots:

R= R(L) = L | 2 (, )(, )

Zfor all L17


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Thus for an integral lattice L the roots are the vectors of norm 1 and 2 in L, which are called theshort and long roots respectively. IfL is even then there are no roots of norm 1 so the root systemofL in this case is R(L) = { L | 2 = 2}.

LetLbe an even lattice. The subgroup W(L) =< r|

R > 0 for all C}. We will also write > 0 to indicate that is a positive root. Similarly, thenegative roots < 0 are the set of roots R ={ R| (, ) < 0 for all C}. The simpleroots are those roots that are perpendicular to the faces of Cand that have product at most 0with all elements ofC. So R is simple if it is not of the form = j + k, wherej, k 1 and, R. (Contrary to what is common, here the simple roots are negative. With this conventiona root is a simple root iff it has product at most 0 with all other simple roots and it is a simpleroot iff it has product at most 0 with all elements ofC.)Denote by Q = Q(R) =ZR L the root lattice with dual weight lattice P =P(R) =Q = { V|(, ) Z for all Q}(={ V|(, ) Zfor all R+}). Let P+ ={ P|(, )N for all R+}be the cone of dominant weights.To each lattice L (not necessarily positive definite) with a fixed Weyl chamber C we can at-tach a Coxeter-Dynkin diagram which is determined by the Gram matrix of the simple roots. Toeach simple root we associate a vertex, and two vertices corresponding to distinct simple rootsi, j are joined by(i, j) lines. IfL is even and positive definite then two vertices are alwaysjoined by 0 or 1 lines, in this case we say that the diagram is simply laced. The only possibleirreducible simply laced diagrams are those of the types An (n1), Dn (n4), E6, E7 andE8(for a proof see[12] section 2.7). So the diagram of an even and positive definite lattice is a unionof diagrams of these types. The diagrams are shown in Figure 3.1 below. Hence ifL is a positivedefinite even lattice then the root lattice Q = ZR(L) is an orthogonal sum of lattices isomorphic toone of the lattices An, Dn, E6, E7 and E8. In the case of a hyperbolic lattice the Coxeter diagrammay contain multiple bonds.

An

Dn

E6

E7

E8

Figure 3.1: Irreducible simply laced root diagrams of finite type.

The Coxeter number h of an irreducible root system is the number of roots divided by the di-mension. The Coxeter number of a component of a positive definite lattice L is defined as theCoxeter number of the lattice generated by the roots of that component. In the following tablewe list the above root lattices with their Coxeter number and the order of their dual quotient group.

For a positive definite lattice L with a chosen Weyl chamber C the Weyl vector L Ris the vector that has inner product1 with all simple roots ofC. It is given by = 12 >0.This does indeed satisfy (, i) = 1 for all simple i:Let i be a simple root. It is well known that in this caseri(R+ \ {i}) =R+ \ {i} ([12]

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Table 3.1: Root latticesRoot system Coxeter number Determinant

An n+ 1 n+ 1Dn 2n 2 4E6 12 3E7 18 2E8 30 1

Proposition 5.6). Hence

ri() = ri

12

R+\{i}

12

i

=

1

2

R+\{i}

+1

2i

= +i

= + (, i)i.

Clearly, the Weyl vector is in the Weyl chamber C. IfL is not positive definite there may ormay not be a Weyl vector, that is, a vector that has product1 with all simple roots.

Assume from now on that L is a positive definite even lattice. For a component R of the Coxeterdiagram ofL the orbit of roots ofL under the Weyl group W(L) has a unique representative 0that is in the Weyl chamber C. This root is called thehighest root. If1,...,j are simple roots of

Rthen0=j

i=1kii, where the ki are positive integers called the weights of the roots i and

ji=1ki = h 1. Equivalently, 0 =

ji=0kii, with k0 = 1 and

ji=0ki = h. The diagram ob-

tained by adding the highest root is the extended Coxeter diagram ofL. It is the Coxeter diagram

of a semi-definite lattice L. We write Rn for the extended Coxeter diagram corresponding to theCoxeter diagram Rn. The extended diagrams for the diagrams in Figure3.1are shown below inFigure3.2 (see [12]section 4.7). The numbers next to the nodes are the coefficients ki. Any nodewith coefficient 1 is called a special vertex, its removal would give back the diagram correspondingto (a component of) the Coxeter diagram of the positive definite lattice.

Example 1. The lattice D +n .The root lattice of type Dn (n 4) is given by

Dn = {(x1,...,xn) Zn | x1+...+xn 2Z}.

The determinant of the Gram matrix ofDn is 4 for alln. Coset representatives for Dn/Dn are:

[0] = (0, 0, ..., 0), with norm 0,

[1] = (1

2,1

2,...,

1

2,1

2), with norm

n

4,

[2] = (0, 0, ..., 1), with norm 1,

[3] = (1

2,1

2,...,

1

2, 1

2), with norm

n

4.

Letn 2N, := [1] and define the latticeD+n =Dn ( + Dn). D+n is a sublattice of the duallattice Dn of index 2, since 2 Dn for n 2N. Also, since2 = n4 , D+n is even iffn 8N. Ifn8N then D+n is the unique even unimodular lattice (up to isomorphism) that contains Dn asa sublattice of rank 2 (D+n is isomorphic to Dn ([3] +Dn)). Furthermore,D+8 =E8. However,D

+16 2E8 since the norm 2 vectors in D

+16 generate the index 2 sublattice D16. Indeed, D16 is

irreducible while 2E8 is reducible.

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1 1A1

1 1 1 1 1 1

1

An(n

2)

2 2 2 2 2

1

1

1

1

Dn(n4)

1 2 3 2 1

2

1

E6

1 2 3 4 3 2

2

1E7

2 3 4 5 6 4 2

3

1E8

Figure 3.2: Extended root diagrams labelled with the integers ki.

The construction ofD +n is an example of gluing lattices.In general, starting with integral lattices L1,...,Lk (not necessarily distinct) of total dimensionn one can construct an n-dimensional integral lattice L that has as a sublattice the direct sumL1 ... Lk. A vector ofL is of the form l = l1+ ... + lk withli Li (not necessarily in Li). Let{

ni |n = 1, . . . , det(Li)} be a set of standard representatives for the cosets ofLi in Li , they arecalled the glue vectors for Li. In this context the group Li /Li is also called the glue group. The

glue vectors are usually chosen to be of minimal length in their coset. The latticeL is generatedbyL1 ... Lk and a set of vectors{gj} of the form

gj =gj1+...+gjk, (3.1)

wheregji {ni } is one of the standard representatives ofLi /Li for all i. These vectors are alsocalled glue vectors. They must have integral product with each other and be closed under additionmodulo L1 ... Lk. The additive group formed by the glue vectors is called theglue code. It canhappen that for a glue vector g j only one of the gji in3.1 is nonzero. In this case the componentLi has self-glue. The lattice D +n is constructed by self-gluing Dn. Since the sublattice Q = ZR(L)of an integral lattice L is a direct sum of root lattices, the root lattices are a particularly good

choice for the lattices Li. The classification of Niemeier lattices described in section 5.3 is a goodexample of the use of gluing theory.

3.2 Covering radius and holes

LetLbe a positive definite lattice and letV =LR be the ambient Euclidean space. The coveringradiusR ofL is the smallest R >0 such that the spheres of radius R centered at the points ofLwill cover V. It is given by

R= supxV

inflL

(x l)2.

For each pointl Lits Voronoi cellV(l) consists of those points in Vthat are at least as close tol as to any other l

L. Thus

V(l) = {x V| (x l)2 (x l)2 for all l L}.

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The Voronoi cells are compact convex polytopes whose union is V. Their interiors are disjoint,but they do have faces in common. All Voronoi cells of a lattice L are congruent and have volumeequal to

det L. We will sometimes callV(0) the Voronoi cell ofL. For any irreducible root lattice

Q =ZR(L) the Voronoi cell is the union of the images of the fundamental simplex (or standardalcove) A=

{

Q

R

|0

(, )

1 for all

R+

}under the (finite) Weyl group W(R(L))

(see[6], Chapter 21, Theorem 5).The vertices of the Voronoi cells are called holes. The radiusR() of a hole is its distance fromLso it is given by

R()2 = inf{(l )2 | l L}.The holes with maximum radius are called the deep holes, their radius is equal to the coveringradius ofL. The other holes are called shallow holes. Theverticesof a hole ofL are the elementsof the set

L() = {l L | (l )2 =R()},i.e. the lattice points l such that is a vertex of the Voronoi cell V(l).Thepacking radiusofL is defined to be the largest r >0 such that balls with radius r centered atthe lattice points do not overlap. The covering radius is the circumradius ofV (0), i,e. the radius

of the smallest circumscribed sphere. The packing radius is the inradius ofV(0), i.e. the radius ofthe largest inscribed sphere.

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Chapter 4

The Algorithm of Vinberg

In this chapter we describe the algorithm of Vinberg. This is an algorithm to determine a fun-damental domain D for a discrete group W < O+(n, 1) generated by reflections. The idea of the

algorithm is as follows: Pick a pointcHn

. The set of mirrorsHc ={H| r W, (, c) = 0}that containc divide Hn into convex regions. Let Dc be the closure of one of those componentsthat containc. It is a fundamental domain for the subgroup Wc < Wgenerated by the reflectionswhose mirrors contain c. Now there is a unique fundamental domain D ofW that contains c andis contained in Dc. The polyhedron D is bounded by the mirrors inHc. The other mirrors thatbound D are found by moving away from the point c inside Dc while checking which mirrors onemeets. That is, the mirrors are enumerated by increasing distance from c.The precise formulation of the algorithm is as follows: Pick a point cHn. Let Rc ={V =Rn,1 | 2 >0, (c, ) 0 and r W}. Form a sequence1, 2, . . . Rc according to the followingrules:

1. Pick 1,...,k Rc such that

Dc =

k

i=1

Hi .

2. forl > k pick l Rc that satisfies

(l, i) 0 for all i < l,

and such as to minimize the distance d(c, Hl) from the pointc to the hyperplane Hl .

ThenD=

i

Hi

is a fundamental domain for W (see[19],3).Now suppose that is a discrete subgroup of motions ofH

n

withW the subgroup generated byreflections that has fundamental domain D. Denote by Sym(D) the symmetry group ofD. Thenthe group decomposes into a semidirect product

=W H,

whereH Sym(D).

Below we will give a formulation of Vinbergs algorithm for the case that W corresponds to aneven lattice and also give a proof for this case. This is also the formulation of Vinbergs algorithmwe will use later.

In general, to obtain a hyperbolic reflection group from a lattice start with a hyperbolic lat-

tice L V = Rn,1, n 1. The automorphism group ofL is the discrete subgroup Aut(L) = {gO(V) | g(L) =L} < GL(V). In the case thatLis an integral lattice Aut(L) =GL(n+1,Z)O(V).

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As before V+ ={Rn,1 |2 0} and the hyperboloid model is used for the hy-perbolic spaceHn. Denote the mirror of a root R(L) byH = { V| (, ) = 0}. Choose aconnected component C=C Hn Hn \R H, also called the Weyl chamber. As before, wedenote by D and D the closure ofC resp. C. DefineW(L) =< r| simple root>. This is thesame situation as in Chapter 2 so again W= W, D is a fundamental domain for the action ofWon Hn and Whas a Coxeter representation.

Now let L be an even lattice. Then a fundamental domain D and its simple roots for the ac-tion ofW =W(L) on Hn can be found by Vinbergs algorithm as follows:Choose a controlling vector c P+(= P D). Define for R the height of with respectto c by ht() =(c, ) N, it can be interpreted as a measure for the distance between Hand c. Let Wc W be the subgroup generated by the reflections whose mirrors contain c, i.e.Wc =< r| R and ht() = 0 >. This is a finitely generated group. In particular ifc2 < 0then R c is of finite type and if c2 = 0 then R c is of finite or affine type. Let Dc bethe closure of the connected component of Hn \ {|rWc} H that contains c (possibly in theclosure). There is a unique fundamental domain D ofW that is contained in Dc and its simpleroots can be found as follows:

1. A positive root with ht() = 0 is a simple root ofD if and only if it is a simple root ofDc.

2. For < 0 with ht() 1: is a simple root if and only if (, ) 0 for all simple < 0with ht() ht()

2 .

Proof. (this proof follows the one in [10])

1. Clear

2. Since (, ) 0 holds for all simple roots , it is clear that the implicationholds.Conversely, it is also true that is a simple root if (, ) 0 for all simple roots. We needto prove that it suffices to only check this condition for all simple such that ht() ht()2 .So let n1ht()

2 +

m

i=1 niht(i).Thus n1 = 1 and ht(i) 1, xk / L.Lemma 4. Ifx L is a primitive norm zero vector then there is a vectory L withy2 = 0 and(x, y) = 1.Proof. Let x L be a primitive norm zero vector. SinceL is unimodular, we have L = L andthus there is an y L with (x, y) = 1. Ify 2 = 0 then y already satisfies the requirements in theLemma. Now suppose y2 = 0, say y 2 =k 2Z. Then (y+ k2 x)2 =y2 + 2 k2 (x, y) =k k= 0 and(y+ k2 x, x) = (x, y) = 1. So y + k2 x is an element ofL that satisfies the requirements.

Thus a primitive norm zero vector x L is contained in a rank 2 sublattice U. Since N andUare both unimodular, N U is also unimodular and has signature (25, 1). HenceL = N Uand the orthogonal complement N ofUis a lattice of type IIn. It follows that there are bijective

correspondences between the sets:1. n-dimensional even unimodular lattices (up to isomorphism),

2. sublattices ofL isomorphic to U(up to the action of Aut(L)),

3. primitive norm zero vectors inL (up to the action of Aut(L)).

Instead of the basis{x, y} ofUwith symmetric bilinear form (, ) that satisfies: x2 =y2 = 0 and(x, y) =1 we will also use the basis (x1, x2) for R1,1, where x1 = 12 x y and x2 = 12 x+y. Sox21= 1, x

22= 1, and (x1, x2) = 0. Note that this is not a lattice basis forU.

If N is a lattice of type IIn that corresponds to a norm zero vector x L then x/x= Nwithx the one-dimensional singular lattice generated by x. Also, x=N 0 where we write 0for the one-dimensional singular lattice. IfNhas roots then the Coxeter diagram ofN

0 is the

Coxeter diagram ofN with all irreducible components R replaced by the extended diagrams R.Now assume that Nhas roots and let L = N Uwith coordinates (,m,n) (with respect to thebasis{x, y} of U) where N,m,n Z and (,m,n)2 = 2 2mn. Then z = (0, 0, 1) is aprimitive norm zero vector that corresponds to the n-dimensional even unimodular latticeN. Theroots in z=N 0 are the vectors of the form (, 0, n) with R(N) and n Z. Suppose thatR(N) =

iRi with the Ri irreducible components ofR(N). Let{ji| j = 1, . . . , rkRi} be a set

of simple roots for the components Ri and let 0i be the corresponding highest roots. As a set ofsimple roots forz we can pick the vectors

fji = (ji , 0, 0), for j 1, and

f0i = (0i , 0, 1).

So the set of vectors {fji| j = 0, . . . , rk Ri} form the extended Coxeter diagram Riand furthermorejkjfji =z , where the kj are the weights of the vertices ofRi. If we apply Vinbergs algorithmwith z as a controlling vector we find a unique fundamental domain D ofL that contains z andsuch that all the vectors fji that form the extended Coxeter diagrams

Ri are simple roots ofD.On the other hand, suppose that D is a fundamental domain of L and that R is a connectedextended Coxeter diagram in the Coxeter diagram ofD. Then there is a primitive norm zerovector x in L that is in D and such that (x, i) = 0 for all simple roots i ofR:Indeed, setx =

ikii with i the simple roots of

Rand the ki their weights. Then x2 = 0 and

(x, i) = 0 for all i. Furthermore, if is a simple root ofD then (x, ) 0 since all i haveinner product at most zero with .It follows that there also is a bijective correspondence between the sets:

1. maximal disjoint sets of extended Coxeter diagrams in the Coxeter diagram ofD such thatno two diagrams in the set are joined to each other, and

2. primitive norm zero vectors ofD that have at least one root perpendicular to them.

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5.2 The lattices II9,1 and II17,1

ConsiderRn+1,1 with quadratic formx2 =n+1

i=1 x2i x2n+2. Denote by {i}n+2i=1 the standard basis

ofRn+1,1. Consider IIn+1,1 with n 8N. Let r = ( 12 , ..., 12 ) Rn+1,1. Then x Rn+1,1 is anelement of IIn+1,1 if:

1. x r Z, i.e.n+1i=1 xi xn+2 2Z, and2. all xi Z or all xi 12 Z.

Write R R(IIn+1,1). The group of orthochronous automorphisms of IIn+1,1 (i.e. the group ofautomorphisms that do not interchange the positive and negative time cones) is the semidirectproduct of the group of reflectionsWof IIn+1,1 with the symmetry group of the Weyl chamber D

determined byW. The polyhedron Dcan be determined by applying Vinbergs algorithm. Choosethe controlling vector c= (0, ..., 0, 2) Rn+1,1. Then c2 = 4 and c P = { Rn+1,1 | (, ) Zfor all R} since(c, x) = 2xn+2 Zfor all x IIn+1,1. Now

R

c = =

n+2

i=1 aii R an+2= 0=

Rn+1,1

an+2= 0, ai Zfor all i, i

ai 2Zand

i

a2i = 2

= Dn+1.

Use for Dn+1 the basis:

ei = i i+1, fori = 1,...,n,en+1 = n+n+1.

So here we have chosen the controlling vector c that corresponds with the sublattice Dn+1 ofIIn+1,1. There are however many more possibilities for the controlling vector. Another good

choice would be the norm zero vector (0,..., 0, 1, 1) that corresponds to the sublattice n E8. If itexists, the Weyl vector is another possible controlling vector. In this case all the simple rootshave height 1 (and all roots of height 1 are simple). For n = 8, 16 or 24 there in fact exists a Weylvector n+1 given by:

9 = (0, 1, . . . , 8, 38),

17 = (0, 1, . . . , 16, 46),

25 = (0, 1, . . . , 24, 70).

This is shown in Chapter 27 of[6]. There it is also remarked that for n = 32, 40,... there exists novector inRn+1,1 having constant inner product with all simple roots of the lattice of type IIn+1,1.

Example 2. II9,1

Now assumen= 8. As above we pick the controlling vector c= (0, . . . , 0, 2) R9,1

. SoRc= D9and we use the basis{e1, . . . , e9} as described above. Let =iaii be a root of height 1, i.e.(c, ) = 2a10= 1. Thusa10= 12 . This vector is accepted as a simple root by Vinbergs algorithmif it satisfies:

ai ai+1 0, fori = 1,..., 8a8+a9 0

Since is a root it also satisfies 2 =9

i=1a2i 14 = 2. Furthermore, sincea10 = 12 , ai+ 12 Z

for all i. Then the only possibility is that all ai {12 , 12}. It follows from the criteria above thatifai = 12 for all i then

ai = 1

2 ai+1= 1

2 fori = 1,..., 8,

ai+1 = 12 ai = 1

2 fori = 1,..., 8.

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From a8+a9 0 it then follows that a1,...,a8 are equal to12 and from r Zit follows thata9=

12

. So there is just one vector e10of height 1 that satisfies the criteria in Vinbergs algorithm,namely

e10=

1

2

8

i=1 i+1

2

9+1

2

10.

The vectors ei are a basis forRn+1,1. The dual basis is:

fi = 1+...+i i10, for i = 1, ..., 7

f8 = 1

2

8i=1

i 12

9 92

10

f9 = 1

2

9i=1

i 72

10

f10 = 210Suppose that is a root of height greater than one. Since the f

iform a basis can be written

as =ibifi. If is a root that satisfies the criteria of Vinbergs algorithm then (, ei) 0for i = 1, ..., 10. From this it follows that all bi 0. Since also f2i 0 for all i and thus also(fi, fj) 0 for all i, j then

2 =

i

j

bibj(fi, fj) 0.

So there can be no root that satisfies the criteria, the sequence of simple roots determined byVinbergs algorithm breaks off at the 10th root.The Coxeter diagram corresponding to the simple roots e1,...,e10is the diagram E10in Figure5.1.

e1 e2 e3 e4 e5 e6 e7

e9

e8 e10

Figure 5.1: E10.

The diagramE10contains the elliptic subdiagramA9and det(E10) = 1 (calculation using Lemma5.1 in [20]: det(E10) = det(A3)det(E7) 4cos2( 3 )det(A2)det(E6) = 4 2 4 14 3 3 = 1),so the signature of the Gram matrix ofE10is (9, 1) and it is the diagram of a Coxeter polyhedronD H9. Removing the leftmost node in the diagram gives the parabolic subdiagram E8, removingany other node gives elliptic subdiagrams. So D has one vertex at infinity corresponding with E8.The diagram has no symmetries so the group of orthochronous automorphisms of II 9,1 is thereflection group determined by E10, W(E10).

Example 3. II17,1Now suppose n= 16. Let c= (0, . . . , 0, 2)R17,1 be the controlling vector and{e1, . . . , e17} theabove described basis forD17= Rc. Suppose that =

iaiiis a root of height 1 that satisfies

the criteria of Vinbergs algorithm. Then ht() = 2a18 = 1 a18 = 12 and thus17

i=1a2i = 2

14

.

Becausea18= 12

and is in II17,1 all ai 12 Z. But then17

i=1a2i 4 14 so there can be no root

of height 1.Suppose now that is a root of height 2. Then a18 = 1 and thus ai Z for all i. Furthermore,2 =

17i+1ai 1 = 2 implies that

17i+1ai = 3 and thus the only possibility is that three of the ai

(i= 1,..., 17) are 1 and the others are zero. Vinbergs algorithm imposes the following conditionson the ai:

ai ai+1 0 fori = 1, ..., 16a16+a17 0

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and thus for ai {1, 0, 1}:

ai = 1 ai+1= 1 for i = 1,..., 16ai = 1 ai1= 1 fori= 2,..., 17.

The first property together with a16+a170 implies that ai= 1 for all i. The second propertyimplies that the only possible root is

e18= 1 2 3 18Continuing with Vinbergs algorithm we now consider roots of height 3. Then a18 =

32

and

2 =17

i=1a2i 94 = 2 so

17i=1a

2i = 4

14

. Since a18= 32

all ai 12 Z and it follows that the onlypossibility is that all ai are12 . From the criteria imposed by Vinbergs algorithm it then followsthat there is just one root of height 3 accepted, namely

e19= 12

17

i=1i+

3

218.

The Coxeter diagram determined by the simple roots e1,...,e19 is shown in Figure 5.2.

e1 e2 e3

e18

e4 e5 e6 e7 e8 e9 e10 e11 e12 e13 e14 e15

e16

e17 e19

Figure 5.2: Coxeter diagram of II17,1.

This is in fact the Coxeter diagram of the lattice II17,1 which we will now prove by showing thatthe sequence of simple roots determined by Vinbergs algorithm breaks off at e19. Suppose thatthere is one more simple root with ht()

4. Then is an element of the convex polyhedral

coneD= { V =R17,1 | (, ei) 0 for all i= 1,..., 19}. A rayR


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same sign. SoL has a one-dimensional intersection with the 2-dimensional faces R0i+ R0jsuch that node i is to the left of the middle node 9 and node j is to the right of node 9. Now thediagram obtained by deleting two such nodes is either positive definite of rank 17 or nonnegativedefinite and degenerate of rank 18. It follows from Proposition 1 that they correspond to (possiblyinfinitely distant) vertices ofD and hence that the extremal rays ofD are all inside of the closure

of V+. Since D is the convex hull of these extremal rays any element of D is in the closure ofV+ and thus cannot be a root. Hence there is no more root that satisfies the criteria of Vinbergsalgorithm.We can also determine directly from the diagram that it corresponds to a reflection group ofH17,1 with a fundamental domain D of finite volume. Write G for the Gram matrix that cor-responds to the diagram above, which we denote by S. Then det(G) = det(E10)det(E8)4cos2( 3 )det(

E8)det(E8) = 0 but S is not parabolic. Furthermore, S contains the elliptic subdia-gram 2E8+ A1 of rank 17. Thus the signature ofG is (17, 1, 1) andSis the diagram of a Coxeterpolyhedron D H17. Removing the middle node gives the parabolic subdiagram 2 E8 of rank 16,so this subdiagram corresponds to a vertex qat infinity. Removing the leftmost and the rightmostnode also gives a parabolic subdiagram of rank 16, namely D16. So this subdiagram also corre-sponds to a vertex at infinity. All the other vertices are obtained by removing one node left of the

middle vertex and one node right of the middle vertex, the subdiagrams that are obtained are allelliptic. So D Hn is a polyhedron with finite volume and two vertices at infinity. Topologically,D is a pyramid with vertex at q, constructed on the direct product of two 9-simplexes.

The only symmetry of the diagram is a reflection through the middle. So Sym(D) = Z2 andthe group of orthochronous automorphisms of II17,1 is the semidirect product ofW(S) =< rei|i= 1,..., 19> withZ2.

5.3 Niemeier lattices

In 1935 Witt found more than 10 of the 24-dimensional even unimodular lattices, now calledNiemeier lattices. In 1965 Leech found such a lattice without roots, now called the Leech lattice.

The list of 24-dimensional even unimodular lattices was completed by Niemeier in 1967. It turnedout that there exist twenty-four Niemeier lattices. Niemeiers proof was later simplified by BorisVenkov ([6], Chapter 18). His proof consists of two parts. First, using modular forms, he deter-mines a list of possible root systems for a Niemeier lattice. These are the twenty-four root systemslisted in Table5.1. Then by a case by case verification it is shown that for each root system thereexists a unique even unimodular lattice. We will discuss this proof in this section.

First of all, the list of possible root systems is determined by showing that a Niemeier latticeNhas to satisfy:

1. R(N) = or rank R(N) = 24,2. All irreducible components ofR(N) have the same Coxeter numberh,

3.|R(N)| = 24h.To show this the next proposition is crucial. It is proved using the theory of modular forms,

see[6], chapter 18.

Proposition 2 (Venkov 1980). IfNis a Niemeier lattice andx R24 thenR(N)

(, x)2 = 1

12(x, x)|R(N)| (5.2)

Corollary 1. IfNis a Niemeier lattice then eitherR(N) = or rankR(N)=24.Proof. Suppose that rank R(N)< 24. Then there is a nonzero vectory R24 that is orthogonalto all R(N). If we now set x = y in 5.2then it follows that 0 = 112 (y, y)|R(N)|. So |R(N)| = 0,i.e. R(N) = .

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Table 5.1: Niemeier lattices listed by their root systems, h is the Coxeter number and V is thenumber of Leech lattice points around the corresponding deep hole (see section 6.5).

Root system h V Root system h V

D24 46 25 2A9+D6 10 27

D16+E8 30 26 4D6 10 283E8 30 27 3A8 9 27A24 25 25 2A7+ 2D5 8 28

2D12 22 26 4A6 7 28A17+E7 18 26 4A5+D4 6 29

D10+ 2E7 18 27 6D4 6 30A15+D9 16 26 6A4 5 30

3D8 14 27 8A3 4 322A12 13 26 12A2 3 36

A11+D7+E6 12 27 24A1 2 484E6 12 28 0 -

Corollary 2. If Nis a Niemeier lattice then all irreducible components of R(N) have the sameCoxeter numberh and|R(N)| = 24h.Proof. If we set x = R(N) in equation5.2 then it follows that

R(N)(, )2 =

1

6|R(N)|. (5.3)

Denote by R the irreducible component of R(N) that contains . Then if / R clearly(, ) = 0. So only R contribute to the sum on the left side of the equation above. Since Ris a simply laced root system for , R we have (, ) {0, 1, 2} and (, ) = 2 if andonly if = . Let(R) ={ R| (, ) = 1}. Then it follows that

R(N)

(, )2 = 2 22 + 2(R). (5.4)

The Coxeter number h has the following property: If R is a simply laced root system then thenumber of elements ofR not orthogonal to a fixed R is equal to 4h 6 ([4], chapter VI, prop.32). Hence 2 + 2(R) = 4h(R) 6 and thus (R) = 2h(R) 4. So combining this with5.3and5.4 we get

2 22 + 2(2h(R) 4) = 16|R(N)|.

Hence|R(N)| = 24h(R) and since R(N) is chosen arbitrarily both claims in the corollarynow follow.

Proposition 3. The root system of a Niemeier lattice is one of those in Table5.1.

Proof. Suppose R:= R(N) = . Then R is isometric to an orthogonal sum of root lattices of thetypes Ai, Dj andEk. Say

R=24i=1

piAi+24

j=4

qjDj +8

k=6

rkEk.

Since all components of R have the same Coxeter number and h(An) = n+ 1, h(Dn) = 2n 2, h(E6) = 12, h(E7) = 18 and h(E8) = 30 it follows that R is of type pAi + qDj +rEk. ByCorollary 1, rank R = 24 and thus pi + qj + rk = 24. So we have the following possibilities forR:

1. R= pAi: Then it follows from pi = 24 that the possibilities are:A24, 2A12, 3A8, 4A6, 6A4, 8A3, 12A2 and 24A1.

2. R= qDj : Then it follows from qj = 24 that the possibilities are:D24, 2D12, 3D8, 4D6 and 6D4.

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3. R= rEk: Then it follows from rk= 24 that the possibilities are: 4E6 and 3E8.

4. R= pAi+qDj (p, q= 0): From h(Dj) = 2j 2 =h(Ai) =i + 1 it follows that i= 2j 3.Hencep(2j 3) +qj = 24 with j 4 and this equation has the following solutions:(j, p, q) = (9, 1, 1), (6, 2, 1), (5, 2, 2) and (4, 4, 1).

So in this case the possibilities forRare: A15+D9, 2A9+D6, 2A7+ 2D5 and 4A5+D4.5. R= pAi+qDj + rEk with r = 0 and qand p not both zero:

(a) Ifk = 6 then h = 12 and thus j = 7 and i = 11 (ifp, q= 0). So 11p+ 7q+ 6r = 24.Now r can be one, two or three and it easily follows that there is only a solution forr= 1, namely (p, q) = (1, 1). So this gives R = A11+D7+E6.

(b) Ifk= 7 then h= 18 and thus j = 10 and i = 17 (ifp, q= 0). There are two solutionsfor 17p+ 10q+ 7r = 24, namely (p, q, r) = (1, 0, 1) or (p, q, r) = (0, 1, 2). Hence we getthe following two possibilities for R: A17+E7 and D10+ 2E7.

(c) If k = 8 then h = 30 and thus j = 16 and i = 29 (if p, q= 0). Now i = 29 is notpossible, sop = 0 and 16q+ 8r= 24. This yields as only solution R = D16+E8.

We see that the possibilities that are found are indeed exactly those listed in Table 5.1.

The next step is to determine the existence and uniqueness of a lattice of type II 24 for all theroot systems in Table 5.1. We postpone the proof of the existence and uniqueness of the latticewith empty root system, the Leech lattice, to Chapter 6.Let Q = ZR be a root lattice generated by one of the nonempty root systems in Table 5.1. IfR= 3E8 thenQ is unimodular and we are finished. In the other cases Q is not unimodular so wethen need to determine a glue code and show that it is unique. Recall that for a lattice L the gluegroup is L/L. So for R =

iniRi (with the Ri irreducible) the glue vectors are elements of the

group

T(R) =P(R)/Q(R) =

i

(P(Ri)/Q(Ri))ni .

Hence the glue code is an additive subgroup A < T(R). As coset representatives for T(Ri) wepick the wni P(Ri) that satisfy (wni, ) {0, 1} for all >0 in Ri. It is well known that thesevectors form a complete set of coset representatives, they correspond to the special vertices ofRi.Furthermore, for gji T(Ri), let l(gji ) be the norm of the corresponding coset representative wni.Finally for a glue vector gj = (gj1, ...g

jk)T(R) (with k the number of irreducible components of

R) the function l is extended to be:

l(gj) =k

i=1

l(gji ).

Since the glue code must give a unimodular lattice, A has to satisfy|A|2 = |T(R)|. Furthermore,since the lattice must be even and with root lattice R all g A \ {0} must have the norm l(g)equal to an even integer that is > 2. We call a subgroupA that satisfies these two conditions evenand self-dual.IfRi is an irreducible root system then the group G(Ri) := Aut(Ri)/W(Ri) acts transitively onT(Ri). This group G(Ri) is the symmetry group of the Coxeter diagram ofRi. For R = niRi thecorresponding symmetry group of the Coxeter diagram ofR acting onT(R) is the wreath productG(Ri)

ni Sni . ForR =

iniRi it is

G(R) =i

(G(Ri)ni Sni) .

Clearly, the norm l(g) corresponding to g A is invariant under the action ofG(R). Hence theset of even and self-dual subgroups A < T(R) isG(R)-invariant. The Niemeier lattice constructedfrom the glue code A is the lattice N =< R , A > generated by the root system R and theglue code A. Conversely, given a Niemeier lattice N with root system R a glue code for thislattice is A = N/Q(R(N)) < P(R(N))/Q(R(N)). It follows that there is a natural one-to-one

correspondence between Niemeier lattices with nonempty root system isomorphic to R (up toisomorphism) and even self-dual subgroups A < T(R) (up to the action ofG(R)).

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Example 4. Let R = 4E6. We have T(E6) = Z/3Z ={0, 1} and G(E6) = Z/2Z ={1, }with (1) =1. Furthermore, l(0) = 0 and l(1) = 4/3. It follows thatT(R) = (Z/3Z)4 andG(R) = (Z/2Z)4S4. So a Niemeier lattice with root system 4E6 corresponds to an even selfdualsubgroup A 2 have three coordinates equal to 1 or

1 and the other one equal to 0. We can

assume that x = (1, 1, 1, 0) A. This is an element of order 3 so we need to add more elementsto get a subgroup of order 9. Because the sum of two elements has to be even and with norm> 2 it follows that the only elements that can be added are of the form y = (y1, y2, y3, y4) with{y1, y2, y3} ={0, 1, 1} and y4 =1. Any element of this form can be changed into any otherelement of this form by an element ofG(R) that leaves x invariant. So without loss of generalitywe can assume that y = (0, 1, 1, 1) A. NowA =< x, y > is an even self-dual subgroup of order9 and this group is unique op to action ofG(R). The glue code A is in fact equal to the tetracode(see for example [6], Chapter 3).

For most of the possible root systems the proof of the existence and uniqueness of a Niemeierlattice with this root system comes down to an easy verification like the one in the example above.Particularly easy cases are the ones R = D24 and R = E8+D16 for which the corresponding

Niemeier lattices are isomorphic to D

+

24 respectively E8 D+

16.The casesR= 24A1and R= 12A2are not that easy. They depend on the existence and uniquenessof the (extended) binary and ternary Golay codes.

Example 5. LetR= 24A1. We haveT(A1) = Z/2Z = {0, 1},G(A1) = 1 ,l(0) = 0 andl(1) = 1/2.So T(R) = (Z/2Z)24 and G(R) = S24. Furthermore, forx T(R) we have l(x) = 12 wt(x). Herewt(x) is the (Hamming) weight of the codeword x that is defined to be the number of nonzerocoordinates of x. It follows that any x in an even self-dual subgroup A < T(R) must satisfywt(x) 8. So an even self-dual subgroupA < T(R) corresponds to a binary self-dual code withminimum distance 8. The unique such code is the (extended) binary Golay code, see [11] Chapter10 for a proof of the existence and uniqueness of this code.

Example 6. Let R = 12A2. We have T(A2) = Z/3Z ={0, 1}, G(A2) = Z/2Z ={1, } with(1) =1, l(0) = 0 and l(1) = 2/3. So T(R) = (Z/3Z)

12

and G(R) = (Z/2Z)12

S12.Furthermore, for x T(R), l(x) = 23 wt(x). It follows that any glue vector x in an even self-dualsubgroupA < T(R) must satisfy wt(x) 6. So an even self-dual subgroup A < T(R) correspondsto a self-dual code in F123 with minimum distance 6. The (extended) ternary Golay code is theunique such code, again see [11]Chapter 10.

As noted in the beginning of this chapter, another way to prove the correctness of the clas-sification of the Niemeier lattices is by using the Minkowski-Siegel mass formula ([6], Chapter16). Yet another proof can be given using the list of deep holes of the Leech lattice as deter-mined in [6] Chapter 23. As we will see in the next chapter, the Leech lattice corresponds tothe Coxeter diagram of the unique lattice of type II 25,1 and deep holes of the Leech lattice corre-spond to subdiagrams of this Coxeter diagram that are unions of the extended Coxeter diagramsAn(n 1), Dn(n 4), E6, E7, and E8. But as seen in section 5.1 these subdiagrams correspondexactly to the Niemeier lattices with roots. Indeed, the diagrams found in the enumeration of thedeep holes of the Leech lattice correspond exactly to the root systems of the Niemeier lattices withroots. The determination of the deep holes is done by extensive computations so this provides byno means a more straightforward proof of the classification. However, in the casesR = 24A1 andR= 12A2 a corresponding deep hole can be constructed quite easily as is shown at the end of thenext chapter.

We end this section with some results on Niemeier lattices we need later. They are all takenfrom [3]section 2.

Lemma 5. Let R be a simply laced irreducible root system and = 12

>0 its Weyl vector.Then2 = 1

12nh(h+ 1), whereh is the Coxeter number ofR andn its rank.

Proof. Leti be a simple root. Then (i) = 1 and it thus follows that = ii where thei are the fundamental weights, i.e. the dual basis for the basis of simple roots. So if =

icii

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(whereci < 0) then 2 = (

icii,

ii) =

ici. The Weyl vectors R for the simply lacedirreducible root systems are:

An = 1

2(n1+ 2(n 1)2+. . .+i(n i+ 1)i+. . .+nn) ,

Dn = (n 1)1+ (2n 3)2+. . .+ (in i(i+ 1)2 )i+. . .+ n(n 1)4 (n1+n) ,E6 = (81+ 112+ 153+ 214+ 155+ 86) ,E7 =

1

2(341+ 492+ 663+ 964+ 755+ 526+ 277) ,

E8 = (461+ 682+ 913+ 1354+ 1105+ 846+ 577+ 298) ,see for example [4] (note that we have a different sign convention). Furthermore, the Coxeternumbers are h(An) =n+ 1, h(Dn) = 2n 2, h(E6) = 12, h(E7) = 18, and h(E8) = 30. Now

2An = 1

2

ni=1

i(n i+ 1) = 12

n

ni=1

i n

i=1

i(i 1)

= 12 n n(n+ 1)2 13 n(n2 1) = 112 n(n+ 1)(n+ 2),2Dn =

n2i=1

in i(i+ 1)

2

+

n(n 1)2

=nn2i=1

i 12

n1i=1

i(i 1) + n(n 1)2

= n(n 2)(n 2 + 1)

2 1

2 1

3(n 1)((n 1)2 1) + n(n 1)

2

= 1

3n3 1

2n2 +

1

6n=

1

12n(2n 2)(2n 1)

2E6 = 8 + 11 + 15 + 21 + 15 + 8 = 78 = 1

12 6 12 13,

2E7 = 1

2

(34 + 49 + 66 + 96 + 75 + 52 + 27) = 399

2

= 1

127

18

19,

2E8 = 46 + 68 + 91 + 135 + 110 + 84 + 57 + 29 = 620 = 1

12 8 30 31,

where we have used the easily verifiable identitiesn

i=1i= 12

n(n+1) andn

i=1i(i1) = 13 n(n21).We see that the Weyl vectors do indeed satisfy 2 = 112 nh(h+ 1).

Note that this formula for the norm of the Weyl vector was not discovered by just calculating2 like in the proof above. It can be proved using the strange formula of Freudenthal and de Vriesthat was proven by them in[8].

Lemma 6 (Venkov 1980). Let N be a Niemeier lattice with roots and its Weyl vector. Then2 = 2h(h+ 1)

Proof. By Corollary 1 rank R(N) = 24, say R(N) =ki=1Ri where the Ri are irreducible root

systems andki=1ni = 24 with ni = rank Ri. Also, by Corollary 2 all irreducible components ofR(N) have the same Coxeter number h= 1

24|R(N)|. Furthermore, for all irreducible components

Ri of R(N) with Weyl vector Ri it follows from Lemma 5 that 2Ri

= 112 nih(h + 1). So 2 =k

i=12Ri

=k

i=11

12nih(h+ 1) = 2h(h+ 1).

Lemma 7. IfNis a Niemeier lattice and N thenR(N)

(, )2 = 2h2.

Proof. By Proposition 2 and Corollary 2 we have:

R(N)(, )2 =

1

122|R(N)|

= 2h2.

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Lemma 8. If is the Weyl vector of a Niemeier latticeN then lies inN.

Proof. We will show that if N then (, ) Z. BecauseN is unimodular it then follows that N. So let N. Then

(2, )2 = (>0

, )2

= (>0

(, ))2

>0

(, )2 mod 2

= 2h by Lemma 7

0 mod 2 since N is even.

So (2, )2 is an even integer. Since (2, ) =>0(, ) Z it follows that (2, ) 2Z andhence (, ) Z.Lemma 9. LetN be a Niemeier lattice with roots. Denote byQ = ZR(N) the root lattice ofNand byP the corresponding dual weight lattice. For all P we have

h

2 2

1 +

1

h

,

and the for which equality holds form a complete set of representatives for the glue group P/Q.

Proof. Since 2 = 2h(h+ 1), for all Pwe have:

h

2

21 +1

h =

h

2

h2

= 2 2h

(, )

= 1

h

>0

(, )2 >0

(, )

by Lemma 7

= 1

h

>0

(, ) [(, ) 1] 0 since (, ) Zfor all > 0.

All terms in the above sum are in fact 0 so there is an equality if and only if (, ) {0, 1}for all > 0. It is well known that the P for which this holds form a complete set of cosetrepresentatives for P /Q.

The lemma implies that the covering radius ofP, and hence also ofN, is at least 2(1 + 1/h). Itis not necessarily equal to this. For example, the covering radius ofN= 3E8is

3>

2(1 + 1/30).

So the covering radius of a Niemeier lattice with roots is always greater than

2. Later on we willsee that the covering radius of the Leech lattice is exactly

2.

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Chapter 6

The Leech Lattice

The Leech lattice is the unique Niemeier lattice without roots. It was discovered by Leech in 1965.Soon after this discovery, Leech conjectured that the lattice had covering radius

2 because there

were several known holes of this radius. He failed to find a proof. Parker later noticed that theknown holes all seemed to correspond to a Niemeier lattice. Inspired by this, Conway, Parker andSloane found all the holes of this radius ([6], Chapter 23). There turned out to be 23 classes ofdeep holes corresponding in a natural way with the 23 Niemeier lattices with roots. Using the factthat the covering radius of the Leech lattice is

2 Conway later proved that the lattice of type

II25,1 has a Weyl vector and that its Coxeter diagram can be identified with the Leech lattice ([6],Chapter 27).Most of the proofs of these results involved long calculations and case by case verifications. In hispaper from 1985 on the Leech lattice ([3]), Richard Borcherds gave new more conceptual proofs ofthe existence and uniqueness of the Leech lattice and of the fact that it has covering radius

2.

He also gave a uniform proof of the correctness of the holy constructions of the Leech latticewhich are described in[6], Chapter 24. In this chapter we present these proofs.

We will first give the proof of the existence and uniqueness of the Leech lattice and then turn tothe covering radius and the deep holes of the Leech lattice. Although this seems like the mostnatural order one must note that the proof we give in section 6.2 of the uniqueness of the Leechlattice does depend on the fact that the Leech lattice has covering radius

2. This will be proved

in section 6.4 without using any results from section 6.2.

6.1 The existence of the Leech lattice

Here we prove the existence of a Niemeier lattice with no roots following section 4 in [3]. Theexistence is proved by showing that given any Niemeier lattice with roots we can construct anotherNiemeier lattice with half as many roots.

LetNbe a Niemeier lattice with Weyl vector = 12 >0and Coxeter numberh. LetL= NUbe the lattice of type II25,1 with coordinates (,m,n), where N, m, m Z and (,m,n)2 =2 2mn. Letz = (0, 0, 1). Thenz is the primitive norm zero vector corresponding to the NiemeierlatticeN. The roots in z, that is the elements in the set{ R(L) | (, z) = 0}, form the affineroot system associated with N.Letz = (,h,h +1). It follows from Lemma 7 that z N. Furthermore, by Lemma 5 it has norm0. Hence z is also a primitive norm zero vector in II25,1 and therefore corresponds to a Niemeierlattice, say N. Note thatz and z are in the same connected component of the set of non zeronorm zero vectors since ((1 t)z+ tz)2 = 2(1 t)t(z, z) = 2h(1 t)t


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Then has to be of the form (, 0, n) with R(N). Thus

(, z) = ((, 0, n), (,h,h+ 1))= (, ) nh= 0.

But for R(N) we have 1 |(, )| h 1 and thus (, ) =nh is not possible.Suppose now that (z, ) = 1. Then is of the form (, 1, n) with2 =2 2n= 2. Furthermore,

(, z) = ((, 1, n), (,h,h+ 1))= (, ) (h+ 1) nh= 0,

and thus h

2=

2

h2 2

h(, ) +2

= 2h(h+ 1)

h2 2

h(h+ 1 +nh) + 2n+ 2

= 2.

This is in contradiction with Lemma 9 so it follows that (, z) 2.Lemma 11. The Coxeter numberh of the Niemeier latticeN is at most 12 h.

Proof. Ifh = 0 then the inequality holds so we can assume h= 0. So N is a Niemeier latticewith roots. The roots R(L) that are in z form the affine root system associated with R(N).LetRbe an irreducible component of this root system and pick a set of simple roots0, 1, . . . , rforR with (i, z) 0. Herer is the rank of the irreducible component ofR(N) corresponding toR. Denote by ki the weights of the roots i. Then

ri=0kii = z

andr

i=0ki = h. By Lemma

10, (z, i) 2 for i = 0, . . . r and thus

(z, z) = (

r

i=0

kii, c) 2h.

Since also(z, z) = ((,h,h+ 1), (0, 0, 1)) = h,

it follows thath 12

h.

Theorem 2. There exists a Niemeier lattice with no roots.

Proof. Start with an arbitrary Niemeier lattice Nwith Coxeter number h= 0 . By Lemma 11we can find a Niemeier lattice N with Coxeter number 12 h. By repeating this process we willeventually find a Niemeier lattice with Coxeter number equal to zero and hence with no roots.

In fact, there are no roots inz and this gives a direct construction of a Niemeier lattice withno roots after one step. The proof of this will be given in Section 6.5.

6.2 The uniqueness of the Leech lattice

Again denote by a Niemeier lattice without roots. We will show that is unique following theproof in section 6 of [3]. let L = U be the lattice of type II25,1 with the same coordinatesas before and let D be the fundamental domain of L that contains the controlling vector z =(0, 0, 1) L. Denote by V+ the open cone of vectors with negative norm that contains z in itsclosure. In the proof of the next theorem we use the fact that the covering radius of is

2. This

will be proved later.

Theorem 3. The simple roots of the fundamental domain D ofL are just the simple roots =(, 1, 1

22 1) of height one with respect to z.

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Proof. Obviously there can be no roots of height zero since has no roots. By Vinbergs algorithmit then follows that all roots= (, 1,

12

2 1) of height one are simple roots. Now suppose thatthese are not all roots and let = (,m, 1

2m(2 1)) be a simple root of height m 2. Because

is a simple root, it satisfies (, ) 0 for all simple roots = (, 1, 12 2 1) of height one.Also, since has covering radius

2 there is a vector

such that (

/m)2

2. But then

(, ) = (, ) m( 12

2 1) 12m

(2 2)

= m+ 1

m+ (, ) m

22 1

2m2

= m+ 1

m m

2

m

2 m+ 1

m m= 1

m >0,

which is a contradiction with (, ) 0.

Corollary 3. The Leech lattice is the unique (up to isomorphism) Niemeier lattice with no roots.

Proof. A Niemeier lattice without roots corresponds (up to isomorphism) to a primitive normzero vectorw inL (up to the action of Aut(L)) with no roots perpendicular to it. So it suffices toshow that any two such vectors are conjugate under Aut(L). Hence if the fundamental domain DofW


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equal to 224. Indeed, since has no roots we can setN2= 0 in equation (5.1). So

(z) = E6 65520691

= 1 +65520

691 (2073q2 + 176896q3 + 4197825q4 +. . .)

= 1 + 196560q2 + 16773120q3 + 398034000q4 +. . . ,

and thus

N0+N4

2 +

N62

+N8

48 = 1 +

196560

2 +

16773120

2 +

398034000

48 = 16777216 = 224.

Hence each equivalence class contains a short vector. This can be either the norm 0 vector, or anopposite pair of vectors of norm 4 or 6, or a collection of 24 mutually orthogonal pairs of vectorsof norm 8.

Corollary 4. The distance between any two vertices of a hole of the Leech lattice is at most2

2.

Proof. (following the proof in [10]). Suppose that vi and vj are two vertices of a hole c such that(vi vj)2 10. By Theorem 4 the equivalence class ofvi vj contains a short vector . Thenvi vj = 2 for a and thus

(vi vj 2)2 =2 8.

Hence the distance between the vectors v i := vi andv j :=vj+ in is at most 2

2 and theyhave the same midpoint (vi+ v

j)/2 = (vi+ vj)/2 asvi andvj . It then easily follows that either v

i

orvj is closer to c thenvi andvj are (also see Figure6.1):

Suppose that

vi

vj

2 , c vi+v

j

2 0. (6.1)Figure 6.1:

vi vj

vi

vj

c

Then

(vi c)2 =

vi vj2

2+

c vi+vj

2

2, by Pythagoras

>

vi vj

2

2+

c v

i+v

j

2

2, since (vi vj)2 >(vi vj)2

(vi c)2, by6.1.

As before, denote by a Niemeier lattice without roots and let L = Ube the lattice of typeII25,1. Also, let z = (0, 0, 1) and let D be the fundamental domain of the reflection group W(L)that contains z. As seen in Theorem 3, the simple roots are the roots = (, 1,

12

2 1) with

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(this does depend on the fact that the covering radius of is 2, which we will prove inthe next section without assuming that all simple roots have height 1). We can therefore identifythe simple roots ofL with the vectors of and thus the Coxeter diagram ofL is indexed by thevectors of . The Gram matrix G = (g) of the simple roots , is given by

g = (, )

= (, ) ( 12

2 1) ( 12

2 1)

= 2 12

( )2.Two points , of are joined by 0, 1, 2, . . . edges if the norm of their difference is 4, 6, 8, . . .. ByCorollary 4, vertices , (c) of a hole c always satisfy ( )2 8. Thus the correspondingnodes in the hole diagram are always joined by 0 , 1, or 2 edges.Now embed R in R25 by identifying R with the hyperplane H ={(x1, . . . , x25) R25 |x25 = 0} in R25.Suppose that R(c)

2 and let c be the point on the line perpendicular to

H that goes through c that has distance

2 from all vertices of the hole (see Figure6.3). So ifR(c) =

2 then c= c. IfR(c)>

2 we define c =c. If the radius of a holec is 2 then for

, : ( )2 = 4 2( c, c),so two nodes , of the hole diagram are joined by 0, 1 or 2 edges if ( c, c) = 0, 1 or2respectively (see Figure6.2).

c

c c

8

2

2

4

2

2

6

2 2

Figure 6.2: Conditions for two vertices , of a hole c of radius 2 to be joined by 0, 1 or 2edges in the hole diagram.

c

c

1 2

34

H

2

2

Figure 6.3: Construction of the point c.

6.4 Covering radius

In this section we prove that the covering radius of a Niemeier lattice with no roots is 2. Thiswas first proved by Conway et al ([6], Chapter 23) by explicit calculation, the proof below is due

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to Borcherds[3]. As before, let L = Ube the lattice of type II25,1. Also, let z = (0, 0, 1) andlet D be the fundamental domain of the reflection group W(L) that contains z .

Proposition 4. Any connected extended Coxeter subdiagram of is contained in a subdiagramof which is a disjoint union of extended Coxeter diagrams of total rank24 (so this subdiagram

has a total of24 + l nodes, where l is the number of connected extended Coxeter diagrams in thedisjoint union).

Proof. LetXbe an extended Coxeter diagram in . Suppose thatv0, v1, . . . , vr are the nodesof this diagram and leti = (vi, 1,

12

v2i 1) be the corresponding simple roots ofL. They determinea primitive norm zero vector w =

ri=0kii (withki the weights of the extended diagram) that

is contained in D. Hencew corresponds to a Niemeier lattice Nwith roots. By Corollary 1 R(N)has rank 24 and by Corollary 2 all its irreducible components have the same Coxeter number

h=r

i=0

ki = (z, w).

This last equality holds because w = ri=0kii and (z, i) = 1 for all i.HenceR w is an affine root system of rank 24. The proposition will follow if we show that allsimple roots of this root system have height 1 with respect to z and thus correspond to vectorsof the Leech lattice. So suppose that 0, 1, . . . , s are the simple roots of another connectedcomponent of the Coxeter diagram ofw with weights k i so that

si=0k

ii = w. Then

h= (z, w) =s

i=0

ki(z, i).

Since this component has the same Coxeter number h as X, alsos

i=0ki = h. Furthermore,

because has no roots it follows immediately that there are no simple roots i with (z, i) = 0.So we have (z, i)

1 and thus the above formula implies that (z, i) =

1 for alli. Hence all

simple roots in R w have height 1.

When changing from the hyperboloid model of hyperbolic space to the upper half-space modelthe rational lines of primitive norm zero vectors in L are mapped bijectively to the points of Q as follows: Letx be a norm zero vector ofL representing a rational line in L Q. Ifx= (0, 0, 1) then x is mapped toin Q . Otherwise, x = (,m,2/2m) withm = 0 andx is mapped to /m in Q .

Lemma 12. The reflection in the simple root = (, 1, 12

2 1) ofD acts on Q asinversion in the sphere of radius

2 with center.

Proof. Letbe a point of Q that corresponds to the rational line in L Q generated by thenorm zero vector x = (, 1,

1

2 2

). Let= (, 1,

1

2 2

1) be a simple root. Then(x, ) = (, ) ( 1

22 1) 1

22

= 1 12

( )2,

and thus (x, ) = 0 ( )2 = 2.

Theorem 5. The covering radius of is

2.

Proof. Let c Q be a hole of . Denote by R(c) the radius of this hole given by R(c)2 =inf{( c)2 | }. By Corollary 4 we know that the distance between any two vertices ofc isat most 22. So the nodes of the hole diagram ofc are joined by 0, 1 or 2 edges. Now there arethe following two possibilities:

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The hole diagram contains no extended diagram:In this case the hole diagram contains no double bonds because if it did it would containthe extended diagram A1. So the diagram is simply laced and it follows that it can containonly simply laced elliptic diagrams, i.e. the diagrams in Figure 1. Indeed, otherwise onecould delete nodes from the diagram until in the next step such an elliptic diagram would be

obtained and thus there would be an extended diagram contained in the hole diagram. Wenow show that a hole whose diagram contains only elliptic diagrams has radius R to be the lattice generated by and c. Then [ : ] = [ : ] = 3and together with the definition of this implies that is an even unimodular lattice in R24.Furthermore, as = it follows that has to be the Niemeier lattice with R() = 12A2because this is the only root system listed in Table 5.1 with h = 3.

In general, suppose that is a primitive vector of norm 2 h2 with hN. Then we canconstruct a Niemeier lattice with Coxeter number equal to h as follows:

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Let= { | (,/h) Z},

and let =< ,/h > be the lattice generated by and /h. This is a Niemeier lattice withCoxeter number equal to h. If there is only one root system in Table 5.1 with this Coxeter number

then this would show the existence of a Niemeier lattice with this roo

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