level-two structure of simply-laced coxeter groups

a

ps

-lacedtones,ertex

ow to

spport

Journal of Algebra 285 (2005) 29–57

www.elsevier.com/locate/jalgebr

Level-two structure of simply-laced Coxeter grou

Mark Reeder1

Boston College, Chestnut Hill, MA 02467, USA

Received 11 September 2003

Available online 12 January 2005

Communicated by Georgia Benkart

Abstract

Let X be a graph, with the corresponding simply-laced Coxeter groupW . ThenW acts naturallyon the latticeL spanned by the vertices ofX, preserving a quadratic form. We give conditions onX

for the form to be nonsingular modulo two, and study the images ofW → O(L/2kL). 2004 Elsevier Inc. All rights reserved.

Introduction

This paper investigates the tower of 2-power congruence subgroups in a simplyCoxeter group, but the story begins with a puzzle for children. We have a pile of sand a graphX with n vertices. At most one stone may be placed on a vertex, so a vhas one of two states: stoned or unstoned. We move by selecting a vertexv having an oddnumber of stoned neighbors, and then change the state ofv. Given an initial configurationof stones onX, we try to reduce the total number of stones as much as possible. Hdetermine this minimal number of stones from the initial configuration?

A configuration of stones is an element in theF2-vector spaceV spanned by the verticeof X. Forv ∈ V , letq(v) be the number of vertices plus the number of edges in the suof v, modulo two. Thenq is a quadratic form onV (see Section 1), and we letO(F2)

denote the subgroup ofGLn(F2) preservingq.

E-mail address:[email protected] Partially supported by National Science Foundation grant DMS-0207231.

0021-8693/$ – see front matter 2004 Elsevier Inc. All rights reserved.doi:10.1016/j.jalgebra.2004.11.010

30 M. Reeder / Journal of Algebra 285 (2005) 29–57

ns

le

se of

singlen-

-

é

t.graphs

in). We

on

on

rst,ted

us,

yinss of a

The moves are linear maps onV preservingq, and are the images of simple reflectiounder the natural homomorphism

ρ : W → O(F2),

whereW is the (simply-laced) Coxeter group havingX as Coxeter diagram. Our puzzcan, and henceforth will be rephrased as follows: Find the orbits ofW onV , determine theorbit of a given vector, and find a vector in each orbit which is minimal, in the senhaving the fewest number of nonzero coefficients in terms of the vertex basis.

We see at once that there are at least two nonzero minimal vectors, namely astoned vertex (contained in an orbit withq = 1) or two nonadjacent stoned vertices (cotained in an orbit withq = 0). If q separates the nonzeroW -orbits onV , then the puzzleis solved: there are two nonzero orbits, determined byq = 0 or q = 1, and a minimal vector can be determined by evaluatingq on the initial vector. So, when doesq separate thenonzeroW -orbits onV ?

Suppose thatq is nonsingular andρ is surjective. Results of Arf–Witt and Dieudonn(see Section 2) imply the following: Ifn is even, thenq separates orbits. Ifn is odd thereis just one additional orbit, consisting of a singleW -invariant vector which is easy to spoThis leads us to the main question addressed in the first part of this paper: For whichX is q nonsingular andρ surjective?

If W is a finite irreducible Weyl group, that is, ifX is a Dynkin diagram of typeADE,the answer is easily checked in each case (and also follows from the results herefind thatq is nonsingular andρ is surjective exactly for typesA1, A2, A4, A5, E6, E7, E8.In fact, it is well known that the mapρ gives isomorphisms

S2/ ± 1→ O1(F2), S3 → O−2 (F2), S5 → O−

4 (F2), S6 → O5(F2),

W(E6) → O−6 (F2), W(E7)/ ± 1→ O7(F2), W(E8)/ ± 1→ O+

8 (F2).

Here,On(F2) or O±n (F2) denotes the orthogonal group of a nonsingular quadratic form

Fn2 which forn even is split (+) or nonsplit (−).

The inverses of the three nontrivial typeA isomorphisms are given by the permutatiaction of the orthogonal group on, respectively, the vectors withq = 1, vectors withq = 0and− type hyperplanes inF5

2. Counting arguments (see [1, pp. 242–243]) show thatρ issurjective in typeE.

In the finite case, theW -orbits onV have several interpretations; we mention two. Fiif we identify vertices inX with simple co-roots in the corresponding simply-connecLie groupG, then a configuration of stones is an involution in a maximal torus ofG, andthe moves are conjugation by simple reflections in the Weyl groupW of G. Our puzzleamounts, for finiteW , to determining the conjugacy class of a given involution. Thin the seven cases above, we have two conjugacy classes of involutions given byq = 1andq = 0, and an additional central involution inA5, E7. We remark that the conjugacclasses of all finite order elements inG were classified by Kac, in terms of coefficientsthe highest root (see [4, Chapter 10]), but it is not always easy to determine the clagiven element.

M. Reeder / Journal of Algebra 285 (2005) 29–57 31

spon-le

ald the

dSec-

ularode

itsns,

vento even

.

per-

hasbolic

The second interpretation arises from recent work on the local Langlands corredence [3, Section 13]. Here, the vectors inV parametrize certain finite sets of irreducibrepresentations of ap-adic groupG, and theW -orbits inV correspond to certain rationclasses of tori inG. The present paper arose in this context, while trying to understanparticularly nice example ofE8 in terms of its graph, without resorting to counting.

In this paper we consider, in places, an arbitrary graphX, but we mostly restrict tothe case whereX is a tree. We say the graphX is nonsingularif the quadratic formq isnonsingular onV .

In Section 4, we give simple graph-theoretic conditions forX to be nonsingular, annonsingular trees are characterized in terms of “sprouting” and “pruning”. Then intion 7, we prove:

1. Theorem. If X is a nonsingular tree, not of typeAn, then the mapρ :W → O(F2) issurjective.

Thus, we find thatE6, E7, E8 are rather the norms than the exceptions, for nonsingtrees; unlikeA1, A2, A4, A5, they are not “low-dimensional accidents”. The branch nmakes all the difference.

Theorem 1 leads us to consider the kernel ofρ, which is only interesting for infiniteW .More generally, we consider the congruence subgroups

Wk = ker[W → O

(Z/2kZ

)].

For k > 1 the groupsWk are torsion-free. It follows easily from properties of the Tcone that the torsion inW1 consists of a finite number of conjugacy classes of involutiocorresponding to subgraphs of typeE8, and certain subgraphs of typeE7 (see Section 8).

The quotientsWk/Wk+1 are elementary abelian 2-groups, and we show, for “most” enonsingular trees, that the rank is as large as possible. (Our arguments apply onlygraphs.) To state the result, letO ′(Z2) denote the kernel of the 2-adic spinor norm

δQ2 :O(Z2) → Q×2 /Q×2

2 .

ThenW ⊂ O ′(Z2) for every nonsingular graphX, sinceδQ2 = 1 on the simple reflectionsIn Sections 9–12, we prove our second main result:

2. Theorem. AssumeX is a nonsingular even tree containing a nonsingular even hybolic subtree. ThenW is 2-adically dense inO ′(Z2). Equivalently,

Wk/Wk+1

so′(F2) if k = 1,2,

so(F2) if k 3,

whereso(F2) is the Lie algebra ofO(F2) andso′(F2) is the commutator subalgebra.

Here, a tree ishyperbolicif its Coxeter group is infinite, and every proper subtreefinite or affine Coxeter group. In fact, there are only two nonsingular even hyper


that1,

ntye

y

e

trees, namelyE10 andT3,3,4 (see Section 4). To prove Theorem 2 we use Kac’s result±Aut(X)W = O(Z) whenX is hyperbolic, along with strong approximation, Theoremand the structure of the adjoint representation ofO(F2).

1. Graphs and quadratic forms

In this paper, a graphX has a finite vertex setS = S(X), and edges are two-elemesubsets ofS. Our graphs have no loops, or multiple edges. Ifi, j is an edge we saverticesi, j are adjacent, or are neighbors, and writei—j . The degree of a vertex is thnumber of edges containing it. Given an ordering onS, let A = [aij ] be the adjacencmatrix ofX, defined by

aij =

1 if i—j,

0 otherwise.

If X andY are two graphs, thenX +Y denotes the disjoint union ofX andY . If J ⊂ S,thefull subgraph on Jis the graph[J ] with vertex setJ and all edges inX between verticesin J .

Given a graphX with vertex setS, let L be a lattice of rankn with basisαi : i ∈ S,and let〈 , 〉 be the symmetric bilinear pairing onL with matrix 2I − A (for some orderingof the basisαi). Note that〈λ,λ〉 ⊂ 2Z for all λ ∈ L. Let q be the quadratic form onLdefined byq(λ) = 1

2〈λ,λ〉. We also writeq for the base extension ofq to R ⊗ L, for anycommutative ringR.

Many coefficient rings appear later in the paper, but until further notice we takeR = F2.Let V = F2 ⊗ L. It has a basisei = 1⊗ αi. If x = ∑

xiei ∈ V , then

q(x) =∑

i

x2i +

∑i<ji—j

xixj ∈ F2.

This yields the description ofq given in the introduction. We can visualizex as a binarycoloring of the vertices ofX, wherei is colored• if xi = 1, and colored if xi = 0. If [x]denotes the full subgraph ofX on the• vertices, thenq(x) is the Euler characteristic of th1-complex[x], modulo two. If[x] has no cycles, then

q(x) ≡ c(x) mod 2,

wherec(x) is the number of connected components of[x].The associated symplectic formf :V ⊗ V → F2 is given by

f (x, y) = q(x + y) + q(x) + q(y).

The bilinear formf has matrix[f (ei, ej )] ≡ A mod 2. We write


s-

in

ts

ker2 X = x ∈ V : f (x,V ) = 0

,

ker2 q = x ∈ ker2 X: q(x) = 0

.

Note that ker2 X = kerA|V . We can visualize ker2 X as the set of binary vertex coloringof X in which every vertex has an even number of• neighbors. Sincef induces a nondegenerate symplectic form onV/ker2 X, we have

dimker2 X ≡ dimV mod 2.

A vectorx ∈ V for whichq(x) = 0 is calledq-isotropic. We define

V (0) := x ∈ V : x = 0, q(x) = 0

, V (1) :=

x ∈ V : q(x) = 1.

The formq and the graphX are callednonsingularif ker2 q = 0.

2. Orthogonal groups over F2

Let O(V ) denote the automorphism group of the quadraticF2 vector spaceV . Manyarguments in this paper depend on the parity ofn. We collect some known facts neededeach case.

2.1. Lemma. Supposen = 2m + 1 is odd. Then the formq is nonsingular if and only ifker2 X = 0, u has two elements, andq(u) = 0. In this case, we have

(1) V (1) = u + V (0).(2) The groupO(V ) has four orbits inV , namely,

0, u, V (0), V (1).

Proof. The restriction ofq to ker2 X is a linear functional from ker2 X to F2. By definition,this functional is injective if and only ifq is nonsingular. Since dimker2 X is odd, the firstassertion is immediate.

Supposeq is nonsingular, and letu be the nonzero element of ker2 X. For each cosex, x+u ∈ V/ker2 X, we haveq(x+u) = q(x)+1, hence (1) holds. Assertion (2) followfrom the Arf–Witt theorem [2, p. 41].

If n = 2m, there are two equivalence classes of quadratic forms onV , according tothe maximal dimension of a subspace on which bothq and f vanish identically. Thisdimension ism − d , whered ∈ 0,1 is called thedefectof q.

2.2. Lemma. Supposen = 2m is even. Thenq is nonsingular iffker2 X = 0, that is, iffdetA is odd. In this case, the following hold.


theo-

he

(1) The defect is given by

d ≡ D2 − 1

8mod 2,

whereD = det[2I − A].(2) The groupO(V ) has three orbits inV , namely,

0, V (0), V (1).

Proof. Since dimker2 X is even, the linear functionalq : ker2 X → F2 is injective iffker2 X = 0, hence the first assertion.

From the classification of quadratic forms overZ2 [6, 5.2.5,6] it follows that

Z2 ⊗ L (m − d)

[0 11 0

]⊥ d

[2 11 2

],

as quadratic spaces. This implies (1). Assertion (2) again follows from the Arf–Wittrem.

Return now to arbitraryn. A transvectionis an element ofO(V ) of the form x →x + f (x, y)y for somey ∈ V (1). It follows from Lemmas 2.1(2) and 2.2(2) that ttransvections form a single conjugacy class inO(V ).

The next result is our main tool in proving surjectivity.

2.3. Lemma. SupposeX is a nonsingular graph, not isomorphic toA2 + A2. Let G be asubgroup ofO(V ) containing a transvection. ThenG = O(V ) if and only ifG is transitiveonV (1).

Proof. If G contains one transvection, and is transitive onV (1), then G contains alltransvections. We will haveG = O(V ) if the transvections generateO(V ). It is known[2, Proposition 14] that, for nonsingularq, the transvections do indeed generateO(V ),except ifn = 4 andd = 0.

There are four nonsingular graphs with four vertices, havingD = −27,−3,5,9. FromLemma 2.2(1), the respective defects ared = 1,1,1,0, the latter coming fromD = 9 forA2 + A2.

3. Nonsingularity conditions for graphs

Let X be a graph with vertex setS = 1, . . . , n, and quadratic formq onV = F2⊗L, asabove. In this section we translate the nonsingularity ofq into conditions on the graphX.

Let

det(X) := detA =∑

ε(σ )a1,σ1a2,σ2 · · ·an,σn ∈ Z,

σ∈Sn


g-

x,

f

r

whereε(σ ) is the sign of the permutationσ ∈ Sn. Theσ -term is nonzero iffσ belongs tothe setA(X) of those permutations that move every vertex to one of its neighbors, so

det(X) =∑

σ∈A(X)

ε(σ ).

Let Z(X) be the set ofn-vertex subgraphsU ⊂ X whose components are either sements— or cycles. Letz(U) be the number of components ofU which are cycles. Theorbits ofσ ∈ A(X) on S define an elementU(σ) ∈ Z(X), andU(σ) = U(σ ′) iff σ ′ is ob-tained fromσ by inverting somek-cycles inσ , for k 3. This implies thatε(σ ) = ε(σ ′),and that

det(X) =∑

U∈Z(X)

ε(U)2z(U), (3a)

whereε(U) is the sign of any permutationσ ∈A(X) with U(σ) = U .An n-vertex subgraphY of X is called amaximal matchingif Y containsn

2 connectedcomponents each of which is a segment, and ifn is odd, one additional isolated vertedenotedj (Y ). Let M(X) be the set of all maximal matchings inX.

3.1. Lemma. If n = 2m then X is nonsingular if and only ifX has an odd number omaximal matchings.

Proof. If n = 2m, then a maximal matching is an elementY ∈ Z(X) with z(Y ) = 0, so theclaim follows from (3a).

For anyj ∈ S, let Xj be the full subgraph ofX supported onS − j. If n is odd, thesegments in a maximal matchingY of X form a maximal matching inXj(Y ). Consider thevector

u :=∑j∈S

det(Xj )ej =∑

Y∈M(X)

ej (Y ). (3b)

3.2. Lemma. If n = 2m + 1, then the following hold.

(1) u ∈ ker2 X.(2) u = 0 if and only ifker2 X = 0, u.(3) X is nonsingular if and only ifq(u) = 1.

Proof. We first show thatu ∈ ker2 X. We must show that everyi ∈ S has an even numbeof neighborsj with detXj odd. LetM(i) be the set of all maximal matchingsY in X suchthatj (Y )—i. ThenM(i) is the disjoint union of the sets of maximal matchings inXj , forj—i. From the even case just proved, we get∣∣M(i)

∣∣ ≡∑

det(Xj ) mod 2.
j—i


reee

e

nds

the

Hence it suffices to show that|M(i)| is even. To this end we construct a fixed-point finvolution onM(i). Let Y ∈ M(i), and letj = j (Y ). Sincei = j , there is a unique edgin Y meetingi, sayi, j ′. Replace this edge byi, j, and keep the remaining edges inY .This gives a new maximal matchingY ′ which again belongs toM(i), sincej (Y ′) = j ′—i.Clearly Y ′ = Y . Repeating the procedure withY ′ will give Y again. This completes thproof of (1).

For (2), note that each detXi is a minor ofX. Henceu = 0 iff dim ker2 X = 1. Finally,if q(u) = 1, then (2) holds, soX is nonsingular. The converse is clear.3.3. Definition. We call the vectoru defined in (3b) thekernel vectorof an odd graphX,and the verticesi in the support ofu (i.e., those with detXi = 0) thekernel verticesof X.

3.4. Example. If X is a Dynkin diagram of finite type, then the kernel vector correspoto a central involution (see Introduction).

3.5. Example. Consider the complete graphKn. By induction, we find there are 1·3 · · · (2m − 1) maximal matchings inK2m, and 1· 3 · · · (2m + 1) maximal matchings inK2m+1. Thus,K2m is nonsingular for allm, by 3.1. ForX = K2m+1, we haveXj K2m

for all j , so

u =2m+1∑j=1

ej ,

and

q(u) ≡ 2m + 1+(

2m + 1

2

)= (m + 1)(2m + 1) ≡ m + 1 mod 2.

Thus, 3.2 shows thatK4m+1 is nonsingular, whileK4m+3 is singular.

3.6. Example. We indicate the kernel vector in the following graphs by using• for thekernel vertices. It shows that the first is singular, and the second is nonsingular.

• • • •

• • •••

If the nonsingular graphs with 2m vertices are known, one can use 3.2 to determinenonsingular graphs with 2m + 1 vertices, by determining whether eachXj is singular ornot, thereby constructing the kernel vectoru, and then evaluatingq(u).


ectedix

better.nsin-

numberpen ifting”,

leaf,ucting

ce no

so we

uting.

For example, there are four nonsingular graphs with four vertices, and 21 conngraphs with five vertices. Examining theXj ’s in each of the latter, one finds exactly snonsingular connected graphs with five vertices. We leave this as an exercise.

This method does not seem practical for larger graphs. For trees, one can doLemma 4.8 below contains a much simpler constructive procedure for finding all nogular trees.

4. Trees

In the previous section we have seen that odd complete graphs can have an oddof maximal matchings and still be singular. In this section, we find that this cannot hapX is a tree. This is a by-product of the construction of all nonsingular trees by “sprouas will be explained.

We adopt the standard terminology: Atree is a connected graph without cycles. Aleafin a graphX is a vertex contained in a unique edge ofX. A branch nodeis a vertex ofdegree at least three.

The first step is a variation on the well known result “trees have leaves”.

4.1. Lemma. If X is a tree then at least one of the following holds.

(1) X consists of a single vertex.(2) There is a vertex inX adjacent to two or more leaves.(3) There is a leaf inX adjacent to a vertex of degree two.

Proof. Suppose (1)–(3) all fail to hold. Then every vertex is adjacent to at most oneand every leaf is adjacent to a branch node. We will get a contradiction by constra cycle inX. Pick a leafv0, and letv1 be a branch node adjacent tov0. Proceed awayfrom v1 on an edge other thanv0, v1. The next vertex cannot be a leaf, sincev1 is alreadyadjacent to the leafv0. If the next vertex has degree 2 then proceed on a new edge. Sinleaf is adjacent to a degree two vertex, we eventually arrive at new branch nodev2. Sincedegv2 3, andv2 is adjacent to at most one leaf, we can exitv2 on a new edge whichdoes not end in a leaf. In this way we visit an unlimited number of branch nodes,eventually visit the same branch node twice.

Note that ifX is any nonsingular tree, then 4.1(2) cannot hold, for ifi, j are leavesadjacent to the same vertex, thenei + ej ∈ ker2 X andq(ei + ej ) = 0. This can also beseen from (3a), sinceZ(X) is empty.

4.2. Definition. If the treeX is obtained by attachingi—j— to some vertexk in a treeX′,we sayX is obtained fromX′ by sproutingat k, andX′ is obtained fromX by pruningat k.

Our eventual aim is to show how all nonsingular trees may be obtained by sproWe begin with even trees.


al

s are

l

n

nts

es

4.3. Lemma. A treeX with 2m vertices is nonsingular if and only if it has a maximmatching, in which case the maximal matching is unique, anddet(X) = (−1)m. The setof nonsingular even trees is preserved under sprouting and pruning. All such treeobtained by starting with the segment— and sprouting at arbitrary vertices.

Proof. If X has no maximal matchings, then it is singular by 3.1. SupposeX has a maximamatching. Then 4.1(1), (2) cannot hold, soX has a subgraphi—j—k with no other edgesin X meetingi or j . Prune atk to obtain a new graphX′. Since any maximal matching iX must containi, j, we have

det(X) = −det(X′).

The lemma follows by induction. 4.4. Example. The nonsingular even trees withn 8 areA2, A4, A6, E6, E8,

T3,3,4 :=

,

,

.

To prepare for odd trees, we first consider the support of vectors in ker2 X. Recall fromSection 1 that ker2 X consists of those binary vertex colorings ofX in which every vertexhas an even number of• neighbors.

4.5. Lemma. SupposeX is any tree, and0 = x ∈ ker2 X. Then the connected componeof [x] are single vertices. At least one vertex in[x] is a leaf inX.

Proof. Any componentc of [x] is a tree. Ifc has more than one vertex, thenc has a leafi,adjacent to a unique vertexj in c. Hencej is the only• neighbor ofi, which contradictsx ∈ ker2 X.

For the second assertion supposeX hasn vertices, and the assertion is true for trewith fewer thann vertices. Writex = ∑n

i=1 xiei . Choose any leafi in X, and letj be theneighbor ofi. If xi = 1 we have found the desired leaf. Ifxi = 0, remove the vertexi andthe edgei, j to obtain the treeXi . We havex ∈ ker2 Xi . Note thatxj = 0 sincei had aneven number of (hence zero)• neighbors inx. By induction, there is a leaf in Xi withx = 0. Sincexj = x, we must havej = , so is also a leaf inX.

ForX odd, we can apply 4.5 to the kernel vector

u =n∑

det(Xj )ej ∈ ker2 X

j=1


3,r of

cause

dsingle

s

yx,

l

he

al

t

and find thatq(u) is the number of nonsingularXj ’s modulo two. By the uniqueness in 4.this implies the following uniform nonsingularity condition for trees with any numbevertices.

4.6. Lemma. Let X be a tree. ThenX is nonsingular if and only ifX contains an oddnumber of maximal matchings.

The construction of odd nonsingular trees by sprouting has an extra wrinkle, beone must also consider singular trees with|ker2X| = 2.

4.7. Lemma. If X is an odd tree, then the following are equivalent:

(1) |ker2X| = 2.(2) The kernel vectoru is nonzero.(3) There exists a maximal matching inX.

Proof. We have seen in 3.2 that (1) and (2) are equivalent for any odd graph. Ifu = 0,then someXi is nonsingular, so has a maximal matchingYi , by 4.3. Adding the vertexito Yi gives a maximal matching inX. Conversely, ifY is a maximal matching inX, thenremoving the isolated vertexj = j (Y ) gives a maximal matching inXj , so det(Xj ) = 0,sou = 0. 4.8. Lemma. The set of odd treesX with |ker2X| = 2 is preserved under sprouting anpruning. Every such tree is obtained by a sequence of sproutings, starting with avertex. SupposeX is obtained fromX′ by sproutingi—j— at the vertexk. Theni is akernel vertex inX if and only ifk is a kernel vertex inX′. If this holds thenX is nonsingularif and only ifX′ is singular, andu = u′+ei . Otherwise(i.e., ifk is not a kernel vertex inX′),X is singular if and only ifX′ is singular, andu = u′. (Hereu andu′ are the kernel vectorof X, X′.)

Proof. We first claim that ifn > 1 and|ker2X| = 2 then 4.1(3) holds. For if not, then b4.1(2) there are at least three distinct leavesi, i′, j with i, i′ adjacent to the same verteand a leafj ′ (which may be one ofi, i′) adjacent the same vertex asj . Thenei + ei′ andej + ej ′ are two linearly independent vectors in ker2 X.

Now, if X′ is obtained by pruning the sprouti—j— from 4.1(3), thenX has a maximamatching if and only ifX′ does. This proves the first two sentences in the lemma.

If k is a kernel vertex inX′, there exists a maximal matchingY in X′k . When we add

the edgej, k to X we get an even treeX′′ which is nonsingular, since it contains tmaximal matchingY ′′ = Y ∪ j, k. Hencei is a kernel vertex inX. Conversely, ifi is akernel vertex inX, thenX′′ = Xi is nonsingular andj, k belongs to the unique maximmatching inX′′. Removing this edge gives a maximal matching inXk , hencek is a kernelvertex inX′.

If k is a kernel vertex inX′, the number of maximal matchings inX is, on accounof Y ′′, one more than the number of maximal matchings inX′. HenceX is nonsingular ifand only ifX′ is singular.


al

phs

ple,

s,

If k is not a kernel vertex inX′ then every maximal matching inX is obtained by addingi, j to a maximal matching ofX′. HenceX andX′ have the same number of maximmatchings. 4.9. Example. We illustrate the method of sprouting/pruning with the family of graTp,q,r . Herep,q, r 2 andTp,q,r is the graph withn = p + q + r − 2 vertices consistingof subgraphsAp, Aq , Ar each joined at one end in a vertex of degree three. For examE8 = T2,3,5, andT3,3,4 was shown in 4.4 above.

Assume first thatn is even. Ifp, q, r are all even, we can pruneTp,q,r down toT2,2,2 =D4, which is singular, henceTp,q,r is singular. If, sayp is even andq, r are odd, we canprune down toT2,3,3 = E6 which is nonsingular, soTp,q,r is nonsingular.

Suppose now thatn is odd. Ifp = 2k+1,q = 2+1, r = 2m+1, thenTp,q,r is obtainedfrom the singular graphT3,3,3 = E6 by k − 1+ − 1+m− 1 sproutings at kernel verticeso T2k+1,2+1,2m+1 is nonsingular if and only ifk + + m is even. Ifp = 2k, q = 2,r = 2m+1, thenTp,q,r is obtained from the singular graphT2,2,r = Dr+2 by k −1+ −1sproutings at kernel vertices, soT2k,2,2m+1 is nonsingular if and only ifk + is odd.

5. Coxeter groups

Given a graphX with vertex setS, let W = W(X) be the group with generatorsσi :i ∈ S, and relations

σ 2i = 1,

σiσjσi = σjσiσj , if i—j,

σiσj = σjσi, otherwise.

The groupW acts linearly onL by

σi(λ) = λ − 〈λ,αi〉αi, (5a)

preserving the form〈 , 〉.

5.1. Lemma. Let k be any field, letVk = k ⊗ L, and letV 0k be the radical of the form on

Vk induced by〈 , 〉. ThenV 0k coincides with the space of invariants ofW in Vk . If X is

connected thenW acts irreducibly onVk/V 0k .

Proof. The first assertion is clear from (5a). Supposex ∈ Vk − V 0k . Thenσi(x) = x for

somei. Hence(1 − σi)(x) = cαi , for somec ∈ k×. If i—j thenσiσj (αi) = 1αj . If X isconnected, this shows there are no proper subspaces ofVk/V 0.
k


-

f

In the next two sectionsk = F2, andV is the quadratic space overF2 constructed froma graphX, as in Section 1. The induced action ofW on V preservesq, and gives a homomorphism

ρ :W → O(V ).

Eachsi := ρ(σi) is a transvection

si(x) = x + f (x, ei)ei .

Visually, if i has an odd number of• neighbors inx, thensi(x) is obtained fromx bychanging the state ofi. If i has an even number of• neighbors inx, thensi(x) = x.

By 2.1, the homomorphismρ is surjective if and only ifW is transitive onV (1), andX = A2 + A2. If i—j thensisj (ei) = ej , so allei in a connected component ofX belongto the sameW -orbit. Since eachei ∈ V (1), the surjectivity ofρ amounts to havingV (1) =Wei for some (any)i.

6. Surjectivity for complete graphs

The caseX = Kn is particularly simple. Recall from 3.4 thatKn is nonsingular whennis even orn ≡ 1 mod 4, andu = ∑n

i=1 ei in the latter case. If 0= x ∈ V , then[x] Kr forsomer n, so

q(x) ≡ r + r

2(r − 1) = r

2(r + 1) =

0 if r = 4j,4j − 1,

1 if r = 4j + 1,4j + 2.

Every• vertex inx hasr − 1 • neighbors and every vertex inx hasr • neighbors. Ifr iseven we can only alter the• vertices; in this case ifi is a• vertex, we have[si(x)] = Kr−1.Likewise if r is odd, we have[si(x)] = Kr+1. It follows that the orbit decomposition oV (1) underW is

V (1) = U1 U5 · · · U4+1,

whereUr = x ∈ V (1): [x] Kr or Kr+1, and = n−24 . By 2.1, this proves

6.1. Proposition. For nonsingular complete graphsKn, the reduction mapρ :W → O(V )

is surjective if and only ifn = 1,2,4,5.

7. Surjectivity for trees

In this sectionX is a tree. We abbreviatewx := ρ(w)x. The vectorx = ∑ni=1 ei belongs

to V (1). As a first step toward surjectivity, we note thatx ∈ Wei for anyi. Indeed, choosea leafi, so thatsix = x − ei . The treeX is replaced by the treeXi , andx is replaced by itsanaloguex − ei . Repeating, we findx ∈ Wei for anyi.


-

-

,

ranche to thelike

moves

nches,he

branche

t

her

Now let x ∈ V be arbitrary and nonzero. Recall thatc(x) denotes the number of components of[x]. Applying the previous argument to each component of[x] proves thefollowing.

7.1. Lemma. Letx ∈ V be nonzero. Then there isw ∈ W such that the components of[wx]are isolated vertices. Moreover,c(wx) = c(x).

The key case for proving surjectivity is the treeTp,q,r whose nonsingularity was discussed in 4.9.

7.2. Lemma. SupposeX = Tp,q,r is nonsingular. Thenρ :W → O(V ) is surjective.

Proof. Let x ∈ V (1). By 7.1, we may assume that[x] consists ofc(x) isolated verticesandc(x) is odd. It suffices to findw ∈ W so thatc(wx) < c(x). For this, it is enough toachieve the following “triad” configuration in the neighborhood of the branch node.

· · · • • · · ·•...

For, the branch reflection will then reduce the number of components by two.Sincesisj (ei) = ej if i—j , we can move stones along branches as follows

· · · • · · · → · · · • · · · .

With these moves, we first “pack down” the stones in each branch. That is, if any bhas stones, we move the stone closest to the leaf onto the leaf, the next closest stonpenultimate spot away from the leaf, and so on, until each branch with stones looks

• • · · ·

and no stones on any branch can be moved towards the leaf on that branch. Thesedo not changec(x).

If there is a stone on the branch vertex, and we cannot move it onto one of the braour configuration isW -invariant, contradictingx ∈ V (1). Hence we can ensure that tbranch vertex has no stone, and this move also does not changec(x).

If all three branches now have stones, we can move those stones closest to thevertex and achieve the triad. If only one branchA has stones, thenA has at least threstones, and sinceX = Dn (by the nonsingularity assumption), some other branchB has atleast two vertices. We move one stone fromA onto the leafb of B. Sinceb is not adjacento the branch vertex, we can move another stone fromA onto the leaf of the third branchC.This takes us back to the previous case.

The remaining possibility is that some branchA has at least two stones, some otbranchB has at least one stone, and the third branchC has no stones. If a stone inB


as

n

at

ng

e

tn

ase.

prevents us from moving a stone fromA ontoC, then the branch neighborhood looksfollows.

• •

Since no stones can be moved towards the leaves ofA or B, this vector is againW -in-variant. This contradiction completes the proof.

Now we can prove the more general result (Theorem 1 of Introduction).

7.3. Theorem. SupposeX is a nonsingular tree, not of typeAn. Thenρ :W → O(V ) issurjective.

Proof. By 7.2 we may assumeX is not of the formTp,q,r . We first supposeX has an evennumbern = 2m of vertices, and argue by induction onm.

By 4.2,X is obtained by sproutingi—j— at some vertex of a nonsingular treeX′ withn − 2 vertices. LetW ′ andV ′ be the analogues ofW , V for X′. NoteX′ is not of typeAn−2, sinceX is not of typeTp,q,r . Henceρ′ :W ′ → O(V ′) is surjective, by the inductiohypothesis.

Let x ∈ V (1), so thatc(x) is odd; we assumec(x) 3. It suffices to findy ∈ Wx suchthatc(y) < c(x).

We may assume, by 7.1, that the components of[x] are isolated vertices. If[x] ⊂ X′ weare done by induction. By moving a stone fromj to i, if necessary, we may assume thx = ei + x′, where[x′] ⊂ X′, andx′ = 0.

SinceX′ = An−2, there is a vertexk in X′ of degree 3. We choosek as near aspossible toj . Hencek is the branch node in a subgraph of typeD4 with neighboringverticesa, b, c, anda, say, is on the path fromk to j . Now c(x′) is even, sox′ ∈ V ′(0).Also eb + ec ∈ V ′(0), becauseX′ is a tree. By surjectivity forW ′, there isw′ ∈ W ′ suchthatw′(x′) = eb + ec, sow′(x) = ei + eb + ec. Hence we can achieve the triad by movithe stone oni along the path towardk. This completes the proof in the even case.

Now supposeX is an odd nonsingular tree, with kernel vectoru = ∑det(Xi)ei . By 4.5,

there is a leafi in X such thatXi is nonsingular. Note thatXi is an even tree, not of typAn−1.

Let x ∈ V (1), and assume the components of[x] are single vertices. Ifxi = 0 then[x] ⊂ Xi and we are reduced to the even case. Hence we may assumexi = 1 andxj = 0,wherej is the neighbor ofi. Note thatq(x + ei) = 0. SinceXi = An−1, there are at leastwo leavesa, b in Xi , other thanj , andq(ea + eb) = 0. From the surjectivity in the evecase, there isw ∈ W(Xi) such thatw(x + ei) = ea + eb, sow(x) = ei + ea + eb. Now j

cannot be adjacent inX to any leaf buti, sinceX is nonsingular. Hencej is not adjacent toa or b. It follows thatsisjw(x) = ej + ea + eb and we are again reduced to the even cThis completes the proof in the odd case.


e

d

lu-

d

8. Involutions at level one

We turn now to the higher level congruence subgroupsWk of W . These groups artorsion-free fork 2. Here, we analyze the torsion inW1.

For any latticeL, and subgroupΓ ⊂ GL(L), define

Γk := ker[GL(L) → GL

(L/2kL

)], k 1.

The quotientΓk/Γk+1 has a Lie algebra structure overF2, induced by the commutator, anthe map

∂k :Γk/Γk+1 → End(L/2L), ∂k(γ ) = 2−k(γ − I ),

is a Lie algebra isomorphism. Ifγ ∈ Γk thenγ 2 ∈ Γk+1 and

∂k+1(γ 2) =

∂k(γ ) + ∂k(γ )2 if k = 1,

∂k(γ ) if k > 1.(8a)

8.1. Lemma. Γ2 is torsion-free, and the torsion elements ofΓ1 are involutions.

Proof. If γ ∈ Γ1 has orderab, with a = 2c andb odd, then∂k(γa) has odd order in the

abelian group End(L/2L), for all k 1. Henceγ a ∈ Γk for all k, soγ a = I andb = 1.If γ ∈ Γk with k 2, then∂k(γ ) = ∂k+c(γ

a) = 0, soγ ∈ Γk+1, again forcingγ = I . Ifγ ∈ Γ1, thenγ 2 is a torsion element inΓ2, henceγ 2 = I .

An involution γ ∈ Γ1 is of the formγ = I + 2D, whereD = ∂1(γ ) andD2 = −D. Itgives a splitting

L = DL ⊕ (I + D)L,

andDL, (I + D)L are the−1,+1 eigenspaces ofγ , respectively.Thus, forΓ = GL(L), taking the−1,+1 eigenspaces gives a bijection between invo

tions inΓ1 and ordered pairs(L′,L′′) of sublattices ofL such thatL = L′ ⊕ L′′.If Γ = O(L) is the orthogonal group of a symmetric form〈 , 〉 :L × L → Z, the involu-

tions inΓ1 correspond to orthogonal pairs(L′,L′′).SupposeΓ = W is the Coxeter group obtained from a connected graphX, andL is the

associated root lattice, with symmetric form〈 , 〉 as in Section 1. ForJ ⊂ S, let W(J) bethe subgroup ofW generated byσj : j ∈ J , and letLJ be theZ-span ofαj : j ∈ J . Bya theorem of Tits (cf. [5, Proposition 3.12]), every finite subgroup ofW can be conjugateinto a finite subgroupW(J), for someJ ⊂ S. It is easy to see that

Wk ∩ W(J) =∏i

Wk ∩ W(Ji),

where[J1], . . . , [Jc] are the connected components of[J ].


o

s

To classify the level one involutionsw ∈ W1, we may therefore assumew ∈ W1∩W(J),where[J ] is connected withW(J) finite, and moreover thatw cannot be conjugated intW(I), for I J . Thenw must be the unique elementwJ ∈ W(J) acting by−1 onLJ . Inparticular,[J ] must have one of typesA1, D2m, E7, E8. We will see that the first two casecannot occur in nonsingular trees.

If i /∈ J , we have

wJ αi = αi +∑j∈J

nijαj ,

with all nij 0. It follows thatLJ is the whole−1 eigenspace ofwJ in L. Hence we havean orthogonal splitting

L = LJ ⊕ L⊥J ,

andL⊥J = λ ∈ L: wJ λ = λ. If J ⊂ K ⊂ S, then replacingW by W(K) shows thatLJ

must be an orthogonal summand ofLK . If X has more than one vertex, thenJ cannot havetypeA1, sinceA1 is not an orthogonal summand ofA2.

For anyJ ⊂ S such that〈 , 〉 is nonsingular onQ ⊗ LJ , there exists, for eachk ∈ S \ J ,a vectorωk ∈ Q ⊗ LJ defined by

〈ωk,αj 〉 = −〈αk,αj 〉, for all j ∈ J.

For example, ifk is not adjacent to any vertex inJ thenωk = 0.

8.2. Lemma. LetJ ⊂ S be such that〈 , 〉 is nonsingular onQ ⊗LJ . ThenLJ is an orthog-onal summand ofL if and only ifωk ∈ LJ for all k ∈ S \ J .

Proof. If L = LJ ⊥ U , then for eachk ∈ S \ J write αk asαk = λk + uk , with λk ∈ LJ

anduk ∈ U . Then−λk satisfies the equations definingωk , soωk = −λk ∈ LJ . Conversely,if ωk ∈ LJ for all k ∈ S \ J , then setuk = αk + ωk , and letU = ∑

k∈S\J Zuk . ClearlyU ⊆ L⊥

J , and it suffices to show equality. Sinceαk = uk −ωk , we haveL = LJ +U . Writeλ ∈ L⊥

J asλ = ω + u, with ω ∈ LJ andu ∈ U . Then 0= 〈αj ,λ〉 = 〈αj ,ω〉 for all j ∈ J ,soω ∈ LJ ∩ L⊥

J = 0, sinceLJ is nonsingular overQ. 8.3. Lemma. If J E8, thenLJ is an orthogonal summand ofL, andwJ ∈ W1.

Proof. This is immediate from 8.2, sinceE8 is a unimodular lattice. 8.4. Lemma. SupposeX is a nonsingular tree. IfJ D2m, m 2, thenLJ is not anorthogonal summand ofL.


n,

Proof. Number the vertices ofJ as

1 2 3 · · · (2m − 2) (2m − 1)

2m.

For j ∈ J , defineλj ∈ Q ⊗ LJ by

〈λj ,αi〉 =−1 if i = j,

0 if i = j,

and letL+J be theZ-lattice spanned byλj : j ∈ J . ThenL+

J containsLJ andL+J /LJ

Z/2Z × Z/2Z. In the latter quotient, we have the relations

λ1 = λ3 = · · · = λ2m−3 = λ2m−1 + λ2m, λ2 = λ4 = · · · = λ2m−2 = 0.

For anyk ∈ S \ J , we have

ωk =∑j∈Jk

λj ,

whereJk = j ∈ J : k − j.SinceX is nonsingular, the vertices 2m − 1,2m cannot both be leaves inX, by the

remark prior to 4.2. Hence there existsk ∈ S \ J such that at least one of2m − 1,2mbelongs toJk . But if ωk ∈ LJ , the above relations then force both 2m − 1 and 2m to bein Jk . Hence there is a 4-cycle inX, with vertices2m−2,2m−1, k,2m. This contradictsour assumption thatX is a tree.

I do not know if it is necessary to assume thatX is a tree in Lemma 8.4.Now takeJ E7, and letuJ ∈ V be the kernel vector of[J ]. Visually,

uJ = • ••

.

Let J0 be the set of• vertices in the above subgraph. These are the kernel vertices inJ . Ifn is odd we can compareuJ with the kernel vectoru = ∑

det(Xi)ei of X.

8.5. Lemma. If J E7 thenLJ is an orthogonal summand ofL if and only ifuJ ∈ ker2 X.If this holds andX is nonsingular thenn is odd anduJ = u.

Proof. Let λj ∈ Q ⊗ LJ , j ∈ J , andJk be as in the proof of 8.4. Viewed as an involutiouJ generates the center of the simply connected Lie groupE7, so we have

∑cjλj ∈ LJ

if and only if∑

j∈J0cj is even. Henceωk ∈ LJ exactly when|Jk ∩ J0| is even. This holds

for all k ∈ S \ J iff uJ ∈ ker2 X. The last assertion is an immediate consequence.


olu-

e

nown

section

f

p

Now 8.1–8.5 yield the following.

8.6. Proposition. SupposeX is a nonsingular tree. Then every conjugacy class of invtions inW1 contains a commuting productw = ∏

J wJ , whereJ runs over full subgraphsof X of typeE7 or E8. If some factor of typeE7 occurs thenn is odd and there are threverticesi, j , k in X suchu = ei + ej + ek andJ0 = i, j, k for every factorwJ of typeE7occuring in an involutionw ∈ W1.

We illustrate 8.6 withX = T2,4,5, labelled as shown.

1 2 3 4 5 6 7 8

9

This graph is nonsingular, withu = e6 + e8 + e9. One involution inW1 comes from theuniqueE8 subdiagram. There are twoE7 subdiagrams, but onlyJ = S − 1,2 gives aninvolution. Explicitly, the vectors

θ7 = α2 + 2α3 + 3α4 + 4α5 + 3α6 + 2α7 + α8 + 2α9,

θ8 = 2α1 + 4α2 + 6α3 + 8α4 + 10α5 + 7α6 + 4α7 + α8 + 5α9

belong toL⊥J and L⊥

E8, respectively. Sinceα2 has coefficient= 1 in θ7 and α8 has

coefficient= 1 in θ8, we have orthogonal decompositions

L = LJ ⊕ Zα1, θ7 = LE8 ⊕ Zθ8

andwJ , wE8 represent the two conjugacy classes of involutions inW1.

9. Density in orthogonal groups

This is the only section where it is not essential to havep = 2. We aim to prove a versioof p-adic density for certain hyperbolic Coxeter groups. This is a combination of knresults, and will be applied in the 2-adic case to more general Coxeter groups. Thistakes place in characteristic zero, and we will recycle some of our earlier notation.

We begin a freeZ-moduleL of rankn, and a symmetric bilinear mapf :L × L → Zsuch thatf (x, x) ∈ 2Z for all x ∈ L. Let D = detf , and setq(x) = 1

2f (x, x). We assumen 5 and thatf is nondegenerate overQ and indefinite overR.

For any integral domainR, let GL(R) denote the group ofR-linear automorphisms oR ⊗ L, O(R) the subgroup ofGL(R) preserving the extension ofq to R ⊗ L, and letSO(R) = g ∈ O(R): detg = 1. Any ring homomorphismR → R′ induces natural grouhomomorphisms

GL(R) → GL(R′), O(R) → O

(R′), SO(R) → SO

(R′).


If x ∈ R ⊗ L andq(x) is a unit inR, we have the “reflection”rx ∈ O(R), defined by

rx(y) = y − q(x)−1f (x, y)x.

Let F be the quotient field ofR, and assumeF has characteristic zero. ThenO(F) isgenerated by reflections [2, Proposition 8], and the spinor norm

δF :O(F) → F×/F×2

is a group homomorphism determined by the ruleδF (rx) = q(x). More generally, ifg ∈O(F) is an involution andEg is the −1-eigenspace ofg in F ⊗ L, then δF (g) is thediscriminant of the restriction off to Eg .

We set

O ′(R) = O(R) ∩ kerδF , SO′(R) = SO(R) ∩ kerδF .

The groupSO′(F ) is the image of Spin(F ) under the two-fold cover Spin(F ) → SO(F ).If F = Qp, the spinor norm is surjective, and we have an exact sequence [8, III.3.2]

1→ ±1 → Spin(F ) → SO(F )δF−→ F×/F×2 → 1.

From the commutative diagram

O(Q)

δQ

O(Qp)

δQp

Q×/Q×2 Q×p /Q×2

p

(9a)

it follows that

O ′(Q) ⊂ O ′(Qp), O ′(Z) ⊂ O ′(Zp).

Fix a primep such thatq is nonsingular onFp ⊗ L.

9.1. Lemma. The image of SO(Zp) underδQpis Z×

p /Z×2p .

Proof. This follows from the results in [6, 5.5].Let Q denote the finite adeles ofQ, and let

SO′(Q) =

g = (gp) ∈ SO(Q

): gp ∈ SO′(Qp) for all p

.


of

p

et

From strong approximation for Spin/Q, it follows that the diagonal embeddingSO′(Q) → SO′(Q) has dense image [7]. Hence, for any primep, and integerk 1, wehave

SO′(Q) = SO′(Q)U ′

kKp,

whereU ′k = ker[SO′(Zp) → SO(Z/pkZ)], and

Kp =∏ =p

SO′(Z).

It follows easily that

SO′(Zp) = SO′(Z)U ′k. (9b)

(Write g ∈ SO′(Zp) in the formg = γ uκ , with γ ∈ SO′(Q), u ∈ U ′k , κ ∈ Kp . Thenγ is

integral at all primes, hence belongs toSO′(Z), andγ κ = 1 for all = p, sog = γ u.)Sinceq is not identically zero onFp ⊗ L, there isλ ∈ Zp ⊗ L such thatq(λ) ∈ Z×

p .From this and (9b), it follows that

O ′(Zp) = O ′(Z)U ′k ⇔ O ′(Z)

det−→ ±1 is surjective. (9c)

This holds if there isλ ∈ L with q(λ) = 1.Suppose now that(L,q) arises from a graphX, as in Section 1. LetΩ be the subgroup

of O(Z) generated by Aut(X) and−I . ThenΩ normalizesW , and we have the subgrou

ΩW ⊆ O(Z).

In fact,W ⊂ O ′(Z) sinceδQ(σi) = 1, so

Ω ′W ⊆ O ′(Z),

whereΩ ′ = kerδQ ∩ Ω .The graphX is calledhyperbolic if W(J) is finite or affine for every proper subs

J ⊂ S, butW is not itself finite or affine. This implies that the quadratic formq on R ⊗ L

has signature(n − 1,1) [1, p. 141].

9.2. Lemma. If X is hyperbolic, thenO ′(Z) = Ω ′W .

Proof. This follows from [5, Corollary 5.10b] and the fact thatW ⊂ O ′(Z). Since det(W) = ±1, 9.2 and (9c) yield the following.

9.3. Corollary. Supposen 5, and X is hyperbolic. ThenO ′(Zp) = Ω ′WU ′k for every

k 1.


lic

n. Foration of

r

Actually, we will only need two examples of 9.3.

9.4. Example. Let X = E10(= T2,3,7). ThenL is the unique even unimodular hyperbolattice in dimension 10. There are no diagram symmetries, soΩ = ±I . The discrimi-nant is−1, soδQ(−I ) = −1, soΩ ′ = 1, andO ′(Z) = W . Hence for allk 1, we haveO ′(Zp) = WU ′

k.

9.5. Example. Let X = T3,3,4. HereΩ = ±I,±ω, whereω is the nontrivial diagramsymmetry. The−1 eigenspace ofω in Q ⊗ L is a hyperbolic plane, so

δQ(−I ) = −3, δQ(ω) = −1.

AgainΩ ′ = 1, soO ′(Zp) = WU ′k for all k 1.

10. Adjoint representation over F2

This section has nothing to do with graphs, and its assertions are surely knowlack of an adequate reference, we recall here the structure of the adjoint representorthogonal groups in characteristic 2. We assume from now on thatn is even.

Let F be a field of characteristic 2, with dual numbersF [ε] = F [x]/(x2). Let V be avector space overF of dimensionn = 2m, and letq be a nonsingular quadratic form onV ,with associated bilinear formf (x, y) = q(x + y) + q(x) + q(y).

ForA ∈ End(V ), we haveI + εA ∈ O(F [ε]) iff

q(v) = q(v + εAv) = q(v) + ε2q(Av) + εf (v,Av) = q(v) + εf (v,Av),

so the scheme-theoretic Lie algebra ofO(F) is

so(F ) := A ∈ End(V ): f (v,Av) = 0 for all v ∈ V

. (10a)

In particular, the Lie algebra depends only onf , not onq. Choose a basis forV such thatthe matrix off has the form

[ 0 QtQ 0

]for some invertible matrixQ ∈ GLm(F). (It will be

convenient to avoid a specific choice forQ.)Let

sm(F) = B ∈ glm(F): txBx = 0 for all x ∈ Fm

.

One can check thatsm(F) is the set of symmetric matrices overF with zero diagonal. FoM ∈ glm(F), let M∗ = Q−1(tM)Q. A matrix calculation with (10a) shows that

so(F ) =A =

[A1 A2A A∗

]: A1 ∈ glm(F) andQA2,

tQA3 ∈ sm(F)

.

3 1


r

ee

Define thehalf-traceτ : so(F ) → F by τ(A) := tr(A1), and set

so′(F ) := kerτ =A =

[A1 A2A3 A∗

1

]∈ so(F ): A1 ∈ slm(F)

.

10.1. Lemma. so′(F ) is the commutator subalgebra ofso(F ).

Proof. We make two preliminary remarks. First, ifS,T ∈ sm(F), then tr(ST ) = 0. Sec-ond, forX ∈ glm(F), we have

tQX ∈ sm(F) ⇔ XQ−1 ∈ sm(F).

Now, for A,B ∈ so(F ), we have

τ([A,B]) = tr

([A1,B1] + A2B3 + A3B2) = tr(A2B3) + tr(A3B2).

Combining our two observations, we have

tr(A2B3) = tr(QA2 · B3Q

−1) = 0, tr(A3B2) = tr(A3Q

−1 · QB2) = 0.

Hence the commutator subalgebra ofso(F ) is contained inso′(F ).Conversely, sinceslm(F) is the commutator subalgebra ofglm(F), it suffices to prove

that[ 0 0

A3 0

] ∈ [so(F ), so(F )] for all A3 ∈ Q−1sm(F), along with a similar assertion fo[ 0 A20 0

]. Since [[

0 0Y 0

],

[X 00 X∗

]]=

[0 0

YX + X∗Y 0

],

it suffices to findY ∈ Q−1sm(F) making the map

φY :glm(F) → Q−1s(F ), φY (X) = YX + X∗Y

surjective. Note thatX ∈ kerφY iff (QY)X + (tX)(QY) = 0.If m = 2 is even, we can chooseY ∈ Q−1sm(F) of rankm, so thatQY is the matrix of

a nondegenerate symplectic form onF 2. Then kerφY = sp2(F ) has dimension 22 + .Since dimsm(F) = (2 − 1), this shows thatφY surjective.

If m = 2 + 1, we can takeY of rank m − 1. Then kerφY sp2(F ) ⊕ F 2, againimplying thatφY is surjective.

The proof for[ 0 A2

0 0

]is similar.

10.2. Lemma. If F is a subfield of an algebraic closureF2, thenF · I and so′(F ) arethe only properO(F)-invariant subspaces ofso(F ). Hence, ifm is odd, we have thirreducible O(F)-decompositionso(F ) = F · I ⊕ so′(F ), and if m is even, we havF · I ⊂ so′(F ) ⊂ so(F ) indecomposable.


pin

stf

Proof. It is clear from 10.1 thatso′(F ) is indeed anO(F) invariant subspace ofso(F ). Forthe moment, letF = F2. Then we can choose a basis ofV so thatq(x) = ∑

xixm+i . Thediagonal matrices inO(F) have the formt = diag(t1, . . . , tm, t−1

1 , . . . , t−1m ) and comprise a

maximal torusT , which acts diagonalizably on anyO(F)-invariant subspaceU ⊆ so(F ).Sincen is even and 6, the roots ofT in so(F ) form a single orbit under the Weyl grouof T . Hence ifU does not consist of diagonal matrices, then all roots must appearU .The calculation [

1 10 1

][0 01 0

][1 10 1

]=

[1 11 1

]shows thatU contains all matrices of the form

[D 00 D∗

], whereD ∈ glm(F) is a diagonal

matrix of trace zero. Henceso′(F ) ⊂ U .If U consists of diagonal matrices, then the calculation[

1 10 1

][s 00 t

][1 10 1

]=

[s s + t

0 t

]implies thatU = F · I . This proves the lemma forF = F2, and shows that the higheweights of the composition factors ofso(F2) take values 0,1 on the simple co-roots oO(F2). By Steinberg’s theorem [9, 1.3] each composition factor ofso(F2) remains irre-ducible underO(F) for any subfieldF ⊂ F2. Since anyO(F)-invariant subspace ofso(F )

remains invariant after extending scalars, the lemma is proved.

11. Higher levels

In this section we assumeL has even rankn = 2m andq is nonsingular onF2 ⊗ L. Asa quadratic space overZ2, we have (see Section 2)

Z2 ⊗ L (m − d)

[0 11 0

]⊥ d

[2 11 2

], (11a)

whered ∈ 0,1 is the defect ofq.Let Uk = ker[SO(Z2) → SO(Z/2kZ)]. The map∂k(u) = 2−k(u − I ) (see Section 8) is

an injection

∂k :Uk/Uk+1 → so(F2) (11b)

whose image is anO(F2)-invariant subspace ofso(F2).

11.1. Lemma. The map(11b) is surjective for allk 1.

Proof. First supposem = 1, so thatso(F2) = 0, I . If d = 0, let

u =[

s 00 s−1

], s = 1+ 2kz, z ∈ Z×

2 .


r

],

Thenu ∈ Uk and∂k(u) = I . If d = 1, let

u = 1

s2 − s + 1

[1− s2 2s − s2

s2 − 2s 1− 2s

], s = 2kz, z ∈ Z×

2 .

Thenu ∈ Uk and∂k(u) = I .For m > 1, the decomposition (11a) shows that the image of∂k contains nonscala

diagonal matrices inso(F2) \ so′(F2). The lemma now follows from 10.2.11.2. Lemma. The spinor normδQ2 is trivial on U3.

Proof. If k 2 andu ∈ Uk , thenu2 ∈ Uk+1, and from (8a) we have

∂k+1(u2) = ∂k(u). (11c)

From 11.1 it follows that the squaring map

Uk/Uk+1 → Uk+1/Uk+2

is surjective fork 2. Hence, givenu ∈ U3, there are elementsuk ∈ Uk , k 2, such that

u = u22u4 = u2

2u23u5 = · · · .

But δQ2 takes values in the finite groupQ×2 /Q×2

2 , and is 2-adically continuous [6, 1.6.5soδQ2 is trivial on squares and onUk for somek. HenceδQ2(u) = 1.

The values ofδQ2 on U1,2 can be expressed in terms of the half-traceτ : so(F2) → F2(see Section 10) and the isomorphism

ε1 × ε2 : Z×2 /Z×2

2 → F2 × F2, ε1(x) ≡ x − 1

2mod 2, ε2(x) ≡ x2 − 1

8mod 2.

11.3. Lemma. For k = 1,2 we haveεk δQ2 = τ ∂k onUk .

Proof. Again start withm = 1. Ford = 0, we have

SO(Z2) =[

a 00 a−1

]: a ∈ Z×

2

, δQ2

([a 00 a−1

])= a,

and the claim is immediate. Ford = 1, one checks that

SO(Z2) =[

a −b

b a + b

]: a, b ∈ Z2, a2 + ab + b2 = 1

,

δQ2

([a −b

b a + b

])= 2− a − 2b.


d

reof,

-

Note that if[

a −bb a+b

] ∈ Uk , thenb ∈ 2k+1Z2. The claim now follows from straightforwarcalculations.

Now letm > 1. The groupUk has the triangular decomposition

Uk = U−k TkU

+k ,

where the subgroupsU±k are generated by subgroups of root groups, or products the

andT = T ∩Uk , for T a maximally split torus ofSO(Q2). The root groups lift to Spin(Q2),henceU±

k ⊂ SO′(Z2). It remains to verify 11.3 onTk . ButT is a product of special orthogonal groups on two-dimensional spaces, so the result follows from the casem = 1.

Recall thatU ′k = Uk ∩ kerδQ2.

11.4. Lemma. We have

∂k

(U ′

k/U ′k+1

) =

so′(F2) if 1 k 2,

so(F2) if k 3.

Proof. If k 3 this is immediate from 11.1, 11.2. Assumek ∈ 1,2. Then

∂k

(U ′

k/U ′k+1

) ⊆ so′(F2)

by 11.3. Ifm = 1 thenso′(F2) = 0, so assumem > 1.From 11.1, we have

[so′(F2) : ∂k

(U ′

k

)] = 1

2

[∂k(Uk) : ∂k

(U ′

k

)].

The map∂k induces an exact sequence

1→ Uk+1/U ′k+1 → Uk/U ′

k → ∂k(Uk)/∂k

(U ′

k

) → 1,

so

[so′(F2) : ∂k

(U ′

k

)] = [Uk : U ′k]

2[Uk+1 : U ′k+1]

.

From 11.1, 11.3, we find that

[Uk : U ′

k

] =

4 if k = 1,

2 if k = 2,

1 if k 3.

The result follows.


ps

w that

at

-

Recall thatO ′(Z) = O(Z) ∩ kerδQ. From (9c), we have

O ′(Z2) = O ′(Z)U ′k, for all k 1,

as long as 1∈ q(L). We can now prove the following density criterion for subgrouof O ′(Z).

11.5. Proposition. Assume1 ∈ q(L), and n 6. Let H be a subgroup ofO ′(Z), andsetHk = H ∩ U ′

k . ThenH is dense inO ′(Z2) (in the 2-adic topology) if and only if thefollowing three conditions hold.

(1) The compositionH → O(Z) → O(F2) is surjective.(2) The image∂k(Hk) ⊆ so(F2) contains a nonscalar matrix for1 k 3.(3) The homomorphismτ ∂3 :U3 → F2 is nontrivial onH3.

Proof. The necessity of (1)–(3) is clear. We assume that (1)–(3) hold, and must shoO ′(Z2) = HU ′

k for all k 1. The latter is true fork = 1, by (1), so assumek 2.In view of (1), we have a containment ofO(F2)-invariant subspaces

∂k(Hk) ⊆ ∂k

(U ′

k

). (11d)

It suffices to show equality in (11d) for allk 2.For 2 k 3, condition (2) and 10.2 imply thatso′(F2) ⊂ ∂k(Hk). Now 11.4 implies

equality in (11d), where fork = 3 we also invoke condition (3).From (11c) we have∂k(Hk) ⊆ ∂k+1(Hk+1) for k 2. We have already proved th

∂3(H3) = so(F2), so we have equality in (11d) for allk 4.

12. Density for W

Let X be a nonsingular even tree withn = 2m vertices, and letL and(W,S) the as-sociated quadratic lattice and Coxeter group, as in Section 1. Then 1∈ q(L), since, forexample,q(α1) = 1. LetJ ⊂ S be an even subset, with corresponding sublatticeLJ , suchthat [J ] is connected and nonsingular. We can labelS = 1, . . . , n so that no edge is contained in1, . . . ,m or m+ 1, . . . , n, and so thatJ = j + 1, . . . ,m,m+ 1, . . . ,2m− j,for some 1 j m. With respect to the basisα1, . . . , αn, the matrix of〈 , 〉 onL has theform

2In −

0 0 ∗ ∗0 0 QJ ∗∗ tQJ 0 0

,

∗ ∗ 0 0


e thate even

where[ 0 QJ

tQJ 0

]is the adjacency matrix of[J ]. Any w ∈ W(J) has matrix onL of the form

w =[

Ij 0 0∗ w1 ∗0 0 Ij

],

wherew1 is the matrix ofw onLJ . If w ∈ Wk(J ) for k 1, then∂k(w) has the form

∂k(w) =[0 0 0

∗ ∂k(w1) ∗0 0 0

],

and∂k(w1) belongs to the Lie algebrasoJ (F2) with respect to[ 0 QJ

tQJ 0

]. Therefore

τ(∂k(w)

) = τJ

(∂k(w1)

),

whereτJ is the half-trace onsoJ (F2).It follows that if conditions (2), (3) of 11.5 hold forH = W(J) andsoJ (F2), then they

hold for H = W andso(F2) as well. We know that 11.5(2), (3) hold ifJ has typeE10 orT3,3,4, by 9.4 and 9.5. In 7.3 we verified 11.5(1) for nonsingularX containing a branchnode. Thus, we have proved Theorem 2 of Introduction:

12.1. Theorem. Let X be a nonsingular even tree, containing a subtree of typeT3,3,4 orE10. ThenW is dense inO ′(Z2).

The nonsingular even trees to which 12.1 does not apply are few, in the sensthey can be easily listed, by considering all possible sproutings on small trees. Thnonsingular trees which do not containT3,3,4 or E10 consist of the familyX2m obtained bym − 1 sproutings at a single vertex inA2 (soX2 = A2, X4 = A4, X6 = E6, . . .), and theseven trees

,

,

,

.


e, RI,

References

[1] N. Bourbaki, Lie Groups and Lie Algebras, Springer-Verlag, Berlin, 2002 (Chapters 4–6).[2] J. Dieudonné, Sur les Groupes Classiques, Hermann, Paris, 1967.[3] S. DeBacker, M. Reeder, Depth-zero supercuspidalL-packets and their stability, preprint, 2004.[4] S. Helgason, Differential Geometry, Lie Groups and Symmetric Spaces, Academic Press, 1978.[5] V. Kac, Infinite Dimensional Lie Algebras, third ed., Cambridge Univ. Press, 1995.[6] Y. Kitaoka, Arithmetic of Quadratic Forms, Cambridge Univ. Press, 1993.[7] M. Kneser, Strong Approximation, Proc. Sympos. Pure Math., vol. IX, Amer. Math. Soc., Providenc

1966.[8] J.-P. Serre, Galois Cohomology, Springer-Verlag, 2002.[9] R. Steinberg, Representations of algebraic groups, Nagoya Math. J. 22 (1963) 33–56.

level-two structure of simply-laced coxeter groups

Documents