Table Of ContentREFLECTION GROUPS OF THE QUADRATIC FORM
−5x2+x2+...+x2
0 1 n
ALICEMARK
2
1
0 Let V be a an (n+1)-dimensional real vector space with basis v ,...,v , and
0 n
2 let L be the integer lattice generated by the same basis. Given a quadratic form of
l the form
u
f(x)=−px2+x2+...+x2
J 0 1 n
9 with p a positive integer, the group Θ of integral automorphisms is the group of
1 symmetries of L preserving that form and mapping each connected component of
the set {x:f(x)<0} to itself. This group splits as a semidirect product
]
R Θ=Γ(cid:111)H
G
whereΓisgeneratedbyreflectionsandH isagroupofsymmetriesofanassociated
.
h polytopeinhyperbolicn-space[1]. WesayLisreflectiveifH isafinitegroup. For
t a fixed p, we wish to know for which values of n the lattice L is reflective. Vinberg
a
m answeredthatquestionforp=1,2in[1]and[3],andMcLeodanswereditforp=3
in [2]. Here we answer it for p=5.
[
FirstwefindgeneratorsforΓusingVinberg’salgorithm. Thenweshowthatthe
1 algorithm terminates after not very many steps when 2 ≤ n ≤ 8. Finally we show
v
that when n≥9, Γ is not finitely generated.
2
Descriptionsofthealgorithmabound(see[1]and[2]andothers),soIwon’tgive
2
(cid:80)
8 a complete one here. Since the algorithm finds roots, the vectors ei = kjvj it
4 finds are subject to the crystallographic condition. In the case p = 5 that means
. (e ,e )=1,2,5, or 10, and if (e ,e )=5 or 10 then 5(cid:45)k and 5|k for j (cid:54)=0.
7 i i i i 0 j
0
Proposition 1. The first several vectors that Vinberg’s algorithm produces are
2
listed in Table 1.
1
:
v
Xi (eik,02ei) ei (ei,ei) i n
1 v +2v +v +v +v 2 n+3 ≥4
r 2 0 1 2 3 4
a v +v +v +v +v +v +v +v 2 n+4 ≥7
0 1 2 3 4 5 6 7
4 2v +5v 5 n+1 ≥2
5 0 1
1 v +2v +v +v 1 n+3 3
1 0 1 2 3
v +v +v +v +v +v +v 1 n+4 6
0 1 2 3 4 5 6
9 3v +5v +5v 5 n+2 ≥2
5 0 1 2
Table 1. Vectors found with Vinberg’s Algorithm. The labels i
arechosenforconveniencetolaterargumentsratherthantheorder
in which the algorithm finds them.
1
2 ALICEMARK
Proof. The batches labeled 1 ,1, and 4 are empty because (e,e) = 5 or 10, and
10 5 10
there is no way to write (e,e)+5k2 =15, 20 or 30 as a sum of squares of integers
0
all divisible by 5.
The batch labeled 1 consists of vectors e=(cid:80)n k v where
2 i=0 n n
n
(cid:88)
k2 =(e,e)+5k2 =7
i 0
i=1
7 may be written as a sum of squares in two ways. This batch contains one vector
if n ≥ 4, and two if n ≥ 7. These two vectors have inner product 0, so we keep
both of them.
The batch labeled 4 consists of vectors e=(cid:80)n k v where
5 i=0 i i
n
(cid:88)
k2 =(e,e)+5k2 =25
i 0
i=1
and 5|k for all i > 0. The vector 2v +5v has inner product 0 and −5 with the
i 0 1
vectors in the previous nonempty batch so we keep it for all n≥2.
The batch labeled 9 is empty since (e,e)=10 and (e,e)+5k2 =55 cannot be
10 0
written as a sum of squares of integers all divisible by 5.
The batch labeled 1 consists of vectors e=(cid:80)n k v where
1 i=0 i i
n
(cid:88)
k2 =(e,e)+5k2 =6
i 0
i=1
Therearetwowaystowrite6asasumofsquares. Oneoftheseproducesthevector
v +2v +v +v . Thishasinnerproduct0withthevectorinbatch 4,sowekeepit
0 1 2 3 5
whenn=3,butithaspositiveinnerproductwiththevectorv +2v +v +v +v
0 1 2 3 4
in batch 1, so we throw it out when n≥4.
2
The other way to write 6 as a sum of squares produces the vector v +v +v +
0 2 3
v +v +v . Thishasinnerproduct0withthevectorv +2v +v +v +v inbatch
4 5 6 0 1 2 3 4
1 and inner product −5 with the vector in batch 4, so we keep it when n = 6. It
2 5
has positive inner product with the vector v +v +v +v +v +v +v +v in
0 1 2 3 4 5 6 7
batch 1, so we throw it out when n≥7.
2
The batch labeled 16 is empty since (e,e)=10 and (e,e)+5k2 =90 cannot be
10 0
written as a sum of squares of integers all divisible by 5.
The batch labeled 9 consists of vectors e=(cid:80)n k v where
5 i=0 i i
n
(cid:88)
k2 =(e,e)+5k2 =50
i 0
i=1
and 5|k for all i > 0. There is exactly one vector satisfying this. 3v +5v +5v
i 0 1 2
has non-positive inner product with all the vectors in previous batches, so we keep
it for all n≥2.
(cid:3)
Proposition 2. The diagrams in figure 1 all describe acute angled polytopes of
finite volume.
We will prove this using Vinberg’s criterion for finite volume. Our notation
follows [1], and the criterion we use is Proposition 2 in that paper. A list of the
affine Coxeter diagrams can also be found there in Table 2.
REFLECTION GROUPS OF THE QUADRATIC FORM −5x2+x2+...+x2 3
0 1 n
Figure 1. Coxeter diagrams for the hyperbolic reflection groups
associated to the form −5x2+x2+...+x2
0 1 n
Ifconditions1and2ofthefinitevolumecriterionaremet, thenanyfacetofthe
polyhedral angle K that passes out of the cone C intersects the boundary of the
S
cone in a line in V corresponding to a cusp of the polytope.
Theonlycocompacthyperbolicdiagramsoccurringassubdiagramsofthegraphs
infigure1aredottedlineedges,andtheonlymaximalaffinesubdiagramshaverank
n−1.
Listed in table 2 are the index sets for the affine subdiagrams and the types
of their maximal extensions, which shows that condition 1 of the finite volume
criterion is satisfied for all of the diagrams in figure 1.
It remains to show that condition 2 of the finite volume criterion is met. There
are 4 different dotted line components which appear. We will show that each
satisfies condition 2.
A sufficient condition for a dotted line subgraph to satisfy condition 2 is that
there exist a set of vertices T ⊂ I such that the diagram with vertex set T is a
spherical subdiagram of rank at least n−1, and there are no edges between the
vertices of S and the vertices of T (this is a corollary of Proposition 2 in [1]). To
see why it is true, compute the dimension of (S∪T)⊥.
Thetwodottedlinesubgraphsthatappearonlywhenn=6,7,8canbeshownto
satisfycondition2usingthisfact. ThefirstofthesehasvertexsetS ={n+2,n+3}.
Let T ={2,3,...,n−1,n+1}. When n=6, T has type D . When n=7, T has
5
type E . When n = 8, T has type E . In each case this is a spherical diagram of
6 7
rank n−1 with no edges joining any vertex in T to any vertex in S.
4 ALICEMARK
n index set maximal diagram type
2 {3,4} A(cid:101)1
(cid:27)
{3,6}
3 A(cid:101)2
{4,5} 1
(cid:27)
{3,4,7}
4 {5,6} A(cid:101)1C(cid:101)2
5 {1,2,3,4,8} A(cid:101)4
(cid:27)
{3,4,5,8}
{6,7} A(cid:101)1B(cid:101)3
(cid:27)
{1,2,3,4,9}
6 {6,10} A(cid:101)1A(cid:101)4
(cid:27)
{3,4,5,6,9}
{7,8} A(cid:101)1B(cid:101)4
(cid:27)
{1,2,3,4,10}
7 {6,7,11} A(cid:101)4C(cid:101)2
(cid:27)
{3,4,5,6,7,10}
{8,9} A(cid:101)1B(cid:101)5
(cid:27)
{1,2,3,4,11}
8 {6,7,8,12} A(cid:101)4B(cid:101)3
(cid:27)
{3,4,5,6,7,8,11}
{9,10} A(cid:101)1B(cid:101)6
Table 2. Decomposition of the maximal affine subdiagrams into
their components
The second of these has vertex set S ={n+2,n+4}. Let T ={1,3,4,...,n−
1,n+1}. Once again T has type D when n = 6, E when n = 7, and E when
5 6 7
n=8.
The remaining two dotted line subgraphs appear for all n, and they cannot be
checkedusingthesamequickmethodastheothertwo. Thefirstofthesehasvertex
set S ={1,n+1}. The two associated vectors are
e =−v +v
1 1 2
e =2v +5v
n+1 0 1
so a vector v ∈K has the form
S
n
(cid:88)
v =av +2a(v +v )+ k v
0 1 2 i i
i=3
Since K ⊂K, (v,e )≤0 for all e . A consequence of this is that 2a≥k ≥...≥
S i i 3
k ≥0.
n
When n=2, (v,e )≤0⇒a≥0, and (v,e )≤0⇒a≤0, so a=0.
2 4
REFLECTION GROUPS OF THE QUADRATIC FORM −5x2+x2+...+x2 5
0 1 n
Figure2. Coxeterdiagramofthepolyhedronindimension9with
the first four vectors found by Vinberg’s algorithm
When n ≥ 3, (v,e ) ≤ 0 ⇒ k +k ≤ −a. Since 2a ≥ 0, a ≥ 0 so −a ≤ 0.
n+3 3 4
Also, since k ≥k ≥0, k +k ≥0. Therefore a=0, and so if v ∈K , v =0.
3 4 3 4 S
The other dotted line subgraph that appears for all n is the one with vertex set
S ={2,n+2}. The associated vectors are
e =−v +v
2 2 3
e =3v +5v +5v
n+2 0 1 2
A vector v ∈K has the form
s
n
(cid:88)
av +bv +(3a−b)(v +v )+ k v
0 1 2 3 i i
i=4
Since K ⊂K, (v,e )≤0 for all e . In particular this holds for i=1,...n, and we
S i i
have b≥3a−b≥k ≥...≥k ≥0. Since b≥0 and 3a−b≥0, a≥0.
4 n
When n = 2, (v,e ) = 0 ⇒ b = 3a, so v = av +3av +2av . Then (v,e ) ≤
2 0 1 2 3
0⇒a≤0, so it must be true that a=0 in which case v =0.
When n≥3, (v,e )≤0⇒a+k ≤0. Since k ≥0, a≤0. Therefore a=0,
n+3 4 4
and so if v ∈K , v =0.
S
This concludes the proof that the diagrams in figure 1 all describe hyperbolic
polytopes of finite volume.
Proposition 3. There are no finitely generated reflection groups associated to the
quadratic form −5x2+x2+...+x2 in n-dimensions for n≥9.
0 1 n
Proof. The computation is very similar to the one McLeod does in the p = 3 case
[2].
When n ≥ 9, Vinberg’s algorithm finds the same first 4 vectors as when n = 8.
The diagram for the polytope with these four vectors is shown in Figure 2. Let
S ={3,4,5,6,7,8,n+3,n+4}. LetG bethesubgraphwithvertexsetS. G has
S S
type D(cid:101)7. Removing all vertices with an edge to GS leaves just GS if n=9, and if
n > 9 leaves also the subgraph Y with vertex set {10,...,n}. Y is of type B
n−9
(B = A , B = I (4)). In order for the algorithm to eventually terminate, there
1 1 2 2
must be some set T with S ⊂T ⊂{1,...,n+4} such that G has rank n−1 and
T
G is a component of T.
S
SinceG wouldbeacomponentofG , anyvertexiinT\S couldhavenoedge
S T
toanyvertexinS. Thevectorcorrespondingtosuchavertexwouldhavezeroinner
6 ALICEMARK
product with the vectors labeled by elements of S. This imposes some conditions
on e = (cid:80)n k v . The fact that (e ,e ) = 0 for j = 3,4,5,6,7,8 implies that
i (cid:96)=0 (cid:96) (cid:96) i j
k =k =...=k . Also (e ,e )=0, so
3 4 9 i n+3
5k =2k +k +2k
0 1 2 3
and (e ,e )=0, so
i n+4
5k =k +k +5k
0 1 2 3
Combine these two equations to get k1 =3k3 and k0 = k2+58k3, so that
n
e = k2+8k3v +3k v +k v +k (v +...+v )+ (cid:88) k v
i 5 0 3 1 2 2 3 3 9 j j
j=10
Then
n
(cid:88)
5(e ,e )=4(k −2k )2+5 k2
i i 2 3 j
j=10
By the crystallographic condition, 5(e ,e ) must be 5,10,25, or 50. Subtracting
i i
4 times a square from any of these cannot yield a non-negative multiple of 5.
Therefore if this equation is to be satisfied, it must be that k −2k =0, meaning
2 3
e has the form
i
n
(cid:88)
e =2k v +3k v +2k v +k (v +...+v )+ k2
i 3 0 3 1 3 2 3 3 9 i
i=10
and
n
(cid:88)
(e ,e )= k2
i i i
i=10
When n = 9, this means Vinberg’s algorithm won’t find any vectors with no edge
to G .
S
Suppose n > 9. If (e ,e ) = 5 or 10, k is divisible by 5 for i > 0, so k2 ≥ 25.
i i i 10
But then if k (cid:54)= 0 already the norm is larger than 10, so (e ,e ) cannot be 5 or
10 i i
10. If (e ,e )=1, k =1 and k =0 for i>10. If (e ,e )=2, k =k =1 and
i i 10 i i i 10 11
k =0 for i>11. There is a family of vectors that the algorithm could find:
i
2av +3av +2av +a(v +...+v )+v
0 1 2 3 9 10
2av +3av +2av +a(v +...+v )+v +v
0 1 2 3 9 10 11
Any two of these have positive inner product with each other for any choice of
values for a, so Vinberg’s algorithm will only ever find one of them. Even if that
one has an edge to the subdiagram Y, making a B(cid:101)n−9, the combined rank with
G is only n−2 which is not enough to satisfy condition 1 of the finite volume
S
criterion.
(cid:3)
References
[1] Vinberg, E`.B.: The groups of units of certain quadratic forms. Math USSR Sbornik Vol. 16
(1972),No.1.17-35.
[2] McLeod, John: Hyperbolic reflection groups associated to the quadratic forms −3x2+x2+
0 1
...+x2.GeomDedicata(2011)152:1-16.
n
[3] Vinberg, E`. B. & Kaplinskaja, I.M.: On The Groups O18,1(Z) And O19,1(Z). Soviet Math.
Dokl.Vol.19(1978),No.1.194-197.
