Hyperbolic Unit Groups and Quaternion Algebras

0.2 MB·English

by  S. O. Juriaans

Hyperbolic Unit Groups and Quaternion Algebras
S. O. Juriaans, I. B. S. Passi∗, A. C. Souza Filho
9
January 14, 2009
0
0
2
Abstract
n
a WeclassifythequadraticextensionsK =Q[√d]andthefinitegroupsGforwhichthegroup
J ring oK[G] of G over the ring oK of integers of K has the property that the group 1(oK[G])
4 of units of augmentation 1 is hyperbolic. We also construct units in theZ-order (oUK) of the
1 quaternion algebra (K)=`−1,−1´,when it is a division algebra. H
H K

]
MathematicsSubjectClassification. Primary[16U60]. Secondary[16S34,20F67]
A
[keywords]HyperbolicGroups,QuaternionAlgebras,FreeGroups,GroupRings,Units
R
.
h 1 Introduction
t
a
m
The finite groups G for which the unit group (Z[G]) of the integral group ring Z[G] is hyper-
U
[ bolic, in the sense of M. Gromov [8], have been characterized in [13]. The main aim of this paper

is to examine the hyperbolicity of the group (o [G]) of units of augmentation 1 in the group
v3 ring oK[G] of G over the ring oK of integers ofU1a qKuadratic extension K =Q[√d] of the field Q of
1 rational numbers, where d is a square-free integer = 1. Our main result (Theorem 4.7) provides a
6 complete characterizationof such group rings o [G6 ].
K
1
2 In the integral case the hyperbolic unit groups are either finite, hence have zero end, or have
9. two or infinitely many ends (see [4, Theorem I.8.32] and [13]); in fact, in this case, the hyperbolic
0 boundaryiseitherempty,orconsistsoftwopoints,orisaCantorset. Inparticular,the hyperbolic
7 boundary is not a (connected) manifold. However, in the case we study here, it turns out that
0
when the unit group is hyperbolic and non-abelian it has one end, and the hyperbolic boundary is
:
v a compact manifold of constant positive curvature. (See Remark after Theorem 4.7.)
i
X Ourinvestigationnaturallyleadsustostudyunitsintheorder (o )ofthestandardquaternion
K
ar algebra H(K) = −1K,−1 , when this algebra is a division algebraH. We construct units, here called
Pell and Gauss units, using solutions of certain diophantine quadratic equations. In particular,
we exhibit units o(cid:0)f norm(cid:1) 1 in (o ); this construction, when combined with the deep work
in [5], helps to provide a s−et of gHenerQa[√to−r7s]for the full unit group ( (o )).
U H Q[√−7]
∗ Senior Scientist, INSA.
1
The work reported in this paper corresponds to the first chapter of the third author’s PhD
thesis [18], where analogous questions about finite semi-groups, see [10], and RA-loops, see [14],
have also been studied.
2 Preliminaries
Let Γ be a finitely generated group with identity e and S a finite symmetric set of generators
of Γ, e S. Consider the Cayley graph = (Γ,S) of Γ with respect to the generating set S and
6∈ G G
d = d the corresponding metric (see [4, chap. 1.1]). The induced metric on the vertex set Γ of
S
(Γ,S) is then the word metric: for γ , γ Γ, d(γ , γ ) equals the least non-negative integer n
1 2 1 2
sGuch that γ1−1γ2 = s1s2...sn, si ∈ S . Reca∈ll that in a metric space (X, d), the Gromov product
(y.z) of elements y, z X with respect to a given element x X is defined to be
x
∈ ∈
1
(y z) = (d(y,x)+d(z,x) d(y,z)),
x
· 2 −
andthatthemetricspaceX issaidtobehyperbolicifthereexistsδ 0suchthatforallw, x, y, z
≥ ∈
X,
(x.y) min (x.z) , (y.z) δ.
w w w
≥ { }−
The group Γ is said to be hyperbolic if the Cayley graph with the metric d is a hyperbolic
S
G
metric space. This is a well-defined notionwhichdepends only onthe groupΓ, andis independent
of the chosen generating set S (see [8]).
A map f : X Y between topological spaces is said to be proper if f 1(C) X is compact
−
→ ⊆
wheneverC Y iscompact. ForametricspaceX,twopropermaps(rays)r , r :[0, [ X are
1 2
⊆ ∞−→
definedtobe equivalentif,foreachcompactsetC X,thereexistsn N suchthatr ([n, [),i=
i
⊂ ∈ ∞
1, 2, are in the same path component of X C. Denote by end(r) the equivalence class of the ray
\
r, by End(X) the set of the equivalence classes end(r), and by End(X) the cardinality of the set
| |
End(X). The cardinality End( , d ) for the Cayley graph ( , d ) of Γ does not depend on the
S S
| G | G
generating set S; we thus have the notion of the number of ends of the finitely generated group Γ
(see [4], [8]).
We next recall some standard results from the theory of hyperbolic groups:
1. Let Γ be a group. If Γ is hyperbolic, then Z2֒ Γ, where Z2 denotes the free Abelian group
6→
of rank 2. [4, Corollary III.Γ3.10(2)]
2. An infinite hyperbolic group contains an element of infinite order. [4, Proposition III.Γ2.22]
3. If Γ is hyperbolic, then there exists n = n(Γ) N such that H n for every torsion
∈ | | ≤
subgroup H <Γ. [4, Theorem III.Γ3.2] and [7, Chapter 8, Corollaire 36]
These results will be used freely in the sequel. In view of (1) above, the following observation is
quite useful.
2
Lemma 2.1. Let A be a unital ring whose aditive group is torsion free, and let θ , θ A be two
1 2
∈
2-nilpotent commuting elements which are Z-linearly independent. Then (A) contains a subgroup
U
isomorphic to Z2.
Proof. Set u=1+θ and v =1+θ . It is clear that u, v (A) and both have infinite order. If
1 2
∈U
1 = w u v then there exists i, j Z 0 , such that, ui = w = vj. Since ui = 1+iθ and
1
6 ∈ h i∩h i ∈ \{ }
vj =1+jθ ,itfollowsthatiθ jθ =0andhence θ , θ isZ-linearlydependent,acontradiction.
2 1 2 1 2
− { }
Hence Z2 u, v (A).
≃h i⊆U
LetC denote the cyclic groupofordern, S the symmetric groupofdegree 3,D the dihedral
n 3 4
groupof order 8, and Q the split extension C ⋊C . Let K be an algebraicnumber field and o
12 3 4 K
its ring of integers. The analysis of the implication for torsion subgroups G of a hyperbolic unit
group (Z[Γ]) leading to [13, Theorem 3] is easily seen to remain valid for torsion subgroups of
hyperbUolic unit groups (o [Γ]). We thus have the following:
K
U
Theorem 2.2. A torsion group G of a hyperbolic unit group (o [Γ]) is isomorphic to one of the
K
U
following groups:
1. C , C , C , an Abelian group of exponent dividing 4 or 6;
5 8 12
2. a Hamiltonian 2-group;
3. S , D , Q , C ⋊C .
3 4 12 4 4
We denote by (K) = (a,b) the generalized quaternion algebra over K:
H K
(K) = K[i, j : i2 = a, j2 = b, ji = ij =: k]. The set 1, i, j, k is a K-basis of (K).
H − { } H
Such an algebra is a totally definite quaternion algebra if the field K is totally real and a, b are
totally negative. If a, b o , then the set (o ), consisting of the o -linear combinations of the
K K K
elements 1, i, j and k, ∈is an o -algebra. WHe denote by N the norm map (K) K, sending
K
H →
x=x +x i+x j+x k to N(x)=x2 ax2 bx2+abx2.
1 i j k 1− i − j k
Let d = 1 be a square-free integer, K = Q[√d]. Let us recall the basic facts about the ring of
integers o6 (see, for example, [11], or [16]). Set
K
√d, if d 2 or 3 (mod 4)
ϑ= ≡
((1+√d)/2, if d 1 (mod 4).
≡
Then o =Z[ϑ] and the elements 1, ϑ constitute a Z-basis of o . If d<0, then
K K
1, ϑ , if d= 1,
{± ± } −
(o )= 1, ϑ, ϑ2 , if d= 3, (1)
K 
U {± ± ± } −
 1 , otherwise.
{± }
If d>0, then there exists a unique unit ǫ>1, called the fundamental unit, such that
(o )= ǫ . (2)
K
U ±h i
3
We need the following:
Proposition 2.3. Let K =Q[√d], with d=1 a square-free integer, be a quadratic extension of Q,
and u (o ). Then ui 1 (mod 2), whe6re
K
∈U ≡
1 if d 1 (mod 8),
≡
i= 2 if d 2, 3 (mod 4),

≡
3 if d 5 (mod 8).
≡

Proof. The assertionfollows immediately on considering the prime factorization of the ideal 2o ,
K
see [3, Theorem 1, p.236].
3 Abelian groups with hyperbolic unit groups
Proposition 3.1. Let R be a unitary commutative ring, C = g . Then u = a + (1 a)g,
2
h i −
a R 0, 1 is a non-trivial unit in (R[C ]) if, and only if, 2a 1 (R).
1 2
∈ \{ } U − ∈U
Proof. Let C = g and suppose that u = a+(1 a)g, a R 0, 1 is a non-trivial unit in
2
h i − ∈ \{ }
R[C ] having augmentation 1. Let ρ : R[C ] M (R) be the regular representation. Clearly
2 2 2
→
a 1 a
ρ(u)= − . Since u is a unit, it follows that 2a 1=detρ(u) (R).
1 a a − ∈U
(cid:18) − (cid:19)
Conversely, let a R 0,1 be such that e = 2a 1 (R). It is then easy to see
∈ \ { } − ∈ U
that u = a + (1 a)g is a non-trivial unit in R[C ] with inverse v = ae 1 + (1 ae 1)g.
2 − −
− −
[9, Proposition I]
Proposition 3.2. The unit group (o [C ]) is trivial if, and only if, K = Q or an imaginary
K 2
U
quadratic extension of Q, i.e., d<0.
Proof. It is clear from the description (1) of the unit group of o that the equation
K
2a 1=u, a o 0, 1 , u (o ) (3)
k K
− ∈ \{ } ∈U
does not have a solution when K =Q or d<0.
Suppose d > 1 and ǫ is the fundamental unit in o . In this case we have (o ) = ǫ . By
K K
Proposition 2.3, ǫi 1+2o for some i 1, 2, 3 . Consequently the equationU(3) has a±sholiution
K
and so, by Proposit∈ion 3.1, (o [C ]) is∈no{n-trivia}l.
K 2
U
4
Theorem 3.3. Let o be the ring of integers of a real quadratic extension K = Q[√d], d > 1 a
K
square-free integer, ǫ>1 the fundamental unit of o and C = g . Then
K 2
h i
1+ǫn 1 ǫn
U1(oK[C2])∼=hgi×h 2 + −2 gi∼=C2×Z,
where n is the order of ǫ mod 2o .
K
Proof. Letu (o [C ])beanon-trivialunit. Then,thereexistsa o suchthat,2a 1= ǫm
1 K 2 K
for some non-z∈erUo integer m. Since n is the order of ǫ mod 2o , m=∈nq with q Z. We−thus h±ave
K
∈
u=a+(1 a)g
−
= 1 ǫm + 1 ǫmg
±2 ∓2
= 1 ǫnq + 1 ǫnqg
±2 ∓2
= 1+ǫn + 1 ǫng q, or g 1+ǫn + 1 ǫng q.
2 −2 2 −2
Hen(cid:0)ce U1(oK[C2])(cid:1)∼=hgi×(cid:0)h1+2ǫn + 1−2ǫng(cid:1)i∼=C2×Z.
As an immediate consequence of the preceding analysis, we have:
Corollary 3.4. If K is a quadratic extension of Q, then (o [C ]) is a hyperbolic group.
1 K 2
U
Corollary 3.5. Let G be a non-cyclic elementary Abelian 2-group. Then (o [G]) is hyperbolic
1 K
if, and only if, o is imaginary. U
K
Proof. Suppose o is real. Since G is notcyclic, there existg,h G, g = h, o(g)=o(h)=2. By
K
T1he.orTemhe3re.3fo,rUe1(oK(o[hg)i])co∼=ntCai2n×sZan∼=AUb1(eoliKan[hhgir]o)u.pSionfcerahngki∩2h,hiso=∈it{1is},nUo16t(ohKyp[hegrib])o∩licU.1(CooKn[vhheris])el=y,
1 K
i{f }o is imaginaUry, then, proceeding by induction on the order G of G, we can conclude that
K
(o [G]) is trivial, and hence is hyperbolic. | |
1 K
U
For an Abelian group G, we denote by r(G) its torsion-free rank. In order to study the hy-
perbolicity of (o [G]), it is enough to determine the torsion-free rank r( (o [G])). Since
1 K 1 K
iUm(aoKgi[nGa]r)y∼=extUeUn(osKio)n,×thUe1n(orK([G(o]),[Gwe]))h=aver(r(U(o1(o[KG[]G))],))w=herre(aUs(iofKK[Gis])a) −rearl(UqUu(oaKdr)a)t.icIefxKtenissioann,
K 1 K
then r( (o ))=1, and therUefore U
K
U
r( (o [G]))=r( (o [G])) 1.
1 K K
U U −
We note that
Q[Cn]∼=⊕ Q[ζd],
dn
X|
5
where ζ is a primitive dth root of unity, and therefore, for any algebraic number field L,
d
L[Cn]∼=⊕ L⊗QQ[ζd].
dn
X|
We say that two groups are commensurable with each other when they contain finite index
subgroups isomorphic to each other. Since the unit group (o [C ]) is commensurable with (Λ),
L n
whereΛ= o ,weessentiallyneedtocomputeUthetorsion-freerankofo foUrthe
needed case⊕s. d|n L⊗Q[ζd] K⊗Q[ζd]
P
Proposition 3.6. Let K = Q[√d], with d a square-free integer = 1. The table below shows the
torsion-free rank of the groups (o [C ]), n 2, 3, 4, 5, 6, 8 . 6
1 K n
U ∈{ }
n r( (o [C ])) n r( (o [C ]))
1 K n 1 K n
U U
1 if d<0,d= 3
0 if d<0 6 −
2 3 0 if d= 3
1 if d>1 −
1 if d>1
1 if d< 1 6 if d<0
−
4 0 if d= 1 5 2 if d=5
−
2 if d>1 6 if d Z+ 1,5
∈ \{ }
4 if d< 1
2 if d< 3 −
− 1 if d= 1
6 0 if d= 3 8 −
− 4 if d=2
3 if d>1
5 if d>2
In all the cases, the computation is elementary and we omit the details.
Theorem 3.7. If K =Q[√d], with d a square-free integer =1, then
6
1. (o [C ]) is hyperbolic;
1 K 3
U
2. (o [C ]) is hyperbolic if, and only if, d<0;
1 K 4
U
3. for an Abelian group G of exponent dividing n>2, the group (o [G]) is hyperbolic if, and
1 K
U
only if, n=4 and d= 1, or n=6 and d= 3 ;
− −
4. (o [C ]) is hyperbolic if, and only if, d= 1;
1 K 8
U −
5. (o [C ]) is not hyperbolic.
1 K 5
U
6
Proof. The Proposition 3.6 gives us the torsion-free rank
r:=r( (o [C ]))
1 K n
U
for n 2, 3, 4, 5, 8 . The group (o [C ]) is hyperbolic if, andonly if, r 0, 1 . Thus, itonly
1 K n
∈{ } U ∈{ }
remains to consider the case (3).
Suppose n = 6 and (o [G]) is hyperbolic. We, hence, have r 0,1 . If G is cyclic, then,
1 K
U ∈ { }
by Proposition 3.6, we have d = 3. If G is not cyclic, then G = Cl Cm, l, m 1. Since
o [C ]֒ o [G], it follows that d−= 3. ∼ 2 × 3 ≥
K 3 K
→ −
Conversely, if n = 6 and d = 3 then, proceeding by induction on G, it can be proved that
(o [G]) is hyperbolic. − | |
1 K
U
The case n=4 can be handled similarly.
Proposition3.8. IfK =Q[√d],withdsquare-freeinteger=1,then (o [C ])isnothyperbolic.
1 K 12
6 U
Psiornosofo. S[Cinc]e֒K[Co12[]C∼=]Kan⊗dQo[Q[[CC1]2]֒] ∼=oK[⊗CQ]Q.[CT3h×ereCfo4r])e,∼=r(K[C(o3 ×[CC4]]),)we hra(ve(tohe[Cim]m))e+r-
K 3 K 12 K 4 K 12 1 K 12 1 K 3
r( (o [C ]))→. → U ≥ U
1 K 4
U
Suppose (o [C ])ishyperbolic. Then,sincer( (o [C ]))<2,wehave,bytheProposition
1 K 12 1 K 12
U U
3.6, d 3, 1 . We also have
∈{− − }
K[C C ]=(K[C ])[C ]=(K K[√ 3])[C ]=
3× 4 ∼ 3 4 ∼ ⊕ − 4 ∼
K[C ] (K[√ 3])[C ]=2K K[√ 1] 2K[√ 3] K[√ 3+√ 1].
4 ⊕ − 4 ∼ ⊕ − ⊕ − ⊕ − −
Set L = Q[√ 3+√ 1] and suppose d = 3. Then o [C ] ֒ 4o 2o and r( (o )) = 1.
K 12 K L L
Thus r( (o [C −]))=2,−and we have a contra−diction. → ⊕ U
K 12
U
Analogously,for d= 1, o [C ]֒ 3o 3o and so r( (o [C ]))=3. Since the extensions
K 12 K L K 12
are non-real, we have th−at r( (o [→C ])) =⊕ r( (o [C ])U) 2, and, hence, we again have a
1 K 12 K 12
U U ≥
contradiction.
We conclude that (o [C ]) is not hyperbolic.
1 K 12
U
4 Non-Abelian groups with hyperbolic unit groups
Theorem 2.2 classifies the finite non-Abelian groups G for which the unit group (Z[G])
1
U
is hyperbolic. These groups are: S , D , Q ,C ⋊ C , and the Hamiltonian 2-group, where
3 4 12 4 4
Q =C ⋊C , with C acting non-trivially on C , and also on C (see [13]).
12 3 4 4 3 4
7
E. Jespers, in [12], classified the finite groups G which have a normal non-Abelian free com-
plement in (Z[G]). The group algebra Q[G] of these groups has at most one matrix Wedderburn
U
component which must be isomorphic to M (Q).
2
Lemma4.1. LetGbeagroupandK aquadraticextension. IfM (K)is aWedderburn component
2
of K[G] then Z2 ֒ (o [G]). In particular, (o [G]) is not hyperbolic.
1 K 1 K
→U U
Proof. The ring Γ =M (o ) is a Z-order in M (K) and
2 K 2
X = e , e √d Γ
12 12
{ }⊂
is a set of commuting nilpotent elements of index 2, where e denotes the elementary matrix. The
ij
set 1,√d is a linearly independent set over Q, and hence so is X. Therefore, by Lemma 2.1,
Z2 ֒{ (Γ}) (o [G]), and so, (o [G]) is not hyperbolic.
1 1 K 1 K
→U ⊂U U
Corollary 4.2. If G S , D , Q , C ⋊C then (o [G]) is not hyperbolic.
3 4 12 4 4 1 K
∈{ } U
Proof. We have that K[G]∼=K⊗Q(Q[G]). For each of the groups under consideration,M2(Q) is
a Wedderburn component of Q[G]; it therefore follows that M (K) is a Wedderburn component of
2
K[G]. The preceding lemma implies that (o [G]) is not hyperbolic.
1 K
U
IfH isanon-AbelianHamiltonian2-group,thenH =E Q ,whereE isanelementaryAbelian
8
×
2-groupand Q is the quaterniongroupof order 8. Since Q contains a cyclic subgroupof order4,
8 8
it follows, by Theorem 3.7, that if (o [Q ]) is hyperbolic, then o is not real.
1 K 8 K
U
Proposition 4.3. If G is a Hamiltonian 2-group of order greater than 8, then (o [G]) is not
1 K
U
hyperbolic.
Proof. Let G = E Q with E elementary Abelian of order 2n > 1. We then have K[G] =
8
K[E×Q8] ∼= K ⊗Q(×Q[E×Q8]) ∼= K ⊗Q(Q[E])[Q8] ∼= K ⊗Q(2nQ)[Q8] ∼= (2nK)[Q8]. If d = −1
it is well known that KQ has a Wedderburn component isomorphic to M (K) and hence, by
8 2
Lemma 4.1, (o Q ) is not hyperbolic. If d < 1, then, by Proposition 3.6, r( (o [C ])) = 1.
1 K 8 1 K 4
Since C is aUsubgroup of Q , it follows that (−(2no )[C ]) embeds into (o [UG]). Thus, since
4 8 1 K 4 1 K
( o [C ]) has rank 2n 2, (o [G]) isUnot hyperbolic. U
U1 2n K 4 ≥ U1 K
Q
In view of the above Proposition, it follows that Q is the only Hamiltonian 2-group for which
8
(o [G]) can possibly be hyperbolic, and in this case o is the ring of integers of an imaginary
1 K K
U
extension. By Lemma 4.1, K[Q ] can not have a matrix ring as a Wedderburn component. Since
8
Q[Q8] ∼= 4Q⊕H(Q), we have K[Q8] ∼= K ⊗Q(4Q⊕H(Q)) ∼= 4K ⊕H(K); hence K[Q8] must be
a direct sum of division rings, or equivalently, has no non-zero nilpotent elements. In particular,
(K) is a division ring.
H
8
Theorem 4.4. Let K = Q[√d], with d square-free integer = 1. Then K[Q ] is a direct sum of
8
6
division rings if, and only if, one of the following holds:
(i) d 1 (mod 8);
≡
(ii) d 2, or 3 (mod 4), or d 5 (mod 8), and d>0.
≡ ≡
Proof. The assertion follows from [1, Theorem 2.3] ; [3, Theorem 1, p.236] and [17, Theorem
3.2].
Corollary 4.5. If K =Q[√d], where dis anegative square-free integer, then thegroup (o [Q ])
1 K 8
U
is not hyperbolic if d 1 (mod 8).
6≡
Let H : C ]0, [ be the upper half-space model of three-dimensional hyperbolic space and
× ∞
Iso(H) its group of isometries. In the quaternion algebra := ( 1, 1) over R, with its usual
H H − −
basis, we may identify H with the subset z+rj : z C,r R+ . The group PSL(2, C) acts on
{ ∈ ∈ }
H in the following way:
ϕ: PSL(2, C) H H
× −→
a b
(M, P) P :=(aP +b)(cP +d) 1,
7→ c d −
(cid:18) (cid:19)
where (cP +d) 1 is calculated in . Explicitly, MP =M(z+rj)=z +r j, with
− ∗ ∗
H
(az+b)(cz+d)+acr2 r
z = , and r = .
∗ ∗
cz+d2+ c2r2 cz+d2+ c2r2
| | | | | | | |
Let K be an algebraic number field and o its ring of integers. Let
K
SL ( (o )):= x (o ):N(x)=1 ,
1 K K
H { ∈H }
where N is the norm in (K). Clearly the groups ( (o )) and (o ) SL ( (o )) are com-
K K 1 K
H U H U × H
mensurable. Consider the subfield F = K[i] (K) which is a maximal subfield in (K). The
⊂ H H
inner automorphism σ,
σ : (K) (K)
H −→ H
x jxj 1,
−
7→
fixes F. The algebra (K)=F Fj is a crossed product and embeds into M (C) as follows:
2
H ⊕
Ψ: (K) ֒ M (C)
2
H →
x y (4)
x+yj .
7→ σ(y) σ(x)
(cid:18) − (cid:19)
This embedding enables us to viewSL ( (o )) andSL ( (K)) assubgroupsofSL(2, C)and
1 K 1
H H
hence SL ( (K)) acts on H.
1
H
9
Proposition 4.6. Let K = Q[√d], d 1 (mod 8) a square-free negative integer, and o its ring
K
of integers. Then ( (o )) and (o ≡[Q ]) are hyperbolic groups.
K K 8
U H U
Proof. ObservethatSL ( (o ))actsonthespaceHand,hence,isadiscretesubgroupofSL (C)
1 K 2
(see[6,Theorem10.1.2,p.4H46]). ThequotientspaceY :=H/SL ( (o ))isaRiemannianmanifold
1 K
of constant curvature −1 and, since H is simply connected, we hHave that SL1(H(oK)) ∼= π1(Y).
Since d 1 (mod 8), (K) is a division ring and, therefore, co-compactand Y is compact (see [6,
Theorem≡10.1.2, itemH(3)]). Hence SL ( (o )) is hyperbolic (see [2, Example 2.25.5]). Since
1 K
( (o )) and (o ) SL ( (o )) areHcommensurable and (o ) = 1, 1 , it follows that
K K 1 K K
UUU((HoHK(o)K∼=))C2is, whyeUpceornbcollui×cd.eSthinacteHUU(o(oKK[Q[Q8]8)])is∼=hyUp(eorKbo)l×ic.U(oK)×UU(oK)×{U−(oK)}×U(H(oK)) and
Combining the results in the present and the preceding section, we have the following main
result.
Theorem 4.7. Let K = Q[√d], with d square-free integer = 1, and G a finite group. Then
(o [G]) is hyperbolic if, and only if, G is one of the groups l6isted below and o (or K) is deter-
1 K K
U
mined by the corresponding value of d:
1. G C , C and d arbitrary;
2 3
∈{ }
2. G is an Abelian group of exponent dividing n for:
n=2 and d<0, or n=4 and d= 1, or n=6 and d= 3.
− −
3. G=C and d<0.
4
4. G=C and d= 1.
8
−
5. G=Q and d<0 and d 1 (mod 8).
8
≡
RtheemEaurckli.deIfanthsepghreoruepoUf(doiKm[eQn8s]i)onis2h,ypaenrdboElincdtsh(en(tohe[Qhyp])e)rbhoalsicobnoeuneldeamryen∂t(U(s(eoeK[[2Q,8E])x)a∼=mSp2le,
K 8
U
2.25.5]). Note that if (Z[G]) is an infinite non-Abelian hyperbolic group, then ∂( (Z[G])) is
U U
totally disconnected and is a Cantor set. So, in this case, (Z[G]) has infinitely many ends and
also is a virtually free group,( [13,Theorem2]and [8, 3]). UHowever,if (o [G]) is a non-Abelian
K
hyperbolic group, then (o [G]) is an infinite group w§hich is not virtuUally free, has one end and
K
U
∂( (Z[G])) is a smooth manifold.
U
10