Two dimensional Complex Kleinian Groups With Four Complex Lines in General Position in its Limit Set

0.18 MB·English

by  Waldemar Barrera

TWO DIMENSIONAL COMPLEX KLEINIAN GROUPS WITH
FOUR COMPLEX LINES IN GENERAL POSITION IN ITS
LIMIT SET
0
1
0 W. BARRERA,A. CANO & J. P.NAVARRETE
2

n
Abstract. In this article we provide an algebraic characterization of
a
those groups of PSL(3,C) whose limit set in the Kulkarni sense has,
J
exactly, four lines in general position. Also we show that, for this class
8
of groups, the equicontinuity set of the group is the largest open set
2
where thegroup acts discontinuously and agrees with thediscontinuity

set of thegroup.
]
S
D
.
h
Introduction
t
a
m In a recent article, see [1], we have proven that for a complex Kleinian
[ group without proper invariant subspaces and ”enough” lines in the Kulka-

rni’slimit set, it holds that its discontinuity set agrees with the equicontinu-
1
ity set of the group, is the largest open set where the group acts discontinu-
v
2 ously and is a holomorphy domain. Such result enable us to understand the
2 relationship, in the two dimensional case and for a ”large class” of groups,
2
bettwen the different notions of limit sets which are usually studied as well
5
. as its geometry, see [3]. This article is a step to extent the results in [1] to
1
the case when the groups has invariant subspaces and ”enough” lines in the
0
0 limit set. More precisely we prove:
1
:
v Theorem 0.1. Let Γ ⊂ PSL(3,C) be a discrete group. The limit set, in the
i
X Kulkarni sense, of Γ has exactly four lines in general position if and only if
r Γ has a hyperbolic toral group, see section 2, whose index is at most 8.
a
Theorem 0.2. Let Γ ⊂ PSL(3,C) be a toral group. Thus the discontinuity
set in the Kulkarni sense agrees with the equicontinuity region and is given
by:
Ω(Γ)= Hǫ1 ×Hǫ2,
ǫ1,ǫ[2=±1
where H+1 and H−1 are the upper half and lower half plane. Moreover Ω(Γ)
is the largest open set on which Γ acts properly discontinuously.
1991 Mathematics Subject Classification. Primary: 32Q45, 37F45; Secondary 22E40,
57R30.
Key words and phrases. kleinian groups, projective complex plane, discrete groups,
limit set.
Research partially supported bygrants from CNPq.
1
2 W.BARRERA,A.CANO&J.P.NAVARRETE
This article is organized as follows: in section 1 we introduce some terms
and notations which will be used along the text. In section 2 we construct
some examples of groups with four lines in general position. Finally in sec-
tion 3, we present the proof of theorem 0.1.
The authors are grateful to Professor Jos´e Seade for stimulating conver-
sations. Part of this research was done while the authors were visiting the
IMATE-UNAM campus Cuernavaca and the FMAT of the UADY. Also,
during this time, the second author was in a postdoctoral year at IMPA,
and they are grateful to these institutions and its people, for their support
and hospitality.
1. Preliminaries and Notations
1.1. Projective Geometry. We recall that the complex projective plane
P2 is
C
P2 := (C3\{0})/C∗,
C
where C∗ acts on C3\{0} by the usual scalar multiplication. This is a com-
pact connected complex 2-dimensional manifold. Let [ ] : C3 \{0} → P2
C
be the quotient map. If β = {e ,e ,e } is the standard basis of C3, we
1 2 3
will write [e ] = e and if w = (w ,w ,w ) ∈ C3 \{0} then we will write
j j 1 2 3
[w] = [w :w : w ]. Also, ℓ ⊂ P2 is said to bea complex line if [ℓ]−1∪{0} is
1 2 3 C
a complex linear subspace of dimension 2. Given p,q ∈ P2 distinct points,
C
there is a unique complex line passing through p and q, such line will be
←→
denoted by p,q.
Consider the action of Z (viewed as the cubic roots of the unity) on
3
SL(3,C) given by the usual scalar multiplication, then
PSL(3,C) = SL(3,C)/Z
3
is a Lie group whose elements are called projective transformations. Let
[[ ]] : SL(3,C) → PSL(3,C) be the quotient map, γ ∈ PSL(3,C) and
γ ∈ GL(3,C), we will say that γ˜ is a lift of γ if there is a cubic root τ of
Det(γ) such that [[τγ]] = γ, also, we will use the notation (γ ) to denote
ij
elements in SL(3,C). Onecan show that PSL(3,C) is a Lie group that acts
e
transitively, effectively andbybiholomorphismsonP2 by[[γ]]([w]) = [γ(w)],
e C
where w ∈ C3\{0} and γ ∈ SL (C).
3
1.2. Complex Kleinian Groups. Let Γ ⊂ PSL(3,C) be a subgroup. We
define(following Kulkarni,see[5]): thesetL (Γ)as theclosureof thepoints
0
in P2 with infinite isotropy group. The set L (Γ) as the closure of the set of
C 1
cluster points of Γz where z runs over P2 \L (Γ). Recall that q is a cluster
C 0
point for ΓK, where K ⊂ P2 is a non-empty set, if there is a sequence
C
(k ) ⊂ K and a sequence of distinct elements (γ ) ⊂ Γ such that
m m∈N m m∈N
γ (k ) // q. ThesetL (Γ)as theclosureof cluster points of ΓK where
m m m→∞ 2
TWO DIMENSIONAL COMPLEX KLEINIAN GROUPS WITH FOUR COMPLEX LINES IN GENERAL POSITION IN
K runs over all the compact sets in P2 \(L (Γ)∪L (Γ)). The Limit Set in
C 0 1
the sense of Kulkarni for Γ is defined as:
Λ(Γ) = L (Γ)∪L (Γ)∪L (Γ).
0 1 2
The Discontinuity Region in the sense of Kulkarni of Γ is defined as:
Ω(Γ) =P2 \Λ(Γ).
C
We will say that Γ is a Complex Kleinian Group if Ω(Γ) 6= ∅.
Lemma 1.1. ( See [2]) Let Γ ⊂ PSL (C) be a subgroup, p ∈ P2 such that
3 C
Γp = p and ℓ a complex line not containing p. Define Π = Π : Γ −→
p,ℓ
Bihol(ℓ) given by Π(g)(x) = π(g(x)) where π = π : P2 − {p} −→ ℓ is
p,ℓ C
←→
given by π(x) = x,p∩ℓ, then:
(i) π is a holomorphic function.
(ii) Π is a group morphism.
(iii) If Ker(Π) is finite and Π(Γ) is discrete, then Γ acts discontinu-
←→
ously on Ω = ( z,p) − {p}. Here Ω(Π(Γ)) denotes the
z∈Ω(Π(Γ))
discontinuity set of Π(Γ).
S
(iv) If Γ is discrete, Π(Γ) is non-discrete and ℓ is invariant, then Γ acts
←→
discontinuously on Ω = z,p−(ℓ∪{p}).
z∈Eq(Π(Γ))
Lemma 1.2. Let Σ ⊂ PSL(2,CS) be a non discrete group, then:
(i) The set P1 \Eq(Σ) is either, empty, one points, two points, a circle
C
or P1.
C
(ii) If C is an invariant closet set which contains at least 2 points, then
P1 \Eq(Σ) ⊂ΣC.
C
(iii) The set P1 \Eq(Σ) is the closure of the loxodromic fixed points.
C
1.3. Counting Lines.
Definition 1.3. Let Ω ⊂ P2 be a non-empty open set. Let us define:
C
(i) The lines in general position outside Ω as:
LG(Ω)= L⊂Gr (P2)|The lines in L are in general position & L⊂P2 \Ω ;
1 C C
(ii) Tnhe number of lines in general position outside Ω a[s: o
LiG(Ω) = max({card(L) :L ∈ LG(Ω)}),
where card(C) denotes the number of elements contained in C.
(iii) Given L ∈LG(Ω) and v ∈ L, we will say that v is a vertex for L
if there are ℓ ,ℓ ∈ L distinct lines and an infinite set C ⊂ Gr (P2)
1 2 1 C
S
such that ℓ ∩ℓ ∩( C) = {v} and C ⊂ P2 \Ω.
1 2 C
Proposition 1.4. Let Γ ⊂ PSL(3C) be a complex Kleinian group. If
T S
LiG(Ω(Γ)) = 4, then for each L ∈ GL(Ω) with card(L) = 4, it falls out
that:
(i) The array of lines L contains exactly two vertexes;
(ii) For every vertex v of L), it follows that Isot(v,Γ) is a subgroup of
Γ with finite index.
4 W.BARRERA,A.CANO&J.P.NAVARRETE
2. Toral Groups
Definition 2.1. Let A∈ SL(2,Z), then A is said to be a Hyperbolic Toral
Automorphism if none of the eigenvalues of A lies on the unit circle.
Theorem 2.2. Let A∈ SL(2,Z) be an hyperbolic toral automorphism then:
(i) The eigenvalues of A are irrational numbers.
(ii) It holds
{x ∈ R2 : An(x)−x ∈ Z×Z for some n ∈ N}= Q×Q.
Definition 2.3. Set S :Q×Q → N which is given by:
S(x) = min{n ∈ N :nx ∈ Z×Z}.
Also define Per : SL(2,Z)×Q×Q → N by
Per(A,x) = min{n ∈ N : Bn(x)−x ∈ Z×Z}.
Finally define φ:SL(2,Z)×Q×Q×Z → Q×Q by
l−1 Bj(x) if l > 0;
j=0
φ(B,x,l) = 0 if l = 0; .

P
 − lj=1B−j(x) if l < 0.
The following straightforward lPemmas will be usefull

Lemma 2.4. Let B ∈ SL(2,Z), ν ∈ Q2 and r,s,l ∈ N with 0 < r,s <
Per(B,ν) . If l = KPer(B,ν)+r and K = n˜S(ν)+t, where K,r,n˜,t ∈ N
are given by the division theorem,
δ = −Br(φ(B,B−KPer(B,ν)(ν)−ν),l);
1
δ = −BPer(B,ν)−r(φ(B,B−Per(B,ν)(ν)−ν,l));
2
δ = S(ν)BPer(B,ν)−R(φ(B,ν,l));
3
δ = K−1φ(B,BiPer(B,ν)(ν)−ν,Per(B,ν));
4 i=0
δ = φ(B,BKPer(B,ν)(ν)−ν,r);
5
P
δ = n˜S(ν)φ(B,ν,Per(B,ν));
6
0 if r+s < Per(B,ν)
δ =
7 φ(B,BPer(B,ν)(ν)−ν,r+s−Per(B,ν)) D.O.F.
(cid:26)
r+sBj(ν) if r+s < Per(B,ν)
δ = jr ,
8 ( PjPrer(B,ν)−1Bj(ν)+φ(B,ν,r+s−Per(B,ν)) D.O.F.
thus δ1,δ2P,δ3,δ4,δ5,δ6,δ7 ∈ Z2 and
φ(B,ν,−l)= δ +δ +δ +(S(ν)−1)BPer(B,ν)−r(φ(B,ν,l));
1 2 3
(2.1) φ(B,ν,l) = δ +δ +δ +tφ(B,ν,Per(B,ν))+φ(B,ν,r);
4 5 6
Brφ(B,ν,s+1) = δ +δ
7 8
Lemma 2.5. Let (a ),(b ) ⊂ C be sequences, then:
m m
(i) If (a ) and (b ) diverges, then the accumulation points of
m m
{[a : b : 1] :m ∈ N}
m m
←−→
lies on e ,e ;
1 2
TWO DIMENSIONAL COMPLEX KLEINIAN GROUPS WITH FOUR COMPLEX LINES IN GENERAL POSITION IN
(ii) If (a ) converges and (b ) diverges, then [a : b :1] // [e ];
m m m m m→∞ 2
(iii) If (a ) diverges and (b ) converges, then [a : b :1] // [e ];
m m m m m→∞ 1
(iv) If k = [a : b : 1] // [z : 0 : 1], where z 6= 0, then there is a
m m m m→∞
subsequence of (k ), denoted (k˜ = [a˜ : ˜b : 1]), such that (a˜ )
m m m m m
and (˜b ) are convergent.
m
Definition 2.6. Let A,B ∈ SL(2,Z), ν ∈ Q×Q, b ∈M(1×2,Z), k,l ∈ Z,
thus we define:
AkBl b+φ(ν,l)
hk : l : b :νi = .
0 1
(cid:18) (cid:19)
From Lemma 2.4, it follows easily
Corollary 2.7. Let A,B ∈SL(2,Z), ν ∈ Q×Q, b ∈ M(1×2,Z), k,l ∈ Z,
then there are m ,...,m ∈ {0,...,S(ν) − 1} and B ∈ Z2 such
0 Per(B,ν)−1
that:
AkBl B + Per(B,ν)−1m Bj(ν)
hk :l :b : νi= j=0 j ,
0 1 !
P
Proposition 2.8. Let A,B ∈ SL(2,Z) be such that the group generated
by A, B is isomorphic to Z×Z and each element in < A,B > \{Id} is a
hyperbolic toral automorphism, also let ν ∈ Q×Q be such that A(ν)−ν ∈
Z×Z. Then
Γ = {hk : l :b : νi| k,l ∈Z, b ∈ M(1×2,Z)}
A,B,ν
is a complex Kleinian group. Moreover Ω(Γ ) is projectively equivalent
A,B,ν
to:
Hǫ1 ×Hǫ2.
ǫ1,ǫ2[∈{±1}
Proof. Let a = hk : l : b : νi,b = hk : l : b : νi ∈ Γ . Thus an easy
1 1 1 2 2 2 A,B,ν
calculation shows:
ab−1 = hk −k : l −l : b +b +b i
1 2 1 2 1 3 4
where
b = −Ak1Bl1(A−k2B−l2(b )+B−l2φ(A−k2(ν)−ν,l ));
3 2 2
b = Bl1(φ(Ak1(ν)−ν,−l )).
4 2
Since b ,b ∈ Z×Z, it follows that Γ is a group.
3 4 A,B,ν
Now, since A,B ∈ SL(2,Z) are commuting hyperbolic toral automor-
phism, it follows that there is Tˆ ∈ SL(2,R) such that TˆATˆ−1,TˆBTˆ−1 are
diagonal matrices. Set
Tˆ 0 α 0 β 0
T = ; TAT−1 = ; TBT−1 = ;
0 1 0 α−1 0 β−1
(cid:18) (cid:19) (cid:18) (cid:19) (cid:18) (cid:19)
6 W.BARRERA,A.CANO&J.P.NAVARRETE
where α,β ∈ R \ {±1} and Tˆ(1,0),Tˆ(0,1) ∈ R2. Moreover, given b ∈
M(1×2,Z) and k,l ∈ Z and taking ν = (ν ,ν ), by Corollary 2.7 there are
1 2
m ,...,m ∈ {0,...,S(ν)−1} and b ,b ∈ Z such that
0 Per(B,ν)−1 1 2
(2.2) Thk:l:b:νiT−1= α0kβm 0α−kβ−m PP2i2i==11xyii((bbii++ννiiPPPjPj==ee0r0rBB((νν))−−11mmjjββ−j)j) .
 0 0 1 
Claim 1. Let z 6= 0, if [z : 0 :1] lies on L (TΓ T−1), then z ∈ R. Let us
1 A,B,ν
assume that Im(z 6= 0). Thus there are w = [a : b : 1] and (γ ) ⊂ Γ a
m A,B,ν
sequenceofdistincelements inΓ−1 suchthatTγ T−1(w) // x. From
A,B,ν m m→∞
equation (2.2), it follows that for each m ∈ N there are n ,k ,b ,b ∈ Z
m m 1m 2m
and {l }PerB(ν)−1 ∈ {0,...,S(ν)−1} such that γ (w) = [a :b : 1] where
j j=0 m m m
2 PerB(ν)−1
a = αkmβnma+ x b +ν l βj ;
m i im i j
 
i=1 j=0
X X
 
2 PerB(ν)−1
b = α−kmβ−nmb+ y b +ν l β−j .
m i im i j
 
i=1 j=0
X X
By Lemma 2.5, we can assume thata // z and b  // 0. Since
m m→∞ m m→∞
Im(a ) = αkmβnmIm(a) → Im(z) 6= 0. We conclude that (k ) and (n )
m m m
are eventually constant. In consequence we conclude that (p ) and (p )
1m 2m
are eventually constant. Thus (γ ) is eventually constant. Which is contra-
m
diction.
Observe that by a similar argument, we can show the claim in the case
x = [0 :z :1] ∈ L (TΓ T−1).
1 A,B,ν
Now let γ ∈ TΓ T−1 induced by the linear map:
A,B,ν
α 0 0
(2.3) 0 α−1 0 ,
 
0 0 1
then is straighfoward to checkthat ←e−,→e ∪←e−,→e ⊂ Λ(TΓ T−1).
1 3 2 3 A,B,ν
On the other hand, from equation (2.2), we conclude that ℓ = h{e ,e }i,
1 2 3
ℓ = h{e ,e }i, e and e , are TΓ T−1-invariant. Thus, taking , π =
2 1 3 1 2 A,B,ν i
π we can define Π : Γ → Bihol(l ). Thus, from Equation 2.2, we
ei,ℓi i 0 i
conclude that Π (TΓ T−1) leaves e , k ∈ {1,2}\{i}, and
j A,B,ν k
[{rα e +se |r,s ∈ R}\{0}]
1 k 3
invariant, moreover, it contains loxodromic and parabolic elements. Thus
Lemma 1.2, yields
ℓ \Eq(Π (TΓ T−1)) = [{rα e +se |r,s ∈ R}\{0}].
j j A,B,ν 1 k 3
TWO DIMENSIONAL COMPLEX KLEINIAN GROUPS WITH FOUR COMPLEX LINES IN GENERAL POSITION IN
Thus a straightforward calculation shows
TΓ T−1(ℓ ∪ℓ ) = P2 \ ←e−→,p ⊂ Λ(TΓ T−1).
A,B,ν 1 2 C j A,B,ν
j∈[{1,2}p∈[R(ℓj)
Thus Ω = Hǫ1 ×Hǫ2 is an open TΓ T−1-invariant set, with
ǫ1,ǫ2∈{±1} A,B,ν
LiG(Ω) = 4. InconsequenceTheorem3.5in[1],yieldsΩ ⊂ Eq(TΓ T−1)) ⊂
S A,B,ν
Ω(Γ). Which clearly concludes the proof. (cid:3)
By means of similar arguments the following proposition can be showed.
Proposition 2.9. Let A ∈ SL(2,Z) be an hyperbolic toral automorphism,
then the following set is a discrete group of PSL(3,C)
Ak b
Γ = |b ∈ M(1×2,Z), k ∈ Zk ∈ Z,
A 0 1
(cid:26)(cid:18) (cid:19) (cid:27)
Moreover Ω(Γ ) is projectively equivalent to Hǫ1 ×Hǫ2.
A,B,ν ǫ1,ǫ2∈{±1}
Definition 2.10. A subgroup Γ ⊂ PSL(3,C)Sis said to be a Hyperbolic
Toral Group if Γ is conjugated to the group described either in proposition
2.9 or the one in proposition 2.8.
3. Four lines Groups
Trough this section Γ ⊂ PSL(3,C) is a complex Kleinian group with
LiG(Ω(Γ)) = 4, L ∈ LG(Ω(Γ)) with card(L) = 4, the vertex of L are
←−→ ←−→
e ,e , ℓ = e ,e , ℓ = e ,e , Γ = Stab(e ,Γ)∩Stab(e ,Γ), π = π and
1 2 1 2 3 2 1 3 0 1 2 i ei,ℓi
Π = Π .
i ei,ℓi
Lemma 3.1. Either Π (Γ ) or Π (Γ ) contains loxodromic elements.
1 0 2 0
Proof. On the contrary, let us assume that Π (Γ ), j ∈ {1,2}, does not
j 0
contains loxodromic elements. Thus each element γ ∈ Γ has a lift γ˜ ∈
GL(3,C) which is given by:
γ 0 γ
11 13
γ˜ = 0 γ γ
22 23
 
0 0 1
 
where |γ | = |γ |= 1. A straightforward calculation shows that Eq(Γ ) =
11 22 0
P2 \←e−,→e . Thus Eq(Γ) = Eq(Γ ). Which is a contradition. (cid:3)
C 1 2 0
Lemma3.2. IfΠ (Γ )containsaloxodromic elementthen Fix(τ)
i0 0 τ∈Πi0(Γ0)
contains a single point.
T
Proof. Without loss of generality we may assume that i = 2. Now, if F =
0 2
Fix(τ)contains morethanonepoint,wededucethatF = {e ,z}
τ∈Π2(Γ0) 2 1
for some z ∈ ℓ \{e }. By conjugating by a projective transformation, if it
T 2 1
8 W.BARRERA,A.CANO&J.P.NAVARRETE
is necessary, we may assume that z = e . Thus each element γ ∈ Γ has a
3 0
lift γ˜ ∈ SL(3,C) which is given by:
γ 0 0
11
γ˜ = 0 γ γ
22 23
 
0 0 1
where abc = 1. In consequence ℓ and e are invariant under the action of
1 1
Γ . By Lemma 1.1, W = P2 \(ℓ ∪{e }) is a discontinuity region for Γ
0 C 1 1 0
which is contained in Eq(Γ ). In consequence Lin(ΓW) < ∞, which is a
0
contradiction, since ΓW ⊂ Γ(Eq(Γ ) = Eq(Γ). (cid:3)
0
Lemma 3.3. The groups Π (Γ ) and Π (Γ ) contains loxodromic elements.
1 0 2 0
Proof. By Lemma 3.1, either Π (Γ ) or Π (Γ ) contains loxodromic ele-
1 0 2 0
ments. Without loss of generality let us assume that Π (Γ ) contains a
1 0
loxodromic element. If Π (Γ ) does not contains loxodromic elements, ev-
2 0
ery element γ ∈ Γ has a lift γ ∈ GL(3,C) which is given by:
0
γ 0 γ
11 13
γ˜ = 0 γ γ
22 23
 
0 0 1
where |γ| = 1. In consequence there are γ,τ ∈ Γ such that Π (γ) and
0 1
Π (τ) are loxodromic elements with Fix(Π (γ)) 6= Fix(Π (τ)), Π (γ) is
1 1 1 2
either parabolic or the identity and Π (τ) is either parabolic or the identity.
2
An easy calculation shows Π (τγτ−1γ−1) is parabolic and Π (τγτ−1γ−1) is
1 1
the identity. In consequence κ = τγτ−1γ−1 has a lift κ˜ ∈ SL(3,C) given by:
1 0 0
γ˜ = 0 1 κ .
23
 
0 0 1
Finally, let γ ∈ Γ be such that Π (γ ) is aloxodromic element. By con-
0 0 1 0
jugating with a projective transformation, if it is necessary, we may as-
sume that Fix(Π (γ )) = {e ,e }. Also, by taking the Inverse of γ, if it
1 0 2 3
is necessary, we may assume that e is an attracting point for Π (γ ). In
2 1 0
consequence, if γ˜ = (γ ∈ SL(3,C)) is a lift of γ, we conclude that:
0 ij
1 0 0
γ˜mκ˜γ˜−m = 0 1 γmκ . // Id,
 22 23  m→∞
0 0 1
which is a contradiction, since Γ is discrete.  (cid:3)
Lemma 3.4. There is an element γ ∈ Γ such that Π (γ ) and Π (γ ) are
0 0 1 0 2 0
loxodromic.
Proof. If this is not the case, there are γ ,γ ∈ Γ such that Π (γ ) is
1 2 0 j k
loxodromic if j = k and either the identity or parabolic in other case. Is
straightforward to check that Π (γ γ ), j ∈ {1,2}, is loxodromic. Which is
j 1 2
a contradiction. (cid:3)
TWO DIMENSIONAL COMPLEX KLEINIAN GROUPS WITH FOUR COMPLEX LINES IN GENERAL POSITION IN
From now on γ will denote a fixed element in Γ such that Π (γ ), j ∈
L 0 j L
{1,2}, isloxodromic. Also, byconjugatingwithaprojective transformation,
if it is necessary, we may assume that γ has a lift γ˜ = (γ ) which is a
L L Lij
diagonal matrix.
Lemma 3.5. There is an element γ ∈ Γ such that Π (τ ) and Π (τ ) are
0 1 0 2 0
parabolic elements.
Proof. Let j ∈ {1,2}, then there is an element γ such that Π (γ ) is loxo-
j j j
dromicand Fix(Π (γ )) 6= Fix(Π (γ )). Set κ = γ γ γ−1γ−1, then Π (κ )
j j j L j L j L j i j
is parabolic if i = j and is either the identity or parabolic in other case.
Thus the only interesting case is Π (κ ) = Id and Π (κ ) = Id. But in such
1 2 2 1
case a simple calculation shows that Π (κ κ ), j ∈ {1,2}, is parabolic. (cid:3)
j 1 2
From now on γ will denote a fixed element in Γ with a lift (γ ), such
P 0 Pij
that Π (γ ), j ∈ {1,2}, is parabolic.
j 0
Lemma 3.6. If |γ |<| γ |, then |γ |>| γ |.
L11 L33 L22 L33
Proof. On the contrary, let us assume that |γ |< |γ |. Then a straight-
L22 L33
forward calculations shows that:
1 0 γm γ γ−m
L11 P13 L33
κ = 0 1 γm γ γ−m .
m  L22 P23 L33 
0 0 1
is a lift of γ−mγ γm. And clearly [[κ ]] // Id, which is a contradiction,
L P L m m→∞
since Γ is discrete. (cid:3)
0
Lemma 3.7. The sets P2 \Eq(Π (Γ )) and P2 \Eq(Π (Γ )) are circles.
C 1 0 C 2 0
Proof. Since the vertex of L are e ,e it follows that
1 2
C = Π {ℓ ∈ Gr (P2)|e ∈ ℓ,ℓ ⊂ Λ(Γ)} ,
j j 1 C j
is a closet, infinite and(cid:16)Π[(Γ )-invariant set. Thus Lemm(cid:17)a 1.2 yields the
j 0
result.
(cid:3)
Lemma 3.8. Up to conjugacy Γ leaves P3 invariant.
0 R
Proof. By Lemma 3.5 there is an element γ ∈ Γ such that Π (γ ) and
0 0 1 0
Π (γ ) are loxodromic elements. Thus after conjugating with a projec-
2 0
tive transformation, if it is necessary, we may assume that Fix(Π (γ )) =
1 0
{[e ],[e ]} and that Fix(Π (γ )) = {[e ],[e ]}. In consequence γ has a lift
2 3 2 0 1 3 0
γ˜ ∈ SL(3,C) given by:
0
γ 0 0
11
γ˜ = 0 γ 0 ,
0 22
 
0 0 γ
22
where γ γ γ = 1. Thus there are α ,α ∈ C∗ such that:
11 22 33 1 2
ℓ \Eq(Π (Γ ) = [{rα e +se |r,s ∈ R}\{0}];
1 1 0 1 2 3
10 W.BARRERA,A.CANO&J.P.NAVARRETE
ℓ \Eq(Π (Γ ) = [{rα e +se |r,s ∈ R}\{0}].
2 2 0 2 1 3
Let η ∈ PSL(3,C) be the element induced by the linear map:
α 0 0
2
η˜= 0 α 0 .
1
 
0 0 1
 
Thus a straightforward calculation shows that
Π (η−1Γ η)[{re +se |r,s ∈ R}\{0}] = [{re +se |r,s ∈ R}\{0}];
1 0 2 3 2 3
Π (η−1Γ η)[{re +se |r,s ∈ R}\{0}] = [{re +se |r,s ∈ R}\{0}].
2 0 1 3 1 3
In consequence P2 is η−1Γ η-invariant. (cid:3)
R 0
From now on we will assume that P2 is Γ -invariant. Also let us define:
R 0
Par(Γ ) = {γ ∈ Γ |Π (γ),j ∈ {1,2}, is either parabolic or the identity}.
0 0 j
Lemma 3.9. The set Par(Γ ) is a group isomorphic to Z×Z.
0
Proof. Clearly Par(Γ ) is a group. Moreover Par(Γ ) can be lifted to a
0 0
^
group Par(Γ )⊂ SL(3,C) where each element has the form:
0
1 0 a
0 1 b .
 
0 0 1
 
where a,b ∈ R. Also observe that the group morphism Lat :P^ar(Γ )→ R2
0
givenbyLat((γ )) = (γ ,γ )enableustoshowthatPar(Γ )isisomorphic
ij 13 23 0
to a lattice in R2. Thus to get the claim, will be enought to show that there
^
aretwoelements inLat(Par(Γ ))whichareR-linearly independent. Clearly
0
1 0 γ γ γ−1
L11 P13 L33
κ = 0 1 γ γ γ−1 .
1  L22 P23 L33 
0 0 1
 
is a lift in P^ar(Γ ) of γ−1γ γL. To conclude observe that the system of
0 L P
linear equations
rLa(κ )+sLa(γ ) = 0
1 P
has determinant γ γ γ−1 (γ −γ )6= 0. Which conclude the proof.
P23 P13 L33 L11 L22
(cid:3)
Set
R(ℓ ) = [{rα e +se |r,s ∈ R}\{0}];
1 1 2 3
R(ℓ ) = [{rα e +se |r,s ∈ R}\{0}];
2 1 1 3
Proposition 3.10. The equicontinuity set of Γ is given by:
Eq(Γ) = Hǫ1 ×Hǫ2.
ǫ1,ǫ2[∈{±1}