Table Of ContentDYNAMIC OF ABELIAN SUBGROUPS OF GL(n, C):
A STRUCTURE’S THEOREM
5
0
0
ADLENEAYADIAND HABIBMARZOUGUI
2
n
a Abstract. Inthispaper,wecharacterizethedynamicofeveryabelian
J subgroups G of GL(n, K), K = R or C. We show that there exists a
0 G-invariant, dense open set U in Kn saturated by minimal orbits with
1 Kn−U a union of at most n G-invariant vectorial subspaces of Kn of
dimension n−1orn−2onK. Asaconsequence,G hasheight atmost
] n and in particular it admits a minimal set in Kn−{0}.
S
D
.
h
t
a 1. Introduction
m
[ Let K = R or C, GL(n, K) be the group of all invertible square matrices
of order n 1 with entries in K, and let be an abelian subgroup of GL(n,
1 ≥ G
K). Thereis anaturallinear action GL(n, K) Kn Kn :(A, v) Av.
v
× −→ 7−→
6 For a vector v Kn, we consider the orbit of through v:
∈ G
3 (v) = Av, A Kn. A subset E Kn is called -invariant if
1 G { ∈ G} ⊂ ⊂ G
A(E) E for any A ; that is E is a union of orbits.
1 ⊂ ∈ G
0
In [3], Kulikov studied the problem of the existence of minimal sets in
5
0 Rn 0 . He constructed an example of discrete subgroup of SL(2,R)
− { }
/ whose linear action on R2 is without minimal set in R2 0 . F. Dal’bot
h
−{ }
t and A.N. Starkov touched in [2] the question of the existence of an infinitely
a
m generated subgroup of SL(2, R) with all orbits dense in R2.
In [1], we studied in the viewpoint closure of orbits the dynamic of a class
:
v of abelian subgroupsof GL(n, R); those containing an element A which
i ∈ G
X satisfies the condition (⋆): all eigenspaces of A are of dimension 1 on C.
r
a This work considers the general case: the study of the dynamic of every
abelian subgroup of GL(n, K). The purpose here is to develop in this gen-
eral situation a setting of a structure’s Theorem analogous to a structure’s
Theorem (in [4]) for foliations on closed manifolds.
Before stating our main results, we introduce the following notions for
groups:
2000 Mathematics Subject Classification. 37C85.
Key words and phrases. linear action,orbit, locally denseorbit,minimal, group, mini-
mal set.
This work is supported by the research unit: syst`emes dynamiques et combinatoire:
99UR15-15.
1
2 ADLENEAYADIANDHABIBMARZOUGUI
A -invariant subset E of Kn is called a minimal set of if every orbit
G G
contained in E is dense in it (this definition it equivalent to say that E is
closed in Kn, non empty, -invariant and has no proper subset with these
G
properties). If V is a -invariant open set in Kn, a minimal set in V is a
G
minimal set of restricted to V. We say that an orbit O in V is called
G
minimal in V if O V is a minimal set in V.
∩
We call class of an orbit L of the set cl(L) of orbits O of such that
G G
O = L. If L is an orbit which is minimal in a -invariant, open set V then
G
cl(L) = L V.
∩
An orbit L of is said to be at level 1 if, L is minimal in Kn 0 .
G − { }
Inductively, we say that L is at level p, p 1 if every orbit O L cl(L)
≥ ⊂ −
is at level < p with at least one orbit at level k for every k < p. The
upper bound of levels of orbits of is called the height of , denoted it
G G
by ht( ). For example, if ht( ) is finite, say p this means that p is the
G G
supremum of k N such that there exist orbits γ , γ , ..., γ of such that
1 2 k
∈ G
γ γ ... γ with γ = γ , 1 i= j k.
1 2 k i j
⊂ ⊂ ⊂ 6 ≤ 6 ≤
The main result of this paper is the following structure’s Theorem:
Structure’s Theorem. Let be an abelian subgroup of GL(n, K)
G
(K = R or C). Then there exists a -invariant, dense open set U in Kn with
G
the following properties:
i) Every orbit of U is minimal in U.
ii) Kn U is a union of at most n -invariants vectorial subspaces of Kn
− G
of dimension n 1 or n 2 on K.
− −
Remark 1.1. If is an abelian subgroup of GL(n,R) containing an el-
G
ement A which satisfies the condition (⋆) we know that all orbits of
∈ G
U are homeomorphic (cf. [1]). If does not contain an element which
G
satisfies the condition (⋆) and if n 3, this property is not true in general:
≥
we give a counterexample for n = 3, 4, 5 (see Examples 6.1, 6.2 and 6.3)
of abelian subgroups which contains two non homeomorphic orbits in any
G
opensetinR3(resp. R4; C5). Ifn = 2, thepropertyremainstrue: agroup
G
does not contain an element which satisfies the condition (⋆) is a subgroup
of homotheties. Therefore, if U = R2 0 and u, v U, there exists
− { } ∈
A GL(2, R) such that Au = v and thus A( (u)) = (v).
∈ G G
Corollary 1.2. Let be an abelian subgroup of GL(n,K). Then has
G G
height at most n.
Ontheotherhand,if weremove 0 whichisfixedby (i.e. (0) = 0 ),
{ } G G { }
does there exist a minimal set in Kn 0 ? we gave in [1] a positive answer
−{ }
for this question which is here a consequence of Corollary 1.2:
DYNAMIC OF ABELIAN SUBGROUPS OF GL(n, C): A STRUCTURE’S THEOREM 3
Corollary 1.3. ([1]) If is an abelian subgroup of GL(n,K), it admits a
G
minimal set in Kn 0 .
−{ }
Corollary 1.4. If has a locally dense orbit O in Kn and C a connected
G
component of U meeting O then:
i) O is dense in C
ii) Every orbit in U meeting C is dense in it.
Corollary 1.5. Let be an abelian subgroup of GL(n,K). If has a
G G
dense orbit in Kn then every orbit in U is dense in Kn.
Thispaperisorganizedasfollows: Insection2wegivesomenotationsand
technical lemmas. In section 3, we prove the main theorem for a subgroup
of S (K). The proof of the structure’s Theorem is done in section 4. In
n
section 5, we prove Corollaries 1.2, 1.3, 1.4, 1.5. In section 6, some examples
are given.
2. Notations and Lemmas
In this paper, we denote by:
- K = R or C.
- T (K) the subgroup of GL(n, K) of lower triangular matrices,
n
- S (K) the subgroup of T (K) of matrices B = (b ) with b =
n n i,j 1≤i,j≤n i,i
µ , i= 1,...,n.
B
Every element B S (K) is written in the form
n
∈
µ 0 .. 0
B
B = b2,1 µB . . = B(1) 0 ,
. . . 0 (cid:18) L µ (cid:19)
B B
bn,1 .. bn,n−1 µB
µ 0 .. 0
B
with B(1) = b2,1 µB . . S (K), µ K and
. . . 0 ∈ n−1 B ∈
bn−1,1 .. bn−1,n−1 µB
L = (b ,....,b ).
B n,1 n,n−1
For this section, is an abelian subgroup of S (K).
n
G
Denote by:
- G(1) = B(1), B .
{ ∈G}
- = (e ,....,e ) the canonical basis of Kn, e = (0,...,0,1,0,...,0) Kn
1 n i
C ∈
(1 is the ith coordinate of e ).
i
- = (B µ I )e Kn, 1 i n 1, B
B n i
F { − ∈ ≤ ≤ − ∈ G}
- I is the identity matrix of Kn.
n
- rang( ) the rang of . We have rang( ) n 1.
F F F ≤ −
4 ADLENEAYADIANDHABIBMARZOUGUI
For every x = (x ,...,x ) Kn, we let x(1) = (x ,...,x ) Kn−1 and
1 n 1 n−1
∈ ∈
(1) = (B(1) µ I )e(1), 1 k n 2, B . Wehavex= (x(1),x )
F { − B n−1 k ≤ ≤ − ∈ G} n
(1)
and e = (e ,0) , k = 1,...,n 2.
k k −
We start with the following lemmas:
Lemma2.1. Underthenotationabove, ifrang( ) = n 1andif v , ..., v
1 n−1
F −
generate , then rang( (1)) = n 2 with (n 2) vectors of (v(1),..,v(1) )
F F − − 1 n−1
generate .
F
Proof. Let A ,...,A such that v = (A µ I )e , k = 1,...,n 1.
1 n−1 ∈ G k k− Ak n ik −
(1)
A 0 (1) (1) (1)
We let A = k and v = (A µ I )e . Then
k (cid:18) L µ (cid:19) k k − Ak n−1 ik
k Ak
(1) (1) (1)
v = (v ,L e ) vect(e ,...,e ) and v vect(e ,...,e ) where
k k k ik ∈ 2 n k ∈ 2 n−1
vect(e ,...,e ) (resp. vect(e ,...,e )) is the vectorial subspace generated
2 n 2 n−1
(1) (1) (1)
by e ,...,e (resp. e ,...,e ). Thus, rang(v ,v ,...,v ) = r n 2.
2 n 2 n−1 1 2 n−1 ≤ −
If r < n 2 there exist 1 k n 2 and α ,...,α ,α ,....α K
1 k−1 k+1 n−1
− ≤ ≤ − ∈
(1) (1) (1) (1) (1)
such that v =α v +...+α v +α v +...+α v .
k 1 1 k−1 k−1 k+1 k−1 n−1 n−1
(1) (1) (1) (1)
Lets to prove that v ,..,v ,v ,..,v are linearly independents:
{ 1 k−1 k+1 n−1}
Suppose the contrary ; that is there exists 1 j < k n 1 such that
≤ ≤ −
(1) (1)
v = β v . Then: v β v = βe andv α v =
j i i j− i i n k− i i
1≤iP≤n−1 1≤iP≤n−1 1≤iP≤n−2
i6=k,i6=j i6=k,i6=j i6=k
(1) (1)
αe , where β = L e β L e and
n j ij − s s is
1≤sP≤n−2
s6=k,s6=j
(1) (1)
α = L e α L e . Thus,
k ik − s s is
1≤sP≤n−2
s6=k
rang(v ,...,v ) = rang(v ,..,v ,βe ,v ,..,v ,αe ,v ,..,v )< n 1.
1 n−1 1 j−1 n j+1 k−1 n k+1 n−1
−
Thereforerang( ) = rang(v ,...,v ) < n 1, acontradiction. We deduce
1 n−1
F −
that r = n 2 and then rang( (1)) = n 2. (cid:3)
− F −
(1)
B 0
Lemma 2.2. Let u, v K⋆ Kn−1 and let B = m ,
m
∈ × (cid:18) L µ (cid:19) ∈ G
m m
m N, B(1) S (K) such that lim B u= v. Suppose that rang( ) =
m n−1 m
∈ ∈ m→+∞ F
(1)
n 1. If (Bm )m∈N is bounded then (Bm)m∈N is bounded.
−
Proof. It suffices to show that (Lm)m∈N is bounded.
By Lemma 2.1, it follows that rang( (1)) = n 2 and there exist (n 2)
F − −
(1) (1) (1) (1)
vectors of v ,..,v which generate , say v ,...,v . We have v =
1 n−1 F 1 n−2 k
DYNAMIC OF ABELIAN SUBGROUPS OF GL(n, C): A STRUCTURE’S THEOREM 5
(1)
(1) (1) (1) A 0
(A λ I )e , v = (A λ I )e , and A = k ,
k − k n ik k k − k n−1 ik k (cid:18) T λ (cid:19) ∈ G
k k
k = 1,...,n 2. Since A B = B A then
k m m k
−
L (A(1) λ I )= T (B(1) µ I )
m k − k n−1 k m − m n−1
Take α = T (B(1) µ I )e , k = 1,...,n 2 and m N.
k,m k m − m n−1 ik − ∈
(1) (1)
Byhypothesis,(Bm )m∈N isboundedthen(Bm µmIn−1)m∈N isbounded
−
and therefore (αk,m)m∈N is bounded for every k = 1,...,n 2.
−
In other part, we have :
(1)
L (A λ I )e = α
m 1 − 1 n−1 i1 1,m
..................................
..................................
(1)
L (A λ I )e = α
m n−2− n−2 n−1 in−2 n−2,m
(1) (1)
TakeL =(b ,...,b )andv = (A λ I )e = (0,a ,...,a ).
m 1,m n−1,m k k − k n−1 ik k,2 k,n−1
Thus,
a b +...+a b = α
1,2 2,m 1,n−1 n−1,m 1,m
..................................
..................................
a b +...+a b = α
n−2,2 2,m n−2,n−1 n−1,m n−2,m
b
2,m
.
ThissystemcanbewrittenintheformMX = Y ,withX = ,
m m m .
bn−1,m
α a .. .. a
1,m 1,2 1,n−1
. .. .. .. ..
Y = and M = .
m . .. .. .. ..
α2,m an−2,2 .. .. an−2,n−1
Since (v(1)) are independents then M is invertible, so M−1Y =
k 1≤k≤n−2 m
Xm. Since(αk,m)m∈N isbounded,(Ym)m∈N isboundedandtherefore(Xm)m∈N
is bounded i.e. (bk,m)m∈N is bounded for k = 2,..,n 1.
−
It remains to prove that (b1,m)m∈N is bounded :
Since lim B u= v we have
m
m→+∞
lim (b u +...+b u +µ u ) = v
1,m 1 n−1,m n−1 m n n
m→+∞
where u = (u ,...,u ) and v = (v ,...,v ) K⋆ Kn−1. Or u = 0 then
1 n 1 n 1
∈ × 6
1
lim b = lim (v b u ... b u µ u ).
1,m n 2,m 2 n−1,m n−1 m n
m→+∞ m→+∞u1 − − − −
Since (bk,m)m∈N is bounded for k = 2,..,n 1, then (b1,m)m∈N is also
bounded. We deduce that (Lm)m∈N is bounded−. (cid:3)
6 ADLENEAYADIANDHABIBMARZOUGUI
Lemma 2.3. Let r = rang( ), 1 r n 1and v ,...,v are the generator
1 r
F ≤ ≤ −
of . For every u Kn, let H be the vectorial subspace of Kn generated by
u
F ∈
u,v ,...,v . Then H is -invariant.
1 r u
G
In particular, the subspace F = H generated by v ,...,v is -invariant.
0 1 r
G
Proof. Let w = (z ,...,z ) H and B with eigenvalue µ. We have
1 n u
∈ ∈ G
n−1
Bw = µw+(B µI )w and (B µI )w = z (B µI )e .
n n i n i
− − −
iP=1
Since v ,...,v generate , then for k = 1,..,n 1, we have (B µI )e =
1 r n k
F − −
r n−1 r
β v H . Therefore: Bw = µw+ z β v H . (cid:3)
k,i i u k k,i i u
∈ ∈
iP=1 kP=1 iP=1
3. Proof of the structure’s Theorem for subgroups of S (K)
n
Let be an abelian subgroup of S (K) and H a -invariant vectorial
n
G G
subspace of Kn. Recall that = (e ,...,e ) is the canonical basis of Kn. Let
1 n
C
B and ϕ the automorphism of Kn with matrix in is B. Let be a
B H
∈ G C C
basis of H and Denote by BH the matrix of the automorphism restriction
(ϕ ) in and = BH, B .
B /H H H
C G { ∈ G}
The main result of this section is the following:
Proposition 3.1. Let be an abelian subgroup of S (K). For every u, v
n
G ∈
K⋆ Kn−1 and for every sequence (Bm)m∈N of such that lim Bmu= v,
× G m→+∞
there exist a -invariant subspace H of Kn and a basis = u,w ,...,w
H 1 p
G C { }
of H such that:
i) is a subgroup of S (K).
H p+1
G
ii) ((Bm)H)m∈N is bounded.
Remark 3.2. In the proposition above, the restriction to a -invariant
G
vectorial subspace H is necessary as shown in Exemple 6.4: there exists a
subgroup of GL(4,R) and an umbounded sequence ( m)m∈N in with
G B G
lim B u= v, for u, v R⋆ R3.
m
m→+∞ ∈ ×
Proof. The proof is by induction on n.
For n = 1, we have B = λ K and u, v K⋆. The condition
m m
∈ ∈
lim Bmu = v shows that (λm)m∈N is bounded. Then i) and ii) are satis-
m→+∞
fied for H = K and = (u).
E
Suppose the proposition is true until the order n 1.
−
Let be an abelian subgroup of S (K). Let u, v K⋆ Kn−1 and
n
G ∈ ×
(Bm)m∈N a sequence of such that lim Bmu= v.
G m→+∞
Every B is written as
∈G
B(1) 0
B = ,
(cid:18) LB µB (cid:19)
DYNAMIC OF ABELIAN SUBGROUPS OF GL(n, C): A STRUCTURE’S THEOREM 7
(1)
where B(1) S (K). Denote by B = Bm 0 , with B(1)
n−1 m m
∈ (cid:18) L µ (cid:19) ∈
m m
S (K), L = (bm ,...,bm ) and by (1) = B(1), B . One can
n−1 m n,1 n,n−1 G { ∈ G}
check that (1) is an abelian subgroup of S (K).
n−1
G
For every z = (z ,...,z ) Kn, denote by z(1) = (z ,...,z ).
1 n 1 n−1
∈
We let u = (x ,...,x ) K⋆ Kn−1, v = (y ,...,y ) Kn, u(1) =
1 n 1 n
∈ × ∈
(x ,...,x ) and v(1) = (y ,...,y ).
1 n−1 1 n−1
As lim B u= v then lim B(1)u(1) = v(1).
m m
m→+∞ m→+∞
By recurrence hypothesis applied to (1) on Kn−1, there exists a (1)-
G G
invariantvectorialsubspaceH ofKn−1andabasis = (u(1),w(1),...,w(1))
1 CH1 1 p
of H such that:
1
i) (1) is a subgroup of S (K) .
GH1 p+1
ii) ((Bm(1))H1)m∈N is bounded.
We distinguish two cases:
Case 1: dim(H ) < n 1.
1
−
We let H′ = z = (z(1),z ) :z(1) H , z K .
n 1 n
{ ∈ ∈ }
H′ is a -invariant vectorial subspace of Kn: if z = (z(1),z ) H′ and
n
G ∈
B , then Bz = B(1)z(1), L z(1) +µ z . As, B(1)z(1) H then
B B n 1
∈ G ∈
Bz H′. (cid:0) (cid:1)
∈ (1)
Let wk = (wk ,0), k = 1,..,p and we let CH′ = (u,w1,..,wp,en). CH′ is
p
a basis of H′: if α,α ,...,α ,β K such that αu+βe + α w = 0 then
1 p n i i
∈
iP=1
p
αu(1) + α w(1) = 0, so, α = α = ... = α = 0 and thus β = 0. Therefore
i i 1 p
iP=1
dim(H′)= 1+dim(H )< n.
1
Denote by H′ = BH′, B . We will to show that H′ is a subgroup
G { ∈ G} G p
of S (K): indeed, if B and k = 1,...p, we have B(1)w(1) = α w(1).
p+2 ∈ G k k,i i
iP=k
p
So, Bw = (B(1)w(1), L w(1)) = ( α w(1), L w(1)) = (L w(1))e +
k k B k k,i i B k B k n
iP=k
p
α w . Moreover, Be = µ e . It follows that BH′ S (K).
k,i i n B n p+2
∈
iP=k
In the basis H′, u (resp. v) has coordinate uH′ = (1,0,...,0) (resp.
vH′ = (v1,...,vp+C2)). We have lim BmH′uH′ = vH′. So, lim µm = v1.
m→+∞ m→+∞
As lim B u = v then lim µ = y1 = 0. It follows that v = 0 and
m→+∞ m m→+∞ m x1 6 1 6
therefore uH′, vH′ K⋆ Kp+1.
∈ ×
We can apply the recurrence hypothesis to H′ on Kp+2, so there exists a
G
H′-invariant vectorial subspaceH” and a basis H” = (uH′,w”1,...,w”q) of
G C
H” which satisfies the assertions i) and ii). As H” is H′-invariant then it
G
8 ADLENEAYADIANDHABIBMARZOUGUI
is a fortiori -invariant. Indeed if B , v H” and ϕ the automorphism
B
G ∈ G ∈
of Kn with matrix in is B, then ϕ(v) = (ϕ/H′)/H”(v) H” with matrix
in is (BH′)H”. C ∈
C
Case 2: dim(H ) = n 1
1
−
In this case, H = Kn−1 and then (B(1))H1 = B(1) is bounded by hypoth-
1 m m
esis.
One distinguish three cases:
Let = (B µ I )e , 1 k n 1, B and r = rang( ).
B n k
F { − ≤ ≤ − ∈ G} F
Case 2. a): r = n 1
−
By Lemma 2.2, (Bm)m∈N is bounded. The assertions i) and ii) of the
proposition follow by taking H = Kn and = (u,e ,...,e ) a basis of Kn.
2 n
C
Case 2. b): 1 r < n 1
≤ −
Let v ,...,v generate . Let H (resp. F) be the vectorial subspace of
1 r u
F
Kn generated by (u,v ,...,v ) (resp. (v ,...,v )). By Lemma 2.3, H (resp.
1 r 1 r u
(2) (2)
F) are -invariant. Let (w ,...,w ) be a basis of F such that for every
G 1 r
B , BF is lower triangular. Let w = (0,w(2)), k = 1,...,r. Then
∈ G k k
= (u,w ,...,w ) is a basis of H . For every B , BHu is lower
CHu 1 r u ∈ G
triangular. Then is a subgroup of S (K). Moreover, u, v K⋆ Kr.
GHu r+1 ∈ ×
Since dim(H ) = r +1 < n, then the proposition follows by applying the
u
recurrence hypothesis on .
GHu
Case 2. c): r = 0
In this case, for every B , (B µ I )e = 0, k = 1,...,n. Then, B =
B n k
∈ G −
µ I and therefore is an abelian subgroup of homotheties of GL(n,K).
B n
G
By taking H = Kn and = (u,e ,...,e ) a basis of Kn, the assertion i) and
2 n
C
ii) of proposition are satisfied:
We have B = µ I , m N and lim λ u = v then lim µ = y1.
m m n ∈ m→+∞ m m→+∞ m x1
Hence (Bm)m∈N is bounded. (cid:3)
Corollary 3.3. Let be an abelian subgroup of S (K) (K = R or C). For
n
G
every u, v K⋆ Kn−1 and for every sequence (Bm)m∈N of such that
∈ × G
lim B u= v, we have lim B−1v = u.
m m
m→+∞ m→+∞
Proof. Let u, v Kn and (Bm)m∈N a sequence in such that lim Bmu =
∈ G m→+∞
v. By Proposition 3.1, there exists a -invariant vectorial subspace H of Kn
G
containing u, v and a basis CH = (u,w1,...,wp) of H such that (BmH)m∈N is
bounded.
DYNAMIC OF ABELIAN SUBGROUPS OF GL(n, C): A STRUCTURE’S THEOREM 9
Denote by u = (u ,...,u ) (resp. v = (v ,...,v )) where u ,...,u
H 1 n H 1 n 1 n
(resp. v ,...,v ) are the coordinate of u (resp. v) in . Denote by the
1 n H
C k k
norm on GL(n,K) defined by : B(x) = sup (kBxk). We have
k x∈Kn−{0} kxk
(BH)−1v u (BH)−1 v BHu .
k m H − H k≤k m k k H − m H k
The matrix N = BH µ I is nilpotent of order p+1. Then
m m − m p+1
p
(BH)−1 = 1 ( 1)k 1 Nk. Therefore :
m µm − (µm)k m
kP=0
p
1
(BH)−1 N k
k m k≤ µ k+1 k mk
Xk=0| m|
Since (BmH)m∈N is bounded and ml→im+∞Bmu = v, u, v ∈ K⋆ × Kn−1,
then (µ1m)m∈N is bounded. Therefore, (Nm)m∈N and then ((BmH)−1)m∈N is
bounded. Since lim BHu = v ,wededucethat lim (BH)−1v = u .
m H H m H H
m→+∞ m→+∞
Hence, lim B−1v = u. (cid:3)
m
m→+∞
4. Proof of the structure’s Theorem : General Case
In fact, we will prove a slightly strong result; that is:
Theorem 4.1. Let be an abelian subgroup of GL(n,K) and let E = x =
i
G {
(x ,...,x ) Kn : x = 0 . Then there exists P GL(n, K) and finitely
1 n i
∈ } ∈
many -invariant subspaces H with H = P(E ) if K = C, and P(E ) or
G k k ik ik
P(E E ), if K = R, 1 k r, 1 i ,...,i n such that satisfies
ik ∩ ik+1 ≤ ≤ ≤ 1 r ≤ G
the property :
P r
if u, v U = Kn Hk and (Bm)m∈N is a sequence of such that
∈ − k∪=1 G
lim B u= v then lim B−1v = u.
m m
m→+∞ m→+∞
The proof of the structure’s Theorem is completed as follows:
r
From Theorem 4.1, we let U = Kn H . It is clear that U is a -
k
−k∪=1 G
invariant denseopen set in Kn andthe property implies in particular that
P
every orbit of U is minimal in U.
Case K = C
The proof uses induction on n.
The Theorem is true for n = 1: take P = IC, H1 = 0 and U = C⋆. So
{ }
the property is satisfied.
P
Suppose the theorem is true until the order n 1. Let be an abelian
− G
subgroup of GL(n,C). We distinguish two cases:
Case 1: Every element of has only one eigenvalue
G
10 ADLENEAYADIANDHABIBMARZOUGUI
In this case, there exists P GL(n,C) such that ′ = P−1 P is a sub-
∈ G G
groupof S (C). By taking H = P(E )andU = Cn H = P C⋆ Cn−1 ,
n 1 1 1
− ×
then, by Corollary 3.3, the property follows. (cid:0) (cid:1)
P
Case 2: There exists A having at least two complex eigenvalues
∈ G
Inthiscaseifλ ,..., λ betheeigenvalues ofAwithorderofmultiplicities
1 p
n1,...,np respectively andEk = Ker(A λkIn)nk bethecharacteristic space
−
of A associated to λ then 1 n < n. The space E (1 k p)
k k k
≤ ≤ ≤
is -invariant: indeed, if B G and x Ek then (A λkIn)nkB(x) =
G ∈ ∈ −
B(A−λkIn)nk(x) = 0. Denote by GEk = {BEk, B ∈ G}. Then GEk is an
abelian subgroup of GL(n ,C), k = 1,..,p.
k
Using the recurrence hypothesis on , there exist P GL(n ,C) and
GEk k ∈ k
finitely many subspaces Hk = P (E ),...,Hk = P (E ) E , where
1 k k,j1 rk k k,jrk ⊂ k
1 j ,...,j n , and for i = 1,...,r , E = x = (x ,...,x )
≤ 1 rk ≤ k k k,ji { k,1 k,nk ∈
Cnk : x = 0 such that the property is satisfied; if u , v U =
k,ji } P k k ∈ k
Cnk−ir=∪k1Hik and(Bm(k))m∈N beasequenceinGk suchthatml→im+∞Bm(k)uk = vk
then lim (B(k))−1v = u .
m k k
m→+∞
p
Sincedim(E ) = n ,k = 1,..,p,and E = Cn,welet = (e ,...,e )
k k k Ck 1,k 1,nk
kL=1
p
a basis of E , k = 1,..,p. Hence, ′ = is a basis of Cn. Let
k k
C k∪=1C
Q GL(n,C) the matrix of base change from to ′. Then for every
∈ C C
B , we have:
∈ G
BE1 0 0
Q−1BQ = 0 . 0
0 0 BEp
P 0 0
1
where BEk GL(nk,C). Take P = 0 . 0 , i = 1,..,rk, k =
∈
0 0 P
p
1,..,p. WeletF = x = (x ,...,x ;...;x ,...,x ) Cn : x = 0 ,
k,ji { 1,1 1,n1 p,1 p,np ∈ k,ji }
p rk
Kk = QP(F ) and U = Cn ( Kk).
i k,ji − i
kS=1 iS=1
Let u, v U and a sequence (Bm)m∈N such that lim Bmu = v.
∈ ⊂ G m→+∞
BE1 0 0
m
Then lim Q−1BmQ(Q−1u) = Q−1v. TakeBm′ = Q−1BmQ = 0 . 0 ,
m→+∞ 0 0 BEp
m
u′ = Q−1u = u′ +...+u′ and v′ = Q−1v = v′ +....+v′, where u′,v′
1 p 1 p k k ∈
p rk
E , k = 1,...,p. We have u′, v′ U = Cn P(F ) . Hence,
k ∈ − (cid:18) k,ji (cid:19)
kS=1 iS=1
u′,v′ U , k = 1,...,p. From lim B u = v we have lim B′ u′ = v′.
k k ∈ k m→+∞ m m→+∞ m
