DENSE ORBITS FOR ABELIAN SUBGROUPS OF GL (n, C)

24 Pages·2012·0.18 MB·English

Available at: http://www.ictp.it/~pub off IC/2005/088
(cid:0)
United Nations Educational Scienti(cid:12)c and Cultural Organization
and
International Atomic Energy Agency
THE ABDUS SALAM INTERNATIONAL CENTRE FOR THEORETICAL PHYSICS
DENSE ORBITS FOR ABELIAN SUBGROUPS OF GL(n;C)
Adlene Ayadi1
Department of Mathematics, Faculty of Sciences of Gafsa, Gafsa, Tunisia
and
The Abdus Salam International Centre for Theoretical Physics, Trieste, Italy
and
Habib Marzougui2
Department of Mathematics, Faculty of Sciences of Bizerte, Zarzouna 7021, Tunisia
and
The Abdus Salam International Centre for Theoretical Physics, Trieste, Italy.
Abstract
In this paper, we give a characterization of existence of dense orbit for any abelian subgroup
G of GL(n;C). We prove in particular that if G has a dense orbit then the closure G is a vector
space of dimension n. If G is (cid:12)nitely generated, this characterization is explicit.
MIRAMARE { TRIESTE
September 2005
[email protected]
2Regular Associate of ICTP. [email protected]
1. Introduction
The existence of dense orbit is one of the main topics of topological dynamical. Examples of
smooth (cid:13)ows with a dense orbit were constructed, for example, in [6] and [7]. For linear groups,
examples of non abelian subgroups of GL(n;C) which have a dense orbit were constructed in
[2](see also [3]): Dalb’bo and Starkov gave an example of an in(cid:12)nitely generated subgroup of
SL(2;R) with a dense orbit in R2. Here we are interested in abelian linear groups. Notice that
if G is a subgroup of isometries, it has no dense orbit (since every orbit of G is contained in a
sphere). Moregenerally, everyequicontinuousabeliansubgroupof GL(n;C) hasnodenseorbit.
Let GL(n;C) be the group of all invertible square matrix of order n 1 with coe(cid:14)cients in C,
(cid:21)
and let G be an abelian subgroupof GL(n;C). There is a natural linear action GL(n;C) Cn :
(cid:2)
Cn. (A;v) Av. For a vector v Cn, denote by G(v) = Av; A G Cn the orbit
(cid:0)! 7(cid:0)! 2 f 2 g (cid:26)
of G through v. A subset E Cn is called G-invariant if A(E) E for any A G; that is
(cid:26) (cid:26) 2
E is a union of orbits.
We say that G(v) is dense in Cn if G(v) = Cn. If all orbits of G (except 0) are dense in Cn,
G is called minimal. If U is an open G-invariant set, the orbit G(v) U is called minimal
(cid:26)
in U if G(v) U = G(w) U for every w G(v) U.
\ \ 2 \
There is no minimal abelian subgroup G of GL(n;C) (n 2); indeed, since G is abelian,
(cid:21)
there exists a common eigenvector v for every element of G and so, G(v) is not dense in Cn.
For n= 1, a subgroup G of C can be minimal (see Examples 6:3 and 7:5).
(cid:3)
So, the question to investigate is the following:
When does an abelian subgroup of GL(n;C) have a dense orbit?
In this paper, we will give a complete answer to the above question for any abelian subgroup
of GL(n;C).
Notice that in [1], the authors gave a global dynamic of every abelian subgroup of GL(n;C).
This paper can be viewed as a continuation of that work.
To state ourmain results, we need to introduce the following notations and de(cid:12)nitions for the
sequel.
Denote by:
- M (C) the set of all square matrix of order n 1 with coe(cid:14)cients in C.
n
(cid:21)
- T (C) the set of all lower triangular matrix of M (C) having one eigenvalue.
n n
- T (C)= T (C) GL(n;C) (i.e. the subset of matrix of T (C) having a non nul eigenvalue).
(cid:3)n n \ n
- D (C) the set of diagonal matrix of M (C).
n n
r
Let r N and n ,...,n N such that n = n. Denote by:
(cid:3) 1 r (cid:3) i
2 2
i=1
A 0 0 P
1
- K (C) = A= 0 : 0 M (C) : A T (C); k = 1;::;r :
n;r 8 0 1 2 n k 2 nk 9
0 0 A
< r =
- K(cid:3)n;r(C) = K:n;r(C@)\GL(n;C) A ;
2
In particular: If r = 1 then K (C) = T (C) and K (C)= T (C). If r =n then K (C) =
n;r n (cid:3)n;r (cid:3)n n;r
D (C).
n
We let:
u u
1 k;1
: :
- U = 8>u= 0 : 1 2 Cn : uk = 0 : 12 C(cid:3) (cid:2)Cnk(cid:0)1; 1 (cid:20) k (cid:20) r 9>
> >
< B u C B u C =
B r C B k;nk C
> @ A @ 1 A >
> e >
: 1;1 0 ;
: 0 1
- u0 = 0 : 12 Cn where ek;1 = : 2Cnk, 1(cid:20) k (cid:20) r,
B : C
BB er;1 CC BB 0 CC
@ A B C
e(k) @ A
1
Denote by e(k) = 0 : 1 Cn where e(k) = 02 Cnj if j 6= k for 1 j r.
: 2 j ek;1 if j = k (cid:20) (cid:20)
(cid:26)
B e(k) C
B r C
@ A
Let A = eB1,....,A = eBp GL(n;C), where B ,..,B M (C). We say that A ,....,A
1 p 1 p n 1 p
2 2
satisfyproperty ifforevery(s ;:::s ; t ;::;t ) Zp+r 0 : rank(v ;::;v ; w ;::;w ) = 2n+1;
1 p 1 r 1 p 1 r
D 2 (cid:0)f g
Re(Bku0) 0Cn
0 1 0 1
where v = Im(B u ) , w = 2(cid:25)e(l) R2n Z, 1 k p ; 1 l r.
k k 0 l
2 (cid:2) (cid:20) (cid:20) (cid:20) (cid:20)
B C B C
B C B C
B s C B t C
k l
B C B C
For a vecto@r v Cn, weAwrite v =@Re(v)+AiIm(v) where Re(v) and Im(v) Rn.
2 2
Consider the exponential map exp: M (C) GL(n;C);M exp(M) =eM.
n
! 7(cid:0)!
We prove (Proposition 6:1), that there exists P GL(n;C) such that PGP 1 is an abelian
(cid:0)
2
subgroup of K (C).
(cid:3)n;r
We let g = exp 1(G) P 1(K (C))P and g = Bu; B g , u Cn. One has
(cid:0) (cid:0) n;r u
\ f 2 g 2
exp(g) = G (see Corollary 3:2(cid:2)). Finally, take (cid:3)v0 = Pu0.
Our principal results can be stated as follows:
Theorem 1.1. Let G be an abelian subgroup of GL(n;C). The following are equivalent:
i) G has a dense orbit in Cn
ii) The orbit G(v ) is dense in Cn
0
iii) g is an additive subgroup dense in Cn
v0
Corollary 1.2. If G has a dense orbit then G is a vector subspace of M (C) of dimension n.
n
For a (cid:12)nitely generated subgroup G, the Theorem can be stated as follows:
Theorem 1.3. Let G be an abelian subgroup of GL(n;C) generated by A ,..,A and let B ,
1 p 1
..., B g such that eB1 = A ,..., eBp = A . The following are equivalent:
p 1 p
2
3
i) G has a dense orbit in Cn
ii) A ,..., A satisfy property
1 p
D
p r
iii)g = Z(B v )+2i(cid:25) ZPe(k) is an additive subgroup dense in Cn.
v0 k 0
k=1 k=1
P P
Corollary 1.4. If G is of (cid:12)nite type p with p 2n r it has no dense orbit.
(cid:20) (cid:0)
Corollary 1.5. If G is of (cid:12)nite type p with p n it has no dense orbit.
(cid:20)
Remark 1.6. 1). Corollary 1:5 is not true in general if p> n as can be shown in Example 7:7.
2) If G is not (cid:12)nitely generated but contains (cid:12)nitely matrix that satisfy property then G
D
has a dense orbit (by Theorem 1:3). Conversely, if G is not (cid:12)nitely generated and have a dense
orbit, G does can not contain any (cid:12)nitely matrix that satisfy property (see Example 6:3).
D
2. Preliminary results
We begin by proving several results concerning matrix exponential map. Denote by =
0
B
(e ;:::;e ) the canonical basis of Cn and by I the identity matrix.
1 n n
Proposition 2.1. exp(K (C))= K (C)
n;r (cid:3)n;r
Lemma 2.2. Let B M (C) having one eigenvalue (cid:22). Then:
n
2
i) Ker(B (cid:22)I )= Ker(eB e(cid:22)I ):
n n
(cid:0) (cid:0)
ii) If eB T (C) then B T (C).
2 (cid:3)n 2 n
Proof. We (cid:12)rst prove: Ker(eB e(cid:22)I ) Ker(etB et(cid:22)I )) for every t R:
n n
(cid:0) (cid:26) (cid:0) 2
Let v Ker(eB e(cid:22)I ). Then emBv = em(cid:22)v for every m Z.
n
2 (cid:0) 2
Take N = B (cid:22)I and P (t) =<(etN I )v; e >, where <; > is the scalar product on Cn.
n i n i
(cid:0) (cid:0)
Since B has only one eigenvalue then N is nilpotent of order p. We have:
k=p(cid:0)1tkNk
e t(cid:22)(etB et(cid:22)I ) = etN I =
(cid:0) n n
(cid:0) (cid:0) k!
k=1
X
Then P is a polynomial of degree less then p 1. We have P (m) = 0 for every m N and
i i
(cid:0) 2
therefore P (t) = 0 for every t R and every i = 1;::;n. We have then (etB et(cid:22)I )v = 0.
i n
2 (cid:0)
Hence v Ker(etB et(cid:22)I ).
n
2 (cid:0)
i) Ker(eB e(cid:22)I ) Ker(B (cid:22)I ) :
n n
(cid:0) (cid:26) (cid:0)
Let v Ker(eB e(cid:22)I ). Since Ker(eB e(cid:22)I ) Ker(etB et(cid:22)I )) then (cid:26) (t) =0 for every
n n n v
2 (cid:0) (cid:0) (cid:26) (cid:0)
t R, where (cid:26) : t (etB et(cid:22)I )v. Thus @(cid:26)v(t) = (BetB (cid:22)et(cid:22)I )v = 0 for every t R.
2 v ! (cid:0) n @t (cid:0) n 2
In particular for t= 0, we have (B (cid:22)I )v = 0. Therefore v Ker(B (cid:22)I )
n n
(cid:0) 2 (cid:0)
Ker(B (cid:22)I ) Ker(eB e(cid:22)I ) : Let v Ker(B (cid:22)I ). We have Bv = (cid:22)v and then :
n n n
(cid:0) (cid:26) (cid:0) 2 (cid:0)
+ +
eBv = 1Bkv = 1(cid:22)kv = e(cid:22)v; So v Ker(eB e(cid:22)I )
k! k! 2 (cid:0) n
k=0 k=0
P P 4
ii) We proceed by induction on n:
for n = 1 it is clear. Suppose the property is true until (n 1). Let B M (C), having
n
(cid:0) 2
only one eigenvalue (cid:22) such that eB T (C). Then e Ker(eB e(cid:22)I ). By i) we have
2 (cid:3)n n 2 (cid:0) n
e Ker(B (cid:22)I ). It follows that:
n n
2 (cid:0)
B(1) 0
B = ; with B(1) M (C) and L M (C):
LB (cid:22) 2 n(cid:0)1 B 2 1;n(cid:0)1
(cid:18) (cid:19)
Since (cid:22) is the only eigenvalue of B then (cid:22) is also the only eigenvalue of B(1). Moreover, we
can check that :
eB(1) 0
eB = ; with L M (C):
(cid:18) LeB e(cid:22) (cid:19) eB 2 1;n(cid:0)1
Since eB T (C) then eB(1) T (C). By induction hypothesis, we have B(1) T (C)
2 (cid:3)n 2 (cid:3)n(cid:0)1 2 n(cid:0)1
and then B T (C). (cid:3)
n
2
Proof of Proposition 2:1
a) Let’s show (cid:12)rst that exp(T (C)) = T (C):
n (cid:3)n
exp(T (C)) T (C) let B T (C) then eB GL(n;C). Since B is lower triangular with
n (cid:26) (cid:3)n 2 n 2
one eigenvalue (cid:21) then eB is lower triangular with one eigenvalue e(cid:21); that is: B T (C).
2 (cid:3)n
T (C) exp(T (C)): let A T (C) with eigenvalue (cid:21). It su(cid:14)ces to show that there exists
(cid:3)n (cid:26) n 2 (cid:3)n
B M (C) having one eigenvalue such that A= eB since by Lemma 2:2;ii), such B T (C).
n n
2 2
Let J be the Jordan normal form of A; that is: there exists P GL(n;C) such that:
2
(cid:21) 0 :: 0
J 0 0
1 1 :: :: 0
PAP(cid:0)1 = J = 0 0 :: 0 1 2 T(cid:3)n(C), where Jk = 0 0 :: :: 0 1 2 T(cid:3)nk(C) if nk (cid:21) 2
0 0 J
r B 0 0 1 (cid:21) C
@ A B C
and Jk = ((cid:21)) if nk = 1. @ A
Since (cid:21)= 0, let (cid:22) C such that e(cid:22) = (cid:21).
6 2
J 0 0
10
We let J = 0 :: 0 T (C), where
0 n
0 12
0 0 J
r0
(cid:22) 0@ :: 0 A
1 :: :: 0
Jk0 = 0 0 :: :: 0 1 2 Tnk(C) if nk (cid:21) 2 and Jk = ((cid:22)) if nk = 1.
B 0 0 1 (cid:22) C
B C
By Le@mma 2:2;i) weAhave dim(Ker(eJk0 (cid:0)e(cid:22)Ink)) = dim(Ker(Jk0 (cid:0)(cid:22)Ink)) =1. Hence, there
exists Qk 2 GL(nk;C) such that QkeJk0Q(cid:0)k1 = Jk, k = 1::;r.
Q 0 0
1
If Q = 0 :: 0 then we get QeJ0Q(cid:0)1 = J.
0 1
0 0 Q
r
Take B@= P 1QJ Q 1AP, we obtain (cid:22) is the unique eigenvalue of B and A = eB.
(cid:0) 0 (cid:0)
b) exp(K (C)) = K (C). Let B K (C). We have:
n;r (cid:3)n;r 2 n;r
B 0 0
1
B = 0 :: 0 where B T (C):
0 1 k 2 nk
0 0 B
r
@ A5
Then
eB1 0 0
eB = 0 0 :: 0 1 and eBk 2 T(cid:3)nk(C)
0 0 eBr
So, eB K (C). @ A
2 (cid:3)n;r
Conversely, if A K (C) then:
2 (cid:3)n;r
A 0 0
1
A = 0 :: 0 where A T (C):
0 1 k 2 (cid:3)nk
0 0 A
r
@ A
By a), there exists Bk 2 T(cid:3)nk(C) such that eBk = Ak for every k = 1;::;r. Hence A = eB
B 0 0
1
where B = 0 :: 0 K (C):
n;r
0 1 2
0 0 B
r
@ A
Proposition 2.3. If A; B K (C) such that eAeB = eBeA then AB =BA.
n;r
2
Proof. a) Let’s show (cid:12)rst the case where A;B T (C).
n
2
Let A; B T (C) such that eAeB = eBeA.
n
2
We have: emAemB = emBemA for every m N.
2
We let N = A (cid:21)I and M = B (cid:22)I , where (cid:21) (resp. (cid:22)) is the only eigenvalue of A (resp.
n n
(cid:0) (cid:0)
of B). We have Nn = Mn = 0.
We let P : R C de(cid:12)ned by P (t) = < (etNetM etMetN)e ; e >, 1 i;j n, where
i;j i;j i j
! (cid:0) (cid:20) (cid:20)
<; > is the scalar product.
P is a polynomial of degree less then 2n.
i;j
By above, we have P (m) = 0 for every m N, 1 i;j n. Then P = 0 for every
i;j i;j
2 (cid:20) (cid:20)
1 i; j n, so etNetM = etMetN and therefore etAetB = etBetA, t R.
(cid:20) (cid:20) 2
Consider the map f : R T (C) de(cid:12)ned by f(t)= etAetBe tAe tB.
! (cid:3)n (cid:0) (cid:0)
By a), we have f(t) = I for every t R. Moreover since etAetB = etBetA and etAA = AetA
n
2
then:
@f(t)
= A+etABe tA etBAe tB B =0
(cid:0) (cid:0)
@t (cid:0) (cid:0)
and
@2f(t)
= AetABe tA etABAe tA BetBAe tB +etBABe tB =0;
(cid:0) (cid:0) (cid:0) (cid:0)
@t2 (cid:0) (cid:0)
for every t R.
2
For t= 0, we obtain AB = BA.
b) Let A; B K (C) such that eAeB = eBeA. We have:
n;r
2
A 0 ::: 0 B 0 ::: 0
1 1
0 : : : 0 : : :
A = 0 1 and B =0 1
: :: : 0 : :: : 0
B 0 ::: 0 Ar C B 0 ::: 0 Br C
B C B C
@ A @ A
where A ;B T (C), k = 1;::;r.
k k 2 nk
6
Then
eA1 0 ::: 0 eB1 0 ::: 0
0 : : : 0 : : :
eA = 0 1 and eB = 0 1
: :: : 0 : :: : 0
B 0 ::: 0 eAr C B 0 ::: 0 eBr C
B C B C
@ A @ A
Since eAeB = eBeA then eAkeBk = eBkeAk for every k = 1;::;r.
By a) we have A B = B A for every k = 1;::;r. Hence AB = BA. (cid:3)
k k k k
Proposition 2.4. Let A; B T (C) such that AB = BA. If eA = eB then there exists k Z
n
2 2
such that A = B+2ik(cid:25)I .
n
Lemma 2.5. Let M T (C) be nilpotent such that eM = I then M = 0.
n n
2
Proof. Take P (t) =< (etM I )e ; e > for t R and 1 i;j n. Where <; > is the scalar
i;j n i j
(cid:0) 2 (cid:20) (cid:20)
product on Cn. P (t) is a polynomial since Mn = 0.
i;j
Since P (m) = 0 for every m Z then P (t) = 0 for every t R and 1 i; j n.
i;j i;j
2 2 (cid:20) (cid:20)
Hence etM I = 0 for every t R.
n
(cid:0) 2
Take (cid:26) : t etM I . Then @(cid:26)(t) = MetM = 0. For t= 0, we have M =0. (cid:3)
7(cid:0)! (cid:0) n @t
Proof. (of Proposition 2:4 : ) Let A, B T (C) such that AB = BA and eA =eB.
n
2
Denote by (cid:21) (resp. (cid:22)) the unique eigenvaleue of A (resp. B). If e(cid:21) = e(cid:22) then (cid:21) = (cid:22)+2ik(cid:25)
for some k Z.
2
Take N = A (cid:21)In and N0 = B (cid:22)In. We have e(cid:21)eN = e(cid:22)eN0. Then eN = eN0. Since
(cid:0) (cid:0)
AB = BA, then NN0 = N0N and so eN(cid:0)N0 = In. By Lemma 2:5, N = N0 and therefore
A = B+2ik(cid:25)I . (cid:3)
n
The following proposition was proved in his book [5]:
Proposition 2.6. [5] Let A M (C). Then if no two eigenvalues of A have a di(cid:11)erence of the
n
2
form 2i(cid:25)k, k Z 0 , then exp :M (C) GL(n;C) is a local di(cid:11)eomorphism at A.
n
2 (cid:0)f g (cid:0)!
Corollary 2.7. The restriction exp=Kn;r(C) : Kn;r(C) (cid:0)! GL(n;C) is an open map.
Proof. Let A T (C). Then by Proposition 2:6, exp is a local di(cid:11)eomorphism at A. Therefore,
n
2
exp=Tn(C) : Tn(C) (cid:0)! GL(n;C) is open. Now, if A2 Kn;r(C) then:
exp=Tn1(A1) 0 ::::: 0
0 : : :
exp=Kn;r(C)(A) = 0 : : : 0 1
BB 0 :::: 0 exp=Tnr(Ar) CC
@ 7 A
where
A 0 :::: 0
1
0 : : :
A =0 : : : 0 1; Ak 2 Tnk(C); k = 1;::;r
B 0 ::: 0 Ar C
B C
Denote by @ A
(cid:31) : K (C) T (C) ::: T (C)
n;r (cid:0)! n1 (cid:2) (cid:2) nr
A 0 :::: 0
1
0 : : :
A= 0 : : : 0 1 7(cid:0)! (cid:31)(A) = (A1; ::::; Ar )
B 0 ::: 0 Ar C
B C
@ A
exp : T (C) ::: T (C) GL(n ;C) ::: GL(n ;C)
n1 (cid:2) (cid:2) nr (cid:0)! 1 (cid:2) (cid:2) r
(A1; ::::; Ar ) 7(cid:0)! exp=Tn1(C)(A1); ::::; exp=Tnr(C)(Ar)
g
Therefore, exp is an open map since ea(cid:16)ch exp=Tnk(C) is open. It follows(cid:17)that since
exp=Kn;r(C) = (cid:31)(cid:0)1 (cid:14)exp(cid:14)(cid:31) and (cid:31) is a isomorphism that exp=Kn;r(C) is an open map. (cid:3)
g
g
3. Some properties of subgroups of K (C)
(cid:3)n;r
In this section, G is an abelian subgroup of K (C) and then g = exp 1(G) K (C).
(cid:3)n;r (cid:0) \ n;r
B 0 :::: 0
1
0 : : :
Every B 2 Kn;r(C) has the form: B = 0 : : : 0 1 where Bk 2 Tnk(C). Denote
B 0 ::: 0 Br C
B C
by: @ A
G = B : B G . G is an abelian subgroupof T (C). We let g =exp 1(G ) T (C).
k f k 2 g k (cid:3)nk k (cid:0) k \ nk
Then every B g has the form :
2
B 0 :::: 0
1
0 : : :
B = 0 : : : 0 1 where Bk 2 gk :
B 0 ::: 0 Br C
B C
B 0 @:::: 0 A eB1 0 :::: 0
1
0 : : : 0 : : :
Indeed, if B = 0 1 then eB = 0 1 G: Then eBk
: : : 0 : : : 0 2 2
B 0 ::: 0 Br C B 0 ::: 0 eBr C
B C B C
Gk and so, Bk g@k. A @ A
2
Corollary 3.1. g is a subgroup of K (C). In particular, g is an additif subgroup of Cn for
n;r v
every v Cn.
2
Proof. Let A; B g. Then eAeB = eBeA (since G is abelian). By Proposition 2:3, AB = BA
2
then eA+B = eBeA G and A+B K (C). It follows that A+B g. (cid:3)
n;r
2 2 2
8
Denote by: (G) = B K (C); BA= AB for every A G
n;r
C f 2 2 g
(G) is a vector subspace of K (C) and we have G (G) since G is abelian.
n;r
C (cid:26) C
Corollary 3.2. i) exp(g) = G
ii) exp( (G)) = (G) GL(n;C).
C C \
iii) (g) = (G). In particular g (G) and all matrix of g commute.
C C (cid:26) C
Proof. Let G be an abelian subgroup of K (C):
(cid:3)n;r
i) We have exp(G) G by de(cid:12)nition.
(cid:26)
G exp(g): Let A G. Since G T (C), then by Proposition 2:1 there exists B K (C)
(cid:26) 2 (cid:26) (cid:3)n 2 n;r
such that eB = A. Hence B exp 1(G) K (C) = g and then A exp(g). It follows that
(cid:0) n;r
2 \ 2
exp(g) = G.
ii) Let A = eB exp( (G)) where B (G). Moreover if C G then BC = CB and
2 C 2 C 2
therefore CeB = eBC. i.e. AC = CA. It follows that A (G) GL(n;C) since A GL(n;C).
2 C \ 2
Conversely, let A (G) GL(n;C). Then A K (C). By Proposition 2:1 there exists
2 C \ 2 (cid:3)n;r
B K (C) such that eB = A. Moreover, if C G then AC = CA. Hence CeB = eBC then
n;r
2 2
eCeB = eBeC. Since B and C G K (C) then by Proposition 2:3 we have BC = CB.
n;r
2 (cid:26)
Therefore B (G). It follows that A exp( (G)).
2 C 2 C
iii) Let B (G) and A g. Then eA G and eAB = BeA. So, eAeB = eBeA. Since
2 C 2 2
A;B K (C) then by Proposition 2:3, AB =BA. So, (G) (g).
n;r
2 C (cid:26) C
Conversely, let B (g) and A G. By i) there exists C g such that eC = A then
2 C 2 2
BC = CB. Hence BeC = eCB. i.e. BA = AB. It follows that B (G). We have
2 C
g (g) = (G) and the elements of g commute. (cid:3)
(cid:26) C C
Proposition 3.3. U is a G-invariant dense open set in Cn.
Proof. It is clear that U is a dense open set in Cn. Let’s show that U is G-invariant:
v v
1 k;1
: :
Let v 2 U and A 2 G. Then v = 0 : 1 2 Cn, where vk = 0 : 1 2 C(cid:3) (cid:2) Cnk(cid:0)1,
B vr C B vk;r C
B C B C
1 k r and @ A @ A
(cid:20) (cid:20)
(cid:22) 0 ::: 0
A1 0 ::: 0 Ak
(k)
0 : : : a : : :
A = 0 : : : 0 1 where Ak = 0 :2:;1 : : : 12T(cid:3)nk(C);
BB@ 0 :::: 0 Ar CCA BBB an(kk);1 :: an(kk);nk(cid:0)1 (cid:22)Ak CCC
@ A
k = 1;::;r. We have:
9
A v v w
1 1 k;1 k;1
: : :
Av = 0 : 1 where Akvk = Ak0 : 1 = 0 : 1 with wk;1 = (cid:22)Akvk;1. Since
vk;1 = 0 tBB@henArwvrk;1CCA= 0. Therefore AkvkBB@ Uvkk;nakndCCAAv BB@Uw. k;nk CCA (cid:3)
6 6 2 2
4. Parametrization
4.1. Parametrisation of subgroup of T (C). In this section, G is a subgroup of T (C)
(cid:3)n (cid:3)n
and then g = exp 1(G) T (C).
(cid:0) n
\
Denote by: F = vect((B (cid:22) I )e : B G; 1 i n 1 )
G B n i
(cid:0) 2 (cid:20) (cid:20) (cid:0)
(resp: F = vect((B (cid:22) I )e : B g; 1 i n 1 )) the vector subspace of Cn gen-
g B n i
(cid:0) 2 (cid:20) (cid:20) (cid:0)
erated by ((B (cid:22) I )e ; B G; 1 i n 1 )
B n i
(cid:0) 2 (cid:20) (cid:20) (cid:0)
(resp: ((B (cid:22) I )e ; B g; 1 i n 1 )) where (cid:22) is the only eigenvalue of B. We
B n i B
(cid:0) 2 (cid:20) (cid:20) (cid:0)
have rank(F ) n 1.
G
(cid:20) (cid:0)
x x
1 1
: v(1) :
For a vector v = 0 : 1 2 Cn, we write v = xn where v(1) = 0 : 1 2 Cn(cid:0)1. In
(cid:18) (cid:19)
B xn C B xn 1 C
B C B (cid:0) C
e@(1) A @ A
particular, e = k ; k = 1;::;n 1.
k
0 (cid:0)
(cid:18) (cid:19)
(cid:22) 0 :::: 0
B
b : :: :
Let B = 0 2::;1 :: : 0 1 2 Tn(C). We can write:
B bn;1 ::::: bn;n 1 (cid:22)B C
B (cid:0) C
@ A (cid:22) 0 :: 0
B
B(1) 0 b : : :
B = LB (cid:22)B ,whereB(1) = 0 2::;1 :: : 0 12 Tn(cid:0)1(C),LB = (bn;1:::;bn;n(cid:0)1)2
(cid:18) (cid:19)
B bn 1;1 ::: bn 1;n 2 (cid:22)B C
B (cid:0) (cid:0) (cid:0) C
M1;n 1(C). @ A
(cid:0)
Denote by:
- G(1) = B(1); B G . Then, G(1) is an abelian subgroup of T (C).
f 2 g (cid:3)n(cid:0)1
- F(1) = (B(1) (cid:22) I )e(1); B G; k = 1::;n 2 .
G f (cid:0) B n(cid:0)1 k 2 (cid:0) g
Proposition4.1. LetGbeanabeliansubgroupofT (C). Ifrank(F ) = n 1(resp. rank(F ) =
(cid:3)n G (cid:0) g
n 1 ) then :
(cid:0)
there exists an injective linear map ’ : Cn T (C) (resp. : Cn T (C) ) such
n n
(cid:0)! (cid:0)!
that:
i) (G) ’(Cn) (resp. (g) (Cn) )
C (cid:26) C (cid:26)
ii) For every v Cn, ’(v)e = v (resp. (v)e = v ).
1 1
2
Proof. Let’s prove the Proposition for rank(F ).
G
10