DYNAMIC 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