Available at: http://publications.ictp.it IC/2009/062

United Nations Educational, Scientific and Cultural Organizationand

International Atomic Energy Agency

THE ABDUS SALAM INTERNATIONAL CENTRE FOR THEORETICAL PHYSICS

ON THE DYNAMIC OF ABELIAN GROUPSOF AFFINE MAPS ON Cn AND Rn

Adlene Ayadi1

Department of Mathematics, Faculty of Sciences of Gafsa, Gafsa, Tunisia,

Habib Marzougui2

Department of Mathematics, Faculty of Sciences of Bizerte, Zarzouna 7021, Tunisiaand

The Abdus Salam International Centre for Theoretical Physics, Trieste, Italy

and

Yahya N’dao3

Department of Mathematics, Faculty of Sciences of Gafsa, Gafsa, Tunisia.

Abstract

In this paper, we characterize the dynamic of any Abelian group G of affine maps on Kn (K = Ror C). We show that there exist a G-invariant affine subspace E of Kn and (if E 6= Kn), at mostn G-invariant affine subspaces H1, ..., Hp (1 ≤ p ≤ n) of Kn of dimension n − 1 or n − 2 over Ksuch that, every orbit in E (resp. U = Kn\

p⋃k=1

Hk) is minimal in E (resp. U). As a consequence,G has height at most n, in particular it has a minimal set in Kn. Moreover, if G has a dense orbit,all orbits in U are dense in Kn.

MIRAMARE – TRIESTE

August 2009

1adleneso@yahoo.fr2Regular Associate of ICTP. hmarzoug@ictp.it3yahiandao@yahoo.fr

1. Introduction

Let GL(n, K) be the group of all invertible square matrices of order n ≥ 1 with entries in K(K = R or C). A map f : Kn −→ Kn is called an affine map if there exist A ∈ GL(n, K) anda ∈ Kn such that f(x) = Ax + a, x ∈ Kn. We denote f = (A, a). Denote by GA(n, K) the set ofall affine maps. For composition of maps, GA(n, K) is a group.

Let G be an Abelian affine subgroup of GA(n, K). For a vector v ∈ Kn, we consider the orbitof G through v: G(v) = {f(v) : f ∈ G} ⊂ Kn. A subset E ⊂ Kn is called G-invariant if f(E) ⊂ Efor any f ∈ G; that is E is a union of orbits. Before stating our main results, we introduce thefollowing notions:

A subset H of Kn is called an affine subspace of Kn if there exist a vector subspace H of Kn

and a ∈ Kn such that H = H + a. For a ∈ Kn, denote by Ta : Kn −→ Kn; x 7−→ x + a thetranslation map by vector a, so H = Ta(H). We say that H has dimension p (0 ≤ p ≤ n), noteddim(H) = p, if H has dimension p and we denote by cod(H) = n − dim(H) the codimension ofH. Denote by FixG = {x ∈ Kn : f(x) = x for every f ∈ G}. If FixG 6= ∅, it is a G-invariantclosed affine subspace of Kn.

A subset E of Kn is called a minimal set of G if E is closed in Kn, non empty, G-invariantand has no proper subset with these properties. It is equivalent to say that E is a G-invariant setsuch that every orbit contained in E is dense in it.

If V is a G-invariant open set in Kn, we say that E is a minimal set in V if it is a minimal setof the restriction G/V of G to V . An orbit O ⊂ V is called minimal in V if O ∩ V is a minimalset in V . For example, a closed orbit in V is minimal in V . In particular, if FixG 6= ∅, everypoint in FixG is minimal in Kn. When F is a G-invariant affine subspace of Kn, an orbit O in Fis called minimal in F if O ∩ V is a minimal set of G/F .

We call class of an orbit O of G the subset cl(O) reunion of orbits γ of G such that γ = O(where γ is the closure of γ). In particular, if V is a G-invariant open set and O is a minimalorbit in V then cl(O) = O ∩ V . (See the proof of Corollary 1.2).

An orbit O of G is said to be at level 0 if O is minimal in Kn. Inductively, we say that O is atlevel p, p ≥ 1 if every orbit γ ⊂ O\cl(O) is at level < p with at least one orbit at level k for everyk < p. The upper bound of levels of orbits of G is called the height of G and denoted by ht(G).For example, if ht(G) is finite, say p this means that p is the supremum of k ∈ N such that thereexist orbits γ0, γ1, ..., γk of G with γ0 ⊂ γ1 ⊂ ... ⊂ γk and γi 6= γj , for every 0 ≤ i, j ≤ k, i 6= j.

The dynamic of matrix groups is an area of a very active research, (see for instance, [4], [6], [1]).

In [1], the first two authors established a structure’s theorem for Abelian subgroup of GL(n, K);

which gives a dynamical decomposition of the phase space Kn into regions saturated by minimal

orbits. For groups of affine maps on Kn, their dynamics were recently initiated for some classes

in different point of view, (see for instance, [8], [3],[7],[5]). The purpose here is to give analogous2

of that theorem for affine Abelian subgroup of GA(n, K) and to improve in the same time the

linear case. Our main results are the following:

Theorem 1.1. Let G be an Abelian affine subgroup of GA(n, K) (K = R or C). Then thereexist a G-invariant affine subspace E of Kn and (if E 6= Kn), an integer 1 ≤ rG ≤ cod(E) andG-invariant affine subspaces H1,..., HrG of Kn of dimension n − 1 or n − 2 over K such thatFixG ⊂ E ⊂

rG⋂k=1

Hk and with the following properties:

(i) Every orbit in E is minimal in E.(ii) If U = Kn\

rG⋃k=1

Hk, every orbit in U is minimal in U .

Under the hypothesis of Theorem 1.1, we have the following corollaries:

Corollary 1.2. ht(G) ≤ cod(E).

Corollary 1.3. (i) G admits a minimal set in Kn (contained in E).(ii) If FixG 6= ∅ then G admits a minimal set in Kn\FixG (resp. in Kn\E.)

Corollary 1.4. If cod(E) = 1 then U = Kn\E and every orbit in U is minimal in it.

Corollary 1.5. If G has a locally dense orbit in Cn then every orbit in U is dense in Cn.

Corollary 1.6. (i) If G has a locally dense orbit O in Rn and C is a connected component of Umeeting O then every orbit meeting C is dense in it.

(ii) If G has a dense orbit in Rn then every orbit in U is dense in Rn.

Remark 1. i) If E = Kn, every orbit in Kn is minimal in it.ii) If n = 1 and G 6= {id} (id is the identity map), then FixG is either ∅ (in this case, G is a

subgroup of translations), or reduced to a point u (in this case, G is conjugated by translationto a subgroup of homotheties and every orbit in K\{u} is minimal in K\{u}. (See Proposition5.6).

iii) If E 6= Kn then for every 1 ≤ k ≤ rG , dimHk = n− 1 (resp. n− 1 or n− 2) if K = C (resp.K = R); this follows from the proof of Theorem 1.1.

iv) If K = C, the open set U = Kn\rG⋃

k=1

Hk is connected, but may be not connected if K = R.(See Example 5.5 and also Example 4.6).

Remark 2. i) Every orbit O in E is at level 0 (indeed, by Theorem 1.1, O is minimal in E andas E is closed in Kn, hence O is minimal in Kn).

ii) The orbits in U are not homeomorphic in general. (See Example 5.2).3

iii) Corollary 1.5 is not true in general in Rn. (See Example 4.6.)

This paper is organized as follows: In section 2, we introduce a notation and some preliminary

lemmas that will be used throughout the remainder of the paper and we prove Theorem 1.1 in

the linear case. The proof of Theorem 1.1 in the affine case and Corollaries 1.2, 1.3, 1.4, 1.5

and 1.6 are done in section 3. In section 4, we give some conditions for topological conjugate an

Abelian affine group to its linear. Section 5 is devoted to the application of some affine subgroups

which illustrate on one hand the dynamic behavior of their orbits and on the other, the difference

between the dynamic of affine and linear groups.

2. The linear case

2.1. Notation. Denote by K∗ = K\{0} and N∗ = N\{0}. Let n ∈ N∗ be fixed. For any integer1 ≤ m ≤ n, denote by

T∗m(K) =

A =

µA 0 . 0a2,1 . . .. . . 0

am,1 . am,m−1 µA

: µA ∈ K

∗

the subgroup of GL(m, K) of lower triangular matrices.

2.1.1. Notation for K = C. Let r ∈ N∗ and r0 ∈ N such that 0 ≤ r0 ≤ r ≤ n and we letη = (n1, ..., nr) ∈ (N∗)r where

r∑k=1

nk = n.

Denote by

- K∗η,r(C) =

A =

A1 0 . 00 . . .. . . 00 . 0 Ar

: Ak ∈ T∗nk(C), 1 ≤ k ≤ r

- K∗η,r0,r(C) =

A1 0 . . . 00 . . .. . .. . Ar0 . . .. . . Ar0+1 . .. . . . . 00 . . . 0 Ar

∈ K∗η,r(C) : |µAk | = 1, for r0 < k ≤ r

.

It is obvious that K∗η,r(C) is a group and that K∗η,r0,r(C) is a subgroup of K∗η,r(C). If r0 = rthere exists, for every 1 ≤ k ≤ r, a matrix A ∈ K∗η,r,r(C) such that |µAk | 6= 1. If r0 = 0 then, forevery A ∈ K∗η,0,r(C), we have |µAk | = 1, for every 1 ≤ k ≤ r.

Let G be an Abelian subgroup of K∗η,r0,r(C). Then every matrix A ∈ G is written in the

form A =

A1 0 . . 00 . . . .. . Ar0 . 0. . . . .0 . . 0 Ar

where Ak ∈ T∗nk(C) and µAk is the only eigenvalue of Ak,

1 ≤ k ≤ r.4

Denote by

J = {k ∈ {1, ..., r} : |µAk | = 1, for every A ∈ G}

and the number of element of J by rG = card(J). We have r0 ≤ rG ≤ r.We let Ek =

{(x1, ..., xr) ∈

r∏i=1

Cni : xi = 0, for every i 6= k}

. Ek is a G-invariant vector

subspace of Cn of dimension nk, 1 ≤ k ≤ r.Denote by

E(G) =

⊕k∈J

Ek if rG 6= 0, r,

Cn if rG = r,

{0} if rG = 0.and

H(G) =

⊕k/∈J

Ek if rG 6= 0, r,

{0} if rG = r,Cn if rG = 0.

We have the decomposition Cn =⊕r

k=1Ek = H(G)⊕

E(G).

We let p =rG∑

k=1

nk. Then E(G) = {0Cp} × Cq and H(G) = Cp × {0Cq}, q = n − p.

2.1.2. Notation for K = R. Let r, s, r0, s0 ∈ N such that 0 ≤ r0 ≤ r, 0 ≤ s0 ≤ s. If rs 6= 0, welet η = (n1, .., nr;m1, ..,ms) ∈ (N∗)r+s where n =

r∑k=1

nk + 2s∑

l=1

ml. If r = 0 (resp. s = 0), we let

η = (m1, ..,ms) (resp. (n1, .., nr)). In particular, we have r + 2s ≤ n.Denote by

- S =

{(α β−β α

): α, β ∈ R

}.

- Bm(R) =

C 0 . 0C2,1 . . 0

. . . 0Cm,1 . Cm,m−1 C

∈ M2m(R) : C, Ci,j ∈ S,

2 ≤ i ≤ m,1 ≤ j ≤ m − 1

- B∗m(R) = Bm(R) ∩ GL(2m, R).

- K∗η,r,s(R) =

A1 0 . . . 00 . . .. . .. . Ar . . .. . . B1 . .. . . . . 00 . . . 0 Bs

∈ Mn(R) :Ak ∈ T∗nk(R), 1 ≤ k ≤ r,Bl ∈ B∗ml(R), 1 ≤ l ≤ s.

.

- K∗η,r0,s0,r,s(R) ={

M =

(M0 00 M1

)∈ GL(n, R)

}

where5

M0 =

A1 0 . . . 00 . . .. . .. . Ar0 . . .. . . B1 . .. . . . . 00 . . . 0 Bs0

∈ K∗η0,r0,s0(R),

M1 =

Ar0+1 0 . . . 00 . . .. . .. . Ar . . .. . . Bs0+1 . .. . . . . 00 . . . 0 Bs

∈ K∗η1,r−r0,s−s0(R),

η0 =

(n1, .., nr0 ;m1, ..,ms0) if r0s0 6= 0,(m1, ..,ms0) if r0 = 0,

(nr0+1, .., nr) if s0 = 0

and

η1 =

(nr0+1, .., nr;ms0+1, ..,ms) if r0s0 6= 0,(ms0+1, ..,ms) if r0 = 0,

(nr0+1, .., nr) if s0 = 0

Let G be an Abelian subgroup of K∗η,r0,s0,r,s(R). Then every matrix M ∈ G is written

in the form M =

(M0 00 M1

)∈ GL(n, R), where

M0 =

A1 0 . . . 00 . . .. . .. . Ar0 . . .. . . B1 . .. . . . . 00 . . . 0 Bs0

∈ K∗η0,r0,s0(R)

and

M1 =

Ar0+1 0 . . . 00 . . .. . .. . Ar . . .. . . Bs0+1 . .. . . . . 00 . . . 0 Bs

∈ K∗η1,r−r0,s−s0(R).

Denote by

- J = {k ∈ {1, ..., r} : |µAk | = 1, for every M ∈ G}- rG = card(J). We have r0 ≤ rG.- J ′ = {l ∈ {1, ..., s} : |µBl | = 1, for every M ∈ G}- sG = card(J

′). We have s0 ≤ sG.6

We let

Ek =

(X,Y ) ∈ Rn : X = (x1, .., xr0 , y1, .., ys0);Y = (xr0+1, .., xr, ys0+1, .., ys),xi ∈ Rni , yj ∈ R2mj ,

xi = yj = 0,for every i 6= k,

1 ≤ j ≤ s, 1 ≤ i ≤ r

.

Ek is a G-invariant vector subspace of Rn of dimension nk, 1 ≤ k ≤ r.

E′l =

(X,Y ) ∈ Rn : X = (x1, .., xr0 , y1, .., ys0), Y = (xr0+1, .., xr, ys0+1, .., ys),xi ∈ Rni , yj ∈ R2mj

xi = yj = 0,for every j 6= l,

1 ≤ i ≤ r, 1 ≤ j ≤ s

.

E′l is a G- invariant vector subspace of Rn of dimension 2ml, 1 ≤ l ≤ s.

Denote by

E(G) =

⊕k∈J

Ek⊕ ⊕

l∈J ′E′l if rG 6= 0, r and sG 6= 0, s,

Rn if rG = r and sG = s,

{0} if rG = 0 and sG = sand

H(G) =

⊕k/∈J

Ek⊕ ⊕

l /∈J ′E′l if rG 6= 0, r, sG 6= 0, s

{0} if rG = r and sG = sRn if rG = 0 and sG = s

We have the decomposition Rn =⊕r

k=1Ek⊕s

l=1E′l = H(G)

⊕E(G).

We let p =rG∑

k=1

nk + 2sG∑l=1

ml and q = n − p. We have 0 ≤ rG ≤ r.Then E(G) = {0Rp} × Rq and H(G) = Rp × {0Rq}.

Finally, throughout the paper, we denote by B0 = (e1, ..., en) the canonical basis of Kn.

2.2. Some results for Abelian subgroups of K∗η,r0,r(C) (resp. K∗η,r0,s0,r,s(R).

Proposition 2.1. Let G be an Abelian subgroup of K∗η,r0,r(C) and let U ′ =r0∏

k=1

(C∗×Cnk−1)×Cq

where q = n −r0∑

k=1

nk. Then for all v, w ∈ U ′ and any sequence (Am)m∈N ⊂ G such that

limm−→+∞

Amv = w, we have limm−→+∞

A−1m w = v. In particular every orbit in U′ is minimal in U ′.

To prove Proposition 2.1, we need the following lemmas.

Lemma 2.2. ([1], Corollary 3.3) Suppose G is an Abelian subgroup of T∗n(K) (K = R or C ).

Then for all v, w ∈ K∗ × Kn−1 and any sequence (Am)m∈N ⊂ G such that limm−→+∞

Amv = w,

we have limm−→+∞

A−1m w = v.

7

Lemma 2.3. Suppose G is an Abelian subgroup of T∗n(K) (K = R or C) such that for every A ∈ G,|µA| = 1. Then for all v, w ∈ Kn and any sequence (Am)m∈N ⊂ G such that lim

m−→+∞Amv = w,

we have limm−→+∞

A−1m w = v.

Proof. The proof is done by induction on n.

For n = 1, G is a subgroup of K∗ and the statement is obvious.

Suppose the Lemma is true up to dimension n− 1 and let G be an Abelian subgroup of T∗n(K)so that for every A ∈ G, |µA| = 1. Let v = (x1, ..., xn) ∈ Kn, w = (y1, ..., yn) ∈ Kn and(Am)m∈N ⊂ G be a sequence such that lim

m−→+∞Amv = w. Then lim

m−→+∞µAmx1 = y1. If x1 6= 0,

so y1 6= 0 (since |µAm | = 1 for every m ∈ N), hence v, w ∈ K∗ × Kn−1. By Lemma 2.2, we havelim

m−→+∞A−1m w = v. Now, if x1 = 0, then y1 = 0 and so v,w ∈ F = {0} × Kn−1. As F is a

G-invariant vector subspace of Kn of dimension n− 1 and since limm−→+∞

(Am/F )u = v, we obtain,

by applying the induction hypothesis to the restriction G/F , limm−→+∞

(Am/F

)−1w = v and so

limm−→+∞

A−1m w = v. �

Proof of Proposition 2.1. Suppose G is a subgroup of K∗η,r0,r(C). Let

v =

v1..vr

∈ U ′ and w =

w1..

wr

∈ U ′. Then vk, wk ∈ C∗ × Cnk−1 (resp. Cnk) for

1 ≤ k ≤ r0 (resp. r0 < k ≤ r).Let (Am)m∈N ⊂ G be a sequence such that lim

m−→+∞Amv = w. Write Am in the form

Am =

Am,1 0 . . 00 . . . .. . Am,r0 . 0. . . . .0 . . 0 Am,r

where Am,k ∈ T∗nk(C) for 1 ≤ k ≤ r and |µAm,k | = 1, for r0 < k ≤ r. Then limm−→+∞Am,kvk = wkfor every 1 ≤ k ≤ r. It follows from Lemma 2.2 (if 1 ≤ k ≤ r0) and from Lemma 2.3 (if r0 < k ≤ r),that lim

m−→+∞A−1m,kwk = vk, for every 1 ≤ k ≤ r. Hence, limm−→+∞A

−1m w = v. �

Analogous of Proposition 2.1 for the real case is the following:

Proposition 2.4. Let G be an Abelian subgroup of K∗η,r0,s0,r,s(R) and let

U ′ =r0∏

k=1

(R∗ × Rnk−1) ×s0∏

l=1

(R2\{0}) × R2ml−2 × Rq,

where q = n − (r0∑

k=1

nk + 2s0∑

l=1

ml). Then for all v, w ∈ U ′ and any sequence (Am)m∈N ⊂ G such

that limm−→+∞

Amv = w, we have limm−→+∞

A−1m w = v. In particular every orbit in U′ is minimal in

U ′.8

To prove Proposition 2.4, we need the following lemmas.

Lemma 2.5. Suppose G is an Abelian subgroup of B∗s(R). Then for all v, w ∈(R2\{0}

)×R2s−2

and any sequence (Bm)m∈N ⊂ G such that limm−→+∞

Bmv = w, we have limm−→+∞

B−1m w = v.

Proof. Take uk =e2k−1+ie2k

2 ∈ C2s, 1 ≤ k ≤ s. Then C0 = (u1, ....., us; u1, ....., us) is a basis ofC2s. Denote by P the matrix of basis change of B0 to C0. For every matrix

B =

C 0 . 0C2,1 . . .

. . . 0Cs,1 . Cs,s−1 C

∈ G, P

−1BP is written in the form :

P−1BP =

(B′ 00 B′

)where B′ ∈ T∗s(C).

Let v = (x1, y1, ...., xs, ys) with (x1, y1) ∈ R2\{(0, 0)}, w ∈(R2\{0}

)× R2s−2 and a sequence

(Bm)m ⊂ G such that limm−→+∞

Bmv = w. Then

P−1v = (x1 + iy1, ..., xs + iys; x1 − iy1, ..., xs − iys) ∈(C∗ × Cs−1

)2

and therefore P−1v, P−1w ∈(C∗ × Cs−1

)2. From lim

m−→+∞Bmv = w, we have

limm−→+∞

P−1BmP (P−1v) = P−1w.

By applying Lemma 2.2 to the subgroup P−1GP of K∗η,2(C) where η = (s, s), we obtain

limm−→+∞

(P−1BmP )−1(P−1w) = P−1v

and therefore limm−→+∞

B−1m w = v. �

Lemma 2.6. Suppose G is an Abelian subgroup of B∗s(R) such that for every B ∈ G, |µB | = 1.Then for all v, w ∈ R2s and any sequence (Bm)m∈N ⊂ G such that lim

m−→+∞Bmv = w, we have

limm−→+∞

B−1m w = v.

Proof. Using Lemma 2.3, the proof is similar to the proof of Lemma 2.5. �

Proof of Proposition 2.4. Using Lemma 2.2, Lemma 2.3, Lemma 2.5 and Lemma 2.6, the proof

is similar to the proof of Proposition 2.1. �

Corollary 2.7. Let G be an Abelian subgroup of K∗η,r0,r(C) (resp. K∗η,r0,s0,r,s(R)). We let q =n −

r0∑k=1

nk (resp.q = n − (r0∑

k=1

nk + 2s0∑

l=1

ml)). Then for all v, w ∈ E(G) = {0Kn−q} × Kq and any

sequence (Am)m∈N ⊂ G such that limm−→+∞

Amv = w, we have limm−→+∞

A−1m w = v. In particular,

every orbit in E(G) is minimal in it.

9

Proof. As E(G) = {0Kn−q}×Kq is a G-invariant vector subspace of Kn, we have limm−→+∞

(Am/E(G)

)v =

w. By applying to the restriction G/E(G), Proposition 2.1 (if K = C) and Proposition 2.4 (if

K = R), we obtain limm−→+∞

(Am/E(G)

)−1w = v and therefore lim

m−→+∞A−1m w = v. �

2.3. Some results for Abelian subgroups of GL(n, K).

Proposition 2.8. Let G be an Abelian subgroup of GL(n, C) (resp. GL(n, R)). Then there exists

P ∈ GL(n, C) (resp. P ∈ GL(n, R)) such that G′ = P−1GP is a subgroup of K∗η,rG,r(C) (resp.K∗η,rG,sG,r,s(R)) for some r ∈ N and η ∈ (N)r (resp. r, s ∈ N and η ∈ (N)r+s).

To prove Proposition 2.8, we need the following lemmas.

Lemma 2.9. ([2], Proposition 2.3) Let G be an Abelian subgroup of GL(n, C). Then there exists

P ∈ GL(n, C) such that P−1GP is a subgroup of K∗η,r(C), for some 1 ≤ r ≤ n and η ∈ (N∗)r.

Analogous of Lemma 2.9 for the real case is the following:

Lemma 2.10. Let G be an Abelian subgroup of GL(n, R). Then there exists P ∈ GL(n, R) suchthat P−1GP is a subgroup of K∗η,r,s(R), for some 0 ≤ r, s ≤ n and η ∈ {N∗}r+s.

Proof. We apply Lemma 2.9 for G considered as an Abelian subgroup of GL(n, C); there exists

Q ∈ GL(n, C) such that G′ = Q−1GQ is a subgroup of K∗η,r′(C), for some 1 ≤ r′ ≤ n and

η′ = (n′1, ..., n′r′) ∈ (N∗)r

′

. We have Cn =r′⊕

k=1

E′k, where Ek are G′-invariant vector subspaces

of Cn of dimension n′k, defined in 2.1.1. There exists a basis C′ = (C′1, ..., C′r′) of Cn whereC′k = (e′k,1, ..., e′k,n′

k) is a basis of E′k such that if Q is the matrix of basis change of B0 to C′, we

have for every matrix M ∈ G:

M ′ = Q−1MQ =

AM1 0 ... 00 . . .. . . 00 ... 0 AMr′

,

where M ′/Ek = AMk , for every 1 ≤ k ≤ r′ and AMk ∈ T∗nk(C).

Denote by

- µAMk

the eigenvalue of AMk ,

- I = {k ∈ {1, ..., n} : µAMk

∈ R, for every M ∈ G};- I ′ = {1, ..., n}\I;

10

Then for every k ∈ I ′, there exists M ∈ G such that µAMk

∈ C\R. Since M ∈ GL(n, R),the conjugate µAM

kis also an eigenvalue of M , hence µAM

k= µAMik

for some ik ∈ I ′. Onecan suppose that I = {1, ..., r′} and I ′ = {k1, ..., ks, ik1 , ..., iks}. There exists a basis C1 =(C′1, ..., C′r ;B′k1, ...,B′ks) of Cn where B′ki = (e′ki,1, e′ki,1..., e

′ki,n′ki

, e′ki,n′ki), 1 ≤ i ≤ s. We let R be the

matrix of basis change of B0 to C1. Then for every M ∈ G, R−1MR is written in the form:

R−1MR =

AM1 0 ... .. .. 00 . . ... ... .. . AMr′ . ... .. . . BMk1 . .. ... .. . . 00 .. .. .. . BMks

where BMki =

(AMki 0

0 AMki

)∈ GL(2n′ki , C), 1 ≤ i ≤ s

By the change of basis used at the begining of the proof of Lemma 2.5, there exists, for 1 ≤ i ≤ s,Pki ∈ GL(2n′ki , C) such that P

−1ki

BkiPki ∈ Bn′ki (R). We let C = (C1, ..., Cr′ ,Bk1, ...,Bks) whereCk = (Re(e

′k,1), ..., Re(e

′k,n′

k)), 1 ≤ k ≤ r and Bki = (Re(e′ki,1), Im(e

′ki,1

)..., Re(e′ki,n′ki), Im(e′ki,n′ki

)), 1 ≤i ≤ s. (For w ∈ Cn′k , Re(w) and Im(w) denote the real and imaginary part of w, respectively,so that w = Re(w) + i.Im(w)). Take P to be the matrix of basis change of B0 to C, then for allM ∈ G, we have P−1MP ∈ Kη′,r′,s(R). �

Notice that this type of proof can be found in [9].

Proof of Proposition 2.8. Suppose that G is an Abelian subgroup of GL(n, C). By Lemma 2.9,

there exists Q ∈ GL(n, C) such that G′ = Q−1GQ is a subgroup of K∗η,r′(C) for some 1 ≤ r′ ≤ nand η′ ∈ {N∗}r′ . So for every M ∈ G, we have:

Q−1MQ =

A1 0 ... 00 . . .. . . 00 ... 0 Ar′

Therefore Cn =r′⊕

k=1

Ek where Ek is a G′-invariant vector subspace of Cn of dimension n′k, such

that for every M ′ ∈ G′ we have M ′/Ek = AMk , k = 1, .., r

′. Let C′ = (C′1, ..., C′r′) be a basis ofCn where C′k = (e′k,1, ..., e′k,n′

k) is a basis of Ek. Let Q be the matrix of change basis of B0 to C′.

Denote by I = {1, ..., r} and recall that J = {k ∈ I : |µAk | = 1, for every A ∈ G′}. Then Jhas the form J = {k1, ..., kr0} and I\J = {kr0+1, ..., kr}. Let C′′ = (C′′1 , C′′2 ) be a basis of Cn

where C′′1 = (C′k1 , ..., C′kr0 ), C′′2 = (C′kr0+1, ...., C

′kr

) and let P be the matrix of basis change of B0 toC.

Therefore if G is an Abelian subgroup of GL(n, R) then, by Lemma 2.10 and by repeating the

same proof above, the conclusion follows. �11

For the sequel, let G be an Abelian subgroup of GL(n, K). We let P ∈ GL(n, K) such thatP−1GP is a subgroup of K∗η,rG,r(C) (resp. K∗η,rG,sG,r,s(R)) and G′ = P−1GP . Then E(G′) ={0Kp} × Kq and H(G′) = Kp × {0Kq}.

Denote by

H(G) = P (H(G′)) = P (Kp × {0Kq})and

E(G) = P (E(G′)) = P ({0Kp} × Kq) .

A matrix A ∈ GL(n, K), is called hyperbolic if it has no eigenvalue of modulus 1. It is calledelliptic if all its eigenvalues are of modulus 1.

Corollary 2.11. Let G be an Abelian subgroup of GL(n, K). Under the notations above, we have:

i) Kn = H(G)⊕E(G) and H(G) and E(G) are G-invariant vector subspaces in Kn of dimensionp and q, respectively. Moreover, all elements of G/E(G) are elliptic.

ii) Every orbit in E(G) is minimal in it. In particular, if all elements of G are elliptic then

E(G) = Kn and so every orbit in Kn is minimal in it.

iii) FixG ⊂ E(G).iv) For every 1 ≤ k ≤ rG (resp. 1 ≤ k ≤ rG, 1 ≤ l ≤ sG), there exists A(k) ∈ G such that

A(k)/H(G) is written in the form

A(k)/H(G) =

Ak,1 0 ... 00 . . .. . . 00 ... 0 Ak,rG

(resp.

Ak,1 0 ... .. .. 00 . . ... ... .. . Ak,rG . ... .. . . B1 . .. ... .. . . 00 .. .. .. . BsG

)

where Ak,j =

µAk,j 0 ... 0

a(k)2,1 . . .

. . . 0

a(k)nj ,1

... a(k)nj ,nj−1 µAk,j

∈ T

∗nj(C), Bl ∈ B∗ml(R), such that |µAk,k | 6= 1

(resp. |µAk,k | 6= 1 and |µBl,l | 6= 1).

Proof. By Proposition 2.8, there exists P ∈ GL(n, K) such that G′ = P−1GP is a subgroup ofK∗η,rG,r(C), if K = C, and of K∗η,rG,sG,r,s(R)) if K = R.

Assertion i) is obvious since Kn = H(G′)⊕

E(G′).12

Assertion ii) follows from Corollary 2.7.

Assertion iii) is clear.

Proof of iv): Suppose the contrary, that is, there exists 1 ≤ k ≤ rG (resp. 1 ≤ k ≤ rG, 1 ≤ l ≤sG), such that for every A ∈ G, we have A/H(G) is written in the form

A/H(G) =

A1 0 ... 00 . . .. . . 00 ... 0 ArG

(resp.

A1 0 ... .. .. 00 . . ... ... .. . ArG . ... .. . . B1 . .. ... .. . . 00 .. .. .. . BsG

)

where Aj =

µAj 0 ... 0a2,1 . . .. . . 0

anj ,1 ... anj ,nj−1 µAj

∈ T

∗nj(C)

(resp. Bl ∈ B∗ml(R)), such that |µAk | = 1 (resp. |µAk | = |µBl | = 1).So by construction, Ek ⊂ H(G) ∩ E(G), a contradiction. �

2.4. Theorem 1.1 in the linear case.

The following theorem is the analogue of Theorem 1.1 in the linear case:

Theorem 1.1’. Let G be an Abelian subgroup of GL(n, K) (K = R or C). Then:

(i) Every orbit in E(G) is minimal in it.

(ii) If H(G) 6= {0} there exist G-invariant vector subspaces H1,..., HrG of H(G) of dimensionp− 1 or p− 2 over K such that if U = Kn\

rG⋃k=1

(Hk ⊕E(G)), every orbit in U is minimalin it.

In particular, we obtain the structure Theorem given in [1].

Proof. Statement (i) follows from Corollary 2.7.

Proof of statement (ii): Case K = C. By Proposition 2.8, there exists P ∈ GL(n, C) such thatP−1GP is an Abelian subgroup of K∗η,rG,r(C), for some r ∈ N and η ∈ (N∗)r. We let G′ = P−1GP .Then we have, Kn = H(G

′

) ⊕ E(G′). We let

H ′k =

{x = (x1, ..., xrG , 0, .., 0) ∈

rG∏

i=1

Cni × {0Cq} : xk ∈ {0} × Cnk−1}

13

and Hk = P (H′k), 1 ≤ k ≤ rG. We have H

′

k (resp. Hk) is a vector subspace of H(G′

) (resp.

H(G)) of dimension p − 1. Take U = Cn\(

rG⋃k=1

Hk ⊕ E(G))

. Then

U = P(Cn\(H ′k ⊕ E(G′)

)= P

(rG∏

k=1

(C∗ × Cnk−1) × Cq)

.

The statement (ii) follows from Proposition 2.1.

Case K = R. By Proposition 2.8, there exists P ∈ GL(n, R) such that P−1GP is an Abeliansubgroup of K∗η,rG,sG,r,s(R) for some r, s ∈ N and η ∈ (N∗)r+s. We let G′ = P−1GP . For1 ≤ k ≤ rG, 1 ≤ l ≤ sG, we let

H ′k =

(x1, ..., xrG , y1, ..., ysG , 0, .., 0) ∈

rG∏

i=1

Rni ×sG∏

j=1

R2mj

× {0Rq} : xk ∈ {0} × Rnk−1

and Hk = P (H′k).

For rG + 1 ≤ l ≤ r, sG + 1 ≤ k ≤ s, we let

H ′′l =

(x1, ..., xrG , y1, ..., ysG , 0, .., 0) ∈

rG∏

i=1

Rni ×sG∏

j=1

R2mj

× {0Rq} : yl ∈ (R2\{0}) × R2ml−2

and Hl = P (H′′l ). Then Hk is a vector subspace of H(G) of dimension p − 1. Take

U = Rn\((

rG⋃

k=1

Hk ⊕ E(G))

∪(

sG⋃

l=1

Hl ⊕ E(G)))

.

Then

U = P

(rG∏

k=1

(R∗ × Rnk−1) ×sG∏

l=1

((R2\{0}) × R2ml−2

)× Rq

).

The statement (ii) follows from Proposition 2.4. �

3. The affine case

Let f = (A, a) ∈ GA(n, K); f(x) = Ax + a, for x ∈ Kn, where A ∈ GL(n, K) and a ∈ Kn. Wecall A the linear part of f . Define the map

Φ : GA(n, K) −→ GL(n + 1, K)

f = (A, a) 7−→(

A a0 1

)

We have the following composition formula(

A a0 1

)(B b0 1

)=

(AB Ab + a0 1

).

Then Φ is a homomorphism of groups.

Let G be an Abelian affine subgroup of GA(n, K) and denote by LG the set of the linear partsof all elements of G. Then Φ(G) is an Abelian subgroup of GL(n + 1, K) and LG is a subgroup ofGL(n, K).

14

Denote by FixLG = {x ∈ Kn : Ax = x for every A ∈ LG}. One can easily check thatFixG = FixLG + x0 where x0 ∈ FixG. Then FixLG is an LG-invariant closed vector subspaceof Kn and we have always 0 ∈ FixLG . Let π : Kn −→ Kn × {1} be the map definedby π(y) = (y, 1). Clearly π is a homeomorphism and we have for every x ∈ Kn, π(G(x)) =Φ(G)(x, 1) = G(x) × {1} ⊂ Kn × {1}.

3.1. Some lemmas.

Lemma 3.1. Let u1, ..., un ∈ Kn where uk = (xk,1, ..., xk,n) with xk,k 6= 0, for every 1 ≤ k ≤ n.Then (Zu1 + .... + Zun) ∩ (K∗)n 6= ∅.

Proof. The proof is done by induction on n.

For n = 1, we have u1 = x1,1 6= 0, so u1 ∈ Zu1 ∩ K∗.Suppose Lemma 3.1 is true up to n − 1, n ≥ 2. Let u1, ..., un ∈ Kn where uk = (xk,1, ..., xk,n)

with xk,k 6= 0, for every 1 ≤ k ≤ n. Denote by u′k = (xk,1, ..., xk,n−1). Using inductionhypothesis, we have (Zu′1 + ....+Zu

′n−1)∩ (K∗)n−1 6= ∅. So, there exist m1, ..,mn−1 ∈ Z such that

m1u′1+...+mn−1u

′n−1 ∈ (K∗)n−1. Then

n−1∑k=1

mkxk,i 6= 0 for every 1 ≤ i ≤ n−1. Denote by I = {i ∈

{1, ..., n} : xn,i 6= 0}. Then I 6= ∅ (since xn,n 6= 0). Let mn ∈ Z\{− 1xn,i

n−1∑k=1

mkxk,i : i ∈ I}

.

Thenn−1∑k=1

mkxk,i + mnxn,i 6= 0 for every i ∈ I. If i /∈ I, then xn,i = 0 and we haven−1∑k=1

mkxk,i +

mnxn,i =n−1∑k=1

mkxk,i 6= 0. Therefore, m1u1 + .... + mnun ∈ (K∗)n. �

Lemma 3.2. Let G be an Abelian affine subgroup of GA(n, K). Then there exists A ∈ LGsuch that the restriction A/H(LG) is hyperbolic. In particular, A/H(LG) − Ip is invertible, wherep = dim(H(LG)) and Ip is the identity matrix.

Proof. i) Case K = C: We let G = LG. By Proposition 2.8, one can suppose that G is asubgroup of K∗η,rG,r(C). Then by Corollary 2.11, H(G) = Cp ×{0Cq} is G-invariant and for every1 ≤ k ≤ rG, there exists A(k) ∈ G such that A(k)/H(LG) is written in the form

A(k)/H(G) =

Ak,1 0 ... 00 . . .. . . 00 ... 0 Ak,rG

,

where Ak,j =

µAk,1 0 ... 0

a(k)2,1 . . .

. . . 0

a(k)nj ,1

... a(k)nj ,nj−1 µAk,j

∈ T

∗nj(C), such that |µAk,k | 6= 1.

15

We let uk = (log |µAk,1 |, ..., log |µAk,rG |) ∈ RrG . As log |µAk,k | 6= 0 for every 1 ≤ k ≤ rG,

then by Lemma 3.1, (Zu1 + .... + ZurG) ∩ (R∗)rG 6= ∅. So there exist m1, ...,mrG ∈ Z suchthat m1u1 + .... + mrGurG ∈ (R∗)rG . Hence, for every 1 ≤ j ≤ rG,

rG∏k=1

|µmkAk,j | 6= 1. We let

A = (A(1))m1 × ..× (A(rG))mrG . ThenrG∏

k=1

µmkAk,j , 1 ≤ j ≤ rG are the eigenvalues of A/H(G), whichare distinct of 1.

ii) Case K = R: The proof is analogous to the above proof. �

Proposition 3.3. Let G be an Abelian affine subgroup of GA(n, K). Then there exists h =(Q, d) ∈ GA(n, K) such that G̃ = h−1 ◦ G ◦ h is an Abelian subgroup of GA(n, K) having thefollowing properties:

(i) LeG = Q−1LGQ and is a subgroup of K∗η,r0,r(C) (resp. K∗η,r0,s0,r,s(R)) if K = C (resp.R) where r0 = rL eG and s0 = sL eG . In particular, H(LeG) = K

p × {0Kq} and E(LeG) ={0Kp} × Kq, where p = dim(H(LeG)) and p + q = n.

(ii) If g = (B, b) ∈ G̃, then b ∈ E(LeG).(iii) H(LG) = Q(H(LeG)) and E(LG) = Q(E(LeG)).

Proof. We let p = dim(H(LG)) and q = dim(E(LG)).• Suppose that pq 6= 0. By Proposition 2.8, there exists P ∈ GL(n, K) such that P−1LGP

is a subgroup of K∗η,r0,r(C) (resp. K∗η,r0,s0,r,s(R)) if K = C (resp. K = R) where r0 = rLG and

s0 = sLG . We let P1 =

(P 00 1

). By Lemma 3.2, there exists A ∈ LG such that A/H(LG) − Ip is

invertible. We let f = (A, a) ∈ G, a ∈ Kn, P−1a =(

a′1a′2

)∈ Rp × Rq and A′ = P−1AP . Then

A′ =

(A′1 00 A′2

), where A′1 = P

−1AP/H(P−1LGP ), A′2 = A/E(P−1LGP ). We have

P−11 Φ(f)P1 =

A′1 0 a′1

0 A′2 a′2

0 0 1

.

We let

P2 =

A′1 − Ip 0 a′10 Iq 00 0 1

.

Since A/H(LG) − Ip is invertible, so A′1 − Ip is invertible and so does P2. We let P3 = P1P−12 .Then P3 =

(Q d0 1

)where Q = PQ−11 , d = −PQ−11

(a′10

)and Q1 =

(A′1 − Ip 0

0 Iq

). We

let h = (Q, d) = Φ−1(P3) and G̃ = h−1 ◦G ◦h. Then h ∈ GA(n, K) and G̃ is an Abelian subgroupof GA(n, K).

Proof of (i) and (ii). Let’s prove that LeG is a subgroup of K∗η,r0,r(C) (resp. K∗η,r0,s0,r,s(R)) ifK = C (resp. R):

16

Let g̃ = h−1 ◦ g ◦ h ∈ G̃ where g = (B, b) ∈ G and take B′ = P−1BP . ThenB′ =

(B′1 00 B′2

)∈ K∗η,r0,r(C) (resp. K∗η,r0,s0,r,s(R)), where

B′1 = P−1BP/H(P−1LGP ) and B

′2 = P

−1BP/E(P−1LGP ). We have

Φ(g̃) = Φ(h−1 ◦ g ◦ h) = P−13 Φ(g)P3 = P2P−11 Φ(g)P1P−12 .

As Φ(g) =

(B b0 1

), then P−11 Φ(g)P1 =

(P−1BP P−1b

0 1

)=

B′1 0 b′1

0 B′2 b′2

0 0 1

, where

P−1b =

(b′1b′2

). As LG is Abelian, then AB = BA and so A′B′ = B′A′. Hence, we have

(A′1 − Ip)b′1 = (B′1 − Ip)a′1 and (A′2 − Iq)b′2 = (B′2 − Iq)a′2. So we obtain

Φ(g̃) = P2(P−11 Φ(g)P1)P

−12

=

B′1 0 −(B′1 − Ip)a′1 + (A′1 − Ip)b′1

0 B′2 b′2

0 0 1

=

B′1 0 00 B′2 b

′2

0 0 1

. (1)

Hence g̃ = (P−1BP, b′) where b′ =

(0b′2

)∈ E(LeG).

On the other hand, we have Φ(g̃) = P−13 Φ(g)P3 =

Q−1BQ Q−1(Bd + b − d)

0 1

. Hence

P−1BP = Q−1BQ and b′ = Q−1(Bd+b−d). So LeG = P−1LGP = Q−1LGQ, B′1 = Q−1BQ/H(L eG), B′2 =

P−1BP/E(L eG) and g̃ = (Q−1BQ, b′). This proves Assertions (i) and (ii).

• Suppose q = 0. Then H(LG) = Kn and r0 = s0 = 0. Assertions (i) and (ii) follow from theprevious proof by taking A′ = P−1AP and P2 =

(A′ − In P−1a

0 1

). So we obtain h = (Q, d)

where Q = (A − In)−1P and d = −(A − In)−1a

• Suppose p = 0. Then E(LG) = Kn and r0 = r, s0 = s. Assertions (i) and (ii) follow bytaking h = (P, 0).

Assertion (iii) results obviously from (i). �

Lemma 3.4. Under the hypothesis of Proposition 3.3, we have:

(i) H(Φ(G̃)) = Kp × {0Kq+1} and E(Φ(G̃)) = {0Kp} × Kq+1,(ii) Fix(G̃) ⊂ E(LeG).

Proof. (i): Let g = (B, b) ∈ G and g̃ = h−1 ◦ g ◦ h ∈ G̃. By Proposition 3.3, i), one hasH(LeG) = Kp × {0Kq} and E(LeG) = {0Kp} × Kq.

17

• Suppose pq 6= 0. Then by (1), we have

Φ(g̃) =

B′1 0 00 B′2 b

′2

0 0 1

,

where B′1 = Q−1BQ/H(L eG) and B

′2 = Q

−1BQ/E(L eG). So {0Kp} × Kq+1 is Φ(G̃)-invariant and the

restriction Φ(g̃)/{0Kp}×Kq+1 has all eigenvalues of modulus 1, for every g ∈ G. Hence E(Φ(G̃)) ={0Kp} × Kq+1. By the same proof, we also have H(Φ(G̃)) = Kp × {0Kq+1}.

• Suppose q = 0. Then by the proof of Lemma 3.3, we have Φ(g̃) =(

B′1 00 1

), for every

g̃ ∈ G̃. Hence the restriction Φ(g̃)/{0Kn}×K has 1 as eigenvalues, for every g ∈ G and so E(Φ(G̃)) ={0Kn} × K. By the same proof, we also have H(Φ(G̃)) = Kn × {0K}.

• Suppose p = 0. Then by the proof of Lemma 3.3, we have Φ(g̃) =(

B′2 b′2

0 1

)for every g̃ ∈ G̃.

Hence Φ(g̃) has all eigenvalues of modulus 1, for every g ∈ G and so E(Φ(G̃)) = Kq+1 = Kn+1.By the same proof, we also have H(Φ(G̃)) = Kn+1.

(ii) We have Fix(Φ(G̃)

)∩ (Kn × {1}) = Fix(G̃) × {1}. Then by (i) and Proposition 3.3,

iii), we have E(Φ(G̃)) = {0Kp} × Kq+1 = E(LeG) × K. Hence Fix(Φ(G̃)

)∩ (Kn × {1})) ⊂

E(Φ(G̃)) ∩ (Kn × {1})) = E(LeG) × {1} and therefore Fix(G̃) ⊂ E(LeG). �

Lemma 3.5. Let G and G̃ = h−1 ◦ G ◦ h be as in Proposition 3.3. Let G = Φ(G̃). Denote byH1, ...,HrG the G-invariant vector subspaces of H(G) of dimension p− 1 or p− 2 as in Theorem1.1’, where p = dim(H(G)). Let H̃k = Hk ⊕ E(LeG). Then:

i) H̃k is a G̃-invariant vector subspace of Kn of dimension n − 1 or n − 2,ii) E(LeG) is G̃-invariant and every orbit of G̃/E(LeG) is minimal in E(LeG).iii) If U ′ is a G-invariant open subset of Kn+1 and if every orbit of G/U ′ is minimal in

U ′, then every orbit of G̃/π−1(U ′ ∩ (Kn × {1})) is minimal in π−1(U ′ ∩ (Kn × {1})).

Proof. Let’s first prove (∗) : π(H̃k) = (Hk ⊕ E(G)) ∩ (Kn × {1})), 1 ≤ k ≤ rG:By Lemma 3.4, i), H(G) = Kp×{0Kq+1} and E(G) = {0Kp}×Kq+1, hence we have Hk⊕E(G) =

Hk ⊕ ({0Kp} × Kq+1) and so

(Hk ⊕ E(G)) ∩ (Kn × {1}) = Hk ⊕ ({0Kp} × Kq × {1}).

As we have, by Proposition 3.3, i), E(LeG) = {0Kp} × Kq, then, π(H̃k) = π(Hk ⊕ E(LeG)) =(Hk ⊕ E(G)) ∩ (Kn × {1})).

i): It is clear that H̃k is a vector subspace of Kn of dimension n − 1 or n − 2. Let’s show

that H̃k is G̃-invariant: Let x ∈ H̃k and f = (A, a) ∈ G̃. Then by (∗), π(x) = (x, 1) ∈(Hk ⊕ E(G)) ∩ (Kn × {1}) .

18

Since Hk ⊕ E(G) and Kn × {1} are G-invariant, we have

Φ(f)(x, 1) ∈ (Hk ⊕ E(G))) ∩ (Kn × {1})

As Φ(f)(x, 1) =

(Ax + a

1

)= π(f(x)) then by (∗), f(x) ∈ H̃k.

ii): It is obvious that E(LeG) is G̃-invariant. Now, let x ∈ E(LeG) and y ∈ G̃(x). By Lemma 3.4,i), we have E(G) = {0Kp} × Kq+1 = E(LeG) × K. We have (x, 1) ∈ E(G) and (y, 1) ∈ G(x, 1).By Theorem 1.1’, every orbit of G/E(G) is minimal in E(Φ(G̃)). Thus G(x, 1) = G(y, 1). HenceG̃(x) × {1} = G̃(y) × {1} and so G̃(x) = G̃(y), this proves ii).

iii): Suppose that U ′ is a G-invariant open subset of Kn+1 and we let U = π−1(U ′∩(Kn×{1})).If x ∈ U and y ∈ G̃(x)∩U , then we have π(x) = (x, 1) ∈ U ′ and π(y) = (y, 1) ∈ G(x, 1) ∩U ′.By hypothesis, we have G(x, 1) ∩ U ′ = G(y, 1) ∩ U ′. Therefore

π−1(G(x, 1) ∩ U ′ ∩ (Kn × {1})

)= G̃(x) ∩ U

and

π−1(G(y, 1) ∩ U ′ ∩ (Kn × {1})

)= G̃(y) ∩ U.

So G̃(x) ∩ U = G̃(y) ∩ U . This proves that every orbit of G̃/U is minimal in U . �

3.2. Proof of Theorem 1.1. Let G be an Abelian subgroup of GA(n, K). By Proposition 3.3,there exists h ∈ GA(n, K) such that G̃ = h−1 ◦ G ◦ h is an Abelian subgroup of GA(n, K) andsatisfying H(LeG) = Kp × {0Kq} and E(LeG) = {0Kp} × Kq, p + q = n. Let G̃ = Φ(G̃). Then G̃ isan Abelian subgroup of GL(n + 1, K). By Lemma 3.4, ii), we have H(G̃) = Kp × {0Kq+1} andE(G̃) = {0Kp} × Kq+1.

If E(LeG) 6= Kn then H(LeG) 6= {0} and so p 6= 0 and H(G̃) 6= {0}. By Theorem 1.1’, wehave Kn+1 = H(G̃) ⊕ E(G̃) and there exist G̃-invariant vector subspaces H̃1,..., H̃r eG of H(G̃) of

dimension p− 1 or p− 2 such that if U ′ = Kn+1\r eG⋃

k=1

(H̃k ⊕E(G̃)), every orbit of G̃/U ′ is minimalin U ′.

We let Fk = H̃k ⊕ E(LeG). By Lemma 3.5, i), Fk is a G̃-invariant vector subspace of Kn ofdimension n− 1 or n− 2. We let E = h(E(LeG)) and Hk = h(Fk). Then E and Hk are G-invariantaffine subspaces of Kn of dimension n − 1 or n − 2, with E ⊂ Hk for every 1 ≤ k ≤ rr eG .

Since G̃ = h−1 ◦ G ◦ h and Fix(G̃) ⊂ E(LeG) (Lemma 3.4, ii)), we have FixG = h(FixG̃) ⊂h(E(LeG)). On the other hand, for h = (Q, d), we have h(E(LeG)) = Q(E(LeG))+d = E(LG)+d = Eand so FixG ⊂ E .

(i): By Lemma 3.5, ii), every orbit of G̃/E(LeG) is minimal in E(LeG), then every orbit of G/Eis minimal in E .

19

(ii): Let Ũ = Kn\r eG⋃

k=1

F̃k. Then by Theorem 1.1’, every orbit of G̃/U ′ is minimal in U ′ where

U ′ = Kn+1\r eG⋃

k=1

(H̃k ⊕ E(G̃)). Let Ũ = π−1(U ′ ∩ (Kn × {1})). By Lemma 3.5, iii), every orbit of

G̃/Ũ is minimal in Ũ . Hence for U = h(Ũ ), every orbit of G/U is minimal in U . �

From the proof of Theorem 1.1, we obtain the following Corollary.

Corollary 3.6. i) If all elements of LG are elliptic then E = Kn and so every orbit in Kn isminimal in Kn.

ii) E = h(E(LeG)

)and dim(E) = dimE(LG).

iii) Let Ũ =r0∏

k=1

(C∗ × Cnk−1) × Cq (resp.r0∏

k=1

(R∗ × Rnk−1) ×s0∏l=1

(R2\{0}) × R2ml−2) × Rq if

K = C (resp. R), where r0 = rL eG and s0 = sL eG . Then every orbit of G̃/ eU is minimal in Ũ .

3.3. Proof of Corollaries 1.2, 1.3, 1.4, 1.5 and 1.6.

Proof of Corollary 1.2. We proceed inductively on n. If n = 1, G is a subgroup of K∗.If E 6= K, then (by Theorem 1.1), E = {a} and so every orbit O in K\{a} is minimal in it.Therefore, O is at level ≤ 1 = cod(E). If E = K, then every orbit in K is minimal in K, hence Ois at level 0 = cod(E).

Now, let L be an orbit in Kn, n ≥ 1. If L ⊂ Kn\U then L ⊂ Hk for some 0 ≤ k ≤ rG . SinceHk is G-invariant and of dimension n− 1 or n− 2 and E ⊂ Hk (Theorem 1.1), so using inductionhypothesis, then L is at level p ≤ dimHk ≤ cod(E) − 1. Suppose L ⊂ U . Then by Theorem1.1, L is minimal in U and therefore cl(L) = L ∩ U : indeed, if γ ⊂ L ∩ U is an orbit of G thenγ ∩ U = L ∩ U . Since U is open, γ = L and therefore L ∩ U ⊂ cl(L). Conversely, let γ ⊂ cl(L)be an orbit. As L ⊂ U and U is open and G-invariant then γ ⊂ U , so cl(L) ⊂ L ∩ U .

So, for every orbit O ⊂ L\cl(L), we have O ⊂ Kn\U and therefore O ⊂ Hk for some 0 ≤ k ≤ rGwhere Hk is of dimension n− 1 or n− 2. By inductive hypothesis applied to the restriction G/Hk ,the orbit O of G is at level ≤ dim(Hk ≤ cod(E) − 1. Hence O is at level ≤ cod(E) − 1. It followsthat L is at level ≤ cod(E). �

Proof of Corollary 1.3. i) By Theorem 1.1’, i), every orbit O ⊂ E is minimal in E . As E isclosed in Kn, hence O ∩ E ⊂ E is a minimal set in Kn.

ii) Since FixG (resp. E) is a G-invariant closed subset of Kn, U = Kn\FixG (resp. Kn\E) isopen and G-invariant. Let γ0, ..., γl of G/U (l ≥ 0) such that γ0U ⊂ ... ⊂ γlU where γiU = γi ∩ Uand γi

U 6= γjU , 0 ≤ i 6= j ≤ l. So γ0 ⊂ ... ⊂ γl with γi 6= γj, 0 ≤ i 6= j ≤ l. By Corollary 1.2,ht(G) ≤ cod(E), hence l ≤ cod(E). This proves that G/U has a minimal set in U . �

Proof of Corollary 1.4. If cod(E) = 1 then E 6= Kn and by Theorem 1.1, we have rG = 1. Sothere exists a G-invariant affine subspace H1 of Kn of dimension n− 1 or n− 2 such that E ⊂ H1.

20

Hence E = H1 and so dimH1 = dimE = n − 1. Therefore U = Kn\E and the Corollary followsby Theorem 1.1. �

Proof of Corollary 1.5. Let O be a locally dense orbit in Cn, that is◦O 6= ∅. Then O ⊂ U

and O∩U is a nonempty closed subset in U . Let’s show that O∩U is open in U : Let v ∈ O∩U .Since O is minimal in U , we have O ∩ U = G(v) ∩ U . So,

◦G(v) =

◦O 6= ∅ and therefore

v ∈◦

G(v) ∩ U ⊂ O ∩ U . As U is connected, then O ∩ U = U and therefore, O = Cn (becauseU = Cn). Now for every orbit L ⊂ U , L ⊂ U ∩ O, we have L ∩ U = O ∩ U and thereforeL = O = Cn. �

Proof of Corollary 1.6 (i): Suppose that O is a locally dense orbit in Rn (i.e.◦O 6= ∅). Then

O ⊂ U . Let C be a connected component of U meeting O. Then O ∩ C is a nonempty closedsubset in C. Let’s show that O ∩ C is open in C: let v ∈ O ∩ C. Since O is minimal in U thenO ∩ U = G(v) ∩ U . So,

◦G(v) =

◦O 6= ∅. Then v ∈

◦G(v) ∩ C ⊂ O ∩ C. As C is connected, then

O ∩ C = C. Let’s show that every orbit meeting C is dense in C: if O′ is an orbit meeting Cthen O′ ⊂ U . Since O is dense in C then O′ ⊂ O and therefore O′ ∩ C = O ∩ C = C. �

(ii): Let O be a dense orbit in Rn (i.e. O = Rn). Then O ⊂ U . Now for every orbit L ⊂ U , wehave L ⊂ U ∩ O and then L ∩ U = O ∩ U . It follows that L = O = Cn. �

4. Criterion of conjugation affine subgroups to linear

In this section, we will give some criterion of topological conjugation by translation Abelian

affine groups to linear groups. In this case, the dynamic of the two groups are then similar.

Two subgroups G and L of GA(n, K) are said to be topologically conjugate (resp. T -conjugate)if there is a homeomorphism h (resp. a translation Tu of K

n of vector u) such that h ◦ G = L ◦ h(resp. Tu ◦ G = L ◦ Tu).

Proposition 4.1. Let G be an Abelian affine subgroup of GA(n, K) (K = R or C).(i) If FixG 6= ∅ then G is T -conjugate to LG.(ii) If FixG = ∅ then G is not topologically conjugate to any subgroup of GL(n, K).

Proof. i) Suppose FixG 6= ∅ and let u ∈ FixG. Let f = (A, a) ∈ G. We have f(u) = u and forevery x ∈ Kn:

T−u ◦ f ◦ Tu(x) = T−u ◦ f(x + u) = T−u (A(x + u) + a)

= T−u (Ax + f(u)) = T−u(Ax + u) = Ax.

Hence LG = T−u ◦ G ◦ Tu and statement i) follows.21

ii) Suppose FixG = ∅ and G is topologically conjugate by a homeomorphism h to a subgroupL of GL(n, K): h ◦ G ◦ h−1 = L. Since 0 ∈ FixL, we have G(h−1(0)) = h−1(L(0)) = h−1(0), soh−1(0) ∈ FixG, this contradicts FixG = ∅. �

Proposition 4.2. Let G be an Abelian affine subgroup of GA(n, K) (K = R or C). If FixLG ={0} then FixG 6= ∅.

We need the following lemma.

Lemma 4.3. Fix(Φ(G)) 6= {0}.

Proof. We let G = Φ(G). Then G is an Abelian subgroup of GL(n + 1, K). Let P ∈ GL(n + 1, C)such that G′ = P−1GP is a subgroup of K∗η,r(C), for some 1 ≤ r ≤ n + 1 given in Proposition2.8. We can consider G as a subgroup of GL(n + 1, C). Denote by tG the transpose of G, that

is, if A = (ai,j)1≤i,j≤n ∈ G then tA = (aj,i)1≤i,j≤n+1 ∈ tG. Then en+1 ∈ Fix(tG) and so(tP )en+1 ∈ Fix(tG′). It follows that for every B′ ∈ G′, we have tB′

((tP )en+1)

)= (tP )en+1. We

let

B′ =

B′1 0 . 00 . . .. . . 00 . 0 B′r

, where B

′k ∈ T∗nk(C), 1 ≤ k ≤ r. Then we have

tB′ =

tB′1 0 . 00 . . .. . . 00 . 0 tB′r

and so µB′r = 1. It follows that Fix(G′) 6= {0} and thus FixG 6= {0}. �

Proof of Proposition 4.2. By Lemma 4.3, there exist (x0, λ0) ∈ Fix(Φ(G)), (x0, λ0) 6= (0, 0),where x0 ∈ Kn, λ0 ∈ K. Then for every f = (A, a) ∈ G, we have Φ(f)(x0, λ0) = (x0, λ0),hence Ax0 + λ0a = x0. Suppose FixLG = {0}. We have λ0 6= 0: indeed, if λ0 = 0, thenx0 6= 0 and Ax0 = x0. So x0 ∈ FixLG\{0}, a contradiction. Now take x = 1λ0 x. We havef(x) = 1λ0 Ax0 + a =

1λx0 = x, hence, x ∈ FixG and therefore FixG 6= ∅. �

Corollary 4.4. Let G be an Abelian affine subgroup of GA(n, K) (K = R or C). If LG containsan element A having no nonzero fixed point, then FixG 6= ∅ and in particular G is T -conjugatedto LG.

Proof. The hypothesis implies that FixLG = {0}, so by Proposition 4.2, FixG 6= ∅ and byProposition 4.1, G is T -conjugated to LG . �

22

Remark 3. Let G be an Abelian affine subgroup of GA(n, C). If FixG = ∅ then rG ≤ n − 1.

Proof. If FixG = ∅ then by Proposition 4.2, FixLG 6= {0}. As FixLG ⊂ E(LG) (Corollary 3.6,ii)), then dim(E) = dimE(LG) and so dimE ≥ dimFixLG ≥ 1, hence codE ≤ n − 1 and byTheorem 1.1, rG ≤ codE ≤ n − 1. �

Example 4.5. Let G be the subgroup of GA(2, R) generated by f = (A, a) and g = (B, b), where

A =

(2 00 1

), a =

(11

)and B =

(3 00 1

), b =

(22

).

Then G is Abelian and since f has no fixed point, FixG = ∅.Let’s show that E = {−1} × R, which implies that cod(E) = 1. Hence, by Corollary 1.4, we

have U = R2\E = (R\{−1}) × R and so every orbit in U is minimal in it.

The construction of h is given in the proof of Proposition 3.3, so we obtain Q = I2 and

d =

(−10

), hence h = (I2, d) = Td. Denote by G̃ = h−1 ◦ G ◦ h. We have E(LeG) = {0} ×R. By

Corollary 3.6, ii), we have E = h(E(LeG)), so E = {−1} × R. �

Example 4.6. Let G be the subgroup of GL(1, R) generated by:

f = e√

6IdR and g = e√

3IdR.

Then G is Abelian and has a locally dense orbit. Indeed;

By Lemma 2.2, we have U = R∗. Let’s show that for every x ∈ R∗+, G(x) is dense in R: Wehave G(x) = {en

√6em

√3x : n,m ∈ Z} ⊂ R∗+. Let ϕ : R∗+ −→ R be the homeomorphism defined

by ϕ(y) = 1√3

log y. Then ϕ(G(y)) = H = Z +√

2Z + logx. Since√

2 /∈ Q, H is dense in R, soG(x) is dense in R∗+. It follows that every orbit in R

∗− is also dense in it.

However, no orbit is dense in R because R∗+ (resp. R∗−) is G-invariant and R

∗+ 6= R (resp.

R∗− 6= R).

5. Application to some affine subgroups

5.1. LG is a group of unipotent matrices. Denote by Unip(n, K) the group of unipotentmatrices of order n ≥ 1 and with entries in K:

Unip(n, K) =

1 0 ..... 0a2,1 . ... .. .. . 0

an,1 ... an,n−1 1

: ai,j ∈ K, 1 ≤ j < i ≤ n

.

and by

UnipA(n, K) = {(A, a) ∈ GA(n, K) : A ∈ Unip(n, K), a ∈ Kn}23

For n = 3, UnipA(n, R) is called the Heisenberg group.

Proposition 5.1. Suppose G is an Abelian subgroup of UnipA(n, K). Then every orbit in Kn isminimal in Kn.

Proof. This follows from Corollary 3.6, i) since E = Kn. �

Example 5.2. Let G be the group generated by f = (A, a) and g = (B, b), where

A =

1 0 20 1 00 0 1

, B =

1 2 20 1 00 0 1

and a = b =

100

Then G is Abelian and all orbits in R3 are minimal in R3 but not homeomorphic. More precisely:i) If u ∈ R × Q × Q∗ , the orbit G(u) is closed in R3.ii) If u ∈ R × (R\Q) × Q∗, the orbit G(u) is dense in a straight line.

Proof. Denote by G = Φ(G). Then G is generated by

Φ(f) =

1 0 2 10 1 0 00 0 1 00 0 0 1

and Φ(g) =

1 2 2 10 1 0 00 0 1 00 0 0 1

Let u =

xyz

∈ R3. One can check that

G(u, 1) ={(

x + 2my + 2(m + n)z + m + n, y, z, 1)

: n,m ∈ Z}

.

As G(u, 1) = G(u) × {1}, we obtain

G(u) ={(

x + (2y + 2z + 1)m + (2z + 1)n, y, z)

: n, m ∈ Z}

i) If u ∈ R×Q×Q∗ then (2y +2z +1)Z+(2z +1)Z is closed in R and therefore G(u) is closedin R3.

ii) If u ∈ R × (R\Q) × Q∗ then (2y + 2z + 1)Z + (2z + 1)Z is dense in R and therefore G(u) isdense in a straight line. �

24

5.2. LG is a group of diagonal matrices. Denote by Dn(K) the group of diagonal matrices oforder n ≥ 1 with entries in K: Dn(K) = {A ∈ GL(n, K) : A is diagonal}

and by DnA(K) = {(A, a) ∈ GA(n, K) : A ∈ Dn(K), a ∈ Kn}.

Proposition 5.3. Suppose G is an Abelian subgroup of DnA(K). Then there exists h ∈ GA(n, K)such that E = h({0Kr0} × Kn−r0) and U = h ((K∗)r0 × Kn−r0). Hence every orbit in U (resp. E)is minimal in U (resp. E).

Proof. Let h ∈ GA(n, K) and G̃ = h−1 ◦ G ◦ h are as in Proposition 3.3. Then LeG = Q−1LGQ is asubgroup of K∗η,r0,r(C) (resp. K∗η,r0,s0,r,s(R)) if K = C (resp. R)) composed of diagonal matrices(since Q is diagonal), where r0 = rL eG and s0 = sL eG . Hence every element A ∈ LeG is written inthe form

A =

λ1 0 ... ... ... 00 . . . . .. . λr0 . . .. . . eiθr0+1 . .. . . . . 00 . . . 0 eiθn

,

where λ1, ..., λr0 ∈ K; θr0+1, ..., θn ∈ R. So E(LeG) = {0Kr0} × Kn−r0 and E = h(E(LeG)

).

By Corollary 3.6, iii), every orbit of G̃/(K∗)r0×Kn−r0 is minimal in (K∗)r0 × Kn−r0, hence forU = h ((K∗)r0 × Kn−r0), every orbit of G/U is minimal in U . �

Proposition 5.4. For every 1 ≤ p ≤ n, 1 ≤ r ≤ p there is an Abelian affine subgroup G ofGA(n, C) having an affine G-invariant subspace E of Kn with cod(E) = p and r G-invariant affinesubspaces H1, ..., Hr of Kn of dimension n − 1 or n − 2 over K satisfying the conclusion ofTheorem 1.1.

Proof. Let G be the affine subgroup of DnA(C) generated by f = (A, a) and g = (B, b), where

A =

D1,r 0

0 In−r

, B =

D2,r 0

0 In−r

, a = b = en

and D1,r,D2,r ∈ Dr(R). By Proposition 5.3, U =(

r∏k=1

C∗)× Cn−r. So, Cn\U has r G-invariant

affine subspaces satisfying the conclusion of Theorem 1.1. �

Example 5.5. Let G be the subgroup of GA(2, K) defined by

G ={

f = (A, a) ∈ GA(2, K) : A =(

λ 00 1

), a =

(λ − 1

0

), λ ∈ K∗

}.

Then G is Abelian and we have:25

i) E = {−1}×K, hence U = K2\({−1}×K) and so every orbit in U (resp. E) is minimal in it.ii) U is connected if K = C and not connected if K = R.

Proof. i) Denote by d =

(−10

)and h = (I2, d) = Td. We let G̃ = h−1 ◦ G ◦ h. We have

E(LeG) = {0} × K. By Corollary 3.6, ii), we have E = h(E(LeG)), so E = {−1} × K. Thencod(E) = 1 and by Corollary 1.4, U = K2\E = K2\({−1} × K).

ii) If K = C, U = C2\({−1} × C) is connected. If K = R, U = R2\({−1} × R) has twoconnected components.

5.3. LG is a subgroup of H∗n(K). We let H∗n(K) = {A = λIn : λ ∈ K∗}and H∗nA(K) = {f = (λIn, a) : λ ∈ K∗, a ∈ Kn}.

Proposition 5.6. Suppose G is an Abelian subgroup of H∗nA(K). Then:i) If FixG = ∅, G is a group of translations and every orbit in Kn is minimal in Kn.ii) If FixG 6= ∅, it is a single-point {u} and G is T -conjugated to a subgroup of H∗n(K). In

particular, every orbit in Kn \ {u} is minimal in it.

Proof. i) Suppose FixG = ∅. Then by Corollary 4.4, every element of LG has nonzero fixed point,hence if f = (λIn, a) ∈ G, we have λ = 1, hence f(x) = x + a, for every x ∈ Kn. Therefore, G isa group of translations. In this case, we have E(LG) = Kn. So by Theorem 1.1, i), every orbit inKn is minimal in Kn.

ii) Suppose FixG 6= ∅. By Corollary 4.4, G is T -conjugated to LG. As LG is a subgroup ofH∗n(K), so G is T -conjugated to a subgroup of H∗n(K). In this case, to prove that every orbitof G/Kn \ {u} is minimal in it, it suffices to prove that every orbit of LG/Kn\{0} is minimal inKn\{0}:

Let v, w ∈ Kn\{0} such that w ∈ LG(v). Then there exists a sequence (hm)m∈N in LG :hm(x) = λmx, x ∈ Kn such that lim

m−→+∞hm(v) = w. As v, w ∈ Kn\{0}, then lim

m−→+∞λm 6= 0 and

so limm−→+∞

h−1m (w) = v. �

5.4. Groups of affine isometries of Kn. Denote by

- SL(n, K) the group of matrices A ∈ GL(n, K) having determinant 1, called the speciallinear group over K.

- O(n, R) the group of orthogonal matrices, called the orthogonal group. It is a subgroup

of GL(n, C).26

Recall that a matrix A ∈ GL(n, R) is called orthogonal if it preserves the inner product,namely if < Ax, Ay >=< x, y > for all x, y ∈ Rn. ( denote the usual inner product on Rn,< x, y >=

∑i xiyi.) Another equivalent definition is that A is orthogonal if

tAA = In. (Here,

tA is the transpose of A).

- SO(n, R) = O(n, R) ∩ SL(n, R), called the special orthogonal group, it is a subgroup ofO(n, R).

Geometrically, elements of O(n, R) are either rotations or combinations of rotations and re-

flections. The elements of SO(n, R) are just the rotations.

- U(n, C) the group of unitary matrices, called the unitary group.

It is a subgroup of GL(n, C). Recall that a matrix A ∈ GL(n, R) is called unitary if it preservesthe inner product, namely if < Ax, Ay >=< x, y > for all x, y ∈ Cn. ( denote the innerproduct on Cn, < x, y >=

∑i xiyi.) Another equivalent definition is that A is unitary if

tAA = In. (Here,tA is the adjoint of A.)

- SU(n, C) = U(n, C) ∩ SL(n, C) the special unitary group.

The corresponding affine groups to the linear above groups are the following:

- SOA(n, R) = {f = (R, a) ∈ GA(n, R) : R ∈ SO(n, R), a ∈ Rn}.- SUA(n, C) = {f = (R, a) ∈ GA(n, C) : R ∈ SU(n, C), a ∈ Cn}.

Proposition 5.7. Let G be an Abelian subgroup of SUA(n, C) ( resp. SOA(n, R)). Then(i) Every orbit in Cn (resp. Rn) is minimal in it.

(ii) If FixG 6= ∅ then every orbit in Cn is bounded.

Proof. Let G be an Abelian subgroup of SUA(n, C) (resp. SOA(n, R)).Assertion i) results from Corollary 3.6, i).

Assertion ii) results from Proposition 4.1, i) and from the fact that every orbit in LG =SO(n, R) (resp. SU(n, C)) is bounded. �

Example 5.8. Let

G ={

f = (A, a) ∈ GA(2, C) : A =(

eiθ 00 1

), a =

(0λ

), λ ∈ C, θ ∈ R

}.

Then G is Abelian and every orbit in C2 is closed in C2.

Indeed, we have E(LG) = C2. For every u = (x, y) ∈ C2, G(u) = Su × C, where Su = {z ∈ C :|z| = |x|}. In particular, G(u) is closed in C2.

27

Example 5.9. Let G ={f = (λ, a) ∈ GA(1, C) : λ = eiθ, a = 1 − eiθ, θ ∈ R

}. Then G is an

Abelian subgroup of GA(1, C) and every orbit in C is dense in a circle.

Indeed, G is clearly Abelian. For every u ∈ C∗, we have G(u) = {1 + eiθ(u − 1) : θ ∈ R} ⊂1 + Su−1 where Su−1 = {v ∈ C : |v| = |u − 1|}. Hence, G(u) = 1 + Su−1 which is a circle.

Acknowledgments. This work was done within the framework of the Associateship Scheme

of the Abdus Salam International Centre for Theoretical Physics, Trieste, Italy. H.M. thanks

ICTP for hospitality where part of this work was done.

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