Control Systems on Three-Dimensional Lie Groups

J Dyn Control Syst DOI 10.1007/s10883-014-9212-0

Control Systems on Three-Dimensional Lie Groups: Equivalence and Controllability Rory Biggs · Claudiu C. Remsing

Received: 21 December 2012 / Revised: 24 September 2013 © Springer Science+Business Media New York 2014

Abstract We consider left-invariant control affine systems, evolving on three-dimensional matrix Lie groups. Equivalence and controllability are investigated. All full-rank systems are classified, under detached feedback equivalence. A representative is identified for each equivalence class. The controllability nature of these representatives is determined. Keywords Left-invariant control system · Detached feedback equivalence · Matrix Lie group Mathematics Subject Classifications (2010) 49J15 · 22E30 · 93B17 · 93B05

1 Introduction Invariant control systems constitute a notable class of smooth nonlinear control systems evolving on (real, finite-dimensional) Lie groups. Such systems form a natural geometric framework for various (variational) problems in mathematical physics, mechanics, elasticity, and dynamical systems (cf. [4, 14, 20, 21]). In the last few decades, substantial work on applied nonlinear control has drawn attention to invariant control affine systems evolving on matrix Lie groups of low dimension (see, e.g. [22, 29, 30, 32, 36, 37] and the references therein). Two natural equivalence relations for (smooth) control systems are state space equivalence and feedback equivalence [19, 33]. State space equivalence is equivalence up to coordinate changes in the state space; it establishes a one-to-one correspondence between

R. Biggs · C. C. Remsing () Department of Mathematics (Pure and Applied), Rhodes University, PO Box 94, 6140 Grahamstown, South Africa e-mail: [email protected] R. Biggs e-mail: [email protected]

Rory Biggs and Claudiu C. Remsing

the trajectories of the equivalent systems. This equivalence relation is very strong. Consequently, any general classification appears to be very difficult, if not impossible. However, in lower dimensions, some reasonable classifications of certain classes of systems have recently been obtained [1, 5, 13]. Feedback equivalence, a weaker equivalence relation, additionally permits feedback transformations of the controls. A classification of controlaffine systems with two inputs evolving on three-dimensional manifolds was obtained in [34]. In the context of left-invariant control systems, feedback equivalence may be specialized by requiring that the feedback transformations are constant over the state space. Such transformations are exactly those that are compatible with the Lie group structure (see, e.g. [6]). This is called detached feedback equivalence. It turns out that two full-rank systems are detached feedback equivalent if and only if there exists a Lie group isomorphism relating their dynamics [12]. Several classes of left-invariant control affine systems have recently been classified [2, 3, 8–10]. In this paper, we consider left-invariant control affine systems evolving on threedimensional matrix Lie groups. For each (connected) matrix Lie group G, we classify, under detached feedback equivalence, all full-rank systems evolving on G. First, the group d Aut (G) of linearized group automorphisms is determined. Class representatives are then constructed (by considering the action of an automorphism on a typical system). For each representative, algebraic classifying conditions are given. Finally, we determine whether or not each representative system is controllable. A classification under detached feedback equivalence is identical to one under local detached feedback equivalence whenever d Aut (G) equals the group Aut (g) of Lie algebra automorphisms. If this is the case, then the classification result is recalled [8–10], but a proof sketch is provided for the sake completeness. The paper concludes with a few remarks. Tables summarizing the main results are appended.

2 Preliminaries 2.1 Invariant Control Affine Systems An -input left-invariant control affine system takes the form g˙ = g (1, u) = g (A + u1 B1 + · · · + u B ) ,

g ∈ G, u ∈ R .

Here, G is a (real, finite-dimensional) connected Lie group with Lie algebra g and A, B1 , . . . , B ∈ g. Also, the parametrization map (1, ·) : R → g is an injective affine map (i.e. B1 , . . . , B are linearly independent). The “product” g (1, u) is to be understood as T1 Lg · (1, u), where Lg : G → G, h → gh is the left translation by g. (When G is a matrix Lie group, this product is simply a matrix multiplication.) Note that the dynamics : G×R → T G are invariant under left translations, i.e. (g, u) = g (1, u). is completely determined by the specification of its state space G and its parametrization map (1, ·). When G is fixed, we specify by simply writing : A + u 1 B1 + · · · + u B . The trace of a system is the affine subspace = A + 0 = A + B1 , . . . , B of g. (Here, 0 = B1 , . . . , B is the subspace of g spanned by B1 , . . . , B .) A system is called homogeneous if A ∈ 0 , and inhomogeneous otherwise. Also, is said to have full rank if its trace generates the whole Lie algebra (i.e. Lie () = g). We shall find

Control Systems on 3D Lie Groups: Equivalence and Controllability

it convenient to refer to an -input system as an (, 0)-system when it is homogeneous and as an (, 1)-system, otherwise. Remark 1 We have the following characterization of the full-rank condition for systems on G when dim G = 3. No (1, 0)-system has full rank. A (1, 1)-system has full rank if and only if A, B1 , and [A, B1 ] are linearly independent. A (2, 0)-system has full rank if and only if B1 , B2 , and [B1 , B2 ] are linearly independent. Any (2, 1)-system or (3, 0)-system has full rank. The admissible controls are piecewise continuous maps u(·) : [0, T ] → R . A trajectory for an admissible control u(·) is an absolutely continuous curve g(·) : [0, T ] → G such that g(t) ˙ = g(t) (1, u(t)) for almost every t ∈ [0, T ]. We say that a system is controllable if for any g0 , g1 ∈ G, there exists a trajectory g(·) : [0, T ] → G such that g(0) = g0 and g(T ) = g1 . For more details about invariant control systems, see, e.g. [4, 20, 23, 31, 35]. The attainable set (from the identity 1 ∈ G ) is the set A of all terminal points g(T ) of all trajectories g(·) : [0, T ] → G starting at 1. A system is controllable if and only if A = G. If is controllable, then it has full rank. Moreover, if is homogeneous or if G is compact, then full rank implies controllability. We recall some sufficient conditions for a full-rank system to be controllable: Lie ( 0 ) = g; the identity 1 is in the interior of A; there exists C ∈ such that t → exp(tC) is periodic [23]. Furthermore, if G is simply connected and completely solvable, then the condition Lie ( 0 ) = g is necessary for controllability [35]. Let and be left-invariant control affine systems on G and G , respectively. and are called detached feedback equivalent if there exists a diffeomorphism : G × R → G × R , (g, u) → (φ(g), ϕ(u)) such that Tg φ · (g, u) = (φ(g), ϕ(u)) for g ∈ G and u ∈ R . Equivalent systems and have the same trajectories (in the sense that φ maps trajectories to trajectories). Hence, if is controllable, then so is . In this paper, we consider only detached feedback equivalence, which will be simply referred to as equivalence. Proposition 1 ([12]) Suppose that and have full rank. Then and are equivalent if and only if there exists a Lie group isomorphism φ : G → G such that T1 φ · = . Proof (sketch) Suppose that and are equivalent. By composing φ with an appropriate left translation, we may assume φ(1) = 1 . Hence, T1 φ · (1, u) = (1 , ϕ(u)) and so T1 φ · = . Moreover, as the elements (1, u) ∈ g, u ∈ R generate g and the push forward (by φ) of the left-invariant vector fields u = (·, u) are leftinvariant vector fields, it follows that φ is a group isomorphism (see, e.g. [6]). On the other hand, suppose that there exists a group isomorphism φ : G → G such that T1 φ · = . Then, there exists a unique affine isomorphism ϕ : R → R such that T1 φ · (1, u) = (1 , ϕ(u)). By left-invariance (and the fact that φ is a homomorphism), it then follows that Tg φ · (g, u) = (φ(g), ϕ(u)). The following result, useful in classifying homogeneous systems, is easy to prove. Proposition 2 Suppose that {i : i ∈ I } is an exhaustive collection of equivalence class i be a system representatives for ( − 1, 1)-systems on a Lie group G. For each i , let


with trace i = i . Then, any (, 0)-system (on G) is equivalent to at least one element i : i ∈ I }. of { 2.2 Three-Dimensional Lie Groups The classification of three-dimensional real Lie algebras is well known. We prefer to use an adaptation of the Bianchi-Behr enumeration (see, e.g. [24–26]). There are 11 types of threedimensional real Lie algebras (in fact, nine algebras and two parametrized infinite families of algebras). In terms of an (appropriate) ordered basis (E1 , E2 , E3 ), the commutation operation is given by [E2 , E3 ] = n1 E1 − aE2 [E3 , E1 ] = aE1 + n2 E2 [E1 , E2 ] = n3 E3 . The structure parameters a, n1 , n2 , n3 for each type are given in Table 1. A classification of the three-dimensional (real connected) Lie groups can be found in [27]. Let G be a three-dimensional (real connected) Lie group with Lie algebra g. 1. If g ∼ = 3g1 , then G is isomorphic to R3 , R2 × T, R × T, or T3 . 2. If g ∼ = g2.1 ⊕ g1 , then G is isomorphic to Aff (R)0 × R or Aff (R)0 × T. ∼ g3.1 , then G is isomorphic to the Heisenberg group H3 or the Lie group 3. If g = H∗3 = H3 /Z(H3 (Z)), where Z(H3 (Z)) is the group of integer points in the centre Z (H3 ) ∼ = R of H3 . 4. If g ∼ = g3.2 , g3.3 , g03.4 , ga3.4 , or ga3.5 , then G is isomorphic to the simply connected Lie group G3.2 , G3.3 , G03.4 = SE (1, 1), Ga3.4 , or Ga3.5 , respectively. (The centres of these groups are trivial.) 5. If g ∼ = g03.5 , then G is isomorphic to the Euclidean group SE (2), the n-fold covering (2). SEn (2) of SE1 (2) = SE (2), or the universal covering group SE 6. If g ∼ = g3.6 , then G is isomorphic to the pseudo-orthogonal group SO (2, 1)0 , the A. Here, A2 ∼ n-fold covering An of SO (2, 1)0 , or the universal covering group = SL (2, R). Table 1 Bianchi-Behr classification Type

Bianchi

a

3g1

I

0

0

0

0

R3 , R2 × T, R × T2 , T3

g2.1 ⊕ g1

III

1

1

−1

0

Aff (R)0 × R, Aff (R)0 × T

g3.1

II

0

1

0

0

H3 , H∗3

g3.2

IV

1

1

0

0

G3.2

g3.3

V

1

0

0

0

G3.3

g03.4

VI0

0

1

−1

0

SE (1, 1)

ga3.4

VIa

a>0 a =1

1

−1

0

Ga3.4

g03.5 ga3.5

VII0

0

1

1

0

(2) SE (2), SEn (2), SE

VIIa

a>0

1

1

0

Ga3.5

g3.6

VIII

0

1

1

−1

g3.7

IX

0

1

1

1

n1

n2

n3

Connected groups

SO (2, 1), SL (2, R), An , A SU (2), SO (3)


7. If g ∼ = g3.7 , then G is isomorphic to either the unitary group SU (2) or the orthogonal group SO (3). A are not matrix Lie groups. Among these Lie groups, only H∗3 , An , n ≥ 3, and For each three-dimensional Lie algebra, a standard computation gives the group of automorphisms (see, e.g. [16, 17, 28]). Suppose that G is a connected Lie group with Lie algebra g and suppose that q : G → G is a universal covering. Then d Aut (G) = {T1 φ : φ ∈ Aut (G)} is a subgroup of Aut (g). Moreover, if ψ ∈ Aut (g), then ψ ∈ d Aut (G) if G) is the unique automorphism such that and only if φ(ker q) = ker q, where φ ∈ Aut ( T1 φ = (T1 q)−1 · ψ · T1 q (cf. [18]).

3 Solvable Groups 3.1 Type 3g1 (Abelian Groups) We consider first the simply connected group R3 . The group of automorphisms d Aut (R3 ) = Aut (3g1 ) is GL (3, R). Any full-rank system on R3 is equivalent to (2,1) : E1 + u1 E2 + u2 E3

or

(3,0) : u1 E1 + u2 E2 + u3 E3 .

As R3 is completely solvable, a system (on R3 ) is controllable if and only if Lie ( 0 ) = 3g1 . Thus, of the above two systems, only (3,0) is controllable. Next, we consider R2 × T. We have ⎫ ⎧⎡ x ⎤ ⎬ ⎨ e 0 0 R2 × T = ⎣ 0 ey 0 ⎦ : x, y, z ∈ R ⎭ ⎩ 0 0 eiz ⎧⎡ ⎫ ⎤ ⎨ x 0 0 ⎬ 3g1 = ⎣ 0 y 0 ⎦ : x, y, z ∈ R . ⎩ ⎭ 0 0 iz Let E1 = diag (1, 0, 0), E2 = diag (0, 1, 0), and E3 = diag (0, 0, i). Proposition 3 The group of automorphisms d Aut (R2 × T) is given by ⎧⎡ ⎫ ⎤ ⎨ u x 0 ⎬ ⎣ v y 0 ⎦ : u, v, w, x, y, z ∈ R, uy − xv = 0, σ = ±1 . ⎩ ⎭ w z σ Proof We have a universal covering q : R3 → R2 × T, diag (ex , ey , ez ) → diag (ex , ey , eiz ) with ker q = diag (1, 1, e2πZ ). Let ⎤ ⎡ a1 a2 a3 ψ = ⎣ b1 b2 b3 ⎦ ∈ Aut (3g1 ). c1 c2 c3 Then ψ ∈ d Aut (R2 × T) if and only if φ(ker q) = ker q, where φ ∈ Aut (R3 ) and T1 φ = (T1 q)−1 · ψ · T1 q. Thus, ψ ∈ d Aut (R2 × T) if and only if 2πZa3 = {0}, 2πZb3 = {0}, and 2πZc3 = 2πZ.


Theorem 1 Any full-rank system on R2 × T is equivalent to exactly one of the following systems: (2,1) E3 ∈ / 0 1,α : α E3 + u1 E1 + u2 E2 (2,1)

2

: E1 + u1 E2 + u2 E3

E3 ∈ 0

(3,0) : u1 E1 + u2 E2 + u3 E3 . Here, α > 0 parametrizes a family of class representatives, each different value corresponding to a distinct non-equivalent representative. Proof No (1, 1)-system or (2, 0)-system has full rank. Any (3, 0)-system is clearly (3,0) . Let be a (2, 1)-system with trace = 3 a E + equivalent to i i i=1 3 3 0 . Then = a E + a E + . Suppose that E b E , c E ∈ 3 1 1 2 2 i=1 i i i=1 i i b1 E1 + b2 E2 , E3 . Hence, ⎤ ⎡ a1 b1 0 ψ = ⎣ a2 b2 0 ⎦ 0 0 1 is an automorphism such that ψ · 2(2,1) = . (Here, 2(2,1) is the trace of 2(2,1) .) Suppose that E3 ∈ / 0 . The equation ⎡ ⎤ a1 b1 c1 v1 v2 v3 ⎣ a2 b2 c2 ⎦ = [ 1 0 0 ] a3 b3 c3 has a unique solution for v1 , v2 , v3 . Moreover v3 = 0. (If v3 = 0, then E3 ∈ 0 .) Hence ⎤ ⎡ 1 0 0 ⎦ 0 1 0 ψ =⎣ v1 v2 sgn (v ) 3 v3 sgn (v3 ) v3 sgn (v3 ) is an automorphism such that 1 E3 + b1 E1 + b2 E2 , c1 E1 + c2 E2 v3 sgn (v3 )

ψ · = a1 E1 + a2 E2 + = αE3 + E1 , E2

(2,1)

for some α > 0, i.e. ψ · = 1,α . (2,1)

Suppose that ψ ∈ d Aut (R2 × T). As ψ · αE3 = ±αE3 , it follows that 1,α equivalent to (2,1)

and 2

(2,1) 1,α

only if α =

cannot be equivalent.

α .

is

2(2,1) ,

(2,1) Similarly, as ψ · αE3 ∈ / it follows that 1,α (2,1) Clearly, (3,0) cannot be equivalent to either 1,α or

2(2,1) . (2,1) Theorem 2 The systems 1,α and (3,0) are controllable, whereas 2(2,1) is not.

Proof Clearly, (3,0) is controllable. The curve t → exp(tαE3 ) is a periodic tra(2,1) jectory of 1,α , and so this systems is controllable. Any trajectory g(·) : t → (2,1)

diag (ex(t) , ey(t) , eiz(t) ), g(0) = 1 of 2 satisfies x(t) = t for t ≥ 0. (Hence, states corresponding to x < 0 are not attainable from identity.) Thus, 2(2,1) is not controllable.


Corollary A (2, 1)-system on R2 × T is controllable if and only if E3 ∈ / 0. Corollary A full-rank system on R2 × T is controllable if and only if there exists A ∈ such that t → exp(tA) is periodic. We now investigate the group R × T2 . We have R × T2 = diag (ex , eiy , eiz ) : x, y, z ∈ R 3g1 = {diag (x, iy, iz) : x, y, z ∈ R} Let E1 = diag (1, 0, 0), E2 = diag (0, i, 0), E 3 = diag (0, 0, i), and let GL (2, Z) denote the (modular) group g ∈ Z2×2 : det g = ±1 . Proposition 4 The group of automorphisms d Aut (R × T2 ) is given by ⎧⎡ ⎫ ⎤ ⎨ x 0 0 ⎬ ⎣ y ⎦ : x, y, z ∈ R, x = 0, M ∈ GL (2, Z) . M ⎩ ⎭ z Proof We have a universal covering q : R3 → R × T2 , diag (ex , ey , ez ) → diag (ex , eiy , eiz ) with ker q = diag (1, e2πZ , e2πZ ). Let ⎤ ⎡ a1 a2 a3 ψ = ⎣ b1 b2 b3 ⎦ ∈ Aut (3g1 ). c1 c2 c3 Now, ψ ∈ d Aut (R × T2 ) if and only if φ(ker q) = ker q, where φ ∈ Aut (R3 ) and T1 φ = (T1 q)−1 · ψ · T1 q. Suppose that ψ ∈ d Aut (R × T2 ). Then, 2πZa2 + 2πZa3 = {0}, 2πZb2 + 2πZb3 = 2πZ, and 2πZc2 + 2πZc3 = 2πZ. Hence, a2 = a3 = 0, and by the Bézout identity, gcd (b2 , b3 ) = gcd (c2 , c3 ) = 1. If ψ ∈ d Aut (R × T2 ), then so is −1 ψ b2 c3 − b3 c2 divides both b2 and b3 . Hence, b2 c3 − b3 c2 = ±1. Thus, . Therefore, b2 b3 b2 b3 ∈ GL (2, Z). Conversely, suppose that a2 = a3 = 0 and ∈ GL (2, Z). c2 c3 c2 c3 Then, gcd (b2 , b3 ) = gcd (c2 , c3 ) = 1, and so φ(ker q) = ker q. Henceforth, (E1∗ , E2∗ , E3∗ ) denotes the dual basis of (E1 , E2 , E3 ). Theorem 3 Any full-rank system on R × T2 is equivalent to one of the following systems: E1∗ ( 0 ) = {0}

(2,1)

1,αγ : αE3 + u1 E1 + u2 (E2 + γ E3 ) (2,1)

2

E1∗ ( 0 ) = {0}

: E1 + u1 E2 + u2 E3

(3,0) : u1 E1 + u2 E2 + u3 E3 . Here, α > 0 and γ ∈ [0, 1) parametrize a family of class representatives, some of which may be equivalent. (2,1)

(2,1)

Remark 2 It turns out that 1,αγ is equivalent to 1,α γ if and only if there exists a1 a2 a1 a4 −a2 a3 4γ ∈ GL (2, Z) such that a1 + a2 γ = 0, γ = aa31 +a +a2 γ , and α = a1 +a2 γ α · a3 a4


Proof No (1, 1)-system or (2, 0)-system has full rank. Any (3, 0)-system is clearly (3,0) . Let be a (2, 1)-system with trace = 3 a E + equivalent to i i i=1 3 3 ∗ ( 0 ) = {0}. Then, = a E + E , E . . Suppose that E b E , c E 1 1 2 3 i=1 i i i=1 i i 1 (2,1)

Hence, ψ = diag ( a11 , 1, 1) is an automorphism such that ψ · = 2 . Suppose ∗ 0 may assume b1 = 0. Hence, = a2 E2 + a3 E3 + that E1( ) = {0}. Then, we E1 + b2 E2 + b3 E3 , c2 E2 + c3 E3 . Therefore, ⎡ ⎤ ⎡ ⎤ 1 0 0 1 0 0 ψ = ⎣ −b2 1 0 ⎦ or ψ = ⎣ −b3 0 1 ⎦ −b3 0 1 −b2 1 0 is an automorphism such that ψ · = a3 E3 + E1 , E2 + c3 E3 . Then, ⎤ ⎡ 1 0 0 ⎦, k ∈ Z 0 ψ = ⎣ 0 1 0 k sgn (a3 ) is an automorphism such that

ψ · ψ · = sgn (a3 )a3 E3 + E1 , E2 + (sgn (a3 )c3 + k)E3 = αE3 + E1 , E2 + γ E3

with α > 0 and γ ∈ [0, 1). (2,1)

(2,1)

Theorem 4 The systems 1,αγ and (3,0) are controllable, whereas 2

is not.

Proof Clearly (3,0) is controllable. The curve t → exp(tαE3 ) is a periodic tra(2,1) jectory of 1,αγ , and so this system is controllable. Any trajectory g(·) : t → (2,1)

satisfies x(t) = t for t ≥ 0. (Therediag (ex(t) , eiy(t) , eiz(t) ), g(0) = 1 of 2 fore, states corresponding to x < 0 are not attainable from identity.) Thus, 2(2,1) is not controllable. Corollary A (2, 1)-system on R × T2 is controllable if and only if E1∗ ( 0 ) = 0. Corollary A full-rank system on R × T2 is controllable if and only if there exists A ∈ such that t → exp(tA) is periodic. Lastly, we consider the group T3 . We have T3 = diag (eix , eiy , eiz ) : x, y, z ∈ R 3g1 = {diag (ix, iy, iz) : x, y, z ∈ R} . Let E1 = diag (i, 0, 0), E2 = diag(0, i, 0), E3 = diag (0, 0, i), and let GL (3, Z) denote the group g ∈ Z3×3 : det g = ±1 . Proposition 5 d Aut (T3 ) = GL (3, Z). Theorem 5 Any full-rank system on T3 is equivalent to one of (2,1) αγ : αE1 + u1 (γ1 E1 + E2 ) + u2 (γ2 E1 + E3 )

(3,0) : u1 E1 + u2 E2 + u3 E3 .


Here, α > 0 and γ1 , γ2 ∈ [0, 1) parametrize a family of class representatives, some of which may be equivalent. (Proofs are omitted as they are similar to those above.) Remark 3 As T3 is compact, all full-rank systems on T3 are controllable. 3.2 Type g2.1 ⊕ g1 We consider first the simply connected group Aff (R)0 × R. We have ⎫ ⎧⎡ ⎤ ⎬ ⎨ 1 0 0 Aff (R)0 × R = ⎣ z ex 0 ⎦ : x, y, z ∈ R ⎭ ⎩ 0 0 ey ⎧⎡ ⎫ ⎤ ⎨ 0 0 0 ⎬ g2.1 ⊕ g1 = ⎣ z x 0 ⎦ = M(x, y, z) : x, y, z ∈ R . ⎩ ⎭ 0 0 y Aff (R)0 × R is completely solvable and simply connected. Hence,

Proposition 6 A system on Aff (R)0 × R with trace = A + 0 is controllable if and only if Lie ( 0 ) = g2.1 ⊕ g1 . Let E1 = M(1, 0, 0), E2 = M(0, 1, 0), and E3 = M(0, 0, 1). The only nonzero commutator is [E1 , E3 ] = E3 . The centre of Aff (R)0 × R is {exp(yE2 ) : y ∈ R}. Remark 4 The basis F1 = 12 (E2 + E3 ), F2 = 12 (E2 − E3 ), F3 = 2E1 has nonzero commutator relations [F2 , F3 ] = F1 − F2 , [F3 , F1 ] = F1 − F2 , as in the Bianchi-Behr classification. With respect to the ordered basis (E1 , E2 , E3 ), the group of automorphisms d Aut (Aff (R)0 × R) = Aut (g2.1 ⊕ g1 ) takes the form ⎧⎡ ⎫ ⎤ ⎨ 1 0 0 ⎬ ⎣ x u 0 ⎦ : x, y, u, v ∈ R, uv = 0 . ⎩ ⎭ y 0 v Theorem 6 ([10]) Any full-rank system on Aff (R)0 × R is equivalent to exactly one of the following systems: (1,1)

1

: E2 + E3 + uE1

(1,1) 2,β : βE1 + u(E2 + E3 )

E1∗ ( 0 ) = {0} E1∗ ( 0 ) = {0}

(2,0) : u1 E1 + u2 (E2 + E3 ) : E 3 + u 1 E1 + u 2 E2

E2 ∈ 0 , E1∗ ( 0 ) = {0}

2(2,1) : E2 + u1 E3 + u2 E1

E3 ∈ 0 , E1∗ ( 0 ) = {0}

(2,1)

1

(2,1)

3

: E2 + u1 E1 + u2 (E2 + E3 )

(2,1) : βE1 + u1 E2 + u2 E3 4,β

(3,0) : u1 E1 + u2 E2 + u3 E3 .

E2 , E3 ∈ / 0 , E1∗ ( 0 ) = {0} E1∗ ( 0 ) = {0}


Here, β = 0 parametrizes families of class representatives, each different value corresponding to a distinct non-equivalent representative. (2,1)

Remark 5 Among the above systems on Aff (R)0 × R, only (2,0) , 3 controllable.

and (3,0) are

Proof (sketch) Let = A + 0 be the trace of a system on Aff (R)0 × R. Suppose that is a (1, 1)-system. Also, suppose that E1∗ ( 0 ) = {0}. Then, = a2 E2 + a3 E3 + E1 + b2 E2 + b3 E3 with a2 a3 = 0 (as has full rank). Thus, ⎡ ⎤ 1 0 0 ψ = ⎣ b2 a2 0 ⎦ b3 0 a3 is an automorphism such that ψ · 1 = . Likewise, if E1∗ ( 0 ) = {0}, then there exists (1,1) an automorphism ψ such that ψ · 2,β = . None of the representatives are equivalent. (1,1)

(1,1)

E2 and E3 are eigenvectors of every automorphism. Hence, 2,β (1,1) 1 .

E1∗ (ψ

cannot be equivalent (1,1)

(1,1)

to Also, · βE1 ) = β for any automorphism ψ. Thus, 2,β and 2,β are equivalent only if β = β . For (2, 0)-systems, the result then follows by Proposition 2. Suppose that is a (2, 1)-system. Furthermore, suppose that E1∗ ( 0 ) = {0}, E2 ∈ / 0, 0 and E3 ∈ / . Then, = a2 E2 + E1 + b2 E2 + b3 E3 , E2 + c3 E3 with a2 c3 = 0. Thus, ⎤ ⎡ 1 0 0 ψ = ⎣ b2 a2 0 ⎦ b3 0 a2 c3 is an automorphism such that ψ · 3 = . Similarly, if E1∗ ( 0 ) = {0} and E2 ∈ 0 0 or E3 ∈ , respectively, then there exists an automorphism ψ such that ψ · 1(2,1) = (2,1) = , respectively. Suppose that E1∗ ( 0 ) = {0}. Then, 0 = E2 , E3 and so or ψ · 2 = 4,β for some β = 0. Likewise, none of these representatives are equivalent. Clearly, any (3, 0)-system is equivalent to (3,0) . (2,1)

Next, we consider the group Aff (R)0 × T. We have ⎧⎡ ⎫ ⎤ ⎨ 1 0 0 ⎬ Aff (R)0 × T = ⎣ z ex 0 ⎦ : x, y, z ∈ R ⎩ ⎭ 0 0 eiy ⎫ ⎧⎡ ⎤ ⎬ ⎨ 0 0 0 g2.1 ⊕ g1 = ⎣ z x 0 ⎦ = M(x, y, z) : x, y, z ∈ R . ⎭ ⎩ 0 0 iy Let E1 = M(1, 0, 0), E2 = M(0, 1, 0), and E3 = M(0, 0, 1). Proposition 7 The group of automorphisms d Aut (Aff (R)0 × T) is given by ⎧⎡ ⎫ ⎤ ⎨ 1 0 0 ⎬ ⎣ x σ 0 ⎦ : x, y, v ∈ R, v = 0, σ = ±1 . ⎩ ⎭ y 0 v


Proof We have a universal covering q : Aff (R)0 × R → Aff (R)0 × T, ⎤ ⎡ ⎤ ⎡ 1 0 0 1 0 0 q : ⎣ z ex 0 ⎦ → ⎣ z ex 0 ⎦ 0 0 ey 0 0 eiy with ker q = diag (1, 1, exp(2πZ)). Now, an automorphism ⎡ ⎤ 1 0 0 ψ = ⎣ x u 0 ⎦ ∈ Aut (g2.1 ⊕ g1 ) y 0 v is in d Aut (Aff (R)0 ×T) if and only if φ(ker q) = ker q, where φ ∈ Aut (Aff (R)0 ×R) and T1 φ = (T1 q)−1 · ψ · T1 q. So ψ ∈ d Aut (Aff (R)0 × T) if and only if φ(exp(2πZE2 )) = exp(2πZE2 ), i.e. 2πZu = 2πZ. Proposition 8 Any full-rank (1, 1)-system on Aff (R)0 × T is equivalent to exactly one of the following systems: (1,1)

: αE2 + E3 + uE1

E1∗ ( 0 ) = {0}

(1,1)

: βE1 + u(E2 + E3 )

E1∗ ( 0 ) = {0}.

1,α 2,β

Here, α > 0 and β = 0 parametrize families of class representatives, each different value corresponding to a distinct non-equivalent representative. 3 . Suppose that Proof Let be a (1, 1)-system with trace = 3i=1 ai Ei + b E i i i=1 ∗ 0 E1 ( ) = {0}. Then, = a2 E2 + a3 E3 + E1 + b2 E2 + b3 E3 . Full rank implies that a2 a3 = 0. Therefore, ⎤ ⎡ 1 0 0 ⎥ ⎢ ψ = ⎣ −b2 sgn (a2 ) sgn (a2 ) 0 ⎦ b

− a3

3

0

1 a3

(1,1) , where α = is an automorphism such that ψ · = a2 sgn (a2 )E2 + E3 + E1 = 1,α a2 sgn (a2 ) > 0. Suppose that E1∗ ( 0 ) = {0}. Then, = 3i=1 ai Ei + b2 E2 + b3 E3 . Full rank implies that a1 b2 b3 = 0. Thus, ⎤ ⎡ 1 0 0 a 2 ⎥ ⎢ ψ = ⎣ − a1 1 0 ⎦ − aa31 bb23 0 bb23 (1,1) . is an automorphism such that ψ · = a1 E1 + E2 + E3 = 2,a 1 ∗ 0 For any ψ ∈ d Aut (Aff (R)0 × T), we have that E1 ( ) = {0} if and only if E1∗ (ψ · (1,1) (1,1) 0 ) = {0}. Therefore, 1,α cannot be equivalent to 2,β . It is a simple matter to verify (1,1) (1,1) (1,1) (1,1) that 1,α and 1,α are equivalent only if α = α . Similarly, 2,β and 2,β are equivalent only if β = β .

By Proposition 2, it then follows that Corollary Any full-rank (2, 0)-system on Aff (R)0 × T is equivalent to (2,0) : u1 E1 + u2 (E2 + E3 ).


Proposition 9 Any full-rank (2, 1)-system on Aff (R)0 × T is equivalent to exactly one of the following systems: (2,1)

1

E2 ∈ 0 , E1∗ ( 0 ) = {0}

: E3 + u1 E1 + u2 E2

E3 ∈ 0 , E1∗ ( 0 ) = {0}

(2,1) 2,α : αE2 + u1 E3 + u2 E1 (2,1)

: αE2 + u1 E1 + u2 (E2 + E3 )

(2,1)

: βE1 + u1 E2 + u2 E3

3,α 4,β

E2 , E3 ∈ / 0 , E1∗ ( 0 ) = {0} E1∗ ( 0 ) = {0}.

Here, α > 0 and β = 0 parametrize families of class representatives, each different value corresponding to a distinct non-equivalent representative. Proof Let be a (2, 1)-system with trace . Suppose that E1∗ ( 0 ) = {0} and E2 ∈ 0 . Then, = a1 E1 + a3 E 3 + b1 E1 + b3 E3 , E2 . As E1∗ ( 0 ) = {0}, b1 = 0. Thus, = a3 E3 + E1 + b3 E3 , E2 . Hence, ⎤ 1 0 0 ψ =⎣ 0 1 0 ⎦ b3 0 a3 ⎡

(2,1)

is an automorphism such that ψ · 1 = . Suppose that E1∗ ( 0 ) = {0} and E3 ∈ 0 . Then, = a1 E1 + a2 E2 + ∗ 0 b 1 E1 + b2 E2 , E 3 . As E1 ( ) = {0} we have that b1 = 0 and so = a2 E2 + E1 + b2 E2 , E3 . Therefore, ⎤ 1 0 0 ψ = ⎣ b2 sgn (a2 ) 0 ⎦ 0 0 1 ⎡

is an automorphism such that ψ · 2,α = , where α = a2 sgn (a2 ). / 0 , and E3 ∈ / 0 . Then, = a2 E2 + a3 E3 + Suppose that E1∗ ( 0 ) = {0}, E2 ∈ E 1 + b2 E2 + b3 E3 , c2 E2 + c3 E 3 . Now, c2 = 0 and c3 = 0. Hence, = a2 E2 + E1 + b2 E2 + b3 E3 , E2 + c3 E3 and (2,1)

⎤ 1 0 0 ⎦ 0 ψ = ⎣ b2 sgn (a2 ) b3 0 c3 sgn (a2 ) ⎡

is an automorphism such that ψ · 3,α = , where α = a2 sgn (a2 ) > 0. Lastly, suppose that E1∗ ( 0 ) = {0}. Then, = βE1 + E2 , E3 for some β = 0. It is easy to verify that no two class representatives are equivalent. (2,1)


In summary, Theorem 7 Any full-rank system on Aff (R)0 × T is equivalent to exactly one of the following systems: (1,1)

: αE2 + E3 + uE1

E1∗ ( 0 ) = {0}

(1,1)

: βE1 + u(E2 + E3 )

E1∗ ( 0 ) = {0}

1,α 2,β

(2,0) : u1 E1 + u2 (E2 + E3 ) 1(2,1) : E3 + u1 E1 + u2 E2

E2 ∈ 0 , E1∗ ( 0 ) = {0}

(2,1) : αE2 + u1 E3 + u2 E1 2,α

E3 ∈ 0 , E1∗ ( 0 ) = {0}

(2,1)

: αE2 + u1 E1 + u2 (E2 + E3 )

(2,1)

: βE1 + u1 E2 + u2 E3

3,α 4,β

E2 , E3 ∈ / 0 , E1∗ ( 0 ) = {0} E1∗ ( 0 ) = {0}

(3,0) : u1 E1 + u2 E2 + u3 E3 . Here, α > 0 and β = 0 parametrize families of class representatives, each different value corresponding to a distinct non-equivalent representative. (2,1)

(2,1)

Theorem 8 Among the above systems on Aff (R)0 × T, only (2,0) , 2,α , 3,α (3,0) are controllable.

and

(2,1)

Proof For (2,0) , 3,α , and (3,0) , we have Lie ( 0 ) = g2.1 ⊕ g1 , and so these systems (2,1) are controllable. The system 2,α has a periodic trajectory t → exp(tαE2 ) and so is controllable. We now show that the remaining systems are not controllable. Suppose that

⎡

⎤ 1 0 0 g(t) = ⎣ z(t) ex(t) 0 ⎦ 0 0 eiy(t) ˙ = is a trajectory of a system : u1 E1 + u2 E2 + u3 E3 . Then, the Cauchy problem g(t) g(t) (u1 (t)E1 + u2 (t)E2 + u3 (t)E3 ), g(0) = 1 takes the form x(0) = 0

x(t) ˙ = u1 (t)

y(0) ∈ 2πZ

y(t) ˙ = u2 (t) z˙ (t) = u3 (t)

e

x(t)

z(0) = 0.

(1,1) and 1(2,1) , we have z˙ (t) = ex(t) > 0 and so z(t) > 0. Correspondingly, for 1,α (1,1) (2,1) Likewise, for 2,β and 4,β , we have that x(t) ˙ = β = 0 and so either x(t) < 0 or x(t) > 0. Hence, these systems are not controllable (as in each case some states are not attainable from the identity).

Corollary A full-rank system on Aff (R)0 × T is controllable if and only if it is homogeneous or there exists A ∈ such that t → exp(tA) is periodic.


3.3 Types g3.1 , g3.2 , g3.3 , g03.4 , and ga3.4 We consider the completely solvable simply connected Lie groups H3 , G3.2 , G3.3 , SE (1, 1) = G03.4 , and Ga3.4 . Of these groups, only the Heisenberg group H3 has a nontrivial centre. However, the quotient H∗3 = H3 /Z(H3 (Z)) is not a matrix Lie group. Controllability on these groups can be characterized as follows. Proposition 10 Any system on G ∈ {H3 , G3.2 , G3.3 , SE (1, 1), Ga3.4 } is controllable if and only if Lie ( 0 ) = g. We have the following matrix representations: ⎡ ⎤ ⎡ ⎤ 1 y x 0 y x H3 : ⎣ 0 1 z ⎦ h3 : ⎣ 0 0 z ⎦ 0 0 1 0 0 0 ⎤ ⎡ ⎡ ⎤ 1 0 0 0 0 0 0 ⎦ g3.2 : ⎣ y z 0 ⎦ G3.2 : ⎣ y ez x −zez ez x −z z ⎤ ⎡ ⎡ ⎤ 1 0 0 0 0 0 z g3.3 : ⎣ y z 0 ⎦ G3.3 : ⎣ y e 0 ⎦ x 0 ez x 0 z ⎡ ⎤ ⎡ ⎤ 1 0 0 0 0 0 SE (1, 1) : ⎣ x cosh z − sinh z ⎦ se (1, 1) : ⎣ x 0 −z ⎦ y − sinh z cosh z y −z 0 ⎡ ⎤ ⎡ ⎤ 1 0 0 0 0 0 Ga3.4 : ⎣ x eaz cosh z −eaz sinh z ⎦ ga3.4 : ⎣ x az −z ⎦ . az az a>0 a>0 y −e sinh z e cosh z y −z az a =1

a =1

As each group G is simply connected, we have d Aut (G) = Aut (g) in each case. The corresponding groups of automorphisms (with respect to the ordered bases) are given by ⎧⎡ ⎫ ⎤ ⎨ yw − vz x u ⎬ 0 y v ⎦ : x, y, z, u, v, w ∈ R, yw − vz = 0 Aut (h3 ) = ⎣ ⎩ ⎭ 0 z w ⎧⎡ ⎫ ⎤ ⎨ u x y ⎬ Aut (g3.2 ) = ⎣ 0 u z ⎦ : x, y, z, u ∈ R, u = 0 ⎩ ⎭ 0 0 1 ⎫ ⎧⎡ ⎤ ⎬ ⎨ x y z Aut (g3.3 ) = ⎣ u v w ⎦ : x, y, z, u, v, w ∈ R, xv − yu = 0 ⎭ ⎩ 0 0 1 ⎧⎡ ⎫ ⎤ ⎨ x y u ⎬ 2 2 Aut (se (1, 1)) = ⎣ σy σ x v ⎦ : x, y, u, v ∈ R, x − y = 0, σ = ±1 ⎩ ⎭ 0 0 σ ⎧⎡ ⎫ ⎤ ⎨ x y u ⎬ Aut (ga3.4 ) = ⎣ y x v ⎦ : x, y, u, v ∈ R, x 2 − y 2 = 0 . ⎩ ⎭ 0 0 1


For each of these five types, a classification of systems is obtained. Theorem 9 ([9]) Any full-rank system on H3 is equivalent to exactly one of the following systems: (1,1) : E2 + uE3 (2,0) : u1 E2 + u2 E3 (2,1)

: E1 + u1 E2 + u2 E3

E1 ∈ / 0

2(2,1) : E3 + u1 E1 + u2 E2

E1 ∈ 0

1

(3,0) : u1 E1 + u2 E2 + u3 E3 . (2,1)

Remark 6 Only (2,0) , 1 ately from Proposition 10.)

, and (3,0) are controllable. (This follows almost immedi-

3 Proof (sketch) Let be a (1, 1)-system with trace = 3i=1 ai Ei + i=1 bi Ei . Then, ⎡ ⎤ a2 b3 − a3 b2 a1 b1 0 a2 b2 ⎦ ψ =⎣ 0 a3 b3 is an automorphism such that ψ · (1,1) = . The (2, 0) case then follows by Proposition 2. Clearly, any (3, 0)-system is equivalent to (3,0) . 0 0 Let be a (2, with trace = A + . Suppose that E1 ∈ . Then, 1)-system 3 3 = i=1 ai Ei + E1 , i=1 bi Ei . As in the (1, 1) case, there then exists an automorphism (2,1) = . On the other hand, suppose that E1 ∈ / 0 . Let = 3i=1 ai Ei + 2 ψsuch that ψ· 3 3 i=1 bi Ei , i=1 ci Ei . Then ⎡ ⎤ ⎡ ⎤ v1 v2 v3 a1 b1 c1 v1 v2 v3 ⎣ a2 b2 c2 ⎦ = 1 0 0 ψ =⎣ 0 1 0 ⎦ 0 0 v1 a3 b3 c3 is an automorphism such that ψ · = 1(2,1) . As E1 is an eigenvector of every (2,1) (2,1) and 2 are not equivalent. automorphism, it follows that 1 Theorem 10 ([9]) Any full-rank system on G3.2 is equivalent to exactly one of the following systems: 1(1,1) : E2 + uE3 E3∗ ( 0 ) = {0} (1,1)

2,β : βE3 + uE2

E3∗ ( 0 ) = {0}

(2,0) : u1 E2 + u2 E3 1(2,1) : E1 + u1 E2 + u2 E3

E1 ∈ / 0 , E3∗ ( 0 ) = {0}

2(2,1) : E2 + u1 E1 + u2 E3

E1 ∈ 0 , E3∗ ( 0 ) = {0}

(2,1)

3,β : βE3 + u1 E1 + u2 E2

E3∗ ( 0 ) = {0}

(3,0) : u1 E1 + u2 E2 + u3 E3 . Here, β = 0 parametrizes families of class representatives, each different value corresponding to a distinct non-equivalent representative.

Rory Biggs and Claudiu C. Remsing (2,1)

Remark 7 Only (2,0) , 1

, and (3,0) are controllable.

Proof (sketch) Let be a (1, 1)-system with trace = A + 0 . Suppose that E3∗ ( 0 ) = {0}. Then, = a1 E1 + a2 E2 + b1 E1 + b2 E2 + E3 with a2 = 0. Hence ⎤ ⎡ a2 a1 b1 ψ = ⎣ 0 a2 b2 ⎦ 0 0 1 = . Similarly, if E3∗ ( 0 ) = {0}, then there exists is an automorphism such that ψ · 1 (1,1) an automorphism ψ such that ψ · 2,β = for some β = 0. The (2, 0) case follows by Proposition 2, and the (3, 0) case is trivial. Let be a (2, 1)-system with trace = A + 0 . Suppose that E3∗ ( 0 ) = {0} and / 0 . Then, = a1 E1 + b1 E1 + E2 , c1 E1 + E3 with a1 = 0. Hence, E1 ∈ ⎡ ⎤ a1 a1 b1 c1 ψ = ⎣ 0 a1 0 ⎦ 0 0 1 (1,1)

= . Similarly, if E3∗ ( 0 ) = {0} and E1 ∈ 0 , then is an automorphism such that ψ ·2 (2,1) = . Lastly, suppose that E3∗ ( 0 ) = there exists an automorphism ψ such that ψ · 1 (2,1) {0}. Then, = 3,β for some β = 0. It is a simple matter to show that none of the representatives are equivalent. (2,1)

Theorem 11 ([9]) Any full-rank system on G3.3 is equivalent to exactly one of the following systems: (2,1)

1

: E 1 + u 1 E2 + u 2 E3

(2,1) 2,β : βE3 + u1 E1 + u2 E2

E3∗ ( 0 ) = {0} E3∗ ( 0 ) = {0}

(3,0) : u1 E1 + u2 E2 + u3 E3 Here, β = 0 parametrizes a family of class representatives, each different value corresponding to a distinct non-equivalent representative. Remark 8 Only (3,0) is controllable. Proof (sketch) It is easy to show that there are no full-rank (1, 1)-systems or (2, 0)-systems. The (3, 0) case is trivial. Let be a (2, 1)-system with trace = A + 0 . Suppose that E3∗ ( 0 ) = {0}. Then, = a1 E1 + a2 E2 + b1 E1 + b2 E2 , c1 E1 + c2 E2 + E3 with a1 b2 − a2 b1 = 0. Thus, ⎡ ⎤ a1 b1 c1 ψ = ⎣ a2 b2 c2 ⎦ 0 0 1 = . Suppose that E3∗ ( 0 ) = {0}. Then, 0 = is a automorphism such that ψ · 1 (2,1) E1 , E2 and so = 2,β for some β = 0. No two of these representatives are equivalent. (2,1)


We shall use a different basis for se (1, 1), resp. ga3.4 ; with respect to this basis, the automorphism groups take a simpler form. More precisely, let F1 = E3

F2 = E1 + E2

F3 = −E1 + E2 .

This basis has commutator relations [F2 , F3 ] = 0

[F3 , F1 ] = −(a + 1)F3

[F1 , F2 ] = (a − 1)F2 .

(Here, a = 0 corresponds to se (1, 1)). With respect to (F1 , F2 , F3 ), the respective automorphism groups take the form ⎧⎡ ⎫ ⎤ ⎡ ⎤ −1 0 0 ⎨ 1 0 0 ⎬ Aut (se (1, 1)) = ⎣ x u 0 ⎦ , ⎣ x 0 u ⎦ : x, y, u, v ∈ R, uv = 0 ⎩ ⎭ y 0 v y v 0 ⎫ ⎧⎡ ⎤ ⎬ ⎨ 1 0 0 Aut (ga3.4 ) = ⎣ x u 0 ⎦ : x, y, u, v ∈ R, uv = 0 . ⎭ ⎩ y 0 v We have Aut (ga3.4 ) = Aut (g2.1 ⊕ g1 ) and

⎧ ⎡ ⎤⎫ −1 0 0 ⎬ ⎨ Aut (se (1, 1)) = Aut (ga3.4 ) 1, ⎣ 0 0 1 ⎦ . ⎩ ⎭ 0 1 0

It follows that the classifications of systems on Aff (R)0 × R (Theorem 6), on SE (1, 1), and on Ga3.4 are very similar. (Hence, we shall omit proofs here.) Theorem 12 ([10]) Any full-rank system on SE (1, 1) is equivalent to exactly one of the following systems: F1∗ ( 0 ) = {0}

1(1,1) : F2 + F3 + uF1 (1,1)

2,α

F1∗ ( 0 ) = {0}

: αF1 + u(F2 + F3 )

(2,0) : u1 F1 + u2 (F2 + F3 ) 1(2,1) : F3 + u1 F1 + u2 F2 (2,1)

: F2 + u1 F1 + u2 (F2 + F3 )

(2,1)

: αF1 + u1 F2 + u2 F3

2

3,α

F2 ∈ 0 or F3 ∈ 0 , F1∗ ( 0 ) = {0} F2 , F3 ∈ / 0 , F1∗ ( 0 ) = {0} F1∗ ( 0 ) = {0}

(3,0) : u1 F1 + u2 F2 + u3 F3 . Here, α > 0 parametrizes families of class representatives, each different value corresponding to a distinct non-equivalent representative. Remark 9 Only (2,0) , 2(2,1) , and (3,0) are controllable. These three systems are, respectively, equivalent to (2,0) : u1 E2 + u2 E3 (2,1) : E1 + u1 E2 + u2 E3 2 (3,0) : u1 E1 + u2 E2 + u3 E3 .


Theorem 13 ([10]) Any full-rank system on Ga3.4 is equivalent to exactly one of the following systems: (1,1)

: F2 + F3 + uF1

F1∗ ( 0 ) = {0}

(1,1)

: βF1 + u(F2 + F3 )

F1∗ ( 0 ) = {0}

1

2,β

(2,0) : u1 F1 + u2 (F2 + F3 ) : F3 + u1 F1 + u2 F2

F2 ∈ 0 , F1∗ ( 0 ) = {0}

2(2,1) : F2 + u1 F3 + u2 F1

F3 ∈ 0 , F1∗ ( 0 ) = {0}

(2,1)

1

(2,1)

3

: F2 + u1 F1 + u2 (F2 + F3 )

F2 , F3 ∈ / 0 , F1∗ ( 0 ) = {0} F1∗ ( 0 ) = {0}

(2,1) : βF1 + u1 F2 + u2 F3 4,β

(3,0) : u1 F1 + u2 F2 + u3 F3 . Here, β = 0 parametrizes families of class representatives, each different value corresponding to a distinct non-equivalent representative. (2,1)

Remark 10 Only (2,0) , 3 respectively, equivalent to

, and (3,0) are controllable. These three systems are,

(2,0) : u1 E2 + u2 E3 (2,1) : E1 + u1 E2 + u2 E3 3 (3,0) : u1 E1 + u2 E2 + u3 E3 . 3.4 Type g03.5 (The Euclidean Group and its Covers) (2) of the Euclidean group SE (2) is a simply connected The universal covering group SE matrix Lie group ⎫ ⎧⎡ ⎤ 1 0 0 0 ⎪ ⎪ ⎪ ⎪ ⎬ ⎨⎢ ⎥ x cos z − sin z 0 ⎥=m (2) = ⎢ (x, y, z) : x, y, z ∈ R SE ⎦ ⎣ ⎪ ⎪ ⎪ ⎪ y sin z cos z 0z ⎭ ⎩ 0 0 0 e with Lie algebra ⎧⎡ 0 ⎪ ⎪ ⎨⎢ x se (2) = ⎢ ⎣y ⎪ ⎪ ⎩ 0

0 0 z 0

0 −z 0 0

⎫ ⎤ 0 ⎪ ⎪ ⎬ 0⎥ ⎥ = M(x, y, z) : x, y, z ∈ R . 0⎦ ⎪ ⎪ ⎭ z

(2) may also be represented as Remark 11 SE ⎡ z ⎤ 0 0 e (2) : ⎣ x ez cos z −ez sin z ⎦ SE y ez sin z ez cos z

⎡

⎤ z 0 0 se (2) : ⎣ x z −z ⎦ . y z z

However, the 4 × 4 representation is more suitable for the subsequent computations and extends easily to n-fold coverings of SE (2).


0, 0), E2 = M(0, 1, 0), (2) is m (0, 0, 2πZ). Let E1 = M(1, The centre of SE and E3 = M(0, 0, 1). With respect to (E1 , E2 , E3 ), the group of automorphisms (2)) = Aut ( se (2)) takes the form d Aut (SE ⎧⎡ ⎫ ⎤ x y u ⎨ ⎬ ⎣ −σy σ x v ⎦ : x, y, u, v ∈ R, x 2 + y 2 = 0, σ = ±1 . ⎩ ⎭ 0 0 σ The n-fold covering SEn (2) of SE (2) ∼ = SE1 (2) is given by ⎧⎡ 1 ⎪ ⎪ ⎨⎢ x SEn (2) = ⎢ ⎣y ⎪ ⎪ ⎩ 0 ⎧⎡ 0 ⎪ ⎪ ⎨⎢ x ⎢ sen (2) = ⎣ y ⎪ ⎪ ⎩ 0

⎫ ⎤ 0 0 0 ⎪ ⎪ ⎬ cos z − sin z 0 ⎥ ⎥ = m(x, y, z) : x, y, z ∈ R sin z cos z 0 ⎦ ⎪ ⎪ iz ⎭ 0 0 en ⎫ ⎤ 0 0 0 ⎪ ⎪ ⎬ 0 −z 0 ⎥ ⎥ = M(x, y, z) : x, y, z ∈ R . z 0 0⎦ ⎪ ⎪ ⎭ 0 0 iz n

Indeed, we have a universal covering morphism (2) → SEn (2), qn : SE

(x, y, z) → m(x, y, z) m

(0, 0, 2nπZ). Let (M(1, 0, 0), M(0, 1, 0), M(0, 0, 1)) be the ordered with ker qn = m basis for sen (2). By a slight abuse of notation, we shall also denote this basis by (E1 , E2 , E3 ). Proposition 11 d Aut (SEn (2)) = Aut (sen (2)). Proof Let ⎡

⎤ x y u ψ = ⎣ −σy σ x v ⎦ ∈ Aut (sen (2)) 0 0 σ (2)), T1 φ = (T1 qn )−1 · ψ · T1 qn . Then, ψ ∈ d Aut (SEn (2)) if and and let φ ∈ Aut (SE only if φ(ker qn ) = ker qn . Now, ker qn = exp(2nπZE3 ). (Here, E3 ∈ se (2).) Hence φ(ker qn ) = {exp(2unπzE1 + 2vnπzE2 + 2σ nπzE3 ) : z ∈ Z} where exp(2unπzE1 + 2vnπzE2 + 2σ nπzE3 ) is ⎡

⎤ 1 0 0 0 ⎢ σ (−v + v cos(2σ nπz) + u sin(2σ nπz)) cos(2σ nπz) − sin(2σ nπz) 0 ⎥ ⎢ ⎥. ⎣ σ (u − u cos(2σ nπz) + v sin(2σ nπz)) sin(2σ nπz) cos(2σ nπz) 0 ⎦ 0 0 0 e2σ nπz Therefore, φ(ker qn ) = ker qn .


(2)) is equivalent to exactly Theorem 14 ([10]) Any full-rank system on SEn (2) (resp. SE one of the following systems: E3∗ ( 0 ) = {0}

1(1,1) : E2 + uE3 (1,1)

2,α

E3∗ ( 0 ) = {0}

: αE3 + uE2

(2,0) : u1 E2 + u2 E3 (2,1)

1

: E1 + u1 E2 + u2 E3

(2,1) 2,α : αE3 + u1 E1 + u2 E2

E3∗ ( 0 ) = {0} E3∗ ( 0 ) = {0}

(3,0) : u1 E1 + u2 E2 + u3 E3 . Here, α > 0 parametrizes families of class representatives, each different value corresponding to a distinct non-equivalent representative. Proof (sketch) Let be a (1, 1)-system with trace = A + 0 . Suppose that E3∗ ( 0 ) = {0}. Then, = a1 E1 + a2 E2 + b1 E1 + b2 E2 + E3 . Thus, ⎡ ⎤ a2 a1 b1 ψ = ⎣ −a1 a2 b2 ⎦ 0 0 1 = . Suppose that E3∗ ( 0 ) = {0}. Then, = is an automorphism such that ψ · 1 a1 E1 + a2 E2 + a3 E3 + b1 E1 + b2 E2 with a3 = 0. Let α = a3 sgn(a3 ). Hence, ⎤ ⎡ a1 b2 sgn(a3 ) b1 α a2 ⎦ ψ = ⎣ −b1 sgn(a3 ) b2 α 0 0 sgn(a3 ) (1,1)

(1,1) = . The (2, 0) case then follows by Proposition 2. is an automorphism such that ψ · 2,α Let be a (2, 1)-system with trace = A + 0 . Suppose that E3∗ ( 0 ) = {0}. Then, = a1 E1 + a2 E2 + b1 E1 + b2 E2 , c1 E1 + c2 E2 + E3 . Thus ⎡ ⎤ v2 b2 v2 b1 c1 v1 a2 b2 −b1 ψ = ⎣ −v2 b1 v2 b2 c2 ⎦ = b1 b2 v2 a1 0 0 1

= . Suppose that E3∗ ( 0 ) = {0}. Then, = is an automorphism such that ψ · 1 a3 E3 + E1 , E2 . Thus, ψ = diag (1, sgn(a3 ), sgn(a3 )) is an automorphism such that ψ · (2,1) = 2,α for some α > 0. Clearly, any (3, 0)-system is equivalent to (3,0) . It is easy to show that no two class representatives are equivalent. (2,1)

SEn (2) decomposes as a semidirect product V K of (a vector space) V ∼ = R2 and a compact subgroup K, where V = {m(x, y, 0) : x, y ∈ R}

and

K = {m(0, 0, z) : z ∈ R}.

Indeed, m(x, y, z) = m(x, y, 0) m(0, 0, z) and V is normal in SEn (2). Proposition 12 (cf. [15, 35]) Any full-rank system on SEn (2) is controllable. (2) is slightly different. The situation for SE


(2), only (1,1) , (2,0) , (2,1) and (3,0) are controllable. Theorem 15 On SE 1 1 (2,1) Proof For (2,0) , 1 and (3,0) , we have Lie ( 0 ) = se (2); so these systems are all (1,1) (0, 0, 2π). Now, m (0, 0, 2π) controllable. For 1 , we have that exp(2(E1 +πE3 )) = m is in the unique connected subgroup O0 with Lie algebra Lie ( 0 ). This is sufficient for controllability of 1(1,1) (see [23]). We show that the remaining two (families of) systems are not controllable. Suppose that (x(t), y(t), z(t)) is a trajectory of a system : u1 E1 + u2 E2 + u3 E3 . Then, the g(t) = m Cauchy problem g(t) ˙ = g(t) (u1 (t)E1 + u2 (t)E2 + u3 (t)E3 ), g(0) = 1 takes the form

x(t) ˙ = u1 (t) cos z(t) − u2 (t) sin z(t)

x(0) = 0

y(t) ˙ = u2 (t) cos z(t) + u1 (t) sin z(t)

y(0) = 0

z˙ (t) = u3 (t)

z(0) = 0.

(1,1) (2,1) and 2,α , we have z˙ (t) = α > 0 and so z(t) > 0 for t > 0. Thus, For both 2,α states corresponding to m(x, y, z), z < 0 are not attainable from identity. Hence, these two (families of) systems are not controllable.

(2) is controllable if and only if E ∗ ( 0 ) = {0}. Corollary A full-rank system on SE 3 3.5 Type ga3.5 The simply connected matrix Lie group

Ga3.5

⎫ ⎧⎡ ⎤ 0 0 ⎬ ⎨ 1 = ⎣ x eaz cos z −eaz sin z ⎦ = m(x, y, z) : x, y, z ∈ R ⎭ ⎩ y eaz sin z eaz cos z

has Lie algebra

ga3.5

⎧⎡ ⎫ ⎤ ⎨ 0 0 0 ⎬ = ⎣ x az −z ⎦ = M(x, y, z) : x, y, z ∈ R . ⎩ ⎭ y z az

The centre of Ga3.5 is trivial. Let E1 = M(1, 0, 0), E2 = M(0, 1, 0), and E3 = M(0, 0, 1). The group of automorphisms d Aut (Ga3.5 ) = Aut (ga3.5 ) is given by ⎧⎡ ⎫ ⎤ ⎨ x y u ⎬ ⎣ −y x v ⎦ : x, y, u, v ∈ R . ⎩ ⎭ 0 0 1 Note that Aut (se (2)) = Aut (ga3.5 ) {1, diag (1, −1, −1)}. Thus, the classifications of systems on SE (2) and Ga3.5 , respectively, are very similar. (Accordingly, the proof of the following theorem is omitted.)


Theorem 16 ([10]) Any full-rank system on Ga3.5 is equivalent to exactly one of the following systems: E3∗ ( 0 ) = {0}

1(1,1) : E2 + uE3 (1,1)

2,β

E3∗ ( 0 ) = {0}

: βE3 + uE2

(2,0) : u1 E2 + u2 E3 (2,1)

1

E3∗ ( 0 ) = {0}

: E1 + u1 E2 + u2 E3

E3∗ ( 0 ) = {0}

(2,1) 2,β : βE3 + u1 E1 + u2 E2

(3,0) : u1 E1 + u2 E2 + u3 E3 . Here, β = 0 parametrizes families of class representatives, each different value corresponding to a distinct non-equivalent representative. (1,1)

Theorem 17 Among the above systems on Ga3.5 , only 1 are controllable.

(2,1)

, (2,0) , 1

and (3,0)

Proof For (2,0) , (2,1) and (3,0) , we have that Lie ( 0 ) = ga3.4 , and so these systems are controllable. (1,1) (2,1) We show that 2,β and 2,β are not controllable. Suppose that g(t) = m(x(t), y(t), z(t)) is a trajectory of a system : u1 E1 + u2 E2 + u3 E3 . Then, the Cauchy problem g(t) ˙ = g(t) (u1 (t)E1 + u2 (t)E2 + u3 (t)E3 ), g(0) = 1 takes the form x(t) ˙ = eaz(t) (u1 (t) cos z(t) − u2 (t) sin z(t))

x(0) = 0

y(t) ˙ = eaz(t) (u2 (t) cos z(t) + u1 (t) sin z(t))

y(0) = 0

z˙ (t) = u3 (t)

z(0) = 0.

(1,1)

(2,1)

For both 2,β and 2,β , we have z˙ (t) = β = 0 and so z(t) > 0 or z(t) < 0 for t > 0. Thus, states corresponding to m(x, y, z), z < 0 (resp. m(x, y, z), z > 0) are not attainable from identity. (1,1) Consider the system 1 . We find it more convenient to show that the equivalent system : E1 + uE3 is controllable; we show that 1 ∈ int A. (It is sufficient to construct a family of trajectories whose terminal points cover an open neighbourhood of the identity). The above Cauchy problem (for ) takes the form x(t) ˙ = eaz(t) cos z(t)

x(0) = 0

y(t) ˙ = e

y(0) = 0

az(t)

z˙ (t) = u(t)

sin z(t)

z(0) = 0.

This may be interpreted as follows. Let γ (·) : [0, T ] → R2 be a curve given by γ (t) = (x(t), y(t)). We have γ˙ (t) = eaz(t) (cos z(t), sin z(t)) and γ˙ (t) = eaz(t) . A trajectory from 1 to 1 corresponds to a curve γ (·) satisfying γ (0) = (0, 0), γ˙ (0) = (1, 0) and γ (T ) = (0, 0), γ˙ (T ) = (1, 0) for some T > 0. We can construct a “figure eight” trajectory satisfying these properties. Let ρ+ = exp(E1 + π2 E3 ) and ρ− = exp(E1 − π2 E3 ). Let φ : R3 → Ga3.5 be the smooth mapping 3 E1 +zE3 e (t, s, z) → etE1 ρ− e(s+1)E1 ρ− eτ1 E1 ρ− eτ2 E1 ρ+

Control Systems on 3D Lie Groups: Equivalence and Controllability Fig. 1 Trajectory of from 1 to 1

where τ1 =

−4e−

aπ 2

2aπ

+ 2e 2 (π + a(2 + aπ)) >0 1 + a2 π

τ2 =

4a + 4e

3aπ 2

+ 2 1 + a 2 eaπ π 1 + a2 π

> 0.

Explicitly, φ(t, s, z) is given by (computed with M ATHEMATICA) ⎡ ⎢ ⎢ ⎢ ⎣

1

e−

aπ 2

−a+ 1+a 2 (−2+t)z+eaz (a cos z+sin z) (1+a 2 )z " ! aπ aπ e 2 − 1+a 2 (−1+s)z+e 2 +az (− cos z+a sin z)

(1+a 2 )z

0

0

⎤

eaz cos z −eaz sin z ⎥ ⎥ ⎥. ⎦ az az e sin z e cos z

Then, φ(t, s, z) ∈ A for t ≥ 0, s ≥ −1, and z ∈ R. Direct computation shows that φ(1, 1, 0) = 1. The corresponding trajectory γ (·) is shown in Fig. 1 (with a = 15 ). A standard computation shows that T(1,1,0) φ has full rank. Hence, by the inverse function theorem, φ map some open neighbourhood of (1, 1, 0) to some open neighbourhood of 1 . Hence, 1 ∈ int A. Corollary A full-rank system on Ga3.5 is controllable if and only if E3∗ ( 0 ) = {0}. 4 Semisimple Groups 4.1 Type g3.6 (The Pseudo-Orthogonal Group and its Covers) There are only two connected matrix Lie groups with Lie algebra g3.6 up to isomorphism, namely the pseudo-orthogonal group SO (2, 1)0 and its double cover SL (2, R). The pseudo-orthogonal group SO (2, 1) = {g ∈ R3×3 : g J g = J, det g = 1}


has two connected components. Here, J = diag(1, 1, −1). The identity component of SO (2, 1) is SO (2, 1)0 = {g ∈ SO (2, 1) : g33 > 0} where g = gij (for g ∈ SO (2, 1)). Its Lie algebra is given by so (2, 1) = {A ∈ R3×3 : A J + J A = 0} ⎧⎡ ⎫ ⎤ ⎨ 0 z y ⎬ = ⎣ −z 0 x ⎦ : x, y, z ∈ R . ⎩ ⎭ y x 0 The special linear group SL (2, R) = {g ∈ R2×2 : det g = 1}

has Lie algebra #$ sl (2, R) =

y−z x 2 2 y+z x 2 −2

%

& : x, y, z ∈ R .

By a slight abuse of notation, we shall denote the (natural) basis for each of these Lie algebras by (E1 , E2 , E3 ). The group of automorphisms Aut (so (2, 1)) is precisely SO (2, 1). Also, the group Int (so (2, 1)) of inner automorphisms of so (2, 1) is precisely SO (2, 1)0 (see, e.g. [27]). (The same holds true for sl (2, R).) Note that Aut (so (2, 1)) is generated by Int (so (2, 1)) and the automorphism ⎡

⎤ −1 0 0 ς = ⎣ 0 1 0 ⎦. 0 0 −1 It is not difficult to show that there exists an automorphism φ ∈ Aut (SO (2, 1)0 ) (resp. φ ∈ Aut (SL (2, R))) such that T1 φ = ς. Accordingly, the following result holds true. Proposition 13 (cf. [13]) d Aut (SO (2, 1)0 ) = Aut (so (2, 1)). Likewise, d Aut (SL (2, R)) = Aut (sl (2, R)). The (Lorentzian) product on so (2, 1) is given by AB = a1 b1 +a2 b2 −a3 b3 . (Here, A = 3i=1 ai Ei and B = 3i=1 bi Ei ). Consider the level sets Hα = {A ∈ so (2, 1)\{0} : A A = α}. Hα is a hyperboloid of two sheets when α < 0, a hyperboloid of one sheet when α > 0, and a (punctured) cone when α = 0. As is preserved by automorphisms, each level set Hα is also preserved. Moreover, Aut (so (2, 1)) acts transitively on each invariant level set Hα , α ∈ R. The sign σ () of an affine subspace (of dimension one or two) is defined as follows. For = A + B (resp. = A + B1 , B2 ), the sign of is given by σ ) = sgn(B B)

()) ' '( ( B B1 B1 B2 ( ( resp. σ () = sgn (( 1 . B1 B2 B2 B2 (

As is preserved by automorphisms, a straightforward computation shows that σ (ψ ·) = σ () for ψ ∈ Aut(so (2, 1)). (The sign of an affine subspace is well defined as it is


independent of parametrization.) It turns out that σ () = −1 if the direction space 0 intersects the inside of the cone H0 , σ () = 0 if 0 is tangent to H0 , and σ () = 1 otherwise. If σ () = 0, then the critical point C () (at which an inhomogeneous affine subspace of nonzero sign is tangent to a level set Hα ; see [8]) is given by AB B B B B1 B1 B1 B2 −1 A B1 C () = A − B1 B2 . B1 B2 B2 B2 A B2 C () = A −

Critical points behave well under the action of automorphisms, i.e. ψ · C () = C (ψ · ) for any automorphism ψ. (The critical point of is well defined as it is independent of parametrization.) Let κ() = C () C (). Then is tangent to the hyperboloid Hκ() . We have κ(ψ · ) = κ() for any automorphism ψ. We define σ () and κ() likewise for sl (2, R). Theorem 18 ([8]) Any full-rank system on SO (2, 1)0 (resp. SL (2, R)) is equivalent to exactly one of the following systems: (1,1)

1

: E3 + u(E2 + E3 )

(1,1) 2,α (1,1) 3,α (1,1) 4,α (2,0) 1 (2,0) 2 1(2,1) (2,1) 2,α (2,1) 3,α (3,0)

: αE2 + uE3

κ() = α ,

: αE1 + uE2

κ() = α 2 ,

: αE3 + uE2

κ() = −α ,

σ () = 0 σ () = −1

2

σ () = 1 2

σ () = 1

: u1 E1 + u2 E2

σ () = 1

: u1 E2 + u2 E3

σ () = −1

: E3 + u1 E1 + u2 (E2 + E3 )

σ () = 0

: αE1 + u1 E2 + u2 E3

κ() = α ,

σ () = −1

: αE3 + u1 E1 + u2 E2

κ() = −α 2 ,

σ () = 1

2

: u1 E1 + u2 E2 + u3 E3 .

Here, α > 0 parametrizes families of class representatives, each different value corresponding to a distinct non-equivalent representative. Proof (sketch) For the sake of convenience, we will use ψ2 (θ ) and ψ3 (θ ) to denote the automorphisms ⎡ ⎤ ⎡ ⎤ cosh θ 0 sinh θ cos θ sin θ 0 ⎣ 0 1 0 ⎦ and ⎣ − sin θ cos θ 0 ⎦ sinh θ 0 cosh θ 0 0 1 respectively. Let be a (1, 1)-system with trace = A + 0 . (If σ () = 0, then κ() = 0.) Suppose that σ () = 0. Then, 0 = B for some B ∈ H0 . Therefore, there exists an automorphism ψ such that ψ · = a1 E1 + a2 E2 + E2 + E3 . We may assume A = −1 for some A ∈ ψ · . Thus, there exists an automorphism ψ such that A ψ · ψ · = E3 + sin θ E1 + cos θ E2 + E3 . Consequently, ψ3 (−θ ) · ψ · ψ · = 1(1,1) . Suppose that σ () = −1 and κ() = α 2 . Then, 0 = B for some B ∈ H−1 . Hence,


there exists an automorphism ψ such that ψ · = α sin θ E1 +α cos θ E2 +E3 . Therefore, (1,1) ψ3 (−θ ) · ψ · = αE2 + E3 = 2,α . Suppose that σ () = 1 and κ() = α 2 . Then, 0 = B for some B ∈ H1 . Hence, there exists an automorphism ψ such that (1,1) ψ · = α cosh θ E1 + α sinh θ E3 + E2 . Thus, ψ2 (−θ ) · ψ · ψ · = 3,α . Suppose that σ () = 0 and κ() = −α 2 . Then, it turns out that σ () = 1. Likewise, it follows (1,1) that there exists an automorphism ψ such that ψ · = 4,α . Let be a (2, 1)-affine subspace with trace = A+ 0 . Suppose that σ () = 0. Then, 0 = r cos θ E1 + r sin θ E2 , c1 E1 + c2 E2 + E3 . Hence, ψ3 (θ ) · 0 = E1 , c2 E2 + E3 with c2 = ±1. Then ψ = diag(sgn(c2 ), sgn(c2 ), 1) is an automorphism such that ψ · A = −1 for some A ∈ ψ3 (θ ) · = a3 E3 + E1 , E2 + E3 . We may assume that A ψ · ψ3 (θ ) · . Therefore, there exists an automorphism ψ such that ψ · ψ · ψ3 (θ ) · = E3 + B1 , B2 . Likewise, it follows that there exists an automorphism ψ such that ψ · B1 , B2 = E1 , E2 + E3 and ψ · E3 = E3 . So ψ · ψ · ψ · ψ3 (θ ) · = 1(2,1) . Suppose that σ () = 0 and κ() = α 2 . Now, κ(2,α ) = α 2 . Thus, there exists an automorphism ψ such that ψ ·C () = C (2,α ). Hence, as ψ ·C () = C (ψ ·), we get that ψ · and 2,α are both equal to the tangent plane of Hα2 at ψ · C () and are therefore identical. (2,1) . Suppose that σ () = 0 and κ() = −α 2 . A similar argument then So, ψ · = 2,α (2,1)

shows that is equivalent to 3,α . The (2, 0)-case follows by Proposition 2. The (3, 0) case is trivial. As the quantities σ () and κ() are invariant under the action of any automorphism, no two representative systems are equivalent. Proposition 14 Let A ∈ sl (2, R) (resp. A ∈ so (2, 1)). Then, A A < 0 if and only if t → exp(tA) is periodic. Proof Let A ∈ sl (2, R) and suppose that A A < 0. Then, there exists an automorphism ψ such that ψ · A = αE3 . We have $ exp(αtE3 ) =

cos αt2 − sin αt2 sin αt2

cos αt2

% .

Thus, t → exp(t ψ · A) is periodic. Let φ ∈ Aut (SL (2, R)) and T1 φ = ψ. Then φ(exp(t A)) = exp(t ψ · A) , and so t → exp(t A) is also periodic. Suppose that A A > 0 (resp. A A = 0). Then, there exists an automorphism ψ such that ψ · A = αE1 (resp. ψ · A = E1 + E3 ). We have $ exp(αtE1 ) =

αt

e2 0 αt 0 e− 2

$

% and

exp(t (E1 + E3 )) =

1 + 12 t − 12 t 1 2t

1 − 12 t

% .

Thus, neither t → exp(αtE1 ) nor t → exp(t (E1 + E3 )) is periodic. Hence, t → exp(tA) is not periodic. For A ∈ so (2, 1), a very similar argument holds. (Alternatively, the double covering q : SL (2, R) → SO (2, 1)0 can be used.) (1,1) Theorem 19 Among the above systems on SO (2, 1)0 (resp. SL (2, R)), only 3,α is not controllable.


(1,1) Proof Suppose that g(t) = gij (t) is a trajectory of 3,α in SO (2, 1)0 . Consider the Cauchy problem g(t) ˙ = g(t) (αE1 + uE2 ), g(0) = 1. In particular, we have g˙ 32 (t) = αg33 (t) > 0 and g32 (0) = 0. Thus, g32 (t) > 0 for t > 0 and so the states ⎤ ⎡ 1 0 0 ⎣ 0 cosh z sinh z ⎦ , z < 0 0 sinh z cosh z (1,1)

are not attainable (from identity). Hence, the system 3,α

is not controllable (on

(1,1) 3,α

SO (2, 1)0 ). Likewise, is not controllable on SL (2, R). (As SL (2, R) is a double cover of SO (2, 1)0 , if the system is controllable on SL (2, R) , then it is controllable on SO (2, 1)0 ; see, e.g. [6].) The homogeneous systems are controllable as they have full rank. Each of the remaining systems is controllable. Indeed, for each of these systems, there exists A ∈ such that (1,1) (1,1) A A < 0 (e.g. for 1 the element A = E3 suffices, whereas for 2,α , the element A = αE2 + 2αE3 suffices). Thus, by Proposition 14, t → exp(t A) is periodic, and so these systems are controllable.

Corollary A full-rank system on SO (2, 1)0 (resp. SL (2, R)) is controllable if and only if it is homogeneous or there exists A ∈ such that A A < 0. 4.2 Type g3.7 (The Orthogonal Groups) There are two connected Lie groups (up to isomorphism) with Lie algebra g3.7 . They are the group of rotations SO (3) and its simply connected double cover SU (2). We have SU (2) = g ∈ C2×2 : gg † = 1, det g = 1 + * i 1 2x 2 (iz + y) : x, y, z ∈ R su (2) = 1 (iz − y) − 2i x 2 SO (3) = g ∈ R3×3 : gg = 1, det g = 1 ⎧⎡ ⎫ ⎤ ⎨ 0 −z y ⎬ so (3) = ⎣ z 0 −x ⎦ : x, y, z ∈ R . ⎩ ⎭ −y x 0 For both Lie algebras, let (E1 , E2 , E3 ) denote the corresponding (natural) basis. With respect to these bases, Aut (su (2)) = SO (3) and Aut (so (3)) = SO (3). As all automorphisms are inner automorphisms, d Aut (SO (3)) = SO (3). (Likewise, d Aut (SU (2)) = SO (3).) 3so (3) is given by A • B = a1 b1 + a2 b2 + a3 b3 . (Here, A = 3The dot product • on a E and B = i i i=1 √ i=1 bi Ei .) The level sets Sα = {A ∈ so (3) : A • A = α} are spheres of radius α. The spheres Sα are preserved by automorphisms. Moreover, Aut (so (3)) acts transitively on each sphere Sα . The critical point C• () (at which an inhomogeneous affine subspace is tangent to a sphere Sα ; see [8]) is given by A•B B C• () = A − B •B B1 • B1 B1 • B2 −1 A • B1 C• () = A − B1 B2 . B1 • B2 B2 • B2 A • B2


Critical points behave well under the action of automorphisms, i.e. ψ · C• () = C• (ψ · ) for any automorphism ψ. (The critical point of is well defined as it is independent of parametrization.) The critical point C• () and invariant level sets Sα are likewise defined for su (2). Theorem 20 ([8]) Any full-rank system on SO (3) (resp. SU (2)) is equivalent to exactly one of the following systems: C• () • C• () = α 2

α(1,1) : αE1 + E2 (2,0)

: u1 E1 + u2 E2

α(2,1) (3,0)

: αE1 + u1 E2 + u2 E3

C• () • C• () = α 2

: u1 E1 + u2 E2 + u3 E3 .

Here, α > 0 parametrizes families of class representatives, each different value corresponding to a distinct non-equivalent representative. Remark 12 As SO (3) and SU (2) are compact, all these systems are controllable. Proof (sketch) Let be a (1, 1)-system with trace . There √ exists an automorphism ψ such that ψ · = α cos θ E1 + α sin θ E3 + E2 , where α = C• () • C• (). Hence, ⎡

⎤ cos θ 0 sin θ 1 0 ⎦ ψ =⎣ 0 − sin θ 0 cos θ

is an automorphism such that ψ · ψ · = α . Let be a (2, 1)-system with trace . We have C• () • C• () = α 2 for some α > 0. As C• (1,α ) • C• (1,α ) = α 2 , there exists an automorphism ψ such that ψ · C• () = C• (1,α ). Hence, ψ · and 1,α are both equal to the tangent plane of Sα2 at ψ · C• (), and are therefore identical. The (2, 0) case follows by Proposition 2; the (3, 0) case is trivial. As the quantity C• () • C• () is invariant under the action of any automorphism, no two representative systems are equivalent. (1,1)

5 Conclusion In this paper, we classified the full-rank left-invariant control affine systems evolving on all three-dimensional matrix Lie groups. For each group, a specific characterization of controllability was identified; see Table 2. Accordingly, a classification of all controllable systems has been obtained; see Tables 3, 4 and 5. A classification of full-rank systems under local detached feedback equivalence is identical to a classification under (global) detached feedback equivalence whenever every Lie algebra automorphism is the linearization of a group automorphism. For the threedimensional matrix Lie groups, this is predominantly the case. The exceptions are R2 × T, R × T2 , T3 , and Aff (R)0 × T. Given a Lie algebra g, it is natural to consider controllability first on the simply connected (universal covering) Lie group G. It is well known that controllability on G implies controllability on any locally isomorphic (connected) group G ∼ G/N. Moreover, =


controllability on G depends, essentially, just on the trace of the system and the Lie algeG/N, additional systems may be bra g. However, on a locally isomorphic group G ∼ = controllable (this is usually attributed to certain “geometric” properties of G, such as the existence of a periodic one-parameter subgroup). On R2 × T, R × T3 , T3 , Aff (R)0 × T, and SEn (2) all additional controllable systems admit a periodic trajectory. For SO (2, 1)0 (resp. SL (2, R)), controllability of systems is also attributed to either the system being homogeneous or the existence of a periodic trajectory. The classification of controllable systems has a natural extension to invariant optimal control problems. To each invariant optimal control problem (with quadratic cost), we associate a cost-extended system. Each cost-extended system corresponds to a family of invariant optimal control problems; by specification of the boundary data, the associated problem is uniquely determined. Cost-extended systems, and their equivalence, have been considered in [7, 11]. A classification of cost extended systems may be founded on a classification of invariant control affine systems. This is a topic for future research.

Appendix

Table 2 Characterization of controllability for full-rank systems Type

Group

Characterization

3g1

R3

= 3g1

R2 × T

∃A ∈ , t → exp(tA) is periodic

R × T2

∃A ∈ , t → exp(tA) is periodic

T3

All are controllable

Aff (R)0 × R

Lie ( 0 ) = g

Aff (R)0 × T

is homogeneous or

g2.1 ⊕ g1

∃A ∈ , t → exp(tA) is periodic g3.1

H3

g3.2 , g3.3 , g03.4 , ga3.4

Lie ( 0 ) = g Lie ( 0 ) = g

(2) SE

E3∗ ( 0 ) = {0}

SEn (2)


ga3.5

Ga3.5

E3∗ ( 0 ) = {0}

g3.6

SL (2, R), SO (2, 1)0

is homogeneous or

g03.5

∃A ∈ , A A < 0 g3.7

SU (2), SO (3)


Rory Biggs and Claudiu C. Remsing Table 3 Classification of controllable systems on (completely) solvable groups Type

Group G

d Aut (G)

3g1

R3

GL (3, R)

R2 × T

R × T2 T3

⎡

u x 0

Trace E1 , E2 , E3

⎤

⎢ ⎥ ⎣v y 0 ⎦ w z σ ⎡ ⎤ x 0 0 ⎢ ⎥ ⎣ y ⎦ GL (2, Z) z

Aff(R)0 × R

Aff(R)0 × T

g3.1

H3

G3.2

E2 + E1 , E2 + E3 E1 , E2 , E3

⎤

E1 , E2 + E3 αE2 + E3 , E1 αE2 + E1 , E2 + E3 E1 , E2 , E3

yw − vz x u

⎢ ⎣0 0 ⎡

g3.2

E1 , E2 + E3

1 0 0 ⎢ ⎥ ⎣x σ 0⎦ y 0 v ⎡

u x y

E1 , E2 , E3

E1 , E2 , E3

⎤ 1 0 0 ⎢ ⎥ ⎣x u 0⎦ y 0 v ⎡

αE3 + E1 , E2 + γ E3

αE1 + γ1 E1 + E2 , γ2 E1 + E3

GL (3, Z)

⎡ g2.1 ⊕ g1

αE3 + E1 , E2 E1 , E2 , E3

⎤

⎥ y v ⎦ z w

⎤

⎢ ⎥ ⎣0 u z ⎦ 0 0 1

E1 + E2 , E3 E1 , E2 , E3 E2 , E3 E1 + E2 , E3 E1 , E2 , E3

⎡ g3.3

G3.3

⎤ x y z ⎢ ⎥ ⎣u v w⎦ 0 0 1

E2 , E3

⎡ g03.4

SE (1, 1)

⎤ x y u ⎢ ⎥ ⎣ σy σ x v ⎦ 0 0 σ ⎡

ga3.4

Ga3.4

x y u

⎤

⎢ ⎥ ⎣y x v⎦ 0 0 1

E1 , E2 , E3

E2 , E3 E1 + E2 , E3 E1 , E2 , E3 E2 , E3 E1 + E2 , E3 E1 , E2 , E3

Control Systems on 3D Lie Groups: Equivalence and Controllability Table 4 Classification of controllable systems on solvable groups (cont.) Type

g03.5

Group G

(2) SE

d Aut (G)

Trace

⎡

E2 + E3

⎤ x y u ⎢ ⎥ ⎣−σy σ x v ⎦ 0

0 σ

E2 , E3 E1 + E2 , E3 E1 , E2 , E3 E2 + E3

⎡ SEn (2)

⎤

x y u ⎢ ⎥ ⎣−σy σ x v ⎦ 0

0 σ

αE3 + E2 E2 , E3 E1 + E2 , E3 αE3 + E1 , E2 E1 , E2 , E3

⎡ ga3.5

Ga3.5

x y u

⎤

⎢ ⎥ ⎣−y x v ⎦ 0 0 1

E2 + E3 E2 , E3 E1 + E2 , E3 E1 , E2 , E3

Table 5 Classification of controllable systems on semisimple groups Type

Group G

d Aut (G)

g3.6

SL (2, R)

MJ M = J

E3 + E2 + E3

SO (2, 1)0

J = diag (1, 1, −1)

αE2 + E3 αE3 + E2

det M = 1

Trace

E1 , E2 E2 , E3 E3 + E1 , E2 + E3 αE1 + E2 , E3 αE3 + E1 , E2 E1 , E2 , E3

g3.7

SU (2) SO (3)

MM = I I = diag(1, 1, 1) det M = 1

αE1 + E2 E1 , E2 αE1 + E2 , E3 E1 , E2 , E3


References 1. Adams RM, Biggs R, Remsing CC. Equivalence of control systems on the Euclidean group SE(2). Control Cybernet. 2012;41:513–24. 2. Adams RM, Biggs R, Remsing CC. On the equivalence of control systems on the orthogonal group SO(4). In: Karimi HR, editor. Recent researches in automatic control, systems science and communications. Athens: WSEAS; 2012, pp. 54–9. 3. Adams RM, Biggs R, Remsing CC. Control systems on the orthogonal group SO (4). Commun Math. 2013;21:107–28. 4. Agrachev A. A., Sachkov Y. L. Control theory from the geometric viewpoint. Berlin: Springer; 2004. 5. Barrett DI, Biggs R, Remsing CC. Affine subspaces of the Lie algebra se (1, 1). (Submitted). 6. Biggs R, Remsing CC. A category of control systems. An S¸tiint¸ Univ Ovidius Constant¸a. 2012;20:355– 68. 7. Biggs R, Remsing CC. On the equivalence of cost-extended control systems on Lie groups. In: Karimi HR, editor. Recent researches in automatic control, systems science and communications. Athens: WSEAS; 2012, pp. 60–5. 8. Biggs R, Remsing CC. Control affine systems on semisimple three-dimensional Lie groups. An S¸tiint¸ Univ AI Cuza Ias¸i Ser Mat. 2013;59:399–414. 9. Biggs R, Remsing CC. Control affine systems on solvable three-dimensional Lie groups. I Arch Math (Brno). 2013;49:187–97. 10. Biggs R, Remsing CC. Control affine systems on solvable three-dimensional Lie groups, II. Note Mat. 2013;33:19–31. 11. Biggs R, Remsing CC. Cost-extended control systems on Lie groups. Mediterr J Math 2014;11:193–215. 12. Biggs R, Remsing CC. On the equivalence of control systems on Lie groups. (Submitted). 13. Biggs R, Remsing CC. Equivalence of control systems on the pseudo-orthogonal group SO (2, 1)0 . (Submitted). 14. Bloch AM. Nonholonomic mechanics and control. New York: Springer; 2003. 15. Bonnard B, Jurdjevic V, Kupka I, Sallet G. Transitivity of families of invariant vector fields on the semidirect product of Lie groups. Trans Amer Math Soc. 1982;271:525–35. 16. Ha KY, Lee JB. Left invariant metrics and curvatures on simply connected three-dimensional Lie groups. Math Nachr. 2009;282:868–98. 17. Harvey A. Automorphisms of the Bianchi model Lie groups. J Math Phys. 1979;20:251–3. 18. Hilgert J, Neeb K-H. Structure and geometry of Lie groups. New York: Springer; 2012. 19. Jakubczyk B. Equivalence and invariants of nonlinear control systems. In: Sussmann HJ, editor. Nonlinear controllability and optimal control. New York: Marcel Dekker; 1990, pp. 177–218. 20. Jurdjevic V. Geometric control theory. Cambridge: Cambridge University Press; 1997. 21. Jurdjevic V. Optimal control on Lie groups and integrable Hamiltonian systems. Regul Chaotic Dyn. 2011;16:514–35. 22. Jurdjevic V, Monroy-Pérez F. Variational problems on Lie groups and their homogeneous spaces: elastic curves, tops, and constrained geodesic problems. In: Anzaldo-Meneses A, Bonnard B, Gauthier JP, Monroy-Pérez F, editors. Contemporary trends in nonlinear geometric control theory and its applications. Singapore: World Scientific; 2002, pp. 3–51. 23. Jurdjevic V, Sussmann HJ. Control systems on Lie groups. J Diff Equations. 1972;12:313–29. 24. Krasiński A, Behr CG, Schücking E, Estabrook FB, Wahlquist HD, Ellis GFR, Jantzen R, Kundt W. The Bianchi classification in the Schücking-Behr approach. Gen Relativ Gravit. 2003;35:475–89. 25. MacCallum MAH. On the classification of the real four-dimensional Lie algebras. In: Harvey A, editor. On Einstein’s path: essays in honour of E. Schücking. New York: Springer; 1999, pp. 299–317. 26. Mubarakzyanov GM. On solvable Lie algebras (Russian). Izv Vyssh Uchebn Zaved Mat. 1963;1963(1(32)):114–23. 27. Onishchik AL, Vinberg EB. Lie groups and Lie algebras III. New York: Springer; 1994. 28. Popovych RO, Boyco VM, Nesterenko MO, Lutfullin MW. Realizations of real low-dimensional Lie algebras. J Phys A Math Gen. 2003;36:7337–60. 29. Puta M, Birtea P, L˘azureanu C, Pop C, Tudoran R. Control, integrability and stability in some concrete mechanical problems on matrix Lie groups. Quad Sem Top Alg e Diff Univ di Roma La Sapienza; 1998. 30. Puta M. Optimal control problems on matrix Lie groups. In: Szenthe J, editor. New developments in differential geometry. Dordrecht: Kluwer; 1999, pp. 361–73. 31. Remsing CC. Optimal control and Hamilton-Poisson formalism. Int J Pure Appl Math. 2010;59:11–7. 32. Remsing CC. Control and integrability on SO (3). In: Ao SI, et al., editors. Proceedings WCE 2010, London, U.K. Hong Kong: IAENG; 2010, pp. 1705–10.

Control Systems on 3D Lie Groups: Equivalence and Controllability 33. Respondek W, Tall IA. Feedback equivalence of nonlinear control systems: a survey on formal approach. Chaos in automatic control, control eng. London: CRC (Taylor & Francis); 2006, pp. 137–262. 34. Respondek W, Zhitomirskii M. Feedback classification of nonlinear control systems on 3-manifolds. Math Control Signals Syst. 1995;8:299–333. 35. Sachkov YL. Control theory on Lie groups. J Math Sci. 2009;156:381–439. 36. Sachkov YL. Maxwell strata in the Euler elastic problem. J Dyn Control Syst. 2008;14:169–234. 37. Sachkov YL. Conjugate points in the Euler elastic problem. J Dyn Control Syst. 2008;14:409–39.