Weak condition for the existence of control sets with a nonempty interior for linear control systems on nilpotent groups

Abstract

In this paper, we show that for a linear control system on a nilpotent Lie group, the Lie algebra rank condition is enough to assure the existence of a control set with a nonempty interior, as soon as one impose a compactness assumption on the generalized kernel of the drift. Moreover, this control set is unique and contains the singularities of the drift in its closure.

1 Introduction

In the study of control systems, the control sets play an important role since they contain several dynamical properties of the systems, such as equilibrium points, recurrent points, periodic and bounded orbits, etc (see [6, Chapter 4]). In particular, exact controllability holds in the interior of such control sets.

On the other hand, due to the work of Jouan [8], any control-affine system with complete vector fields acting transitively on a connected manifold is equivalent to a linear control system on a Lie group or on a homogeneous space, as soon as the Lie algebra they generate is finite dimensional. Therefore, linear control systems appear as a classifying family for more general control-affine systems on connected manifolds, showing that the understanding of such systems is worth pursuing.

In this paper, we show that for linear control systems on nilpotent groups, if the generalized kernel of the drift of a linear control system is compact (see Definition 2.2), the Lie algebra rank condition is enough to assure the existence and uniqueness of a control set with a nonempty interior. This is a remarkable fact, since the Lie algebra rank condition is the weakest necessary condition one could ask to assure the existence of control set with a nonempty interior.

The paper is divided as follows: In Sect. 2, we introduce the basic concepts of linear vector fields and linear control systems. In this section, we also study a family of diffeomorphisms induced by a linear vector field and prove some related results. The understanding of such family proves to be central to the main results of the paper. In Sect. 3, we prove our main result, namely that a linear control system whose drift has a compact generalized kernel admits a unique control set with a nonempty interior. We start by analyzing the regular case. By using the diffeomorphisms induced by the drift of the system, we first show the existence of control sets with a nonempty interior. In sequence, we show that all of the previous control sets have nonempty intersections and, hence, coincide. For the general case, we consider a fibration between nilpotent groups, whose fiber is compact and given by the generalized kernel of the drift. Using general results, we are then able to show that the preimage of the control set obtained for the regular case is in fact a control set for the initial system, implying its existence and uniqueness. In Sect. 4, we provide two beautiful examples of non-regular systems on the Heisenberg group whose dynamics are drastically different. Both systems satisfy the Lie algebra rank condition; however, one of them is controllable, and the other has no control sets with a nonempty interior but only one-point control sets.

1.1 Notations

Let N be a connected Lie group with Lie algebra $\mathfrak {n}$. By $\exp :\mathfrak {n}\rightarrow N$, we denote the exponential map of N. For any element $x\in N$, we denote by $L_x$ and $R_x$ the left and right translations of N by x, respectively. The center of N is denote by Z(N). By $\textrm{Aut}(N)$ and $\textrm{Aut}(\mathfrak {n})$ we denote, respectively, the group of automorphisms of N and $\mathfrak {n}$. If $f:N\rightarrow N$ is a differentiable map, its differential at a point $x\in N$ is denoted by $(df)_x$. The set of fixed points of f is the set

$$\begin{aligned} {\text {fix}}(f):=\{x\in X, f(x)=x\}. \end{aligned}$$

If X is a complete vector field on N, with associated flow $\{\psi _S\}_{S\in \mathbb {R}}$, then

$$\begin{aligned} {\text {fix}}(\psi ):=\bigcap _{S\in \mathbb {R}}{\text {fix}}(\psi _S), \end{aligned}$$

is the set of singularities of X.

2 Background

In this section, we give some background needed to understand the problem at hand. For more on the subject, the reader can consult, for instance, Ayala and Tirao [5],Da Silva [7], Jouan [8], Jouan [9].

2.1 Linear vector fields

Let N be a connected nilpotent Lie group, and $\mathfrak {n}$ its associated nilpotent Lie algebra, which we identify with the set of right-invariant vector fields on N.

Definition 2.1

A linear vector field $\mathcal {X}$ on N is a complete vector field whose associated flow $\{\varphi _S\}_{S\in \mathbb {R}}$ is a 1-parameter group of automorphisms, that is, $\varphi _S\in \textrm{Aut}(N)$ for all $S\in \mathbb {R}$.

Associated with any linear vector field $\mathcal {X}$, there is a derivation A of $\mathfrak {n}$ defined by $A(Y):=[\mathcal {X}, Y]$. The relation between A and $\mathcal {X}$ is given by the equations

$$\begin{aligned} (d\varphi _{S})_{e}=\textrm{e}^{SA}\hspace{.5cm} \text{ and } \hspace{.5cm} \varphi _S(\exp X)=\exp (\textrm{e}^{SA}X), \end{aligned}$$

(1)

where $\textrm{e}^{tA}$ is the matrix exponential of A.

Definition 2.2

Let $\mathcal {X}$ be a linear vector field on N with associated derivation A. The generalized kernel of $\mathcal {X}$ is the connected subgroup $N_0$ whose associated Lie algebra is given by

$$\begin{aligned} \mathfrak {n}_0:=\{X\in \mathfrak {n}; \;A^kX=0, \text{ for } \text{ some } k\in \mathbb {N}\}. \end{aligned}$$

We say that $\mathcal {X}$ is regular if $N_0=\{e\}$ or, equivalently, if $\det A\ne 0$.

The next result associates a diffeomorphism with any automorphism with only the origin as a fixed point. Since it is a slightly generalization of [1, Proposition 2.7], we omit its proof.

Proposition 2.3

If $\psi \in \textrm{Aut}(N)$ is such that ${\text {fix}}((d\psi )_e)=\{0\}$, the map

$$\begin{aligned} f_{\psi }: N\rightarrow N, \hspace{1cm} f_{\psi }(x):=x\psi (x^{-1}), \end{aligned}$$

is a surjective local diffeomorphism. Moreover, $f_{\psi }$ is injective if and only if ${\text {fix}}(\psi )=\{e\}$.

In what follows, we specialize on the previous result for the automorphisms $\varphi _S$ associated with a linear vector field $\mathcal {X}$.

Proposition 2.4

Let $\mathcal {X}$ be a linear vector field on N, and consider the maps $f_S:=f_{\varphi _S}$, where $\{\varphi _S\}_{S\in \mathbb {R}}$ is the flow of $\mathcal {X}$. It holds

• 1. If A is the derivation associated with $\mathcal {X}$, then

  $$\begin{aligned} {\text {fix}}(\varphi )=\exp \left( \ker A\right) . \end{aligned}$$

  Moreover, if $N_0$ is a compact subgroup, then $N_0={\text {fix}}(\varphi )\subset Z(N)$.

• 2. If $\mathcal {X}$ is regular, then N is simply connected;

• 3. $f_{S_0+S_1}(x)=f_{S_0}(x)\varphi _{S_0}(f_{S_1}(x))=f_{S_1}(x)\varphi _{S_1}(f_{S_0}(x))$, for any $S_1, S_2\in \mathbb {R}$, $x\in N$;

• 4. There exists a discrete subset $\mathcal {Z}$ of $\mathbb {R}$ such that $f_S$ is a diffeomorphism for any $S\in \mathbb {R}{\setminus }\mathcal {Z}$

Proof

1. The inclusion $\exp \ker A\subset {\text {fix}}(\varphi )$ is straightforward and hence we will omit its proof.

Let $x\in {\text {fix}}(\varphi )$. Since the exponential is a covering map, there exists $X\in \mathfrak {n}$ such that $x=\exp X$. In particular,

$$\begin{aligned}{} & {} \forall S\in \mathbb {R}, \hspace{.5cm}x=\exp (X)=\varphi _S(\exp X)=\exp (\textrm{e}^{SA}X)\hspace{.5cm}\\{} & {} \quad \implies \textrm{e}^{SA}X\in \exp ^{-1}(\exp X)=\exp ^{-1}(x). \end{aligned}$$

However, the fact that $\exp ^{-1}(x)$ is discrete for covering maps, allows us to conclude that

$$\begin{aligned} \forall S\in \mathbb {R}, \hspace{.5cm} \textrm{e}^{SA}X=X\hspace{.5cm}\iff \hspace{.5cm}X\in \ker A, \end{aligned}$$

showing that ${\text {fix}}(\varphi )\subset \exp \ker A$, and proving the first assertion. For the second assertion, by [11, Theorem 1.6], any compact Lie subgroup of N is central, that is, $N_0\subset Z(N)$. Since, up to isomorphism, any compact, connected and abelian Lie group is a torus, we conclude that $\textrm{Aut}(N_0)$ is discrete. In particular, $\{\varphi _S|_{N_0}\}_{S\in \mathbb {R}}\subset \textrm{Aut}(N_0)$ implies

$$\begin{aligned} \forall S\in \mathbb {R},\hspace{.5cm} \varphi _S|_{N_0}=I_{N_0}\hspace{.5cm}\implies \hspace{.5cm}N_0\subset {\text {fix}}(\varphi ). \end{aligned}$$

Since the opposite inclusion is direct, the result follows.

2. Let $\widetilde{N}$ be the simply connected covering of N and $\pi :\widetilde{N}\rightarrow N$ be the canonical projection between them. Consider the lifting $\widetilde{\mathcal {X}}$ of $\mathcal {X}$ to $\widetilde{N}$, whose associated derivation is also A (see [5, Theorem 2.2]). In particular, the respective flows satisfy

$$\begin{aligned} \forall S\in \mathbb {R}, \hspace{.5cm}\pi \circ \widetilde{\varphi }_S=\varphi _S\circ \pi , \end{aligned}$$

implying that $\widetilde{\varphi }_S(\ker \pi )=\ker \pi$ for all $S\in \mathbb {R}$. Since $\ker \pi$ is a discrete subgroup, we conclude that

$$\begin{aligned} \ker \pi \subset {\text {fix}}(\widetilde{\varphi }). \end{aligned}$$

On the other hand, the fact that $\widetilde{N}$ is a simply connected nilpotent Lie group implies that the exponential map is a diffeomorphism, and hence, any $x\in \widetilde{N}$ can be uniquely written as $x=\textrm{e}^X$ for some $X\in \mathfrak {n}$. As a consequence, we get using equation (1) that

$$\begin{aligned} \forall S\in \mathbb {R}, \hspace{.3cm}\widetilde{\varphi }_S(x)=x\hspace{.3cm}\iff \hspace{.3cm}\forall S\in \mathbb {R}, \hspace{.3cm}\textrm{e}^{S A}X=X, \hspace{.3cm}\iff \hspace{.3cm} X\in \ker A, \end{aligned}$$

and since $\det A\ne 0$, we conclude that $\ker \pi =\{e\}$, showing that $\pi$ is a diffeomorphism and hence N is simply connected.

3. It follows directly from the definition of $f_S$. In fact,

$$\begin{aligned} f_{S_0+S_1}(x)=x\varphi _{S_0+S_1}(x^{-1})=x\varphi _{S_0}(x^{-1})\varphi _{S_0}(x)\varphi _{S_0}(\varphi _{S_1}(x^{-1}))=f_{S_0}(x)\varphi _{S_0}(f_{S_1}(x)). \end{aligned}$$

4. Let us consider $\pm i\mu _j, j=1, \ldots , m$ be the pure imaginary eigenvalues of A and define the discrete subset of $\mathbb {R}$ given by

$$\begin{aligned} \mathcal {Z}:=\left\{ \frac{2k\pi }{\mu _j}, j=1, \ldots , m, k\in \mathbb {Z}\right\} , \end{aligned}$$

where $\mathcal {Z}:=\{0\}$ if A has no pure imaginary eigenvalue. By Proposition 2.3, $f_S$ is a diffeomorphism as soon as ${\text {fix}}(\textrm{e}^{S A})=\{0\}$ and ${\text {fix}}(\varphi _S)=\{e\}$. Moreover, since N is simply connected, relation (1) implies that

$$\begin{aligned} \varphi _S(x)=x\hspace{.5cm}\iff \hspace{.5cm}\textrm{e}^{SA}X=X,\hspace{.5cm} \text{ with } \hspace{.5cm} x=\textrm{e}^X, \end{aligned}$$

and hence, $f_S$ is a diffeomorphism as soon as $\det (I_{\mathfrak {n}}-\textrm{e}^{SA})\ne 0$ or equivalently, if $S\in \mathbb {R}{\setminus }\mathcal {Z},$ concluding the proof. $\square$

2.2 Linear control systems

Let N be a connected Lie group with Lie algebra $\mathfrak {n}$ identified with the set of right-invariant vector fields on N.

A linear control system (abrev. LCS) on N is defined by the family of ODEs,

$$\begin{aligned} \dot{x}(t)=\mathcal {X}(x(t))+\sum _{j=1}^mu_j(t)Y^j(x(t)), \end{aligned}$$

($\Sigma _N$)

where the drift $\mathcal {X}$ is a linear vector field, $Y^1, \ldots , Y^m\in \mathfrak {n}$ and $\textbf{u}=(u_1, \ldots , u_m)\in \mathcal {U}$. Here, $\mathcal {U}$ is the set of admissible control functions given by

$$\begin{aligned} \mathcal {U}:=\{\textbf{u}:\mathbb {R}\rightarrow \mathbb {R}^m; \;\textbf{u}\; \text{ is } \text{ a } \text{ piecewise } \text{ constant } \text{ function } \text{ with } u(\mathbb {R})\subset \Omega \}, \end{aligned}$$

where $\Omega \subset \mathbb {R}^m$ is a compact and convex subset containing the origin in its interior. We say that the LCS is regular if its drift is regular.

For any $x\in N$ and $\textbf{u}\in \mathcal {U}$, the solution $S\mapsto \phi (S, x, \textbf{u})$ of $\Sigma _N$ is complete and satisfies

$$\begin{aligned} \phi _{S, \textbf{u}}\circ R_x=R_{\varphi _{S}(x)}\circ \phi _{S, \textbf{u}}, \hspace{.5cm} \text{ for } \text{ any } \hspace{.5cm}S\in \mathbb {R}, x\in N, \end{aligned}$$

(2)

where $\phi _{S, \textbf{u}}(x):=\phi (S, x, \textbf{u})$, for any $x\in N$, and $\{\varphi _S\}_{S\in \mathbb {R}}$ is the flow of $\mathcal {X}$.

For any $x\in N$ and $S>0$, we define respectively

$$\begin{aligned} \mathcal {O}^{+}_{S}(x):=\{\phi (S,x, \textbf{u}), \textbf{u}\in \mathcal {U}\}\hspace{.5cm} \text{ and } \hspace{.5cm}\mathcal {O}^{+}(x):=\bigcup _{S>0}\mathcal {O}^{+}_{S}(x), \end{aligned}$$

as the set of points reachable from x in time $S>0$ and the reachable set of x. We say that $\Sigma _{N}$ satisfy the Lie algebra rank condition (abrev. LARC) if $\mathfrak {g}$ is the smallest $\mathcal {D}$-invariant Lie subalgebra containing $Y^1, \ldots , Y^m$. In particular, if the LARC is satisfied, the sets

$$\begin{aligned} \mathcal {O}^{+}_{\le S}(x):=\bigcup _{0<s\le S}\mathcal {O}^{+}_{s}(x), \hspace{.5cm}S>0, \end{aligned}$$

have a nonempty interior. Moreover, ${\text {int}}\mathcal {O}^{+}_{\le S}(x)$ (resp. ${\text {int}}\mathcal {O}^{+}(x)$) is dense in $\mathcal {O}^{+}_{\le S}(x)$ (resp. $\mathcal {O}^{+}(x)$) for all $S>0$, $x\in N$.

The next proposition, [9, Proposition 2], states the main properties of the reachable sets of a LCS.

Proposition 2.5

It holds:

• 1. $\mathcal {O}^{+}_{\le S}(e)=\mathcal {O}^{+}_{S}(e)$ for all $S>0$;

• 2. $\mathcal {O}^{+}_{S_1}(e)\subset \mathcal {O}^{+}_{S_2}(e)$ for all $0<S_1\le S_2$;

• 3. $\mathcal {O}^{+}_{S}(x)=\mathcal {O}^{+}_{S}(e)\varphi _S(x)$ for all $S>0$ and $x\in N$;

• 4. $\mathcal {O}^{+}_{S_1+S_2}(e)=\mathcal {O}^{+}_{S_1}(e)\varphi _{S_1}\left( \mathcal {O}^{+}_{S_2}(e)\right) =\mathcal {O}^{+}_{S_2}(e)\varphi _{S_2}\left( \mathcal {O}^{+}_{S_1}(e)\right)$, for all $S_1, S_2>0$.

Remark 2.6

Under the LARC, [10, Chapter 3, Theorem 2] implies that

$$\begin{aligned} \textrm{cl}\left( \mathcal {O}^+_{\le S}(e)\right) =\textrm{cl}\left( \textrm{int}\left( \mathcal {O}^+_{\le S}(e)\right) \right) , \hspace{.5cm}\forall S>0. \end{aligned}$$

Therefore, the previous proposition and the fact that translations are diffeomorphisms imply that

$$\begin{aligned} {\text {cl}}\left( {\text {int}}\mathcal {O}^{+}_{S}(x)\right){} & {} ={\text {cl}}\left( {\text {int}}\left( \mathcal {O}^{+}_{S}(e)\varphi _S(x)\right) \right) ={\text {cl}}\left( {\text {int}}\left( \mathcal {O}^{+}_{S}(e)\right) \right) \varphi _S(x)\\{} & {} =\textrm{cl}\left( \textrm{int}\left( \mathcal {O}^+_{\le S}(e)\right) \right) \varphi _S(x)=\textrm{cl}\left( \mathcal {O}^+_{\le S}(e)\right) \varphi _S(x) \\{} & {} =\textrm{cl}\left( \mathcal {O}^+_{S}(e)\right) \varphi _S(x)=\textrm{cl}\left( \mathcal {O}^+_{S}(e)\varphi _S(x)\right) =\textrm{cl}\left( \mathcal {O}^+_{S}(x)\right) , \end{aligned}$$

that is,

$$\begin{aligned} {\text {cl}}\left( {\text {int}}\mathcal {O}^{+}_{S}(x)\right) ={\text {cl}}\left( \mathcal {O}^{+}_{S}(x)\right) , \hspace{.5cm}\forall S>0, x\in N. \end{aligned}$$

Next, we define the concept of control sets.

Definition 2.7

A subset D of N is called a control set of the system $\Sigma _N$ if it satisfies the following properties:

• (i) (Weak invariance) For every $x\in N$, there exists a $\textbf{u}\in \mathcal {U}$ such that $\phi (\mathbb {R}^+,x,\textbf{u})\subset D$;

• (ii) (Approximate controllability) $D\subset \textrm{cl}\left( \mathcal {O}^{+}(x)\right)$ for every $x\in D$;

• (iii) (Maximality) D is maximal, with relation to set inclusion, satisfying properties (i) and (ii).

In particular, when the whole state space N is a control set, $\Sigma _N$ is said to be controllable.

The next technical lemma gives us a condition for a point to belong to the interior of a control set.

Lemma 2.8

Let $\Sigma _N$ be a LCS on N satisfying the LARC and $x_0\in N$. If $x_0\in {\text {int}}\mathcal {O}^+(x_0)$, there exists a control set D such that $x_0\in {\text {int}}D$.

Proof

Let us denote by $\mathcal {O}^-(x_0)$ the set of points reachable from $x_0$ for the reversed-time system. Since $\Sigma _N$ satisfies LARC, ${\text {int}}\mathcal {O}^-(x_0)$ is dense in $\mathcal {O}^-(x_0)$ implying that

$$\begin{aligned} D_0:={\text {int}}\mathcal {O}^+(x_0)\cap {\text {int}}\mathcal {O}^-(x_0), \end{aligned}$$

is a nonempty open subset containing $x_0$. By definition, for any point $y\in D_0$, there exist times $S_1, S_2>0$ and control functions $\textbf{u}_1, \textbf{u}_2\in \mathcal {U}$ such that

$$\begin{aligned} \phi (S_1, x, \textbf{u}_1)=y\hspace{.5cm} \text{ and } \hspace{.5cm}\phi (S_2, y, \textbf{u}_2)=x. \end{aligned}$$

Therefore, conditions (i) and (ii) of Definition 2.7 are satisfied and, by [6, Proposition 3.2.5], there exists a control set D such that $D_0\subset D$, concluding the proof. $\square$

Remark 2.9

It is worth mentioning that several results for control sets of LCSs were obtained in [4] under the (much stronger) condition of local controllability around the identity element of the group. Such a condition, although sufficient, is not necessary for the existence of a control set with a nonempty interior. In fact, examples of LCSs with the identity element on the boundary of a control set with a nonempty interior can be found in [3].

Let us finish this section with some comments on conjugations of linear systems.

Suppose $\widehat{N}$ is another nilpotent Lie group, and

$$\begin{aligned}{} & {} \dot{z}(s) = \widehat{\mathcal {X}}(z(s)) + \sum _{j=1}^{m}u_j(s)\widehat{Y}^j(z(s)), \hspace{.5cm} \textbf{u} = (u_1,\ldots , u_m)\in \mathcal {U}.{} & {} \hspace{1cm}\left( \Sigma _{\widehat{N}}\right) \end{aligned}$$

a LCS on $\widehat{N}$. If $\psi :N\rightarrow \widehat{N}$ is a surjective homomorphism, we say that $\Sigma _N$ and $\Sigma _{\widehat{N}}$ are $\psi$-conjugated if their respective vector fields are $\psi$-conjugated, that is,

$$\begin{aligned} \psi _*\circ \mathcal {X}=\widehat{\mathcal {X}}\circ \psi \hspace{.5cm} \text{ and } \hspace{.5cm}\;\;\;\psi _*\circ Y^j=\widehat{Y}^j\circ \psi , \hspace{.5cm} j=1,\ldots m. \end{aligned}$$

If such $\psi$ exists, we say that $\Sigma _N$ and $\Sigma _{\widehat{N}}$ are $\psi$-conjugated. If $\psi$ is an isomorphism, $\Sigma _N$ and $\Sigma _{\widehat{N}}$ are said to be equivalent. The next result, whose prove we omit, relates control sets of conjugated systems.

Proposition 2.10

Let $\Sigma _N$ and $\Sigma _{\widehat{N}}$ be $\psi$-conjugated LCSs satisfying the LARC. It holds:

• 1. If D is a control set of $\Sigma _N$, there exists a control set $\widehat{D}$ of $\Sigma _{\widehat{N}}$ such that $\psi (D)\subset \widehat{D}$;

• 2. If for some $y_0\in {\text {int}}\widehat{D}$ it holds that $\psi ^{-1}(y_0)\subset {\text {int}}D$, then $D=\psi ^{-1}(\widehat{D})$.

Proof

1. Since $\Sigma _N$ and $\Sigma _{\widehat{N}}$ are $\psi$-conjugated, their solutions satisfy

$$\begin{aligned} \psi (\phi (t, x, \textbf{u}))=\widehat{\phi }(t, \psi (x), \textbf{u}), \hspace{.5cm}\forall t\in \mathbb {R}, \textbf{u}\in \mathcal {U}. \end{aligned}$$

In particular, if D is a control set of $\Sigma _N$, the image $\psi (D)$ satisfy properties 1 and 2 in Definition 2.7, implying the existence of a control set $\widehat{D}$ of $\Sigma _{\widehat{N}}$ such that $\psi (D)\subset \widehat{D}$, showing item 1.

2. Let $y_0\in {\text {int}}\widehat{D}$ such that $\psi ^{-1}(y_0)\subset {\text {int}}D$. Since both systems satisfy LARC, exact controllability holds in ${\text {int}}D$ and in ${\text {int}}\widehat{D}$. Therefore, by the conjugation property, exact controllability holds also in $\psi ^{-1}({\text {int}}\widehat{D})$, which implies that $\psi ^{-1}({\text {int}}\widehat{D})\subset {\text {int}}D$. Continuity, together with item 1 implies the result. $\square$

Remark 2.11

Let us assume that H is a connected, closed, normal of N. If N is a normal subgroup, the homogeneous space $N':=N/H$ is a connected nilpotent Lie group whose canonical projection $\pi :N\rightarrow N'$ is a homomorphism. Following [8, Proposition 4], if $\mathcal {X}$ is a linear vector field on N and H is invariant by the flow of $\mathcal {X}$, the map $\mathcal {X}'$ defined by the relation

$$\begin{aligned} \pi _*\circ \mathcal {X}=\mathcal {X}'\circ \pi , \end{aligned}$$

is a well-defined linear vector field on $N'$. Consequently, any LCS $\Sigma _N$ on N induces a LCS $\Sigma _{N'}$ on $N'$ that is $\pi$-conjugated to $\Sigma _{N}$. Furthermore, if A and $A'$ are the associated derivation of $\mathcal {X}$ and $\mathcal {X}'$, respectively, the conjugation property implies that

$$\begin{aligned} (d\pi )_{e}\circ A=A'\circ (d\pi )_{e}. \end{aligned}$$

In particular, if $\mathfrak {h}$ is the Lie algebra of H, the previous relation implies that

$$\begin{aligned} (d\pi )_{e}X\in \ker A'\hspace{.5cm}\iff \hspace{.5cm} AX\in \ker (d\pi )_e=\mathfrak {h}. \end{aligned}$$

(3)

3 Control sets of LCSs

In this section, we show that the LARC is enough to assure the existence of control sets with nonempty interiors for LCSs whose associated drift has compact set of singularities. Moreover, we show that such a control set is unique and contains the identity element in its boundary.

We start with a technical lemma which will help us ahead.

Lemma 3.1

Let $\gamma :[0, 1]\rightarrow N$ be a continuous path and assume that for any $t\in [0, 1]$, there exists a control set $D_{\gamma (t)}$ of $\Sigma _N$ such that $\gamma (t)\in {\text {int}}D_{\gamma (t)}$. Then, $D_{\gamma (t)}=D_{\gamma (0)}$ for all $t\in [0, 1]$.

Proof

For any $t\in [0, 1]$, there exists, by continuity, $\varepsilon >0$ such that $\gamma \left( [t, t+\varepsilon )\right) \subset {\text {int}}D_{\gamma (t)}$. Consequently, if

$$\begin{aligned} T_0:=\max \{T\in [0, 1]; \;D_{\gamma (t)}=D_{\gamma (0)}, \forall t\in [0, T]\}, \end{aligned}$$

then necessarily $T_0=1$, which shows the result. $\square$

3.1 The regular case

In this subsection, we prove the existence of a unique control set with nonempty interior for regular systems.

Theorem 3.2

A regular LCS on a nilpotent connected Lie group N satisfying the LARC admits a unique control set with a nonempty interior.

Proof

Let $\Sigma _N$ be a regular LCS on N with drift $\mathcal {X}$. Denote as previously by $\{\varphi _S\}_{S\in \mathbb {R}}$ the flow of $\mathcal {X}$. By Proposition 2.3, there exists an open and dense set $I:=\mathbb {R}{\setminus } \mathcal {Z}$ such that

$$\begin{aligned} \forall S\in I, \hspace{.5cm }x\in N\mapsto f_S(x)=x\varphi _S(x^{-1})\hspace{.5cm} \text{ is } \text{ a } \text{ diffeomorphism. } \end{aligned}$$

The rest of the proof is divided into the next four steps.

Step 1: For any $S\in I$ and $x\in f_S^{-1}\left( {\text {int}}\mathcal {O}_S^+(e)\right)$, there exists a control set $D_{S, x}$ satisfying $x\in {\text {int}}D_{S, x}$.

By the LARC, it holds that

$$\begin{aligned} \forall S>0, \hspace{1cm} {\text {int}}\mathcal {O}^+_{S}(e)\ne \emptyset . \end{aligned}$$

Therefore, for all $S\in I$ and $y\in {\text {int}}\mathcal {O}^+_{S}(e)$, there exists $x\in N$ such that $f_{S}(x)=y$. Writing $y=\phi (S, e, \textbf{u})$ we get

$$\begin{aligned}{} & {} x\varphi _{S}(x^{-1})=f_{S}(x)=y=\phi (S, e, \textbf{u})\hspace{.5cm}\\{} & {} \quad \iff x=\phi (S, e, \textbf{u})\varphi _{S}(x)\in {\text {int}}\mathcal {O}^+_{S}(e)\varphi _{S}(x)={\text {int}}\mathcal {O}^+_{S}(x), \end{aligned}$$

showing, in particular, that $x\in {\text {int}}\mathcal {O}^+(x)$. By Lemma 2.8, there exist a control set $D_{S, x}$ satisfying $x\in {\text {int}}D_{S, x}$, showing step 1.

Step 2: For any $S\in I$, there exists a control set $D_S$ satisfying

$$\begin{aligned} f_S^{-1}\left( {\text {int}}\mathcal {O}_S^+(e)\right) \subset {\text {int}}D_S. \end{aligned}$$

By the previous step, it is enough to show that the control sets $D_{S, x}$ are independent of $x\in f_S^{-1}\left( {\text {int}}\mathcal {O}_S^+(e)\right)$.

However, the fact that ${\text {int}}\mathcal {O}_S^+(e)$ is connected ([10, Chapter 3, Theorem 3]) and that, for any $S\in I$, the map $f_S$ is a diffeomorphism, imply that the set $f_S^{-1}\left( {\text {int}}\mathcal {O}_S^+(e)\right)$ is open and connected, and hence path connected. Consequently, for any $x, y\in f_S^{-1}\left( {\text {int}}\mathcal {O}_S^+(e)\right)$, there exists a continuous path $\gamma :[0, 1]\rightarrow N$ such that $\gamma (0)=x$, $\gamma (1)=y$ and $\gamma (t)\in f_{S}^{-1}\left( {\text {int}}\mathcal {O}_S^+(e)\right)$ for any $t\in [0, 1]$. By Step 1, $\gamma (t)\in {\text {int}}D_{S, \gamma (t)}$ which by Lemma 3.1 implies necessarily $D_{S, x}=D_{S, y}$, and hence

$$\begin{aligned} D_S=D_{S, x}, \hspace{.5cm}\forall x\in f_S^{-1}\left( {\text {int}}\mathcal {O}_S^+(e)\right) , \end{aligned}$$

showing the step 2.

Step 3: There exists a control set D such that

$$\begin{aligned} \forall S\in I, \hspace{1cm}f_S^{-1}\left( {\text {int}}\mathcal {O}_S^+(e)\right) \subset {\text {int}}D. \end{aligned}$$

By the previous step, we only have to show that $D_{S_0}=D_{S_1}$ for any $S_0, S_1\in I$. Assume first that $[S_0, S_1]\subset I$ and let $x\in {\text {int}}\mathcal {O}_{S_0}^+(e)$. Define the curve

$$\begin{aligned} \gamma :[0, 1]\rightarrow N, \hspace{1cm} \gamma (t):=f^{-1}_{tS_1+(1-t)S_0}(x). \end{aligned}$$

Since for all $t\in (0, 1)$ we have that $tS_1+(1-t)S_0\in I$, we get that

$$\begin{aligned} \gamma (t){} & {} =f_{tS_1+(1-t)S_0}^{-1}(x)\in f_{tS_1+(1-t)S_0}^{-1}\left( {\text {int}}\mathcal {O}_{S_0}^+(e)\right) \subset f_{tS_1+(1-t)S_0}^{-1}\\{} & {} \quad \left( {\text {int}}\mathcal {O}_{tS_1+(1-t)S_0}^+(e)\right) \subset {\text {int}}D_{tS_1+(1-t)S_0}, \end{aligned}$$

where for the first inclusion we used Proposition 2.5. By Lemma 3.1, we conclude that $D_{S_0}=D_{S_1}$, showing that control sets $D_S$ depend only on the connected components of I.

Let $(0, \varepsilon )$ be a connected component of I, and denote by D the control set satisfying

$$\begin{aligned} \forall S\in (0, \varepsilon ), \hspace{1cm} f_S^{-1}\left( {\text {int}}\mathcal {O}_S^+(e)\right) \subset D. \end{aligned}$$

Take an arbitrary $S\in I$ and choose $n\in \mathbb {N}$ large enough, such that $S<n\varepsilon$, and consider

$$\begin{aligned} x\in f_{\frac{S}{n}}^{-1}\left( {\text {int}}\mathcal {O}^+_{\frac{S}{n}}(e)\right) \subset {\text {int}}D. \end{aligned}$$

Using Proposition 2.4, we get

$$\begin{aligned}{} & {} f_{2\frac{S}{n}}(x)=f_{\frac{S}{n}}(x)\varphi _{\frac{S}{n}}\left( f_{\frac{S}{n}}(x)\right) \in {\text {int}}\mathcal {O}^+_{\frac{S}{n}}(e)\varphi _{\frac{S}{n}}\left( {\text {int}}\mathcal {O}^+_{\frac{S}{n}}(e)\right) ={\text {int}}\mathcal {O}^+_{2\frac{S}{n}}(e), \\{} & {} f_{3\frac{S}{n}}(x)=f_{\frac{S}{n}}(x)\varphi _{\frac{S}{n}}\left( f_{2\frac{S}{n}}(x)\right) \in {\text {int}}\mathcal {O}^+_{\frac{S}{n}}(e)\varphi _{\frac{S}{n}}\left( {\text {int}}\mathcal {O}^+_{2\frac{S}{n}}(e)\right) ={\text {int}}\mathcal {O}^+_{3\frac{S}{n}}(e), \end{aligned}$$

and inductively,

$$\begin{aligned} f_S(x)= & {} f_{n\frac{S}{n}}(x)=f_{\frac{S}{n}}(x)\varphi _{\frac{S}{n}}\left( f_{(n-1)\frac{S}{n}}(x)\right) \in {\text {int}}\mathcal {O}^+_{\frac{S}{n}}(e)\varphi _{\frac{S}{n}} \left( {\text {int}}\mathcal {O}^+_{(n-1)\frac{S}{n}}(e)\right) \\= & {} {\text {int}}\mathcal {O}^+_{n\frac{S}{n}}(e)={\text {int}}\mathcal {O}^+_S(e). \end{aligned}$$

Consequently,

$$\begin{aligned} x\in f_{S}^{-1}\left( {\text {int}}\mathcal {O}^+_S(e)\right) \subset {\text {int}}D_S, \end{aligned}$$

which implies $D_{S}=D$ for any $S\in I$, showing step 3.

Step 4: D is the only control set of $\Sigma _N$.

Since control sets are either disjoint or coincide (see [6, Chapter 3]), it is enough to show that any arbitrary control set $D'$ with a nonempty interior satisfies

$$\begin{aligned} {\text {int}}D'\cap f_{S}^{-1}\left( {\text {int}}\mathcal {O}^+_S(e)\right) \ne \emptyset , \hspace{.5cm} \text{ for } \text{ some } \hspace{.5cm}S\in I. \end{aligned}$$

Let us consider $x\in {\text {int}}D'$ and $S_0>0$. Since ${\text {int}}\mathcal {O}^+_{S_0}(x)$ is dense in $\mathcal {O}^+_{S_0}(x)$ (see Remark 2.6), we get

$$\begin{aligned} {\text {int}}\mathcal {O}^+_{S_0}(x)\cap {\text {int}}D'\ne \emptyset . \end{aligned}$$

For y in this intersection, there exists, by exact controllability, $S_1>0$ and $\textbf{u}\in \mathcal {U}$ such that $\phi (S_1, y, \textbf{u})=x$ and hence

$$\begin{aligned} x=\phi (S_1, y, \textbf{u})\in \phi _{S_1, \textbf{u}}({\text {int}}\mathcal {O}^+_{S_0}(x))\subset {\text {int}}\mathcal {O}^+_{S_0+S_1}(x)={\text {int}}\mathcal {O}^+_{S_0+S_1}(e)\varphi _{S_0+S_1}(x), \end{aligned}$$

implying that

$$\begin{aligned} x\varphi _{S_0+S_1}(x^{-1})\in {\text {int}}\mathcal {O}^+_{S_0+S_1}(e). \end{aligned}$$

On the other hand, the fact that I is dense in $(0, +\infty )$ assures, by continuity, the existence of $S\in I$, with $S>S_0+S_1$, such that

$$\begin{aligned} x\varphi _S(x^{-1})\in {\text {int}}\mathcal {O}^+_{S_0+S_1}(e)\subset {\text {int}}\mathcal {O}^+_S(e), \end{aligned}$$

and so,

$$\begin{aligned} f_S(x)=x\varphi _S(x^{-1})\in {\text {int}}\mathcal {O}^+_S(e)\hspace{.5cm}\iff \hspace{.5cm} x\in f_S^{-1}({\text {int}}\mathcal {O}^+_S(e)), \end{aligned}$$

which implies step 4 and concludes the proof of the theorem. $\square$

As a direct corollary, we get the following.

Corollary 3.3

Under the assumptions of the previous theorem, the only control set D of a LCS $\Sigma _N$ on N satisfies $e\in \textrm{cl}(D)$.

Proof

In fact, since $f_S(e)=e$, for any $S\in I$, we get that

$$\begin{aligned} e{} & {} =f_S^{-1}(e)\in f_S^{-1}\left( \textrm{cl}\left( \mathcal {O}_S^+(e)\right) \right) =f_S^{-1}\left( \textrm{cl}\left( {\text {int}}\mathcal {O}_S^+(e)\right) \right) \\{} & {} =\textrm{cl}\left( f_S^{-1}\left( {\text {int}}\mathcal {O}_S^+(e)\right) \right) \subset \textrm{cl}(D). \end{aligned}$$

$\square$

3.2 The general case

In this section, we show the result for more general nilpotent Lie groups, under a compactness assumption on the generalized kernel of the drift.

Theorem 3.4

A LCS on a nilpotent Lie group N satisfying the LARC and whose associated drift has compact generalized kernel admits a unique control set with nonempty interior. Moreover, its closure contains the singularities of the drift.

Proof

Let $\Sigma _N$ be a LCS whose drift $\mathcal {X}$ has compact generalized kernel $N_0$. By Proposition 2.4, it holds that $N_0={\text {fix}}(\varphi )\subset Z(N)$. Consequently, $\widehat{N}:=N / N_0$ is a nilpotent Lie group. Moreover, the fact that $N_0$ is invariant by the flow of $\mathcal {X}$ assures the existence of a LCS $\Sigma _{\widehat{N}}$ on $\widehat{N}$ that is $\pi$-conjugated to $\Sigma _N$, where $\pi :N\rightarrow \widehat{N}$ is the canonical projection (see [8, Proposition 4]). Let us denote by $\widehat{\mathcal {X}}$ the drift of $\Sigma _{\widehat{N}}$, by $\{\widehat{\varphi }_S\}_{S\in \mathbb {R}}$ the flow of $\widehat{X}$ and by $\widehat{A}$ the associated derivation. If A is the derivation associated with $\mathcal {X}$, Eq. (3) implies that

$$\begin{aligned} (d\pi )_eX\in \ker \widehat{A}\hspace{.5cm}\iff \hspace{.5cm} AX\in \mathfrak {n}_0\hspace{.5cm}\iff \hspace{.5cm} A^k(AX)=0, \text{ for } \text{ some } k\in \mathbb {N}, \end{aligned}$$

and hence $X\in \mathfrak {n}_0$. Consequently, $\ker \widehat{A}=\{0\}$, implying that $\Sigma _{\widehat{N}}$ is regular. By Theorem 3.2, the LCS $\Sigma _{\widehat{N}}$ admits a unique control set $\widehat{D}$ with nonempty interior.

Let $x\in \pi ^{-1}({\text {int}}\widehat{D})$ and consider, as in step 4 of the previous result, $S>0$ such that $\pi (x)\in {\text {int}}\widehat{\mathcal {O}}_S^+(\pi (x))$, where $\widehat{\mathcal {O}}_S^+(\pi (x))$ is the set of points reachable from $\pi (x)$ in time S of the system $\Sigma _{\widehat{N}}$. By the conjugation property,

$$\begin{aligned} \pi ^{-1}\left( {\text {int}}\widehat{\mathcal {O}}_S^+(\pi (x))\right) =\bigcup _{g\in H}{\text {int}}\mathcal {O}_S^+(xg)=\bigcup _{g\in H}{\text {int}}\mathcal {O}_S^+(x)g, \end{aligned}$$

and hence, there exists $g\in H$ satisfying $xg\in {\text {int}}\mathcal {O}_S^+(x)$. Moreover, the fact that the elements of finite order are dense in a compact group, together with the openness of ${\text {int}}\mathcal {O}_S^+(x)$, allows us to assume w.l.o.g. that

$$\begin{aligned} xg\in {\text {int}}\mathcal {O}_S^+(x), \hspace{.5cm} \text{ with } \hspace{.5cm}g^n=e, \hspace{.5cm} \text{ for } \text{ some } \hspace{.5cm}n\in \mathbb {N}. \end{aligned}$$

Let then $\textbf{u}\in \mathcal {U}$ such that $\phi (S, x, \textbf{u})=xg$ and extend it S-periodically. By concatenation, one gets that

$$\begin{aligned} \forall k\in \mathbb {N}, \hspace{.5cm}\phi (kS, x, \textbf{u})=xg^k\hspace{.5cm}\implies \hspace{.5cm}x=\phi (nS, x, \textbf{u})\in {\text {int}}\mathcal {O}_{nS}^+(x)\subset {\text {int}}\mathcal {O}^+(x), \end{aligned}$$

which, by Lemma 2.8, assures the existence of a control set $D_x$ with $x\in {\text {int}}D_x$.

Since the previous holds for any $x\in \pi ^{-1}({\text {int}}\widehat{D})$, for any $h\in H$ there exists a control set $D_{xh}$ with $xh\in {\text {int}}D_{xh}$. However, the fact that H is a connected Lie group (see Proposition 2.4) implies that, for any $h_1, h_2\in H$, there exists a continuous curve $\gamma :[0, 1]\rightarrow H$ such that $\gamma (0)=h_1$, $\gamma (1)=h_2$ and

$$\begin{aligned} x\gamma (t)\in xH=\pi ^{-1}(\pi (x))\subset \pi ^{-1}({\text {int}}\widehat{D}),\hspace{.5cm}\forall t\in [0, 1]. \end{aligned}$$

Therefore, for any $t\in [0, 1]$, there exists a control set $D_{x\gamma (t)}$ with $x\gamma (t)\in {\text {int}}D_{x\gamma (t)}$, which by Lemma 3.1 implies that $D=D_{x\gamma (t)}$ for all $t\in [0, 1]$. Therefore, D is a control set of $\Sigma _N$ with nonempty interior, such that

$$\begin{aligned} xH=\pi ^{-1}(\pi (x))\subset {\text {int}}D\hspace{.5cm} \text{ and } \hspace{.5cm} \pi (x)\in {\text {int}}\widehat{D}, \end{aligned}$$

which by Proposition 2.10 implies that $D=\pi ^{-1}(\widehat{D})$. Uniqueness follows direct from the previous equality. Also, by Corollary 3.3 and the fact that $\pi$ is a closed map, we get that

$$\begin{aligned} \pi (H)\subset \textrm{cl}\left( \widehat{D}\right) \hspace{.5cm}\implies \hspace{.5cm} H\subset \pi ^{-1}\left( \textrm{cl}\left( \widehat{D}\right) \right) =\textrm{cl}\left( \pi ^{-1}(\widehat{D})\right) =\textrm{cl}\left( D\right) , \end{aligned}$$

concluding the proof. $\square$

4 LCSs on the Heisenberg group

In this section, we give some examples of non-regular LCSs on the Heisenberg group.

The Heisenberg group $\mathbb {H}$ is by definition $\mathbb {H}:=(\mathbb {R}^3, *)$ where the product is given by

$$\begin{aligned} (x_1, y_1, z_1)*(x_2, y_2, z_2):=\left( x_1+x_2, y_1+y_2, z_1+z_2+\frac{1}{2}(x_2y_1-x_1y_2)\right) . \end{aligned}$$

Its Lie algebra $\mathfrak {h}$ is $\mathfrak {h}:=(\mathbb {R}^3, [\cdot , \cdot ])$ where

$$\begin{aligned}{}[(\alpha _1, \beta _1, \gamma _1), (\alpha _2, \beta _2, \gamma _2)]:=\left( 0, 0, \alpha _2\beta _1-\alpha _1\beta _2\right) . \end{aligned}$$

Example 4.1

Non-regular LCS on $\mathbb {H}$ with an infinite number of one-point control sets

The control-affine system

$$\begin{aligned}{} & {} \left\{ \begin{array}{l} \dot{x}(S)=y(S)\\ \dot{y}(S)=\textbf{u}(S)\\ \dot{z}(S)=\frac{1}{2}x(S)\textbf{u}(S) \end{array}\right. , \hspace{.5cm} \text{ where } \Omega :=[\rho , \nu ], \hspace{.5cm}\rho<0<\nu ,{} & {} \hspace{1cm}\left( \Sigma _{\mathbb {H}}\right) \end{aligned}$$

is a LCS on $\mathbb {H}$ whose associated derivation and right-invariant vector field are given, respectively, as

$$\begin{aligned} A=\left( \begin{array}{lll} 0 &{} \quad 1 &{} \quad 0 \\ 0 &{} \quad 0 &{} \quad 0 \\ 0 &{} \quad 0 &{} \quad 0 \end{array}\right) \hspace{.5cm} \text{ and } \hspace{.5cm}Y(x, y, z)=\left( 0, 1, \frac{1}{2}x\right) . \end{aligned}$$

In particular, if $Y=Y(0, 0, 0)$, we have that

$$\begin{aligned} AY=(1, 0, 0)\quad \text{ and } \quad [AY, Y]=(0, 0, 1)\quad \implies \quad \mathfrak {h}=\textrm{span}\{Y, AY, [AY, Y]\}, \end{aligned}$$

and $\Sigma _{\mathbb {H}}$ satisfies the LARC. Moreover, $\det A=0$ and the system is non-regular.

The plane $Q=\{(x, 0, z), x, z\in \mathbb {R}\}$ is the set of singularities of the drift. In particular, any point in Q has to be contained in a control set of $\Sigma _{\mathbb {H}}$. Our aim here is to show that any point in Q is in fact a control set of $\Sigma _{\mathbb {H}}$.

In order to show the previous, let us consider the diffeomorphism,

$$\begin{aligned} f:\mathbb {H}\rightarrow \mathbb {R}^3, \hspace{1cm}f(x, y, z)=\left( x, y, z-\frac{1}{6}xy\right) , \end{aligned}$$

we get that $\Sigma _{\mathbb {H}}$ is equivalent to the system¹

$$\begin{aligned}&\left\{ \begin{array}{l} \dot{x}(S)=y(S)\\ \dot{y}(S)=\textbf{u}(S)\\ \dot{z}(S)=\frac{1}{6}(2x(S)\textbf{u}(S)-y(S)^2) \end{array}\right. , \end{aligned}$$

($\Sigma _{\mathbb {R}^3}$)

whose solution for a constant control function $\textbf{u}\equiv u$ is

$$\begin{aligned} \phi (S, P_0, \textbf{u})=\left( x_0+y_0S+\frac{uS^2}{2}, y_0+Su, z_0+\frac{1}{6}(2x_0u-y_0^2)S\right) . \end{aligned}$$

Let us consider the function

$$\begin{aligned} F:\mathbb {R}^3\rightarrow \mathbb {R}, \hspace{1cm}F(x, y, z):=6z\sigma +y(y^2-2x\sigma ), \end{aligned}$$

where $\sigma <\rho$ is a fixed real number. For any $u\in \Omega$ and $P_0=(x_0, y_0, z_0)$, it holds that:

If $u=0$, then

$$\begin{aligned} F(\phi (S, P_0, 0))= & {} F\left( x_0+Sy_0, y_0, z_0-\frac{y_0^2}{6}S\right) \\= & {} 6\left( z_0-\frac{y_0^2}{6}S\right) \sigma +y_0(y_0^2-2(x_0+Sy_0)\sigma ) \\= & {} F(P_0)-3Sy_0^2\sigma \ge F(P_0),\hspace{.5cm} \text{ since } \hspace{.5cm} \sigma<\rho <0, \end{aligned}$$

with equality iff $y_0=0$.

On the other hand, if $u\ne 0$, then

$$\begin{aligned} g_u(S):= & {} F(\phi (S, P_0, u))=6\left( z_0+\frac{1}{6}(2x_0u-y_0^2)S\right) \sigma \\{} & {} +(y_0+uS)\left( (y_0+uS)^2-2\left( x_0+y_0t+\frac{uS^2}{2}\right) \sigma \right) \\= & {} (y_0+uS)^3+\sigma \left( 6z_0-2x_0y_0-3y_0^2S-3y_0uS^2-u^2S^3\right) \\= & {} \frac{u-\sigma }{u}(y_0+uS)^3+\sigma \left( 6z_0-2x_0y_0+\frac{1}{u}y_0^3\right) . \end{aligned}$$

Derivation in S gives us that

$$\begin{aligned} g_u'(S)=3(u-\sigma )(y_0+uS)^2\ge 0, \end{aligned}$$

showing that, for all $S>0$,

$$\begin{aligned} F(\phi (S, P_0, u))\ge F(P_0), \end{aligned}$$

with equality iff $y_0=0$. By concatenation, we conclude that

$$\begin{aligned} F(\phi (S, P_0, \textbf{u}))> F(P_0), \hspace{.5cm} \text{ if } \hspace{.5cm}y_0\ne 0\hspace{.5cm} \text{ or } \hspace{.5cm}\textbf{u}\not \equiv 0, \end{aligned}$$

implying that the control sets of $\Sigma _{\mathbb {R}^3}$ are the singletons $\{(x_0, 0, z_0)\}$.

Example 4.2

Controllable non-regular LCS on $\mathbb {H}$

Let us consider the linear control system

$$\begin{aligned}{} & {} \left\{ \begin{array}{l} \dot{x}(S)=-y(S)\\ \dot{y}(S)=x(S)+\textbf{u}(S)\\ \dot{z}(S)=\frac{1}{2}x(S)\textbf{u}(S) \end{array}\right. , \hspace{.5cm} \text{ where } \Omega :=[\rho , \nu ], \hspace{.5cm}\rho<0<\nu ,{} & {} \hspace{1cm}\left( \Sigma _{\mathbb {H}}\right) \end{aligned}$$

is a LCS on $\mathbb {H}$ whose associated derivation and right-invariant vector field are given, respectively, as

$$\begin{aligned} A=\left( \begin{array}{lll} 0 &{} \quad -1 &{}\quad 0 \\ 1 &{} \quad 0 &{} \quad 0 \\ 0 &{} \quad 0 &{} \quad 0 \end{array}\right) \hspace{.5cm} \text{ and } \hspace{.5cm}Y(x, y, z)=\left( 0, 1, \frac{1}{2}x\right) . \end{aligned}$$

In particular, for $Y=Y(0, 0, 0)$, it holds that

$$\begin{aligned} AY=(-1, 0, 0)\quad \text{ and } \quad [AY, Y]=\left( 0, 0, 1\right) \quad \implies \quad \mathfrak {h}=\textrm{span}\{Y, AY, [AY, Y]\}, \end{aligned}$$

showing that $\Sigma _{\mathbb {H}}$ satisfies the LARC. In particular, $\det A=0$ and the system is non-regular.

The solutions of $\Sigma _{\mathbb {H}}$ for a constant control function $\textbf{u}\equiv u$ is

$$\begin{aligned} \phi (S, P_0, \textbf{u})= & {} \bigg ((x_0+u)\cos S-y_0\sin S-u, (x_0+u)\sin S+y_0\cos S, z_0 \\{} & {} -\frac{u}{2}\left( (x_0+u)\sin S+y_0(\cos S-1)\right) +\frac{u^2}{2}S\bigg ). \end{aligned}$$

Let us note that the induced system on $\mathbb {R}^2$

$$\begin{aligned}{} & {} \left\{ \begin{array}{l} \dot{x}(S)=-y(S)\\ \dot{y}(S)=x(S)+\textbf{u}(S) \end{array}\right. ,{} & {} \end{aligned}$$

($\Sigma _{\mathbb {R}^2}$)

is a LCS whose associated derivation has a pair of pure imaginary eigenvalues. Since $\Sigma _{\mathbb {R}^2}$ also satisfies the LARC, it is controllable (see for instance [6, Chapter 3]). Therefore, $\Sigma _{\mathbb {H}}$ is controllable, as soon as $\{0\}\times \mathbb {R}$ is controllable. In order to show the previous, we construct explicitly a trajectory connecting two arbitrary points $(0, z_1), (0, z_2)\in \{0\}\times \mathbb {R}$ with $z_1<z_2$ as follows:

• From $(0, z_1)$ to $(0, z_2)$:

Let $\delta >0$ such that $(-\delta , \delta )\subset (\rho , \nu )$. Take $\epsilon >0$ satisfying $3\epsilon =z_2-z_1$ and consider $S_0>0$ such that

$$\begin{aligned} \overline{\mathcal {O}_{S_0}^+(0, z_1)}\subset B_{\epsilon }(0, z_1)\hspace{.5cm} \text{ and } \hspace{.5cm}\overline{\mathcal {O}_{S_0}^-(0, z_2)}\subset B_{\epsilon }(0, z_2), \end{aligned}$$

which exists by the continuity of the solutions and the compactness of $[\rho , \nu ]$. Moreover, the LARC implies that $\pi (\mathcal {O}_{S_0}^+(0, z_1))$ and $\pi (\mathcal {O}_{S_0}^-(0, z_2))$ are open neighborhoods of the origin in $\mathbb {R}^2$, where $\pi : (v, z)\in \mathbb {H}\rightarrow v\in \mathbb {R}^2$ is the projection onto the two first components. Since $\Sigma _{\mathbb {R}^2}$ is controllable, there exists $u\in (-\delta , 0)$ and $\textbf{u}_1, \textbf{u}_2\in \mathcal {U}$ such that

$$\begin{aligned} \phi (S_0, (0, z_1), \textbf{u}_1)=(v(u), \bar{z}_1) \hspace{.5cm} \text{ and } \hspace{.5cm} \phi (S_0, (v(u), \bar{z}_2), \textbf{u}_1)= (0, z_2), \end{aligned}$$

and by our choices $\bar{z}_1<\bar{z}_2$, where $v(u):=(-u, 0)$ is an equilibrium point of $\Sigma _{\mathbb {R}^2}$. On the other hand,

$$\begin{aligned} S_1=\frac{\bar{z}_2-\bar{z}_1}{p(u)}>0\quad \implies \quad \phi (S_1, (v(u), \bar{z}_1), u)=(v(u), \bar{z}_1+S_1p(u))=(v(u), \bar{z}_2), \end{aligned}$$

and so, by concatenation, we get

$$\begin{aligned} \phi (S_0, \phi (S_1, \phi (S_0, (0, z_1), \textbf{u}_1), u), \textbf{u}_2)=(0, z_2). \end{aligned}$$

• From $(0, z_2)$ to $(0, z_1)$:

Let us fix $\alpha \in (0, \nu )$ and define $v^*:=-\alpha \left( 1, \pi \right)$. By the controllability of the induced system $\Sigma _{\mathbb {R}^2}$, there exist $\mathbf{u_1}, \mathbf{u_2}, \textbf{u}^*\in \mathcal {U}$, $S_1, S_2, S^*>0$ and $\bar{z}_1, \bar{z}_2, z^*\in \mathbb {R}$ satisfying

$$\begin{aligned}{} & {} \phi (S_2, (0, z_2), \textbf{u}_2)=(v^*, \bar{z}_2), \hspace{.5cm}\phi (S_1, (v(\alpha ), \bar{z}_1), \textbf{u}_1)=(0, z_1)\hspace{.5cm} \text{ and } \\{} & {} \quad \phi (S^*, (v^*, 0), \textbf{u}^*)=(v(\alpha ), z^*), \end{aligned}$$

where as previously $v(\alpha )=(-\alpha , 0)$. A trajectory connecting $(0, z_2)$ to $(0, z_1)$ is construct by concatenation as follows:

• (i) With control $\textbf{u}_2$ and time $S_2$ connect $(0, z_2)$ with $(v^*, \bar{z}_2)$;

• (ii) Using the constant control $\textbf{u}\equiv \alpha$ and $\tau _0=\pi$ we have that

  $$\begin{aligned} \phi \left( \pi , (v^*, \bar{z}_2), \alpha \right) =\left( -\alpha \left( 1, -\pi \right) , \bar{z}_2-\alpha ^2\frac{\pi }{2}\right) ; \end{aligned}$$

• (iii) Since $v^*$ and $-\alpha (1, -\pi )$ have the same norm, there exists $\tau _1>0$ such that $R_{\tau _1}\left( -\alpha (1, -\pi )\right) =v^*$, where $R_{\tau _1}$ is the counter-clockwise rotation of $\tau _1$-degrees. Hence,

  $$\begin{aligned} \phi \left( \tau _1, \phi \left( \tau _0, (v^*, \bar{z}_2), \alpha \right) , 0\right){} & {} =\phi \left( \tau _1, \left( -\alpha (1, -\pi ), \bar{z}_2-\alpha ^2\frac{\pi }{2}\right) , 0\right) \\{} & {} =\left( R_{\tau _1}\left( -\alpha (1, -\pi )\right) , \bar{z}_2-\alpha ^2\frac{\pi }{2}\right) \\ {}{} & {} =\left( v^*, \bar{z}_2-\alpha ^2\frac{\pi }{2}\right) , \end{aligned}$$

  which gives us a periodic trajectory for the induced system on the plane (see Fig. 1a)

• (iv) Repeat items (ii) and (iii) $n_0$-times to obtain a trajectory connecting $(v^*, \bar{z}_2)$ to the point $(v^*, \hat{z}_2)$, where

  $$\begin{aligned} \hat{z}_2:= \bar{z}_2 -n_0\alpha ^2\frac{\pi }{2}\hspace{.5cm} \text{ satisfies } \hspace{.5cm}\hat{z}_2\le \bar{z}_1-z^*; \end{aligned}$$

• (v) Now, with control $\textbf{u}^*$ and time $S^*$ we have that

  $$\begin{aligned} \phi (S^*, (v^*, \hat{z}_2), \textbf{u}^*){} & {} =\phi (S^*, (v^*,0), \textbf{u}^*)+(0, \hat{z}_2)=(v(\alpha ), z^*)+(0, \hat{z}_2)\\{} & {} =(v(\alpha ), z^*+\hat{z}_2); \end{aligned}$$

• (vi) With time $S_0:=\frac{2}{\alpha ^2}(\bar{z}_1-(z^*+\hat{z}_2))\ge 0$ and control $u\equiv \rho$, we get that

  $$\begin{aligned} \phi (S_0, (v(\alpha ), z^*+\hat{z}_2), \rho )=\left( v(\alpha ), S_0\frac{\alpha ^2}{2}+(z^*+\hat{z}_2)\right) =(v(\alpha ), \bar{z}_1); \end{aligned}$$

• (vii) Now, with control $\textbf{u}_1$ and time $S_1$ we get that $\phi (S_1, (v(\alpha ), \bar{z}_1), \textbf{u}_1)=(0, z_1)$, which gives us the desired trajectory (see Fig. 1b).

Fig. 1

a Periodic orbit on the plane. b Trajectory connecting (0, $z_{2}$) to (0, $z_{1}$)

Remark 4.3

It is important to note that in both examples, the LCS induced on $\mathbb {R}^2$ given by the first two components is controllable. However, the differences in the dynamics of the systems on $\mathbb {H}$ are quite remarkable, showing that one cannot expect to find a condition assuring the existence of control sets with nonempty interiors for non-regular systems.

5 Remarks/future work

In the search for control sets with nonempty interiors, the LARC is the weakest condition one could impose on the system in order to have reachable sets with nonempty interiors. As a consequence, any kind of generalization of the previous results has to be concentrated on the properties of the derivation A. Since the results presented here heavily depend on the regularity of A and invertible derivations only appear on nilpotent groups, for more general Lie groups as state spaces, one would have to decompose the system into a nilpotent component, where the induced system is regular and a compact component.

One suitable candidate for the previous split appears in the proof of [2, Theorem 4.1] where the compact assumption is made on the connected Lie group with associated Lie algebra given by the sum of the generalized eigenspaces of A with zero real parts. However, for such a split, the system that appears on the nilpotent part is not linear but (conjugated to an) affine. Therefore, our approach to a generalization of the previous results would first prove the same results for affine systems on nilpotent groups.