In this section, we give some examples of boundedness and unboundedness control sets on some decomposable Lie groups. But first, we explain how to find the face of the drift 𝓧 when it is induced by an inner derivation.

Here, exp_{G}:𝔤 → *G* is the usual exponential map. Hence, *X* defines by conjugation a 1-parameter group of inner automorphism as follows

$$\begin{array}{}{\displaystyle {\phi}_{t}(g)={X}_{t}(e)\phantom{\rule{thinmathspace}{0ex}}g\phantom{\rule{thinmathspace}{0ex}}{X}_{-t}(e),g\in G,\text{\hspace{0.17em}and\hspace{0.17em}}{\phi}_{t}\in \mathrm{A}\mathrm{u}\mathrm{t}(G)\text{\hspace{0.17em}for any\hspace{0.17em}}t\in \mathbb{R}.}\end{array}$$

Therefore, it is possible to compute the linear vector field as

$$\begin{array}{}{\displaystyle \mathcal{X}(g)=(\frac{d}{dt}{)}_{t=0}{\phi}_{t}(g).}\end{array}$$

The associated derivation 𝓓 : 𝔤 → 𝔤 is 𝓓(*Y*) = −[*X*, *Y*], *Y* ∈ 𝔤.

Recall that any derivation on a semisimple Lie group is inner. This property has interest for us in the compact case. On the other hand, in [13] we built an algorithm which provides an effective means to compute the Lie algebra *∂* 𝔤 that we use in this section.

#### Example 5.2

*Consider the solvable affine group*

$$\begin{array}{}{\displaystyle \left\{G=\left(\begin{array}{cc}x& y\\ 0& 1\end{array}\right):x>0\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}and\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}y\in \mathbb{R}\right\}}\end{array}$$

*with Lie algebra* 𝔤 = *Span* {*X*, *Y*} *and* [*X*, *Y*] = *Y*. *An easy computation shows that* *∂* 𝔤 *is given just by inner derivation with the shape*

$$\begin{array}{}{\displaystyle \mathrm{\partial}\mathfrak{g}\mathfrak{=}\left\{\mathcal{D}\mathcal{=}\left(\begin{array}{cc}0& 0\\ a& b\end{array}\right):a,b\in \mathbb{R}\right\}.}\end{array}$$

*From Remark 5.1 the linear vector fiel* 𝓧 *associated to* 𝓓 *is given by*

$$\begin{array}{}{\displaystyle {\mathcal{X}}_{(x,y)}=\left(\begin{array}{cc}0& a(x-1)+by\\ 0& 0\end{array}\right).}\end{array}$$

Let *Σ* be the transitive linear system on *G* defined by

$$\begin{array}{}{\displaystyle \dot{g}(t)=\mathcal{X}(g(t))+u(t)X(g(t)),u\in \mathcal{U}}\end{array}$$

where, 𝓓 = *ad*(*Y*) comes from *a* = −1 and *b* = 0. Since *ad*(*Y*)*X* = −*Y* then *Span* {*X*,𝓓*X*} = 𝔤. So, *Σ* satisfies the *ad*-rank condition, 𝓐 is open and of course, *Σ* satisfies also LARC. Moreover, *G* is solvable thus the control set 𝓒_{e} is the only one with nonempty interior. It turns out that,

$$\begin{array}{}{\displaystyle {\mathfrak{g}}^{+}={\mathfrak{g}}^{-}=0\Rightarrow {\mathfrak{g}}^{0}=\mathfrak{g}.}\end{array}$$

Thus,

$$\begin{array}{}{\displaystyle {G}^{+,0}=G\subset \mathcal{A}\text{\hspace{0.17em}and\hspace{0.17em}}{G}^{-,0}=G\subset {\mathcal{A}}^{\ast}\Rightarrow {\mathcal{C}}_{e}=G.}\end{array}$$

To conclude, the system is controllable from the identity. This fact is completely concordant with Theorem 3 in [11]. Actually, it is shown there that a transitive system in a canonical form, like *Σ*, is controllable if and only if *b* = 0.

#### Example 5.3

*Let* 𝔤 = ℝ*X* + ℝ*Y* + ℝ*Z* *be the Lie algebra of the connected and simply connected Heisenberg Lie group* *G*

$$\begin{array}{}{\displaystyle G=\left\{g=\left(\begin{array}{ccc}1& x& z\\ 0& 1& y\\ 0& 0& 1\end{array}\right):(x,y,z)\in {\mathbb{R}}^{3}\right\}}\end{array}$$

*of dimension three*. *The generators of* 𝔤 *are provided by*

$$\begin{array}{}{\displaystyle X=\frac{\mathrm{\partial}}{\mathrm{\partial}x},Y=\frac{\mathrm{\partial}}{\mathrm{\partial}y}+x\frac{\mathrm{\partial}}{\mathrm{\partial}z}\text{\hspace{0.17em}and\hspace{0.17em}}Z=\frac{\mathrm{\partial}}{\mathrm{\partial}z}.}\end{array}$$

The only one non-vanishing Lie bracket is [*X*, *Y*] = *Z*. Any derivation 𝓓 is represented by a 6 real parameters matrix in the basis {*X*, *Y*,*Z*} as follow

$$\begin{array}{}{\displaystyle \mathrm{\partial}\mathfrak{g}\mathcal{=}\left\{\left(\begin{array}{ccc}a& b& 0\\ c& d& 0\\ e& f& a+d\end{array}\right):a,b,c,d,e,f\in \mathbb{R}\right\}.}\end{array}$$

Consider the linear system *Σ* with derivation 𝓓 determined by its coefficients *a* = *d* = −1, *b* = 1, *c* = −1, *e* = *f* = 0 and the control vectors *X* and *Z*,

$$\begin{array}{}{\displaystyle \dot{g}(t)=\mathcal{X}(g(t))+{u}_{1}(t)X(g(t))+{u}_{2}(t)Z(g(t)),u\in \mathcal{U},\text{\hspace{0.17em}with\hspace{0.17em}}\mathit{\Omega}=\left[-1,1\right].}\end{array}$$

We have, *Spec*_{Ly}(𝓓) = {−1, −2}. So, 𝔤^{−,0} = 𝔤^{−} = 𝔤 and 𝔤^{+,0} = 0 are both ideals of 𝔤. On the other hand,

$$\begin{array}{}{\displaystyle Span\left\{X,Z,\mathcal{D}\mathcal{(}X)=X-Y\right\}=\mathfrak{g}.}\end{array}$$

Since 𝓓 is a hyperbolic derivation, Theorem 3.7 shows that the existent control set 𝓒_{e} is bounded.

#### Example 5.4

*On the rotational group S*0(3, ℝ) *with Lie algebra* 𝔰𝔬 (3, ℝ) *the skew*-*symmetric real matrix of order three*

$$\begin{array}{}{\displaystyle \mathfrak{g}=Span\left\{X,Y,Z\right\}}\end{array}$$

*consider the system*

$$\begin{array}{}{\displaystyle \dot{g}(t)=\mathcal{X}(g(t))+{u}_{1}(t)X(g(t))+{u}_{2}(t)Y(g(t)),u\in \mathcal{U},\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}with\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\mathit{\Omega}=\mathbb{R},}\end{array}$$

*where* 𝓧 = *ad*(*X*). *Since* *Σ* *satisfies LARC*, *the control set is bounded and coincides with the group*. *The system is controllable from the identity*.

#### Example 5.5

*Take the linear system* *Σ* *on the Heisenberg group* *G* *like in Example* 2, *but with different dynamic behavior*

$$\begin{array}{}{\displaystyle \dot{g}(t)=\mathcal{X}(g(t))+{u}_{1}(t)(X-Y)(g(t))+{u}_{2}(t)(X+Y+Z)(g(t)),u\in \mathcal{U},\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}with\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\mathit{\Omega}=\left[-1,1\right]}\end{array}$$

*where the derivation* 𝓓 *is furnished by* *a* = 1, *d* = −1 *and* *b* = *c* = *e* = *f* = 0. *Hence*, 𝓓(*X* − *Y*) = *X* + *Y*. *Thus*,

$$\begin{array}{}{\displaystyle Span\left\{X-Y,Z,X+Y\right\}=\mathfrak{g}}\end{array}$$

*an* 𝓐 *is an open set*.

If we restrict *Σ* to the plane ℝ^{2} = *Span* {*X*, *Y*} we get a classical linear system on the vector space *Σ*_{ℝ2}

$$\begin{array}{}{\displaystyle \left(\begin{array}{c}\dot{x}\\ \dot{y}\end{array}\right)=\left(\begin{array}{cc}1& 0\\ 0& -1\end{array}\right)\left(\begin{array}{c}x\\ y\end{array}\right)+\left(\begin{array}{c}1\\ 1\end{array}\right)u:u\in \mathcal{U}\text{\hspace{0.17em}with\hspace{0.17em}}\mathit{\Omega}=\left[-1,1\right]}\end{array}$$

which satisfies Lemma 2.5. Moreover,

$$\begin{array}{}{\displaystyle \mathcal{A}\mathcal{=}\mathbb{R}\times (-1,1)\text{\hspace{0.17em}and\hspace{0.17em}}{\mathcal{A}}^{\ast}\mathcal{=}(-1,1)\times \mathbb{R}.}\end{array}$$

Thus, the control set 𝓒_{e} restricted to the plane is bounded and reads

$$\begin{array}{}{\displaystyle ({\mathcal{C}}_{e}{)}_{{\mathbb{R}}^{2}}=(-1,1)\times \left[-1,1\right],\text{\hspace{0.17em}see\hspace{0.17em}}\left[4\right].}\end{array}$$

However, 𝓒_{e} can not be bounded. Despite the fact that 𝔤^{+,0} = *Span* {*X*, *Z*} and 𝔤^{−,0} = *Span* {*Y*, *Z*} are ideals, the derivation 𝓓 = *diag*(1, −1, 0) is just hyperbolic on the plane not on *G*. Actually,

$$\begin{array}{}{\displaystyle {\mathcal{C}}_{e}=(-1,1)\times \left[-1,1\right]\times (1,1,1)\mathbb{R}.}\end{array}$$

#### Example 5.6

*Let us consider the nilpotent Lie group* *G* *with Lie algebra*

$$\begin{array}{}{\displaystyle \phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\mathfrak{g}=\mathbb{R}{X}_{1}+\mathbb{R}{X}_{2}+\mathbb{R}{X}_{3}+\mathbb{R}{X}_{4},\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}and\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}the\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}rules}\\ \\ {\displaystyle \left[{X}_{4},{X}_{2}\right]={X}_{4},\left[{X}_{3},{X}_{2}\right]={X}_{4}+{X}_{2},\left[{X}_{1},{X}_{2}\right]={X}_{3}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}and\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\left[\text{\hspace{0.17em}}{X}_{1},{X}_{3}\right]={X}_{4}.}\end{array}$$

Let *Σ* be a linear system with an arbitrary derivation 𝓓 ∈ *∂* 𝔤 such that the reachable set 𝓐 of *Σ* is open. Hence, the control set 𝓒_{e} is unbounded. In fact, a straightforward computation shows that the Lie algebra of 𝔤-derivations is five dimensional and reads as

$$\begin{array}{}{\displaystyle \mathrm{\partial}\mathfrak{g}\mathfrak{=}\left\{\left(\begin{array}{cccc}a& -a& 0& 0\\ 0& 0& 0& 0\\ b& c& a+b& 0\\ d& e& b+c+d& 2a+b\end{array}\right):a,b,c,d,e\in \mathbb{R}\right\}.}\end{array}$$

Since the underlying topological space of *G* is the connected and simply connected manifold ℝ^{4}, Theorem 3.5 applies. However, 0 ∈ *Spec*_{Ly}(𝓓) for any 𝓓 ∈ *∂* 𝔤. Thus, no hyperbolic derivation exits, ending the proof.

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