Krener's theorem

Let ${\displaystyle {\ }{\dot {q}}=f(q,u)}$  be a smooth control system, where ${\displaystyle {\ q}}$  belongs to a finite-dimensional manifold ${\displaystyle \ M}$  and ${\displaystyle \ u}$  belongs to a control set ${\displaystyle \ U}$ . Consider the family of vector fields ${\displaystyle {\mathcal {F}}=\{f(\cdot ,u)\mid u\in U\}}$ .

Let ${\displaystyle \ \mathrm {Lie} \,{\mathcal {F}}}$  be the Lie algebra generated by ${\displaystyle {\mathcal {F}}}$  with respect to the Lie bracket of vector fields. Given ${\displaystyle \ q\in M}$ , if the vector space ${\displaystyle \ \mathrm {Lie} _{q}\,{\mathcal {F}}=\{g(q)\mid g\in \mathrm {Lie} \,{\mathcal {F}}\}}$  is equal to ${\displaystyle \ T_{q}M}$ , then ${\displaystyle \ q}$  belongs to the closure of the interior of the attainable set from ${\displaystyle \ q}$ .

Even if ${\displaystyle \mathrm {Lie} _{q}\,{\mathcal {F}}}$  is different from ${\displaystyle \ T_{q}M}$ , the attainable set from ${\displaystyle \ q}$  has nonempty interior in the orbit topology, as it follows from Krener's theorem applied to the control system restricted to the orbit through ${\displaystyle \ q}$ .

When all the vector fields in ${\displaystyle \ {\mathcal {F}}}$  are analytic, ${\displaystyle \ \mathrm {Lie} _{q}\,{\mathcal {F}}=T_{q}M}$  if and only if ${\displaystyle \ q}$  belongs to the closure of the interior of the attainable set from ${\displaystyle \ q}$ . This is a consequence of Krener's theorem and of the orbit theorem.

As a corollary of Krener's theorem one can prove that if the system is bracket-generating and if the attainable set from ${\displaystyle \ q\in M}$  is dense in ${\displaystyle \ M}$ , then the attainable set from ${\displaystyle \ q}$  is actually equal to ${\displaystyle \ M}$ .

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