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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