Krener's theorem

In mathematics, Krener's theorem is a result attributed to Arthur J. Krener in geometric control theory about the topological properties of attainable sets of finite-dimensional control systems. It states that any attainable set of a bracket-generating system has nonempty interior or, equivalently, that any attainable set has nonempty interior in the topology of the corresponding orbit. Heuristically, Krener's theorem prohibits attainable sets from being hairy.

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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