Let $X_1,\dots,X_n$ be complete vector fields on $\mathbb R^n$ and suppose that $(X_1(p),\dots,X_n(p))$ is a basis for all $p \in \mathbb R^n$.

Question: Is it possible to choose a cube $C$ around the origin of $\mathbb R^n$ such that there is for every $p \in C$ a piecewise smooth curve $\alpha \subset C$ which connects $p$ with $0$ where the smooth parts of the curve are given by the flows of the vector fields $\pm X_1,\dots,\pm X_n$? (With other words: is it possible to travel from $0$ to $p$ following only the integral curves of the given vector fields in a bounded domain?)

For $n=2$ this is pretty clear; w.l.o.g. $X_1=\partial/\partial x_1$ and the image of $x_1=0$ under the flow $X_2$ fills all of $\mathbb R^2$, since the flow lines of $X_2$ intersect $x_1=0$ transversally. Now it is easy to find a $C$ and $\alpha$ for a $p \in C$. But I can not generalize this for $n$ arbitrary.

Edit: Instead of demanding $C$ to be a cube, one could also ask if there is a open neighbourhood of the origin with the desired properties.

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    $\begingroup$ Maybe looking for the keyword "control theory" might be useful. $\endgroup$ – Qfwfq Feb 18 '12 at 9:12
  • $\begingroup$ You cannot assume that $X_1 = \partial/\partial x_1$, because it might be nonconstant. $\endgroup$ – Matthias Ludewig Feb 18 '12 at 11:38
  • $\begingroup$ Since $X_1$ has no zeros, I can assume after a change of the coordinate system, that $X_1$ is constant. $\endgroup$ – Panagiotis Konstantis Feb 18 '12 at 12:38
  • $\begingroup$ (Title re-edited. "Linearly independent[...]" is the right spelling) $\endgroup$ – Qfwfq Feb 18 '12 at 23:02

For a suitable nbd of the origin $U$, yes, even following the $n$ flows in a prescribed order. Assuming $X_1,\dots,X_n$ locally Lipschitz continuous, the corresponding flows $\phi_1(t,x),\dots, \phi_n(t,x)$ are $C^1$ maps (in the pair) so the map $\Phi$ $$(t _ 1,\dots,t_ n)\mapsto \phi _ n (t _ n, \phi_{n-1}( t _ {n-1}, \dots \phi _1 (t_1, 0))\dots)$$ is of class $C^1$ on a nbd of $0\in\mathbb{R}^n$. Since for $(t,x)\to 0$ in $ \mathbb{R} \times\mathbb{R}^n$, $$ \phi_j(t,x)= x + tX_j(0) + o\big((t,x) \big) $$ it follows that the differential of $\Phi$ at $0\in\mathbb{R}^n$ is the linear map $$(t _ 1,\dots,t_ n)\mapsto t_1X_1(0)+\dots t_n X_n(0)$$ which is invertible because by assumption $X_1(0),\dots, X_n(0)$ is a base. By the implicit function theorem, $\Phi$ is a local diffeo between a cube $Q$ and a nbd $U$ of $0$. The images of the broken lines in $Q$ with edges parallel to the axes is what you want and reach any point of $U$.

(edit) Also, given any connected open subset $\Omega\subset \mathbb{R}^n$, any two points $p$ and $q$ of $\Omega$ can be joined by a continuous curve made by finitely many flow-lines of the fields $X_1,\dots,X_n$, just because the set of points that may be reached this way starting from $p$ is a non-empty open and closed subset of $\Omega$, thanks to the preceding local fact.

However, there could be problems to reach some point if $\Omega$ is not open, e.g a closed cube. Take e.g. in $n=2$ and two constant fields $X_1:=(2,1)$, $X_2:=(1,2)$. Then, the vertex $(r,r)$ of the square $Q:=[−r,r]\times[−r,r]$ is easily joined to the origin by a two-edges broken flow-line, but the vertex $(r,−r)$ can't be reached by any such broken line in $Q,$ in any finite number of steps.

  • $\begingroup$ Thanks! That is a nice answer! Though I think it is nevertheless true for the cube-version. $\endgroup$ – Panagiotis Konstantis Feb 18 '12 at 19:01
  • $\begingroup$ I've added some hints for the case of nbd's of prescribed shape. $\endgroup$ – Pietro Majer Feb 18 '12 at 22:28
  • $\begingroup$ Thanks again, I had only open nbd's in mind and indeed there could be some problem with closed ones... $\endgroup$ – Panagiotis Konstantis Feb 19 '12 at 0:59

Let my first vector field just be $(1,0,0)$. The second is $(0,1,e^x)$. The third can be anything, say $(0,0,1)$.

I claim that one cannot make a square with the first two fields, a necessary component of a cube. Obviously two of the edges will be straight lines that keep $y$ and $z$ fixed and vary $x$. The other edges will also be straight lines, but tilted at different angles in a way that cannot form a square.

More rigorously, since the other flow does not change $x$, $x$ will go between two values, say $x_1$ and $x_2$. $y$ and $z$ similarly go between two values, say $y_a,y_b$ and $z_a,z_b$.

On the first edge of flow two, we must have $(z_a-z_b)=e^{x_1}(y_a-y_b)$. On the second, $(z_a-z_b)=e^{x_2}(y_a-y_b)$. This is impossible.

  • $\begingroup$ Sorry, I don't understand your answer. Please correct me, if I'm wrong: The flow for $(0,1,e^x)$ starting from the origin is given by $t\mapsto(0,t,t)$. Choose for $C$ the unit cube around the origin. Then for an arbitrary $(x,y,z) \in C$ first I would flow along the flow of $(0,1,e^x)$ ending at $(0,y,y)$, then I would adjust the other coordinates by flowing along the other vector fields, which will not change the y--coordinate. I'm always staying in the the unit cube. What is my mistake? $\endgroup$ – Panagiotis Konstantis Feb 18 '12 at 9:48
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    $\begingroup$ Ah. I misunderstood your question to imply something very different. Pietro's answer is correct, mine is not. $\endgroup$ – Will Sawin Feb 18 '12 at 18:38

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