An invariant distribution

control theorydifferential-geometrydynamical systems

I am studying differential geometry, and I am focusing on distributions. I am in particular focusing on the concept of invariant distributions.

If we consider a nonlinear system:

$\dot{x}=f(x)$

I have seen that a distribution is invariant with respect to $f$ if:

$\left [ f,\Delta \right ]\subset \Delta $

i.e any $\tau (x)\in \Delta (x)$ we have:

$\left [ f,\tau \right ](x)\in \Delta (x)$

where $\left [ f,\tau \right ]$ is the Lie Bracket operation.

I am not sure about the meaning of this, I just copied from the notes of my professor as it is, but it shoud mean that if I have any vector field $\tau$, if I do the Lie Bracket with the vector field $f$, the generator vecotr field will always be in the distribution. So, it should mean that it is impossible to go out of the distribution if I do a Lie Bracket with $f$. (this is just what I think, I do not have a confirm).

then, studyng on the notes of my professor, I have that if the distribution is invariant with respect to $f$ and involutive, it is possible to define a change of coordinates:

$\Phi (x)=\begin{bmatrix}
\Phi _1(x)\\
\lambda _1(x)\\
…\\
\lambda _k(x)
\end{bmatrix}$

where the functions $\lambda (x)$ are $n-k$ and such that:

$\frac{d\Phi _2(x)}{dx}\Delta =0$

where $\Phi _2(x)$ are the $n-k$ functions $\lambda (x)$.
So, $\Phi _2(x)$ is defining foliations.

After this, the notes say that after this change of coordinates, it is possible to express the system as follows:

$\dot{z_1}=f_1(z_1,z_2)$

$\dot{z_2}=f_2(z_2)$

but how did he came to this solution?

Moreover, it says that this system puts in evidence an important propery of the system, which is the fact that the system can be considered as composed by two different subsystems, and that the evolution of two states that belong to the same foliation, evolve together in foliations, which I do not understand why, and also what it means.

Can somebody help me?

Best Answer

Let $x\in \mathbb{R}^n$ and $U$ be an open neighbourhood of $x.$ Let $\Delta$ be a smooth $k$-dimensional distribution on $\mathbb{R}^n.$ When $\Delta$ is involutive, we have, by Frobenius' theorem, that it is locally completely integrable. That means that there exists a coordinate transformation where the immersed submanifolds tangent to $\Delta$ are "flattened" in the new coordinates.

Let us use that change of coordinates. Let the coordinate transformation be denoted $\Phi: U\to V.$ Define our new set of coordinates $$\begin{pmatrix}z_1(x) \\ \vdots \\z_n(x)\end{pmatrix} = z(x) = \Phi(x).$$ The sets tangent to $\Delta$ are immersed submanifolds given in the new coordinates $z$ by fixing the $n-k$ functions $z_{k+1}(x), \dots, z_n(x)$ to any constant. These are your $\lambda$ functions.

It helps to move to these new coordinates $z$ where $\Delta$ is flattened. Notice that the $\Phi$-related distribution of $\Delta$ is generated by the vector fields

$$\partial_{z_1},\dots,\partial_{z_k}.$$

Let us call this distribution (defined on the open set $V$) $\bar{\Delta}.$ Let us also denote the $\Phi$-related vector field of $f$ as $\bar{f}.$

All of this discussion has ignored the important property connecting $f$ and $\Delta.$ Now let us talk about that. Since $\Delta$ is involutive we also have $\bar{\Delta}$ is involutive. Further, if $[f, \Delta] \subseteq \Delta$ we have that $[\bar{f}, \bar{\Delta}] \subseteq \bar{\Delta}$ Recognize that since $\bar{\Delta}$ is generated by the constant, standard vector fields $\partial_{z_1}, \dots, \partial_{z_k}$ we can say

$$ \begin{aligned}\\ [\bar{f}, \partial_{z_1}] &= \sum_{\ell=1}^{k} c_{1,\ell} \partial_{z_\ell}\\ &~\vdots\\ [\bar{f}, \partial_{z_k}] &= \sum_{\ell=1}^{k} c_{k,\ell} \partial_{z_\ell} \end{aligned}$$

where $c_{i,j}$ are smooth functions on $V.$ At this point if you write $\bar{f}$ as a smooth function combination of the constant vector fields $\partial_{z_1},\ldots,\partial_{z_n}$ and combine with the above equation, what can you say about the coefficients that multiply the vector fields $\partial_{z_{k+1}},\dots, \partial_{z_n}$? Direct computation should verify that those coefficients cannot be functions of $z_{1}$ through til $z_k.$