I am working in geometry control field, fall last week on this exercice and I can't figure it out. I have a distribution $\mathscr{D}$ with $rank(\mathscr{D})=m+1$ in $\mathbb{R}^n$ with $n\leq 2m+1$. I know that there exists an involutive sub-distribution $\mathscr{L}\subset\mathscr{D}$ with rank $m$. I also know that the growth vector is $(m+1,n)$. (So I guess $\mathscr{D}+[\mathscr{D},\mathscr{D}]=T\mathbb{R}^n$, am I right?).


What I need is to find a local normal form $\varphi$ of $\mathscr{D}$ so that $\varphi\_\*\mathscr{D}=span\{f'\_1,\dots,f'\_n\}$ assuming $\mathscr{D}=span\{f\_1,\dots,f\_n\}$.


So far, I guess I can use the fact that $\mathscr{L}$ is involutive to write it $\mathscr{L}=span\{\frac{\partial}{\partial x^1},\dots,\frac{\partial}{\partial x^m}\}$ and then add a vector to build $\mathscr{D}$ but I don't know how. I guess Fröbenius can provide some help but can't find how. Someone has an idea ?


Thanks in advance.



Finding a local normal form regarding distribution rank properties
In what follows I am assuming (but I'm pretty sure) that your integrable sub-distribution $\mathscr{L}$ is supposed to be the same distribution as $\mathscr{F}$ that you mention later in your question but please correct me if I've misunderstood. Additionally, I'll assume we're working locally, i.e. on sufficiently small open subsets of a fixed point $p$ in $\mathbb{R}^{2m+1}$ (in order to use Frobenius theorem). Oh, and I'll be using the Einstein summation convention, too.


You are right that the Frobenius theorem is key here. Indeed, let $z, y^1,\ldots, y^m$ be the first integrals of $\mathscr{L}$ (i.e. $X(z)=X(y^i)=0$ for all $1\leq i\leq m$ and for all $X\in\mathscr{L}$). These help define a set of coordinates on $\mathbb{R}^{2m+1}$ on $(z, y^1,\ldots,y^m,x^1,\ldots, x^m)$ so that
$$ 
\mathscr{L}=\text{span}\{\partial\_{x^1},\ldots,\partial\_{x^m}\},
$$
as you had already noticed. Now, since the first derived system, call it $\mathscr{D}'$, (informally defined as $\mathscr{D}'=\mathscr{D}+\left[\mathscr{D},\mathscr{D}\right]$), then the growth vector of $(m+1,n)$ does indeed guarantee that $\mathscr{D}'=T\mathbb{R}^n$. Since $\mathscr{L}$ has rank $m$, this means that
$$
\mathscr{D}=\mathscr{L}\oplus\text{span}\{Z\}
$$
for some vector field $Z$ with no components from the directions in $\mathscr{L}$. In particular, since $\text{rank}\,\mathscr{D}'=2m+1$ and $\mathscr{L}$ is integrable, then the vector fields defined by $Y\_i=[\partial\_{x^i},Z]$ for all $1\leq i\leq m$ must not be in the distribution $\mathscr{D}$ (but they are of course in $\mathscr{D}'$). Writting $Z=A\partial\_{z}+B^i\partial\_{y^i}$ then one can make the argument that the first integrals of $\mathscr{L}$ may be written in such a way so that
\begin{equation\*}
\begin{aligned}
Y\_1&=\frac{\partial A}{\partial x^1}\partial\_z+\frac{\partial B^i}{\partial x^1}\partial\_{y^i}=\partial\_{y^1}\\
\,&\vdots\\
Y\_m&=\frac{\partial A}{\partial x^m}\partial\_z+\frac{\partial B^i}{\partial x^m}\partial\_{y^i}=\partial\_{y^m}.
\end{aligned}
\end{equation\*}
This is a system of linear 1st order PDE and we can quickly see a solution to be $A=A(y^1,\ldots,y^m)$ (i.e. no dependence on the $x^i$'s) and $B^i=x^i\tilde{B}^i(y^1,\ldots,y^m)$ for all $1\leq i\leq m$. Thus the distribution $\mathscr{D}$ can be written as
\begin{equation\*}
\mathscr{D}=\text{span}\{A\partial\_z+x^1\tilde{B}^1\partial\_{y^1}+\cdots+x^m\tilde{B}^m\partial\_{y^m},\partial\_{x^1},\ldots,\partial\_{x^m}\}
\end{equation\*}


Edit: If I recall correctly, this is essentially a special case of Robert Bryant's doctoral thesis, so you may also find that as an enlightening resource.


