The original references are:

**W.-L. Chow,** Über Systeme von linearen partiellen Differentialgleichungen erster Ordnung. *Math. Ann.* 117 (1939), 98–105

**P. K. Rashevsky,** Any two points of a totally nonholonomic space may be connected by an admissible line.
*Uch. Zap. Ped. Inst. im. Liebknechta, Ser. Phys. Math.* 2 (1938), 83–94 (in Russian).

There are several different proofs of this result. I learned it from Proposition III.4.1 in:

**N. Th Varopoulos, L. Saloff-Coste, T. Coulhon**, *Analysis and Geometry on Groups*. Cambridge University Press.

The proof is quite concise, but not too difficult. Here is my own version of the proof from that book.

## Chow–Rashevsky theorem

Let $Z$ be a smooth vector field and $Z_{t}$ the local $1$-parameter
family of diffeomorphisms associated with $Z$. Fix $f\in C^{\infty}$
and a point $m$. Then the function $h(t) = f(X_{t}(m))$ is smooth
and $h^{(k)}(0) = (X^{k}f)(m)$. Hence the Taylor series for
$h$ at $t=0$ is given by
\begin{equation}
(1)\qquad \sum_{k=0}^{\infty} X^{k}f(m) \frac{t^{k}}{k!},
\end{equation}
which means
$$
h(t) = \sum_{k=0}^{i} X^{k}f(m) \frac{t^{k}}{k!} + O(t^{i+1})
\qquad {\rm as} \ t\to\infty.
$$
We will use the formal expresion $(e^{tX}f)(m)$ to denote (1).

Let $Z_{1},\ldots,Z_{k}$ be smooth vector fields. Let $f\in C^{\infty}$.
Fix a point $m$ and define
$$
H(t_{1},\ldots,t_{k}) =
f(Z_{1,t_{1}}\circ Z_{2,t_{2}} \circ\cdots \circ Z_{k,t_{k}}(m)).
$$
Note that
$$
\frac{\partial^{m_{1}}}{\partial t_{1}^{m_{1}}}
H(0,t_{2},\ldots,t_{k}) =
(Z_{1}^{m_{1}} f) (Z_{2,t_{2}} \circ\cdots\circ Z_{k,t_{k}}(m)).
$$
Taking then the derivatives with respect to $t_{2},\ldots,t_{k}$
yields
$$
\frac{\partial^{m_{1}+\ldots+m_{k}}}{\partial
t_{1}^{m_{1}}\ldots \partial t_{k}^{m_{k}}}
H(0,\ldots,0) =
(Z_{k}^{m_{k}}\ldots Z_{1}^{m_{1}}f)(m).
$$
Hence the Taylor series for $H$ is given by
$$
\sum_{m_{1}=0}^{\infty} \ldots \sum_{m_{k}=0}^{\infty}
\frac{t_{1}^{m_{1}}\ldots t_{k}^{m_{k}}}{m_{1}!\ldots m_{k}!}
(Z_{k}^{m_{k}}\ldots Z_{1}^{m_{1}}f)(m),
$$
which will be formally denoted by
$$
(e^{t_{k}Z_{k}}\ldots e^{t_{1}Z_{1}} f)(m).
$$
Before we prove the Chow--Rashevsky's theorem we show how to use
the above Taylor's formula to prove the following theorem.

**Theorem.** Let $G$ be a Lie group. Then $$ \exp(tX)\exp(tY) = \exp\Big( t(X+Y) + \frac{t^{2}}{2}[X,Y] +
O(t^{3})\Big). $$

*Proof.* Note that
$\exp(tX)\exp(sY)$ is the same as $Y_{s}\circ X_{t}(e)$
($e$ denotes the neutral element of $G$), because
$s\mapsto \exp(tX)\exp(sY)$ is the integral curve of $Y$
passing through $\exp(tX)$ at $s=0$. Thus the Taylor series for
$f(\exp(tX)\exp(sY))$ is
$e^{tX}e^{sY} f(e)$ and hence the Taylor series for
$h(t) = f(\exp(tX)\exp(tY))$ at $t=0$ is
\begin{eqnarray*}
e^{tX}e^{tY}f(e)
& = &
\Big(1+tX + \frac{t^{2}}{2}X^{2} + O(t^{3}) \Big)
\Big(1+tY + \frac{t^{2}}{2}Y^{2} + O(t^{3}) \Big)f(e) \\
& = &
f(e) + t(X+Y)f(e) +
t^{2}\Big(\frac{X^{2}}{2} + XY + \frac{Y^{2}}{2}\Big)f(e)
+ O(t^{3})
\end{eqnarray*}
Now there is a smooth function $t\mapsto Z(t)$, $Z(0) = 0$ such that
$$
\exp(tX)\exp(tY) = \exp(Z(t))
$$
for small $t$. We can write $Z(t) = tZ_{1}+t^{2}Z_{2} + O(t^{3})$.
Since $f(\exp(tW)) = f(e) + tWf(e) + \frac{t^{2}}{2}W^{2}f(e) + O(t^{3})$
and since obviously $f(A(t) + O(t^{3})) = f(A(t)) + O(t^{3})$,
we have
$$
f(\exp(Z(t)) =
f(\exp(t(Z_{1}+tZ_{2}))) + O(t^{3}).
$$
Fix $s$ and then
$$
f(\exp(t(Z_{1}+sZ_{2}))) =
f(e) + t(Z_{1}+sZ_{2})f(e) +
\frac{t^{2}}{2}(Z_{1}+sZ_{2})^{2}f(e) + O(t^{3}) = A
$$
Now substituting $s=t$ yields
$$
A = f(e) + tZ_{1}f(e) + t^{2}Z_{2}f(e)
+ \frac{t^{2}}{2}Z_{1}^{2}f(e) + O(t^{3})).
$$
Taking coordinate functions as $f$ and comparing the Taylor series yields
$$
Z_{1} = X+Y,\qquad Z_{2} + \frac{Z_{1}^{2}}{2} =
\frac{X^{2}}{2} + XY + \frac{Y^{2}}{2}.
$$
Hence $Z_{2} = \frac{1}{2}[X,Y]$, which implies
$$
Z(t) = t(X+Y) + \frac{t^{2}}{2}[X,Y] + O(t^{3}),
$$
and hence the theorem follows. $\Box$

As an immediate consequence we obtain

**Corollary** $\exp(-tX)\exp(-tY)\exp(tX)\exp(tY) = \exp(t^{2}[X,Y] + O(t^{3})).$

We will see now that the corollary holds for arbitrary smooth
vector fields, not necessarily on the Lie group.

**Corollary** $Y_{t}\circ X_{t}\circ Y_{-t} \circ X_{-t}(m)
= m+ t^{2}[X,Y]_{m} + O(t^{3})$.

*Proof.* The Taylor series for
$h(t) = f(Y_{t}(X_{t}(Y_{-t}(X_{-t}(m)))))$ is
\begin{eqnarray*}
e^{-tX}e^{-tY}e^{tX}e^{tY} f(m)
& = &
(1 - tX + \frac{t^{2}}{2}X^{2} + O(t^{3}))
(1 - tY + \frac{t^{2}}{2}Y^{2} + O(t^{3})) \times \\
& \times &
(1 + tX + \frac{t^{2}}{2}X^{2} + O(t^{3}))
(1 + tY + \frac{t^{2}}{2}Y^{2} + O(t^{3})) f(m) \\
& = &
(1 + t^{2}[X,Y] + O(t^{3})) f(m).
\end{eqnarray*}
Now we can turn to the main subject of the section, namely the connectivity
theorem of Chow and Rashevsky.

**Theorem (Chow-Raschevsky)** Let $\Omega\subset\mathbb{R}^{n}$ be an open domain and let
$X_{1},\ldots,X_{k}$ be smooth vector fields satisfying
H"ormander's condition i.e. for some positive integer $d$ comutators
of length less than or equal to $d$ span the tangent space
$\mathbb{R}^{n}$ at every point of $\Omega$. Then every two points in
$\Omega$ can be connected by an admissible curve. Moreover for any
compact set $K\subset\Omega$ there is a constant $C>0$ such that
\begin{equation}
(2)\qquad \rho(x,y) \leq C|x-y|^{1/d} \qquad
\mbox{for all $x,y\in K$}. \end{equation}

**Remark.** The estimate (2) is due to Nagel, Stein and Waigner.

*Proof.* Let $Y_{1},\ldots,Y_{p}$ be smooth vector fields.
Fix $m\in\Omega$. Define by induction
\begin{eqnarray*}
C_{1}(t) & = & Y_{1,t}(m) \\
C_{p}(t) & = &
C_{p-1}(t)^{-1}\circ Y_{p,-t}\circ C_{p-1}(t) \circ Y_{p,t}(m).
\end{eqnarray*}
Recall that $Y_{j,t}$ denotes the local family of diffeomorpisms associated to
$Y_j$.
Since both $C_{p}(t)$ and $C_{p}(t)^{-1}$ are compositions of diffeomorphisms
$Y_{j,\pm t}$ one easily obtaines that the Taylor series for
$f(C_{p}(t))$ and $f(C_{p}(t)^{-1})$ are given by
$\widetilde{c}_{p}(t)f(m)$ and $\widetilde{c}_{p}(t)^{-1}f(m)$ where
$\widetilde{c}_{p}(t)$ is a formal series defined by induction as follows
\begin{eqnarray*}
\widetilde{c}_{1}(t) & = & e^{tY_{1}} \\
\widetilde{c}_{p}(t) & = & e^{tY_{p}} \widetilde{c}_{p-1}(t)
e^{-tY_{p}} \widetilde{c}_{p-1}(t)^{-1}.
\end{eqnarray*}

It is easy to prove by induction that
\begin{equation}
(3)\qquad
\widetilde{c}_{p}(t) = 1 + t^{p} [Y_{p},[Y_{p-1},[\ldots,Y_{1}]\ldots]
+ O(t^{p+1}),
\end{equation}
and hence
$$
\widetilde{c}_{p}(t)^{-1} = 1 - t^{p} [Y_{p},[Y_{p-1},[\ldots,Y_{1}]\ldots]
+ O(t^{p+1}).
$$
Indeed, for $p=1$, (3) is obvious.
Assume it is true for $p$ and we prove it for $p+1$.
We have
\begin{eqnarray*}
\widetilde{c}_{p+1}(t)
& = &
e^{tY_{p+1}}\widetilde{c}_{p}(t) e^{-tY_{p+1}}\widetilde{c}_{p}(t)^{-1} \\
& = &
e^{tY_{p+1}} (\widetilde{c}_{p}(t) - 1)e^{-tY_{p+1}}
\widetilde{c}(t)^{-1} + \widetilde{c}_{p}(t)^{-1} \\
& = &
(1 + tY_{p+1})(\widetilde{c}_{p}(t)-1)(1 - tY_{p+1})
\widetilde{c}_{p}(t)^{-1} + \widetilde{c}_{p}(t)^{-1} + O(t^{p+2}) \\
& = &
(\widetilde{c}(t)-1)\widetilde{c}_{p}(t)^{-1} +
t^{p+1}[Y_{p+1},[Y_{p},[\ldots,Y_{1}]\ldots] +
\widetilde{c}_{p}(t)^{-1} + O(t^{p+2}) \\
& = &
1 + t^{p+1}[Y_{p+1},[Y_{p},[\ldots,Y_{1}]\ldots] + O(t^{p+2}).
\end{eqnarray*}
The claim is proved.

Hence the Taylor series of $f(C_{p}(t))$ at $t=0$ begins with
$$
f(m) + t^{p}[Y_{p},[Y_{p-1},[\ldots,Y_{1}]\ldots]f(m) + O(t^{p+1})
$$
and the Taylor series of $f(C_{p}(t)^{-1})$ at $t=0$ begins with
$$
f(m) - t^{p}[Y_{p},[Y_{p-1},[\ldots,Y_{1}]\ldots]f(m) + O(t^{p+1}).
$$
Now if $F_{1}$ and $F_{2}$ are two $C^{\infty}$ functions with Taylor series
$F_{1}(t) = a + bt^{p} +\ldots$ and
$F_{1}(t) = a - bt^{p} +\ldots$ then it is easy to see that the function
$$
G(t) = \left\{
\begin{array}{cc}
F_{1}(t^{1/p}) & \mbox{if $t\geq 0$} \\
F_{2}((-t)^{1/p}) & \mbox{if $t<0$}
\end{array}
\right.
$$
is $C^{1}$ in the neighborhood of $0$ and $G'(0)=b$.

Taking $F_{1}(t)=f(C_{p}(t))$ and $F_{2}(t) = f(C_{p}(t)^{-1})$,
where $f$ are all coordinate functions we conclude that the function
$$
\phi(t) = \left\{
\begin{array}{cc}
C_{p}(t^{1/p}) & \mbox{if $t\geq 0$} \\
C_{p}((-t)^{1/p})^{-1} & \mbox{if $t<0$}
\end{array}
\right.
$$
is a $C^1$ path passig through $m$ at $t=0$ with
$\phi'(0) = [Y_{p},[Y_{p-1},[\ldots,Y_{1}]\ldots]$.

Let $V_{1},\ldots,V_{n}$ be a basis of $\mathbb{R}^{n}=T_{m}\Omega$ arising from
H"ormander's condition i.e.,
$$
V_{i} = [X_{i,p_{i}},[X_{i,p_{i}-1},[\ldots,X_{i,1}]\ldots],
$$
where $i=1,2,\ldots,n$, $p_{i}\leq d$ and $X_{i,l}\in\{ X_{1},\ldots,X_{k}\}$.
Let $\phi_{i}(t)$ be a $C^1$ path defined as above for
$Y_{1}=X_{i,1},\ldots,Y_{p_{i}} = X_{i,p_{i}}$. Then $\phi_{i}'(0)=V_{i}$.
Finally define $\Phi$ by
$$
\Phi(\theta) = \phi_{1}(\theta_{1})\circ \cdots \phi_{n}(\theta_{n}),
\qquad \theta = (\theta_{1},\ldots,\theta_{n}).
$$
Then $\Phi$ is a $C^1$ mapping from a neighborhood of $0$ in $\mathbb{R}^{n}$
to $\Omega$. Since $\partial\Phi/\partial\theta_{i}(0)=\phi_{i}'(0)=V_{i}$
we conclude that $\Phi$ is a diffeomorphism in a neighborhood of $0$.
This implies that any point in the neighborhood of $m=\Phi(0)$ can be
connected to $m$ by an admissible curve.

More procisely $\phi_{i}(\theta_{i})$ is a composition of diffeomorphisms
of the form $X_{j,\pm|\theta_{i}|^{1/p_{i}}}$. Hence denoting the composition
by $\prod$ we may write
\begin{equation}
(4)\qquad
\Phi(\theta) = \left( \prod_{i=1}^{n} \prod_{\alpha=1}^{M_{i}}
X_{i,j_{\alpha},\pm|\theta_{i}|^{1/p_{i}}} \right)(m).
\end{equation}
Assume that $|\theta|\leq 1$. For any $x$,
the two points $x$ and $X_{i,j_{\alpha},\pm|\theta_{i}|^{1/p_{i}}(x)}$
can be connected by an admissible curve --- an integral curve of
$X_{i,j_{\alpha}}$ and hence the Carnot--Carath'eodory distance between
these two pints is no more than $|\theta_{i}|^{1/p_{i}}\leq |\theta|^{1/d}$.
Now we can move from $m$ to $\Phi(\theta)$ on such admissible curves and hence
\begin{equation}
(5)\qquad
\rho(\Phi(\theta),m) \leq C_{1}|\theta|^{1/d}
\approx C_{2}|\Phi(\theta)-m|^{1/d},
\end{equation}
where $C_{1}=\sum_{i=1}^{n}M_{i}$ equals the number of integral curver we use to
join $m$ with $\Phi(\theta)$ (see (4)). We employed also the fact
that $|\theta|\approx |\Phi(\theta) - m|$ which is a consequence of the fact
that $\Phi$ is a diffeomorphism.

Since we can connect all the points in a neighborchood of any point it easily
follows that we can connect any two points in $\Omega$.
The estimate (2) follows from (5).
$\Box$