Diffeomorphism in $\mathbb{R}^n$ I am stuck with this question for quite some time now. Please help.
Let $f$ and $g$ be $2$ linearly independent vector fields on $\mathbb{R}^n$. State with reasons if one can always get a diffeomorphism $h$ from an open set $U$ to an open set $V$ so that $Dh_x(f(x))=e_1$ and $ Dh_x(g(x))=e_2$ on $U$?
Thanks for any help.
 A: Hint: Notice that if there exists such a diffeomorphism $h$ then the Lie bracket $[f,g]$ is zero.
Without using Lie brackets, an equivalent viewpoint is to notice that the flows associated to the vector fields $e_1$ and $e_2$ commute; therefore, if a diffeomorphism $h$ exists, the flows associated to $f$ and $g$ commute. To get a contradiction, you may exhibit an example of two linearly independent vector fields whose flows do not commute.
For example, let $X(x,y)=(1,2x)$ a vector field so that integral curves be simple parabolics:

By considering the differential equation $\left\{ \begin{array}{ll} \partial_t \varphi(t,(x_0,y_0))= X(\varphi(t,(x_0,y_0))) \\ \varphi(0,(x_0,y_0))=(x_0,y_0) \end{array} \right.$, one may easily find that the associated flow is $\varphi(t,(x_0,y_0))=(t+x_0,t^2+2x_0t+y_0)$. 
Now let $Y(x,y)=(1,0)$ be another vector field, whose flow is $\phi(t,(x_0,y_0))=(t+x_0,y_0)$.
Finally, you can check that $\phi(r,\varphi(s,(x_0,y_0))) \neq \varphi(s, \phi(r,(x_0,y_0)))$ in general.
Remark: In fact, the converse is true, that is if $[f,g]=0$ then there exists a diffeomorphism $h$ sending $f$ and $g$ on $e_1$ and $e_2$ respectively. For that, show that $\psi : (r,s) \mapsto \varphi^r \circ \phi^s (0)$ (where $\varphi$ and $\phi$ are the flows associated to $f$ and $g$ respectively) is a local diffeomorphism sending $e_1$ and $e_2$ on $f$ and $g$ respectively (using the fact that $\varphi$ and $\phi$ commute).

I add a lemma to justify that the flows associated to $f$ and $g$ must commute:
Lemma: Let $f : U \subset \mathbb{R}^n \to V \subset \mathbb{R}^n$ be a diffeomorphism and let $X_1, X_2$ (resp. $Y_1, Y_2$) be two vector fields over $U$ (resp. $V$) such that $df(x) \cdot X_i(x)= Y_i(f(x))$ for all $x \in U$. If $\varphi_{X_1}$ and $\varphi_{X_2}$ commute then $\varphi_{Y_1}$ and $\varphi_{Y_2}$ commute.
Proof. First, notice that if $\gamma$ is an integral curve for $X_i$, that is $\gamma'(t)=X_i(\gamma(t))$ for all $t \in \mathbb{R}$, then $f \circ \gamma$ is an integral curve for $Y_i$: for all $t \in \mathbb{R}$, $$(f \circ \gamma)'(t)= df(\gamma(t)) \cdot \gamma'(t) = df(\gamma(t)) \cdot X_i(\gamma(t)) = Y_i(f \circ \gamma(t)).$$
Therefore, $f \circ \varphi_{X_i}(t,x_0)= \varphi_{Y_i}(t,f(x_0))$ or $\varphi_{Y_i}(t,y_0)=f \circ \varphi_{X_i} (t,f^{-1}(y_0))$.
By hypothesis, $$\varphi_{X_1}(r,\varphi_{X_2}(s,x_0))= \varphi_{X_2}(s, \varphi_{X_1}(r,x_0))$$ hence $$f \circ \varphi_{X_1}(r,\varphi_{X_2}(s,x_0))= f \circ \varphi_{X_2}(s, \varphi_{X_1}(r,x_0)). \hspace{1cm} (1)$$
But $$f \circ \varphi_{X_1}(r,\varphi_{X_2}(s,x_0))= f \circ \varphi_{X_1}(r, f^{-1} \circ f \circ \varphi_{X_2}(s,f^{-1} \circ f(x_0)))= \varphi_{Y_1}(r, \varphi_{Y_2}(s,f(x_0))),$$
and in the same way $$f \circ \varphi_{X_2}(s,\varphi_{X_1}(r,x_0))= \varphi_{Y_2}(s, \varphi_{Y_1}(r,f(x_0))).$$
Therefore, $(1)$ becomes $$\varphi_{Y_1}(r,\varphi_{Y_2}(s,f(x_0))) = \varphi_{Y_2}(s, \varphi_{Y_1}(r,f(x_0))). \hspace{1cm} \square$$
A: Here is an "undergraduate" approach for the case $n=2$.
Such a map $h$ does not always exist. There are integrability conditions to the effect that the curls of certain auxiliary fields must vanish.
Denote the independent variables by $x$ and $y$, and assume $h$ in the form
$$h:\quad (x,y)\mapsto\bigl(u(x,y),v(x,y)\bigr)\ .$$
Let $f=(f_1,f_2)$, $g=(g_1,g_2)$, and put $J:=(f_1g_2-f_2g_1)\ne0$.
The condition $dh.f=e_1$ for all $(x,y)$ unpacks to
$$u_x f_1+u_y f_2=1,\quad v_x f_1 +v_y f_2=0\ ,$$
and similarly the condition $dh.g=e_2$ for all $(x,y)$ unpacks to
$$u_x g_1+u_y g_2=0,\quad v_xg_1+v_y g_2=1\ .$$
Solving the last four equations for the partial derivatives of $u$ and $v$ we obtain
$$u_x={g_2\over J},\quad u_y=-{g_1\over J};\qquad v_x=-{f_2\over J},\quad v_y={f_1\over J}\ .$$
These equations define the gradients $\nabla u$, $\nabla v$ of the scalar functions $u$ and $v$. It is well known that such functions (locally) exist iff the implied mixed derivatives are equal. This means that a necessary and sufficient condition for the existence of $h$ is that
$$\left({g_2\over J}\right)_y\equiv\left(-{g_1\over J}\right)_x,\qquad \left(-{f_2\over J}\right)_y\equiv\left({f_1\over J}\right)_x\ ,\tag{1}$$
which is the same as requiring that
$${\rm curl}\left({g_2\over J},-{g_1\over J}\right)\equiv0,\qquad {\rm curl}\left(-{f_2\over J},{f_1\over J}\right)\equiv0\ .$$
When these conditions are fulfilled then $u$ and $v$ can be obtained by simple integration. Furthermore, since
$${\rm det}(dh)={\rm det}\left[\matrix{u_x&u_y\cr v_x& v_y\cr}\right]={1\over J}\ne0\ ,$$
it follows that $h$ is locally a diffeomorphism.
That the conditions $(1)$ are not void is exemplified by the following simple example: Put
$$f(x,y):=(1,0), \quad g(x,y):=(x,1)\ .$$
Then $J(x,y)\equiv1$ and therefore $(g_2/J)_y=0$, $(-g_1/J)_x=-1$.
