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I came across the following conclusion in a textbook, but can't really understand it. I would be grateful if anyone could elaborate:

Assume that we have two linearly independent vector fields $V_{1},V_{2}$ on $\mathbb{R}^{3}$. Then,for a function $f$ on $\mathbb{R}^{3}$,

if $[V_{1},V_{2}]=0$ and $V_{1}f=0$, $V_{2}f=0$, it follows from the Frobenius theorem that $f$ is a function of one variable.

I only have a basic understanding of Frobenius theorem, namely, if the vector fields $V_{i}$ are linearly independent and commute, then a PDE system $V_{i}f=0$ is solvable. But how do I get "function of one variable" statement?

Thank you for any help.

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The Frobenius theorem implies that you can find a chart which has $V_1=\frac{\partial}{\partial x^1}$ and $V_2=\frac{\partial}{\partial x^2}$, so locally your $f$, which is a function of $(x^1,x^2,x^3)$, is such that $$\frac{\partial f}{\partial x^1}=\frac{\partial f}{\partial x^2}=0.$$ Of course, this means that (in an appropriate neighborhood of each point) $f$ depends only on $x^3$.

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    $\begingroup$ Thank you for the answer. As a follow up: would the same be true if the commutator of $V_{1}$ and $V_{2}$ was a linear combination of $V_{1}$ and $V_{2}$? $\endgroup$
    – GregVoit
    May 29, 2016 at 9:17
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    $\begingroup$ @GregVoit, in that case $V_1$ and $V_2$ are everywhere tangent to a family of surfaces (the distribution is integrable), and $f$ is constant on each of those surfaces. You can still choose coordinates so that these surfaces are given by $x^3=\text{constant}$ and $f$ will depend just on $x^3$. $\endgroup$ May 29, 2016 at 14:59
  • $\begingroup$ as a follow up question: is it equivalent to say that there is a coordinate system, say $(x_{1},x_{2},x_{3})$, such that $V_{1}=\frac{\partial}{\partial x_{1}}$ and $V_{2}=g(x_{1},x_{2},x_{3})\frac{\partial}{\partial x_{1}}+h(x_{1},x_{2},x_{3})\frac{\partial}{\partial x_{2}}$? @TedShifrin $\endgroup$
    – GregVoit
    Jun 2, 2016 at 14:20
  • $\begingroup$ or does it have to be in the exact form $V_{1}=\frac{\partial}{\partial x_{1}}$ and $V_{2}=\frac{\partial}{\partial x_{2}}$? @TedShifrin $\endgroup$
    – GregVoit
    Jun 2, 2016 at 14:21
  • $\begingroup$ @GregVoit, the former. The latter can happen only when the Lie bracket is $0$. $\endgroup$ Jun 2, 2016 at 17:16

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