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.
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 $x^3$.