Example of two linearly independent, nowhere vanishing vector fields in $\mathbb{R}^{2}$ I knew that two linearly independent and nowhere-vanishing vector fields provide a basis for the tangent space at each point in $\mathbb{R}^{2}$.
Is it necessary that these two vector fields commute? Would you give me an example for these two vector fields?
 A: Let us take the first vector field to be given by $V_{1}=\partial _{x}$.
Any other vector field will be given by $V_{2} = A(x, y)\partial _{x} + B(x, y)\partial _{y}$ . The commutator will be given by
$[V_{1} , V_{2} ] = [\partial _{x} A(x, y)]\partial _{x} + [\partial _{x} B(x, y)]\partial _{y}$
We want this not to vanish. So either one of A, B must depend on x. Set
$V_{2} =x \partial _{x} + \partial _{y}$ .  $V_{1}$ ,  $V_{2}$ are nowhere vanishing, and their commutator is nowhere vanishing as well.
A: Consider the vectors
$$\partial_{\theta}=-y\partial_{x}+x\partial_{y}$$
and
$$\partial_{r}=\frac{x}{\sqrt{x^{2}+y^{2}}}\partial_{x}+\frac{y}{\sqrt{x^{2}+y^{2}}}\partial_{y}$$
Then
$$\partial_{\theta}\partial_{r}-\partial_{r}\partial_{\theta}\neq0$$
A: I think, @Alex answer is not correct. A slight change is required to Alex's answer.
For the vectors,
$$\partial_{\theta}=-y\partial_{x}+x\partial_{y}$$
and
$$\partial_{r}= x\partial_{x}+y\partial_{y},$$
$$\partial_{\theta}\partial_{r}-\partial_{r}\partial_{\theta}\neq0,$$ whereas for the vectors,
$$\partial_{\theta}=-y\partial_{x}+x\partial_{y}$$
and
$$\partial_{r}=\frac{x}{\sqrt{x^{2}+y^{2}}}\partial_{x}+\frac{y}{\sqrt{x^{2}+y^{2}}}\partial_{y},$$
$$\partial_{\theta}\partial_{r}-\partial_{r}\partial_{\theta} = 0.$$
Hence, the first set of vectors is the correct answer, not the second.
