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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?

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    $\begingroup$ What are "these commutators"? No commutators or operators, for that matter, had been mentioned before that sentence. Otherwise. Vector fields $(1,0)$ and $(1,0)$, independent of $x,y$, are surely an example and one can easily write down infinitely many others, pretty much arbitrary fields, right? What is the real question? $\endgroup$ Nov 30, 2012 at 15:57
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    $\begingroup$ $\partial/\partial x +f(x,y)\partial/\partial y$ and $\partial/\partial y$ are linearly independent vector fields for any choice of smooth function $f$. We can choose $f$ such that commutators do not vanish. For $f=0$ commutators vanish. $\endgroup$
    – user10001
    Nov 30, 2012 at 16:04
  • $\begingroup$ @Luboš Motl I read somewhere that these fields cannot be a coordinate basis since the commutator does't vanish (like what we have in QM). But I don't know why. $\endgroup$
    – Fatima
    Nov 30, 2012 at 16:05
  • $\begingroup$ @Fatima: the commutator of those fields certainly vanishes: $[\partial_{x}, \partial_{y}] = 0$ trivially. $\endgroup$ Nov 30, 2012 at 17:04
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    $\begingroup$ I think @Fatima might be asking if it is necessary that the Lie derivative $ \mathcal{L}_X(Y) = [X,Y]$ vanishes in order for them to be a coordinate basis? In which case, no, as long as they're non vanishing and independent at each point, you're fine. $\endgroup$
    – twistor59
    Nov 30, 2012 at 18:04

3 Answers 3

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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.

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  • $\begingroup$ Did you mean $V_2=x\partial_x+\partial_y$ ? I mean your $V_1$ and $V_2$ are not independent everywhere (specifically they are not independent at $x=0$). $\endgroup$
    – user10001
    Nov 30, 2012 at 22:04
  • $\begingroup$ @dushya Yes, you are right. $\endgroup$
    – Fatima
    Dec 1, 2012 at 20:23
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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 $$\partial_{\theta}\partial_{r}-\partial_{r}\partial_{\theta}\neq0$$

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    $\begingroup$ Those fields are not nowhere vanishing. $\endgroup$ Nov 30, 2012 at 16:49
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    $\begingroup$ I think my favourite vector fields would be $\frac{\partial}{\partial x}$ and $\frac{\partial}{\partial y}$ ! $\endgroup$
    – twistor59
    Nov 30, 2012 at 17:59
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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.

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