Is my reasoning about commutators of vectors right?

[JuanC97]

Hello guys, I have a question regarding commutators of vector fields and its pushforwards.

Let me define a clockwise rotation in the plane $\,\phi:\mathbb{R}^2\rightarrow\mathbb{R}^2 \,.\; [\,\partial_x\,,\,\partial_y\,]=0 \,$, $\;(\phi_{*}\partial_x) = \partial_r$ and $\,(\phi_{*}\partial_y) = \partial_\theta$

I'd like to compute $[\, \frac{x\partial_x+\,y\partial_y}{\sqrt{x^2+y^2}} \,,\, \frac{x\partial_y-\,y\partial_x}{\sqrt{x^2+y^2}} \,]$, i.e: $[\, (\cos\hspace{-0.025cm}\theta)\partial_x+(\sin\hspace{-0.025cm}\theta)\partial_y \,,\, -(\sin\hspace{-0.025cm}\theta)\partial_x+(\cos\hspace{-0.025cm}\theta)\partial_y \,]$.

We can identify this result with $[\, \partial_r \,,\, \partial_\theta \,] = [\, \phi_{*}\partial_x \,,\, \phi_{*}\partial_y \,]$ but we know that $[\, \phi_{*}\partial_x \,,\, \phi_{*}\partial_y \,] = \phi_{*}[\, \partial_x \,,\, \partial_y \,]$
hence $[\, \partial_r \,,\, \partial_\theta \,] = 0$. But... I don't know, this seems to contradict some ideas I had before about holonomic bases.

As far as I knew, if one defines a new basis $\{e_{\mu'}\}$ in terms of an holonomic basis $\{\partial_{\mu}\}$ such that $e_{\mu'} = A_{\mu'}^{\mu}\partial_\mu$, one should not expect this new basis to be holonomic but, following the reasoning I just showed you above, one could always identify the action of the matrix A with the action of the pushforward of some map, ending up again with a situation like $[\, e_{\mu'} \,,\, e_{\nu'} \,] = \phi_{*}[\, \partial_\mu \,,\, \partial_\nu \,] = 0$ which would explicitly contradict the original idea about what to expect (or not) about this kind of bases.

Could anyone, please, help me to understand where is the problem in this reasoning (if there's one)?.

 

[Orodruin]

et's define a clockwise rotation in the plane ϕ:R2→R2ϕ:R2→R2 \,\phi:\mathbb{R}^2\rightarrow\mathbb{R}^2 .

We know that [∂x,∂y]=0[∂x,∂y]=0 [\,\partial_x\,,\,\partial_y\,]=0 \,, also (ϕ∗∂x)=∂r(ϕ∗∂x)=∂r \,(\phi_{*}\partial_x) = \partial_r and (ϕ∗∂y)=∂θ(ϕ∗∂y)=∂θ \,(\phi_{*}\partial_y) = \partial_\theta

This does not look like a rotation to me. Rather a non-linear map where the Cartesian coordinates are mapped to the points as if they were polar coordinates.

As far as I knew, if I define a new basis {eμ′}{eμ′}\{e_{\mu'}\} in terms of an old holonomic basis {∂μ}{∂μ}\{\partial_{\mu}\} such that eμ′=Aμμ′∂μeμ′=Aμ′μ∂μ e_{\mu'} = A_{\mu'}^{\mu}\partial_\mu , I should not expect this new basis to be holonomic but, following the reasoning I just showed you, one could always identify the action of the matrix A as the action of the pushforward and one would always end up with a situation like
[eμ′,eν′]=ϕ∗[∂μ,∂ν]=0[eμ′,eν′]=ϕ∗[∂μ,∂ν]=0 [\, e_{\mu'} \,,\, e_{\nu'} \,] = \phi_{*}[\, \partial_\mu \,,\, \partial_\nu \,] = 0 contradicting the little I thought I knew.

You should not expect an arbitrary basis to be holonomic. However, it can be holonomic. In particular, when you choose a basis to be holonomic basis and then express that basis in another holonomic basis, it is obviously not going to change the fact that your chosen basis was holonomic from the beginning.

 

[JuanC97]

This does not look like a rotation to me. Rather a non-linear map where the Cartesian coordinates are mapped to the points as if they were polar coordinates.

Yeah, you're right, I should've called it a "change of coordinates" instead of "a clockwise rotation".

And... about this, what do you think?

hence [∂r,∂θ]=0

 

[Orodruin]

The Lie bracket of any two basis vectors from a holonomic basis is always zero.