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there is this problem that keeps bothering me.

Consider the following local frame of the tangent space of $\mathbb{R}^2\setminus \{0\}$:

$$ E_1 = \frac{x}{\sqrt{x^2+y^2}} \partial_x + \frac{y}{\sqrt{x^2+y^2}}\partial_y, E_2 = \frac{-y}{\sqrt{x^2+y^2}}\partial_x + \frac{x}{\sqrt{x^2+y^2}}\partial_y.$$

Geometrically, one can see this to be the unit normal vectors tangent to the concentric circles and the (outward pointing) normal.

One can show that these vector fields do not commute. Hence, in theory, we cannot find a smooth chart $(U,\phi=(x_1,...,x_n))$ such that locally $E_j = \partial_{x_j}$.

We can find a neighbourhood $U$ such that the following ARE smooth and invertible: $f_1(x,y) = \frac{-1}{2}\log(x^2+y^2)$ and $f_2(x,y) = -\tan^{-1}(\frac{x}{y})$. Thus, they would give us diffeomorphisms, and according to my calculations, they DO induce these vector fields, meaning that they are a chart bringing it to a local coordinate frame... Where did I go wrong in this reasoning/calculation?

Any advice is greatly appreciated.

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I'm not quite sure why you have the negative signs, but in both cases, you're off by a factor of $r$. $-df_1 = \dfrac{dr}r$ and $-df_2=d\theta$, so the dual vector fields are $r\dfrac{\partial}{\partial r}$ and $\dfrac{\partial}{\partial\theta}$, but $E_1=\dfrac{\partial}{\partial r}$ and $E_2=\dfrac1r\dfrac{\partial}{\partial \theta}$.

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  • $\begingroup$ Thank you very much! $\endgroup$ – Jonas Jan 5 at 22:29
  • $\begingroup$ I assume that you can't find $f_1,f_2$ such that they give $E_1$ and $E_2$ then, right? :) $\endgroup$ – Jonas Jan 5 at 22:31
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    $\begingroup$ Yes, you were right. If you look at the differential forms version, $dr$ is exact, but $r\,d\theta$ is not. Scaling by $1/r$ makes them both exact. $\endgroup$ – Ted Shifrin Jan 5 at 22:33
  • $\begingroup$ I have no idea how I could have forgotten about the differential form interpretation... Thanks! $\endgroup$ – Jonas Jan 5 at 22:35
  • $\begingroup$ You're very welcome! :) $\endgroup$ – Ted Shifrin Jan 5 at 22:36

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