# Parallel transport in two different polar coordinates

I need help with the basics of parallel transport. So I will write down what I have done in the plane $\mathbb{R}^2$ with non-cartesian coordinates, mixed with some small questions.

First, I use polar coordinates $(r,\theta)$ and the coordinate frame for the tangent space, which is the plane itself, given by partial derivatives $(\partial_r,\partial_\theta)$.

The operator $\partial_\theta$ generates rotations, in the sense that $e^{\phi\partial_\theta}$ is a rotation by angle $\phi$, independent of $r$. As a tangent vector, $\partial_\theta$ is not a unit vector, however, since $\partial_\theta=x\partial_y-y\partial_x$ so $||\partial_\theta||^2=x^2+y^2=r^2$. The vector $\partial_\theta$ increases in size with increasing $r$. This is in line with the fact that it generates rotation by a fixed angle, so the arc sweeped by such a rotation indeed increases linearly with $r$. [is this reasoning correct?]

The translated vector $\vec{U}$ will have coordinates related to those of the initial vector $\vec{V}$ by $U^\mu=V^\mu-V^\lambda\Gamma^{\mu}_{\nu\lambda}dx^\nu$.

Take a vector $\vec{V}=V^r\partial_r+V^\theta\partial_\theta$ with $V^r=V\cos\phi$ and $V^\theta=V\sin\phi/r$ for some $\phi$. If we transport it along $r$ by some amount $dr$, the new coordinates are $U^r=V^r$ and $U^\theta=V^\theta-V^\theta dr/r$. Therefore, $\Gamma^r_{ra}=0$, $\Gamma^\theta_{rr}=0$, $\Gamma^\theta_{r\theta}=1/r$.

If we transport it along $\theta$ by some amount $d\theta$, the coordinates become $U^r=V^r+V^\theta rd\theta$ and $U^\theta=V^\theta-V^r d\theta/r$. Therefore, $\Gamma^r_{\theta r}=\Gamma^\theta_{\theta \theta}=0$, $\Gamma^r_{\theta \theta}=-r$, $\Gamma^\theta_{\theta r}=1/r$.

Since $\Gamma^\mu_{\nu\lambda}=\Gamma^\mu_{\lambda\nu}$, there is no torsion.

Now, I want to change the coordinate system in the tangent space, to test my understanding. I want to use $(\partial_r,\partial_s)$, where $s=r\theta$ is an arc coordinate. Since $\partial_s=\frac{1}{r}\partial_\theta$, the operator $\partial_s$ has unit norm (it is the versor $\hat\theta$) and also produces rotations, but by a fixed arc instead of a fixed angle. Notice that $\partial_r$ and $\partial_s$ do not commute.

Now $\vec{V}=V^r\partial_r+V^s\partial_s$ with $V^r=V\cos\phi$ and $V^s=V\sin\phi$. When we transport it along $r$, the coordinates do not change at all, so $\Gamma^{\mu}_{r\lambda}=0$. When we transport it along $s$ by some amount $ds$ the coordinates become $U^r=V^r+V^sds/r$ and $U^s=V^s-V^r ds/r$ so $\Gamma^r_{\theta r}=\Gamma^s_{s s}=0$, $\Gamma^r_{s s}=-1/r$, $\Gamma^s_{s r}=1/r$.

This time $\Gamma$ is not symmetric, as $\Gamma^{s}_{rs}=0$ but $\Gamma^{s}_{sr}=1/r$. That means this way of doing it, which can be seen as a different connection on the plane from the usual, has torsion. Is this correct?

I was expecting that things would turn out the same in the end. I mean, I thought I could choose whatever coordinate system I wanted and parallel transport and torsion would be invariant notions. I mean, I DEFINED the final vetor to be identical to the original vector, so... how can there be torsion??

The torsion tensor is defined as $$T(X, Y) = \nabla_X Y - \nabla_Y X - [X, Y].$$ To say $\nabla$ is torsion free is to say that $T \equiv 0$.
As you noted, $[\partial_s, \partial_r] \ne 0$ and so $$\Gamma_{sr}^i \partial_i - \Gamma_{rs}^j \partial_j = T(\partial_s, \partial_r) + [\partial_s, \partial_r] = [\partial_s, \partial_s] \ne 0.$$ and so $\Gamma$ should not be symmetric in the lower two slots just as you have.
The Christoffel symbols $\Gamma$ are coordinate dependent while the torsion tensor $T$ is not. As you have discovered, vanishing $T$ is not equivalent to symmetric $\Gamma$. It is only equivalent to symmetric $\Gamma$ for commuting vector fields, which is true for $\partial_r, \partial_{\theta}$ but not for $\partial_r, \partial_s$.