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 $\partial_r, \partial_s$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.