Marked points on fibers of $TS^2$ can trace out a helix during (parallel) transport.

Question

It is claimed here that a mark on a rod parallelly transported by an observer moving along a geodesic $\gamma(t)$ on a smooth manifold $M$ (with a metric $g$) will trace out a helix, instead of a geodesic.

I am failing to see this on a simple case: Specifically, say that $M=\mathcal{S}^2$ (the unit sphere) $g=d\theta^2 + \sin^2 \theta d\phi^2$ (the standard metric) and $\nabla$ is any metric-compatible connection with Torsion (for example, the one where a vector is to be parallel transported if the angle between the vector and the latitude is kept fixed during the navigation - see here)

Since $\nabla$ is Metric Compatible, a parallelly transported Fiber $T_pM \cong \mathbb{R}^2$ will be a rotation. If we then move along $\partial_{\phi}$, it seems that $\partial_{\theta}$ (i.e. the "rod") should be facing in the same direction the whole time (say towards the North pole).

If that's the case, how is a curve of the form $\theta=constant$ a helix?

Edit: After the wonderful reply by @Jackozee Hakkiuz, it is almost certainly the case that the wikipedia statement is incorrect/incomplete. For starters, helicity seems to occur only when we have at least 3 dimensions. Further assumptions may also be necessary, so if anyone knows the full generality in which a parallelly transported rod traces a helix, feel free to share it with us.

Answer 1

The example doesn't work because the dimension is too low. Metric-compatible parallel transport maps "look like rotations" from the outside, so since parallel-transporting along your geodesics neccessarily fixes the velocity vector, that is enough to fix the whole tangent plane.

Consider $\mathbb R^3$ with the Euclidean metric and with Cartesian coordinates. If a connection is compatible with the metric, it needs to satisfy $$\Gamma^i_{jk} + \Gamma^k_{ji} = 0$$ Moreover, if we ask the coordinate lines to be geodesics, we get the additional restrictions $$\Gamma^i_{11}=\Gamma^i_{22}=\Gamma^i_{33}=0.$$ These two conditions allow $\Gamma^i_{jk}\neq 0$ only if $i,j,k$ are all different. For the connection to be flat, all components of the curvature tensor must vanish: $$R^s_{ljk}=\partial_k\Gamma^s_{jl}-\partial_j\Gamma^s_{kl}+\Gamma^s_{kr}\Gamma^r_{jl}-\Gamma^s_{jr}\Gamma^r_{kl}=0.$$ For simplicity, suppose the connection coefficients are constant, so that we only get the restriction $$\Gamma^s_{kr}\Gamma^r_{jl}=\Gamma^s_{jr}\Gamma^r_{kl}$$ This gets us down to one degree of freedom, say, $\Gamma^1_{23}=-\Gamma^3_{21}=A$. Now observe that the parallel transport of a vector $v$ along a curve $\gamma$ is given by $$\frac{dv^i}{dt} + \Gamma^i_{jk}\dot\gamma^jv^k=0.$$ In particular, note that the $\Gamma$ terms vanish if $\dot\gamma$ doesn't have $y$ component. Now, for a line $\gamma(t)=(a,t,c)$ parallel to the $y$ axis, we have $\dot\gamma=(0,1,0)$, so \begin{align} \frac{dv^1}{dt} &= -Av^3 \\ \frac{dv^2}{dt} &= 0 \\ \frac{dv^3}{dt} &= Av^1 \end{align} Whose solution is, you guessed, a helix!

Edit The main goal of this answer was to see that the statements in the wiki make some sense, at least in particular cases. Of course, this is just a particular space in which we can get a feel for what a helix looks like, but how would you detect "helicity" in an arbitrary manifold?. My guess would be that, in the general case, vectors which are parallel-transported using a connection with torsion will form helices when you compare them to the same vectors parallel-transported using the Levi-Civita connection, but proving this would need more work. Probably this would also work for nonzero curvature.