Parallel Transport along Great Circle of 2-sphere I have come across the topic of parallel transport in an introductory book to General Relativity, and am unsure about what distinguishes parallel transport along a geodesic from an arbitrary path. Is the transported vector have any "special" features when moved across the geodesic?
In particular, I am interested in the example of transport on a 2-sphere. If transporting a vector from points $A$ to $B$ along a circular arc, would that be equivalent to rotating the embedded vectors about the axis of circle (the rotation would be carried out in 3D space). Would this be true just for transport about the great circle, or for any circular path on the sphere's surface? If so, I would appreciate an explanation of why this works or a reference that I could look up.
 A: One definition of a geodesic is that the tangent vector to the geodesic is parallel-transported along the geodesic. In other words, if the curve is a geodesic (a great circle on the two-sphere) and ${\bf t}$ is the tangent at some point, and we parallel transport ${\bf t}$ to another point on the gedesic, the transported vector  will coincide with the tangent vector at the new point. For the "shortest path" and ``parallel-transported tangent vector" properties to coincide, one needs that the connection be the Levi-Civita torsion-free one. 
A: The only special property of parallel transport along a geodesic is that it is angle-preserving.
Let $\gamma:\mathbb{R}\rightarrow M$ be a geodesic, let $T=\dot{\gamma}$ be the tangent vector and let $V$ be a vector field along $\gamma$ that is parallel transported, so $\nabla_TV=0$.
Because $\gamma$ is a geodesic, $T$ is also parallel ($\nabla_TT=0$). The Levi-Civita connection is metric compatible, so $$ \nabla_T(g(T,V))=g(\nabla_TT,V)+g(T,\nabla_TV)=0 \\ \nabla_T(g(T,T))=\nabla_T(g(V,V))=0, $$ which incidentally implies that $$=\nabla_T\cos\alpha=\nabla_T\frac{g(T,V)}{\sqrt{g(T,T)g(V,V)}}=0,$$ so during parallel transport, the angle between $V$ and $T$ will stay constant.
Had $\gamma$ not been a geodesic, the relation $\nabla_TT=0$ would no longer be true, thus parallel transport would not preserve the angle between the transported vector and the curve's tangent vector.

Parallel transport along a closed loop will always results in rotation in Riemannian geometry. Because the metric $g$ is preserved, the parallel transport map $P_\gamma$ along a closed loop must be a $P_\gamma:T_pM\rightarrow T_pM$ linear transformation. Which, because it preserves the metric, must be a linear isometry of $g_p$, but $g_p$ is just a usual inner product on $T_pM$, so $P_\gamma\in SO(T_pM)$ is a rotation.
In particular, this is one reason why the curvature tensor in two dimensions has one independent component only. $R^{ab}_{\ \ \ \ cd}$ can be seen as a map from parallelograms (in $cd$) to infinitesimal rotations (in $ab$) (infinitesimal rotations are elements of $\mathfrak{o}(T_pM)$ and as such are antisymmetric matrices). In 2 dimensions, there is only one linearly independent parallelogram, so the  $cd$ indices have only one possible value and there is only one axis of rotation, so rotations can be described by a single number, hence $R^{ab}_{\ \ \ \ cd}$ is a map from a one dimensional space to another one dimensional space.
