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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.

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  • $\begingroup$ In general, the relationship between parallel transports along different paths depends on the curvature of the manifold. If the manifold is simply connected, then the condition which ensures parallel transport is path-independent is that the curvature vanishes. The 2-sphere is simply connected, and if you are using the metric induced from the usual embedding into $\mathbb{R}^3$ (as the unit sphere) then the curvature is the same everywhere (it's not zero!). The angle between two parallel transported vectors will be something like the solid angle spanned by the two paths. $\endgroup$ – Sean Pohorence Apr 30 '17 at 15:25
  • $\begingroup$ Would Mathematics be a better home for this question? $\endgroup$ – Qmechanic May 1 '17 at 8:14
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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.

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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.

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  • $\begingroup$ Attendum: The fact that this angle stays constant during parallel transport can be used to recover the connection from its geodesics (after all geodesics in riemannian geometry can be defined without the connection). I think it is called Schild's ladder, but checking the wikipedia page made me unsure. With that said, this procedure is detailed in Arnold's The Mathematical Methods of Mechanics (appendix I: Riemannian curvature). $\endgroup$ – Bence Racskó May 1 '17 at 9:21
  • $\begingroup$ You want $P_\gamma\in\mathrm O(T_pM)$, I don't think orientability was assumed. $\endgroup$ – Ryan Unger May 1 '17 at 14:36
  • $\begingroup$ @0celouvskyopoulo7 But if I take a smooth, one-parameter family of loops $\gamma_\epsilon$, all homotopic to the trivial curve, and consider $P_{\gamma_\epsilon}$ in the $\epsilon\rightarrow 0$ limit, I need to recover the identity map, right? Since the connection is smooth, its dependence on paths should also be. But then $P_{\gamma_\epsilon}$ is a smooth one-parameter family of linear isometries that are connected to the identity element of $O(T_pM)$, so they must belong in $SO$. $\endgroup$ – Bence Racskó May 1 '17 at 16:34
  • $\begingroup$ @0celouvskyopoulo7 Well, I forgot to say that $\gamma_0=\text{trivial curve}$. $\endgroup$ – Bence Racskó May 1 '17 at 16:36
  • $\begingroup$ Why should a general curve be nullhomotopic? $\endgroup$ – Ryan Unger May 1 '17 at 18:08

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