I've been trying to understand parallel transport. Many of the descriptions present a mathematical version: $\nabla_V X = 0$.

And/or they present an example involving soldiers (usually Roman) carrying spears around the world. Two soldiers start at the North Pole $[x,y,z]=[0,1,0]$ both pointing their spears towards $[1,0,0]$. One soldier walks forwards to the easternmost point on the Earth's equator $[x,y,z]=[1,0,0]$. His spear stays tangent to the Earth and points southwards when he arrives $[0,-1,0]$. The other solider walks sideways to the equator $[x,y,z]=[0,0,1]$ and his spear still points eastwards $[1,0,0]$ when he arrives. Now both soldiers are on the equator with spears pointing in different directions, one south, the other east. If they walk towards each other, the spears will still point different ways.

The problem is that both soldiers held their spears "parallel" while walking, so there are two seemingly parallel paths between two points which end up with different results. And this is bad, and not just for the exhausted Romans.

Solving this issue involves rotating the spears as the soldiers walk towards each other along the equator, so they point in the same direction when they meet. I would like to see a mathematical description of this.

For my own attempt to describe this problem, I started with a "congruence of curves" of concentric spheres $x^2+y^2+z^2=C$. To find a tangent vector field to these curves, I found the normal vector $[2x,2y,2z]$ and picked a vector field perpendicular to it such as $V=[y+z,-x,-x]$ so that $[2x,2y,2x] \cdot V = 0$.

By my understanding, a tensor $X$ will be parallel transported along this congruence of curves (the sphere) if the covariant derivative of $X$ contracted with the tangent vector field $V$ equals $0$: $\nabla_V X=0$. The tensor $X=[x^2+y^2+z^2,0,0]$ can do this: $$\nabla_V X = (\partial X + \Gamma X) V = (\partial X + 0) V = \begin{bmatrix} 2x & 2y & 2z \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} \begin{bmatrix} y+z \\ -x \\ -x \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\0 \end{bmatrix}$$

This tensor $X$ points the same direction everywhere on the sphere $[1,0,0]$. So this describes a solution to the Romans' concerns. The first soldier just needs to raise his spear as he walks so it always points at $[1,0,0]$. But although it is a solution, I can't (mathematically) see what the problem was in the first place.

On the other hand, a vector field which describes the Romans' spear movements as above, including the equatorial rotation is $Y=[y^2+z^2,-x^2,0]$. But this field does not get parallel transported around the sphere because $\nabla_V Y \ne 0$. This rotation is supposedly how parallel transport fixes the Romans' spear pointing issues, but in this case it doesn't actually seem to be an example of parallel transport.

I also tried doing these calculations in polar coordinates but I got similarly lost. I feel like I'm missing a piece of the puzzle. Any pointers much appreciated.

  • 1
    $\begingroup$ When you are standing at the North Pole, the only direction you can ever take is South. Eastward motion would mean that you are turning in place. In general what you are looking at is an (arbitrary) embedding of the sphere in a three dimensional coordinate system. That is exactly the opposite of what differential geometry is trying to achieve: the description of certain types of metric sets without embedding in some sort of higher dimensional cartesian coordinate system. $\endgroup$
    – CuriousOne
    Sep 8, 2015 at 22:13
  • 2
    $\begingroup$ Two comments and a point of clarification: (1) Your use of directions is a bit confusing -- usually they refer to directions of increasing/decreasing latitude/longitude, rather than directions to fixed points on the surface, so e.g. there is no East direction at the North Pole. (2) Do you mean $\nabla_XV$? (3) "Parallel transport is supposed to solve this issue" -- this is where things get confusing. What issue exactly is there? Parallel transport is what you did in the first paragraphs. It is path dependent. $\endgroup$
    – user10851
    Sep 8, 2015 at 22:14
  • 1
    $\begingroup$ (1) Sorry for the confusing directions. I am imagining myself looking at a sphere from a distance within a 3D coordinate system. (2) I meant $\nabla_V X$, corrected. (3) It gets confusing because I'm confused. So if the Roman soldier problem is parallel transport - then how is it described mathematically? What $X$ in which coordinate system describes their motion which satisfies $\nabla_V X=0$? $\endgroup$
    – Paul
    Sep 8, 2015 at 22:29
  • 3
    $\begingroup$ As a general hint I would suggest that you do not imagine looking at an embedding when dealing with manifolds. All the objects in differential geometry are living in two possible places: 1) Points are living in the set of the manifold. 2) Vectors are living in tangent spaces that are attached to individual points. At no time is it necessary to leave the manifold and to escape into a higher dimensional embedding space. . $\endgroup$
    – CuriousOne
    Sep 8, 2015 at 22:35
  • 1
    $\begingroup$ Thank you for the quick replies. But describing it all in polar coordinates, I understand even less how the $\nabla_V X=0$ equation relates to the Romans. North Pole = $[lat,long]= [\theta,\phi]=[90,0]$. Which directions are the spears pointing when they start? The first soldier walks forwards towards $[0,0]$, the other sideways to $[0,90]$. What would $V$ and $X$ be in this case? Or is each soldier described by a different $X$? $\endgroup$
    – Paul
    Sep 8, 2015 at 22:49

1 Answer 1


The answer to my question is simpler than I suspected. It is fairly easy to describe the movements of both soldiers mathematically.

The first soldier's spear is being transported along $X_1 = [y^2+z^2, -x^2, 0]$. The second soldier's spear along $X_2 = [y^2+z^2, 0, -x^2]$. Both are valid parallel transports relative to the tangent vector field $V=[0, -z, y]$. For example, with the first one:

$$\nabla_{V} X_1 = (\partial X_1 + \Gamma X_1) V = (\partial X_1 + 0) V = \begin{bmatrix} 0 & 2y & 2z \\ -2x & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} \begin{bmatrix} 0 \\ -z \\ y \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\0 \end{bmatrix}$$

But they can both end up at the same point on the sphere pointing in different directions. At the point $[1,0,0]$, $X_1=[0,-1,0]$ and $X_2=[0,0,-1]$.

This isn't possible on a flat surface, or even on a cylinder. On a cylinder, spears which are parallel transported along different paths will point in the same direction when they meet again.

The reason for this is that a sphere is intrinsically curved. If you try to flatten out the sphere (by converting to spherical polar coordinates for instance), then the corresponding parallel transports within spherical polar coordinates will also end up pointing in a different directions. That's because the sphere's $\Gamma$ leads to a non-zero Riemann curvature tensor (a 4 dimensional tensor computed from $\Gamma$). Unlike a cylinder whose Riemann curvature tensor is zero.

I also made a mistake in the original post. The vector field I mentioned $Y=[y^2+z^2,−x^2,0]$ does describe parallel transport (including the equatorial rotation) relative to the tangent vector field $[0,-z,y]$.


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.