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Parallel Transport on the Torus

Because it really is all about the torus, baby

After reading about the torus’s curvature, shape operator, and geodesics, you’re probably asking yourself how parallel transport on the torus affects tangent vectors. Good question!

We start as always with our standard parameterization of the surface:

x (u, v) = \{\begin{cases} x = (c + a \cos v) \cos u \\ y = (c + a \cos v) \sin u \\ z = a \sin v \end{cases}

This parameterization is particularly nice, since it’s an orthogonal patch ( $F = x_{u} \cdot x_{v} = 0$ ). This lets us easily compute the associated frame field E₁,E₂:

\begin{array}{l} E_{1} = \frac{x_{u}}{\sqrt{E}} = (- \sin u, \cos u, 0) \\ E_{2} = \frac{x_{v}}{\sqrt{G}} = (- \cos u \sin v, - \sin u \sin v, \cos v) \end{array}

Let’s check that E₁ and E₂ are orthogonal by computing their dot product. Also, the cross product should be normal to the surface:

\begin{array}{l} E_{1} \cdot E_{2} = \sin u \cos u \sin v - \sin u \cos u \sin v + 0 = 0, check \\ E_{1} \times E_{2} = |\begin{array}{c} \vec{x} & \vec{y} & \vec{z} \\ - \sin u & \cos u & 0 \\ - \cos u \sin v & - \sin v \sin u & \cos v \end{array}| = (\cos u \cos v, \sin u \cos v, \sin^{2} u \sin v + {cos}^{2} u \sin v) = (\cos u \cos v, \sin u \cos v, \sin v), check \end{array}

Taking partial derivatives of E and G, we have an equation that gives the connection form $ω_{12}$ :

ω_{12} = - \frac{{(\sqrt{E})}_{v}}{\sqrt{G}} ⅆ u + \frac{{(\sqrt{G})}_{u}}{\sqrt{E}} ⅆ v = \frac{a \sin v}{a} ⅆ u + 0 = \sin v ⅆ u

$ω_{12}$ tells us how a vector rotates as it is parallel transported. It encodes pretty much everything about the space’s curvature.

What does this tell us about parallel transport on the torus?

First, the $ⅆ u$ term tells us that parallel transport along lines of constant u (longitude lines) doesn’t affect vectors:

[vectors don't rotate during parallel transport along a line of longitude]

We could have predicted this from the symmetry of the torus; along a line of longitude, the neighborhoods to the left and right are mirror images, so there’s no preferred direction for a vector to rotate.

Second, the sinv term tells us that parallel transport along lines of constant v (latitude lines) causes vectors to rotate through the angle $2 π \sin v$ during their journey back to their starting point. Let’s look at some specific cases, starting with parallel transport along the outer equator:

[vectors don't rotate during parallel transport along outer equator]

Here $v = 0$ , so $\sin v$ is also zero, and the vectors don’t rotate at all. This is also true at the inner equator.

At the top of the torus ( $v = \frac{π}{2}$ ), $\sin v = 1$ , so a vector rotates through a full $2 π$ during its journey. Note how the angle between the blue vector and its path (red) changes as the vector is parallel transported:

[vectors rotate 2pi during parallel transport along top of torus]

Something else interesting is happening simultaneously: the vector’s origin is also rotating through $2 π$ . These rotations cancel, leaving the blue vector pointing in the same direction in the embedding space. So someone living on the torus would say the vector rotates as it is parallel transported, while someone living outside the surface would not. (Note that at $v = \frac{π}{2}$ the torus’s Gaussian curvature is zero, so it’s not surprising that vectors parallel transported along that path don’t appear to rotate in the embedding space.)

Parallel transport along other lines of latitude causes vectors to rotate varying amounts ( $2 π \sin v$ ). In the next illustration, four frames aligned with the u and v axes (at $v = 0, \frac{π}{6}, \frac{π}{3}, and \frac{π}{2}$ ) are parallel transported widdershins around the torus. The amount of rotation is a function of their v coördinate.

[vectors rotation rates differ for parallel transport at different latitudes]

By the time these frames return to their starting point, each will have rotated $2 π \sin v$ around its origin. For example, the second line of latitude from the bottom is at $v = \frac{π}{6}$ , so vectors parallel transported along it will have rotated through an angle of $2 π \sin \frac{π}{6} = 2 π \frac{1}{2} = π$ , as indeed they have.

[parallel transported vectors approaching 2pi]

Putting all of this together, here’s how a whole bunch of frames rotate while being parallel transported widdershins along lines of latitude, starting at the red longitude line:

[many vectors parallel transported along lines of latitude]

When creating these images, one thing that surprised me was how quickly the value of $2 π \sin v$ changed near $v = 0$ . Then I remembered that $\frac{ⅆ \sin v}{ⅆ v} = \cos v$ , which has extremes at $v = k π$ , k an integer. So the rate of change of the effect of parallel transport along lines of latitude is most extreme at the outer and inner equators.

Finally, parallel transport on the bottom half of the torus is the same except for direction of rotation, since $\sin v$ is negative there.

[parallel transport along bottom half of torus]

Students of differential geometry may have noticed that $\sin v ⅆ u$ is also the ω₁₂ of the sphere. The difference is that for the sphere, the range of v is $[- \frac{π}{2}, \frac{π}{2}]$ , while for the torus it’s $[- π, π]$ . Indeed, you can see that parallel transport on the outer half of the torus mirrors parallel transport on the sphere.

Bonus: computing the Gaussian curvature from the patch

Given an orthogonal patch, we can directly compute the Gaussian curvature K from E and G:

K = \frac{- 1}{\sqrt{EG}} \{{(\frac{{(\sqrt{G})}_{u}}{\sqrt{E}})}_{u} + {(\frac{{(\sqrt{E})}_{v}}{\sqrt{G}})}_{v}\} = \frac{- 1}{a (c + a \cos v)} {(\frac{- a \sin v}{a})}_{v} = \frac{{(\sin v)}_{v}}{a (c + a \cos v)} = \frac{\cos v}{a (c + a \cos v)}

This agrees with the value for the Gaussian curvature we computed from the shape operator, but unlike that calculation this one doesn’t require a normal to the surface. Thus, if we lived on the torus, we could compute our space’s Gaussian curvature directly from measurements made within our space, and we don’t have to assume the existence of an embedding space. That’s the beauty of differential geometry.

Last updated 18 November 2005
http://www.rdrop.com/~half/math/torus/parallel.transport.xhtml
All contents released into the public domain by Mark L. Irons

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