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I have a question about parallel transport that I'm very confused about and would appreciate some help. The question reads:

What vector field $X$ on the unit 2-sphere in $\mathbb{R}^3$ has rotations around the $z$-axis as flows? The orbits of these flows would be the lines of latitude. Solve the parallel transport equations $\nabla_X(V_i)=0$, for $V_1, V_2$ the elements of a basis of the tangent plane, along the curve of latitude 45 degrees. What do the $V_i$ come back to once one has transported them all the way around the circle?

I am confused by which vector field the question is talking about. Would this field just be the sphere parametrized as:

$$(x,y,z) = (r\cos(\theta)\sin(\varphi), r\sin(\theta)\sin(\varphi), r\cos(\varphi))$$ Where $\varphi$ is constant?

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    $\begingroup$ The curves $\varphi=\text{constant}$ are the lines of latitude to which they refer. But you want the vector field everywhere tangent to those. When you work with the sphere, parametrized as you did in spherical coordinates, what are the natural choices for $V_1$ and $V_2$? $\endgroup$ Commented Apr 2, 2017 at 22:02
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    $\begingroup$ Yikes, way too hard. Just differentiate the parametrization with respect to the independent variables! (You'll find all this in my text you've already looked at, of course.) :) $\endgroup$ Commented Apr 2, 2017 at 22:27
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    $\begingroup$ Well, rotating about the $z$-axis means that we're varying $\theta$, fixing $\phi$, so it must be $S_\theta$ :) Remember that to do the parallel transport around a closed curve, you want to fix a starting (and ending) point and see how the vector field turns when you return to the starting point after one trip around. $\endgroup$ Commented Apr 4, 2017 at 23:15
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    $\begingroup$ They mean to set $\varphi = \pi/4$, but your computation is not correct. $S_\theta$ definitely does not have covariant derivative $0$ along $\varphi=\pi/4$. ... And there are pictures of this exact thing, and a good deal of discussion and derivations, in my text ... :P $\endgroup$ Commented Apr 4, 2017 at 23:25
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    $\begingroup$ Yes, Felicio. See also the remark on p. 71 for an intuitive discussion of this phenomenon. Probably reading and working through the section carefully would be helpful. $\endgroup$ Commented Apr 4, 2017 at 23:43

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This isn't a direct answer to your question about the parallel transport equations, but the standard geometric solution. Assuming a unit sphere, parallel transport around the latitude $L_{\varphi}$ making angle $\varphi$ with the equator can be found by constructing the cone tangent to the sphere along $L_{\varphi}$, then cutting the cone along a generator, rolling it flat, and performing parallel transport of a tangent vector around an arc of a circle.

Elementary geometry shows that the unrolled cone is a sector of a disk of radius $\cot\varphi$ whose central angle is $2\pi\sin\varphi$. Parallel transport around $L_{\varphi}$ in the northern hemisphere rotates a vector clockwise (looking "down" at the tangent plane, toward the center of the sphere) by an angle $2\pi\sin\varphi$. The animation is a full-circle pan around the sphere.

Parallel transport on the sphere

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  • $\begingroup$ Just wanted to say thanks for this great way to think intuitively about parallel transport on a sphere. I had been confusing me for ages why the vector needs to turn. $\endgroup$ Commented Feb 22, 2018 at 17:13
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For points on the equator. you may find all elipsoids that are present on the north hemisphere exert the same integral sum of vector acceleration around a regular surface. their transport should be similar to a circle transport at the equator. it is the case that a transport with some location a (t)i that the integral sum of acceleration is greater than the distance around the world at the equator so it is not similar.

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  • $\begingroup$ a (t) i must be below the equator for some t,i $\endgroup$ Commented Apr 9, 2017 at 18:11

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