Geodesics on S2S^2

I was asked to show the following:

Given a Riemannian metric on S2S^2, let x∈S2x \in S^2 be the north pole, and let V(θ)∈TxS2V(\theta) \in T_xS^2 be unit tangent vectors at xx. Then there exists a constant t>0t > 0 such that expx(tV(θ))\exp_x(tV(\theta)) is the south pole for all θ\theta.

The conclusion in this statement is really not intuitive for me, and I am not sure how to approach this problem. Could someone give a hint on this?

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Presumably the question concerns the round metric on the sphere (for which the conclusion is true), not an arbitrary metric (for which the conclusion is false).
– Andrew D. Hwang
Oct 20 at 22:19

  

 

I was not provided with this condition. But could you explain more about it?
– P. Factor
Oct 20 at 22:22

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1 Answer
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Hints: What are the geodesics of a round unit sphere?

Start at a point xx of a round unit 22-sphere and walk at unit speed. How long does it take to reach the antipode −x-x?

By contrast, consider an arbitrary metric on a 22-sphere. As a physical analogy, imagine the surface of a squash or a gourd, where the center of the stem end is the north pole and the south pole is the center of the blossom end. It’s not obvious that a geodesic starting at the north pole ever reaches the south pole. Certainly, different geodesics between the poles have different lengths.

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I do understand what you said here. I guess I did not state the problem correctly. The assumption on the Riemannian metric is that all the geodesics originating from the north pole go through the same point (maybe call it the south pole), and the conclusion would be that all these geodesics have the same length.
– P. Factor
2 days ago

  

 

Ah, I see. The “usual site culture” would be to post a separate question, possibly linking back to this one, or linking forward from this one to the revised statement. (It would be helpful if, in the new question, you could say something about “available technology”, e.g., whether you know about Jacobi fields or the Gauss-Bonnet theorem.) If you revise this question, however, I’ll delete this answer and my comment.
– Andrew D. Hwang
yesterday