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I've gotten a germ of the proof, which involves using the standard metric on the two sphere $ds^2=d\theta^2 + \sin^2{\theta}d\phi^2$. Thing is, how do I continue from here? How do I use this metric to calculate those interior angles regardless of the lengths of the legs of the triangle?

Mathipulator
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    Do you need to prove for any arbitrary triangle, or would a counter example of the statement ‘interior angles add up to 180’ suffice? – Soham Saha Sep 28 '23 at 05:36
  • Note: perhaps $\mathrm d\phi^2$ is correct? – Anton Vrdoljak Sep 28 '23 at 06:54
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    Not that the metric induces a dot product on the tangent space which can be used to find the angles of the triangle but looking at the initial tangent vectors of the curves forming said triangle – Chris Sep 28 '23 at 07:09
  • Soham, the latter but both would be nice to derive. – Mathipulator Sep 28 '23 at 12:59
  • Are you interested in a proof that does not use the formula you gave for the standard metric on the 2-sphere? I ask because the comment of @Chris would make a very nice answer, although if that does not interest you it might not be worth it. – Lee Mosher Sep 28 '23 at 13:16
  • Let me just add that the formula you gave is not actually the best primitive definition of the metric; instead, the metric is more properly defined using the dot product method. – Lee Mosher Sep 28 '23 at 13:16
  • If all you need is a counterexample, look at the simplest triangles you can think of. Remember that each side is part of a great circle, so it's a matter of choosing the great circles. And it's useful to have guideposts such as the north and south poles. – Deane Sep 28 '23 at 13:18
  • @LeeMosher are you proposing a counter example which does not depend on a choice of metric? I would personally be very interested in that, and I selfishly suggest you write up an answer if so – Chris Sep 28 '23 at 21:38
  • No, I'm proposing a proof that does not depend on spherical coordinates, and it's the same proof proposed by @Chris. – Lee Mosher Sep 29 '23 at 01:34
  • @LeeMosher oh yes ok that makes much more sense. I think it would probably be easier to not use spherical coordinates at all tbh, at least from a calculation standpoint – Chris Sep 29 '23 at 17:02

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