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I'm confused what exactly is happening in the following proof of Lemma 4.3 (Lemma 1.5 and Exercise 4.5 are also attached as they are apparently used in this proof). In particular, what does the integration process actually prove? How does it prove what we want? I would imagine that we would take a general geodesic and show that it looks like an exponential.

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Is there perhaps a more direct way to prove the result that geodesics in compact Lie groups look like exponential maps? I do apologize in advance - I have little background knowledge in this area of math, and am trying to understand this result as it is used in the proof of Bott-Periodicity. These images are taken from Dietmar Salamon's 'Notes on Compact Lie Groups' (link: https://people.math.ethz.ch/~salamon/PREPRINTS/liegroup.pdf)

Arthur
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They're overcomplicating it. Here's another argument that combines three simple facts.

1) the flows of left-invariant fields consists of right translations; the flows of right-invariant fields consist of left translations.

2) for a bi-invariant metric, both left-invariant fields and right-invariant fields are Killing fields (this follows from item (1)).

3) for any pseudo-Riemannian manifold, the integral curves of constant length Killing fields are geodesics (Killing's equation with $\langle X,X\rangle$ constant implies that $\nabla_XX = 0$).

This means that given $g \in G$ and $v \in T_gG$, we have that $g^{-1}v \doteq {\rm d}(L_{g^{-1}})_g(v) \in \mathfrak{g}$. Then $\alpha(t) = \exp(t g^{-1}v)g$ is the geodesic with $\alpha(0) = g$ and $\alpha'(0) = v$.

Ivo Terek
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  • I don't understand point 3. Isn't the length of a Killing field always constant along an integral curve? – Max May 02 '20 at 23:13
  • Yes, and this with Killing's equation says that $\langle \nabla_XX,X\rangle = 0$, but that's not enough to say that $\nabla_XX = 0$. For this you need $\langle \nabla_XX,Y\rangle = 0$ for any $Y$, hence the assumption with this more "global" nature. – Ivo Terek May 02 '20 at 23:19
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    Ah, I wasn't parsing correctly. Your Killing fields are constant length everywhere. Thanks – Max May 02 '20 at 23:38
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    Actually, I understand your confusion. I hope I have improved the wording in the answer enough now. – Ivo Terek May 02 '20 at 23:51
  • Thank you for this answer. For point 3 when you say that integral curves of constant length Killing fields are geodesics, is that an if and only if? Is every geodesic an integral curve of constant length Killing field? This stuff is quite foreign to me, I apologize if my questions are naive. – Arthur May 03 '20 at 01:40
  • Actually, I believe I have verified this to be true. I suppose I am a bit confused with the computation done at the end, and the stuff in brackets in point 3? I think I am missing information regarding some equations that I should take a look at. Thanks in advance. – Arthur May 03 '20 at 01:43
  • @Arthur There is a terse proof of (3) here. https://math.stackexchange.com/questions/1915567/if-x-is-killing-and-p-is-a-critical-point-of-fp-xp-2-then-the-tr – Max May 04 '20 at 20:22
  • Thank you for that @Max – Arthur May 05 '20 at 22:18