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Let $N$ be a submanifold of a Riemannian manifold $M$. Does a geodesic $\gamma$ in $M$ with image contained in $N$ remain a geodesic in $N$? I'm pretty sure this has to be true true because its locally length minimizing properties.

To prove this can I just say that if $\nabla$ is the connection on $N$ and $\overline{\nabla}$ is the connection on $M$, then $\nabla_{\gamma'}\gamma'=(\overline{\nabla}_{\gamma'}\gamma')^{T}=0$ since $\overline{\nabla}_{\gamma'}\gamma'=0$?

Tuo
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  • I'm a bit rusty but that looks basically right. You should probably cite Gauss' formula when splitting the covariate derivative on $M$ into the covariate derivative on $N$ and the second fundamental form. – Neal Nov 04 '18 at 00:50
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    @Neal I'm pretty sure it would also contradict local length minimizing if it wasn't a geodesic still in $S$. – Tuo Nov 04 '18 at 00:52
  • Yes, that is true and your argument is also correct. – Ernie060 Nov 04 '18 at 00:52
  • @TuoTuo I think that's an even cleaner argument – Neal Nov 04 '18 at 00:54
  • @TuoTuo I think you should write up your answers and accept them :) – Neal Nov 04 '18 at 00:59
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    @Neal I didn't think of it until after I asked the question. I think the best phrasing is "if you have a curve that locally minimizes the distance between points along it, but doesn't minimize the distance between those points anymore when you are in a submanifold with the same metric,then you have a problem" – Tuo Nov 04 '18 at 01:16

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