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Let M be a Riemannian homogeneous space, i.e. the isometry group acts transitively. Prove: any geodesic loop (with possible angle at the starting point) is a closed geodesic (smooth at the starting point).

And there is a hint: prove and use the fact that the vector field associated to a one parameter group of isometries is a Jacobi field when restricted to any geodesic.

I just don't know how this hint comes into play..

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  • $\begingroup$ It turns out to be not hard..just apply the first variation formula.. $\endgroup$ – somebody trivial Dec 12 '14 at 0:32
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    $\begingroup$ Could you post your own answer? That will be helping others. Thanks. $\endgroup$ – Xipan Xiao Dec 12 '14 at 19:26
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Here is my own answer:

Say $\gamma(s),s\in[0,a]$ is a geodesic loop starting at $\gamma(0)=p$. Consider any one parameter group of isometries $g_t$, since the isometry group acts transitively, $g_t p$ takes p to any direction in $T_p M$ as $g_t$ varies. Now note that $g_t \gamma$ is a geodesic variation which preserves length, apply the first variation formula for energy:

$$0=E'(0)=-\int_0^a<U(s),\gamma''(s)>ds+ <U,\gamma'>|_0^a=<U(0),\gamma'(a)-\gamma'(0)>.$$

where $U(s)=\frac{d}{dt}|_{t=0}g_t\gamma(s)$ is the variation vector field. $E'(0)=0$ because it preserves length, and $\gamma''(s)=0$ because it is geodesic. Now recall that $U(0)$ can be any vector in $T_pM$, which forces $\gamma'(a)=\gamma'(0)$. That's geometrically what we need to prove.

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