I heard of a result on Riemannian manifolds but I couldn't find a good reference:
For every point $p$ on a complete Riemannin manifold there exists a closed geodesic passing through $p$ with length two times the injective radius at $p$.
Thanks!
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asked Feb 25, 2018 at 11:21
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5$\begingroup$ Do you mean the geodesic to pass through the point $p$? If so, you heard wrongly. If, say a compact Riemannian manifold has negative sectional curvature at every point, then there are only countably many closed geodesics. By Baire category they cannot contain all points of the manifold, so there are points which do not lie on any closed geodesic. $\endgroup$– user1688Commented Feb 25, 2018 at 13:52
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$\begingroup$ @Corbennick Yes, I forgot to mention that the geodesic passes through $p$. I added this in the edited version. $\endgroup$– Mostafa - Free PalestineCommented Feb 25, 2018 at 14:13
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1 Answer
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The statement is not true.
First, a comment on terminology. There are two related notions for a geodesic that passes twice through a point $p$: a periodic geodesic and a geodesic loop at $p$. The former is smooth at all points, while the latter may be non-smooth at $p$. Sometimes periodic geodesics are called closed geodesics.
By shortening one shows that at any point of a complete Riemannian manifold there is a shortest geodesic loop at that point. Similarly, any compact Riemannian manifold contains a periodic geodesic (but not necessarily through every point).
On a non-compact complete manifold a shortest periodic geodesic through a point may not exist. For example, the surface of revolution obtained by rotating the graph of $z=\frac{1}{x}$ has no periodic geodesics, even though its injectivity radius if finite. (Of course, the standard Euclidean space also has no periodic geodesics but its injectivity radius is infinite).
Another good example is the acute cone of revolution smoothed at the tip. The shortest loop from a point not near the tip goes around the tip point, and is never a periodic geodesic.
Now what is true (in a complete Riemannian manifold) is that the injectivity radius is the minimum of two numbers:
The half of the length of a shortest geodesic loop at $p$,
The distance from $p$ to nearest conjugate point at $p$. (This number is called the conjugate radius at $p$ and it is bounded below by $\frac{\pi}{\sqrt{k}}$ where $k$ is an upper section curvature bound of the ambient manifold).
Finally, let me address a possible source of the OP confusion. Recall that the injectivity radius of Riemannian manifold is the infimum of injectivity radii at its points. Klingenberg proved that if the injectivity radius of a compact Riemannian manifold is smaller than $\frac{\pi}{\sqrt{k}}$, then it equals half of the length of a closed periodic geodesic.
Good references are textbooks by P.Petersen and T.Sakai, which are called "Riemannian Geometry".
answered Feb 25, 2018 at 14:11