Let (M,g) be a closed manifold and let $\alpha$ be an element of $G=\pi_1(M,p)$ we can define the norm of $\alpha$ with respect to p as the infinimum riemannian length of a representative of $\alpha$ . people say this norm is realised by a geodesic loop at p my question is why it is realised by a geodesic loop and not by a geodesic it must be something with the regularity here is a simple proof for why i am saying geodesic and not geodesic loop. fix $\tilde{p}$ in the universal cover of $(M,g) $ let c be a representitve of $\alpha$ , $\tilde{c}$ be the lift of c to $\tilde{M}$ with base point $\tilde{p}$ and let $c\prime$ be the unique minimizing geodsic joinging $\tilde{p}$ to $\alpha(\tilde{p})=\tilde{c}(\tilde{p})$ . obviously $\tilde{c}$ and $c\prime$ are homotopic and hence $\pi oc\prime$ and $\pi o\tilde{c}=c$ are homotopic where $\pi : \tilde{M}\rightarrow M$. and I know the the image of a geodesic by a local isometry is a geodesic hence $\pi oc\prime$ is a closed geodesic at p and sure it minimise the length otherwise we will have a contradiction cause it lift $c\prime$ is minimizing .
Geodesic versus geodesic loop
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riemannian-geometry
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1Hi, welcome to this site. I'm sorry to ask, but: What exactly is your *question*? It would be great if you could insert a few more periods and make this one block of text into two or three paragraphs. – 2011-08-01
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0my question is the length of alpha is realised by a closed geodesic or closed geodesic loop ? – 2011-08-01
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1On the other hand, if instead of looking at homotopy classes of *based* maps from $S^1$, one just considers the homotopy classes of maps from $S^1$ into a compact manifold $M$, then such classes *are* represented by closed geodesics. – 2011-08-01