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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 .

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Hi, 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. –  t.b. Aug 1 '11 at 10:05
    
my question is the length of alpha is realised by a closed geodesic or closed geodesic loop ? –  Alfie Aug 1 '11 at 10:17
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On 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. –  Jason DeVito Aug 1 '11 at 16:02
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2 Answers

Are you sure you have the definition of a "closed geodesic" and "geodesic loop" correct?

Usually we have

Definition A closed geodesic $\gamma$ on $(M,g)$ is a smooth image of $\mathbb{S}^1$ that is geodesic.

Definition A geodesic loop $\gamma$ on $(M,g)$ is a smooth image of $[0,1]$, that is geodesic, and such that $\gamma(0) = \gamma(1)$.

See, for example, the Springer EOM.

Then by the argument given in your question statement, you have that a minimizing object is automatically a geodesic loop. But it doesn't have to be a closed geodesic as there may be an angle.

Example Imagine a T-shaped pipe formed by joining two cyclinders at right angles. (To make it closed you can glue the ends of the pipe to a big sphere.) Smooth out the junctions. Let your base point $p$ be the point on the horizontal part of the T that sits directly opposite the vertical leg of the T.

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$\pi \circ c′$ is not smooth at its endpoints. That is, it is not smooth at $p$. Thus $\pi \circ c′$ is not a geodesic. Instead it is a geodesic loop.

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