Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 .

share|improve this question
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
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
add comment

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.

share|improve this answer
add comment

$\pi \circ c′$ 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.

share|improve this answer
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.