Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let $M$ be a (without boundary and not necessarly complete) Riemannian manifold.

A map $c\colon [a,b]\rightarrow M$ is called geodesic of type A iff $c$ is piecewise smooth, parametrized proportional to arclength and for all $t\in[a,b]$ there exists an $\epsilon>0$ such that $L(c|_{[t-\epsilon,t+\epsilon]})=d(c(t-\epsilon),c(t+\epsilon))$. ($L$ is the lengthfunctional and $d$ the distance in $M$.)

A map $c\colon [a,b]\rightarrow M$ is called geodesic of type B iff $c$ is smooth and is autoparallel with respect to the levi-civita-connection of $M$.

Are those two notions always to 100% equivalent? If not, why and under which precondition and/or changes are they?

share|improve this question
    
This should be in most Riemannian geometry textbooks. The first implies the second by first variation of length (or energy), while the second implies the first via existence of convex neighborhoods (With Gauss Lemma being the important ingredient). –  Dan Lee Apr 2 '13 at 2:10
    
Is the usage of the variation of energy or length necessary or can I prove $A\leftrightarrow B$ fully just by using convex neighbourhoods? –  Lisa Apr 2 '13 at 7:07

1 Answer 1

These notions coincide always with the exception that B is in general parameterized by a constant multiple of arc-length.

share|improve this answer
    
A does too, as I said, that it is parametrized proportional to arclength, don't it? Do I have to use variations of the length or energy or is this equivalence provable only by using the exponential map and Riemann normal coordinates? –  Lisa Apr 1 '13 at 13:05
1  
@Lisa: your geodesic type A does parametrize wrt. arc length, but B does not in general, only when the initial (and therefore at each time) velocity c'(a) has unit length. –  Jaap Eldering Apr 1 '13 at 15:04

Your Answer

 
discard

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.