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Let $c$ be a geodesic on a Manifold $M$. Some books define $c$ to be a Geodesic iff $\nabla_{c'}c'=0$. Therefore for every $c(t)$ the Geodesic must be extendable into a smooth vector field on an open set of $c(t)$.

How can I prove this? Is this true for any smooth curve on M or just with geodesics?


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I'm not sure about what means extending a geodesic into a vector field. Geodesics are curves and not vector fields. Could you clarify please? – matgaio May 14 '12 at 21:15
Ok no problem! I mean, one can find a (usual) Vector field X on M, s.t. X(c(t))=c'(t) $\forall t$. That is X is equal to the vector field c'(t) along c, but is also defined an an open set (like a coordinate nbhd.) – Braten May 14 '12 at 21:23
If you are working on borderless manifolds, it will be true for geodesics. You can try the following: for each $c(t)$, consider a totally convex neighborhood of $c(t)$ and a small coordinate chart defined on it in rrder to construct your desired vector field. I think it could work. – matgaio May 14 '12 at 21:27
Why should this definition imply that every $c(t)$ admit such an extension? $\nabla_{c'}c'$ only depends on the values of $c'$ along the curve $c$ (just write down the formula in coordinates) so one does not need to extend $c'$ to make this definition meaningful. – treble May 14 '12 at 21:35
@treble Extension is in tangent field original in the curve, does'nt refer to domain extension of the curve. – Gastón Burrull May 14 '12 at 21:39
up vote 2 down vote accepted

Perhaps you mean to geodesic field when your affine conection is compatible with you riemannian metric. It's always exists and existence follows directly from the differential equations defining the paralel transport of the tangent unitary map.

I dont know a direct counterexample when a smooth curve doesnt have this property, you must construct some curve such that any differentiable tangent field is forced to have some singular points. For example in $S^2$ there isn't a differentiable tangent field, because always you can find a singularity.

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I think the field he is looking for is a field over a neighborhood of $c(t)$ in $M$. The geodesic field is a field on $TM$. – matgaio May 14 '12 at 21:28
Differential equations given us the answer, the problem is about existence and it could be very hard to determine. If you curve have a tangent field not parallel to manifold, differentials equations may be quite hard and probably existence doesnt hold even in the curve (indeed a neighborhood of curve). – Gastón Burrull May 14 '12 at 21:32

Take the vector field and extend it via $d\exp_{c(t)}|_\nu$, where $\nu$ is the normal bundle over $c$.

The tangent bundle of $c$, $Tc$, resides in $TM|_c$. The existence of a metric lets us pick out elements of $TM|_c$ perpendicular to $Tc$. All such perpendicular elements form the normal bundle $\nu(c\subset M)$. The exponential map takes a small neighborhood of the zero section of $\nu$ to a small open neighborhood of $c$.

If we have a vector field $X$ defined on $c$, we can extend it to a vector field on $\nu$, defined by extending it constantly on each fiber (note that $\dim\nu = \dim TM|_c$). Now just push $X$ out to $\exp\nu$ by $d\exp$, which is a diffeomorphism.

Alternately, since $c$ is a submanifold, it has an atlas of coordinate charts which take the form $c\times \mathbb{R}^{n-1}$ (here $n = \dim M$). You can extend $X$ to a neighborhood of $c$ by extending $X$ to a neighborhood of $c$ in each coordinate chart so that it commutes with the projection. (You will need to be a little careful about coordinate changes, though.)

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Ok no problem! I mean, one can find a (usual) Vector field X on M, s.t. X(c(t))=c'(t) ∀t. That is X is equal to the vector field c'(t) along c, but is also defined an an open set (like a coordinate nbhd.)

does not match your question: if $v\in T_pM$ is a tangent vector, then $\nabla_v Y$ may be defined to be $\nabla_X Y$ for some locally defined vector field $X$ as long $X_p=v$. This is done pointwise, if $c(t)$ is a curve there is no need for a locally defined vector field $X$ having $X(c(t))=c'(t)$ for all $t$ (I think such an $X$ doesn't have to exist in an open nhood of $c(I)$ where $c:I\rightarrow M$ is a curve.

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I think the statement of the question is just a little bit confused. I think it is not for all $t$, but just for a neighborhoof ot "t", or, we just have to find a local vector field. – matgaio May 14 '12 at 22:09
I think $\nabla_vY$ makes sense even if $v$ is not $X(p)$ for a local vector field $X$. In the expression in coordinates of the covariant derivative, you have just to know $Y$ locally and $v$ in the point – matgaio May 14 '12 at 22:13
Therefore for every ... must be extendible indicates that the OP uses a definition of $\nabla$ that depends on vector fields. – Blah May 14 '12 at 22:19
You mean $\nabla$ will depend of the choice of $c$ too? – matgaio May 14 '12 at 22:21
No, $\nabla$ is given (the Levi-Civita-Connection) - the question is why you said that the tangent vector $c'(t)$ must be extendible to a nhood of $c(t)$ (for one/all $t$)? – Blah May 15 '12 at 4:46

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