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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?

Regards.

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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?2012-05-14
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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) $\forall t$. 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.)2012-05-14
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    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.2012-05-14
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    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.2012-05-14
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    @treble Extension is in tangent field original in the curve, does'nt refer to domain extension of the curve.2012-05-14

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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 $TM$.2012-05-14
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    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).2012-05-14
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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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Your comment

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.2012-05-14
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    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 point2012-05-14
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    *Therefore for every ... must be extendible* indicates that the OP uses a definition of $\nabla$ that depends on vector fields.2012-05-14
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    You mean $\nabla$ will depend of the choice of $c$ too?2012-05-14
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    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 $t$)?2012-05-15
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    If you fix a $t$, then in a neighborhood we have $c(t)$ embedded. You can suppose it is inside a coordinate chart and via this chart you can extend the vector field $c'(t)$ to $X$, but just locally. It will happen yet $\nabla_{\gamma'}X=0$ over $\gamma$, inside this neighborhood. Perhaps we are thinkink about different things so we are not convincing each other =)2012-05-15