Extending geodesics to vector fields 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.
 A: 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.
A: 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.)
A: 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.
