Vector Field Generating Variation Along Curve I'm learning a proof of the fact that length extremising curves are geodesics of the Levi-Civita connection, and have found something I don't understand. The argument states the following.
Suppose $\gamma$ is an arbitrary curve in a manifold $M$ and $N$ is an arbitrary vector field along $\gamma$ which vanishes at the endpoints. Then there must exist a variation $\tilde{\gamma}$ of $\gamma$ such that $\gamma'$ and $N$ are the pushforwards of the coordinate vector fields on the domain of the variation. 
I can't see how this is true, because it would immediately imply that all vector fields commute (I think!). Am I missing something or is it false? If it's wrong, does anyone know how to prove that statement that length extremizing implies geodesic. 
The relevant lectures notes from which I've taken this argument are here, page 32. Thanks in advance!
 A: The claim is only that if you're given a vector field $N$ along the curve, then you can find a variation of the curve such that the pushforwards of the coordinate vector fields restrict to $\gamma'$ and $N$ along the curve.  This is not the same as being given a vector field on the manifold.  
If the dimension of $M$ is 2, and the variation $\tilde\gamma$ happens to be a smooth embedding, then the pushforwards actually will define vector fields on $M$ in a neighborhood of the curve (minus the endpoints), and these vector fields will commute.  The fact that you can do this for an arbitrary $N$ is no contradiction, because in this case $N$ must be nowhere tangent to the curve, and the values of the bracket along the curve are not determined solely by the values of $N$ and $\gamma'$ there; they depend on how the vector fields are extended, because the coordinate formula for the bracket involves differentiating the coefficients of $\gamma'$ in the direction of $N$.  
In the example @Ted Shifrin gave with $N=fT$, the above argument doesn't apply, because if $\gamma'$ and $N$ are linearly dependent, then $\tilde \gamma$  cannot be an embedding.  And in dimensions higher than 2, you won't ever get vector fields defined on an open subset of $M$; they'll only be defined on the (at most) 2-dimensional image of $\tilde \gamma$.  
A: To answer your point in the third paragraph: No, it only implies that vector fields that are push-forwards of (constant multiples of) coordinate vector fields commute. True enough, two linearly independent vector fields can, in appropriate coordinates, be coordinate vector fields if (and only if) their Lie bracket vanishes. 
