I'm trying to resolve the following:
"Let $M$ be a regular and oriented surface, and $\alpha = \alpha(t)$ a regular parametrized curve in $M$. An let $\mathbf{V}=\alpha'(t)$. We know that if $\mathbf{V}$ is a parrallel vector field (this meaning its covariant derivative is zero) then $\alpha$ is a geodesic curve. Prove that the reciprocal isn't true, and give a necessary and sufficient condition for it to be true."
I think I have proven that if the curve is parametrized by its natural parameter (arc length). $$ \alpha (\text{geodesic}) \Rightarrow K_g =(\vec{N},\dot{\alpha},\ddot{\alpha}) = 0$$ $$\left. \begin{array}{l} \dot{\alpha} \perp \vec{N}\\ \dot{\alpha} \perp \ddot{\alpha} \end{array}\right\} \Rightarrow \ddot{\alpha}\parallel \vec{N}$$ Which is what we wanted, but I still can't disprove what I was asked to (I can't find a counter example) And clearly it isn't a necessary condition, since the equator of a sphere with radius $R$ and center at the origin parametrized as $R(\cos\theta,\sin\theta,0)$ (which isn't naturally parametrized) has a parallel velocity vector.
$|\alpha'|'=0$ is a N&S condition: $$ \alpha' = R\vec{t} \Rightarrow \ \alpha'' = R\vec{t'} \\ $$ Here $\vec{t}$ is the tangent vector, and since it is unitary, we get $$ \vec{t} \perp \vec{t'} \Rightarrow \ \alpha' \perp \alpha'' $$ Then since $K_g = 0 = \frac{1}{R^3}(\vec{N},\alpha',\alpha'')$ and given the perpendicular relation before, the only way this determinat can be zero is if $\alpha'' \parallel \vec{N}$ Which is the same as having a covariant derivate zero.
In the other direction it is easy to see that if geodesic implies covariant derivative equal to $0$: $$ \alpha'' = (|\alpha'|\vec{t})’=|\alpha'|'\vec{t} + |\alpha'|\vec{t'} $$ And for this to be paralle to the perpendicular we need $|\alpha'|'$ to be zero. A counter example to proof can be the $\theta = \theta_0$ of the revolution surface $(f(x)\sin\theta, f(x)\cos\theta, z(x)$