Vector field and integral curve Let M a riemaniann manifold, $V$ a vector field in M, and $\phi_t$ the respective flow
Let $p\in M$, and $\gamma$ the integral curve of $V$. Show that for all $v\in T_pM$
$$\frac{D}{dt}\bigg|_{t=0}(d\phi_t)_pv=\nabla_vV$$
where $\frac{D}{dt}$ is the covariant derivate along $\gamma$ and $\nabla$ is the Levi-Civita conection.
Anyone can help me? Any idea at least?
 A: The answer requires you to unwrap some definitions. Let me give you a starting point.
My suggestion is to look at what the equation is trying to tell you about geometric quantities. You are given an ODE on $M$ by $\dot \gamma(t) = V(\gamma(t))$. This gives you a $1$-parameter family of diffeomorphisms of $M$ by following the flow $\phi_t$. You are particularly interested in the integral curve of this equation passing through $p$. The quantity $d\phi_t(p) v$ the linearization of this flow acting on a vector at the point $p$. You want to differentiate this with respect to time (at $t=0$) and see what you get.
What is this geometrically? this is saying that you look at your ODE, and think of the time $t$ flow $\phi_t$ and look at what this does to initial conditions near $p$. As you wobble the initial condition $p'$ near $p$, you look at how $\phi_t(p')$ varies from $\phi_t(p)$.
Now, there are two ways of measuring what the linearized flow is doing infinitesimally (i.e. how an infinitesimally small "wobble" changes the output). The first is by considering this covariant derivative, as you have on the left of your equation. The second is by considering the Lie derivative of $v$ with respect to $V$. Now, what is the relationship between the two? This is where you use that the Levi-Civita connection is torsion-free! Torsion free means
\[
\nabla_X Y - \nabla_Y X = [X, Y] = L_X Y.
\]
I think this should be enough to get you started, but I am happy to tell you more if you get stuck -- leave me a comment to let me know.
