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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 $v\in T_pM$


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?

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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.

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I understand all except the last, i don´t see how to follow from that, i mean what you mean by Lie derivative of $v$ with respect to $V$.I should prove that $\nabla_V v=0$ or something like that? (considering that $v$ is not a vector field – Dimitri Mar 18 '13 at 14:13
Indeed, I was thinking of $v$ as being a local vector field. You can always extend your vector $v$ to a vector field and then show the result doesn't depend on your extension. (So, no, you can't prove $\nabla_V v = 0$, since that will depend on how you extend $v$.) – Sam Lisi Mar 19 '13 at 23:56
After you've extended the vector field, you want to expand the LHS as $= \nabla_V(d\phi_t \cdot v)|_{t=0} = \nabla_{d\phi_t \cdot v} V|_{t=0} + [V, d\phi_t \cdot v]|_{t=0}$ first using the definition of covariant derivative and then using the fact it is torsion-free. You now want to rewrite this last term as $L_V(d\phi_t v)$ and evaluate it. – Sam Lisi Mar 20 '13 at 0:00
Thanks for the aclaration, but in the last term i think i´m going back where i started, or maybbe i don´t understand how you finish the prove from that – Dimitri Mar 20 '13 at 0:27
You know the total expression needs to be independent of the extension of $v$. However, you can extend $v$ however you want... what if you take an extension of $v$ that is invariant under the flow of $V$? Once you've seen what happens in that case, you can try to understand what happens more generally, though that case is sufficient to prove what you want. – Sam Lisi Mar 20 '13 at 10:26

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