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Let $\tilde{\gamma}$ be a reparametrization of $\gamma$, so that $\tilde{\gamma}(t) = \gamma (\phi (t))$ for some smooth function $\phi$ with $\frac{d\phi}{dt} = 0$ for all values of $t$.

If $v$ is a tangent vector field along $\gamma$, we have that $\tilde{v}(t) = v(\phi (t))$ is one along $\tilde{\gamma}$.

I want to prove that $$\nabla_{\tilde{\gamma}}\tilde{v}=\frac{d\phi}{dt}\nabla_{\gamma}v$$

We have that $\nabla_{\tilde{\gamma}}\tilde{v}$ is the orthogonal projection of $\frac{d\tilde{v}}{dt}$ onto $T_{\tilde{\gamma}(t)}S$, so onto $T_{\gamma (\phi (t))}S$.

We also have that $$\nabla_{\tilde{\gamma}}\tilde{v}=\frac{d\tilde{v}}{dt}-\left (\frac{d\tilde{v}}{dt}\cdot \textbf{N}\right )\textbf{N}$$

Could you give me a hint ho we get to the desired result?

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    $\begingroup$ Hint Apply the chain rule to $\frac{d \tilde v}{dt} = \frac{d}{dt} v(\phi(t))$ and substitute in your formula for orthogonal projection. $\endgroup$ – Travis Jan 11 '16 at 13:30
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    $\begingroup$ Yes, that's right. $\endgroup$ – Travis Jan 11 '16 at 13:40
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    $\begingroup$ We should get $\frac{d\phi}{dt} \nabla_{\ast} \nu$ for an appropriate $\ast$: reverse substituting using the definition of the projection gives that the argument in $\nu$. On the other hand, $\ast$ just tells us the subspace onto which we're projecting (namely, the tangent space to the curve at the point, which does not depend on the parameterization), so we may as well take $\ast$ to be $\gamma$. $\endgroup$ – Travis Jan 11 '16 at 14:07
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    $\begingroup$ Because $\gamma$ and $\tilde \gamma$ are different parameterizations of the same curve $S$, and the tangent space to a point on a curve does not depend on the parameterization. In fact, this is essentially the content of the equation $T_{\tilde \gamma(t)} S = T_{\gamma (\phi(t))} S$ given in the question. $\endgroup$ – Travis Jan 11 '16 at 14:19
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    $\begingroup$ You're welcome, I'm glad you found my comments useful. $\endgroup$ – Travis Jan 11 '16 at 14:48

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