# Orthogonal projection onto the tangent space

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?

• 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. – Travis Jan 11 '16 at 13:30
• Yes, that's right. – Travis Jan 11 '16 at 13:40
• 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&#39;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 $\gamma$. – Travis Jan 11 '16 at 14:07
• 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. – Travis Jan 11 '16 at 14:19
• You're welcome, I'm glad you found my comments useful. – Travis Jan 11 '16 at 14:48