Proving $H(\gamma',\gamma')=d^2(f\circ\gamma)/ds^2$

Question

If $\gamma$ is a geodesic and $f\in\mathscr{T}(M)$, then $H(\gamma',\gamma')=d^2(f\circ\gamma)/ds^2$. H is the hessian.

I tried to apply the Hessian formula proven in the chapter: $H^f(\gamma',\gamma')=\langle D_{\gamma'}(grad f),\gamma'\rangle=\langle \sum\frac{d gradf\circ\gamma}{dt}+\sum gradf^i D_{\alpha'(t)}\partial_i,\gamma'\rangle$

But I do not see how I can get a second derivative out of this. I expect to use the geodesic definition, where $D_{\gamma''}=0$ but that does not show here.

Question:

How do I simplify the expression?

Thanks in advance!

Answer 1

I ended up coming up with an answer.

By the definition of 1-form

$$ \nabla d f (\gamma', \gamma') = \gamma' \gamma' f - \nabla_{\gamma'}\gamma' f $$ then it follows from gamma being a geodesic ie. $\nabla_{\gamma'}\gamma' = 0$$