Proper Way to view vector field component functions with respect to covariant derivative

Question

I follow this lecture https://www.youtube.com/watch?v=2eVWUdcI2ho&list=PLFeEvEPtX_0S6vxxiiNPrJbLu9aK1UVC_&index=8 and at minute 35, the covariant derivative is used to calculate the geodesic equation. The lecturer says, that the term to calculate the second derivative is strictly not defined. He is right, because Gamme dot q, is a function from R to R and not from M to R. So the partial operator is not defined. How does he calculate the second derivative and how to make sense out this. Is it better to think of the vector field component functions as function from M to R? In the lecture they are functions from R to R. Iam grateful for any help!

Answer 1

The point is that although $\frac{\partial}{\partial x^m}\dot\gamma^n$ doesn't exist (since $\dot\gamma^n$ is only defined along the curve $\gamma$), the quantity $$ \left(\dot\gamma^m\frac{\partial}{\partial x^m}\right)\dot\gamma^n $$ does exist, since it really means $$ \frac{d}{dt}\dot\gamma^n(\gamma(t)), $$ i.e. the derivative of $\dot\gamma^n$ along the curve $\gamma$. And this latter is just $\ddot\gamma^n$.

As an alternative way to understand this, it's probably conceptually clearer to say $$ \nabla_{v_\gamma}v_\gamma = \nabla_{v_\gamma}\left(\dot\gamma^n\frac{\partial}{\partial x^n}\right) = v_\gamma(\dot\gamma^n)\frac{\partial}{\partial x^n} + \dot\gamma^n\nabla_{v_\gamma}\frac{\partial}{\partial x^n} $$ and then using $v_\gamma(\dot\gamma^n) = \ddot\gamma^n$ (since $v_\gamma(f) = \dot f$ for any function $f$ by definition).