How to evaluate the partial derivative of a differential?

Question

Suppose $M$ is a finite-dimensional smooth manifold and $f\in C^\infty (M)$. Let us now define function $F:TM\to \mathbb{R}$ by $(p,w)\mapsto \bigr(df(p)\bigr)(w)$ for $p\in M$ and $w\in T_pM$. Now, I want to show that if $\gamma:I\to M$ is a smooth curve and $(x,U)$ is a local chart in $M$ such that $\gamma(I)\subset U$; then $$\frac{\partial F}{\partial x^i}(\gamma(t),\gamma'(t))=\frac{d}{dt}\Bigr(\frac{\partial F}{\partial v^i}\bigr(\gamma(t),\gamma'(t) \bigr) \Bigr),$$ for $i=1,\ldots,$dim($M$); where $\displaystyle v^i:=\frac{\partial }{\partial x^i}$. I do know that this should follow easily from the chain rule, but I simply do not understand the partial derivative of $F$, seems realy weird to me.

Answer 1

I think you meant $v^i=dx^i$, otherwise ($x,v$) do not form a coordinate system. Note that $df$ is a 1-form on $M$; in particular, its restriction to $U$ can be written in the form $\sum_j A_j dx^j=\sum_jA_jv^j$ in the coordinates $(x,v)$ of $TM$ (Here $A_j=(\partial f/\partial x^j)$). Now we may compute both sides of your equation:

$$\frac{\partial F}{\partial x^i}(\gamma(t),\gamma'(t))=\sum_j\frac{\partial}{\partial x^i}\Bigr|_{(\gamma(t),\gamma'(t))} A_j v^j=\sum_j\frac{\partial A_j}{\partial x^i}(\gamma(t))\cdot (x^j\circ \gamma)'(t),$$ now compute the RHS taking into account that $\partial A_j/\partial v^i= 0$ $$\frac d{dt} \Bigr(\frac{\partial F}{\partial v^i}\Bigr)(\gamma(t),\gamma'(t))=\frac{d}{dt}A_i(\gamma(t))=\sum_k\frac{\partial A_i}{\partial x^k}(\gamma(t))\cdot(x^k\circ \gamma)'(t)$$ We used the chain rule in the last step, since $A_i$ is a partial derivative of $f$, we clearly have $\displaystyle\frac{\partial A_i}{\partial x^k}=\frac{\partial A_k}{\partial x^i}$(This is just a restatement of $d^2f=0$).