Let M be a differentiable manifold. On page 26 of Riemannian Geometry by Do Carmo, he makes this statement:
Observe that if $\varphi:M\longrightarrow M$ is a diffeomorphism, $v\in T_pM$ and $f$ is a differentiable function in a neighborhood of $\varphi(p)$, we have
$$(d\varphi(v)f)\varphi(p)=v(f\circ\varphi)(p).$$
Indeed, let $\alpha:(-\varepsilon,\varepsilon)\longrightarrow M$ be a differentiable curve with $\alpha'(0)=v$, $\alpha(0)=p$. Then
$$(d\varphi(v)f)\varphi(p)=\left.\frac{d}{dt}(f\circ\varphi\circ\alpha)\right|_{t=0}=v(f\circ\varphi)(p).$$
I understand what he means by the right hand side. Namely, a vector $v$ in the tangent space at $p$ acting as a derivation on the differentiable function $f\circ\varphi$ and then evaluating the resultant function at $p$. However, I'm unsure if I'm understanding what the object on the left hand side is supposed to be. You take the differential of the map $\varphi$ (not at any point in particular?, I assume there's supposed to be a subscript $p$?) and then act this Jacobian on the vector $v$. This is fine, and it should give you a vector in the tangent space at $\varphi(p)$. You then treat this as a derivation and so it maps $f$ to some other differentiable function which you then evaluate at $\varphi(p)$. Is this correct? If so, I still can't see where the middle step comes from.
Best Answer
Let $g = f\circ \varphi : M \to M $. Then $g$ is differentiable and the chain rules says $$ \mathrm{d}g (p) = \mathrm{d}(f\circ \varphi)(p) = \mathrm{d}f(\varphi(p))\circ\mathrm{d}\varphi(p). $$ If $\alpha$ is a path in $M$ with $\alpha(0)=p$ and $\alpha'(0)=v$, then by definition $$ (g\circ\alpha)'(0) = \mathrm{d}g(\alpha(0))\alpha'(0)=\mathrm{d}(f\circ \varphi)(p)v = v\cdot \left( f\circ \varphi\right), $$ where the last equality is just a notation for the vector $v$ acting on the function $f\circ \varphi$ by evaluating its differential. But if you write $\beta = \varphi\circ \alpha$, which is a path with $\beta(0) = \varphi(p)$ and $\beta'(0) = \mathrm{d}\varphi(p)v$ $$ (f\circ\beta)'(0) = \mathrm{d}f(\beta(0))\beta'(0) = \mathrm{d}f\left(\varphi(p)\right) \left(\mathrm{d}\varphi(p)v\right) = \left(\mathrm{d}\varphi(p)v\right)\cdot f, $$ where the last equality is the vector $\mathrm{d}\varphi(p)v$ acting on the function $f$.
The equality, $g \circ \alpha = f \circ \beta$ shows that $$ v\cdot\left(f\circ \varphi \right) = \left(\mathrm{d}\varphi(p)v \right)\cdot f. $$