classical example that I see a lot and don't understand (this one from Lie groups and Lie algebra):
$T_0\mathbb{R}^n \cong \mathbb{R}^n$ (denote the isomorphism as $\phi:T_0\mathbb{R}^n \rightarrow \mathbb{R}^n$)
$X,Y \in \mathbb{R}^n$
$L_g:\mathbb{R}^n\rightarrow \mathbb{R}^n : x \mapsto g+x$
${L_g}_{*h}(X) = \frac{d}{dt}(g+h+tX)\vert_0 = X$
My question his how do we calculate ${L_g}_{*h}(X)$? Normally ${L_g}_{*h}(X)$ is from $T_0\mathbb{R}^n$ to $T_0\mathbb{R}^n$. So how is this calculated properly?
Is it in fact an abuse of notation? it should actually be ${L_g}_{*h}(\phi^{-1}(X))$
Also $T_0\mathbb{R}^n$ acts on fonctions, but how does this action translate to $\mathbb{R}^n$?
Any help would welcome.
In fact $T\mathbb R^n$ is "really" $\mathbb R^n\times \mathbb R^n$. In the sense that there exists a canonical isomorphism between them. (It's a bit silly but also an important exercise to do.)
To compute ${L_g}_*$ you just have to compute a differential as usual. The fact that $$ L_g(x) = x+g $$shows you that $$ L_g(x + h) = (x+h)+g = (x+g) +h = L_g(x) + h $$ so its differential is just the identity.
Ok, now, I choose a base $(x_1,\dots,x_n)$ for $\mathbb R^n$. Let's take a vector field $$ V = \sum a_i \frac{\partial}{\partial x_i} $$ the isomorphism $T\mathbb R^n\cong \mathbb R^n\times \mathbb R^n $ is given by $$ V\mapsto (a_1,\dots,a_n)=: W. $$ If $f\in C^\infty(M)$ then $$ V(f) = \sum a_i \frac{\partial f}{\partial x_i} = \left\langle (a_1,\dots,a_n), \left(\frac {\partial f}{\partial x_1},\dots,\frac {\partial f}{\partial x_n}\right)\right\rangle = \langle W, df\rangle .$$ And I think this last expression is what you are looking for.