I am reading the John Lee's Introduction to smooth manifolds, Proposition 20.8 and stuck at understanding some statement :
Im struggling with understand the underlined statements.
C.f. There, in the proof, the isomorphism $T_gG \oplus T_X\mathfrak{g} \cong T_{g,X}(G\times \mathfrak{g})$ is, $(v,w) \mapsto d(\iota_G)_g(v) +d(\iota_{\mathfrak{g}})_X(w)$, where $\iota_G : G \to G\times \mathfrak{g}$ sends $G$ to the slice $G \times \{X\}$ and similarly for $\iota_{\mathfrak{g}}$ (c.f. : Tangent Space of Product Manifold ).
Q. And why can we write $\Xi f(w^{i}, x^{i}) = x^jX_j f(w^{i},x^{i})$ locally? And what does the sentence "where each vector field $X_j$ differentiates $f$ only in the $w^{i}$-direction exactly means? And finally, why $x^jX_j f(w^{i},x^{i})$ depends smoothly on $(w^{i},x^{i})$?
As in the proof, let $(\mathfrak{g}, \eta:=(x^{i})) $ be the canonical global coordinates for $\mathfrak{g}$ and let $(U, \psi:=(w^{i}))$ be a smooth local coordinates for $G$. Then $(U\times \mathfrak{g} , \varphi)$, where $\varphi : (g, X:=x^{i}X_i) \mapsto (\psi(g)=(w^{i}(g)),\eta(X)=(x^{i}))$, is a smooth local coordinates for $G\times \mathfrak{g}$. Let $f \in C^{\infty}(U \times \mathfrak{g})$ be an arbitary smooth function. Note that $\Xi f : U \times \mathfrak{g} \to \mathbb{R}$. Then my question is, why the coordinate representation
$$\hat{\Xi f} := \Xi f \circ \varphi^{-1} : \hat{(U\times \mathfrak{g})}:=\varphi(U\times \mathfrak{g}) \subset \mathbb{R}^{k} \times \mathbb{R}^{k} \to \mathbb{R} ,$$
sends an element $((w^{i}), (x^{i}))$ to $x^{j}X_j f(w^{i},x^{i})$ ? Here I think that the notation $f(w^{i},x^{i})$ means the coordinate representation $(f \circ \varphi^{-1})((w^i),(x^{i}))$ (Am I following well the author's argument ?) I can't prove this calculation at all. How can we prove this formally? And how can we deduce the smoothness of $\hat{\Xi f}$ from this form of explicit correspondence?
EDIT (My first attempt) : Let $((w^{i}),(x^{i})) \in \varphi(U\times \mathfrak{g})$.
And let $\varphi^{-1}((w^{i}),(x^{i}))= (g:=\psi^{-1}((w^{i})), X:=x^{j}X_j)$. Then letting $\iota_G|_U : U \to U \times \mathfrak{g}$,
$$ \Xi f \circ \varphi^{-1}((w^{i}), (x^{i})) = \Xi_{(g,X)}f := (X_g, 0)f = (d(\iota_G)_g(X_g) + d(\iota_{\mathfrak{g}})_X(0))f = (d(\iota_G)_g(X_g))f = (d(\iota_G|_U)_g(X_g))f= X_g( f\circ (\iota_G|_U)) = (x^{j}X_j)_g(f\circ \iota_G|_U) \overset{\mathrm{x^{j} \in \mathbb{R}}}{:=} x^{j}(X_j)_{\psi^{-1}((w^{i}))}(f\circ\iota_G|_U).$$
For the fifth equality, we use the identification $T_gU \cong T_gG $ and $T_{(g,X)}(U\times \mathfrak{g}) \cong T_{(g,X)}(G\times \mathfrak{g})$. Note that $f\circ \iota_G|_U \in C^{\infty}(U)$. Since $X_1, \dots X_k$ are smooth vector fields on $G$, each functions $X_j(f\circ \iota_G|_U)$ is smooth on $U$ (c.f. his book Proposition 8.14). In particular, $(X_j(f\circ \iota_G|_U)) \circ \psi^{-1} : \psi(U) \to \mathbb{R}$ is smooth.
Perhaps, from these data, can we show that $\Xi f \circ \varphi^{-1}$ is smooth on $\varphi(U\times \mathfrak{g})$? If so, how formally?
Can anyone help?
