I read that given an open subset $U$ of a manifold $M$, one can give the set of all smooth vector fields over $U$ the structure of a $C^\infty(U)$-module. I checked that defining $(X + Y)(m) := X(m) + Y(m) \quad\forall m \in U$ and $(fX)(m) := f(m)X(m)$ satisfies the module axioms, but I'm having trouble showing that the vector fields on the right side of each of these equations are in fact smooth maps $U \to T(M)$.
Clearly the proof relies on the manifold structure given to $T(M)$ by coordinate maps of the form $$\tilde{\phi} : \pi^{-1}(V) \to \mathbb{R}^{2d}, \qquad \tilde{\phi}(v) = (x_1(\pi(v)),\ldots, x_d(\pi(v)),dx_1(v),\ldots,dx_d(v)),$$
where $(\phi,V)$ is a coordinate system on $M$ with coordinate functions $x_1,\ldots,x_d$ and $\pi$ is the natural projection $T(M) \to M$. I know this much, but I'm getting lost in the details.
Best Answer
(I'll call the induced chart on the tangent bundle $(T \phi, TV)$ for quicker typing).
We have to prove that $T \phi \circ(fX) \circ \phi^{-1} : \phi[V] \subset \Bbb{R}^n \to T \phi[V] \subset \Bbb{R}^n \times \Bbb{R}^n$ is smooth, from the assumption that $X$ and $f$ are smooth, which means that $T \phi \circ X \circ \phi^{-1}$ and $f \circ \phi^{-1}$ are smooth maps (from the open subset $\phi[V] \subset \Bbb{R}^n$ into $\Bbb{R}^n \times \Bbb{R}^n$ and $\Bbb{R}$ respectively).
Well, now we just apply the definition of $T\phi$ and compute the compositions. Given any $\xi \in \phi[V]$, we have that \begin{align} (T \phi \circ (fX) \circ \phi^{-1})(\xi) &= T \phi \left( f(\phi^{-1}(\xi)) \cdot X(\phi^{-1}(\xi)) \right) \end{align} So, the vector $v$ we are considering is $f( \phi^{-1}(\xi)) \cdot X(\phi^{-1}(\xi)) \in T_{\phi^{-1}(\xi)}M$. If we apply $T\phi$ to this vector, the first $n$ entires will be the coordinates of the base point $\phi^{-1}(\xi)$, and the second $n$ entires will be the components of the vector (i.e apply $dx_i$ to this vector $v$). Hence, \begin{align} (T \phi \circ (fX) \circ \phi^{-1})(\xi) &= \left( \phi(\phi^{-1}(\xi)), \dots, dx_i[f( \phi^{-1}(\xi)) \cdot X(\phi^{-1}(\xi))] \dots \right) \in \Bbb{R}^n \times \underbrace{\Bbb{R}\times \dots \times \Bbb{R}}_{n\, \text{times}} \\ &= \left( \xi, \dots, (f \circ \phi^{-1})(\xi) \cdot dx_i (X \circ \phi^{-1})(\xi) \dots \right) \in \Bbb{R}^n \times \underbrace{\Bbb{R}\times \dots \times \Bbb{R}}_{n\, \text{times}} \end{align} (Here, I used the fact that $dx_i$ is linear in each tangent space, so that $dx_i(cv) = c \cdot dx_i(v)$) Compare this with $(T\phi \circ X \circ \phi^{-1})(\xi) = \left( \xi, \dots, dx_i (X \circ \phi^{-1})(\xi) \dots \right) \in \Bbb{R}^n \times \underbrace{\Bbb{R}\times \dots \times \Bbb{R}}_{n\, \text{times}}$.
In other words, $(T \phi \circ (fX) \circ \phi^{-1})(\xi)$ is that $2n$-tuple whose first $n$ entries are simply $(\xi_1, \dots, \xi_n)$, and whose $(n+i)^{th}$ entry is obtained by multiplying the $(n+i)^{th}$ entry of $(T \phi \circ X \circ \phi^{-1})(\xi)$ and $(f \circ \phi^{-1})(\xi)$.
Stated as an equality of functions, If we write the coordinate functions as $(T \phi \circ (fX) \circ \phi^{-1})(\cdot) = (y_1(\cdot), \dots, y_{2n}(\cdot))$ then for $1 \leq i \leq n$, we have that \begin{align} y_i(\cdot) = (\text{id}_{\Bbb{R}^n})_i(\cdot) \quad \text{and} \quad y_{n+i}(\cdot) = (f \circ \phi^{-1})(\cdot) \cdot dx_i[X \circ \phi^{-1}(\cdot)] \end{align} Then, clearly each $y_i, y_{n+i}$ is a smooth function $\phi[V] \to \Bbb{R}$, hence the entire thing is smooth as a map from $\phi[V] \subset \Bbb{R}^n \to T\phi[V] \subset \Bbb{R}^{2n}$, which is what we needed to prove to show that $fX$ is a smooth vector field.
If you follow a similar reasoning for the sum $X+Y$ of vector fields, you'll find that \begin{align} y_i(\cdot) = (\text{id}_{\Bbb{R}^n})_i(\cdot) \quad \text{and} \quad y_{n+i}(\cdot) = dx_i[Y \circ \phi^{-1}(\cdot)] + dx_i[X \circ \phi^{-1}(\cdot)] \end{align} Hence, the sum of smooth vector fields is also smooth.