Let $M$ be a smooth manifold of dimension $n$. Let $U \subseteq M$ be an open subset over which the tangent bundle $TM$ is trivial. Let $X_1,\ldots,X_n : U \to TM$ be a local frame for $TM$ and let $\omega_1,\ldots,\omega_n : U \to T^*M$ be the corresponding coframe. So, by definition, we have: \begin{align*} \omega_i (X_j)= \delta_{ij}. \end{align*}
In principle, any 1-form is a unique $C^\infty(U)$-linear combination of $\omega_1,\ldots,\omega_n$. In particular, that means we can think of the exterior derivative $d : C^\infty(U) \to C^\infty(U;T^*M)$ as a combination of $\omega_1,\ldots,\omega_n$ whose coefficients are operators $C^\infty(U) \to C^\infty(U)$. A bit of thought reveals that actually the coefficient operators should be vector fields. For a calculation, I wanted to know explicitly what these coefficient vector fields are. My presumption was that they would be built out of the vector fields $X_i$ and their Lie brackets $[X_i,X_j]$. However, to my surprise, the Lie brackets do not enter into the picture. The coefficient operator for $\omega_i$ is simply $X_i$. In other words: \begin{align} df &= \sum_{i=1}^n (X_i f) \omega_i && \text{ for all } f \in C^\infty(U). \tag{1} \end{align}
Of course there is a case where this is not surprising. If we have a smooth local coordinate system $(x_1,\ldots,x_n)$ on $U$ and are working with $X_i = \frac{\partial}{\partial x_i}$ and $\omega_i = dx_i$, then this is just the familiar expression: \begin{align*} df = \sum_{i=1}^n \tfrac{\partial f}{\partial x_i} dx_i. \end{align*} On the other hand, I am was quite surprised that this also works with respect to a general frame $(X_i)$ and coframe $(\omega_i)$. This doesn't seem particularly difficult to prove. Choosing local coordinates, we can write \begin{align*} X_i = \sum_j a_{ij}\tfrac{\partial }{\partial x_j} && \text{ and } && \omega_i = \sum_j b_{ij} dx_j \end{align*} where the smooth function coefficients are related by \begin{align*} \sum_{i=1}^n a_{ij} b_{ik} = \delta_{jk}. \end{align*} (Note: In matrix formalism, $A = [a_{ij}]$ and $B=[b_{ij}]$ are smooth functions $U \to \operatorname{GL}(n,\mathbb{R})$ related by $B = (A^t)^{-1}$.) With these notations, for any $f \in C^\infty(U)$, we have \begin{align*} \sum_{i=1}^n (X_i f_i) \omega_i &= \sum_{i,j,k=1}^n a_{ij}b_{ik} \tfrac{\partial f}{\partial x_j} dx_k \\ &= \sum_{j,k=1}^n \left( \sum_{i=1}^n a_{ij}b_{ik} \right) \tfrac{\partial f}{\partial x_j} dx_k \\ &= \sum_{k=1}^n \tfrac{\partial f}{\partial x_k} dx_k \\ &=df \end{align*}
Question: Should (1) surprise me? My intuition was that we should not expect (1) to hold without the additional assumption that $[X_i,X_j]=0$, the vanishing of Lie brackets being the obstruction to integrating the vector fields to get a local coordinate system. However, it seems my intuition is wrong here as nothing more than linear algebra enters in the computation above. We never end up differentiating any of the cooefficient functions. It's just linear algebra with no calculus.
Any insight? I am finding this a bit surprising and would value any comments which remove some or all of the "magic" from (1) for me.
Absolutely not surprising. It is basic linear algebra. For any vector space $W$, any $\lambda\in W^*$ and any basis $\{w_1,\dots, w_m\}$ of $V$ with dual basis $\{\omega^1,\dots,\omega^m\}$ of $W^*$, $\lambda$ has the basis expansion $\lambda=\sum_{i=1}^m\lambda(w_i)\,\omega^i$.
Taking now for example $W=\bigwedge^k(V)$ for some vector space $V$, you’ll find a similar result. For any $\lambda\in\left[\bigwedge^k(V)\right]^*\cong\bigwedge^k(V^*)$, if you’re given a basis $\{v_1,\dots, v_n\}$ of $V$, and you construct the corresponding dual basis $\{\nu^1,\dots,\nu^n\}$ of $V^*$ and the induced bases of the exterior powers, you find that \begin{align} \lambda=\sum_{I\uparrow}\lambda(v_I)\,\nu^I \end{align} (the sum over increasing multiindices).
Finally, in the manifold setting, you’re just doing this one tangent space at a time. So, the equation (1) is not just for differential $1$-forms on manifolds, but has a natural analogue for differential $k$-forms as well.