Let $M$ be a manifold and let $\{\alpha_k\}$ be a set of $1$- forms s.t. $\{\alpha_k(p)\}$ forms a basis for $T^*_pM$.
Let $(x,U)$ be a chart based in $p$ and denote $\partial_i := \frac{\partial}{\partial x^i}$. Now Consider the $X_k=\large{\sum_i \alpha_k(\partial_i)\partial_i}$. By linear algebra $\{X_k(p)\}$ is linearly independent.
Is it necessarily true that there is a neighborhood of $p$ in which $[X_i,X_j]=0$ for all $i,j$?
I know that whenever all $\alpha_k$'s are exact the conclusion holds since we can invert the jacobian and so we have a local diffeomorphism. Is it true in this larger setting?
On the first quadrant in $\mathbb{R}^2$, let $\alpha_1 = y dx$ and $\alpha_2 = x dy $. The form a basis since $x\neq 0 \neq y$ in the first quadrant.
Use the identity chart to compute $X_1$ and $X_2$, getting $X_1 = y \partial_x$ and $X_2 = x \partial_y$.
For any smooth $f$, we compute: \begin{align*} [X_1,X_2]f &= X_1(X_2(f)) - X_2(X_1(f))\\ &= X_1(x f_y) - X_2(y f_x) \\ &= yf_y + yxf_{yx} -(xf_x + xyf_{xy})\\ &= yf_y - xf_x.\end{align*}
So, $[X,Y] =y \partial_y - x\partial_x$ which is nonzero at every point in our manifold.