$\textbf{Theorem:}$ Suppose that $X_1, \dots, X_k$ are vector field on a manifold $M$ and at a point $p \in M$, we have that $X_1(p), \dots, X_k(p) \in T_p(M)$ are linearly independent, then the Lie brackets, $$[X_i, X_j]$$ vanish in a neighborhood of $p$ if and only if there exists a coordinate chart containing $p$ such that in these coordinates $X_i= \frac{\partial}{\partial x^i}$.
$\textbf{Example:}$ Let $\frac{\partial}{\partial x}$ and $\frac{\partial}{\partial y} + x \frac{\partial}{\partial z}$ be vector field in $\mathbb{R}^3$. Then their Lie bracket is $\frac{\partial}{\partial z}$ and the integral curves are $$ (a + t, b ,c)$$ and $$(a, b + s, c +as)$$ respectively.
From this example my notes concluded that we $\textbf{don't get a surface at all that matches these two vector field}$.
This conclusion was drawn from looking at the integral curves.
- What is the relationship between integral curves, vector fields, and forming a surface?
I see the Lie bracket is not equal to zero in this case, but I am not sure how the theorem is related to the example. Does the Lie bracket not vanishing imply that we don't have a surface?