Computing the curvature of a connection

Question

I'm trying to compute the curvature of a connection $\nabla$ on a 2-dimensional real vector bundle $E$ over a 2-dimensional manifold $M$. I'm using the definition $$F_{\nabla}(X,Y,s)= \nabla_X\nabla_Yf-\nabla_Y\nabla_Xf -\nabla_{[X,Y]}f$$ where $X,Y$ are vector field over $M$ and $s$ is a section of $E$. I know that $F_{\nabla}$ is $C^{\infty}(M)$-linear in all components, hence my idea is to compute the curvature just for vector field of the form $\frac{\partial}{\partial x}$ and $\frac{\partial}{\partial y}$, since they locally give a basis for the $C^{\infty}$-module of vector fields. Now, I'm not really familiar with Lie derivative, so I'm thinking of $[X,Y]$ just as an operator, but I think $[\frac{\partial}{\partial x},\frac{\partial}{\partial x}]$, $[\frac{\partial}{\partial x},\frac{\partial}{\partial y}]$, $[\frac{\partial}{\partial y},\frac{\partial}{\partial x}]$ and $[\frac{\partial}{\partial y},\frac{\partial}{\partial y}]$ should all be zero.

It follows that in my computation $\nabla_{[X,Y]}f$ always vanishes. Is it possible? Am I allowed to work locally like that?

Answer 1

Yes, if you use coordinate vector fields, then the Lie brackets all vanish and the formula simplifies to $$F(\partial_i, \partial_j)s = \nabla_{\partial_i}\nabla_{\partial_j}s - \nabla_{\partial_j}\nabla_{\partial_i}s.$$

Answer 2

You can assume that $[\frac{\partial}{\partial x}, \frac{\partial}{\partial x}]$ and the other two are equal to zero at a single point $p \in M$. You cannot assume that they are zero in an open neighbourhood. As you need open neighbourhoods to define the covariant derivates, $\nabla_{[X,Y]}f$ doesn't need to vanish.