Relation between exterior derivative and Lie bracket There is a formula connecting the exterior derivative and the Lie bracket 
$$d\omega (X,Y) = X \omega(Y) - Y \omega(X) -  \omega([X,Y]).$$
What is a good way to remember this? By which I mean, what structure does this reveal? (Or, what essentially is going on here?) In his book, John Lee says, "In a sense, the Lie bracket is dual to the exterior derivative." But that's not really satisfying to me.
 A: I cannot improve on Ted Shifrin's answer describing why the bracket term is necessary. So, I'll only attempt in this answer to elaborate the sense in which the exterior derivative and bracket are dual.
Fix a local frame $(E_a)$ and let $(\theta^a)$ denote its dual coframe, so that $\theta^a (E_b) = \delta^a{}_b$; in particular each such contraction is constant. Then, for frame and coframe elements the exterior derivative formula simplifies to
$$\phantom{(ast)} \qquad d\theta^a(E_b, E_c) = -\theta^a([E_b, E_c]) . \qquad (\ast)$$ But the right-hand side is (up to sign) exactly the structure function $C^a{}_{bc}$ of the frame, and the structure functions tell us everything there is to know about the geometry of the frame $(E_a)$, or equivalently the coframe $(\theta^a)$. Put another way:

The exterior derivatives $d\theta^a$ contain the same information encoded by the Lie bracket but instead expressed in the (dual) language of the dual coframe.

A: Eric, if you take $X=\partial/\partial x^i$, $Y=\partial/\partial x^j$, and $\omega = f dx^k$, then of course the formula gives you what you expect from the first two terms, and the bracket term disappears because $[X,Y]=0$ in this case. Why is the bracket term there in general? It's because $d\omega$ is a tensor field, and therefore must be linear over the $C^\infty$ functions. However, $X(\omega(Y)) - Y(\omega(X))$ is not linear over the $C^\infty$ functions, and it is the bracket term that precisely corrects for that. In general, formulas such as this that one can guess by applying to coordinate vector fields have bracket terms appearing exactly for that reason.
