Relation between exterior derivative and Lie bracket

3.9k Views Asked by At

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.

1

There are 1 best solutions below

1
On BEST ANSWER

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.