Let $(M,g)$ be a Riemannian manifold. From $g$ and a fixed vector field $V$ we can derive the following two differential forms:
- A $1$- form $\alpha(X) = g(V,X)$, i.e. $\alpha = \iota_Vg$.
- A $2$-form $\beta(X,Y) = g(V,[X,Y]) = g(V,\mathcal{L}_XY)$.
Notice that we obviously have $\beta(X,Y) = \alpha([X,Y]).$
Is there a nice expression for the exterior derivative $d\alpha$ and $d\beta$ of these forms? In particular, when are they zero?
I can get expressions in local coordinates, but they are kinda ugly and I cannot relate them to other known quantities.
A literal back-of-the envelope calculation leads me to $$d\alpha(X,Y)=g(\nabla_XV,Y)-g(\nabla_YV,X).$$ Does that help?
EDIT: Here is what I presume, 2 1/2 years later, was on the back of the envelope (with a few extra words).
Since $\nabla$ is $g$-compatible, we have $\nabla g = 0$, and so $$X(g(V,Y)) = \nabla_X(g(V,Y)) = g(\nabla_X V,Y) + g(X,\nabla_X Y).$$ Also, since $\nabla$ is torsion-free, we have $\nabla_X Y - \nabla_Y X = [X,Y]$, so $g(V,[X,Y]) = g(V,\nabla_X Y)-g(V,\nabla_Y X)$.
Using the standard formula for the derivative of a $1$-form, we have \begin{align*} d\alpha(X,Y) &= X(g(V,Y)) - Y(g(V,X))- g(V,[X,Y]) \\ &= g(\nabla_X V,Y) + g(V,\nabla_X Y) - g(\nabla_V, X) - g(V,\nabla_Y X) - \big(g(V,\nabla_X Y)-g(V,\nabla_Y X)\big)\\ &= g(\nabla_X V,Y) - g(\nabla_Y V,X), \end{align*} as required.