What is the skew-symmetric part of the covariant derivative of a one-form?

Question

This is a followup question to here.

Let $E \to M$ be a vector bundle with connection $D := \nabla$. Extend $D$ to $E^*$ and $\text{Hom}(E, E)$. Let $E = TM$ here, and suppose that the torsion is $0$:

$$T(X, Y) = \nabla_XY - \nabla_YX - [X, Y] = 0.$$

Does it follow that $d\theta$, the exterior derivative of a $1$-form $\theta$, is the skew-symmetric part of $D\theta$?

Answer 1

Note that

$$(D\theta)(X, Y) = (\nabla\theta)(X, Y) = (\nabla_X\theta)(Y) = X(\theta(Y)) - \theta(\nabla_XY).$$

So the skew-symmetric part of $(D\theta)(X, Y)$ is

\begin{align*} \frac{1}{2}[(D\theta)(X, Y) - (D\theta)(Y, X)] &= \frac{1}{2}[X(\theta(Y)) - \theta(\nabla_XY) - Y(\theta(X)) + \theta(\nabla_YX)]\\ &= \frac{1}{2}[X(\theta(Y)) - Y(\theta(X)) - \theta(\nabla_XY - \nabla_YX)]\\ &= \frac{1}{2}[X(\theta(Y)) - Y(\theta(X)) - \theta([X, Y])], \end{align*}

which is either $(d\theta)(X, Y)$ or $\frac{1}{2}(d\theta)(X, Y)$, depending on your convention.