 Solution

Let $e(\omega )$ denote exterior multiplication, $e(\omega )\alpha =\omega \wedge \alpha$. For a $q$-form $\alpha$, evaluation of the proposed operator on vector fields $Y_0,\dots ,Y_q$ givessince ${\sum }_i{\omega }_i(Y_a)X_i=Y_a$. This expression is independent of the chosen local basis.

Expand each covariant derivative into the derivative of $\alpha (Y_0,\dots ,{\hat{Y}}_a,\dots ,Y_q)$ minus the terms obtained by differentiating each argument. For every pair $a<b$, those argument terms combine intoThe Levi-Civita connection is a torsion-free connection, so the vector in this expression is $[Y_a,Y_b]$. The result is exactly the defining evaluation formula for the exterior derivative:Therefore $\boxed{d=\sum _{i}e({\omega }_i){\nabla }_{X_i}}$. No orthonormality or compactness is needed for this identity; the dual-frame property and vanishing torsion tensor suffice.