= 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> $Y_0,\ldots,Y_q$ gives
$$
\left(\sum_i\omega_i\wedge\nabla_{X_i}\alpha\right)(Y_0,\ldots,Y_q)
=\sum_{a=0}^q(-1)^a(\nabla_{Y_a}\alpha)(Y_0,\ldots,\widehat Y_a,\ldots,Y_q),
$$
since $\sum_i\omega_i(Y_a)X_i=Y_a$. This expression is independent of the chosen local <basis>.
Expand each <covariant derivative> into the derivative of $\alpha(Y_0,\ldots,\widehat Y_a,\ldots,Y_q)$ minus the terms obtained by differentiating each argument. For every pair $a<b$, those argument terms combine into
$$
(-1)^{a+b}\alpha(\nabla_{Y_a}Y_b-\nabla_{Y_b}Y_a,
Y_0,\ldots,\widehat Y_a,\ldots,\widehat Y_b,\ldots,Y_q).
$$
The <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>:
$$
(d\alpha)(Y_0,\ldots,Y_q)
=\sum_a(-1)^aY_a\bigl(\alpha(Y_0,\ldots,\widehat Y_a,\ldots,Y_q)\bigr)
+\sum_{a<b}(-1)^{a+b}\alpha([Y_a,Y_b],Y_0,\ldots,\widehat Y_a,\ldots,\widehat Y_b,\ldots,Y_q).
$$
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.
Back to article page
