# Tensor Derivation

O’Neill 2.6: from 2.13, we know that $({\cal D}\theta)(X) = {\cal D}(\theta X) - \theta({\cal D} X)$.

Suppose that ${\cal D} \partial_{i} = \sum F_{i}^{j} \partial_{j}$. Thus

${\cal D}(dx^i)(\partial_j) = {\cal D}(dx^i(\partial_j)) - dx^i({\cal D} \partial_j) = -dx^i(\sum F_j^k \partial_k) = -F_j^i$.

And so we must have

${\cal D}(dx^i) = -\sum F_j^i dx^j$

as required.