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.

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