 Solution

Write a local frame as $e_a$, its dual basis as ${ϵ}^a$, and set $D{e}_a=A^b{}_a{e}_b$. The one-form definition gives $D{ϵ}^a=-A^a{}_b{ϵ}^b$. Repeated indices are summed. Every tensor field has a unique local expansionDefine $DT$ by applying the scalar operator to its coefficient and applying $D$ to one frame factor at a time, adding all these terms. In components this isIn each term only the indicated slot is replaced. This formula is real-linear and satisfies the Leibniz rule for a tensor product: differentiating a coefficient product uses the scalar product rule, and the list of differentiated frame factors splits into the two factors' lists.

To check that it is intrinsic, write $e$ as the row of frame fields and $ϵ$ as the column of dual fields. Change frame by $e^{\prime }=eP$, where $P$ is an invertible smooth matrix. ThenDifferentiating the inverse matrix gives $Z(P^{-1})=-P^{-1}(ZP)P^{-1}$; it cancels the extra frame-change terms in the dual factors. The same cancellation in each tensor slot makes the two local formulas agree. Thus they glue to a global tensor derivation.

In particular, for arbitrary tensor types, not necessarily the same type,The resulting type is the sum of their two covariant counts and their two contravariant counts. This also supplies the required real-linearity for every type, including functions as type $(0,0)$.