 Solution

For the Lie derivative of a tensor field, the scalar rule follows because a vector field $X$ differentiates products of smooth functions. On vector fields the Lie bracket of vector fields satisfiesThis follows by applying both sides to an arbitrary smooth function and expanding the two compositions of derivations. Both the scalar and vector-field operators are real-linear, so the extension theorem applies and gives the unique contraction-compatible tensor operator ${\mathcal{L}}_X$.

For a type-$(1,1)$ tensor $A$, regard it as the pointwise endomorphism of $TM$ obtained by contracting its covector slot with a vector. The endomorphism-induced tensor derivation starts withIt is real-linear and obeys $D_A(fY)=fA(Y)=f{D}_{A}Y+(D_{A}f)Y$. The scalar operator zero is a derivation in every dimension, so there is no zero-dimensional obstruction here. Consequently it also extends uniquely. On a differential one-form, the two operators areTheir actions on arbitrary tensor fields follow by the tensor-product rule; $D_A$ adds $A$ in every contravariant slot and subtracts its dual action in every covariant slot.

Functions in this calculation belong to $C^{\infty }(M)$: the printed $C^{\infty }(X)$ in Q2(c) is a typographical error. The contraction $C_2^1$ pairs the sole covector with the new vector $Y$.