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A smooth endomorphism $A$ of the tangent bundle defines the tensor derivation $D_A$ by $D_{A}f=0$ and $D_{A}Y=A(Y)$. On a differential one-form, $D_A\omega =-\omega \circ A$. On a general tensor field, it acts by $A$ in each vector factor and by the negative dual action in each covector factor. The two actions cancel in each contracted pairing, proving compatibility with tensor contraction.

For a smooth vector field $X$, the Lie derivative is the tensor derivation determined by ${\mathcal{L}}_{X}f=X(f)$ and ${\mathcal{L}}_{X}Y=[X,Y]$. The identity $[X,fY]=f[X,Y]+X(f)Y$ permits its unique extension. On a differential one-form,On differential forms it agrees with the usual Lie derivative of a differential form and Cartan's magic formula.

For a decomposable tensor field, consider a selected pairing $\omega (Y)$ in a tensor contraction. Its derivative isby the defining one-form formula. In the Leibniz rule expansion of the uncontracted tensor, the terms differentiating these two selected factors combine into exactly this derivative of the pairing. Every term differentiating another factor passes unchanged through the contraction. Thus, first on local decomposable tensors and then by linearity on every local tensor expansion,This is basis independent; in components the negative covector term and positive vector term for the two contracted slots cancel.

For uniqueness, any extension satisfying the tensor-product rule is local. If $T$ vanishes near $p$, multiply it by a smooth cutoff function $h$ equal to one near $p$ and supported where $T=0$. The identity $D(hT)=(Dh)T+hDT$ implies $DT(p)=0$. Any allowed extension on a differential one-form is forced by differentiating its contraction with every vector field. Its value on every local frame tensor is then forced by the tensor-product rule and its value on scalar coefficients. These local tensors span each tensor bundle, so two extensions agree everywhere.

Together with the preceding construction, this proves existence and uniqueness of the contraction-compatible tensor derivation, with the scalar-rule qualification already stated. No claim that every global tensor field is a finite sum of products of global vector fields is needed.

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 $(0,0)$.

First establish the scalar Leibniz rule, since it is needed for an extension to all tensor fields. Suppose every component of $M$ has positive dimension. Computing $D(fgY)$ first with scalar $fg$ and then as $f(gY)$ givesAt any point, a smooth cutoff function times a coordinate vector field can be chosen nonzero there. Evaluating the displayed identity provesThus the scalar operator is a derivation of an algebra. Derivations of smooth functions are vector fields, so the scalar operator is differentiation along a unique vector field $Z$; no continuity assumption is needed. Indeed, if $h$ vanishes near $p$, take a smooth cutoff function $\chi$ equal to one near $p$ and supported where $h=0$. The identity $D(\chi h)=(D\chi )h+\chi Dh=0$ gives $Dh(p)=0$. Thus $D$ depends only on the local function germ, so local coordinate functions may be extended with cutoffs before applying it. The local identity $h(x)=h(p)+{\sum }_i(x^i-x^i(p))h_i(x)$ gives $Dh(p)={\sum }_{i}D(x^i)(p){\partial }_{i}h(p)$. Smooth local coefficients $D(x^i)$ define $Z$.

The vector-field operator is local as well. If $Y$ vanishes near $p$, choose a cutoff $h$ equal to one near $p$ with support where $Y=0$. Then $D(hY)=hDY+(Dh)Y$ gives $DY(p)=0$. We can therefore work with local frames without presuming a global frame.

For a differential one-form $\omega$, the only possible contraction-compatible definition isThe scalar product rule and the given vector-field rule show that this is linear over $C^{\infty }(M)$ in $Y$, so it defines a one-form. Its coefficients are smooth by evaluating the formula on local frame fields extended with cutoffs. It is real-linear in $\omega$, and direct substitution gives $D(f\omega )=fD\omega +(Df)\omega$.

The original PDF's hint has a transpose error. If $D{e}_i={\sum }_j{a}_{ij}{e}_j$ for $e_i=\partial /\partial x^i$, then evaluation on every $e_j$ forcesnot the untransposed coefficient array printed in the hint. For example, take $Df=0$, $D{e}_1=e_2$ and $D{e}_2=0$ on ${\mathbb{R}}^2$. The correct values are $D(d{x}^1)=0$, $D(d{x}^2)=-d{x}^1$. The printed hint instead makes the derivative of $d{x}^1(e_2)=0$ equal to $-1$.

There is also a genuine zero-dimensional edge case in the hypotheses: on a one-point manifold the vector-field space is zero, so the printed rule imposes no condition on the scalar map. Taking $Df=f$ satisfies that rule but cannot extend to a tensor derivation, since the product rule requires $D1=0$. Thus the extension theorem is valid on positive-dimensional manifolds as above, or in every dimension if the scalar product rule is added explicitly. The next two parts use these precise hypotheses.

The tensor denoted $X\otimes df$ acts as the rank-one endomorphism $Y\mapsto df(Y)X=Y(f)X$. In the paper's covector-first ordering it is written $df\otimes X$, using the canonical interchange of tensor factors; this is the same endomorphism.

On any smooth function $h$,On a vector field $Y$, the Lie bracket of vector fields givessoMultiplication by $f$ preserves the tensor-product Leibniz rule and commutation with tensor contractions, since the latter are linear over smooth functions. Hence the difference is a real-linear contraction-compatible tensor derivation whose scalar operator is zero. Its agreement with $D_{X\otimes df}$ on functions and vector fields determines its action on every tensor by uniqueness:For example, on a differential one-form this gives $-\omega (X)df$. Equivalently ${\mathcal{L}}_{fX}\omega =f{\mathcal{L}}_X\omega +\omega (X)df$, which checks the sign independently. This is the scaled Lie derivative defect identity.

For a smooth function $f$ and vector field $X$,on every tensor field, with $X\otimes df$ interpreted as the endomorphism $Y\mapsto Y(f)X$. Both sides vanish on functions; on vector fields this follows from $[fX,Y]=f[X,Y]-Y(f)X$. Both are contraction-compatible tensor derivations, so agreement on functions and vector fields proves equality on all tensor types. In particular ${\mathcal{L}}_{fX}\omega =f{\mathcal{L}}_X\omega +\omega (X)df$.

A tensor derivation is a real-linear operation preserving every tensor type, obeying the tensor-product Leibniz rule, and commuting with every tensor contraction. A derivation $D$ of smooth functions and a real-linear operator on vector fields satisfying $D(fY)=fDY+(Df)Y$ extend uniquely to a tensor derivation. On a differential one-form it must satisfyThis expression is linear over smooth functions in $Y$. In a local frame $D{e}_a=A^b{}_a{e}_b$, the dual rule is $D{ϵ}^a=-A^a{}_b{ϵ}^b$. Apply the product rule to every coefficient and frame factor to define the extension; these dual signs cancel under contraction. Frame changes agree by differentiating the inverse matrix. Cutoffs prove locality and hence uniqueness from local expansions.