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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.

Use the PDF's covector-first convention: its space ${\mathcal{T}}_l^k$ consists of tensor fields inThis reverses the order in which some texts list tensor type. For a finite-dimensional real vector space $V$, define the tensor contraction first on decomposable tensors:The hats mean omission, with the other factors left in their original order. The formula is a multilinear map of its individual factors, so the universal property of a tensor product gives a unique linear map with this formula.

Apply it with $V=T_{p}M$ at every $p$. Evaluation of a covector on a vector is basis independent: under a change of frame, one factor transforms by a matrix and the other by its inverse transpose, and the matrices cancel in the pairing. Thus the fiber maps agree on overlapping vector bundle trivializations. In a local dual basis, the coefficients of the contracted tensor are finite sums of coefficients with the selected covariant and contravariant indices set equal. They remain smooth. Henceis a globally defined smooth contraction for $k,l\ge 1$. It is also linear over $C^{\infty }(M)$.

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.

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 $Y$.

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 field is a smooth section of a vector bundle formed from tensor products of the tangent bundle and cotangent bundle. With $k$ covector factors and $l$ vector factors it belongs toIts coefficients in any local frame vary smoothly. Different conventions list the two counts in different orders, so the factor order should be specified. Tensor contractions use the canonical evaluation of a covector on a vector and are independent of the frame.

A multilinear operation on vector fields is tensorial when it is linear over smooth functions in every argument. Its value at a point then depends only on the argument vectors at that point. To see this, use a smooth cutoff function to reduce to local fields, expand them in a local frame, and apply linearity over smooth functions to their coefficients. A smoothly valued tensorial operation therefore defines a tensor field. The curvature of an affine connection is tensorial, whereas a covariant derivative differentiates a scalar coefficient in its second argument and is not tensorial there.