Past exam of the mathematics course of the University of Cambridge / 2017 / iii / Paper 115 / 2 / b

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.

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 $(0,0)$.

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.