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For an associative algebra $A$ over a field $k$, a derivation is a $k$-linear map $D:A\to A$ satisfying $D(ab)=D(a)b+aD(b)$. It is the case $M=A$ of a derivation into a bimodule; the order of factors matters when $A$ is noncommutative. Substituting $a=b=1$ gives $D(1)=0$. Such maps form a Lie algebra under the commutator $[D,E]=D\circ E-E\circ D$.

More generally, for a commutative ring $R$, a derivation from a commutative $R$-algebra $B$ into a $B$-module $M$ is an $R$-linear map $D:B\to M$ with $D(r)=0$ for $r\in R$ and $D(b{b}^{\prime })=bD(b^{\prime })+b^{\prime }D(b)$. The second term has this order because the algebra is commutative. Derivations of smooth functions are vector fields provide a geometric example.

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.