Derivation of an algebra (${\mathrm{Der}}_k(A)$)

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.

Every real-linear derivation of an algebra $D:C^{\infty }(M)\to C^{\infty }(M)$ on a smooth manifold has the form $Dh=Z(h)$ for a unique smooth vector field $Z$. The product rule makes $D$ local: multiplying a function that vanishes near a point by a smooth cutoff function supported there gives zero derivative at that point. Locally write $h(x)=h(p)+{\sum }_i(x^i-x^i(p))h_i(x)$, where $h_i(p)={\partial }_{i}h(p)$. Applying $D$ at $p$ gives $Dh(p)={\sum }_{i}D(x^i)(p){\partial }_{i}h(p)$; the smooth coefficients $D(x^i)$ define the claimed vector field. No continuity hypothesis is needed. In dimension zero all such derivations vanish.