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On smooth functions, the Lie derivative of a function satisfiesOn an arbitrary vector field $W$, the Lie derivative of a vector field givesThe middle equality follows from the Jacobi identity for the Lie bracket of vector fields. One can verify that identity without assuming the result being proved: view vector fields as derivations acting on smooth functions, expand their commutators, and cancel the six compositions. Equality as derivations implies equality as vector fields.

For the second identity, let $A,B,C$ be any operators on either smooth functions or vector fields. Associativity of composition givesEach term cancels in the sum. Applying this to $A={\mathcal{L}}_X$, $B={\mathcal{L}}_Y$, $C={\mathcal{L}}_Z$ proves both requested identities on both kinds of arguments:The commutator identity for Lie derivatives expresses that the Lie derivative of a tensor field represents the Lie bracket of vector fields.

The Lie derivative of a function is differentiation along the flow, while the Lie derivative of a vector field compares a vector with its pullback by that flow. Their index-free expressions areHere $X(f)$ is the directional derivative of the smooth function, and $[X,Y]$ is the Lie bracket of vector fields, characterized by $[X,Y](f)=X(Y(f))-Y(X(f))$. To see the second expression directly, if ${\varphi }_t$ is the local flow of $X$, differentiating $(d{\varphi }_t{)}^{-1}Y\circ {\varphi }_t$ at $t=0$ gives $dY(X)-dX(Y)=[X,Y]$. This uses the flow definition of the Lie derivative of a tensor field.