Past exam of the mathematics course of the University of Cambridge / 2016 / iii / Paper 309 / 3

Write $V=T_p\mathcal{M}$ and $V^{\ast }$ for its dual space. A type-$(1,1)$ tensor is equivalently a linear map $A:V\to V$: the corresponding scalar-valued multilinear map is $(X,\eta )\mapsto \eta (A(X))$. Thus the composition definesIt is linear in $X$ because $A$ and $B$ are linear maps, and linear in the covector $\eta$ by evaluation. This proves directly that the composition is a tensor of type $(1,1)$.

Choose a basis $e_{\alpha }$ of the tangent space and its dual basis $e^{\beta }$. With the paper's component order $A(e_{\alpha })=A_{\alpha }{}^{\gamma }{e}_{\gamma }$,Equivalently, in the usual output-first matrix notation, $C^{\beta }{}_{\alpha }=B^{\beta }{}_{\gamma }{A}^{\gamma }{}_{\alpha }$: the matrix product is $BA$, not $AB$. This composition of mixed tensors contracts one output index with one input index. Under the tensor component transformation law, the two Jacobian factors at that contracted index cancel, leaving exactly one factor for each free index.

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.

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.