Past exam of the mathematics course of the University of Cambridge / 2013 / iii / Paper 49 / 1

Let ${\varphi }_s$ be the local flow of the smooth vector field $X$, with ${\varphi }_0=\mathrm{id}$ and $\frac{d}{ds}{\varphi }_s(p)=X_{{\varphi }_s(p)}$. The flow definition of the Lie derivative of a tensor field isFor a tensor field of type $(r,s)$, the mixed tensor pullback transports the tensor at ${\varphi }_s(p)$ back to $p$: it applies $(d{\varphi }_s{)}_p^{-1}$ to every contravariant factor and $(d{\varphi }_s{)}_p^{\ast }$ to every covariant factor. Thus every difference quotient belongs to the same tensor space at $p$. This definition gives another tensor field of the same type and uses no choice of affine connection. The tensor Lie derivative is defined wherever the local flow exists, including points where $X$ vanishes.

For a diffeomorphism, the differential and its inverse preserve the natural pairing of a vector with a covector. Consequently the mixed tensor pullback commutes with every tensor contraction $C$:Differentiating at zero gives ${\mathcal{L}}_X(CT)=C({\mathcal{L}}_{X}T)$. The mixed tensor pullback also preserves tensor products, soThe ordinary product rule for differentiation therefore yieldsThis proves the contraction and Leibniz rule properties for all smooth generators, without needing straightening coordinates.

The suggested flow-box theorem applies locally only where $X\ne 0$. For example $X=x{\partial }_x$ vanishes at $x=0$, so it cannot equal a coordinate basis vector there. Its local flow is nevertheless ${\varphi }_s(x)=e^{s}x$ and ${\varphi }_s^{\ast }dx=e^{s}dx$, giving ${\mathcal{L}}_{X}dx=dx$ even at that zero. This Lie derivative at a zero of its generator illustrates why the general flow proof is needed to cover every point.

For a smooth function, pullback of a smooth function is composition. The chain rule givesFor a vector field $Y$, use any local coordinates. To first order,Multiplying this inverse differential by $Y({\varphi }_s(x))$ givesThusThis is the Lie bracket of vector fields, whose action on a function is $X(Yf)-Y(Xf)$.

Write the covector field as $\omega ={\omega }_{\mu }d{x}^{\mu }$. From the contraction and product rule properties,Take $Y={\partial }_{\mu }$. The Lie bracket of vector fields is $[X,{\partial }_{\mu }]=-({\partial }_{\mu }{X}^{\nu }){\partial }_{\nu }$, soEquivalently, the covariant-factor contribution comes from ${\mathcal{L}}_{X}d{x}^{\mu }=d{X}^{\mu }=({\partial }_{\nu }{X}^{\mu })d{x}^{\nu }$. The formula holds in any coordinate basis and is the one-form case of the coordinate tensor Lie derivative.

Expand $T=T^{\mu }{}_{\nu }{\partial }_{\mu }\otimes d{x}^{\nu }$. Using the tensor Lie derivative of functions, vectors and covectors and its Leibniz rule givesThe contravariant slot has a minus sign and the covariant slot a plus sign.

For the commutator identity for Lie derivatives, put $D=[{\mathcal{L}}_X,{\mathcal{L}}_Y]-{\mathcal{L}}_{[X,Y]}$. The commutator of two tensor derivations is itself a tensor derivation, and $D$ commutes with tensor contractions. On a function, $Df=XYf-YXf-[X,Y]f=0$. On a vector field $Z$,by the Jacobi identity, which follows here by expanding the commutators of the operators $X,Y,Z$ acting on functions. For a type $(1,1)$ tensor, $T(Z)$ is a vector field, and contraction compatibility givesAs this holds for every $Z$, $DT=0$. ThereforeThe derivation argument also establishes the identity for arbitrary tensor types by applying it to covector–vector pairings and then to tensor products.