Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 309 / 1 / a

Let ${\Phi }_s$ be the local flow generated by the vector field $X$. For a covariant tensor, its Lie derivative isFor contravariant indices one equivalently uses the differential of the inverse flow, and the construction extends to every tensor product by the Leibniz rule. It measures the infinitesimal change of $T$ under transport by the flow of $X$.

At a point where $X\ne 0$, the flow-box theorem supplies a transverse hypersurface with coordinates $x^i$. Flow each point on it for parameter $t$ and keep the $x^i$ constant along the flow. In the resulting coordinates,Because the coordinate basis is transported by this flow, the Lie derivative of any tensor is obtained by differentiating its coordinate components:

For a function, pullback by the flow givesIn flow-box coordinates $X={\partial }_t$, write $Y=Y^{\mu }{\partial }_{\mu }$. ThenBoth sides are vector fields defined invariantly, so equality in these coordinates provesin every coordinate system.

For one-forms, the defining pullback or the identitygivesApplying the Leibniz rule to both covariant slots of a type-$(0,2)$ tensor yields