At a point where $X\neq0$, 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,
$$
\boxed{X=\frac{\partial}{\partial t}}.
$$
Because the coordinate basis is transported by this flow, the Lie derivative of any tensor is obtained by differentiating its coordinate components:
$$
\boxed{\mathcal L_XT=\partial_tT^{\mu_1\cdots}{}_{\nu_1\cdots}
\text{the unchanged coordinate basis tensors}}.
$$


Solution

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

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2021/iii/paper-309/1/a/ii/solution