Past exam of the mathematics course of the University of Cambridge / 2017 / iii / Paper 309 / 3

For a smooth vector field $X$, solve the initial-value problem $d{\varphi }_t(p)/dt=X({\varphi }_t(p))$, ${\varphi }_0(p)=p$. Local existence, uniqueness and smooth dependence for ordinary differential equations give a local flow. Uniqueness yields ${\varphi }_{t+s}(p)={\varphi }_t({\varphi }_s(p))$ whenever both sides are defined, and ${\varphi }_{-t}$ is the inverse of ${\varphi }_t$ on the corresponding domains. Thus each local flow map is a diffeomorphism between open subsets.

A group of diffeomorphisms of the entire manifold for every real $t$ requires a complete vector field. This follows, for example, on a compact manifold without boundary: local existence times can be made uniform on a finite cover, and repeated continuation prevents finite-time escape. Without completeness, the source's group is only local. The vector field $x^2{\partial }_x$ on $\mathbb{R}$ has flow $x\mapsto x/(1-tx)$ and blows up at $t=1/x$ for $x>0$, so smoothness alone cannot imply a global one-parameter group.

Define the Lie derivative of a differential form by ${\mathcal{L}}_X\alpha =(d/dt{)}_{t=0}{\varphi }_t^{\ast }\alpha$, using the local flow. The pullback of a differential form respects the wedge product of differential forms and commutes with the exterior derivative. Therefore ${\mathcal{L}}_X$ is a derivation of an algebra of degree zero and commutes with $d$.

Let $D_X={\iota }_{X}d+d{\iota }_X$, where ${\iota }_X$ is the interior product of a differential form. The graded product rules for $d$ and ${\iota }_X$ show that their anticommutator $D_X$ is also a degree-zero derivation of an algebra: the two mixed terms cancel. Since $d^2=0$, it commutes with $d$. On a function $f$, ${\iota }_{X}f=0$ and $D_{X}f={\iota }_{X}df=Xf={\mathcal{L}}_{X}f$. Hence $D_X(df)=d(Xf)={\mathcal{L}}_X(df)$.

In a manifold chart, every differential form is a sum of functions times products $d{x}^{i_1}\wedge \cdots \wedge d{x}^{i_p}$. The two derivations agree on its generators $f$ and $d{x}^i$, and therefore agree on every such form. This proves Cartan's magic formula, including the case $p=0$:

Use Cartan's magic formula. The exterior derivative and interior product of a differential form giveThus $d({\iota }_X\alpha )=-2xdx-2ydy+2zdz$, and the first two components cancel:The rotation in the $xy$-plane leaves $ydx-xdy$ invariant, while the dilation in $z$ doubles the infinitesimal weight of $zdz$, providing a geometric check of this Lie derivative.

Integrating the two vector fields gives their complete flow maps:The first is a simultaneous rotation about the z-axis through angle $t$ and a dilation by $e^t$ in the $z$-direction. The second is an isotropic dilation about the origin. Both satisfy the group law for all real $t$ and have inverse given by time $-t$. In particular ${\varphi }_t$ is not a pure rotation unless restricted to the plane $z=0$.

The Jacobian determinant of ${\varphi }_t$ is $e^t$, while every coordinate and every coordinate differential is multiplied by $e^t$ under ${\psi }_t$. Hence the pullback of a differential form operation givesTo compare directly with Cartan's magic formula, note $d\mu =0$ and computeThe first two differentiated terms vanish because their coordinate differentials repeat; the last contributes $dz\wedge dx\wedge dy=\mu$. AlsoTheir sum is $2\alpha$, as required. Equivalently the divergence of $X$ relative to the standard volume form is one.

An electromagnetic field tensor obeys the homogeneous Maxwell equations $dF=0$. Together with its invariance under the Killing vector field $V$, Cartan's magic formula gives $d({\iota }_{V}F)={\mathcal{L}}_{V}F-{\iota }_{V}dF=0$. The local potential of a closed differential one-form construction, the one-form Poincaré lemma, therefore supplies a scalar field $\Phi$ on a sufficiently small contractible neighborhood withFor an explicit local proof, write ${\iota }_{V}F={\alpha }_i(x)d{x}^i$ on a coordinate ball centered at zero and take $\Phi (x)={\int }_0^1{x}^i{\alpha }_i(sx)ds$. Since $d\alpha =0$, differentiating under the integral gives ${\partial }_j\Phi ={\int }_0^1[{\alpha }_j(sx)+s{x}^i{\partial }_i{\alpha }_j(sx)]ds={\alpha }_j(x)$. There need not be a global potential when this closed one-form has nonzero periods. The homogeneous Maxwell equation is essential; an arbitrary invariant two-form would not suffice.

Use proper time for the massive particle, let $u$ be its tangent, and assume $m\ne 0$. The Killing equation makes $u^a{u}^b{\nabla }_a{V}_b=0$. The Lorentz force equation then impliesConsequently the local conserved quantity isThe sign follows from contracting $F$ in its first argument, exactly as in the source. Adding a constant to $\Phi$ merely shifts the conserved quantity. This is a charged-particle conserved quantity from a Killing symmetry.