Past exam of the mathematics course of the University of Cambridge / 2014 / iii / Paper 50 / 3

Treat $V$ as a derivation. For an arbitrary smooth function $f$ on $\mathcal{N}$, the definition of the differential of a smooth map and the chain rule giveComparison with $({\varphi }_{\ast }V{)}^{\alpha }{\partial }_{\alpha }f$ for all $f$ provesThe Jacobian is an $n\times m$ matrix; it need not be square or invertible.

Evaluate the pullback of a covector on the coordinate basis vector ${\partial }_i$. Its pushforward is $({\partial }_i{y}^{\alpha }){\partial }_{\alpha }$, soEquivalently, pulling back the coordinate one-forms gives ${\varphi }^{\ast }(d{y}^{\alpha })=d(y^{\alpha }\circ \varphi )=({\partial }_i{y}^{\alpha })d{x}^i$. The same Jacobian enters the vector pushforward and the covector pullback, with the different index contractions reflecting their duality.

With vanishing extrinsic curvature, the momentum constraint is identically satisfied and the vacuum Hamiltonian constraint reduces to $\mathcal{R}=0$. To evaluate that Ricci scalar, write ${\psi }_i={\partial }_i\psi$, raise these intermediate indices with ${\delta }^{ij}$, and use ${\gamma }^{ij}={\psi }^{-4}{\delta }^{ij}$. The Levi-Civita connection isSubstitution into the stated curvature convention givesTracing with ${\gamma }^{ij}$ cancels the gradient-square terms:Here the Laplacian and norm on the right are those of the flat Euclidean metric. Thus the time-symmetric conformally flat vacuum initial data constraints becomeThe equivalence uses $\psi \ne 0$. This identity is also the three-dimensional specialization of scalar curvature under conformal rescaling; locally the conformal exponent is $2\log |\psi |$, so either fixed nonzero sign of $\psi$ gives the same metric.

On the punctured region $r>0$, choose a positive constant $a$ and setFor a radial function in three dimensions, $\Delta \psi =r^{-2}{\partial }_r(r^2{\partial }_r\psi )$. Since $r^2{\partial }_r\psi =-a$, this harmonic function has $\Delta \psi =0$ at every $r>0$, is nonconstant and positive, and approaches one at infinity. It defines the requested time-symmetric conformally flat vacuum initial data on that punctured or exterior region. One often writes $a=M/2$, giving the time-symmetric spatial metric of the Schwarzschild metric in isotropic coordinates.

A puncture or inner boundary is necessary if global regularity was intended. A smooth harmonic function on all of ${\mathbb{R}}^3$ that tends to one at infinity must equal one everywhere: applying the maximum principle for harmonic functions to larger and larger balls bounds $|\psi -1|$ by its arbitrarily small boundary values. The nonconstant example is therefore not a smooth solution through $r=0$, and no such globally regular example exists under those stronger assumptions.