Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 309 / 4

For the displayed frame,and all other pairings vanish. Hence the proposed one-forms are the dual coframe. Substitution intoreproduces the metric, proving that the frame is an orthonormal frame with signature $(-,+,+)$.

The exterior derivatives of the coframe areInsert the proposed connection into Cartan's first structure equation and use ${\omega }_{\mu \nu }=-{\omega }_{\nu \mu }$. The three equations giveThus

Using Cartan's second structure equation givesSince ${\Theta }_{\mu \nu }=R_{\mu \nu \rho \sigma }{e}^{\rho }\wedge e^{\sigma }/2$, the independent nonzero lowered Riemann curvature tensor components arewith all others determined by the symmetries of the Riemann curvature tensor.

Contraction in the orthonormal frame yieldsHenceFor $u^{\mu }=(1,0,0)$, pressureless matter has $T_{00}=\rho$ and vanishing spatial components. The spatial Einstein equations $G_{ii}+\Lambda {\eta }_{ii}=0$ give $\Lambda =-1/2$, while the time component gives $1=8\pi \rho$. Thus, for $G_N=1$,If the convention is $G_{ab}+\Lambda g_{ab}=T_{ab}$, the corresponding density is simply $\rho =1$.

Because the metric coefficients are independent of $t$ and $y$,are Killing vector fields. The mapssatisfy ${\varphi }_s\circ {\varphi }_r={\varphi }_{s+r}$. Moreover, $dx$ is unchanged and both $e^{x}dy$ and $dt+\sqrt{2}{e}^{x}dy$ are invariant, so ${\varphi }_s^{\ast }g=g$. Differentiating at $s=0$ produces the third Killing fieldGiven two points, first use ${\varphi }_s$ to match their $x$ coordinates, then translations generated by ${\partial }_y$ and ${\partial }_t$ to match $y$ and $t$. The isometry group therefore acts transitively, so the spacetime is a homogeneous space.