Take both curvature radii to be one. Coordinate ranges depend on the global identification, which a line element alone cannot fix. For the ordinary two-dimensional de Sitter spacetime,This is the unit hyperboloid $X_0^2-X_1^2-X_2^2=-1$ in ambient signature $(+--)$, parametrized by $X_0=\sinh t$, $X_1=\cosh t\cos \chi$, $X_2=\cosh t\sin \chi$. Unwrapping $\chi$ instead gives its universal cover.

For two-dimensional anti-de Sitter spacetime, the embedding $X_0^2+X_1^2-X_2^2=1$ in ambient signature $(++-)$ is parametrized by $X_0=\cosh r\cos t$, $X_1=\cosh r\sin t$, $X_2=\sinh r$. On this hyperboloid $r\in \mathbb{R}$ and $t$ is periodic modulo $2\pi$, creating closed timelike circles. The physically usual universal cover removes that periodicity:Both signs of $r$ are needed for the full two-dimensional spatial line. Taking $r\ge 0$ without an additional construction would omit one of its two ends. The following causal comparison uses the AdS universal cover.

Both models are maximally symmetric Lorentzian manifolds, with constant sectional curvature and no curvature singularities. In the paper's convention their Ricci scalars are respectively $+2$ and $-2$. The de Sitter spatial scale factor $a(t)=\cosh t$ contracts to a nonzero minimum and then expands; $t=0$ is not a big-bang singularity. The AdS metric is static, and ${\partial }_t$ is timelike everywhere because $g_{tt}={\cosh }^{2}r>0$.

The embeddings give a concise classification of all affinely parametrized geodesics. Geodesic acceleration in the ambient space is normal to the hyperboloid, hence proportional to $X$. Differentiating the constraint and setting $X^{\prime 2}=\kappa$, with $\kappa =1,0,-1$ for timelike, null and spacelike curves, givesEvery solution stays in the two-plane spanned by its initial point $P$ and tangent $V$, so every geodesic is a plane section through the ambient origin. For de Sitter, these are $P\cosh s+V\sinh s$ for timelike curves, $P+sV$ for null curves, and $P\cos s+V\sin s$ for spacelike curves. Thus timelike geodesics and null geodesics are complete and nonclosed, while spacelike geodesics are closed circles on the ordinary hyperboloid. These are the geodesics of two-dimensional de Sitter spacetime.

For AdS the roles of the circular and hyperbolic solutions are reversed: timelike curves are $P\cos s+V\sin s$, null curves are $P+sV$, and spacelike curves are $P\cosh s+V\sinh s$. Timelike curves close after proper time $2\pi$ on the hyperboloid, but their lifts on the universal cover continue to ever later time rather than forming closed curves. All three kinds extend for arbitrary affine parameter. This is geodesics of two-dimensional anti-de Sitter spacetime, not an assertion that geodesics reach the conformal boundary in finite physical affine length.

The coordinate first integrals make the behavior concrete. In de Sitter, spatial rotational symmetry gives a conserved $P_{\chi }={\cosh }^{2}td\chi /ds$, and normalization givesConstant-$\chi$ observers are timelike geodesics. Nontrivial null geodesics have affine parameter proportional to $\sinh t$, which diverges at both $t\to \pm \infty$. In AdS the conserved static energy is $E={\cosh }^{2}rdt/ds$, withFor timelike curves, $E\ge 1$ andThey oscillate through the center and never reach spatial infinity; $r=0$ is the $E=1$ geodesic. For null curves, $\lambda =\pm \sinh r/E+\text{constant}$, so both ends are at infinite affine parameter. Spacelike AdS geodesics likewise extend to the two spatial ends.

For the conformal structure, introduce de Sitter conformal timeMultiplying by ${\cos }^2\eta$ gives the conformal cylinder of two-dimensional de Sitter spacetime. The past and future conformal infinities are spacelike circles at $\eta =\mp \pi /2$. Null curves are $\chi =\pm \eta +\text{constant}$ modulo $2\pi$. Although the conformal-time interval is finite, their physical affine parameters and timelike observers' proper times are infinite at its ends. Constant-$\eta$ circles are Cauchy hypersurfaces; global de Sitter is globally hyperbolic.

For AdS putThe conformal strip of two-dimensional anti-de Sitter spacetime has unbounded time and timelike conformal boundaries at $\psi =\pm \pi /2$. Null curves have $t\pm \psi =\text{constant}$. A signal from any finite $r$ reaches the center in coordinate time $|\psi |<\pi /2$, and a null curve crosses the entire conformal strip in time $\pi$. This finite coordinate travel time coexists with infinite physical affine length to the boundary. The covering spacetime has no closed timelike curves, but is not globally hyperbolic: causal curves can arrive from timelike infinity without meeting a proposed initial slice. Field evolution consequently requires boundary conditions as well as initial data. Unwrapping time removes closed timelike curves, not the timelike boundary.

The contrast in observer event horizons follows directly from these null curves. For the complete de Sitter observer $\chi =0$, let $d(\chi ,0)\in [0,\pi ]$ be shortest angular distance. An event can send a signal to this observer before its future endpoint precisely whenEquality is its future observer event horizon. It can receive a signal emitted by the observer after its past endpoint precisely when $d(\chi ,0)<\eta +\pi /2$, whose equality is the past horizon. In the observer's fundamental domain these horizons are null linesTheir intersection encloses the observer's static patch of de Sitter spacetime, the diamond $d(\chi ,0)<\pi /2-|\eta |$. They are observer-dependent cosmological horizons, not curvature singularities or Cauchy horizons. Every inertial observer has the corresponding horizons by de Sitter symmetry.

For the complete AdS static observer at $r=0$, every event at finite $r$ can send a signal to the observer, and can receive one, because time is unbounded and the coordinate distance $|\psi |$ is finite. Thus there is no corresponding global static-observer event horizon. The lapse never vanishes in these coordinates. Restricted accelerated-observer patches can have observer horizons, but those are distinct from the global static model. On the original periodic-time AdS hyperboloid, the more serious issue is closed timelike curves.

The figure identifies the de Sitter spatial edges and displays only a finite time window of the AdS covering strip, whose time continues indefinitely. Solid red curves are null geodesics; the oscillating AdS curve is a timelike geodesic. The shaded de Sitter diamond is the static observer's two-way communication region, not the entire spacetime.

The concise comparison is de Sitter has spacelike conformal infinities and cosmological observer horizons; covering AdS has timelike conformal boundaries and no horizon for an eternal global static observer. Both are constant-curvature and geodesically complete, but their geodesic recurrence, global topology, causal boundaries and initial-value properties differ.

Raise perturbation indices with the background Minkowski metric. To first order, the inverse metric is $g^{ik}={\eta }^{ik}-ϵh^{ik}+O({ϵ}^2)$ and the Levi-Civita connection isQuadratic connection products in the Riemann curvature tensor are of order ${ϵ}^2$. Substitution into the paper's curvature convention givesThe signs follow the convention on the cover sheet; reversing the curvature definition would reverse this expression.

The gauge freedom is the freedom to identify points of the perturbed spacetime with points of the flat background in slightly different coordinates. Under the passive infinitesimal change $x^{\prime i}=x^i+ϵ{\xi }^i(x)$, the metric transformation law gives the linearized coordinate gauge transformationThus a nonzero perturbation can partly represent a coordinate change rather than physical curvature. Substituting this variation into the displayed curvature formula givesEach third derivative has a partner differing only by the order of its commuting partial derivatives. This explicitly proves gauge invariance of the linearized Riemann tensor. The restriction to first order matters: a coordinate change acts on nonzero background curvature as well, but the background here is flat.

Define the trace-reversed metric perturbation and d'Alembert operator byIn four dimensions $\bar{h}=-h$ and $h_{ik}={\bar{h}}_{ik}-\frac{1}{2}{\eta }_{ik}\bar{h}$. Contract the curvature formula using $R_{ik}=R^m{}_{imk}$:up to $O({ϵ}^2)$. Consequently the Einstein tensor isChoose Lorenz gauge in linearized gravity, ${\partial }^k{\bar{h}}_{ik}=0$. It can be imposed locally by solving $\square {\xi }_i={\partial }^k{\bar{h}}_{ik}$, since the gauge transformation changes this divergence by $-\square {\xi }_i$. The remaining coordinate freedom satisfies $\square {\xi }_i=0$. In this gauge,This is the requested Linearized Einstein equations in the stated sign convention. Their divergence requires ${\partial }^k{T}_{ik}=0$ to this order.

For the photon beam, introduce $u=t-x$ and the constant null covector ${\ell }_i=(1,-1,0,0)$, so ${\ell }^i=(1,1,0,0)$. The stress-energy tensor isIt is trace-free and conserved: ${\ell }^i{\partial }_i\rho (u)=0$. A beam-adapted particular solution, with no independently added homogeneous gravitational field, isThe equality follows because its trace is zero, and the Lorenz condition holds because ${\ell }^k{\partial }_{k}H=0$. The field equation becomes a transverse Poisson equation:For the transversely uniform source in the question, one particular solution is $H=4\pi G\rho (u)(y^2+z^2)$. Thus the sourced components can be chosen asBoundary conditions and homogeneous solutions determine additional freedom; the source does not uniquely specify every component in every gauge. In particular, taking $H$ to depend only on $t-x$ would give $\square H=0$ and cannot solve the equation for nonzero $\rho$. The uniform idealization is not an asymptotically flat finite beam, and its quadratic particular solution is a weak-field description only where $|ϵH|\ll 1$.

Finally consider another photon with tangent proportional to ${\ell }^i$. Its tangent remains null because $h_{ik}{\ell }^i{\ell }^k=0$. Moreover $h_{ik}{\ell }^k=0$, and since $\ell$ is constant, the connection contraction isThe geodesic equation is therefore satisfied by a straight affinely parametrized ray along the beam. Hence co-propagating photons experience no deflection from the beam's gravitational field. This is parallel photon beams in linearized gravity; it does not assert absence of forces on counter-propagating photons or effects from unrelated homogeneous gravitational perturbations.