Observer Patches and Cosmological Horizons

Global de Sitter spacetime is geodesically complete and has compact $S^3$ Cauchy surfaces, yet no single geodesic observer can exchange signals with all of it. The apparent tension disappears once three objects are separated: the maximal spacetime, a coordinate patch, and the causal diamond of a chosen worldline. In four-dimensional vacuum de Sitter, the observer’s boundaries are null cosmological horizons inside the physical spacetime; they are not the spacelike components $\mathscr I^\pm$ of the global conformal completion.

Required background. We use The Einstein Equation to fix the four-dimensional vacuum solution with $\Lambda\gt 0$ and de Sitter radius $\ell=\sqrt{3/\Lambda}$.

Helpful background. Pseudo-Riemannian Metrics and Signature distinguishes null horizons from spacelike infinity, Affine and Variational Geodesics describes the central observer, Causality and Global Structure supplies observer-relative futures and pasts, Exact Solutions of Einstein’s Equations supplies the coordinate transformations, and Formulations of General Relativity clarifies the difference between a chart restriction and an evolution domain.

One observer’s diamond in global de Sitter

Use global conformal coordinates for exact four-dimensional de Sitter:

$$d\widehat s^2 =\frac{\ell^2}{\cos^2\eta} \left(-d\eta^2+d\chi^2+\sin^2\chi\,d\omega_2^2\right),$$

with $-\pi/2\lt\eta\lt\pi/2$ and $0\leq\chi\leq\pi$. The conformal factor does not change null curves. Radial light rays therefore obey

$$\frac{d\chi}{d\eta}=\pm1.$$

Let $\gamma$ be the complete timelike geodesic at $\chi=0$. Its ideal endpoints are the points $i^-$ and $i^+$ at the north poles of $\mathscr I^-$ and $\mathscr I^+$. An event $(\eta,\chi)$ can send a signal to some later point on $\gamma$ precisely when

$$\eta+\chi\lt\frac{\pi}{2}.$$

It can receive a signal from some earlier point on $\gamma$ precisely when

$$\eta-\chi\gt-\frac{\pi}{2}.$$

Thus the observer’s causal diamond is

$$I^-(\gamma)\cap I^+(\gamma) =\left\{\chi\lt\frac{\pi}{2}-|\eta|\right\}.$$

Its future and past null boundaries are the observer’s cosmological event horizons. They meet at the equatorial two-sphere $\eta=0$, $\chi=\pi/2$. The rest of the global spacetime still exists; it is simply outside this observer’s two-way causal access. The global completion and the observer-dependent character of the horizons are discussed in Ashtekar, Bonga, and Kesavan 2015, §III.A, pp. 10–11 and Ellis and Uzan 2015, §2.2, pp. 4–5.

Changing the geodesic observer moves the diamond by a de Sitter isometry. There is no unique null hypersurface called “the de Sitter horizon” until the worldline or static region is specified. The spacelike boundaries $\mathscr I^\pm$, by contrast, are global three-dimensional components of the completion.

The static patch

Coordinates $(t,r,\theta,\phi)$ adapted to $\gamma$ put its diamond in the form

$$d\widehat s^2 =-\left(1-\frac{r^2}{\ell^2}\right)dt^2 +\left(1-\frac{r^2}{\ell^2}\right)^{-1}dr^2 +r^2d\omega_2^2, \qquad 0\leq r\lt\ell.$$

The central curve $r=0$ is $\gamma$, and $K=\partial_t$ is timelike in the patch because

$$\widehat g(K,K)=-\left(1-\frac{r^2}{\ell^2}\right)\lt0.$$

At $r=\ell$, $K$ becomes null and the static coefficients are singular. Curvature is not: de Sitter has constant curvature everywhere. Global coordinates extend through the two null branches. The surface $r=\ell$ is therefore simultaneously a coordinate boundary of this static chart, a Killing horizon of $K$, and a causal horizon of the selected observer. Those three descriptions agree here because of exact symmetry; they need not agree in a dynamical asymptotically de Sitter spacetime.

Beyond the horizon the same Killing field is spacelike. More generally, where a de Sitter Killing field has a nonzero boundary value on spacelike $\mathscr I$, that value is spacelike and remains so in a sufficiently small nearby bulk region. The qualification matters: the static generator vanishes at the observer’s ideal endpoints in the conformal completion and can remain timelike along the central worldline all the way to them. There is no Killing field timelike everywhere on global de Sitter. This obstructs a single globally preferred positive energy while permitting $K$ to define a useful observer-patch energy where it is future timelike. Static-patch thermodynamics and quantum states require additional normalization and quantum-field assumptions and belong to Quantum Fields in Curved Spacetime, not to this causal comparison.

The expanding planar patch

Another common chart is

$$d\widehat s^2 =-dt_{\rm p}^2+e^{2t_{\rm p}/\ell}d\mathbf x^2 =\frac{\ell^2}{\eta_{\rm p}^2} \left(-d\eta_{\rm p}^2+d\mathbf x^2\right),$$

where

$$\eta_{\rm p}=-\ell e^{-t_{\rm p}/\ell}, \qquad -\infty\lt\eta_{\rm p}\lt0, \qquad \mathbf x\in\mathbb R^3.$$

The surface $\eta_{\rm p}=0$ is a complete spacelike future infinity with topology $\mathbb R^3$. The limit $\eta_{\rm p}\to-\infty$ is not global $\mathscr I^-$. It is a null past boundary of the expanding Poincaré patch, which is only one half of global de Sitter. Exact global coordinates extend through it.

This distinction matters in inflationary language. A flat FLRW chart can accurately describe an expanding region while omitting the antipodal region and most of past infinity. Its constant-$t_{\rm p}$ slices are noncompact, whereas global constant-$\tau$ slices are compact three-spheres. Neither difference is a local curvature distinction. Ashtekar, Bonga, and Kesavan trace how $\mathbb R^3$ future infinity and the patch’s null past edge arise in Ashtekar, Bonga, and Kesavan 2015, §III.E, pp. 15–18.

A realistic inflationary spacetime is only approximately de Sitter for a finite interval and may have a big-bang boundary or a pre-inflationary phase. One must then compute particle or event horizons from the full scale factor and actual temporal endpoints. Replacing the exact chart’s past horizon by a universal statement about every inflationary model is not justified.

Global boundary, patch boundary, and horizon

The three descriptions answer different questions:

• Global $\mathscr I^+$ is the spacelike endpoint set of the entire completed spacetime. It is not crossed by a physical observer.
• A Poincaré-patch edge is the boundary of a chosen open submanifold of global de Sitter. One branch is a regular null hypersurface in the maximal extension; the other is spacelike conformal infinity.
• An observer horizon is $\partial I^-(\gamma)$ or $\partial I^+(\gamma)$ for a specified complete history. It is invariant after $\gamma$ is chosen but changes when the observer changes.

These distinctions survive away from exact de Sitter. In a dynamical future asymptotically de Sitter spacetime, a complete observer may still have a cosmological event horizon because the remaining conformal time is finite. The horizon need not be Killing, its area need not be constant, and it can depend on the entire future evolution. Neither smoothness nor conformal flatness of $q_{ab}$ is required merely to define the causal boundary. This page assumes exact conformal flatness only for the explicit de Sitter calculations; it does not impose the ABK restriction on the general asymptotic class.

Common pitfalls

Treating the static patch as the maximal spacetime. The divergence of $g_{rr}$ at $r=\ell$ is a chart failure on a regular null horizon. Global de Sitter continues beyond it.

Calling the planar past edge a big bang. In exact de Sitter it is extendible. It becomes a physical singular boundary only after a different cosmological solution and its matter evolution establish that fact.

Assigning one horizon to all observers. Isometrically related observers have equally valid but different horizon locations. The observer history belongs in the definition.

Exercises

For the north-pole observer $\gamma$ in global conformal coordinates, derive the two inequalities defining its causal diamond. Then decide whether the event $p=(\eta,\chi)=(0,2\pi/3)$ can ever send a signal to $\gamma$, and whether the failure signals a spacetime singularity.

Solution

A radial signal from $(\eta,\chi)$ reaches $\chi=0$ no earlier than conformal time $\eta+\chi$. It arrives before the observer’s future endpoint only if $\eta+\chi\lt\pi/2$. Time reversal gives the condition for receiving a signal from the observer: tracing a radial ray backward to $\chi=0$ gives emission time $\eta-\chi$, which must exceed $-\pi/2$. Hence

$$p\in I^-(\gamma)\cap I^+(\gamma) \quad\Longleftrightarrow\quad -\frac{\pi}{2}\lt\eta-\chi \quad\text{and}\quad \eta+\chi\lt\frac{\pi}{2}.$$

For $p=(0,2\pi/3)$, the second inequality fails because $2\pi/3\gt\pi/2$. No signal from $p$ reaches the north-pole observer before $i^+$. The point is nevertheless a regular point of global de Sitter, accessible to other observers. The obstruction is an observer-dependent cosmological horizon, not curvature blowup or a boundary of the maximal spacetime.

References

• Ashtekar, Abhay, Béatrice Bonga, and Aruna Kesavan. “Asymptotics with a Positive Cosmological Constant: I. Basic Framework.” Classical and Quantum Gravity 32, 025004 (2015). arXiv.
• Ellis, George F. R., and Jean-Philippe Uzan. “Causal Structures in Cosmology.” Comptes Rendus Physique 16, 928–947 (2015; arXiv version 2016). arXiv.