de Sitter Spacetime

Global and static de Sitter charts describe one four-dimensional constant-positive-curvature spacetime but different domains. Global coordinates cover the complete hyperboloid with compact $S^3$ slices; a static chart covers only one observer’s causal diamond and ends at cosmological Killing horizons where its time-translation field becomes null.

Required background. The Riemann Curvature Tensor supplies the constant-curvature invariant checks; The Einstein Equation fixes $\Lambda=3/L^2$ in four dimensions.

Helpful background. Killing Fields and Isometry Groups distinguishes the global isometry group from the static chart’s preferred generator.

Hyperboloid and global chart

In five-dimensional Minkowski space with metric

$$ds_5^2=-dX_0^2+dX_1^2+dX_2^2+dX_3^2+dX_4^2,$$

four-dimensional de Sitter spacetime of radius $L>0$ is the connected hyperboloid

$$-X_0^2+X_1^2+X_2^2+X_3^2+X_4^2=L^2.$$

Write

$$X_0=L\sinh(t/L), \qquad (X_1,\ldots,X_4)=L\cosh(t/L)\,\mathbf n,$$

where $t\in\mathbb R$ and $\mathbf n\in S^3$. The induced metric is

$$ds^2=-dt^2+L^2\cosh^2(t/L)d\Omega_3^2.$$

This is a global chart up to the ordinary angular overlaps on $S^3$. The shrinking and expansion of the slice radius do not select a geometrically preferred “center”: the full $O(1,4)$ action is transitive. Reall derives the embedding, ten Killing generators, and global slicing in Reall 2013, §13.1, printed pp. 149–153, Open PDF.

The invariant curvature data are

$$R_{\mu\nu\rho\sigma} =\frac1{L^2}(g_{\mu\rho}g_{\nu\sigma}-g_{\mu\sigma}g_{\nu\rho}), \quad R_{\mu\nu}=\frac3{L^2}g_{\mu\nu}, \quad R=\frac{12}{L^2}.$$

Thus the vacuum Einstein equation holds with $\Lambda=3/L^2$.

The embedding also makes topology visible at the level needed here. Constant global time slices are three-spheres, and the coordinate $t$ runs through all real values. The neck at $t=0$ is a minimum-volume slice, not an edge of spacetime. Antipodal sides of a global slice belong to the same connected hyperboloid, whereas one static observer sees only a proper subset.

Static observer domain

Choose one polar direction in the embedding and set

$$X_0=\sqrt{L^2-r^2}\sinh(t_s/L), \qquad X_4=\sqrt{L^2-r^2}\cosh(t_s/L),$$

with $(X_1,X_2,X_3)=r\mathbf n_2$. For $0\le r<L$ and $t_s\in\mathbb R$,

$$ds^2=-\left(1-\frac{r^2}{L^2}\right)dt_s^2 +\frac{dr^2}{1-r^2/L^2}+r^2d\Omega_2^2.$$

The vector $\partial_{t_s}$ is Killing and timelike only inside this diamond. At $r=L$ it becomes null, while all curvature invariants retain their constant values. The two limiting null hypersurfaces are cosmological Killing horizons for this normalization and observer family, not curvature boundaries. Other geodesic observers obtain other static diamonds related by de Sitter isometries.

The static domain follows directly from the embedding inequalities $X_4^2-X_0^2>0$ and $X_4>0$. They select the intersection of the causal future and past of one geodesic observer. The opposite sign of $X_4$ supplies the antipodal static diamond. Regions with the reversed first inequality are regular but cannot be covered by real $t_s$ in this transformation. The coordinate horizon is the boundary of a particular boost-like flow, not the end of the hyperboloid.

Radial null propagation gives a second check. A tortoise coordinate for the static radial metric diverges as $r$ approaches $L$, even though a global null curve crosses the same hypersurface. Divergence of an adapted coordinate is evidence about the chart, not about affine completeness.

Flat and open slicings

Planar coordinates are related to the embedding by

$$\begin{aligned} X_0&=L\sinh(t_p/L)+\frac{e^{t_p/L}}{2L}|\mathbf x|^2,\\ (X_1,X_2,X_3)&=e^{t_p/L}\mathbf x,\\ X_4&=L\cosh(t_p/L)-\frac{e^{t_p/L}}{2L}|\mathbf x|^2. \end{aligned}$$

Substitution verifies the hyperboloid constraint, while addition gives the domain test

$$X_0+X_4=Le^{t_p/L}>0.$$

Thus the coordinates cover exactly that embedding half and give

$$ds^2=-dt_p^2+e^{2t_p/L}d\mathbf x^2, \qquad t_p\in\mathbb R,\quad \mathbf x\in\mathbb R^3.$$

Each slice is intrinsically flat, despite the positive four-dimensional curvature. Open slicing gives

$$ds^2=-dt_o^2+L^2\sinh^2(t_o/L) \left(d\chi^2+\sinh^2\chi d\Omega_2^2\right),$$

with $t_o>0$ for one expanding region. The apparent scale-factor zero at $t_o=0$ is the tip of that chart, not a curvature singularity. Together these charts demonstrate that “closed,” “flat,” and “open” describe chosen spacelike foliations, not three different local de Sitter geometries. Bičák emphasizes this slicing dependence in Bičák 2000, §1.3, printed pp. 9–10, Open PDF.

The planar chart provides a useful overlap test. Its scale factor has constant Hubble parameter $H=L^{-1}$, and a direct FLRW curvature calculation returns $R=12H^2=12/L^2$, agreeing with the embedding. Yet it covers only half the hyperboloid. Geodesics can leave that half through its null boundary even though planar cosmic time ranges over all real values. An infinite coordinate-time range is not a proof that a chart is global.

Independent checks and limits

The embedding constraint proves that each coordinate form lies on the same hyperboloid; pulling back the ambient metric verifies its coefficients and domain. The constant Riemann tensor independently verifies the field equation and shows why chart boundaries carry no curvature blowup. In a region whose physical size and time are fixed while $L\to\infty$, each form approaches Minkowski geometry and $R\to0$.

This page identifies charts and horizon data only. General causal-boundary theorems, quantum states, cosmological dynamics, and de Sitter thermodynamics belong elsewhere.

The horizon claim is deliberately bounded. It is a Killing horizon of the static generator normalized to unit norm at $r=0$. Calling it an event horizon requires specifying the observer or asymptotic region relative to which a causal past is taken. No surface-gravity or temperature claim is needed to identify the metric and its atlas.

Check your understanding

Why does $g_{t_st_s}=0$ at $r=L$ not contradict the nondegeneracy of de Sitter spacetime?

Solution

The static metric components are singular because the static map uses $\sqrt{L^2-r^2}$ and ceases to be an invertible chart at $r=L$. The embedding hyperboloid remains smooth there, and invariants such as $R=12/L^2$ are finite. The vanishing norm says that the particular Killing vector $\partial_{t_s}$ becomes null. An overlapping global or horizon-penetrating chart crosses the same hypersurface with nondegenerate components.

References

• Bičák, Jiří. “Selected Solutions of Einstein’s Field Equations: Their Role in General Relativity and Astrophysics.” Einstein’s Field Equations and Their Physical Implications, Lecture Notes in Physics 540 (2000), 1–126. arXiv:gr-qc/0004016, §1.3, printed pp. 8–10.
• Reall, Harvey S. General Relativity. Cambridge Part III lecture notes (2013). §13.1, printed pp. 149–160, Open PDF.