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Represent unit-radius de Sitter spacetime by the spacelike unit hyperboloid in five-dimensional Minkowski spacetime. Appending its position vector as the last column of an oriented pseudo-orthonormal frame produces a matrix in $SO(4,1)$. Projection to that last column identifies $SO(4,1)$ with the oriented orthonormal frame bundle, a principal bundle with group $SO(3,1)$. The full bundle allowing both orientations is $O(4,1)$ with group $O(3,1)$. Requiring a time orientation as well selects the corresponding identity components.

For the full unit-radius two-dimensional de Sitter spacetime, the coordinate ranges areOne can use $0\le \chi <2\pi$ with endpoints identified. If the identification is omitted, the same line element describes the universal covering spacetime instead. The periodicity follows from the hyperboloid embeddingThe induced ambient Minkowski metric is precisely the given Lorentzian metric.

Using an affine parameter $s$, the geodesic Lagrangian is $\frac{1}{2}({\dot{t}}^2-{\cosh }^{2}t{\dot{\chi }}^2)$. The nonzero Christoffel symbols areThus the geodesic equations and their first integrals areHere $\epsilon =1$ for unit-speed timelike geodesics, $0$ for null geodesics and $-1$ for unit-speed spacelike geodesics. The conserved $J$ is the geodesic conserved quantity from a Killing vector for the angular symmetry.

All geodesics of two-dimensional de Sitter spacetime can be written especially simply in the embedding. If $P$ is the initial point and $V$ the initial tangent, with $P^2=-1$, $P\cdot V=0$, $V^2=\epsilon$, the geodesic has normal ambient acceleration $X^{\prime \prime }=\epsilon X$. ThereforeThese curves stay in the two-plane spanned by $P$ and $V$, preserve both constraints and have no tangential acceleration. They consequently solve the geodesic equation and exhaust all initial directions.

An isometry can move the chosen point to $P=(0,1,0)$, namely $(t,\chi )=(0,0)$. For future-directed timelike geodesics take $V=(\cosh \alpha ,0,\sinh \alpha )$, obtainingThe continuous angular branch has $|\chi |<\pi /2$, and every real $\alpha$ gives a distinct timelike direction. The two null geodesics have $V=(1,0,\pm 1)$ andFor spacelike geodesics take $V=(\sinh \alpha ,0,\cosh \alpha )$, givingThey are closed with period $2\pi$ in affine arclength and $|t|\le |\alpha |$. In particular, $\alpha =0$ is the equatorial geodesic $t=0$. These three families, with reversals of parametrization allowed, are the requested complete set through the chosen point.

The conformal time substitution givesHenceMultiplying by ${\cos }^{2}T$ gives the regular conformal cylinder of two-dimensional de Sitter spacetime. Adding $T=\pm \pi /2$ completes the conformal compactification; both boundary circles are spacelike. Null geodesics are straight lines $\chi =\pm T$ through the origin. The other families have the useful diagram equationsThe image sketches representative directions from each family; the formulas allow all intermediate directions.

For the complete observer $\chi =0$, let $d(\chi ,0)\in [0,\pi ]$ be the shortest angular distance on the circle. Since light travels one unit of angular distance per unit conformal time, an event at $(T,\chi )$ can signal to the observer at some finite future time precisely whenEquality defines its future observer event horizon. Likewise an event can receive a signal emitted by the observer precisely whenand equality defines the past horizon. In the lift $-\pi \le \chi \le \pi$, their branches are $\chi =\pm (\pi /2-T)$ and $\chi =\pm (T+\pi /2)$. The common communication region is the diamond $|\chi |+|T|<\pi /2$, the observer's static patch of de Sitter spacetime. These are the observer horizons in two-dimensional de Sitter spacetime. Every complete timelike observer has analogous horizons after a de Sitter isometry; they depend on the observer, rather than enclosing an intrinsic black-hole region.

Despite these horizons, each constant-$T$ circle is a Cauchy hypersurface of the full spacetime: any inextendible causal curve crosses it exactly once. The global spacetime is therefore globally hyperbolic and has no Cauchy horizons. The conformal boundaries lie at infinite proper time for complete timelike geodesics and infinite affine distance for complete null geodesics, so their finite height in the diagram is not a physical incompleteness.

For a smooth curve $\gamma (t)$ on a Riemannian surface, its energy of a curve isChoose a local parameterization $\mathbf{X}(u,v)$ and write $\gamma (t)=\mathbf{X}(u(t),v(t))$. With coefficients of the first fundamental formthe Lagrangian isThe Euler-Lagrange equations are equivalentlyAfter multiplying by the inverse metric these become the geodesic equationwhere the Christoffel symbols are determined by $E,F,G$.

If a straight line segment lies in the surface, parameterize it byThen $\ddot{\gamma }=0$, so its acceleration has zero tangential component. The two displayed equations hold directly, and the segment is a geodesic.

For the one-sheeted hyperboloidputTwo distinct ruling lines through $P$ areIndeed, the direction $d_{\pm }$ satisfiesso substitution shows that every point of the line lies in $H$. These are geodesics by the straight-line argument.

A third geodesic is the meridian through $P$. Choose $s_0$ with $\sinh s_0=z_0$; thenIts acceleration is ${\gamma }_m$, which is normal to $H$, so it is a geodesic. If $z_0\ne 0$, these give the required three distinct subsets.

If $z_0=0$, there is also the equatorial circle. Writing $P=(\cos {\theta }_0,\sin {\theta }_0,0)$,Its acceleration is normal to $H$ along $z=0$, so it is a fourth geodesic distinct from the meridian and the two rulings.

Finally, write $\rho =\sqrt{x^2+y^2}=\sqrt{1+z^2}$. Clairaut's relation gives the conserved quantityChoose initial data in $z>0$ withSince $|\sin \psi |\le 1$, every point of the resulting geodesic satisfies $\rho \ge c$, and thereforeContinuity keeps the geodesic in the component $z>0$, and the stated completeness assumption defines it for every real time. The continuum of choices of $c$ supplies infinitely many such geodesics.

The $xy$ block has characteristic polynomial $(2-\lambda {)}^2-a^2$. A convenient orthonormal eigenbasis for the full real symmetric matrix iswith eigenvalues $2-a$, $2+a$, and $c$, respectively. Hence its orthogonal diagonalization isFor repeated eigenvalues, every nonzero combination within the corresponding eigenspace is also an eigenvector; the displayed basis remains valid. Introduce principal axes coordinatesThe quadratic form becomesThe principal-axis reduction of a quadric therefore has three regimes.

For $0<a<2$, all coefficients are positive: the surface is an ellipsoid centred at the origin, withFor $a=2$, the $X$ term disappears. The surface is an elliptic cylinder, unbounded along $e_{+}$, whose cross-section is $4Y^2+c{Z}^2=1$:For $a>2$, there is one negative coefficient, so the surface is a one-sheet hyperboloid:Its waist at $X=0$ is an ellipse with semiaxes $(a+2{)}^{-1/2}$ and $c^{-1/2}$. At other $X$, both waist radii are multiplied by $\sqrt{1+(a-2)X^2}$. There is no intercept on its longitudinal axis. Its asymptotic cone is $(a+2)Y^2+c{Z}^2=(a-2)X^2$. If positive eigenvalues coincide, the corresponding choice of principal axes is not unique, reflecting a rotational symmetry of the surface.

Close the hyperboloid segment with the two disks at $z=\pm 1$, and let $V$ be its interior. The flux through either disk is zero because $A_z=0$. Thus the desired surface integral is the total outward flux through the closed boundary. Sincethe divergence theorem and cylindrical coordinates giveThe extra factor $r$ is the volume integral measure in cylindrical coordinates; omitting it would give the wrong flux.

Writing the coordinates as $(x,y,z)$, the level equation is $x^2+y^2-z^2=\alpha$. Therefore:

The cone vertex is an instance of the double cone singularity. The tips of the two-sheeted hyperboloid are regular because the gradient $(2x,2y,-2z)$ is nonzero there.

A quadric surface in three-dimensional Euclidean space is the zero set of a real polynomial of degree two. Its quadratic part can be written with a symmetric matrix $A$, since the antisymmetric matrix part contributes zero to $x^{T}Ax$. An orthogonal diagonalization and a translation expose its geometry: examples include an ellipsoid, an elliptic cone, and a hyperboloid. This affine geometric description allows singular surfaces such as a cone's vertex.

Unit-radius two-dimensional de Sitter spacetime is the hyperboloid $(X^0{)}^2-(X^1{)}^2-(X^2{)}^2=-1$ in three-dimensional Minkowski spacetime. Its closed coordinates are $X^0=\sinh t$, $X^1=\cosh t\cos \chi$, $X^2=\cosh t\sin \chi$, with $t\in \mathbb{R}$ and $\chi$ periodic modulo $2\pi$. Unwrapping $\chi$ gives its universal covering spacetime instead.