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In constant sectional curvature $\kappa$, a deviation vector orthogonal to a unit timelike geodesic obeys $D^{2}S=\kappa S$, where $D={\nabla }_T$. Therefore $K=g(DS,DS)-\kappa g(S,S)$ is constant, since its derivative is $2g(DS,D^{2}S)-2\kappa g(S,DS)=0$. Its squared length $f=g(S,S)$ satisfies $\ddot{f}=2K+4\kappa f$. In four spacetime dimensions, $\kappa =R/12$.

In a parallel-propagated orthonormal frame, orthogonal geodesic deviation in constant sectional curvature reduces to three scalar equations ${\ddot{S}}^i=\kappa S^i$. For $\kappa >0$, $S=A{e}^{\sqrt{\kappa }\tau }+B{e}^{-\sqrt{\kappa }\tau }$: generic initial data grow exponentially, but the pure decaying mode is a counterexample to universal growth. Initially comoving neighbors have $S=S(0)\cosh (\sqrt{\kappa }\tau )$. For $\kappa <0$, $S=A\cos (\sqrt{-\kappa }\tau )+B\sin (\sqrt{-\kappa }\tau )$, so the deviation is bounded and oscillatory. Simultaneous focusing requires compatible vector initial data and is not automatic for every oscillatory solution.

For a nondegenerate orthonormal coframe in spacetime $E^a$ and independent antisymmetric Lorentz connection 1-form, the action $\int {ϵ}_{abc}(R^{ab}\wedge E^c+\lambda E^a\wedge E^b\wedge E^c)$ has independent coframe and connection variations. Since $\delta R^{ab}=D\delta {\omega }^{ab}$, integration by parts yields a connection equation ${ϵ}_{abc}{T}^c=0$. The three variations of the cubic coframe term give the coframe equation ${ϵ}_{abc}(R^{ab}+3\lambda E^a\wedge E^b)=0$. In three dimensions these are exactly the equations displayed above. The solution has constant sectional curvature $K=-3\lambda$. Omitting the factor of three changes the cosmological normalization.

For a connected smooth Riemannian manifold $(M,g)$ without boundary, let $I(M)$ be the group of all global diffeomorphisms $f$ with $f^{\ast }g=g$, endowed with its compact-open topology. Equivalently these maps preserve the intrinsic distance; the Myers-Steenrod theorem supplies smoothness of distance-preserving bijections and makes $I(M)$ a finite-dimensional Lie group acting smoothly. The group is metric-dependent: a topological manifold can carry a very symmetric metric or one with no continuous symmetries. Connectedness matters here. With infinitely many identical disconnected components, permutations of those components can produce a group which is not a finite-dimensional Lie group.

An isometry is determined by its value and derivative at one point. To prove this, suppose $f(p)=p$ and $d{f}_p=I$. Preservation of the Levi-Civita connection giveswhere the exponential map is defined locally. Thus $f$ is the identity in a normal neighborhood of $p$. The set where its value and derivative agree with the identity is both closed and, by the same argument, open. Connectedness makes it all of $M$. No completeness is needed for this determination argument.

The stabilizer $I(M{)}_p$ consequently acts faithfully and orthogonally on $T_{p}M$, so embeds in $O(n)$. More generally an isometry takes a chosen orthonormal frame at $p$ to one at its image point, providing an injective map into the orthonormal frame bundle. It is a smooth embedding, and the full isometry action is proper. The orbit-stabilizer dimension formula therefore givesIsotropy is compact. If $M$ is compact then $I(M)$ itself is compact, by the same frame description or Arzela-Ascoli applied to isometries and their inverses. Properness also means that the quotient by isometries has well-controlled local orbit structure, even though orbit dimensions may vary.

The infinitesimal equation is the Killing equation. Differentiating $f_t^{\ast }g=g$ at zero givesConversely a vector field satisfying this equation has local flows preserving the metric. A global one-parameter subgroup requires the field to be complete. On a geodesically complete manifold every Killing field is complete: its length is constant along each of its own trajectories, so a finite-time trajectory has bounded speed and stays in a closed bounded ball. The Hopf-Rinow theorem makes that ball compact, and the ordinary differential equation can then be continued. The caveat is real: on the Euclidean interval $(0,1)$, the field ${\partial }_x$ is Killing but its translations are not globally defined for all time. This illustrates Killing fields on incomplete manifolds need not generate global isometries.

The Killing transport identities determine a Killing field from $X(p)$ and the skew-adjoint map $(\nabla X{)}_p$. Along a geodesic the field is a Jacobi field, so uniqueness for its second-order differential equation propagates these initial data. They have $n+n(n-1)/2$ components, reproducing the dimension bound infinitesimally. The important distinction on an incomplete manifold is between all local Killing solutions and the Lie algebra of complete global isometry flows.

In dimension three the bound is six, and rotational isotropy gives stronger information. The Lie subalgebras of $\mathfrak{so}(3)$ have dimensions zero, one or three: identifying the bracket with the cross product, a two-plane is not closed under brackets. Suppose $\dim I(M)=5$. Since an orbit has dimension at most three, its stabilizer would have dimension at least two, and hence exactly three. Its connected isotropy acts as all of $SO(3)$ on the tangent space. But the orbit tangent space is invariant under that action and would have dimension $5-3=2$, impossible because the standard rotational representation is irreducible. Therefore dimension five cannot occur.

If the dimension is six, each orbit has dimension three and each stabilizer dimension three. All orbits are open, so connectedness makes the action transitive. At each point the isotropy is transitive on tangent two-planes, forcing sectional curvature to be independent of the plane; transitivity makes its value independent of the point. Thus a maximally symmetric connected three-manifold has constant sectional curvature. Homogeneous Riemannian manifolds are complete, so its simply connected cover is the round sphere, Euclidean space or hyperbolic space after normalization. Conversely these three simply connected metrics attain dimension six:Here $O^{+}(3,1)$ preserves the chosen sheet of the hyperboloid; it includes orientation-reversing hyperbolic isometries. Constant curvature alone does not imply that a quotient retains six-dimensional global symmetry.

Dimension four likewise forces transitivity: stabilizer dimension three would give a one-dimensional invariant orbit tangent space, again impossible for full rotational isotropy. The stabilizer thus has dimension one and the orbits are three-dimensional. Typical four-dimensional isometry groups occur for $S^2\times \mathbb{R}$ and ${\mathbb{H}}^2\times \mathbb{R}$, where the factor groups have dimensions three and one. The usual Nil metric has three translation symmetries and one rotational isotropy symmetry. In centred Heisenberg coordinates it iswhich exhibits the horizontal rotations. Its distinct vertical and horizontal Ricci eigenvalues bound continuous isotropy by that rotation group, so its full isometry dimension is four. Standard metrics on $\widetilde{S{L}_2(\mathbb{R})}$ and nonround Berger spheres also have dimension four. The standard Sol geometry has a three-dimensional translation group and only finite isotropy, giving dimension three. These examples show how a preferred vertical direction reduces rotational symmetry.

The isometry dimensions in dimension three are precisely $0,1,2,3,4,6$, all realizable. A compact hyperbolic three-manifold gives zero; the product of a circle and a closed hyperbolic surface gives one. For dimension two take $T^2\times (0,1)$ withIts scalar curvature is the constant $-14$, but its Ricci eigenvalues are the distinct constants $-3,-6,-5$ in the two horizontal directions and the vertical direction. Hence connected isotropy is trivial and possible isometries preserve these line distributions. A local isometry preserving them has $t^{\prime }=t+c$ and would require constant dilations $e^{-c}$ and $e^{-2c}$ of the two circle coordinates. A global circle diffeomorphism with such a constant dilation has degree $\pm 1$, forcing $c=0$. The identity component is therefore exactly the two circle translations. A flat three-torus gives dimension three; $S^2\times \mathbb{R}$ gives four; a simply connected constant-curvature model gives six. This construction distinguishes local homogeneity from the smaller global isometry group.

For quotients, let a complete simply connected Riemannian manifold $X$ cover $M=X/\Gamma$. Every isometry of $M$ lifts to an isometry of $X$ which normalizes the deck transformation group. Conversely, every normalizing isometry descends. Two lifts differ by a deck transformation, giving the isometry group of a Riemannian quotient formulaThus one should not replace the isometry group of a quotient by that of its model geometry. For a flat torus, for example,The second factor is finite, whereas translations give the three-dimensional identity component.

There is a useful curvature test for compact manifolds. Contracting and differentiating the Killing equation gives ${\nabla }^{\ast }\nabla X=\mathrm{Ric}(X)$, with the nonnegative rough-Laplacian convention. Integration yields the integrated Bochner identity for Killing fieldsIf Ricci curvature is negative definite, no nonzero Killing field exists. The compact isometry group then has dimension zero and is finite. This explains why closed hyperbolic manifolds have finite global symmetry, despite the six-dimensional isometry group of their universal cover. Compactness is essential to this conclusion.

For a compact flat Riemannian manifold, the same identity makes every Killing field parallel. Parallel fields are exactly the vectors fixed by linear holonomy. ThereforeApplied to the six flat types of Question 1, this gives dimensions three for the torus, one for each of the four nontrivial cyclic types, and zero for the Hantzsche-Wendt manifold. These are identity-component dimensions; disconnected finite symmetries can still be present. Local curvature symmetry, global deck-group constraints, and completeness thus play different roles in determining the isometry group.

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.

Treat the orthonormal coframe in spacetime and the Lorentz connection 1-form as independent variables, with nondegenerate coframe and ${\omega }_{ab}=-{\omega }_{ba}$. Fix the curvature conventionThe second expression is Cartan's first structure equation; the first is Cartan's second structure equation. Use compactly supported variations, or boundary data that make the variation's surface term vanish.

For a connection variation, $\delta R^{ab}=D\delta {\omega }^{ab}$, where $D$ is the covariant exterior derivative. Since $D{ϵ}_{abc}=0$, its graded product rule givesFor a coframe variation, relabeling the dummy indices makes the three variations of the cubic term identical. HenceThe independent variations therefore give ${ϵ}_{abc}{T}^c=0$ and ${ϵ}_{abc}(R^{ab}+3\lambda E^a\wedge E^b)=0$. In three dimensions the alternating symbol identifies an antisymmetric index pair with a single index, so these equations are equivalent toThese are the equations of first-order three-dimensional gravity. The first selects the Levi-Civita connection; the second fixes constant sectional curvature $K=-3\lambda$. The factor of three follows from varying the coefficient exactly as printed.

On a static coordinate patch where $f>0$, choose the Lorentz frame metric ${\eta }_{ab}=\mathrm{diag}(-1,1,1)$ and coframePut $A=f^{\prime }/(2\sqrt{f})$ and $B=\sqrt{f}/r$. Exterior differentiation gives $d{E}^0=-A{E}^0\wedge E^1$, $d{E}^1=0$, and $d{E}^2=B{E}^1\wedge E^2$. Solving the equation of vanishing torsion forms givesHere antisymmetry is in the two raised or two lowered frame indices; mixed time-space components have equal, rather than opposite, signs. This is the Lorentzian connection-form antisymmetry needed to retain the curvature signs.

Using Cartan's second structure equation, the nonzero curvature 2-forms areFor example, $d(A{E}^0)=-(\sqrt{f}{A}^{\prime }+A^2)E^0\wedge E^1=-f^{\prime \prime }{E}^0\wedge E^1/2$, while the $02$ form is the product ${\omega }^0{}_1\wedge {\omega }^{12}=-AB{E}^0\wedge E^2$. Thus the field equations require $f^{\prime \prime }/2=3\lambda$ and $f^{\prime }/(2r)=3\lambda$. Integrating the latter gives $f=C+3\lambda r^2$; the former is then satisfied automatically. The origin condition fixes $C=1$, yieldingAs a sign and normalization check, the Ricci tensor is $R_{\mu \nu }=-6\lambda g_{\mu \nu }$ and the scalar curvature is $R=-18\lambda$. The coframe calculation applies initially in the region with $f>0$; if $\lambda <0$, $f$ vanishes at $r=1/\sqrt{-3\lambda }$, delimiting that static chart without changing the derived function.

At each point, apply the indefinite-signature version of the Gram-Schmidt process to any local coframe, choosing one timelike and three spacelike one-forms. This produces an orthonormal coframe in spacetime $\{{\theta }^a\}$ satisfyingIt is unique only up to a local Lorentz transformation. The torsion-free connection 1-forms are then determined by Cartan's first structure equation and metric compatibility,Cartan's second structure equation gives the curvature 2-formsand contraction gives the Ricci tensor $R_{bd}=R^a{}_{bad}$.

For the displayed metric, takeThen $d{\theta }^0=0$ and $d{\theta }^i=-{\theta }^0\wedge {\theta }^i$. The nonzero connection forms are thereforewith ${\omega }^i{}_j=0$. A second application of the structure equations givesEquivalently,so this is a constant sectional curvature spacetime with $K=1$. In four dimensions,The Vacuum Einstein equations with a cosmological constant are $G_{ab}+\Lambda g_{ab}=0$, and henceThe metric is the contracting flat slicing of de Sitter spacetime with curvature radius one.

Contract the isotropic-curvature formula on $a,c$. In dimension $n$ it givesSubstitution into the contracted Bianchi identity givesSince $n>2$, ${\nabla }_{b}K=0$, so $K$ is constant on each connected component. Thus this is constant sectional curvature, withThis argument is Schur theorem in pseudo-Riemannian geometry.

In the propagated frame $T=e_0$, and the constant sectional curvature formula reduces geodesic deviation toThe stated temporal initial data give $y^0(t)=0$. Writing $\omega =\sqrt{-K}$ for $K<0$ and $\kappa =\sqrt{K}$ for $K>0$, the spatial displacement isThese formulas are valid to first order in the initial separation and relative velocity.