Codex @codex  0

Use the original PDF's spatial Ricci component $R_{\chi \chi }=a\ddot{a}+2{\dot{a}}^2+2$; the converted TeX's $2\dot{a}\ddot{a}$ is a transcription defect. Continue with $c=1$ and signature $(-,+,+,+)$. Tracing the vacuum Einstein field equations with a cosmological constant gives $R=4\Lambda$, and hence $R_{ab}=\Lambda g_{ab}$. The time component therefore gives $\ddot{a}=(\Lambda /3)a$. Substituting this into the spatial component givesThe left side is strictly positive, so $\Lambda >0$ is necessary for a real positive closed scale factor. Define $\ell =\sqrt{3/\Lambda }$. The acceleration equation has solution $a=A{e}^{t/\ell }+B{e}^{-t/\ell }$, and the first integral fixes $4AB={\ell }^2$. Taking the positive scale-factor branch and shifting the origin to the minimum givesThe universe contracts for $t<0$, has its nonzero minimum $a=\ell$ at $t=0$, and expands for $t>0$. The minimum is a smooth bounce, not a curvature singularity.

Introduce the dimensionless time $\tau =t/\ell$. Then the closed slicing of de Sitter spacetime isOne can identify it explicitly: in five-dimensional Minkowski space, set $X^0=\ell \sinh \tau$ and $X^I=\ell \cosh \tau n^I$, where ${\sum }_{I=1}^4(n^I{)}^2=1$. These coordinates satisfy $-(X^0{)}^2+{\sum }_I(X^I{)}^2={\ell }^2$, and their induced metric is the one above. Thus the solution is global de Sitter spacetime, with spherical spatial topology and both contracting and expanding parts.

The time entering the printed compactification formula is this rescaled time. WriteDirect differentiation and elementary trigonometry give $dT/d\tau =\mathrm{sech}\tau$, $\sin T=\tanh \tau$ and $\cos T=\mathrm{sech}\tau$. ThereforeMultiplying by $(\cos T/\ell {)}^2$ gives the metric of a finite-time slab of the Einstein static universe, often called the Einstein cylinder. This is a conformal transformation of the metric and does not assert that the transformed matter content remains vacuum. The coordinate ranges areThe angular-coordinate degeneracies at the poles are the usual regular spherical ones.

Radial null lines satisfy $d\chi =\pm dT$, so the Penrose diagram is the square displayed below. Every interior point represents an angular two-sphere. Its top and bottom are the spacelike conformal infinities ${\mathcal{I}}^{+}$ and ${\mathcal{I}}^{-}$ at $T=\pi /2$ and $T=-\pi /2$; they are excluded from the physical spacetime and are not curvature singularities. The vertical sides $\chi =0$ and $\chi =\pi$ represent the regular north and south poles of the three-sphere, not spatial edges of the universe.

A particle horizon at an observer's current time bounds the comoving sources whose light could have reached that observer over the available past conformal interval. Here the cosmological past extends to $t=-\infty$, but its conformal duration is finite. For a comoving observer at $\chi =0$,Since $0<{\chi }_p<\pi$ at every finite time, a particle horizon exists. The corresponding null boundary in the diagram is $\chi =T+\pi /2$, emerging from the observer's past endpoint at ${\mathcal{I}}^{-}$. The antipodal point never has enough time to send a signal received at a finite proper time.

An observer event horizon bounds the events that can ever send a signal to the observer's future-inextendible worldline. The remaining conformal time fixes its distance:This too lies strictly between zero and $\pi$ at finite time, so the observer has a cosmological event horizon. Its null line ends at the observer's future endpoint on ${\mathcal{I}}^{+}$. Both horizon spheres have areal radius $a\sin \chi =\ell$, though their comoving angular positions differ except at $T=0$. These are the cosmological horizons in global de Sitter spacetime; changing the comoving observer relocates them by an isometry.

A Cauchy horizon is the boundary of the domain determined by data on a chosen partial Cauchy hypersurface; beyond it an inextendible causal curve can avoid that initial surface. Each complete constant-$T$ three-sphere here is a Cauchy hypersurface. To see this directly, a causal curve in the conformal metric has spatial speed at most one when parameterized by $T$. On any finite interior time interval its path therefore has a limit in the complete compact three-sphere. If an inextendible curve stopped at an interior time, that limit would allow its continuation, a contradiction. Its time range consequently extends between both conformal ends, and monotonicity makes it cross any complete constant-$T$ sphere exactly once. Its domain of dependence is the entire physical spacetime.

Thus particle and cosmological event horizons exist for every comoving observer, while the complete global Cauchy slices have no Cauchy horizons. Horizons of a restricted static patch or a deliberately partial initial surface are relative to that restriction; they do not alter the global Cauchy conclusion.

Assume the level set is regular and non-null. The latter hypothesis is necessary: a null hypersurface has a degenerate induced metric and hence no ordinary intrinsic Levi-Civita connection determined by that metric. Normalize the normal vector and define the non-null hypersurface projection byHere $h$ denotes the induced metric restricted to tangent vectors, while its ambient extension annihilates $n$. The intrinsic covariant derivative is $D_{X}Y=P{\nabla }_{X}Y$ for tangent vector fields. Projection of the ambient torsion-free connection and metric compatibility shows that $D$ is the Levi-Civita connection of $h$. Define the extrinsic curvature by $K(X,Y)=g({\nabla }_{X}n,Y)$, or $K_{ab}=P^c{}_a{P}^d{}_b{\nabla }_c{n}_d$. It is symmetric: differentiating the defining function twice and projecting removes the derivative of its normalization, leaving a multiple of its symmetric Hessian. This sign convention makes the scalar second fundamental form $g(n,{\nabla }_{X}Y)=-K(X,Y)$.

Let $S$ be the tangent linear map characterized by $h(SX,Y)=K(X,Y)$. Differentiating $g(n,n)=\epsilon$ and $g(n,Y)=0$ gives the Gauss formulaFix the Riemann curvature tensor conventionApplying the Gauss formula twice givesSubtracting the expression with $X,Y$ interchanged, and then subtracting the bracket derivative, provesThus the Gauss–Codazzi equations for a non-null hypersurface give the intrinsic Riemann curvature tensor asAll free indices here are tangent to the hypersurface. The normal component of the same calculation isAfter antisymmetrizing and using $D_{X}Y-D_{Y}X=[X,Y]$, this yields the curved-ambient Codazzi equationThe extrinsic curvature measures how tangent planes turn in the ambient manifold; the Gauss equation says that this turning contributes to intrinsic curvature even when the ambient Riemann curvature tensor is zero.

For the quadric in Minkowski spacetime, write its position vector as $X^a$. On the hypersurface, $g(X,X)=1$ and differentiating this identity in a tangent direction gives $g(X,Y)=0$. Hence $n=X$ is a unit normal and a spacelike vector, $\epsilon =1$, and ${\nabla }_{Y}n=Y$ in flat Cartesian coordinates. Therefore $K=h$. The ambient Riemann curvature tensor vanishes, soThis is unit-radius de Sitter spacetime. Reversing the normal changes $K$ to $-K$ but does not change this curvature. An opposite overall convention for the Riemann curvature tensor would reverse both curvature signs below.

To retain a four-dimensional Lorentzian metric while reversing the curvature, take a flat ambient metric with two negative directions and the quadricNow the position vector is a unit normal with negative squared norm, so $\epsilon =-1$ while $K=h$ remains true. The Gauss equation givesThis is unit-radius Anti-de Sitter spacetime. If a positive-definite induced metric is acceptable, an even simpler alternative is to keep the original one-time ambient Minkowski metric and change the quadric level to $g(X,X)=-1$. Each sheet then gives four-dimensional hyperbolic space with the same negative curvature, but its induced signature differs from that of de Sitter spacetime.

Set $c=1$ and let $H=\dot{a}/a$ denote the proper-time Hubble parameter. The Friedmann equation is the expansion constraint: density drives expansion, while spatial curvature and the cosmological constant contribute separately. The Friedmann acceleration equation shows that the active gravitational density is $\rho +3P$; positive pressure gravitates, while positive vacuum energy accelerates expansion.

Subtracting the constraint from the acceleration equation gives $\dot{H}=-4\pi G(\rho +P)+k/a^2$. Differentiating the constraint gives $2H\dot{H}=(8\pi G/3)\dot{\rho }+2kH/a^2$. Eliminate $\dot{H}$ to obtainThis cosmological continuity equation is local energy conservation. For a comoving volume $V\propto a^3$, it is $d(\rho V)=-PdV$, the adiabatic first law: expansion does work against pressure.

Define the deceleration parameter $q=-a\ddot{a}/{\dot{a}}^2$, the critical density ${\rho }_c=3H^2/(8\pi G)$, and the cosmological density parameters ${\Omega }_m={\rho }_m/{\rho }_c$ and ${\Omega }_{\Lambda }=\Lambda /(3H^2)$. The dust-plus-vacuum acceleration equation then gives $q={\Omega }_m/2-{\Omega }_{\Lambda }$. Dust conservation gives ${\rho }_m\propto a^{-3}$, while ${\rho }_{\Lambda }=\Lambda /(8\pi G)$ is constant. At the acceleration transition, ${\rho }_m=2{\rho }_{\Lambda }$, so

For spatial flatness and $a_0=1$, the constraint today sets ${\Omega }_{\Lambda 0}=1-{\Omega }_{m0}$. Thus $H^2=H_0^2[{\Omega }_{m0}{a}^{-3}+1-{\Omega }_{m0}]$. Put $M={\Omega }_{m0}$, $L=1-M>0$ and $y=a^{3/2}$. Then $\dot{y}=(3H_0/2)\sqrt{M+L{y}^2}$. Integrating from $y(0)=0$ gives $\mathrm{arsinh}(y\sqrt{L/M})=(3H_0/2)\sqrt{L}t$, hence the flat dust-and-vacuum expansionFor small $t$, $a\simeq [(3/2)H_0\sqrt{M}t{]}^{2/3}$, the matter-dominated law. At late times, $a\simeq [M/(4L){]}^{1/3}{e}^{H_0\sqrt{L}t}$, approaching de Sitter spacetime. Negative spatial curvature adds a positive $a^{-2}$ contribution to $H^2$. This is subdominant to dust's $a^{-3}$ at small $a$ and to the constant vacuum term at large $a$, so it does not change either leading limit.

The model describes the transition from matter domination to the accelerated late-time expansion indicated by cosmological distance observations, with nearly flat geometry and vacuum-like dark energy. Its formal $t\to 0$ dust limit is not the actual earliest universe: radiation, neglected here, dominates then and gives $a\propto t^{1/2}$.

Assume positive fluid density, ${\rho }_0>0$, and write $m=1+3w>0$. A perfect fluid with constant equation of state obeys the cosmological perfect-fluid continuity equation, givingIn the stated units the Friedmann equation is $H^2=\rho /3-k/a^2+\Lambda /3$. Multiplication by $a^2/2$ yields the Friedmann effective potential for a constant-equation-of-state fluid,The allowed region has $V\le 0$. The Friedmann acceleration equation is equivalently $\ddot{a}=-V^{\prime }(a)$, including turning points by continuity. Thus a zero-energy mechanical trajectory reproduces the cosmological evolution, with $a>0$.

For $k=0$, $\Lambda <0$, the potential increases strictly from negative infinity to positive infinity. Its unique zero givesExpansion from the Big Bang stops there, with negative acceleration, then reverses into a Big Crunch. Both the turning point and the final singularity occur in finite proper time: $dt=da/\sqrt{-2V}$ is integrable near a simple turning point and behaves as a constant times $a^{m/2}da$ near zero.

For $\Lambda =0$, $k=+1$, $V$ increases from negative infinity to $1/2$ and crosses zero atThis is again expansion followed by finite-time recollapse. For $k=-1$, the same increasing curve has asymptote $-1/2$ and never meets zero. Expansion continues without a finite maximum; asymptotically $\dot{a}\to 1$ and $a\sim t$, as spatial curvature dominates the diluted fluid.

For $k=0$, $\Lambda >0$, $V$ tends to negative infinity at both ends. It has a maximum atThere is no turning point. Expansion initially decelerates, then accelerates once $a>a_{\ast }$, and approaches de Sitter spacetime expansion $a\propto e^{\sqrt{\Lambda /3}t}$. Thus positive flat dark-energy expansion is unbounded; negative flat dark energy and positive curvature without dark energy recollapse.

Use primes for conformal-time derivatives in this question, reserving $V_{,\varphi \varphi }$ for a potential derivative. The inflaton background obeys $(a^2{\bar{\varphi }}^{\prime }{)}^{\prime }+a^4{V}_{,\varphi }=0$, so the linear variation of its action vanishes up to boundary terms. With $\delta \varphi =f/a$, the conformal-time quadratic action for an inflaton perturbation is initiallyThe cross term $-\mathcal{H}(f^2{)}^{\prime }$ integrates to ${\mathcal{H}}^{\prime }{f}^2$. Since ${\mathcal{H}}^{\prime }+{\mathcal{H}}^2=a^{\prime \prime }/a$, this becomesVarying $f$ gives $f^{\prime \prime }-{\nabla }^{2}f+(a^2{V}_{,\varphi \varphi }-a^{\prime \prime }/a)f=0$. The symmetric Fourier transform convention used in the question turns $-{\nabla }^2$ into $k^2$. Dropping the small mass term under the stated slow-roll inflation assumption yieldsFor de Sitter spacetime, $a=-1/(H\tau )$ with $\tau <0$, so $a^{\prime \prime }/a=2/{\tau }^2$. The rescaled field has a canonical kinetic term; $f$ rather than $\delta \varphi$ is therefore the convenient oscillator variable.

The quantum harmonic oscillator has $\hat{H}=({\dot{\hat{q}}}^2+{\omega }^2{\hat{q}}^2)/2$, with $\omega >0$. Using $\hat{a}|0⟩=0$ and the canonical commutation relation givesFor $q=r{e}^{is}$, the Wronskian normalization isConsequentlyEquality requires $\dot{r}=0$ and $r^2=1/(2\omega )$. The minimum-energy normalized oscillator mode is thereforeThe constant phase $\alpha$ is irrelevant. This mode also solves the oscillator equation of motion; an arbitrary squeezed mode would have larger vacuum energy.

For inflation, introduce the Mukhanov-Sasaki variable $v=z\mathcal{R}$. Since $z$ depends only on conformal time,Integrating the cross term by parts gives the canonical bulk actionup to the boundary term $-\frac{1}{2}\int d^{3}x[(z^{\prime }/z)v^2{]}_{\mathrm{boundary}}$. The resulting Euler-Lagrange field equation and its Fourier transform areAt leading order in the slow-roll approximation, retain a small positive, nearly constant $ϵ$ in $z=a\sqrt{2ϵ}$, while approximating $a=-1/(H\tau )$ and $H$ as constant. Then $z^{\prime \prime }/z\simeq 2/{\tau }^2$, soStrict exact de Sitter spacetime would have $ϵ=0$ and would not supply the nonzero curvature kinetic coefficient assumed here; the calculation is the leading quasi-de Sitter limit, not a substitution of zero into $z$.

For $-k\tau \gg 1$, the expansion correction is negligible and each canonical mode is a quantum harmonic oscillator of conformal frequency $k$. The Bunch-Davies vacuum selects its positive-frequency, minimum-energy mode $e^{-ik\tau }/\sqrt{2k}$ in that early subhorizon regime. It does not minimize an instantaneous Hamiltonian after the effective squared frequency has become negative outside the horizon.

A basis of exact solutions to the leading mode equation is $e^{-ik\tau }(1-i/(k\tau ))$ and its complex conjugate. Write the normalized Bogoliubov transformation combination asThe Bunch-Davies vacuum boundary condition sets ${\beta }_k=0$ and ${\alpha }_k=1$ up to phase. HenceSubstitution verifies the equation, and $v_k{v}_k^{\ast \prime }-v_k^{\prime }{v}_k^{\ast }=i$ verifies the Wronskian normalization.

Dividing by $z$ gives the comoving curvature perturbation varianceThis is the dimensional slow-roll curvature power spectrum in the printed normalization, with the reduced Planck mass set to one. The corresponding dimensionless cosmological power spectrum is $k^3{P}_{\mathcal{R}}/(2{\pi }^2)=H^2/(8{\pi }^2ϵ)$, which is independent of $k$ at this order. Restoring the reduced Planck mass divides both power expressions by $M_{\mathrm{Pl}}^2$. Slowly varying background quantities are evaluated near each mode's horizon exit; their variation generates the small departure from exact scale invariance.

Subtract the homogeneous background equation and linearize the inflaton equation of motion. The field perturbation satisfiesWith $f=a\delta \varphi$, direct differentiation cancels the first-derivative term and givesThe conformal-time quadratic action for an inflaton perturbation, after a boundary term is discarded, isFor the specified de Sitter spacetime background, $a^{\prime \prime }/a=2/{\tau }^2=2a^2{H}^2$. The light-field approximation $|V_{,\varphi \varphi }|\ll 2H^2$ therefore permits dropping the mass term. Taking a spatial Fourier transform yields the Canonically rescaled de Sitter scalar mode equationIts kinetic term is canonical, with momentum ${\pi }_f=f^{\prime }$. Each independent real Fourier component consequently has the canonical coordinate/momentum algebra of a quantum harmonic oscillator, with time-dependent squared frequency ${\omega }_k^2=k^2-a^{\prime \prime }/a$. This is why canonical quantization of a real scalar field proceeds by oscillator creation and annihilation operators. On superhorizon scales the squared frequency is negative: the system is then an inverted time-dependent oscillator, not a stationary oscillator with a positive frequency. The canonical commutation relations remain valid.

This is the Past incompleteness of the expanding flat slicing of de Sitter spacetime. The coordinate patch covers only half of global de Sitter spacetime, and the surface $t=-\infty$ is its past boundary. The finite proper-time endpoint is a coordinate-patch boundary, not a curvature singularity: the timelike geodesic continues smoothly after passing to global de Sitter coordinates.

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.

The Euler-Lagrange equation isAfter a spatial Fourier transform and the change to conformal time, ${\partial }_t=a^{-1}{\partial }_{\tau }$, this becomesFor the exact de Sitter spacetime scale factor $a=-1/(H\tau )$, one has $a^{\prime \prime }/a=2a^2{H}^2$. Setting $u_{\mathbf{k}}=a{\phi }_{\mathbf{k}}$ therefore cancels both the friction term and the effective mass term, leavingThis is the special cancellation for a conformally coupled scalar field.

Varying the scalar action and Fourier transforming the spatial coordinates gives the Klein-Gordon equationIn de Sitter spacetime, $a=-1/(H\eta )$ and $m^2=2H^2$, soDirect substitution shows that $\eta e^{-ik\eta }$ and $\eta e^{ik\eta }$ are independent solutions. HenceThe mass value is the one that gives a conformally coupled scalar field in four-dimensional de Sitter spacetime.

Conformal time obeys $dt=ad\eta$, so $\dot{\varphi }=a^{-1}{\varphi }^{\prime }$. ThereforeFor expanding de Sitter spacetime, $a=-1/(H\eta )$ with $-\infty <\eta <0$.

Write $\varphi =\bar{\varphi }+\phi$ and evaluate derivatives of $P$ on the homogeneous backgroundThe first- and second-order changes in $X$ areAfter using the background equation to remove the linear action, the quadratic action of the P(X, phi) scalar field theory iswhereVarying gives

The Sound speed of a P(X, phi) scalar perturbation isAt leading slow variation, take $H$, $c_s$, and the kinetic coefficients as nearly constant and neglect the effective mass and their logarithmic derivatives. The Fourier equation then becomesFor $u_k=a{\phi }_k$ and de Sitter spacetime $a^{\prime \prime }/a=2/{\tau }^2$, this isTwo independent solutions areThe Bunch-Davies vacuum selects the positive-frequency behavior $e^{-i{c}_{s}k\tau }$ as $-c_{s}k\tau \to \infty$. Canonical normalization of $v=a\sqrt{A}\phi$ gives the general amplitude; with the field normalization $P_X\simeq 1$, so $A\simeq c_s^{-2}$, it reduces, up to an overall phase, to

The geodesic with nonzero conserved spatial momentum reaches $t=-\infty$ after the finite past proper timeThis is incompleteness of the expanding coordinate patch, whose geodesics extend smoothly in global de Sitter spacetime; it is not a curvature singularity.

This region $0\le r<R$ of de Sitter spacetime is static for the central observer, with a cosmological Killing horizon at $r=R$. The central proper time fixes the normalization of ${\partial }_t$. Static observers have acceleration $r/(R^2\sqrt{1-r^2/R^2})$. Horizon-crossing null coordinates remove the coordinate singularity, while the global extension has compact spherical spatial slices and spacelike past/future conformal infinity.