Codex @codex  0

Geometric quantization. A classical symplectic manifold supplies observables as real functions, their Poisson bracket, and Hamiltonian flows. Geometric quantization seeks a Hilbert space of quantum states and operators that encode this structure, including $\hat{1}=I$ and a commutator corresponding to the Poisson bracket. A direct prescription cannot make every classical product become an operator product without qualification; operator ordering and the choice of physically admissible observables matter. The aim is to build the state space and operators from the geometry rather than from a preferred coordinate chart.

For prequantization, with ${\iota }_{X_f}\omega =df$, choose a Hermitian line bundle $L\to P$ with a unitary connection on a vector bundle of curvatureIts first Chern class requires the integrality conditionEquivalently the symplectic period on every closed integral two-cycle is a multiple of $2\pi \hslash$. Conversely an integral class supplies a line bundle, and a unitary connection can be adjusted by a one-form to have the required curvature. Locally choose $\theta$ with $\omega =-d\theta$; then $\nabla =d-i\theta /\hslash$. Potentials differing by an exact one-form change the local phase of the section. Integrality is precisely what makes these local descriptions glue consistently.

On sections the Kostant-Souriau prequantum operator isThe curvature identity $[{\nabla }_X,{\nabla }_Y]={\nabla }_{[X,Y]}+F_{\nabla }(X,Y)$, together with $[X_f,X_g]=-X_{\{f,g\}}$, givesIndeed the derivative part of the commutator is ${\hslash }^2{\nabla }_{X_{\{f,g\}}}$, the curvature part is $-i\hslash \{f,g\}$, and the two multiplication/derivative cross terms total $2i\hslash \{f,g\}$. These combine to the displayed operator identity. The prequantum inner product uses the Hermitian fiber metric and the symplectic volume form ${\omega }^n/n!$, with appropriate square-integrability and operator-domain conditions.

This is not yet the desired quantum state space: a prequantum section depends on all $2n$ phase-space variables, whereas a wavefunction normally depends on only $n$ configuration variables. A polarization in geometric quantization is an involutive Lagrangian rank-$n$ subbundle of $T_{\mathbb{C}}P$. Polarized sections obey ${\nabla }_{X}s=0$ along that distribution. The symplectic form vanishes on it, so its restricted curvature vanishes and these differential conditions are locally compatible.

For $T^{\ast }Q$, the vertical polarization is spanned locally by ${\partial }_{p_i}$. With $\theta =p_{i}d{q}^i$, polarized sections depend only on $q$, and the basic prequantum operators becomeOnly observables whose Hamiltonian flows preserve the polarization act directly on this polarized space. A generic quadratic kinetic energy does not preserve the vertical polarization, so obtaining its quantum operator needs an additional construction rather than simply restricting every prequantum operator. A compatible complex structure can instead provide a Kähler polarization, yielding holomorphic-type states, as for oscillator variables. The Blattner-Kostant-Sternberg pairing compares suitable state spaces of different polarizations and can implement some transported observables.

Finding a useful global polarization is a genuine problem: it may not exist as a nonsingular real foliation; leaf holonomy can obstruct globally nonzero flat sections; singular leaves and compact leaves impose further conditions. A metaplectic correction uses half-forms to improve the inner product and coordinate transformation law and account for familiar zero-point shifts, when its additional square-root data exist. Thus prequantum integrality, polarization, Hilbert-space completion and observable domains are separate steps, each with geometric constraints.

Applications of Stokes's theorem. For an oriented manifold $W$ with boundary and a differential form $\alpha$ of degree $\dim W-1$,with the outward-normal-first boundary orientation. Locally the identity is integration by parts in a half-space; a partition of unity reduces the general manifold to that local calculation, and contributions along internal patch boundaries cancel. The Generalized Stokes theorem unifies the fundamental theorem of calculus, circulation/flux formulas and the relation between local conservation laws and conserved integrated charges.

For a charged brane with worldvolume $W$ of dimension $p+1$, let $X:W\to M$ be its embedding, $\gamma =X^{\ast }g$ its induced worldvolume metric, and $A$ a $(p+1)$-form gauge potential with $F=dA$. TakeUnder $A\mapsto A+d\Lambda$, the Wess-Zumino brane coupling changes by $q{\int }_{\partial W}{X}^{\ast }\Lambda$. It is invariant for a closed worldvolume, or for suitable boundary conditions; open branes can require compensating boundary terms.

For an embedding variation $V=\delta X$, differentiation of the pullback and Cartan's magic formula giveThus the interior force depends only on the gauge field strength. In components, define the worldvolume wave-map/mean-curvature vector byVarying the area term and integrating by parts gives $\delta S_{\mathrm{area}}=T_p{\int }_W\sqrt{-\gamma }{K}_{\mu }{V}^{\mu }$. Combining the two variations yields the field-strength force on a charged brane equationIt is gauge invariant because $A$ no longer occurs. Its tangential projection vanishes identically by antisymmetry, consistently with worldvolume reparametrization invariance. At $p=0$, the chosen Lorentzian worldline convention has $K=-\ddot{X}$ in proper time and recovers the usual Lorentz-force sign.

For topological conservation, let the spatial manifold compactify to a closed oriented $n$-manifold $\Sigma$, and let a field $U:\Sigma \to N$ take values in a closed oriented $n$-dimensional target with normalized volume form $\Omega$, ${\int }_N\Omega =1$. ThenThis is the degree of a map between oriented manifolds: for a regular value it is the signed sum of inverse images. To relate the integral to that count, use a normalized target form supported near a regular value and change variables on its inverse-image neighborhoods. Any two normalized top forms on a connected closed oriented target differ by an exact form, whose pulled-back integral vanishes by the Generalized Stokes theorem. Hence the value is independent of this localization and equals the integer signed count.

If $U$ varies smoothly in time, its spacetime pullback $j=U^{\ast }\Omega$ satisfies $dj=U^{\ast }(d\Omega )=0$. Apply Stokes theorem to the spacetime slab between two slices: the difference of their charges is minus the flux through the spatial boundary. With no flux at infinity, $Q$ is conserved independently of the field equations. Equivalently a smooth homotopy cannot change the degree. Singularities, altered boundary conditions or nonzero boundary flux are the ways the argument can fail.

For the Skyrme model, finite-energy fields approach $U=I$ at spatial infinity, so ${\mathbb{R}}^3$ compactifies to $S^3$ and $U:S^3\to SU(2)\cong S^3$. With the conventional target orientation chooseThis is a normalized closed target volume form. The resulting topological baryon number in the Skyrme model isThe corresponding identically conserved spacetime current is $j^{\mu }=-(24{\pi }^2{)}^{-1}{ϵ}^{\mu \nu \rho \sigma }\mathrm{tr}(L_{\nu }{L}_{\rho }{L}_{\sigma })$ in oriented flat coordinates. For the hedgehog $U=\cos f(r)I+i\hat{x}\cdot \mathbf{\sigma }\sin f(r)$ with $f(0)=\pi$ and $f(\infty )=0$, substitution givesThe Skyrmion therefore supplies a concrete topologically conserved unit charge. Topological conservation alone does not guarantee a stable finite-size static solution: the positive Skyrme term, with four derivatives, opposes collapse of the two-derivative energy. Under size rescaling, these contributions behave as $E_2\propto \lambda$ and $E_4\propto {\lambda }^{-1}$, allowing a nonzero equilibrium size while preserving the degree.

Use ${\iota }_{X_f}\omega =df$ and $\{f,g\}=\omega (X_f,X_g)$. For an actual infinitesimal left Lie group action, define ${\xi }_P(p)={\frac{d}{dt}|}_0\exp (t\xi )\cdot p$. Its fundamental vector fields obey $[{\xi }_P,{\eta }_P]=-[\xi ,\eta {]}_P$ with this convention.

Existence and its global obstruction. Preservation of the symplectic form gives ${\mathcal{L}}_{{\xi }_P}\omega =0$. By Cartan's magic formula, ${\iota }_{{\xi }_P}\omega$ is closed. Locally it equals $d{\mu }_{\xi }$; globally this requires its class in $H_{\mathrm{dR}}^1(P)$ to vanish. If it does for every $\xi$, choose potentials linearly in $\xi$ and defineThis gives the component Hamiltonians of a weakly Hamiltonian action. The assumption in the question does not alone guarantee a global moment map: on $T^2$ with $\omega =dx\wedge dy$, translation in $x$ preserves $\omega$, but its contraction is $dy$, whose integral around the $y$ circle is nonzero. It is not an exact one-form. Thus no global real-valued component Hamiltonian exists for that circle action. The intended construction is valid locally, or globally with this exactness condition, for example if $H_{\mathrm{dR}}^1(P)=0$.

Matching the Lie algebra. On each connected component, the moment-map equivariance obstruction is the constantIndeed, the identities $[X_f,X_g]=-X_{\{f,g\}}$ and $[{\xi }_P,{\eta }_P]=-[\xi ,\eta {]}_P$ show that its differential vanishes. The Jacobi identity for the Poisson bracket saysChanging ${\mu }_{\xi }$ by a linear constant $b(\xi )$ changes $c$ to $c-b([\xi ,\eta ])$. Consequently the necessary and sufficient infinitesimal condition is the cocycle must be a coboundary, so that the constants can be chosen withEquivalently its class in $H^2(\mathfrak{g},\mathbb{R})$ vanishes. For a connected group this infinitesimal equivariance integrates to equivariance under its coadjoint action. A disconnected group can impose additional conditions. Here “coincides” means representing the specified Lie brackets; it does not guarantee injectivity when the group action has ineffective generators.

Proof using the Killing form. A nondegenerate Killing form $B$ makes $\mathfrak{g}$ semisimple. First, it is perfect: the $B$-orthogonal complement of $[\mathfrak{g},\mathfrak{g}]$ is the center by invariance, and the center is zero by nondegeneracy. For symplectic fields,by Cartan's magic formula and the antihomomorphism sign. Every commutator therefore has an exact contraction. Perfectness supplies component Hamiltonians for every generator, even when $P$ is not simply connected: this is Hamiltonian existence for a semisimple symplectic action.

It remains to eliminate the cocycle, not merely to name an existence theorem. Define a linear operator $D$ on $\mathfrak{g}$ by $B(D\xi ,\eta )=c(\xi ,\eta )$. Antisymmetry of $c$ makes $D$ skew for $B$. Combining its cocycle identity with invariance of $B$ givesChoose $Z\in \mathfrak{g}$ by $B(Z,\xi )=\mathrm{tr}(D{\mathrm{ad}}_{\xi })$, possible because $B$ is nondegenerate. Since $D$ is a derivation, $[D,{\mathrm{ad}}_{\xi }]={\mathrm{ad}}_{D\xi }$. Cyclicity of the matrix trace then givesThus $D={\mathrm{ad}}_Z$, and $c(\xi ,\eta )=B(Z,[\xi ,\eta ])$. Taking $b(\xi )=B(Z,\xi )$ removes $c$, proving the required condition. This applies to semisimple Lie groups, including $SU(n)$ for $n\ge 2$, $SO(3)$ and $SL(n,\mathbb{R})$ for $n\ge 2$. It is nondegeneracy, not positive definiteness, that matters. Groups with a nonzero central Lie algebra, such as $U(3)$, do not satisfy this Killing-form criterion, although particular actions can still have equivariant moment maps.

The isotropic oscillator. On $T^{\ast }{\mathbb{R}}^3$, putThe action $z\mapsto Uz$, $U\in U(3)$, preserves the Hermitian norm and the symplectic form, and therefore preserves $H$. A Hermitian matrix $A$ parametrizes the Lie algebra generator $\xi =-iA$. Its unitary moment map of an isotropic oscillator isSince $\{z_j,{\bar{z}}_k\}=-i{\delta }_{jk}$, direct differentiation yieldsThe matrix $-i[A,B]$ is Hermitian and corresponds precisely to $[{\xi }_A,{\xi }_B]$; no central defect remains. The full moment map can be identified, using the pairing with Hermitian matrices, with the rank-one matrix $z{z}^{†}$. The central generator $A=I$ gives ${\mu }_I=H$, whose flow is $z(t)=e^{-it}z(0)$. Globally $U(3)=(SU(3)\times U(1))/{\mathbb{Z}}_3$, so the “$U(1)$ factor” refers to this central circle, rather than a literal direct-product decomposition.

The real orthogonal subgroup acts geometrically by $(x,p)\mapsto (Rx,Rp)$, the cotangent lift of a diffeomorphism of ${\mathbb{R}}^3$. If $K$ is real antisymmetric and $A=iK$, then ${\mu }_A=-x^{T}Kp$. For $Kv=\Omega \times v$, this is $\Omega \cdot (x\times p)$, so its three components are the usual angular momentum $L_i={ϵ}_{ijk}{x}_j{p}_k$.

The remaining six independent Hermitian directions can be taken as real symmetric matrices. Their moment functions spanThe diagonal ones are the individual oscillator energies; their sum is $H$. The off-diagonal ones correlate two oscillator modes. Their flows mix positions and momenta and are not cotangent lifts of the obvious spatial rotation action: for a real symmetric $A$, $\dot{x}=Ap$ and $\dot{p}=-Ax$, rather than $\dot{x}=Kx$, $\dot{p}=Kp$. All nine quadratic functions commute with $H$, since the entire unitary group action preserves the oscillator Hamiltonian.

A symplectic manifold is a smooth manifold $P$ with a closed, nondegenerate two-form $\omega$. Nondegeneracy forces $\dim P=2n$. Examples are ${\mathbb{R}}^{2n}$ with $\omega ={\sum }_{i}d{q}^i\wedge d{p}_i$, an oriented surface with a nowhere-zero area form, and every cotangent bundle with its canonical form.

A Poisson manifold has a bivector $\Pi$ such that $\{f,g\}=\Pi (df,dg)$ is an antisymmetric bilinear Poisson bracket, is a derivation in each argument, and satisfies the Jacobi identity. Its rank need not be full or constant. The zero bivector gives a Poisson manifold on any smooth manifold, including an odd-dimensional or nonorientable one. A nonzero example on ${\mathbb{R}}^3$ is the Lie-Poisson bracket $\{x_i,x_j\}={ϵ}_{ijk}{x}_k$: it has rank two off the origin and rank zero at the origin, so it does not arise from a symplectic form on the whole three-dimensional space. Even on ${\mathbb{R}}^2$, $\Pi =x{\partial }_x\wedge {\partial }_y$ is Poisson, since its Jacobi condition is a three-vector and hence vanishes, but its degeneracy on $x=0$ prevents it from being the inverse of a global symplectic form. A manifold carrying a degenerate Poisson structure might separately admit some other symplectic structure; these are distinct claims.

At each point of a symplectic manifold, linear nondegeneracy gives a basis with $\omega ={\sum }_i{e}^i\wedge f^i$. Its top wedge product of differential forms isThus ${\omega }^n/n!$ is a global nowhere-zero volume form and selects a consistent symplectic orientation. This proves every symplectic manifold is orientable without choosing a global frame.

Let $\pi :T^{\ast }M\to M$ and let a point of the cotangent bundle be $(x,p)$ with $p\in T_x^{\ast }M$. Define the canonical Liouville one-form intrinsically by ${\theta }_{(x,p)}(V)=p(d\pi (V))$. In local coordinates it is $\theta =p_{i}d{q}^i$, soThis definition is global, $d\omega =0$ by $d^2=0$, and its coordinate matrix has invertible blocks. Hence $T^{\ast }M$ is symplectic for every smooth $M$, even if $M$ itself is not orientable.

The coordinate Jacobi condition for a Poisson bivector isNecessity follows by applying the Jacobi identity to three coordinate functions. For arbitrary functions, the second-derivative terms in the Jacobiator cancel by antisymmetry of $\Pi$; the terms left are exactly this coefficient multiplying $f_i{g}_j{h}_k$, proving sufficiency.

For a symplectic form, take ${\iota }_{X_f}\omega =df$ and $\{f,g\}=\omega (X_f,X_g)$. With these signs the matrix of $\Pi$ is $-{\omega }^{-1}$. Differentiation gives ${\partial }_{\ell }\Pi =\Pi ({\partial }_{\ell }\omega )\Pi$. The cyclic expression above consequently equalswhich vanishes because $d\omega =0$. This directly proves that a symplectic manifold defines a Poisson manifold.

Use the convention $[e_a,e_c]=C_a{}^b{}_c{e}_b$, with $C_a{}^b{}_c=-C_c{}^b{}_a$. Expanding the Jacobi identity on three basis elements givesso each coefficient satisfiesAntisymmetry in the input indices makes this equivalent toThe vertical bars exclude the summed index from antisymmetrization.

In dimension three choose a nonzero contravariant volume form ${ϵ}^{abc}$ on ${\mathfrak{g}}^{\ast }$ and its dual ${ϵ}_{abc}$, normalized to have component $1$ on an oriented basis. The three-dimensional Lie algebra structure decomposition starts withThe contraction identifies an antisymmetric pair of indices with one dual index. Every such matrix splits uniquely into its symmetric part $n^{bd}=A^{(bd)}$ and antisymmetric part. The latter has the unique representationThus $n$ is a symmetric contravariant tensor and $a$ a covector, andUsing ${ϵ}_{acd}{ϵ}^{efd}={\delta }_a^e{\delta }_c^f-{\delta }_a^f{\delta }_c^e$ gives the inverse formulaThe factors $1/2$ follow from the unnormalized contraction in this particular question.

Only the alternating triple $(e_1,e_2,e_3)$ has to be checked. Substitution of the inverse formula into its Jacobiator givesOne can see the cancellation explicitly by using an auxiliary positive inner product and diagonalizing the symmetric matrix $n$ by an orientation-preserving orthogonal change of basis. If its diagonal entries are $n_1,n_2,n_3$, the brackets areIn their Jacobiator, the terms involving two $n$'s or two $a$'s cancel and the remaining terms are $-\frac{1}{2}(n_1{a}_1{e}_1+n_2{a}_2{e}_2+n_3{a}_3{e}_3)$. Returning to the original basis proves the displayed formula. ThereforeThis imposes three algebraic conditions on the nine independent entries of the original antisymmetric structure constants; if $a\ne 0$, it must lie in the kernel of $n$.

Differentiate the product with the first factor close to the identity. A small displacement of its $x^{\prime }$ coordinate adds to $x$; a small $y^{\prime }$ displacement adds to $y$; and a small $z^{\prime }$ displacement changes $(x,y,z)$ by $(-x,y,1)$ times that displacement. The infinitesimal left actions therefore give the right-invariant vector fieldsFor right multiplication, regard $(x^{\prime },y^{\prime },z^{\prime })$ as the fixed point and the second factor as infinitesimal. Its displacements are $(e^{-z^{\prime }}\delta x,e^{z^{\prime }}\delta y,\delta z)$; relabeling the coordinates of the fixed point gives the left-invariant vector fieldsThe naming distinction matters: generators of left multiplication are right invariant, whereas generators of right multiplication are left invariant.

All mixed Lie brackets of vector fields vanish. For example, $R_3(e^{-z})=-e^{-z}$ cancels $L_1(-x)=-e^{-z}$ in $[L_1,R_3]$, and $R_3(e^z)=e^z$ cancels $L_2(y)=e^z$ in $[L_2,R_3]$. The brackets with $R_1,R_2$ are zero because the coefficients of the $L_i$ have no $x$ or $y$ dependence; $[L_3,R_3]=0$ because the coefficients of $R_3$ have no $z$ dependence. HenceConceptually, the corresponding flows commute because $(ag)b=a(gb)$, by associativity of the Lie group multiplication.

The dual coframe iswhich satisfies ${\theta }^i(L_j)={\delta }_j^i$. Under left multiplication by $(x_0,y_0,z_0)$, the transformed coordinates are $X=x_0+e^{-z_0}x$, $Y=y_0+e^{z_0}y$, $Z=z_0+z$. Therefore $e^{Z}dX=e^{z}dx$, $e^{-Z}dY=e^{-z}dy$, and $dZ=dz$. Each one-form is a left-invariant differential form, so their sum of squares is the left-invariant metricThis is the standard metric of the Sol Lie group. As a sign check, $[L_3,L_1]=-L_1$, $[L_3,L_2]=L_2$, while $d{\theta }^1={\theta }^3\wedge {\theta }^1$ and $d{\theta }^2=-{\theta }^3\wedge {\theta }^2$, in agreement with the Maurer-Cartan equation.

Hold the metric fixed when varying the gauge potential, put $F=dA$, and take variations with compact support or with appropriate fixed boundary data. The Hodge star operator then givesFor the Abelian Chern-Simons 5-form, the two occurrences of $F$ commute as even-degree differential forms. Since $dF=0$,The factor three comes from all three appearances of $A$, including those hidden in $F=dA$. The bulk variation is thereforeand the Abelian Chern-Simons five-form field equation is

Under an Abelian gauge transformation $A\mapsto A+d\lambda$, the gauge field strength stays fixed and the action changes byThus the local integrand is not gauge invariant, but its change is a boundary term by the Generalized Stokes theorem. With the stated boundary conditions the variational bulk equations are unchanged; they depend only on $F$ and the metric, so they are manifestly gauge invariant. On a spacetime with boundary, gauge transformations nonzero at the boundary require boundary conditions or boundary degrees of freedom; the bulk claim does not assert invariance of every possible boundary action.

For the metric variation, write the kinetic action in components,At fixed covariant components $A_{\mu }$ and $F_{\mu \nu }$,Substituting into the given definition of the stress-energy tensor yieldsThe Chern-Simons 5-form uses only wedge products of differential forms and orientation, with no metric or Hodge star operator. Its bulk metric derivative is zero, so the Chern-Simons term contributes no bulk stress-energy tensor. The displayed tensor is generally not traceless in five dimensions: $T^{\mu }{}_{\mu }=-F_{\rho \sigma }{F}^{\rho \sigma }/4$.

A smooth fiber bundle $\pi :E\to M$ with typical fiber $F$ is locally a product: every base point has a neighborhood $U$ and a local trivialization ${\pi }^{-1}(U)\cong U\times F$ commuting with projection to $U$. Its overlap maps act smoothly on $F$. A principal bundle with structure Lie group $G$ has a free right $G$-action, each fiber is one orbit, and the local trivializations are equivariant for right multiplication on $G$. A smooth section of a fiber bundle is a map $s:M\to E$ with $\pi \circ s={\mathrm{id}}_M$.

For a principal bundle, a global section of a fiber bundle gives the mapThe free transitive action on each fiber makes $\Phi$ bijective. In a local trivialization, write $s(x)=(x,a(x))$; then $\Phi (x,g)=(x,a(x)g)$ and its inverse is $(x,k)\mapsto (x,a(x{)}^{-1}k)$. Both are smooth and $\Phi$ is equivariant. Conversely, a product principal bundle has the section $x\mapsto (x,e)$. This proves a principal bundle is trivial exactly when it has a global section. The argument depends on the free transitive action: an arbitrary bundle section need not trivialize its bundle.

For an $n$-dimensional smooth manifold, the frame bundle has fiber over $x$ consisting of ordered bases of $T_{x}M$, equivalently linear isomorphisms $u:{\mathbb{R}}^n\to T_{x}M$. Right composition by $GL(n,\mathbb{R})$ makes it a principal bundle. If a metric has signature $(p,q)$, its pseudo-orthonormal frame bundle consists of the isometries $u:{\mathbb{R}}^{p,q}\to T_{x}M$, with structure group $O(p,q)$. In the positive-definite case this is the usual orthonormal frame bundle. Choosing space and time orientations reduces the Lorentzian structure group to its corresponding connected subgroup.

For Minkowski spacetime, use a fixed origin and a standard Lorentzian frame. Every pair consisting of a point and a pseudo-orthonormal frame is the image of this reference pair under a unique affine isometry. Thus, for the full groups,Here the first expression denotes the total space of the orthonormal frame bundle, not merely the Lorentz group. Projection sends an affine isometry to the image of the reference point, and the right action of the stabilizer changes its frame.

Realize de Sitter spacetime of radius $\ell$ as ${\eta }_{AB}{X}^A{X}^B={\ell }^2$ in ${\mathbb{R}}^{1,4}$. The ambient orthogonal group is transitive on this hyperboloid. Its point stabilizer acts on the Lorentzian tangent space as $O(1,3)$; choosing a tangent orthonormal frame removes that stabilizer completely. ConsequentlyFor oriented, time-oriented frames, replace these groups by $S{O}_0(1,4)$ and $S{O}_0(1,3)$, and similarly use the proper orthochronous Poincaré group for Minkowski spacetime. The isometries are uniquely determined by their action on a point and its frame, which explains these bundle identifications.

Finally, choose a basis $v_1,\dots ,v_n$ of the Lie algebra $T_{e}G$. The left-invariant vector fields $X_i(g)=(d{L}_g{)}_e{v}_i$ form a basis at every $g$, because $L_g$ is a diffeomorphism. They give a global section of a fiber bundle of the frame bundle, so that bundle is trivial. More explicitly,Thus every Lie group is a parallelizable manifold; connectedness is not needed for this construction.

Use geometrized units $G=c=1$ until converting the clock readings. By the Birkhoff theorem, the spherical vacuum exterior is a portion of the Schwarzschild spacetime with fixed mass $M$:The exterior is not flat and is not radiating merely because the surface moves. When the surface crosses $r=2M$, this exterior must be represented in regular ingoing Eddington-Finkelstein coordinates, rather than treating the Schwarzschild-coordinate divergence as a physical barrier.

The Oppenheimer-Snyder model has a homogeneous pressureless matter interior. Since its initial expansion vanishes but its density is positive, the Friedmann equation requires positive spatial curvature. Choose a comoving radial angle $\chi$ and the boundary's proper time $\tau$, with the stopwatch set to zero at release. The interior FLRW metric and density areThe density law follows from the dust stress-energy tensor and its covariant conservation. The Friedmann equation and the initial rest condition giveIts contracting branch has the parameterizationIndeed $d\tau /d\eta =a$ and $\dot{a}=-\sin \eta /(1+\cos \eta )$ satisfy the displayed equation; $\eta =0$ is the turning point and $\eta =\pi$ is the final curvature singularity.

For homogeneous dust-ball matching, the boundary has areal radius $R=a\sin {\chi }_0$. Matching its mass to the exterior givesThis $M$ is the gravitational mass parameter, not the integral of rest density over the curved proper-volume element. The matching can also be checked directly. The outward angular extrinsic curvature inside is $K_{\theta \theta }=R\cos {\chi }_0$; outside it is $R\sqrt{1-2M/R+{\dot{R}}^2}$. The boundary motion hasso these curvatures agree. The induced metric is $-d{\tau }^2+R^{2}d{\Omega }^2$ on both sides and $K_{\tau \tau }=0$ because the pressure-free boundary is geodesic. No artificial surface stress layer is required.

The observer moving with this boundary meets the event horizon when $R=2M$. From $R=R_0(1+\cos \eta )/2$, the proper-time horizon crossing in homogeneous dust collapse occurs atThe singularity is reached at ${\tau }_s=\pi a_{\max }/2$, so the remaining clock time isThese are finite proper times, even though the external static time diverges at horizon crossing.

Convert the supplied rounded units consistently: $M=10^{38}$ Planck masses and $R_0=6.25\times 10^{39}$ Planck lengths. Thus $2M/R_0=0.032$, ${\chi }_0\simeq 0.17985$, and $a_{\max }\simeq 3.4939\times 10^{40}$ in Planck units. Multiplying geometric times by $5\times 10^{-44}$ seconds givesAt the accuracy justified by the rounded constants, these are about three milliseconds to horizon crossing and about seven microseconds more to the singularity. For $R_0\gg 2M$, expanding ${\eta }_H=\pi -2{\chi }_0$ gives $\Delta \tau \simeq (2/3)a_{\max }{\chi }_0^3\simeq 4M/3$, independently confirming the microsecond scale. The exact expression above retains the finite release-radius correction.

Strictly, this boundary clock reading is when the observer crosses the horizon, not when the horizon first exists anywhere. The event horizon inside an Oppenheimer-Snyder cloud is an outgoing radial null ray. Since $d\tau =ad\eta$, the interior metric is conformal to $-d{\eta }^2+d{\chi }^2+{\sin }^2\chi d{\Omega }^2$, and that ray has $d\chi /d\eta =1$. Tracing it back from $({\eta }_H,{\chi }_0)$ givesFor this cloud it starts at the centre after release, at comoving time ${\tau }_{\mathrm{birth}}\simeq 2.722\times 10^{-3}$ seconds. This answers the literal global-formation interpretation as well; it is not an event on the outer observer's worldline. The distinction matters because an event horizon is defined by the entire future causal escape problem, not by a locally detectable signal of formation.

For the observer at infinity, use the boundary's conserved Killing energy $E$ in the exterior:As $\tau \uparrow {\tau }_H$, $\dot{R}\to -E$ and $R-2M\simeq E({\tau }_H-\tau )$. Consequentlywhere ${\tau }_{\ast }>0$ fixes a harmless dimensionless logarithm. For actual outgoing light signals the relevant retarded time is $u=t-r_{\ast }(R)$. Since $r_{\ast }\simeq 2M\log (R-2M)+O(1)$,Therefore she never receives a signal showing the surface cross the horizon at any finite asymptotic time. Radiation emitted ever closer to crossing arrives ever later and is increasingly redshifted and diluted; photons emitted at or inside the event horizon cannot reach her. The formal crossing time at infinity is infinite, not the finite boundary stopwatch reading. Nor can the earlier global birth of the horizon be observed directly as a local flash from its central birth event.

Use Planck units for the following formulas. The black-hole area theorem is a classical statement with hypotheses. Assume the Einstein field equations, the null energy condition and appropriate global future regularity/predictability. For a clean version of the focusing sketch, take the future horizon generators to remain on a regular horizon and to be future complete in affine parameter. The usual strong asymptotic predictability formulation supplies the global causal control needed to exclude the same focusing pathology; an arbitrary spacetime with naked future breakdown is not covered by the theorem.

Let ${\ell }^a$ be an affinely parametrized generator of a smooth portion of the event horizon, and let $\theta =d\log dA/d\lambda$ be its null expansion. The horizon is a null hypersurface, so its generators have zero null twist. In four dimensions the Null Raychaudhuri equation isThe Einstein field equations give $R_{ab}{\ell }^a{\ell }^b=8\pi T_{ab}{\ell }^a{\ell }^b\ge 0$, since the metric terms vanish when contracted with a null vector. The screen-space null shear term is also nonnegative, so ${\theta }^{\prime }\le -{\theta }^2/2$. If ${\theta }_0<0$ at ${\lambda }_0$, integrating this inequality while the congruence remains regular givesIt focuses within affine distance at most $2/|{\theta }_0|$. The vanishing area element is a focal point to an earlier spacelike horizon cut. Past such a point the null geodesic can no longer generate an achronal boundary, but the event horizon is precisely such a boundary. Future completeness and the global hypotheses exclude its escaping this contradiction by ending prematurely at a pathology. Therefore $\theta \ge 0$ everywhere the smooth horizon description applies.