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Write $\epsilon =n_a{n}^a\in \{-1,+1\}$. The non-null hypersurface projection is $P^a{}_b={\delta }^a{}_b-\epsilon n^a{n}_b$, reproducing both signs in the question. It annihilates the normal and is the identity on tangent vectors to the hypersurface.

Represent $V$ by the tangent to a curve $\lambda$ through $p$. The curve $\varphi \circ \lambda$ lies in $\Sigma$, so its tangent $d{\varphi }_{p}V$ is tangent to $\Sigma$. Orthogonality therefore gives $n_a({\varphi }_{\ast }V{)}^a=0$, andThis uses the stated smooth hypersurface assumption; the dimension relation alone would not make the image of an arbitrary smooth map a regular hypersurface.

Write $h_{ij}={}^{(3)}{g}_{ij}$ for the positive spatial metric tensor and $h^{ij}$ for its inverse. Expanding the 3+1 decomposition gives $g_{00}=N^2-h_{ij}{N}^i{N}^j$, $g_{0i}=-h_{ij}{N}^j$ and $g_{ij}=-h_{ij}$. The inverse isFor example, $g^{00}{g}_{00}+g^{0i}{g}_{i0}=1$, and $g^{00}{g}_{0j}+g^{0i}{g}_{ij}=0$. The spatial block similarly gives $g^{i0}{g}_{0j}+g^{ik}{g}_{kj}={\delta }^i{}_j$. These checks determine all blocks without treating the spatial block alone as the inverse of the four-metric.

Raising the normal covector with this inverse metric tensor yieldsTake $N>0$ so it is future-pointing. Its norm is $n_{\mu }{n}^{\mu }=1$. The spatial projection tensor obeys $P^{\mu }{}_{\nu }{n}^{\nu }=0$, and multiplication givesThus it is an idempotent linear projection onto vectors tangent to the spatial hypersurface. The negative spatial components of the spacetime metric tensor do not change this idempotence.

For an affine variety, every coherent ideal sheaf is quasi-coherent; vanishing of quasi-coherent cohomology on an affine scheme therefore gives $H^1(X,\mathcal{I})=0$.

For the projective-space complement, assume $n\ge 1$ and choose two distinct closed points $P,Q\in X$. Take the ideal sheaf of two closed points $\mathcal{I}={\mathfrak{m}}_P\cap {\mathfrak{m}}_Q$ on $X$. The codimension-two extension of regular functions on a normal variety givesOne can see this directly: on every standard affine chart of ${\mathbb{P}}^n$, a rational function written in lowest terms cannot have a nonconstant denominator, because an irreducible polynomial factor of the denominator would define a pole along a codimension-one hypersurface, and such a hypersurface is not removed by $Z$. The extended function is constant because every global regular function on projective space is constant. Now the short exact sequencesends $k$ diagonally into $k^2$ on global sections. Its cokernel is $k$, and the long exact sequence in sheaf cohomology injects that cokernel into $H^1(X,\mathcal{I})$. ThusThe assumption $n\ge 1$ is necessary: ${\mathbb{P}}^0$ is already affine and has no such example.

Choose local coordinates $y^1,\dots ,y^{D-1}$ on the hypersurface $\Sigma$, and let ${\Phi }_s$ be the local flow of the vector field $X$. The map $F(s,y)={\Phi }_s(p(y))$ follows the integral curves of a vector field starting at $p(y)\in \Sigma$. At $s=0$, its differential maps the $s$ direction to $X$ and the remaining directions to a basis of $T_p\Sigma$. Transversality makes these vectors linearly independent, so the inverse function theorem makes $F$ a local manifold chart.

Set $x^0=s$ and transport the $y^i$ unchanged along each integral curve of a vector field. Then the adapted coordinates satisfyThe flow-box theorem gives this construction locally. The parameter is the flow parameter, and need not be arc length; a global manifold chart would require additional hypotheses.

Take the orientation induced by the future-directed unit normal and positive lapse function $\alpha$. In coordinates adapted to the 3+1 decomposition of spacetime, any component ${ϵ}_{\ell ijk}$ with all four indices spatial vanishes: only three distinct spatial indices are available. The shift vector therefore contributes nothing, andHere $[ijk]$ is the Levi-Civita symbol, and the last equality uses the metric determinant in a 3+1 decomposition, $|g|={\alpha }^2\gamma$. The right side is precisely the spatial metric volume tensor, with component $\sqrt{\gamma }$ at $ijk=123$, the appropriate permutation sign, and zero for repeated indices. HenceMore invariantly, the spatial volume form is the restriction to the hypersurface of the contraction of the spacetime volume form with $n$.

Use signature $(-,+,+,+)$, so the future-directed unit normal obeys $n^{\mu }{n}_{\mu }=-1$. The spatial projection tensor and the induced metric areIndeed $P^{\mu }{}_{\alpha }{n}^{\alpha }=0$ and $P^{\mu }{}_{\rho }{P}^{\rho }{}_{\alpha }=P^{\mu }{}_{\alpha }$: this is an idempotent linear map onto the tangent space of the spatial hypersurface. For a covariant tensor both slots must be projected:Contracting either slot with $n$ supplies the vanishing factor $P^{\mu }{}_{\alpha }{n}^{\alpha }$. Thus both normal contractions vanish, without requiring symmetry of $T$.

Locally write a smooth hypersurface as $\Phi =0$ with $n_a={\nabla }_a\Phi \ne 0$. It is a null hypersurface when $n_a{n}^a=0$ on it. Then $n^a$ is both normal and tangent, and the induced metric is degenerate along $n^a$.

Because covariant derivatives commute on a scalar,The scalar $n^2$ vanishes on the hypersurface, so its derivative annihilates every tangent direction there. Its derivative is consequently proportional to $n_a$: locally $n^2=\Phi q$ for a smooth function $q$, givingThus the normal generates null pregeodesics. Rescale $k^a=\alpha n^a$, choosing $n^b{\nabla }_b\log \alpha =-\kappa$, to get $k^b{\nabla }_b{k}^a=0$. These are the affinely parametrized generators of the null hypersurface. The proportionality need not vanish: setting $n^2=0$ only on the hypersurface does not set its full transverse derivative to zero.

The general-order Cauchy-Kovalevskaya theorem applies to analytic data on a non-characteristic hypersurface. In local coordinates $(t,z)$, a real analytic system of order $m$ in an unknown vector $U$ is in normal form whenwith $F$ real analytic near the initial jet. Analytic data ${\partial }_t^{j}U(0,z)=h_j(z)$ for $0\le j<m$ determine a unique real analytic solution near each initial point. More general systems, including different orders for the components, have the same conclusion after solving for the highest normal derivatives: the corresponding coefficient matrix, or highest-derivative Jacobian for a nonlinear system, must be invertible. This is the system's non-characteristic hypersurface condition. The uniqueness asserted by the Cauchy-Kovalevskaya theorem is initially uniqueness in the real analytic class.

A real analytic coordinate change flattens a real analytic initial hypersurface. Non-characteristicity permits solving for its highest normal derivatives by the real analytic implicit function theorem. Introduce all derivatives through order $m-1$ as additional unknowns. Their normal derivatives are either another jet variable, a tangential first derivative of a jet variable, or the right-hand side of the original equation. This gives a first-order real analytic system. The derivative-compatibility identities have zero initial data and are preserved by the system; equivalently, its real analytic coefficient recurrence reproduces the derivatives of $U$. This explains both reductions, rather than assuming arbitrary real analytic hypersurfaces are already flat.

For the first example, the principal symbol of the Laplace equation at the conormal $d{x}_1$ is $1$, so the plane is a non-characteristic hypersurface. The theorem therefore applies to real analytic $u{|}_{x_1=0}$ and ${\partial }_{1}u{|}_{x_1=0}$:This does not give Hadamard well-posedness in smooth or Sobolev space norms. For $n\ge 2$, the harmonic functionshave zero value data and normal derivative data $e^{-\sqrt{N}}\cos (N{x}_2)$ tending to zero with every fixed tangential derivative. At any fixed $x_1=\delta >0$, however, $u_N(\delta ,0)$ diverges. Thus the elliptic Cauchy problem for a partial differential equation is unstable despite real analytic solvability. The exceptional case $n=1$ is the elementary affine ordinary differential equation and has no such tangential high-frequency instability.

Use $G=c=1$, the Einstein field equations with zero cosmological constant, and metric signature $(-,+,+,+)$. Write $m=m_{a}d{x}^a$ for the differential one-form dual to the axial Killing vector field. To fix the sign of the volume formula, use the component Hodge star operator convention $(\star j{)}_{abc}={ϵ}_{abcd}{j}^d$ and $(\star F{)}_{ab}=\frac{1}{2}{ϵ}_{abcd}{F}^{cd}$. The Killing equation gives $(dm{)}_{ab}=2{\nabla }_a{m}_b$, and its contracted curvature identity givesTo see the curvature step, tracing the Killing equation gives ${\nabla }_b{m}^b=0$. The second covariant derivative of a Killing vector then gives ${\nabla }^b{\nabla }_b{m}_a=-R_{ab}{m}^b$, while commuting the covariant derivatives gives ${\nabla }^b{\nabla }_a{m}_b=R_{ab}{m}^b$. Subtracting these two terms is precisely ${\nabla }^b(dm{)}_{ab}$ above. The component Hodge star operator converts this divergence to $d\star dm$ with the displayed minus sign.

In a vacuum spacetime region $R_{ab}=0$, so $d\star dm=0$. If $S_1$ and $S_2$ are enclosing spacelike submanifolds in the same homology class bounding a vacuum three-dimensional region $W$, Stokes theorem yieldsThus the Komar angular momentum is independent of the enclosing vacuum spacelike two-manifold, provided the surfaces have the same orientation and enclose the same sources and inner boundaries. The vacuum region need not be stationary: an axial Killing vector field suffices.

For a regular filling hypersurface $\Sigma$ with $\partial \Sigma =S$, the Einstein field equations giveThe pullback of $\star m$ to $\Sigma$ vanishes because $m^a$ is tangent to $\Sigma$: the dual three-form measures the normal component, which is zero. ThereforeThis is the stress-energy current from a Killing vector integrated over the slice, with the signs fixed by the stated Hodge star operator convention.

There is a necessary boundary qualification omitted from the printed formulation. If the slice has inner boundaries $S_i$, orient them so $\partial \Sigma =S-{\bigcup }_i{S}_i$. The actual Komar angular momentum with inner boundaries identity isFor a vacuum Kerr black hole, $T_{ab}=0$ on the exterior slice but $J\ne 0$; its horizon supplies precisely the inner boundary contribution. Thus the matter-only formula is false for arbitrary exterior slices. It is valid when a nonsingular filling with no inner boundaries exists, or when all omitted inner charges vanish. The Komar angular momentum independence likewise concerns homologous surfaces in the same vacuum region, not an unrestricted comparison of differently enclosed objects.

The relevant Penrose singularity theorem states: a connected time-oriented four-dimensional globally hyperbolic spacetime with a noncompact Cauchy hypersurface, a nonempty compact orientable boundaryless trapped surface, and the null convergence condition is future null-geodesically incomplete. Under the Einstein field equations, the null energy condition supplies the curvature hypothesis. A trapped surface here is spacelike and has both future normal null expansions strictly negative. The conclusion is an incomplete null geodesic, not necessarily a divergent curvature invariant at an identifiable point.

Suppose, for contradiction, that every future null geodesic is complete. Write $S$ for the trapped surface andThe equality follows because the causal future of a compact set is closed in a globally hyperbolic spacetime. This future horismos is a closed achronal boundary. It is nonempty: the restriction of a Cauchy time function to compact $S$ has a minimum, and a point at that minimum cannot be chronologically preceded by another point of $S$.

Normalize the two future null normal directions along $S$ against a smooth future timelike field, fixing the affine scale continuously. Their initial null expansions are continuous and strictly negative. Compactness of the normalized normal bundle gives a uniform $c>0$ with ${\theta }_{\pm }\le -c$ everywhere on $S$. The preceding null focusing theorem forces a conjugate point to a spacelike surface along each normal generator within affine length $L=2/c$.

By the supplied boundary-generator result, every point of $E^{+}(S)$ is reached by an orthogonal future null geodesic which has no earlier conjugate point. No such boundary generator can remain on the boundary beyond its first focal point. Every point of $E^{+}(S)$ therefore lies inThe parameter set is compact. Future null completeness makes its geodesic flow defined throughout this common finite interval, and smooth dependence on initial data makes its image $K$ compact. Since $E^{+}(S)$ is closed and contained in $K$, it is compact. This is the compactness of the future horismos of a trapped surface step; a merely pointwise finite bound would not suffice without compactness and uniform normalization.

Choose a smooth complete timelike vector field, obtained if necessary by positive rescaling against a complete auxiliary Riemannian metric. Its inextendible integral curves of a vector field meet a chosen smooth Cauchy hypersurface $C$ exactly once. Projection along those curves defines a continuous map ${\pi }_C:E^{+}(S)\to C$. The achronal set property makes this map an injective function: two boundary points on the same timelike integral curve of a vector field would be timelike related.

An achronal boundary is a topological hypersurface without boundary, even at nonsmooth generator junctions; the submanifold property allowed in the question supplies this fact. Thus domain and codomain are both three-dimensional topological manifolds without boundary. Invariance of domain makes ${\pi }_C(E^{+}(S))$ open in $C$. Compactness makes it closed in the Hausdorff space $C$, and it is nonempty. Connectedness of the spacetime gives connectedness of $C$, so this image is all of $C$. That would make $C$ compact, contradicting the noncompact Cauchy hypersurface hypothesis. The completeness assumption is false, proving the theorem.

The noncompactness hypothesis and strict trapping are essential to this version. Global hyperbolicity cannot simply be omitted in the compactness and projection steps, and nonpositive initial expansion with zeros does not supply the uniform focusing bound used here.

Locally write the real analytic hypersurface as $\Sigma =\{F=0\}$ with $dF\ne 0$. The conormal bundle is spanned by $dF$. The surface is a characteristic hypersurface at $p$ precisely when the principal symbol vanishes on that conormal:Only the matrix $(a_{ij}+a_{ji})/2$, which is a symmetric matrix, contributes to this expression.

Take $n^a{n}_a=-1$. The first fundamental form, or induced metric, is the spatial projection tensorWith the convention used in the question, define the second fundamental form byLocally $n_a$ is proportional to the gradient of a defining function for the hypersurface, so $h_a{}^c{h}_b{}^d{\nabla }_{[c}{n}_{d]}=0$ and $K_{ab}=K_{ba}$. Both derivative indices are projected tangentially. If two extensions of $n$ agree on $\Sigma$, their difference vanishes there, as does its tangential derivative, so $K_{ab}$ is independent of the extension.

Only the second-order terms contribute to the principal symbol. For a covector $\xi =({\xi }_t,{\xi }_r,{\xi }_z)$ it isThe conormal to the hypersurface $\Sigma =\{\varphi =0\}$ is $d\varphi =({\varphi }_t,{\varphi }_r,{\varphi }_z)$. After evaluating the coefficient at the prescribed boundary value of $u_z$, the non-characteristic condition is therefore

A pencil is a one-parameter linear family of geometric objects. For linearly independent homogeneous polynomials $F,G$ of equal degree, $sF+tG=0$, indexed by $[s:t]\in {\mathbb{P}}^1$, is a pencil of plane curves or hypersurfaces. Lines through a fixed point give a simple example; a pencil of plane curves specializes the general construction.

A degree-$d$ projective hypersurface $X_d\subseteq {\mathbb{P}}_k^n$ is the closed subscheme cut out by one nonzero homogeneous polynomial of degree $d$.

This is a homogeneous-polynomial hypersurface in projective space.

The tangential component is the orthogonal projection of a vector onto the tangent space. For a plane curve, it is $(w\cdot T)T$; for a hypersurface, it is $w-(w\cdot n)n$. The scalar tangential component of acceleration is $\dot{v}$, which can be positive, zero or negative as the speed increases, remains constant or decreases.