Let ${\eta }_{ab}=g(e_a,e_b)$ be the constant diagonal metric matrix in a local pseudo-orthonormal frame, and define ${\theta }_{ab}={\eta }_{ac}{\theta }^c{}_b$. The metric compatibility of the Levi-Civita connection givesThus the lowered connection matrix is antisymmetric, including in indefinite signature. Ordinary matrix antisymmetry of ${\theta }^a{}_b$ itself is not the corresponding assertion in indefinite signature.

Let ${\omega }^a$ be the dual coframe, and write a vector field as $Y=e_b{\omega }^b(Y)$. ThenSubtract the same expression with $X,Y$ exchanged and subtract ${\omega }^a([X,Y])$. By the definition of the exterior derivative, this gives the components of the torsion tensor:The Levi-Civita connection is torsion-free, so Cartan's first structure equation is

Use $R(X,Y)={\nabla }_X{\nabla }_Y-{\nabla }_Y{\nabla }_X-{\nabla }_{[X,Y]}$ and define its matrix of curvature two-forms by $R(X,Y)e_b=e_a{R}^a{}_b(X,Y)$. Applying two covariant derivatives to $e_b$ yieldsThe first three terms are $d{\theta }^a{}_b(X,Y)$ and the last two are $({\theta }^a{}_c\wedge {\theta }^c{}_b)(X,Y)$. Therefore Cartan's second structure equation isThe associated Riemann curvature tensor components satisfy $R^a{}_b=\frac{1}{2}{R}^a{}_{bcd}{\omega }^c\wedge {\omega }^d$ with this convention.

For a position-dependent change of pseudo-orthonormal frame, write the frame row as $E^{\prime }=E\Lambda$, with ${\Lambda }^T\eta \Lambda =\eta$. The dual coframe obeys ${\omega }^{\prime }={\Lambda }^{-1}\omega$. Applying the Leibniz rule to $\nabla (E\Lambda )$ givesThe connection one-form has an inhomogeneous term because differentiating the moving basis produces $d\Lambda$. The curvature two-form instead transforms homogeneously. Indeed, curvature is tensorial in the vector to which it is applied, so $R(X,Y)(E\Lambda )=(R(X,Y)E)\Lambda =E^{\prime }{\Lambda }^{-1}R(X,Y)\Lambda$. ConsequentlyEquivalently, substituting the transformed connection into $d{\theta }^{\prime }+{\theta }^{\prime }\wedge {\theta }^{\prime }$ and using $d{\Lambda }^{-1}=-{\Lambda }^{-1}(d\Lambda ){\Lambda }^{-1}$ cancels all derivative terms and gives the same result. This transformation law holds for a general invertible frame change as well; the orthogonality condition ensures both frames remain pseudo-orthonormal.

One definition of a principal bundle $P\to M$ uses local trivializations ${\varphi }_i:{\pi }^{-1}(U_i)\to U_i\times G$ whose overlap maps arewith smooth $G$-valued transition functions of a principal bundle satisfying the cocycle condition. This definition displays the global right group action rather than assuming it: if ${\varphi }_i(p)=(x,g_i)$, define ${\varphi }_i(p\cdot h)=(x,g_{i}h)$. On an overlap,so the local definitions agree. The resulting action is smooth, free and transitive on each fibre, and respects the projection.

If $s:M\to P$ is a global section of a fiber bundle, set $F(x,h)=s(x)\cdot h$. The free and transitive fibre action makes $F:M\times G\to P$ bijective. In a local trivialization, write $s(x)=(x,a_i(x))$; then $F(x,h)=(x,a_i(x)h)$ and its inverse is $(x,g)\mapsto (x,a_i(x{)}^{-1}g)$. Both are smooth and respect the action and projection. Conversely a principal bundle isomorphism $M\times G\to P$ sends $x\mapsto (x,e)$ to a global section. Hence

For a pseudo-Riemannian manifold of metric signature $(p,q)$, its orthonormal frame bundle has fibre at $x$ consisting of linear isometries $u:{\mathbb{R}}^{p,q}\to T_{x}M$. Its structure group is $O(p,q)$ and its right action is $u\cdot h=u\circ h$. Two such frames differ by a unique element of $O(p,q)$, proving the required free and transitive fibre action. A global section is exactly a global pseudo-orthonormal frame. Orientation and time orientation, when specified, restrict the structure group and the allowed frames.

Realize unit-radius four-dimensional de Sitter spacetime asIts tangent space is $X^{\perp }$, with Lorentzian metric signature $(3,1)$. If $e_0,e_1,e_2,e_3$ is a pseudo-orthonormal frame at $X$, append $X$ as the final column. The resulting five-by-five matrix preserves $\eta$, and conversely every such matrix supplies a point $X$ and a frame at that point. Thus the full frame bundle is $O(4,1)$, with projection $g\mapsto g{e}_4$ and fibre group $O(3,1)$.

The printed $SO(4,1)$ statement requires the orientation restriction. Choose the tangent orientation by requiring the appended matrix to have determinant one. Thenis precisely the oriented frame bundle of de Sitter spacetime, with right structure group $SO(3,1)$ embedded as $\mathrm{diag}(h,1)$. If future time orientation is also required, use $S{O}_0(4,1)$ and $S{O}_0(3,1)$. Calling $SO(4,1)$ the full bundle without specifying orientation would exclude valid frames of the opposite orientation.

Finally write the global coordinatesDirect differentiation gives the de Sitter spacetime metric $-d{\tau }^2+{\cosh }^2\tau g_{S^3}$. The assumed global frame of a vector bundle on $S^3$ can be orthonormalized smoothly by Gram-Schmidt orthogonalization for the round positive metric. More explicitly, view $S^3$ as the unit quaternions: $Y_1(q)=qi$, $Y_2(q)=qj$, $Y_3(q)=qk$ form a smooth global orthonormal tangent basis. SetThese have inner products $g(E_0,E_0)=-1$, $g(E_i,E_j)={\delta }_{ij}$ and $g(E_0,E_i)=0$. They are smooth everywhere because $\cosh \tau$ never vanishes. Choose the spatial orientation consistently, reversing one $Y_i$ if necessary, and choose $E_0$ future-directed. This global frame on four-dimensional de Sitter spacetime yields a global section of the full frame bundle and also of either restricted bundle. Yes: the de Sitter frame bundle admits a global section and is trivial.

Use ambient quadratic form $Q=-(X^0{)}^2-(X^4{)}^2+(X^1{)}^2+(X^2{)}^2$. The printed sine coordinate has a duplicated superscript: it must be $X^2$, not another $X^1$. With that correction the embedding lies on $Q=-1$. Put $s=\sqrt{1+r^2}$. Differentiating the two timelike coordinates giveswhereas the two spatial coordinates give $(d{X}^1{)}^2+(d{X}^2{)}^2=d{r}^2+r^{2}d{\theta }^2$. Their difference is the induced Lorentzian metricThe polar coordinate $\theta$ collapses at $r=0$; this is a coordinate degeneracy, not a degeneracy of the metric.

For the matrix model of anti-de Sitter three-space, defineThen $\det M=-Q$. This linear identification of ambient space with real two-by-two matrices identifies $Q=-1$ with $SL(2,\mathbb{R})$. Its inverse recovers each $X$ by sums and differences of matrix entries, so the identification is global. The identity corresponds to $X^4=1$ and the other coordinates zero. For a traceless tangent matrix $U$, the induced squared norm isPolarization gives $g_I(U,V)=\frac{1}{2}\mathrm{tr}(UV)$.

The Killing form of ${\mathfrak{sl}}_2(\mathbb{R})$ is $B(U,V)=4\mathrm{tr}(UV)$. This normalization can be checked without assuming it: for $H=\mathrm{diag}(1,-1)$, $E=\left(\begin{array}{cc}0& 1\\ 0& 0\end{array}\right)$ and $F=\left(\begin{array}{cc}0& 0\\ 1& 0\end{array}\right)$, the brackets are $[H,E]=2E$, $[H,F]=-2F$, $[E,F]=H$. Taking traces of their three-dimensional adjoint matrices gives $B(H,H)=8$, $B(E,F)=B(F,E)=4$, and all other basis pairings zero. These agree with $4\mathrm{tr}(UV)$.

Both left and right multiplication by matrices in $SL(2,\mathbb{R})$ preserve $-\det M$, hence its polarized ambient bilinear form and the induced metric. Equality at the identity therefore propagates throughout the group:In particular the Killing-form Einstein metric calculation gives $\mathrm{Ric}=-2g$, consistent with unit-radius three-dimensional Anti-de Sitter spacetime.

Every element of $SO(2,2)$ preserves the ambient quadratic form, hence the hyperboloid and its induced metric, and so acts by isometries. Define the left-right double cover of SO0(2,2) byThis is a group homomorphism, since successive actions multiply the two factors in the same order. It preserves $\det M$; as a transformation of the four-dimensional matrix space it has determinant $(\det A{)}^2(\det B^{-1}{)}^2=1$. Thus it takes values in $SO(2,2)$.

If its action is the identity, applying it to $M=I$ gives $A=B$. Commuting with every matrix then forces $A$ to be scalar. The condition $\det A=1$ leaves only $A=I$ or $A=-I$. HenceThe differential is $(a,b):M\mapsto aM-Mb$. A zero differential implies $a=b$ and that $a$ is scalar; tracelessness makes $a=b=0$. It is therefore an isomorphism onto the target Lie algebra, since both dimensions are six. The image is an open subgroup. Since $SL(2,\mathbb{R})$ is connected, the image is connected and is exactly the identity component $S{O}_0(2,2)$: an open subgroup of a connected group is also closed and must be the whole group. Thus the precise covering statement isIt is not onto the disconnected full $SO(2,2)$. For example, conjugation $M\mapsto DM{D}^{-1}$, $D=\mathrm{diag}(1,-1)$, sends $(X^0,X^4,X^1,X^2)$ to $(-X^0,X^4,X^1,-X^2)$ and belongs to the other component of $SO(2,2)$. It reverses the orientation of the negative two-plane and is not in the image of the connected source.

The Adjoint representation of a Lie group of $SL(2,\mathbb{R})$ preserves its Killing form of metric signature $(2,1)$ and gives a second double cover,Its kernel is the scalar centre; its differential is injective and both dimensions are three, so the same connected-image argument proves surjectivity onto the identity component. The full $SO(2,1)$ again has an additional component. Thus $S{O}_0(2,1)$ is locally the same Lie group geometry as $SL(2,\mathbb{R})$, with its Killing metric, and globally is its central quotient.

With the printed periodic time, the group model has topology $S^1\times {\mathbb{R}}^2$ and closed timelike curves, since varying $t$ at fixed spatial position gives a timelike circle. The causally unwrapped version of Anti-de Sitter spacetime instead has $t\in \mathbb{R}$ and is the universal covering Lie group. It must not be substituted silently for the periodic spacetime specified here.

A Poisson manifold is a smooth manifold with an antisymmetric contravariant tensor field $\pi =\frac{1}{2}{\pi }^{ij}{\partial }_i\wedge {\partial }_j$ such that the bracketsatisfies the Jacobi identity. Bilinearity, antisymmetry and the Leibniz rule follow directly from this formula. The Poisson bivector is allowed to be degenerate.

Apply the Jacobi identity to coordinate functions $x^i,x^j,x^k$. Necessity gives the coordinate Jacobi condition for a Poisson bivector:For sufficiency, expand $\{f,\{g,h\}\}+\{g,\{h,f\}\}+\{h,\{f,g\}\}$. The terms differentiating $\pi$ are $J^{ijk}{f}_i{g}_j{h}_k$. All terms containing second derivatives cancel. For example, those containing $g_{j\ell }$ have coefficient $({\pi }^{i\ell }{\pi }^{jk}+{\pi }^{k\ell }{\pi }^{ij})f_i{h}_k$, which is antisymmetric under $j\leftrightarrow \ell$ after contraction with $f_i{h}_k$, whereas $g_{j\ell }$ is symmetric. The same argument handles the second derivatives of $f$ and $h$. Thus the displayed condition is both necessary and sufficient.

A symplectic manifold has a smooth nondegenerate closed differential form $\Omega$ of degree two. Write $\Omega =\frac{1}{2}{\Omega }_{ij}d{x}^i\wedge d{x}^j$ and choose the explicit inverse convention ${\pi }^{ik}{\Omega }_{kj}={\delta }_j^i$. This defines a smooth antisymmetric bivector. Differentiating the inverse matrix givesSubstitution into the Jacobi coefficient, followed by relabeling summed indices, givessince the parentheses are the components of $d\Omega$. Hence every symplectic manifold carries a Poisson structure. The opposite overall convention for the inverse bracket gives the same conclusion. Nondegeneracy and closure together are crucial; an arbitrary invertible antisymmetric tensor need not satisfy the Jacobi identity.

For a finite-dimensional Lie algebra $\mathfrak{g}$, its dual space ${\mathfrak{g}}^{\ast }$ carries the Lie-Poisson bracketwhere $d{f}_{\ell },d{g}_{\ell }$ are identified with elements of $\mathfrak{g}$. If $[e_i,e_j]=c_{ij}{}^k{e}_k$ and $x_i(\ell )=\ell (e_i)$, this reads $\{x_i,x_j\}=c_{ij}{}^k{x}_k$. The Jacobi coefficient iswhich vanishes by the Lie algebra Jacobi identity.

In particular, the dual of $\mathfrak{so}(3)$ is ${\mathbb{R}}^3$ with the rotational Lie-Poisson structure on R3,The Poisson bivector has rank two away from the origin and rank zero at the origin. The function $x^2+y^2+z^2$ is a Casimir function, since its bracket with every function vanishes. The centred spheres are its two-dimensional symplectic leaves, and the origin is a point leaf. This is a Poisson manifold but not a symplectic manifold: its bivector is not invertible, and an odd-dimensional manifold cannot have a nondegenerate alternating bilinear form.

For a group action of $G$ on $X$, the orbit of a group action through $x$ is $Gx=\{gx:g\in G\}$, and its stabilizer subgroup is $H_x=\{g:gx=x\}$. A transitive group action has one orbit. A simply transitive group action has exactly one element carrying any given point to any other; equivalently it is transitive and every stabilizer subgroup is trivial.

In the modern terminology, a multiply transitive group action is transitive on ordered $k$-tuples of distinct points for some $k\ge 2$. In older geometric usage, the phrase can instead mean transitive but not simply transitive, so that there are several elements carrying one point to another. Under either interpretation the action is transitive, and that alone is what the following argument needs. These meanings should not be confused: a transitive action with nontrivial stabilizers need not be two-transitive.

Fix $x_0\in X$ and put $H=H_{x_0}$. DefineThis is well-defined because every element of $H$ fixes $x_0$. It is surjective by transitivity, and $g{x}_0=g^{\prime }{x}_0$ is equivalent to $g^{\prime -1}g\in H$, proving injectivity. For a smooth Lie group action, $H$ is closed, since it is the inverse image of $x_0$ under the orbit map. The orbit map has constant rank: translations in $G$ and the action diffeomorphisms of $X$ intertwine its differentials. Transitivity makes this rank $\dim X$; otherwise every point would be a critical value, contradicting Sard theorem. Its local submersion charts give smooth local sections and hence a smooth inverse for $F$. ThereforeChoosing $x_1=a{x}_0$ replaces $H$ by $aH{a}^{-1}$. Thus the stabilizer subgroup is determined by the base point, and intrinsically only up to conjugacy. A simply transitive group action gives $H=\{e\}$.

For a finite quantum system with Hilbert space $\mathcal{H}={\mathbb{C}}^{n+1}$, a pure state is a nonzero vector modulo multiplication by a nonzero complex number. Equivalently, it is a unit vector modulo its physically irrelevant phase. Consequently the space of pure states is the space of complex lines,The qualification “pure” is essential: arbitrary quantum states include mixed density matrices and do not form Complex projective space.

The unitary group $U(n+1)$ acts transitively on these lines: choose unit vectors in two given lines, complete each to an orthonormal basis, and map one basis to the other by a unitary transformation. The stabilizer of the line $\mathbb{C}{e}_0$ consists exactly of block matrices $\mathrm{diag}(z,V)$ with $z\in U(1)$ and $V\in U(n)$. Indeed a unitary transformation preserving the line also preserves its orthogonal complement. The homogeneous space description is thereforeIts real dimension is $(n+1{)}^2-1-n^2=2n$. This action is generally not two-transitive: the absolute inner product between unit representatives of two lines is invariant.

The Hopf fibration is the projection $\pi :S^{2n+1}\to {\mathbb{CP}}^n$, $z\mapsto [z]$. The right action $z\cdot \lambda =z\lambda$, $\lambda \in S^1$, is free, and its orbits are exactly the fibres. On the chart $U_i$ where $z_i\ne 0$, write $w_j=z_j/z_i$ for $j\ne i$. A smooth local section isEvery unit vector over this chart has the unique expression $z={\sigma }_i([z])\lambda$, where $\lambda =z_i/|z_i|$. These formulas provide local trivializations ${\pi }^{-1}(U_i)\cong U_i\times S^1$, proving that this is a principal bundle with fibre $S^1$. For $n\ge 1$ it has no global section: a global section would give $S^{2n+1}\cong {\mathbb{CP}}^n\times S^1$, contradicting their fundamental groups. For $n=0$ the base is a point and the bundle is trivial.

For the Lie algebra $\mathfrak{g}$ of a Lie group $G$, let ${\mathrm{ad}}_X(Y)=[X,Y]$. The Killing form is the symmetric bilinear formFor a finite-dimensional real Lie algebra, the Cartan criterion for semisimplicity says that $B$ is nondegenerate precisely when $\mathfrak{g}$ is a semisimple Lie algebra. Thus a semisimple Lie group supplies the required condition. Compactness alone is insufficient: an abelian Lie algebra, including that of a torus, has zero Killing form.

The Jacobi identity gives ${\mathrm{ad}}_{[Z,X]}=[{\mathrm{ad}}_Z,{\mathrm{ad}}_X]$. Cyclic invariance of the trace therefore givesThere is also invariance under the full Adjoint representation of a Lie group: ${\mathrm{ad}}_{{\mathrm{Ad}}_{a}X}={\mathrm{Ad}}_a{\mathrm{ad}}_X{\mathrm{Ad}}_{a^{-1}}$, so $B({\mathrm{Ad}}_{a}X,{\mathrm{Ad}}_{a}Y)=B(X,Y)$.

Define a pseudo-Riemannian metric by translating $B$ from the identity:It is smooth and nondegenerate. Left translations preserve it by construction. Right translation, expressed in left-translated tangent coordinates, acts by ${\mathrm{Ad}}_{b^{-1}}$, so it also preserves the metric. This constructs a bi-invariant pseudo-Riemannian metric.

For left-invariant vector fields, the derivative terms in the Koszul formula vanish. Invariance of $B$ then givesUse the Riemann curvature tensor convention $R(X,Y)Z={\nabla }_X{\nabla }_{Y}Z-{\nabla }_Y{\nabla }_{X}Z-{\nabla }_{[X,Y]}Z$. Substituting the Levi-Civita connection and using the Jacobi identity yieldsThe Ricci tensor is the trace of $X\mapsto R(X,Y)Z$. Since $[[X,Y],Z]={\mathrm{ad}}_Z{\mathrm{ad}}_{Y}X$, it follows thatThis is the Killing-form Einstein metric. Rescaling to $g=cB$, with nonzero constant $c$, gives Einstein metric constant $-1/(4c)$.

For a compact semisimple Lie group, $B$ is negative definite and $-B$ is a positive-definite bi-invariant Riemannian metric. For example, $SU(2)$ has topology $S^3$ and metric signature $(0,3)$ for $B$, while $SO(3)$ has the same Lie algebra and metric signature, but topology ${\mathbb{RP}}^3$. The special linear group $SL(2,\mathbb{R})$ has metric signature $(2,1)$ and topology $S^1\times {\mathbb{R}}^2$; its universal covering Lie group has topology ${\mathbb{R}}^3$ and the same Killing form. The real Lie algebra of $SL(2,\mathbb{C})$ has metric signature $(3,3)$, and the group has topology $S^3\times {\mathbb{R}}^3$. Thus the Killing form distinguishes compact from noncompact directions in these semisimple examples, but does not determine global topology or distinguish different covering groups.

Keep the unit-normal sign explicit: $n^a{n}_a=\epsilon \in \{+1,-1\}$. Since the hypersurface is spacelike its normal is timelike. ConstructChoose the sign of n for the desired time orientation. The pullback of h is the nondegenerate induced metric. The second fundamental form isIts sign reverses when n reverses. It is tangential in both indices and symmetric: the antisymmetric part of the derivative of a normalized gradient vanishes after both tangential projections. Equivalently it is the projected Hessian of f divided by the normalizing factor. This is the normal-change definition of extrinsic curvature.

For tangent vector fields X,Y, let D be the induced Levi-Civita connection. Orthogonality to n gives the derivative decompositionUse the curvature convention of the preceding solution, $R(X,Y)Z=[{\nabla }_X,{\nabla }_Y]Z-{\nabla }_{[X,Y]}Z$. Substitute the decomposition twice and take its tangential component. Terms differentiated normally cancel, while the two normal components contribute the quadratic shape terms:Lowering the first curvature index gives the Gauss equation for a nonnull hypersurfaceAll four displayed indices on the left are tangential.

Contract with $h^{ac}{h}^{bd}$ and use $h^{ab}=g^{ab}-\epsilon n^a{n}^b$ for tangential contractions. The double-normal term vanishes by curvature antisymmetry and each mixed contraction is ambient Ricci curvature. With $K=h^{ab}{K}_{ab}$ this givesThis also records which metric and curvature conventions enter the sign.

In vacuum without a cosmological constant, the ambient Ricci tensor and scalar vanish. A totally umbilic hypersurface has $K_{ab}=\lambda h_{ab}$, so $K=3\lambda$ and $K_{ab}{K}^{ab}=3{\lambda }^2$. Therefore the scalar curvature of an umbilic vacuum hypersurface is $6\epsilon {\lambda }^2$.

To express the requested nonnegative version, take Lorentz signature $(+---)$, so the timelike unit normal has $\epsilon =+1$ and h is the induced negative-definite spatial metric. ThenIf instead one uses signature $(-+++)$ and its positive-definite spatial metric with the same curvature-slot convention, the scalar is $-6{\lambda }^2$. Replacing h by $-h$ reverses its scalar curvature. Thus the printed inequality has a signature/induced-metric convention built into it; the general formula above makes that choice explicit rather than losing the sign. The metric in the separate final question explicitly uses the mostly-plus signature.

An orthonormal tetrad, or vierbein, is a local basis $e_I{}^a$ of tangent vectors together with inverse coframe $e_a{}^I$, satisfying $g_{ab}{e}_I{}^a{e}_J{}^b={\eta }_{IJ}$ and $g_{ab}=e_a{}^I{e}_b{}^J{\eta }_{IJ}$. Capital indices label the locally orthonormal Lorentz frame; lower-case indices label spacetime coordinates. Changing the frame by a local Lorentz transformation leaves the metric unchanged. A spinor description uses a compatible local spin frame and the torsion-free spin connection.

Let fixed flat matrices satisfy $\{{\Gamma }_I,{\Gamma }_J\}=2{\eta }_{IJ}1$. The curved spacetime gamma matrices are ${\gamma }_a=e_a{}^I{\Gamma }_I$ and ${\gamma }^a=e_I{}^a{\Gamma }^I$. ThereforeThe identity matrix is four by four for a four-dimensional Dirac spinor. The combined tensor/spin covariant derivative makes ${\nabla }_a{\gamma }_b=0$, the Clifford version of the tetrad postulate.

Fix the curvature slots by $R_{abcd}=g_{ae}{R}^e{}_{bcd}$, with $[{\nabla }_c,{\nabla }_d]v^a=R^a{}_{bcd}{v}^b$, and $R=g^{ac}{g}^{bd}{R}_{abcd}$. The torsion-free spin connection then has $[{\nabla }_a,{\nabla }_b]\psi =\frac{1}{4}{R}_{abcd}{\gamma }^c{\gamma }^d\psi$. These choices fix the signs in the requested identities.

Write ${\gamma }^{ab}={\gamma }^{[a}{\gamma }^{b]}$ and similarly for four indices. Clifford anticommutation givesOn contraction with the Riemann curvature tensor, the four-form term vanishes by the first Bianchi identity. The first and last two-form terms vanish by pair antisymmetry; the remaining two-form terms are symmetric Ricci contractions against antisymmetric gamma matrices and vanish too. The scalar terms are zero, minus R and minus R respectively. Thus the Riemann tensor Clifford contraction isThis algebraic result is independent of Lorentz signature when the contraction convention is held fixed.

Apply the Dirac operator twice, using covariant constancy of the gamma matrices. Define the rough wave operator by $\square =g^{ab}{\nabla }_a{\nabla }_b$, with the full connection acting on the intermediate covector-spinor derivative. Splitting the product into symmetric and antisymmetric parts givesConsequently the Lichnerowicz spinor-square formula and the assumed massless Dirac equation imply $\boxed{(\square -R/4)\psi =0}$. Torsion would modify the commutator and introduce extra terms, so the torsion-free connection is part of this argument.

A stationary black hole behaves mechanically like an equilibrium thermodynamic system. Its horizon area A, asymptotic energy E, angular momentum J and electric charge Q are linked by the laws of black-hole mechanics. Let ${\chi }^a=t^a+{\Omega }_H{\varphi }^a$ generate its Killing horizon, with the time translation normalized at infinity; its surface gravity is defined by ${\chi }^b{\nabla }_b{\chi }^a=\kappa {\chi }^a$ on that horizon. We use $c=1$ and keep G explicit. The horizon electrostatic potential ${\Phi }_H$ is measured relative to infinity.

The Zeroth law of black-hole mechanics states that kappa is constant on each connected stationary horizon, under the Einstein equations and suitable energy/regularity assumptions. It parallels the zeroth law of thermodynamics, which makes temperature uniform in equilibrium. Kappa need not be constant in a dynamical geometry. For an asymptotically flat static hole, local observers see gravitationally redshifted temperatures; the equilibrium quantity comparable to kappa is the temperature referenced to infinity.

The First law of black-hole mechanics relates nearby stationary solutions:The same leading relation describes a small physical process which perturbs a horizon and subsequently settles to equilibrium, with the appropriate hypotheses on fluxes and final state. The work terms correspond to rotational and electromagnetic work. The structural analogy with the first law of thermodynamics, $dE=TdS+\text{work}$, identifies area as an entropy-like variable and surface gravity as a temperature-like variable. Classically this fixes their product in the first law, rather than their separate normalizations.

The second law of black-hole mechanics, or black-hole area theorem, states that future event-horizon area does not decrease classically. Its essential geometric ingredient is the Raychaudhuri equation for affinely parametrized null generators. In four dimensions a hypersurface-orthogonal congruence obeysThe null energy condition and Einstein equations make the last term nonpositive. If the expansion became negative, the inequality would focus the generators to a conjugate point in finite affine parameter. Under future completeness and predictability assumptions, that would conflict with their remaining on an achronal event horizon. Thus $\vartheta \ge 0$, and the area element, satisfying $d\log dA/d\lambda =\vartheta$, cannot decrease. For a merger the final area is at least the sum of the initial component areas. These assumptions matter: the theorem is a classical statement, not a prohibition of quantum evaporation.

The third law of black-hole mechanics is an unattainability statement: a nonextremal hole cannot be driven to zero surface gravity in a finite physical process satisfying the usual bounded-stress and regularity conditions. This parallels the unattainability form of the third law of thermodynamics. Extremal stationary solutions exist, so the law concerns how an initially nonextremal hole approaches them. The stronger entropy form of Nernst's theorem does not transfer unchanged: extremal holes can have finite area and entropy, depending on their conserved charges.

Quantum field theory turns the mechanical analogy into a physical thermodynamic identification. Hawking radiation is thermal at infinity to leading semiclassical order, withThese give $T_H\delta S_{\mathrm{BH}}=\kappa \delta A/(8\pi G)$ exactly, so the mechanical first law becomes the thermodynamic first law with Bekenstein-Hawking entropy. Scattering through the exterior adds greybody factors to the emitted spectrum but leaves this temperature identification intact. Equilibrium with an external bath and isolated evaporation are different physical states of the field.

During black-hole evaporation, the area can decrease because the semiclassical stress tensor need not satisfy the classical null energy condition. The proposed generalized second law applies instead to $S_{\mathrm{gen}}=S_{\mathrm{BH}}+S_{\mathrm{outside}}$, asserting nondecrease of the combined entropy. A Schwarzschild hole illustrates an unusual thermodynamic property: $T_H\propto E^{-1}$, so its heat capacity is negative. It becomes hotter as it loses energy and is unstable in a simple canonical heat bath, while entropy and energy laws still describe an isolated system. The quantum temperature, area entropy and generalized second law supply the correspondence; the mechanical laws alone do not establish a thermal emission spectrum or a microscopic entropy count.

Assume standard radiation-dominated expansion, negligible chemical potential, one neutrino helicity and one antineutrino helicity, and a stable species $X$ with no later entropy injection. Use natural units and unreduced Planck mass $m_{\mathrm{pl}}\simeq 1.22\times 10^{19}\mathrm{GeV}$. The relativistic energy density is ${\rho }_R={\pi }^2{g}_{\ast }{T}^4/30$. The Friedmann equation then givesFor thermal decoupling in cosmology, equate the interaction rate $G_A^2{T}_D^5$ to $H(T_D)$:This order-of-magnitude estimate assumes the supplied rate law remains applicable there.

A decoupled relativistic species cools as $T_X\propto a^{-1}$, while annihilations heat the coupled bath. Treat the supplied high-temperature count $100$ as the remaining bath's entropy count, approximately equal to its energy count. Cosmological entropy conservation down to ordinary neutrino decoupling givesSubsequent electron-positron annihilation in cosmology heats photons but neither neutrino population. Therefore the two-stage entropy dilution of an early-decoupled neutrino givesThe ordinary Cosmic neutrino background has $T_{\nu }/T_{\gamma }=(4/11{)}^{1/3}$. If the rounded count $100$ instead includes the new two helicities, the remaining bath count is $98.25$; this changes the answer only slightly, to about $0.341$. Keeping the extra species' entropy separate is essential in either convention.

For the two relativistic fermion states, $g_{\ast s,X}=(7/8)2=7/4$. Photon and three standard neutrino species have present entropy degrees of freedom, referred to $T_{\gamma }$,Writing $\theta =T_X/T_{\gamma }$, the fractional entropy of a relativistically decoupled species isThus approximately 1.7% of the thermal particle entropy is in the new neutrinos and antineutrinos. If $100$ counts all states before the new decoupling, the fraction is $1.75\%$. These are the same rounded estimate. The question concerns the radiation/relic thermal entropy budget; it does not supply a model for entropy in later black holes or other gravitational structures.

If $X$ has a small mass but decoupled relativistically, its comoving particle number is still conserved. The relativistic-neutrino number-density ratio givesOnce nonrelativistic, ${\rho }_X\simeq m_X{n}_X$. The density bound on a relativistically decoupled massive relic is thereforeUsing the standard present photon abundance and critical-density normalization, this is approximately ${\Omega }_X{h}^2\simeq m_X/(9\times 10^2\mathrm{eV})$, where $h=H_0/(100\mathrm{km}{\mathrm{s}}^{-1}\mathrm{Mp}{\mathrm{c}}^{-1})$ is the present Hubble normalization, not a magnetic field. A stable relic must not exceed the permitted matter density; for example, if ${\Omega }_M{h}^2=0.15$, even assigning all matter to it gives only an order-$10^2\mathrm{eV}$ mass ceiling.