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Source: released examination PDF, bundle pages 92–95.

We use metric signature $(-,+,+,+)$ and set the speed of light to one. In question 2, $M$ and $Q$ have the geometric normalization of the Reissner-Nordstrom metric, with $M>0$ and $r>0$. Every numbered part is treated, including all three cases in question 2(b).

Since the real conformal factor $\Lambda$ is nowhere zero, ${\Lambda }^2>0$ at every point. For any tangent vector,The sign is unchanged. With metric signature $(-,+,+,+)$, negative, zero and positive values therefore describe the same timelike vectors, null vectors and spacelike vectors for both metric tensors. In particular the two null cones coincide.

Choose the same time orientation for the two Lorentzian metrics. A future-directed causal curve or timelike curve for one has the same character for the other. Hence the causal future, causal past and chronological future of every point agree:This is the preservation of causal structure by a smooth, nonvanishing conformal transformation. It does not assert preservation of proper time or of affine parameter lengths.

Complete solutions to the three questions of the Part C General relativity II course of the University of Oxford paper, sat on 1 June 2021. Original released PDF, bundle pages 63–65; paper pages 1–3. The bundle filename says 2020, but the paper's cover explicitly says Trinity Term 2021. The metric signature is $(-,+,+,+)$ and the speed of light is set to one. All questions and both Roman-numbered parts of question 1(d) are covered.

First use the original null coordinates $u=t+m/\lambda$, $v=t-m/\lambda$. They obey $u-v=2m/\lambda$, and the radial metric tensor becomesApply the given substitutions $u=\tan (U/2)$ and $v=-\cot (V/2)$. The identitiesgive the analytic extension of a spacetimeThe poles of the individual tangent and cotangent substitutions are now ordinary coordinate lines wherever $\cos ((U-V)/2)\ne 0$. Set $T=(U+V)/2$ and $X=(U-V)/2$. Allowing these coordinate functions to continue across every such line givesThis is the universal cover of two-dimensional anti-de Sitter spacetime of radius $m$, times a two-sphere of radius $m$: the simply connected Bertotti-Robinson spacetime. In defining this maximal analytic extension, $T$ ranges over the whole real line without periodic identification. A periodic identification would instead create closed timelike curves and would be an additional global choice.

For clarity, the original patch is embedded by the branches $-\pi <U<\pi$ and $0<V<2\pi$. The inequality $u>v$ is equivalent on these branches to $|U-V|<\pi$, so the original patch occupiesThe first spacetime extension in part (a) occupies the larger band $0<T-X<2\pi$, with $|X|<\pi /2$. Its finite-$v$ Killing horizon is the interior line $T+X=\pi$. The edges $T-X=0,2\pi$ of that band are further regular Poincaré horizons; continuing through them produces the full conformal strip of two-dimensional anti-de Sitter spacetime.

The original coordinate $\lambda$ has a smooth continuation throughout the strip:Consequently the complete set of zero-level Killing horizons isThese are two alternating families of parallel null lines; different families meet only at the excluded conformal boundaries, never inside the spacetime. The continuation of the original Killing vector field isThe field never vanishes in the interior, including on the Killing horizons.

To obtain the Penrose diagram, suppress the two-sphere and multiply by the conformal factor ${\cos }^{2}X/m^2$. The resulting radial metric tensor is $-d{T}^2+d{X}^2$, so the diagram is an infinitely tall strip of width $\pi$, with null lines at 45 degrees and timelike edges $X=\pm \pi /2$. Generically,At the special edge points where the two families of Killing horizons end, the displayed formula is a $0/0$ limit and $\lambda$ has no unique limiting value. Thus the infinity labels on the diagram refer to the open edge segments, not those vertices. Each interior point of the diagram represents a two-sphere of fixed radius $m$; neither edge is a centre where that two-sphere shrinks.

Finally, these timelike edges are conformal boundaries, not omitted regular points of a further spacetime extension. One way to establish maximality is geodesic completeness. The two-dimensional anti-de Sitter spacetime factor embeds locally as $Z\cdot Z=-m^2$ in a flat space with metric signature $(-,-,+)$. Along an affinely parametrized geodesic with constant $ϵ=\dot{Z}\cdot \dot{Z}$, the ambient equation isIts oscillatory, linear or hyperbolic solutions for negative, zero or positive $ϵ$ exist for all real affine parameters and lift to the universal cover. The two-sphere is also geodesically complete, and geodesics of the product split into geodesics of its factors. Hence $\tilde{g}$ is geodesically complete. A proper regular spacetime extension would furnish a geodesic reaching its boundary in finite affine parameter, contradicting that completeness. This proves that the displayed simply connected analytic spacetime is maximal.

Use metric signature $(+,-,\dots ,-)$ and initially set $\hslash =1$. For an internal scalar line with momentum $p$, the Feynman propagator is $i/(p^2-m^2+i0)$. An $r$-leg Feynman vertex, for $r=3,4,6$, contributes $-i{\lambda }_r$, with all incident momenta incoming. Its $1/r!$ in the Lagrangian density is canceled by the permutations of its identical fields. Each vertex conserves momentum and supplies $(2\pi {)}^d{\delta }^{(d)}(\sum p)$. Integrate each independent loop momentum with $d^{d}k/(2\pi {)}^d$, multiply the line and vertex factors, and divide by the Feynman-diagram symmetry factor for labeled external legs. Sum the distinct graphs and external-momentum assignments without counting the same assignment twice.

An unamputated momentum-space correlation function also contains the external propagators. Dividing the generating functional by its zero-source value removes disconnected vacuum bubbles; taking its logarithm selects connected correlation functions. Amputation of external propagators removes the external line factors when computing vertex amplitudes. Translational invariance leaves one overall momentum-conservation delta function for each connected graph.

The loop order is the number of independent graph cycles, equivalently the number of independent momentum integrations left after vertex conservation. A spanning tree through $V$ interaction vertices contains $V-1$ internal edges; hence, with $I$ internal lines,Restoring $\hslash$ in $\exp (iS/\hslash )$ gives $\hslash$ for each internal propagator and ${\hslash }^{-1}$ for each interaction vertex. Thus an amputated connected amplitude hasThe same counting holds for a diagram in $\log Z[J]$ when the external source is coupled as $J\varphi /\hslash$: every external propagator brings $\hslash$, canceled by the source insertion's ${\hslash }^{-1}$. This is Planck-constant loop counting with source normalization. An ordinary connected expectation of $n$ fields, with no such source factors, instead scales as ${\hslash }^{L+n-1}$; multiplying $\log Z$ by $-i\hslash$ to form $W$ shifts its source-kernel counting to ${\hslash }^L$. The normalization must be specified when applying the printed loop rule.

Let $V_r$ count $r$-valent interaction vertices. Counting their incident half-edges gives $2I+n=3V_3+4V_4+6V_6$. Substituting the loop identity yieldsEach coefficient on the right lies between $1/2$ and $2$, proving the vertex-count bounds for bounded scalar valenceThe free two-point term, which has no interaction vertex, separately satisfies the bounds with $V=L=0$ and $n=2$.

A one-particle-irreducible Feynman diagram is connected and remains connected after any single internal line is cut. The left graph below contains two quartic vertices and two parallel internal lines: cutting either line leaves the vertices connected by the other. The right graph contains two cubic vertices joined by one internal line, whose removal disconnects it, so it is a one-particle-reducible Feynman diagram. Both have four external legs.

The quantum effective action generates the one-particle-irreducible vertices. Their ultraviolet divergences and those of their proper divergent subgraphs are removed by local counterterms. Reducible connected graphs are reconstructed by joining these vertices with full propagators; they do not introduce independent local interaction counterterms merely because they contain bridges. This makes the one-particle-irreducible Feynman diagram decomposition useful for organizing renormalization. Overall power counting still has to be supplemented by subtraction of divergent subgraphs.

For a scalar with a two-derivative kinetic term, mass dimensions obeyThe general power-counting criterion used here is that interactions of nonnegative coupling dimension can be included in a perturbatively renormalizable quantum field theory, with all symmetry-allowed required local counterterms included; negative-dimensional couplings generally require an unbounded tower of higher-dimensional counterterms. Indeed, the scalar graph power-counting identity givesfor the superficial degree of divergence. Nonnegative coupling dimensions prevent this degree from increasing indefinitely with the vertex count.

The coupling dimensions areThus all three are power-counting renormalizable in three dimensions; cubic and quartic in four; cubic only in six. Positive-dimensional interactions are super-renormalizable, while the zero-dimensional ones are marginal by classical power counting. Usual kinetic, mass, vacuum and, when no symmetry forbids it, linear counterterms must also be admitted. This conclusion concerns perturbative ultraviolet renormalizability; it does not assert global stability of a cubic scalar potential.

Use the same metric signature as above and define the Minkowski generating functional byHere $\hslash =1$. Integration by parts writes the quadratic action as $S_0=\frac{1}{2}\varphi K\varphi$ with $K=-\square -m^2+i0$; the infinitesimal imaginary part damps the oscillatory Gaussian functional integral and selects the vacuum Feynman propagator. Abbreviating repeated spacetime integrations, completing the square givesTranslation of the regulated integration variables therefore yieldsThe source-independent normalization contains the regulated functional determinant and its oscillatory phase; it cancels in $Z_0[J]/Z_0[0]$. Define ${\Delta }_F=i{K}^{-1}$. The Gaussian evaluation of a free scalar generating functional is equivalently $Z_0[J]/Z_0[0]=\exp [-J{\Delta }_{F}J/2]$.

With Fourier convention $e^{-ip\cdot x}$, $-\square$ acts as $p^2$. The inverse-kernel equation $K{K}^{-1}=I$ then givesThis is the time-ordered free two-point correlation function: differentiating twice with respect to $J$ and multiplying by $1/i^2$ recovers ${\Delta }_F$. It obeys $(\square +m^2){\Delta }_F=-i{\delta }^{(d)}$ in the distributional limit. The $i0$ is essential: the algebraic denominator alone would not distinguish Feynman, advanced and retarded Green functions.

The subspace spanned by $e_4,e_5,e_6$ is the center of a Lie algebra. Each of their tensor squares is therefore invariant. On tensor products the infinitesimal Adjoint representation of a Lie algebra acts by a derivation, $D_x(u\otimes v)=[x,u]\otimes v+u\otimes [x,v]$. Write $u\odot v=u\otimes v+v\otimes u$ and let $S=e_1\odot e_5+e_3\odot e_6-e_2\odot e_4$. The only potentially nonzero checks areThe remaining generators are central, so they act trivially. This proves every claimed tensor invariant.

These tensors are contravariant. To obtain an ordinary metric tensor, take an invertible linear combination and invert it. For arbitrary real $a_4,a_5,a_6$ and $t\ne 0$, setIn the ordered basis $(e_1,e_2,e_3;e_4,e_5,e_6)$ its matrix isSince $K^{T}K=I$, $\det Q=-t^6\ne 0$, andLet ${\lambda }^a$ be the dual left-invariant coframe. The resulting neutral invariant metrics on the free two-step Lie algebra on three generators areHere $({\lambda }^a{)}^2={\lambda }^a\otimes {\lambda }^a$. Infinitesimal invariance of $Q$ implies that its inverse is an adjoint-invariant bilinear form. Its left-translated metric is right-invariant on a connected $G$, giving the desired four-parameter family of bi-invariant metrics. A change of variables removes the central $A$ block from the quadratic form, leaving three off-diagonal two-dimensional blocks. Each has one positive and one negative eigenvalue. Thus the metric signature is $(3,3)$; these are bi-invariant pseudo-Riemannian metrics, not positive-definite ones.

Connectedness, conventionally understood here, matters for the passage from infinitesimal to full group invariance. If $G$ is disconnected, its other components must preserve $Q$ separately. For example the Lie algebra automorphism reversing $e_1,e_4,e_6$ and fixing $e_2,e_3,e_5$ preserves all brackets but sends $S$ to $-S$. In the semidirect product of the simply connected group with this order-two automorphism, none of these nondegenerate $t\ne 0$ tensors is invariant under that component. Thus the Lie-algebra data alone give the stated family on the identity component, rather than an unconditional extension to every disconnected realization.

Define the raw Killing form byFor a finite-dimensional real or complex Lie algebra in characteristic zero, it is nondegenerate exactly when the algebra is a semisimple Lie algebra. For a real compact semisimple algebra it is negative definite. Consequently the positive compact Killing metric is $g=-K$: for example on $\mathfrak{su}(2)$ with $[e_i,e_j]={\epsilon }_{ijk}{e}_k$, $K_{ij}=-2{\delta }_{ij}$ and $-K$ gives a Riemannian metric on $SU(2)$. An overall minus sign in the term “Killing metric” is common here; raw trace $K$ itself is negative definite in this example.

For a Lorentzian metric use $\mathfrak{sl}(2,\mathbb{R})$ with $[h,e]=2e$, $[h,f]=-2f$, $[e,f]=h$. Direct traces givewhose metric signature is $(2,1)$. For a non-Abelian degenerate example, the three-dimensional Heisenberg Lie algebra has $[x,y]=z$ central. Every product of two adjoint maps is zero, so $K=0$.

For any adjoint-invariant nondegenerate metric on a Lie group, the Koszul formula on left-invariant fields simplifies toTake $R(X,Y)Z={\nabla }_X{\nabla }_{Y}Z-{\nabla }_Y{\nabla }_{X}Z-{\nabla }_{[X,Y]}Z$. The Jacobi identity givesTaking the trace of $Z\mapsto R(Z,X)Y=-\mathrm{ad}Y\mathrm{ad}X(Z)/4$ yieldsThus the metric $g=K$ is a Killing-form Einstein metric with Einstein constant $-1/4$. With $g=-K$ the same connection has constant $+1/4$. This proves the Einstein manifold assertion for every semisimple group, including products when the unscaled Killing form is used on all factors.

There is also a different, flat metric-preserving connection. Declare the global left-invariant frame parallel:Its curvature is zero because applying two derivatives to coefficient functions and subtracting their commutator gives zero. Its metric derivative vanishes since $g_{ab}$ is constant in this frame. Its torsion tensor on invariant fields isFor each fixed $c$, antisymmetry gives $T_a{}^b{}_c=C_c{}^b{}_a$, so this is the matrix of $\mathrm{ad}{e}_c$. The torsion contractions of a left-parallel Lie-group connection therefore areThe zero trace follows because a semisimple algebra is perfect: every element is a sum of brackets, and the adjoint of a bracket is a matrix commutator with zero trace. If the metric convention is $g=-K$, the second contraction is $-g_{cd}$. These flat translation connections on a Lie group are not the torsion-free Levi-Civita connection; nonzero torsion permits flatness alongside the curved Killing metric.

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.

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.

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.

General relativity models spacetime by a smooth four-dimensional Lorentzian manifold whose metric tensor governs clock readings, causal cones and free-particle motion. The Einstein equivalence principle motivates local special relativity and universal geodesic free fall for test bodies; it does not by itself select a unique gravitational field equation. We assume a Levi-Civita connection, a metric tensor as the gravitational field with no additional dynamical gravitational variables, and matter with a covariantly conserved stress-energy tensor. Units have $c=1$.

The existence of local inertial frames is a local statement. At any regular event choose an orthonormal frame and Riemann normal coordinates, so that $g_{ab}={\eta }_{ab}$ and ${\Gamma }^a{}_{bc}=0$ at the event, with metric signature $(+---)$. Metric compatibility then gives ${\partial }_c{g}_{ab}=0$ there. The geodesic equation at that point is the special-relativistic inertial equation. Along a timelike geodesic, a parallel-transported orthonormal frame gives Fermi normal coordinates, eliminating the connection along the central worldline. These constructions do not in general eliminate second derivatives of the metric tensor in a neighborhood. The Riemann curvature tensor is built from the derivatives of the connection coefficients as well as their quadratic terms; nonzero Riemann curvature tensor components can remain when all connection coefficients vanish at a chosen point. Conversely, Polar coordinates in Minkowski spacetime can have nonzero connection coefficients but zero Riemann curvature tensor. Connection coefficients alone are therefore not an invariant measure of gravity.

For the tensorial content, consider a two-parameter family of geodesics parametrized by an affine parameter, with tangent $U$ and connecting field $J$. Their coordinate construction gives $[U,J]=0$, and torsion-freeness gives ${\nabla }_{U}J={\nabla }_{J}U$. Since ${\nabla }_{U}U=0$, commuting derivatives in the convention of Question 1 yields geodesic deviation:The left side is covariant relative acceleration, rather than the second coordinate derivative of an arbitrary separation. In an orthonormal frame in free fall with $U=(1,0,0,0)$, its spatial components are $D^2{J}^i/d{\tau }^2=-R^i{}_{00j}{J}^j$. This is a measurable tidal force, and a coordinate transformation cannot make a nonzero Riemann curvature tensor disappear. It is the distinction between gravity at one event and the behavior of an extended freely falling laboratory.

In a static Newtonian limit, $g_{00}=1+2\Phi$ and ${\Gamma }^i{}_{00}={\partial }_i\Phi$, so the test-body acceleration is $-\nabla \Phi$. The curvature-sign convention in Killing derivative identities gives $R^i{}_{00j}\simeq {\partial }_i{\partial }_j\Phi$. Hence relative acceleration is the negative Hessian of the gravitational potential. For the Newtonian potential of a point mass $\Phi =-GM/r$, the Hessian has radial eigenvalue $-2GM/r^3$ and two tangential eigenvalues $GM/r^3$: radial separations stretch and tangential ones compress. Outside the source the Ricci tensor can vanish while the Riemann curvature tensor remains nonzero. Its vacuum tidal part is encoded by the Weyl tensor; vacuum does not mean flatness. Gravitational waves likewise carry Riemann curvature tensor disturbances without a local material source.

To obtain the field equations, make the additional dynamical assumptions explicit: locality and diffeomorphism invariance of general relativity; a symmetric metric tensor field equation using at most second derivatives of the metric tensor; a geometrical left-hand side linear in the Riemann curvature tensor with constant coefficients; and compatibility with local matter conservation. These restrictions exclude higher-curvature or extra-field alternatives. Under them the available rank-two curvature contractions give the ansatzThe contracted Bianchi identity and metric compatibility giveFor this to vanish identically for arbitrary metric tensors, $b=-a/2$. Assuming $a\ne 0$ and absorbing its normalization into the coupling givesThe permitted constant term is the cosmological constant. Its value is not fixed by the equivalence principle. Conservation of $T$ follows for generally covariant matter on its equations of motion: varying the matter action under a compactly supported infinitesimal diffeomorphism, then integrating its term $T^{ab}{\nabla }_a{\xi }_b$ by parts, gives ${\nabla }_a{T}^{ab}=0$. The divergence-free Einstein tensor is consistent with this identity.

The coupling sign must be fixed in the actual curvature-sign convention in Killing derivative identities, with $R_{ab}=R^c{}_{acb}$. For the static weak-field approximation, $R_{00}\simeq -\Delta \Phi$. Set $\Lambda =0$ for the local asymptotically flat Newtonian comparison. In four dimensions the trace equation gives $R=-\kappa T$, so $R_{00}=\kappa (T_{00}-\frac{1}{2}{g}_{00}T)\simeq \kappa \rho /2$ for slowly moving matter. Matching $\Delta \Phi =4\pi G\rho$ forces $\kappa =-8\pi G$. Thus, with this convention,Under the opposite curvature-sign convention in Killing derivative identities the coupling sign reverses, as in Einstein-equation coupling under reversed curvature. This sign statement is compatible with the supplied linearized handout. Local inertial frames remove the connection at an event; tidal Riemann curvature tensor effects remain tensorial, and the field equations require dynamical assumptions beyond the equivalence principle.

Use metric signature $(-,+,+,+)$ and units $c=1$. Write ${\dot{x}}^{\mu }=d{x}^{\mu }/d\lambda$ in this part. Varying the worldline einbein, without first fixing it, givesThe canonical four-momentum is $p_{\mu }=m{g}_{\mu \nu }{\dot{x}}^{\nu }/e$, so the same mass shell constraint is the mass shell $g^{\mu \nu }{p}_{\mu }{p}_{\nu }=-m^2$.

The Euler-Lagrange equation for the trajectory isMultiplying by $e$ and then by the inverse metric, and collecting the metric tensor derivatives into the Christoffel symbols, givesThis is a geodesic equation in a possibly nonaffine parameter.

Under a change of parameter, $e^{\prime }({\lambda }^{\prime })=e(\lambda )d\lambda /d{\lambda }^{\prime }$, so $ed\lambda$ is invariant. On the positive branch, the mass shell constraint identifies this invariant with the proper time increment $ds=\sqrt{-g_{\mu \nu }d{x}^{\mu }d{x}^{\nu }}=ed\lambda$. Choosing ${\lambda }^{\prime }=s$ therefore sets $e^{\prime }=1$. The trajectory equation is then the affinely parametrized geodesic equation and its tangent has norm minus one. This is proper-time gauge for a massive worldline einbein. The mass shell constraint must still be imposed: fixing $e$ in the action before varying would lose it. A fixed parameter interval can retain a proper-time modulus; here changing to the actual proper-time interval is permitted.

Use metric signature $(+---)$ and average over the four initial spin states, summing over final spins and the $N_c=3$ quark color charge states. The electrons carry no color charge, so their initial color charge average is trivial. A final color charge average would not give the inclusive relativistic cross-section requested.

Photon exchange gives the tree amplitude, up to an irrelevant overall phase,Massless spin sums give $\sum u\bar{u}=/p$ and $\sum v\bar{v}=/p$. ThereforeThe color charge multiplicity is ${\sum }_{a,b}|{\delta }_{ab}{|}^2=N_c$, not $N_c^2$. Each trace equals $4(a^{\mu }{b}^{\nu }+a^{\nu }{b}^{\mu }-g^{\mu \nu }a\cdot b)$. Contracting them, the terms proportional to $(p_1\cdot p_2)(q_1\cdot q_2)$ cancel, leaving $32[(p_1\cdot q_1)(p_2\cdot q_2)+(p_1\cdot q_2)(p_2\cdot q_1)]$. Thus