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Let $q^i=e^i-\frac{1}{2}{h}^i{}_j{e}^j$, so $p^0=ϵ/a^2$ and $p^i=(ϵ/a^2)q^i$. The orthonormal tetrad gives ${\delta }_{ij}{e}^i{e}^j=1$. To first order,Hence the time-index Christoffel symbols contract toThe $\mathcal{H}{h}_{ij}$ term cancels the perturbation in the coordinate direction norm. This cancellation is why the local tetrad direction must not be mistaken for the coordinate direction.

Along the photon null geodesic, $d/d\lambda =p^{0}d/d\eta$, andThe time component of the geodesic equation now gives the photon energy redshift from a tensor metric perturbation:The derivative on the left follows the ray; the dot on $h$ is a partial conformal time derivative. The background redshift has already been removed by using comoving energy. The calculation only needs ${\Gamma }^0{}_{00}$ and ${\Gamma }^0{}_{ij}$; the unrelated printed ${\Gamma }^i{}_{j0}$ should contain $\frac{1}{2}{\dot{h}}^i{}_j$, rather than the full derivative.

Let $p_i:M_1\times M_2\to M_i$ be the projections. The derivative of the two projections identifies the tangent bundle withTake the direct sum of the two pullback connections. This is the product affine connection. In product coordinates $(x^a,y^{\alpha })$, its only nonzero Christoffel symbols are the two blocks inherited from the factors; all mixed coefficients are zero.

To make the connection on arbitrary fields explicit, write $V=V^a{\partial }_{x^a}+V^{\alpha }{\partial }_{y^{\alpha }}$ and $W=W^a{\partial }_{x^a}+W^{\alpha }{\partial }_{y^{\alpha }}$, allowing all coefficients to depend on both variables. ThenThe terms $V(W^a)$ and $V(W^{\alpha })$ include derivatives in both factors. Omitting these cross derivatives for fields with variable coefficients would not define a connection.

The formula is real-bilinear, linear over smooth functions in $V$, and obeys the Leibniz rule in $W$. Under a change of product coordinates each block has the factor's connection transformation law, so the local formulas agree; equivalently the pullback connection construction already guarantees this gluing.

A field $X_i$ on one factor has a canonical lift to the product. Opposite-factor lifts commute, and their mixed covariant derivatives vanish. For such lifted fields the coefficient derivatives reduce to their factor derivatives, givingThese values determine the connection uniquely: lifted coordinate frames span locally, and the connection's product rules determine its action on all their smooth linear combinations.

The product Riemannian metric isIt is smooth and symmetric. If $(u_1,u_2)\ne 0$, at least one component is nonzero and contributes a strictly positive squared norm; the other contributes a nonnegative term. Thus $g$ is positive definite.

Let the factor connections be Levi-Civita connections. The torsion form of the product affine connection vanishes on two lifted fields from one factor because it equals the factor torsion; on opposite-factor fields it vanishes because both covariant derivatives and the Lie bracket of vector fields vanish. By tensoriality, torsion vanishes for all fields.

Check metric compatibility on lifted fields as well. When the two paired fields are from one factor, their metric pairing depends only on that factor. Differentiation along that factor gives the usual factor compatibility identity, and differentiation along the other factor gives zero on both sides. When the paired fields are from different factors, their pairing is identically zero and their derivatives stay in their original summands, so both sides are again zero. Since $\nabla g$ has tensoriality, these checks imply $\nabla g=0$ globally.

The Existence and uniqueness of the Levi-Civita connection therefore givesEquivalently, applying the Christoffel symbol formula to the block metric gives the factor coefficients and zero mixed coefficients, because each factor metric is independent of the other coordinates.

At $(p,q)$, choose neighborhoods on which the two factors admit local isometries to open subsets of Euclidean space. Their product is a local isometry of the product Riemannian metric to the standard Euclidean metric on ${\mathbb{R}}^{2n}$. In these product coordinates the metric matrix is constant, so every Christoffel symbol of the Levi-Civita connection is zero. Its Riemann curvature tensor therefore vanishes at every point of that neighborhood. Curvature has tensoriality, so this is independent of the chosen coordinates; the neighborhoods cover the product.

The splitting provides another useful direct check. Expanding the definition on lifted fields, using zero mixed derivatives and commuting opposite-factor fields, gives the curvature splitting for a product connectionTensoriality extends this identity from lifts to arbitrary tangent vectors at a point. Each factor curvature is zero under the stated local-isometry hypothesis. ConsequentlyThis is a local conclusion about the connection; no global identification of either factor or their product with Euclidean space is required.

Use spatial indices $i,j=1,2,3$, set ${\varphi }_i={\partial }_i\varphi$, and assume $\varphi$ is smooth. The inverse metric has $g^{tt}=-c^{-2}{e}^{-2\varphi /c^2}$ and $g^{ij}={\delta }^{ij}$. The Christoffel symbols of the Levi-Civita connection follow from ${\Gamma }^{\alpha }{}_{\beta \gamma }=g^{\alpha \delta }({\partial }_{\beta }{g}_{\gamma \delta }+{\partial }_{\gamma }{g}_{\beta \delta }-{\partial }_{\delta }{g}_{\beta \gamma })/2$:All other components vanish. Although $g_{tt}=-c^2-2\varphi +O(c^{-2})$ diverges, its affine connection has a smooth limit on compact subsets:This is the Newtonian connection from an exponential lapse. It is torsion-free, and its geodesic equation, using $t$ as an affine parameter when $\dot{t}\ne 0$, is $d^2{x}^i/d{t}^2=-{\partial }_i\varphi$. Thus $\varphi$ has the role of a Newtonian gravitational potential. There is no finite limiting nondegenerate Lorentzian metric in these fixed coordinates; $-g(c)/c^2$ tends to $d{t}^2$, a rank-one tensor. The limiting connection preserves $dt$ and the contravariant spatial tensor with components $h^{ij}={\delta }^{ij}$, $h^{t\mu }=0$, which describe the degenerate temporal/spatial structures of this limit.

For the Expanding flat slicing of de Sitter spacetime, take the geodesic LagrangianThe Euler-Lagrange equation for $t$ iswhile each spatial coordinate $x^i$ obeysComparing these equations with the geodesic equation gives the complete list of nonzero Christoffel symbols:

For a local parametrization $\varphi (u,v)$, write the first fundamental form and second fundamental form asThe Gaussian curvature is

To prove the Theorema Egregium, set $g_{ij}=⟨{\varphi }_i,{\varphi }_j⟩$ and decomposeDifferentiating $g_{ij}$ and solving for the tangential coefficients givesso every Christoffel symbol depends only on $E,F,G$ and their first derivatives. Equality of mixed third derivatives yields the Gauss equationThe Riemann curvature tensor is formed from first derivatives and quadratic products of the Christoffel symbols, and hence from the first and second derivatives of $E,F,G$. Consequentlydepends only on the first fundamental form.

Because the Levi-Civita connection is torsion-free, its Christoffel symbols satisfy ${\Gamma }_{\rho \alpha }^{\mu }={\Gamma }_{\alpha \rho }^{\mu }$. Substitute the stated coordinate expression for each term in the cyclic sumThe partial-derivative terms cancel pairwise by the lower-index symmetry of $\Gamma$, and the quadratic terms cancel after relabelling their dummy indices. Thereforewhich is the first Bianchi identity.

Use the geodesic Lagrangianwhere here the overdot on a coordinate means differentiation with respect to the affine parameter. Applying the Euler-Lagrange equation to each coordinate and comparing with the geodesic equationdetermines the Christoffel symbols. To display derivatives of the metric coefficients unambiguously, write ${\lambda }_t={\partial }_t\lambda$, ${\lambda }_r={\partial }_r\lambda$, and similarly for $\mu$. The nonzero symbols, together with those obtained by symmetry in the two lower indices, areThis is the Christoffel symbols of a diagonal spherical spacetime metric.

The coordinate expression for the Lie bracket of vector fields isFor the Levi-Civita connection,The final term vanishes because the Christoffel symbols are symmetric in $\mu ,\nu$ while the vector product in parentheses is antisymmetric. Hence

For a smooth curve $\gamma (t)$ on a Riemannian surface, its energy of a curve isChoose a local parameterization $\mathbf{X}(u,v)$ and write $\gamma (t)=\mathbf{X}(u(t),v(t))$. With coefficients of the first fundamental formthe Lagrangian isThe Euler-Lagrange equations are equivalentlyAfter multiplying by the inverse metric these become the geodesic equationwhere the Christoffel symbols are determined by $E,F,G$.

If a straight line segment lies in the surface, parameterize it byThen $\ddot{\gamma }=0$, so its acceleration has zero tangential component. The two displayed equations hold directly, and the segment is a geodesic.

For the one-sheeted hyperboloidputTwo distinct ruling lines through $P$ areIndeed, the direction $d_{\pm }$ satisfiesso substitution shows that every point of the line lies in $H$. These are geodesics by the straight-line argument.

A third geodesic is the meridian through $P$. Choose $s_0$ with $\sinh s_0=z_0$; thenIts acceleration is ${\gamma }_m$, which is normal to $H$, so it is a geodesic. If $z_0\ne 0$, these give the required three distinct subsets.

If $z_0=0$, there is also the equatorial circle. Writing $P=(\cos {\theta }_0,\sin {\theta }_0,0)$,Its acceleration is normal to $H$ along $z=0$, so it is a fourth geodesic distinct from the meridian and the two rulings.

Finally, write $\rho =\sqrt{x^2+y^2}=\sqrt{1+z^2}$. Clairaut's relation gives the conserved quantityChoose initial data in $z>0$ withSince $|\sin \psi |\le 1$, every point of the resulting geodesic satisfies $\rho \ge c$, and thereforeContinuity keeps the geodesic in the component $z>0$, and the stated completeness assumption defines it for every real time. The continuum of choices of $c$ supplies infinitely many such geodesics.

The Levi-Civita connection is a torsion-free connection. Its torsion form isSince $T=0$,In coordinates this is the symmetry ${\Gamma }_{jk}^i={\Gamma }_{kj}^i$ of the Christoffel symbols.

In the absence of scalar anisotropic stress, the traceless spatial Einstein field equations giveThe photon mass-shell condition $g_{\mu \nu }{P}^{\mu }{P}^{\nu }=0$, together with $p^2=g_{ij}{P}^i{P}^j$ and $ϵ=ap$, gives to first orderUse the time component of the geodesic equation, divide it by $d\eta /d\lambda =P^0$, and substitute the stated Christoffel symbols. Keeping the perturbed ratios $P^i/P^0=(1+\Phi +\Psi ){\hat{p}}^i$ where they multiply the background $\mathcal{H}$ terms makes those terms cancel. Comparing the result with the total derivative of $P^0=ϵa^{-2}(1-\Psi )$ yieldsThe first term is the time-dependent gravitational redshift and the second is the gravitational frequency shift along the photon direction.

The Schwarzschild exterior is vacuum, soand consequently $C_{\alpha \beta \gamma \delta }=R_{\alpha \beta \gamma \delta }$. PutThe relevant Christoffel symbols areWith the conventionone findsSince $f^{\prime \prime }=-4M/r^3$,This is the leading Radial Weyl curvature of Schwarzschild spacetime. The opposite convention for the Riemann tensor reverses the sign but not the physical tidal magnitude.

For the Friedmann-Lemaitre-Robertson-Walker metric in Cartesian spatial coordinates, the only time-dependent components are $g_{ij}=a^2{\delta }_{ij}$. The Christoffel symbol formula therefore givesSubstituting these into the definition of the Ricci tensor givesThus, with no summation over $i$, the requested components are ${\Gamma }^0{}_{ii}=a\dot{a}$, ${\Gamma }^i{}_{0i}={\Gamma }^i{}_{i0}=\dot{a}/a$, and $R_{ii}=a\ddot{a}+2{\dot{a}}^2$.

Writing $f=1-2M/R$, orthogonality from part a givesThe needed Christoffel symbols areThe radial and azimuthal transport equations reduce toHenceWith the stated initial direction,

Forthe nonzero Christoffel symbols needed here areOn a line of constant longitude, $\dot{\varphi }=0$, and the geodesic equation reduces to $\ddot{\theta }=0$ after choosing an affine parameter. Thus every meridian is a geodesic.

For a nonconstant line of constant latitude, $\dot{\theta }=0$ and the $\theta$ equation requiresAway from the coordinate-degenerate poles, this holds only whenso only the equator is a constant-latitude geodesic.

LetThe Euler-Lagrange equation givesSince ${\dot{x}}^{\mu }=Ld{x}^{\mu }/ds$ and $d/d\lambda =Ld/ds$, this becomesExpanding the derivative, multiplying by the inverse metric, and using the Christoffel symbolgives the geodesic equation

At an arbitrary point choose normal coordinates, so the Christoffel symbols vanish there. Torsion freedom and commutation of partial derivatives then give the algebraic first Bianchi identityDifferentiate the coordinate curvature expression and cyclically antisymmetrize. Third derivatives cancel, giving the second Bianchi identityBoth statements are tensorial and hence hold in every coordinate system. Contracting the differential identity, using the curvature symmetries and metric compatibility, yieldsthe contracted Bianchi identity.

Apply the definition to the coordinate basis, for which $[{\partial }_{\mu },{\partial }_{\nu }]=0$ and ${\nabla }_{\mu }{\partial }_{\tau }={\Gamma }^{\rho }{}_{\tau \mu }{\partial }_{\rho }$. Comparing coefficients givesequivalent to the paper's ordering after commuting scalar factors. Although the individual Christoffel symbols are not tensors, the preceding intrinsic definition proves that this complete combination is tensorial.