Curvature: tidal forces, with the mask off

Idea. Locally the cone is a piece of flat paper rolled up, so its curvature is zero; but the paper is missing a wedge, and a loop round the apex notices the missing wedge. Curvature is local; holonomy round a loop that cannot be shrunk is not.

1. With \(g_{rr} = 1\), \(g_{\phi\phi} = r^2\sin^2\alpha\), the geodesic Lagrangian \(\tfrac12(\dot r^2 + r^2\sin^2\alpha\,\dot\phi^2)\) gives \(\ddot r - r\sin^2\alpha\,\dot\phi^2 = 0\) and \(\ddot\phi + (2/r)\dot r\dot\phi = 0\), so \[ \Gamma^r_{\phi\phi} = -r\sin^2\alpha, \qquad \Gamma^\phi_{r\phi} = \Gamma^\phi_{\phi r} = \frac1r , \] and all others vanish. Then \[ R^r{}_{\phi r\phi} = \partial_r\Gamma^r_{\phi\phi} - \partial_\phi\Gamma^r_{r\phi} + \Gamma^r_{r\lambda}\Gamma^\lambda_{\phi\phi} - \Gamma^r_{\phi\lambda}\Gamma^\lambda_{r\phi} = -\sin^2\alpha - 0 + 0 - (-r\sin^2\alpha)\Big(\frac1r\Big) = 0 . \] In two dimensions the whole tensor is \(K(g_{\rho\mu}g_{\sigma\nu} - g_{\rho\nu}g_{\sigma\mu})\) with \(K = R_{r\phi r\phi}/\det g = g_{rr}R^r{}_{\phi r\phi}/\det g = 0\), so every component vanishes.

2. Along \(r = r_0\), with \(\phi\) as parameter, \(\mathrm dV^r/\mathrm d\phi = -\Gamma^r_{\phi\phi}V^\phi = r_0\sin^2\alpha\,V^\phi\) and \(\mathrm dV^\phi/\mathrm d\phi = -\Gamma^\phi_{\phi r}V^r = -V^r/r_0\). In orthonormal components \(A = V^r\), \(B = r_0\sin\alpha\,V^\phi\): \[ \frac{\mathrm dA}{\mathrm d\phi} = \sin\alpha\,B, \qquad \frac{\mathrm dB}{\mathrm d\phi} = -\sin\alpha\,A . \] This is unit 06's oscillator: \((A,B)\) turns clockwise in the \((\hat e_r,\hat e_\phi)\) plane at the constant rate \(\sin\alpha\), and after \(\phi = 2\pi\) it has turned through \(-2\pi\sin\alpha\). The frame itself is back where it began, so the holonomy is a rotation by \[ -2\pi\sin\alpha \;\equiv\; 2\pi(1-\sin\alpha) \pmod{2\pi} . \]

3. \(\alpha = 30^\circ\): \(2\pi(1-\tfrac12) = \pi\), a half-turn, \(180^\circ\). \(\alpha = 60^\circ\): \(2\pi(1 - 0.8660) = 0.8418\) rad \(= 48.23^\circ\). A surface with \(R = 0\) everywhere is locally flat: cut along \(\phi = 0\) and the chart \(\tilde\phi = \phi\sin\alpha\) unrolls it onto a plane with a wedge of angle \(2\pi(1-\sin\alpha)\) removed. A loop that does not enclose the apex bounds a flat disc and has trivial holonomy; the circle encloses the apex, which is not part of the surface, so it bounds nothing on the surface and the holonomy theorem (angle \(=\int K\,\mathrm dA\) over the enclosed region) does not apply to it. This is why the flatness theorem requires simple connectivity. If the apex is put back as a point carrying curvature, the theorem says that point carries \(\int K\,\mathrm dA = 2\pi(1-\sin\alpha)\): a delta function.

Check. Limit \(\alpha = 90^\circ\): the "cone" is the plane in polar coordinates, and the holonomy is \(0\) ✓. A second check: the cone tangent to a sphere along the circle of colatitude \(\theta_0\) has \(\sin\alpha = \cos\theta_0\), and unit 32 found transport round that latitude on the sphere turns a vector through \(2\pi\cos\theta_0\) in magnitude — the same number, because transport along a curve depends only on the metric along an infinitesimal strip round the curve, and the cone and the sphere agree there to first order. Dimensions: angles, nothing to check; \(r_0\) has dropped out, as it must for a cone, which looks the same at every scale.

Idea. Antisymmetry in two slots of a two-dimensional index leaves a one-dimensional space; so the Riemann tensor is a multiple of any nonzero tensor with the same symmetries, and \(g\) supplies one.

1. By S1, \(R_{\rho\sigma\mu\mu} = 0\), so a nonzero component has \(\mu\nu \in \{12, 21\}\); by S2 likewise \(\rho\sigma \in \{12, 21\}\). The four candidates are \(R_{1212}\), \(R_{1221} = -R_{1212}\), \(R_{2112} = -R_{1212}\), \(R_{2121} = R_{1212}\). So the space of arrays obeying S1 and S2 is one-dimensional. S4 holds automatically: exchanging the pairs of \(R_{1212}\) gives \(R_{1212}\), of \(R_{1221}\) gives \(R_{2112}\), equal to it. For S3, with only two index values two of \(\sigma,\mu,\nu\) coincide, and then S1 alone kills the cyclic sum: if \(\sigma = \mu\) it is \((R_{\rho\sigma\sigma\nu} + R_{\rho\sigma\nu\sigma}) + R_{\rho\nu\sigma\sigma} = 0 + 0\); if \(\mu = \nu\) it is \(R_{\rho\sigma\mu\mu} + (R_{\rho\mu\mu\sigma} + R_{\rho\mu\sigma\mu}) = 0\); if \(\nu = \sigma\) it is \((R_{\rho\sigma\mu\sigma} + R_{\rho\sigma\sigma\mu}) + R_{\rho\mu\sigma\sigma} = 0\). So nothing further is imposed.

Now \(T_{\rho\sigma\mu\nu} = g_{\rho\mu}g_{\sigma\nu} - g_{\rho\nu}g_{\sigma\mu}\) obeys S1 and S2 by inspection, and \(T_{1212} = g_{11}g_{22} - g_{12}^2 = \det g > 0\). Both \(R\) and \(T\) lie in the one-dimensional space, \(T \ne 0\), so \(R = fT\) with \(f = R_{1212}/T_{1212} = R_{1212}/\det g\). Since \(R\) and \(T\) are tensors, \(f\) at a point does not depend on the chart; call it \(K\). \(\square\)

2. With \(\varphi = \ln a - \ln y\): \(\varphi_x = 0\), \(\varphi_y = -1/y\). For \(g = e^{2\varphi}\delta\) the Christoffel formula gives \(\Gamma^x_{xx} = \varphi_x\), \(\Gamma^x_{xy} = \varphi_y\), \(\Gamma^x_{yy} = -\varphi_x\), \(\Gamma^y_{yy} = \varphi_y\), \(\Gamma^y_{xy} = \varphi_x\), \(\Gamma^y_{xx} = -\varphi_y\). So \[ \Gamma^x_{xy} = -\frac1y,\quad \Gamma^y_{yy} = -\frac1y,\quad \Gamma^y_{xx} = \frac1y,\quad \Gamma^x_{xx} = \Gamma^x_{yy} = \Gamma^y_{xy} = 0 . \] Term by term: \(\partial_x\Gamma^x_{yy} = 0\); \(\ -\partial_y\Gamma^x_{xy} = -\partial_y(-1/y) = -1/y^2\); \(\ \Gamma^x_{x\lambda}\Gamma^\lambda_{yy} = \Gamma^x_{xx}\Gamma^x_{yy} + \Gamma^x_{xy}\Gamma^y_{yy} = 0 + 1/y^2\); \(\ -\Gamma^x_{y\lambda}\Gamma^\lambda_{xy} = -(\Gamma^x_{yx}\Gamma^x_{xy} + \Gamma^x_{yy}\Gamma^y_{xy}) = -1/y^2\). \[ R^x{}_{yxy} = 0 - \frac1{y^2} + \frac1{y^2} - \frac1{y^2} = -\frac1{y^2}, \qquad R_{xyxy} = g_{xx}R^x{}_{yxy} = -\frac{a^2}{y^4}, \qquad K = \frac{R_{xyxy}}{\det g} = \frac{-a^2/y^4}{a^4/y^4} = -\frac1{a^2} . \]

3. \(\Delta\varphi = \partial_y^2(-\ln y) = 1/y^2\) and \(e^{-2\varphi} = y^2/a^2\), so \(K = -(y^2/a^2)(1/y^2) = -1/a^2\) ✓. The metric components blow up as \(y \to 0\), but \(K\) — a scalar — is the same everywhere, and \(y = 0\) is at infinite distance: \(\int_{y}^{1} a\,\mathrm dy'/y' = a\ln(1/y) \to \infty\). The line \(y = 0\) is not part of the surface; it is where the chart sends infinity.

Check. Dimensions: \(K\) is \(\mathrm L^{-2}\) ✓. Limit \(a \to \infty\): \(K \to 0\) ✓. And \(K\) is constant, as it must be: \((x,y) \mapsto (x+b, y)\) and \((x,y)\mapsto(\lambda x,\lambda y)\) are isometries (they leave \(g\) unchanged), they can carry any point to any other, and an isometry preserves \(K\). The sphere is the same surface with the opposite sign.

Idea. In these coordinates the \(r\)-lines are a family of geodesics fanning out from a point, and \(f\) is literally the rate at which they separate. Curvature is the second derivative of that separation — which is the deviation equation, and which is why \(K = -f''/f\).

1. From \(L = \tfrac12(\dot r^2 + f^2\dot\phi^2)\): \(\ddot r - ff'\dot\phi^2 = 0\) and \(\ddot\phi + 2(f'/f)\dot r\dot\phi = 0\), so \(\Gamma^r_{\phi\phi} = -ff'\), \(\Gamma^\phi_{r\phi} = \Gamma^\phi_{\phi r} = f'/f\). Then \[ R^r{}_{\phi r\phi} = \partial_r\Gamma^r_{\phi\phi} - \partial_\phi\Gamma^r_{r\phi} + \Gamma^r_{r\lambda}\Gamma^\lambda_{\phi\phi} - \Gamma^r_{\phi\lambda}\Gamma^\lambda_{r\phi} = -(f'^2 + ff'') - 0 + 0 - (-ff')\frac{f'}{f} = -ff'' . \] \(R_{r\phi r\phi} = g_{rr}R^r{}_{\phi r\phi} = -ff''\), \(\det g = f^2\), so \(K = -f''/f\). \(\square\)

2. \(f = r\): \(K = 0\), the plane. \(f = a\sin(r/a)\): \(f'' = -f/a^2\), \(K = 1/a^2\), the sphere with \(r\) the distance from the north pole. \(f = a\sinh(r/a)\): \(f'' = f/a^2\), \(K = -1/a^2\), the hyperbolic plane of P2 in polar form. \(f = r\sin\alpha\): \(K = 0\), the cone of P1.

3. For \(\gamma(s) = (r,\phi) = (s, \phi_0)\): \(\ddot r + \Gamma^r_{\phi\phi}\dot\phi^2 = 0\) ✓ and \(\ddot\phi + 2\Gamma^\phi_{r\phi}\dot r\dot\phi = 0\) ✓ since \(\dot\phi = 0\); and \(g(\dot\gamma,\dot\gamma) = 1\). The family parameter is \(\phi_0\), so \(\xi = \partial_\phi\), orthogonal to \(u = \partial_r\), with length \(|\xi| = f\); for two members \(\Delta\phi\) apart, separation \(f(r)\Delta\phi\). By the deck's two-dimensional reduction, \(\mathrm D^2\xi/\mathrm ds^2 = -K\xi\) for \(\xi \perp u\) with \(|u| = 1\). In the orthonormal frame \((\partial_r, f^{-1}\partial_\phi)\), which is parallel along the \(r\)-lines — \(\nabla_r\partial_r = 0\) and \(\nabla_r(f^{-1}\partial_\phi) = -f'f^{-2}\partial_\phi + f^{-1}(f'/f)\partial_\phi = 0\) — \(\xi\) has the single component \(f\), so the equation reads \(f'' = -Kf\). The deviation equation and part 1 are one statement.

4. \(f'' + K_0f = 0\), \(f(0) = 0\), \(f'(0) = 1\) is a linear ODE with constant coefficients and two initial conditions; by unit 01's existence and uniqueness theorem it has exactly one solution, and \(f = \sin(\sqrt{K_0}\,r)/\sqrt{K_0}\) is one. So the metric is \(\mathrm dr^2 + K_0^{-1}\sin^2(\sqrt{K_0}\,r)\,\mathrm d\phi^2\): the sphere of radius \(1/\sqrt{K_0}\). Constant positive curvature determines the metric near a point.

Check. Dimensions: \(f\) is a length, \(f''/f\) is \(\mathrm L^{-2}\) ✓. Near \(r = 0\) every smooth case has \(f = r + O(r^3)\), which is P5's starting point, and the first three examples satisfy it ✓. The cone fails \(f'(0) = 1\) unless \(\alpha = 90^\circ\) — the smoothness condition is exactly what the apex violates.

Idea. A rotation angle is only defined modulo \(2\pi\), and "relative to what" is the whole question. The curvature theorem gives the rotation relative to the starting vector; the Earth-bound observer measures it relative to a frame that has itself gone once round.

1. \(\int_{\text{cap}}K\,\mathrm dA = \frac{1}{R_\oplus^2}\cdot 2\pi R_\oplus^2(1-\cos\theta_0) = 2\pi(1-\sin\lambda)\). For Paris, \(\sin\lambda = 0.75295\), so this is \(1.5523\) rad \(= 88.94^\circ\), anticlockwise seen from above.

2. With unit 32's Christoffel symbols on the latitude circle (parameter \(\phi\)), \(\dot V^\theta = \sin\theta_0\cos\theta_0\,V^\phi\), \(\dot V^\phi = -\cot\theta_0\,V^\theta\). In orthonormal components \(A = R_\oplus V^\theta\) (south), \(B = R_\oplus\sin\theta_0\,V^\phi\) (east): \(\dot A = \cos\theta_0\,B\), \(\dot B = -\cos\theta_0\,A\). So relative to the (south, east) frame — equivalently the (east, north) frame, which differs by a fixed rotation — the vector turns at rate \(\cos\theta_0 = \sin\lambda\), clockwise seen from above, through \[ -2\pi\sin\lambda \text{ per sidereal day}. \] Consistency with part 1: \(-2\pi\sin\lambda = 2\pi(1-\sin\lambda) - 2\pi\), the same rotation modulo a full turn ✓. The extra \(-2\pi\) is the frame's own turn: on a tiny circle round the pole the cap, and so the holonomy, shrink to nothing, yet the east direction still swings once round as it circles the pole.

3. \(360^\circ\times0.75295 = 271.06^\circ\) per sidereal day, clockwise seen from above. A full turn relative to the building takes \(86\,164.09\ \mathrm s/0.75295 = 1.1443\times10^5\) s \(= 31.79\) h — unit 03's Coriolis answer, obtained with no force at all.

Check. At the pole, \(\sin\lambda = 1\): one full turn per sidereal day, the Earth simply turning under an inertial swing plane ✓. At the equator: no turning, and the cap is a hemisphere with \(\int K\,\mathrm dA = 2\pi\), a full turn, invisible ✓. In the southern hemisphere \(\sin\lambda < 0\) and the sense reverses ✓. Dimensions: angles and times only.

Idea. Radial geodesics from a point separate like \(f\), and positive curvature makes them reconverge (the meridians of the deck), so circles are shorter than Euclid says. The deficit is cubic in \(r\) with coefficient \(K/6\): curvature can be measured with a tape measure, from inside, which is Gauss's point.

1. In this form \(K = -f''/f\) is a function of \(r\), and \(K(0)\) is its value at the centre. From \(f'' = -Kf\): \(f''(0) = 0\); \(f''' = -K'f - Kf'\), so \(f'''(0) = -K(0)\). Taylor to third order: \[ f(r) = r - \frac{K(0)}{6}r^3 + o(r^3), \qquad C(r) = 2\pi r\Big(1 - \frac{K(0)}{6}r^2 + o(r^2)\Big). \] Rearranging, \((2\pi r - C)/r^3 = \pi K(0)/3 + o(1)\), which is the limit formula. \(\square\)

2. \(C = 2\pi R_\oplus\sin(r/R_\oplus)\) with \(r/R_\oplus = 0.15696\): \(C = 6257.42\) km against \(2\pi r = 6283.19\) km, a shortfall of \(25.77\) km, or \(0.41\%\). The leading term predicts \(2\pi r\cdot r^2/(6R_\oplus^2) = \pi r^3/(3R_\oplus^2) = 25.80\) km, within \(0.12\%\) of the exact shortfall.

3. \(K_{\text{est}} = \frac{3}{\pi}\cdot\frac{25.768\ \mathrm{km}}{(1000\ \mathrm{km})^3} = 2.4606\times10^{-14}\ \mathrm{m^{-2}}\), against the true \(1/R_\oplus^2 = 2.4637\times10^{-14}\ \mathrm{m^{-2}}\). The implied radius is \(K_{\text{est}}^{-1/2} = 6375\) km, \(4\) km (\(0.06\%\)) too large — the next term of the series, of relative size \(r^2/(20R^2)\) in \(K\), is what is left out.

Check. Dimensions: \(K r^2\) is dimensionless ✓. \(K = 0\): \(C = 2\pi r\) exactly ✓. On the hyperbolic plane \(C = 2\pi a\sinh(r/a) > 2\pi r\), a surplus, as \(K < 0\) requires ✓.

Idea. Three symmetries make \(R\) a symmetric form on planes; the fourth, the first Bianchi identity, removes exactly the totally antisymmetric part, and no more. Count the forms, subtract the antisymmetric tensors.

1. Fix an ordering of the \(N\) pairs \(\rho < \sigma\). By S1 and S2, \(R\) is fixed by its values \(B_{(\rho\sigma),(\mu\nu)} = R_{\rho\sigma\mu\nu}\) on ordered pairs, and any values extend uniquely by the signs. S4 says \(B\) is a symmetric \(N\times N\) matrix, and any symmetric matrix arises. So \(\dim W = N(N+1)/2\).

2. \(Z\) is invariant under cyclic permutations of \((\sigma,\mu,\nu)\) by its form. Swapping \(\sigma\) and \(\mu\): \(Z_{\rho\mu\sigma\nu} = R_{\rho\mu\sigma\nu} + R_{\rho\sigma\nu\mu} + R_{\rho\nu\mu\sigma}\), and by S1 these are \(-R_{\rho\mu\nu\sigma}\), \(-R_{\rho\sigma\mu\nu}\), \(-R_{\rho\nu\sigma\mu}\): so \(Z_{\rho\mu\sigma\nu} = -Z_{\rho\sigma\mu\nu}\). A transposition and the 3-cycles generate all permutations of the last three slots, so \(Z\) is antisymmetric in them. Swapping \(\rho\) and \(\sigma\): \(Z_{\sigma\rho\mu\nu} = R_{\sigma\rho\mu\nu} + R_{\sigma\mu\nu\rho} + R_{\sigma\nu\rho\mu}\). First term, S2: \(-R_{\rho\sigma\mu\nu}\). Second, S4 then S2: \(R_{\sigma\mu\nu\rho} = R_{\nu\rho\sigma\mu} = -R_{\rho\nu\sigma\mu}\). Third, S4 then S1: \(R_{\sigma\nu\rho\mu} = R_{\rho\mu\sigma\nu} = -R_{\rho\mu\nu\sigma}\). Total \(-Z_{\rho\sigma\mu\nu}\). Adding the transposition \((\rho\sigma)\) to the permutations of the last three generates all of \(S_4\). \(\square\)

3. If \(A\) is totally antisymmetric then it obeys S1, S2, and S4 (exchanging the pairs is an even permutation), so \(A \in W\), and its \(Z\) is \(A_{\rho\sigma\mu\nu} + A_{\rho\mu\nu\sigma} + A_{\rho\nu\sigma\mu} = 3A\) (each is a cyclic, even, permutation). So \(R \mapsto Z\) is onto \(\Lambda^4\), of dimension \(\binom n4\). S3 is \(Z = 0\), so the tensors obeying S1–S4 are its kernel, and by rank–nullity \[ \frac{N(N+1)}{2} - \binom n4 = \frac{n(n-1)(n^2-n+2)}{8} - \frac{n(n-1)(n-2)(n-3)}{24} = \frac{n(n-1)}{24}\big[3(n^2-n+2) - (n^2-5n+6)\big] = \frac{n(n-1)\cdot2n(n+1)}{24} = \frac{n^2(n^2-1)}{12}. \] \(n = 2\): \(1\). \(n = 3\): \(6\). \(n = 4\): \(20\). \(\square\)

4. Let \(\mathcal R\) be the space of curvature tensors (dimension \(6\) at \(n = 3\)) and \(\mathrm{Sym}\) the symmetric \((0,2)\)-tensors (dimension \(6\)). \(F : \mathrm{Sym}\to\mathcal R\) and \(\mathrm{Ric} : \mathcal R\to\mathrm{Sym}\) are linear with \(\mathrm{Ric}\circ F = \mathrm{id}\). So \(F\) is injective, hence, the dimensions being equal, bijective; and then \(\mathrm{Ric} = F^{-1}\) is injective too. The Ricci tensor determines the Riemann tensor, by the explicit formula \(R = F(\mathrm{Ric})\). \(\square\) Physically: in 2+1 dimensions, wherever \(R_{\mu\nu} = 0\) — in vacuum, once unit 35 ties Ricci to matter — the whole curvature vanishes. No tides in empty space, no gravitational waves, no field outside a star: only the global deficit angles of P1.

Check. \(n = 1\) gives \(0\) — a curve has no intrinsic curvature ✓. \(n = 4\): \(N = 6\), \(21 - 1 = 20\), which is also the chart-counting shortfall \(100 - 80\) on the deck's opening slide ✓. At \(n = 3\) the count \(\binom 34 = 0\) says S3 is automatic, consistent with part 3.

Idea. The commutator of covariant derivatives is a derivation with no first-order part, so it acts on a product slot by slot. Applied to the metric, which \(\nabla\) cannot see, it forces an antisymmetry.

1. Expand as in rung 3 and subtract the same with \(\mu \leftrightarrow \nu\). The two middle terms become \((\nabla_\nu\omega_\rho)(\nabla_\mu\chi_\sigma) + (\nabla_\mu\omega_\rho)(\nabla_\nu\chi_\sigma)\), symmetric in \(\mu\nu\), and cancel. What remains is \(([\nabla_\mu,\nabla_\nu]\omega_\rho)\chi_\sigma + \omega_\rho[\nabla_\mu,\nabla_\nu]\chi_\sigma = -R^\lambda{}_{\rho\mu\nu}\omega_\lambda\chi_\sigma - R^\lambda{}_{\sigma\mu\nu}\omega_\rho\chi_\lambda\). Both sides are additive in \(T\), and in a chart every \(T\) is a sum of products: \(T = \sum_\beta \omega^{(\beta)}\otimes\mathrm dx^\beta\) with \(\omega^{(\beta)} = T_{\alpha\beta}\,\mathrm dx^\alpha\). So the product case gives the general one. \(\square\)

2. \(\nabla g = 0\), so the left side vanishes: \(0 = -R^\lambda{}_{\rho\mu\nu}g_{\lambda\sigma} - R^\lambda{}_{\sigma\mu\nu}g_{\rho\lambda} = -R_{\sigma\rho\mu\nu} - R_{\rho\sigma\mu\nu}\), which is S2. Then \(R^\rho{}_{\rho\mu\nu} = g^{\rho\sigma}R_{\sigma\rho\mu\nu}\), a symmetric \(g^{\rho\sigma}\) against an antisymmetric pair: \(0\).

3. Lowered, \(M_{\rho\sigma} = -R_{\rho\sigma\mu\nu}\delta a^\mu\delta b^\nu\) is antisymmetric: an element of the Lie algebra \(\mathfrak{so}(g)\) of infinitesimal rotations (or Lorentz transformations, in signature \((-,+,+,+)\)). It must be, because transport preserves \(g\) (metric compatibility), so the holonomy of a loop is in the orthogonal group, and a group element close to the identity is \(1 + M\) with \(g(MV,W) + g(V,MW) = 0\) to first order. S2 is that sentence in components.

Check. On the sphere, \(R^\theta{}_{\theta\theta\phi} = 0\) and \(R^\phi{}_{\phi\theta\phi} = 0\) from the deck's Christoffel symbols directly, as \(R^\rho{}_{\rho\mu\nu} = 0\) demands ✓. For a connection that is not metric the argument fails at the first step of part 2, and \(R^\rho{}_{\rho\mu\nu}\) is then a genuine nonzero antisymmetric \((0,2)\)-tensor — Weyl's 1918 theory, whose failure is a story for another course.

Idea. The second Bianchi identity holds for every metric. Traced twice, it says a particular combination of Ricci and scalar curvature has zero divergence for every metric — so it is the only combination that can be set equal to a conserved source.

1. Contract with \(g^{\lambda\mu}\). Term one: \(\nabla^\mu R_{\rho\sigma\mu\nu}\). Term two: \(g^{\lambda\mu}R_{\sigma\lambda\mu\nu} = -g^{\lambda\mu}R_{\lambda\sigma\mu\nu} = -R_{\sigma\nu}\) (S2, then the definition). Term three: \(g^{\lambda\mu}R_{\lambda\rho\mu\nu} = R_{\rho\nu}\) (definition). With \(\nabla g^{-1} = 0\): \[ \nabla^\mu R_{\rho\sigma\mu\nu} - \nabla_\rho R_{\sigma\nu} + \nabla_\sigma R_{\rho\nu} = 0 . \] Contract with \(g^{\sigma\nu}\). Term one: \(g^{\sigma\nu}R_{\rho\sigma\mu\nu} = g^{\sigma\nu}R_{\sigma\rho\nu\mu} = R_{\rho\mu}\) (S1 and S2 together). Term two: \(-\nabla_\rho R\). Term three: \(\nabla^\nu R_{\rho\nu}\). So \(2\nabla^\mu R_{\rho\mu} - \nabla_\rho R = 0\); by the symmetry of Ricci this is \(2\nabla^\mu R_{\mu\nu} = \nabla_\nu R\). \(\square\)

2. \(\nabla^\mu G_{\mu\nu} = \nabla^\mu R_{\mu\nu} - \tfrac12\nabla^\mu(Rg_{\mu\nu}) = \tfrac12\nabla_\nu R - \tfrac12 g_{\mu\nu}\nabla^\mu R = 0\), using \(\nabla g = 0\). And \(\nabla^\mu(\Lambda g_{\mu\nu}) = \Lambda\nabla^\mu g_{\mu\nu} = 0\). \(\square\)

3. If \(R_{\mu\nu} = \kappa T_{\mu\nu}\) then \(R = \kappa T\) and, by part 1, \(\kappa\nabla_\nu T = \nabla_\nu R = 2\nabla^\mu R_{\mu\nu} = 2\kappa\nabla^\mu T_{\mu\nu} = 0\). So \(T\) is constant on spacetime (on each connected component). For a star of dust in empty space, \(T = -\rho c^2\) is \(0\) outside and \(-\rho c^2 \ne 0\) inside: not constant. The only dust this theory permits is dust of the same density everywhere, which is no star at all. The wrong equation is not wrong in some detail; it contradicts the conservation of energy and momentum for every realistic source.

4. By P2, \(R_{\rho\sigma\mu\nu} = K(g_{\rho\mu}g_{\sigma\nu} - g_{\rho\nu}g_{\sigma\mu})\). Then \(R_{\sigma\nu} = g^{\rho\mu}R_{\rho\sigma\mu\nu} = K(2g_{\sigma\nu} - g_{\sigma\nu}) = Kg_{\sigma\nu}\), \(R = 2K\), and \(G_{\sigma\nu} = Kg_{\sigma\nu} - \tfrac12(2K)g_{\sigma\nu} = 0\). \(\square\)

Check. The sphere: \(R_{\mu\nu} = g_{\mu\nu}/a^2\), \(R = 2/a^2\) constant, so both sides of \(2\nabla^\mu R_{\mu\nu} = \nabla_\nu R\) vanish ✓. Dimensions: every term is \(\mathrm L^{-3}\) ✓. This identity is the only reason the field equation of unit 35 can be consistent at all: it holds before any equation is imposed.

Idea. \(P\) is a machine that builds a curvature tensor with a prescribed Ricci trace. Subtract what it builds from Ricci and the remainder has no trace at all; count what is left.

1. \(g_{\rho[\mu}S_{\nu]\sigma} - g_{\sigma[\mu}S_{\nu]\rho} = \tfrac12\big(g_{\rho\mu}S_{\nu\sigma} - g_{\rho\nu}S_{\mu\sigma} - g_{\sigma\mu}S_{\nu\rho} + g_{\sigma\nu}S_{\mu\rho}\big)\). Contract with \(g^{\rho\mu}\): \(nS_{\nu\sigma}\), \(-S_{\nu\sigma}\), \(-S_{\nu\sigma}\), \(+g_{\sigma\nu}s\). Total \(\tfrac12\big((n-2)S_{\sigma\nu} + sg_{\sigma\nu}\big)\). And \(g^{\rho\mu}\cdot\tfrac12(g_{\rho\mu}g_{\nu\sigma} - g_{\rho\nu}g_{\mu\sigma}) = \tfrac12(n-1)g_{\sigma\nu}\). So \[ g^{\rho\mu}P(S)_{\rho\sigma\mu\nu} = \frac{2}{n-2}\cdot\frac{(n-2)S_{\sigma\nu} + sg_{\sigma\nu}}{2} - \frac{2s}{(n-1)(n-2)}\cdot\frac{(n-1)g_{\sigma\nu}}{2} = S_{\sigma\nu} + \frac{s\,g_{\sigma\nu}}{n-2} - \frac{s\,g_{\sigma\nu}}{n-2} = S_{\sigma\nu}. \] Hence \(\mathrm{Ric}(C) = \mathrm{Ric}(R) - \mathrm{Ric}(P(\mathrm{Ric}(R))) = R_{\sigma\nu} - R_{\sigma\nu} = 0\); the other traces of \(C\) are \(\pm\) this one or zero, by the deck's "only trace" argument, since \(C\) has S1–S4. For that: \(P(S)\) is built from \(h_{\rho\mu}k_{\sigma\nu} + h_{\sigma\nu}k_{\rho\mu} - h_{\rho\nu}k_{\sigma\mu} - h_{\sigma\mu}k_{\rho\nu}\) with \((h,k) = (g,S)\) and \((g,g)\); this has S1, S2, S4 by inspection, and writing out its cyclic sum over \(\sigma\mu\nu\) gives twelve terms that cancel in pairs by the symmetry of \(h\) and \(k\). \(\square\)

2. \(\mathrm{Ric}\circ P = \mathrm{id}\) on symmetric tensors, so \(\mathrm{Ric}\) is onto a space of dimension \(n(n+1)/2\). By rank–nullity and P6, its kernel has dimension \[ \frac{n^2(n^2-1)}{12} - \frac{n(n+1)}{2} = \frac{n(n+1)\big(n(n-1) - 6\big)}{12} = \frac{n(n+1)(n-3)(n+2)}{12}. \] \(n = 3\): \(0\) — so at \(n = 3\) every curvature tensor is \(P(\mathrm{Ric})\), the formula of P6 part 4 (check: at \(n = 3\), \(\tfrac{2}{n-2} = 2\) and \(\tfrac{2}{(n-1)(n-2)} = 1\), which reproduce it term for term). \(n = 4\): \(10\), so \(20 = 10 + 10\). \(n = 5\): \(35\), so \(50 = 15 + 35\).

3. In vacuum \(R_{\mu\nu} = 0\) — the physical input unit 35 supplies, and which the Newtonian limit already shows for the \(00\) component, \(R_{00} = \nabla^2\Phi/c^2 = 0\) outside matter. Then \(P(\mathrm{Ric}) = 0\) and \(C = R\): all of the curvature, tides included, is Weyl. The trace: \(R^i{}_{0i0} = \tfrac{GM}{c^2r^3}(-2 + 1 + 1) = 0\) ✓, the deck's \((-2,1,1)\).

Check. The formula blows up at \(n = 2\), where the Weyl tensor is not defined and the count would give \(2\cdot3\cdot4\cdot(-1)/12 = -2\), a negative dimension: the Ricci map is not onto at \(n = 2\) (it hits only multiples of \(g\), by P8 part 4), which is exactly the step that fails ✓.

Idea. Unit 05 computed the tidal tensor; this unit says it is six components of the Riemann tensor, up to \(c^2\). The ocean's tidal bulges are a measurement of spacetime curvature.

1. \(k = GM_{\text{Moon}}/d^3 = 8.632\times10^{-14}\ \mathrm{s^{-2}}\), and \(k/c^2 = 9.604\times10^{-31}\ \mathrm{m^{-2}}\). \[ R^i{}_{0j0} = \frac{k}{c^2}\begin{pmatrix}-2&0&0\\0&1&0\\0&0&1\end{pmatrix} = \begin{pmatrix}-1.921&0&0\\0&0.960&0\\0&0&0.960\end{pmatrix}\times10^{-30}\ \mathrm{m^{-2}} . \] Radius of curvature \((k/c^2)^{-1/2} = 1.020\times10^{15}\) m \(= 0.108\) light-year.

2. Sub-lunar point, \(\xi = R_\oplus\hat x\): \(\ddot\xi^x = -c^2R^x{}_{0x0}R_\oplus = 2kR_\oplus = 2(8.632\times10^{-14})(6.371\times10^6) = 1.100\times10^{-6}\ \mathrm{m\,s^{-2}}\), positive: away from the Earth's centre, towards the Moon. On the \(y\)-axis: \(\ddot\xi^y = -kR_\oplus = -5.499\times10^{-7}\ \mathrm{m\,s^{-2}}\), towards the centre. Stretch along the Moon line, squeeze across it: two bulges.

3. \(GM_\odot/(1\ \mathrm{AU})^3 = 3.964\times10^{-14}\ \mathrm{s^{-2}}\), so \(2kR_\oplus = 5.051\times10^{-7}\ \mathrm{m\,s^{-2}}\), and the ratio Sun/Moon is \(0.459\). The Sun pulls \(179\) times harder on the Earth as a whole, but its curvature at the Earth is less than half the Moon's, because curvature falls as \(1/d^3\).

Check. Dimensions: \(c^2\cdot\mathrm{m^{-2}}\cdot\mathrm m = \mathrm{m\,s^{-2}}\) ✓. Trace \(-2 + 1 + 1 = 0\): vacuum at the Earth's centre as far as the Moon's field goes ✓. Compare with the Moon's full pull, \(GM/d^2 = 3.32\times10^{-5}\ \mathrm{m\,s^{-2}}\): the tide is \(1/30\) of it, the ratio \(2R_\oplus/d\), since only the difference across the Earth is felt by a freely falling Earth ✓.

Idea. The bare commutator of \(\nabla_X\) and \(\nabla_Y\) detects curvature and also the failure of the fields \(X, Y\) themselves to commute. The bracket term removes the second.

1. \(\nabla_YZ = r^{-2}\partial_\theta\). Then \(\nabla_X(r^{-2}\partial_\theta) = (\partial_r r^{-2})\partial_\theta + r^{-2}\nabla_r\partial_\theta = -2r^{-3}\partial_\theta + r^{-3}\partial_\theta = -r^{-3}\partial_\theta\). And \(\nabla_XZ = \nabla_r\partial_r = 0\), so \(\nabla_Y\nabla_XZ = 0\). The bare commutator is \(-r^{-3}\partial_\theta \ne 0\).

2. \([X,Y]f = \partial_r(r^{-1}\partial_\theta f) - r^{-1}\partial_\theta\partial_r f = -r^{-2}\partial_\theta f\), so \([X,Y] = -r^{-2}\partial_\theta\). Then \(\nabla_{[X,Y]}Z = -r^{-2}\nabla_\theta\partial_r = -r^{-3}\partial_\theta\), and \[ R(X,Y)Z = -r^{-3}\partial_\theta - (-r^{-3}\partial_\theta) = 0 . \]

3. \(X\) and \(Y\) are not coordinate fields — their flows do not commute, because the arc a unit step in \(\theta\) covers depends on \(r\) — and the bare commutator measures that failure as well as any curvature.

Check. Size: \(|r^{-3}\partial_\theta| = r^{-3}\cdot r = r^{-2}\), dimension \(\mathrm L^{-2}\) — that of curvature, which is why the impostor is convincing ✓. With the coordinate fields \(X = \partial_r\), \(Y = \partial_\theta\) instead, the bracket vanishes and the bare commutator is \(\nabla_r(r^{-1}\partial_\theta) - 0 = -r^{-2}\partial_\theta + r^{-2}\partial_\theta = 0\) directly ✓.

Idea. Going round a loop, the zeroth-order effect of \(\Gamma\) cancels (out and back), and what survives is the first-order variation of \(\Gamma V\) integrated against the enclosed signed area — antisymmetric, so only the antisymmetrised derivative survives, which is \(R\).

1. Transport is \(\mathrm dV^\rho = -\Gamma^\rho_{\mu\sigma}V^\sigma\,\mathrm dx^\mu\); integrating round the loop, \(\Delta V^\rho = -\oint\Gamma^\rho_{\mu\sigma}(x)V^\sigma(x)\,\mathrm dx^\mu\). To first order in \(\delta x = x - p\), \(\Gamma(x) = \Gamma + (\partial_\nu\Gamma)\delta x^\nu\) and \(V^\sigma(x) = V^\sigma - \Gamma^\sigma_{\nu\lambda}V^\lambda\delta x^\nu\), everything on the right at \(p\). Multiplying, \[ \Gamma^\rho_{\mu\sigma}(x)V^\sigma(x) = \Gamma^\rho_{\mu\sigma}V^\sigma + X^\rho{}_{\nu\mu}\,\delta x^\nu + O(\delta^2), \qquad X^\rho{}_{\nu\mu} = \big(\partial_\nu\Gamma^\rho_{\mu\sigma} - \Gamma^\rho_{\mu\lambda}\Gamma^\lambda_{\nu\sigma}\big)V^\sigma . \] The constant term gives \(\oint\mathrm dx^\mu = 0\). The \(O(\delta^2)\) terms integrate to \(O(\delta^3)\) since the loop has length \(O(\delta)\). So \(\Delta V^\rho = -X^\rho{}_{\nu\mu}\oint\delta x^\nu\,\mathrm dx^\mu + O(\delta^3)\).

The loop integral, leg by leg (\(\delta x\) is \(t\,\delta a\), then \(\delta a + t\,\delta b\), then \(\delta b + (1-t)\delta a\), then \((1-t)\delta b\), with \(\mathrm dx\) equal to \(\delta a\,\mathrm dt\), \(\delta b\,\mathrm dt\), \(-\delta a\,\mathrm dt\), \(-\delta b\,\mathrm dt\)): \[ \oint\delta x^\nu\mathrm dx^\mu = \tfrac12\delta a^\nu\delta a^\mu + \big(\delta a^\nu + \tfrac12\delta b^\nu\big)\delta b^\mu - \big(\delta b^\nu + \tfrac12\delta a^\nu\big)\delta a^\mu - \tfrac12\delta b^\nu\delta b^\mu = \delta a^\nu\delta b^\mu - \delta b^\nu\delta a^\mu . \] Call it \(S^{\nu\mu}\); it is antisymmetric, so only the antisymmetric part of \(X\) in \(\nu\mu\) survives: \(X^\rho{}_{\nu\mu}S^{\nu\mu} = \tfrac12(X^\rho{}_{\nu\mu} - X^\rho{}_{\mu\nu})S^{\nu\mu}\), and \(X^\rho{}_{\nu\mu} - X^\rho{}_{\mu\nu} = \big(\partial_\nu\Gamma^\rho_{\mu\sigma} - \partial_\mu\Gamma^\rho_{\nu\sigma} + \Gamma^\rho_{\nu\lambda}\Gamma^\lambda_{\mu\sigma} - \Gamma^\rho_{\mu\lambda}\Gamma^\lambda_{\nu\sigma}\big)V^\sigma = R^\rho{}_{\sigma\nu\mu}V^\sigma\). Hence \[ \Delta V^\rho = -\tfrac12R^\rho{}_{\sigma\nu\mu}V^\sigma\big(\delta a^\nu\delta b^\mu - \delta a^\mu\delta b^\nu\big) = -R^\rho{}_{\sigma\nu\mu}V^\sigma\delta a^\nu\delta b^\mu = -R^\rho{}_{\sigma\mu\nu}V^\sigma\delta a^\mu\delta b^\nu, \] using S1 on the second half and renaming dummies. \(\square\)

2. Only \(\mu\nu = \theta\phi\) contributes. With the deck's \(R^\theta{}_{\phi\theta\phi} = \sin^2\theta\) and \(R^\phi{}_{\theta\theta\phi} = -R^\phi{}_{\theta\phi\theta} = -1\): \(\Delta V^\theta = -\sin^2\theta\,V^\phi\,\delta\theta\,\delta\phi\), \(\Delta V^\phi = +V^\theta\,\delta\theta\,\delta\phi\). In orthonormal components \(A = aV^\theta\), \(B = a\sin\theta\,V^\phi\): \[ \Delta A = -\sin\theta\,\delta\theta\,\delta\phi\;B, \qquad \Delta B = +\sin\theta\,\delta\theta\,\delta\phi\;A , \] an anticlockwise rotation in the \((\hat e_\theta,\hat e_\phi)\) plane by \(\epsilon = \sin\theta\,\delta\theta\,\delta\phi\). The area is \(a^2\sin\theta\,\delta\theta\,\delta\phi\) and \(K = 1/a^2\), so \(\epsilon = K\times\text{area}\) ✓. And \(\hat e_\theta\times\hat e_\phi = \hat e_r\), the outward normal, so going first along \(\hat e_\theta\) then \(\hat e_\phi\) is positively oriented, and the rotation is positive — as the deck's octant example found on the large scale.

Check. Dimensions: \(R^\rho{}_{\sigma\mu\nu}\delta a^\mu\delta b^\nu\) is dimensionless in any coordinates, since the two \(\delta\)'s carry the units the two lower indices remove ✓. Flat space: \(R = 0\), no holonomy at second order, and in fact none at all ✓. Reversing the loop swaps \(\delta a\) and \(\delta b\) and flips the sign ✓.