Part E · curved space · spine

Curvature: tidal forces, with the mask off

Two covariant derivatives fail to commute, and the failure is a tensor. That tensor is what no change of coordinates can remove — and, component for component, it is unit 05's tidal tensor.

needs 32, 05~3½ h slides + 7 h problemsproblems: ./mr problems 33

Unit 30 proved a negative: a freely falling frame kills the first derivatives of the metric at a point and cannot kill the second, so gravity is not a force field on flat spacetime. Unit 31 built the space, unit 32 built the derivative. This unit builds the object the first two were for, and then spends its second half showing that the object is not new: it is unit 05's $\partial_i\partial_j\Phi$, wearing the only notation in which it could ever have been a spacetime tensor.

Conventions, fixed here and used to the end of the course

Signature $(-,+,+,+)$ as in unit 26; $x^0 = ct$; Greek indices $\mu,\nu,\rho,\sigma,\lambda$ run over all $n$ coordinates, Latin $i,j,k$ over the spatial ones.
$R^\rho{}_{\sigma\mu\nu} = \partial_\mu\Gamma^\rho_{\nu\sigma} - \partial_\nu\Gamma^\rho_{\mu\sigma} + \Gamma^\rho_{\mu\lambda}\Gamma^\lambda_{\nu\sigma} - \Gamma^\rho_{\nu\lambda}\Gamma^\lambda_{\mu\sigma}$.
$R_{\mu\nu} = R^\rho{}_{\mu\rho\nu}$ and $R = g^{\mu\nu}R_{\mu\nu}$.

With these signs a sphere has positive curvature and positive scalar curvature, and the Einstein equation of unit 35 will read $G_{\mu\nu} = 8\pi G\,T_{\mu\nu}/c^4$ with no stray minus. Other books differ; every published sign convention is one of eight, and the only defence is to state yours and check it on the sphere, which is what the first worked example does.

How to read this deck

One example runs through every section: the round two-sphere of radius $a$. Section 1 computes its curvature, section 2 counts how many numbers that took, section 3 identifies the number as $1/a^2$ and proves no map of it can be flat, section 4 lets two of its geodesics fall together, section 5 takes the traces and finds that in two dimensions the Einstein tensor is identically zero. Proofs are on the slides where they are short, in these notes under Full proof where they are not, and every result not proved here is marked cited with where it is proved.

33 · overview

What this unit is for

1 · the object

$[\nabla_\mu,\nabla_\nu]$ is not zero, and is not a differential operator: it is multiplication by a tensor. That tensor is the Riemann curvature tensor.

2 · how big it is

Four symmetries cut $n^4$ components to $n^2(n^2-1)/12$: twenty in four dimensions, six in three, one in two.

3 · what it measures

The angle a vector turns through when carried round a loop, and — Gauss — a quantity no bending of the surface in space can alter.

4 · what it is

Two freely falling particles accelerate apart at a rate that is curvature. In the Newtonian limit that rate is $\partial_i\partial_j\Phi$.

getting around

A dotted underline is a glossary term: click it. G lists every term, N opens the notes, where the long proofs live. The round sphere of radius $a$, solid-underlined like that, runs through all five sections; its octant triangle and two of its meridians do the work.

section 1

The thing no chart removes

Second derivatives that refuse to commute, and the tensor that measures by how much.

33 · definition

Where to look for it

Unit 32's normal coordinates at $p$ gave $g_{\mu\nu}(p) = \eta_{\mu\nu}$ and $\partial_\lambda g_{\mu\nu}(p) = 0$, hence $\Gamma^\rho_{\mu\nu}(p) = 0$ — and could not be pushed to the second derivatives. Count why:

Order by order in the Taylor expansion about $p$: the derivatives of $g$ to be set to their flat values (purple) against the coefficients a change of chart offers (orange). Enough at orders 0 and 1 — exactly enough at order 1 — and $20$ short at order 2.

At second order the chart runs out of freedom by $100 - 80 = 20$. Twenty numbers at each point cannot be set to zero, and this unit is about those twenty.

the count is a promise, not a proof

It says something survives; it does not say what, nor that the survivor is a tensor. Both have to be produced. The honest route is not to count but to find an operator that is manifestly coordinate-free and see what it contains.

The numbers: $g$ symmetric in $n=4$ gives $10$; its first derivatives $10\times4 = 40$; its seconds $10\times10 = 100$ using Clairaut in the two derivative slots. A chart change $x \mapsto x'$ has $4\times4 = 16$ first coefficients, $4\times10 = 40$ second, $4\times20 = 80$ third.

Where each number comes from. Expand the new chart about $p$ as a Taylor series. The linear coefficients $\partial x'^\mu/\partial x^\nu$ number $n^2 = 16$ and are used to set $g_{\mu\nu}(p) = \eta_{\mu\nu}$ — that takes $10$ of them, leaving $6$, which is exactly $\dim\mathrm{SO}(1,3)$: the residual Lorentz freedom at a point, unit 26's group appearing as the stabiliser of a point of spacetime. The quadratic coefficients are symmetric in their two lower indices, $n\cdot\tfrac{n(n+1)}2 = 40$, and are used to kill the $40$ first derivatives of $g$ — exact fit, which is why normal coordinates exist and why the equivalence principle is true. The cubic coefficients are symmetric in three lower indices, $n\cdot\binom{n+2}{3} = 4\cdot20 = 80$, against $100$ second derivatives. Shortfall $20$.

That $20$ is the same $20$ the counting theorem of section 2 will produce from the symmetries of the Riemann tensor, by a completely different argument. When two independent counts agree it is usually because they are counting the same thing, and here they are: the second derivatives of the metric that no chart removes are precisely the components of the curvature.

Riemann's own route (Habilitationsvortrag, Göttingen 1854, with Gauss in the audience) was a count of this kind, done for general $n$: how many functions a metric carries once the freedom of coordinates is spent. The tensor was written down properly only in his 1861 Paris prize essay, unpublished in his lifetime; Christoffel (1869) and then Ricci and Levi-Civita (1900) gave it the calculus it needed, which Einstein learnt from Grossmann in 1912.

33 · definition

The curvature operator

Mathematics first. Let $\nabla$ be unit 32's Levi-Civita connection on a manifold $M$, and let $\mathfrak X(M)$ be its vector fields.

the curvature operator

$R : \mathfrak X(M)^3 \to \mathfrak X(M)$,

$$R(X,Y)Z := [\nabla_X,\nabla_Y]Z - \nabla_{[X,Y]}Z ,$$

where $[\nabla_X,\nabla_Y]Z = \nabla_X\nabla_YZ - \nabla_Y\nabla_XZ$.

Every term is a vector field, so $R(X,Y)Z$ is one, with no coordinates anywhere. The bracket term is unit 31's Lie bracket, and it is there to remove the part of the failure that is merely the two flows not commuting.

and why not the obvious thing

"How much the surface bends in space" is not available: a spacetime sits in nothing. Any definition of curvature this course can use must be built from $g$ and $\nabla$ alone. Section 3 proves — Gauss — that nothing is lost.

The next slide draws what the commutator compares, and why the bracket term has to be there.

33 · definition

Why a commutator: two routes round a parallelogram

why a commutator, physically

Unit 32 made $\nabla$ the derivative that compares a vector here with a vector transported from there. Carrying $Z$ along $X$ then $Y$, against $Y$ then $X$, compares two transports round the two sides of a parallelogram. The difference is the failure of parallel transport to be path-independent — which is what "curved" is supposed to mean.

If the flows of $X$ and $Y$ do not commute, the two routes do not even end at the same point: the gap is $[X,Y]$, and part of the mismatch in $Z$ is just $Z$ changing across that gap. The term $-\nabla_{[X,Y]}Z$ subtracts exactly that part.

Two routes from $p$, each with $Z$ carried along it. They end a gap $[X,Y]$ apart (exaggerated here), and the two carried copies of $Z$ disagree. Close the gap and what remains of the disagreement is $R(X,Y)Z$, times the area enclosed.

Reading the figure. Going along $X$ for a small parameter $\varepsilon$ and then along $Y$ for $\varepsilon$ lands at a different point from going along $Y$ then $X$; to second order the separation is $\varepsilon^2[X,Y]$, which is unit 31's description of the Lie bracket as the failure of two flows to close a square. The carried vectors differ at order $\varepsilon^2$ too, for two reasons: the endpoints differ, and the transport is path-dependent. The definition of $R$ keeps the second and subtracts the first.

Why the bracket term is needed. Take flat $\mathbb R^2$ in polar coordinates, where the answer must be zero. Set $X = \partial_r$ and $Y = \partial_\theta$; these commute, so the bracket term vanishes and $R$ is the bare commutator. But take instead $X = \partial_r$ and $Y = r^{-1}\partial_\theta$, the orthonormal frame. Now $[X,Y] = -r^{-2}\partial_\theta \ne 0$, and the bare commutator $\nabla_X\nabla_YZ - \nabla_Y\nabla_XZ$ is not zero. The third term is exactly what cancels it. Without it the definition would depend on the frame, which is to say it would not be a definition at all.

33 · definition

It is a tensor, and that is the surprise

theorem · tensoriality

$R(X,Y)Z$ is $C^\infty(M)$-linear in each of $X$, $Y$, $Z$: for every smooth $f$, $R(fX,Y)Z = R(X,fY)Z = R(X,Y)(fZ) = f\,R(X,Y)Z$. Hence $R$ is a $(1,3)$-tensor field: its value at $p$ depends on $X,Y,Z$ only through $X_p, Y_p, Z_p$.

proof, the $Z$ slot

$\nabla_X\nabla_Y(fZ) = \nabla_X\big(f\nabla_YZ + (Yf)Z\big)$ $= f\nabla_X\nabla_YZ + (Xf)\nabla_YZ + (Yf)\nabla_XZ + (XYf)Z$.

Subtract the same with $X \leftrightarrow Y$: the two middle terms are symmetric and cancel, leaving $f[\nabla_X,\nabla_Y]Z + (XYf - YXf)Z$. And $XYf - YXf = [X,Y]f$, which is exactly what $-\nabla_{[X,Y]}(fZ)$ subtracts. $\square$

what would have gone wrong

$\nabla_XZ$ alone is not $C^\infty$-linear in $Z$ — the Leibniz term $(Xf)Z$ is the whole point of a connection. Only in the antisymmetrised pair does it cancel. Curvature is a second derivative of the metric that behaves like a tensor, and nothing weaker does.

full proof in the notes — the $X$ and $Y$ slots, and the lemma that turns $C^\infty$-linearity into pointwise dependence.

Full proof

The $Y$ slot. $\nabla_X\nabla_{fY}Z - \nabla_{fY}\nabla_XZ - \nabla_{[X,fY]}Z$. First term: $\nabla_X(f\nabla_YZ) = f\nabla_X\nabla_YZ + (Xf)\nabla_YZ$. Second: $f\nabla_Y\nabla_XZ$. Third: the bracket obeys $[X,fY] = f[X,Y] + (Xf)Y$ (unit 31), so $\nabla_{[X,fY]}Z = f\nabla_{[X,Y]}Z + (Xf)\nabla_YZ$. The two $(Xf)\nabla_YZ$ terms cancel and $f\,R(X,Y)Z$ remains. The $X$ slot follows from $R(X,Y)Z = -R(Y,X)Z$, which is immediate from the definition since $[Y,X] = -[X,Y]$.

From $C^\infty$-linearity to a tensor

Lemma (the tensor characterisation). If $T : \mathfrak X(M)^k \to \mathfrak X(M)$ is $C^\infty(M)$-linear in each slot, then $T(X_1,\dots,X_k)|_p$ depends only on $X_1|_p,\dots,X_k|_p$. Proof. By multilinearity it is enough to show that $X_1|_p = 0$ forces $T(\dots)|_p = 0$. Write $X_1 = \sum_\mu X_1^\mu\partial_\mu$ in a chart around $p$, with $X_1^\mu(p) = 0$; patch the chart fields to global fields with a bump function $\chi$ equal to $1$ near $p$ (unit 31's partitions of unity are built from exactly these). Then near $p$, $X_1 = \sum_\mu X_1^\mu\,(\chi\partial_\mu)$, so $T(X_1,\dots)|_p = \sum_\mu X_1^\mu(p)\,T(\chi\partial_\mu,\dots)|_p = 0$. $\square$

This lemma is the whole reason the words "is a tensor" can be earned rather than asserted. It is also the cleanest illustration of unit 16's insistence that a tensor is a multilinear map: linearity over functions, not merely over numbers, is exactly the algebraic shadow of "pointwise".

The trap this closes. The Christoffel symbols have three indices and are not a tensor; $\partial_\mu\Gamma^\rho_{\nu\sigma}$ has four and is not a tensor either. The particular combination on the next slide is. Nothing about the index pattern could have told you that — only the invariant definition above could.

33 · components

Components, and the Ricci identity

Take $X = \partial_\mu$, $Y = \partial_\nu$ — coordinate fields, so $[X,Y] = 0$ and the bracket term drops. Writing $R(\partial_\mu,\partial_\nu)\partial_\sigma = R^\rho{}_{\sigma\mu\nu}\,\partial_\rho$ defines the components, and expanding gives

the Riemann curvature tensor in a chart $$R^\rho{}_{\sigma\mu\nu} = \partial_\mu\Gamma^\rho_{\nu\sigma} - \partial_\nu\Gamma^\rho_{\mu\sigma} + \Gamma^\rho_{\mu\lambda}\Gamma^\lambda_{\nu\sigma} - \Gamma^\rho_{\nu\lambda}\Gamma^\lambda_{\mu\sigma}$$

Equivalently, acting on any vector field — the Ricci identity:

Ricci identity $$\begin{aligned}[\nabla_\mu,\nabla_\nu]V^\rho &= R^\rho{}_{\sigma\mu\nu}\,V^\sigma\\ [\nabla_\mu,\nabla_\nu]\omega_\rho &= -R^\sigma{}_{\rho\mu\nu}\,\omega_\sigma\end{aligned}$$

mnemonic, and its limits

Read $\partial\Gamma - \partial\Gamma + \Gamma\Gamma - \Gamma\Gamma$ with the last two lower slots antisymmetrised, and the differentiated $\Gamma$ carrying the other two. It is a mnemonic for these conventions only: swap any sign convention of unit 32 and it changes.

sanity

On a function, $[\nabla_\mu,\nabla_\nu]f = \nabla_\mu\partial_\nu f - \nabla_\nu\partial_\mu f = -(\Gamma^\lambda_{\mu\nu}-\Gamma^\lambda_{\nu\mu})\partial_\lambda f = 0$: torsion-free means scalars never see curvature. Curvature needs something with an index to rotate.

Full proof of the component formula

$\nabla_\nu\partial_\sigma = \Gamma^\lambda_{\nu\sigma}\partial_\lambda$. Apply $\nabla_\mu$: $\nabla_\mu\nabla_\nu\partial_\sigma = (\partial_\mu\Gamma^\lambda_{\nu\sigma})\partial_\lambda + \Gamma^\lambda_{\nu\sigma}\Gamma^\rho_{\mu\lambda}\partial_\rho$. Rename $\lambda\to\rho$ in the first term and read off the $\partial_\rho$ coefficient: $\partial_\mu\Gamma^\rho_{\nu\sigma} + \Gamma^\rho_{\mu\lambda}\Gamma^\lambda_{\nu\sigma}$. Antisymmetrise in $\mu\nu$ and the boxed formula appears. $\square$

Full proof of the Ricci identity

For a vector field, $\nabla_\mu\nabla_\nu V^\rho = \partial_\mu(\nabla_\nu V^\rho) - \Gamma^\lambda_{\mu\nu}\nabla_\lambda V^\rho + \Gamma^\rho_{\mu\lambda}\nabla_\nu V^\lambda$ — note the middle term, which is there because $\nabla_\nu V^\rho$ is a $(1,1)$-tensor and its lower index must be differentiated too. Expand and antisymmetrise in $\mu\nu$. The middle term is symmetric in $\mu\nu$ (torsion-free) and drops out. Every term carrying a derivative of $V$ cancels against its partner, and what is left is $\big(\partial_\mu\Gamma^\rho_{\nu\sigma} - \partial_\nu\Gamma^\rho_{\mu\sigma} + \Gamma^\rho_{\mu\lambda}\Gamma^\lambda_{\nu\sigma} - \Gamma^\rho_{\nu\lambda}\Gamma^\lambda_{\mu\sigma}\big)V^\sigma$. That the derivatives of $V$ cancel is the previous slide's theorem, made visible.

The covector form follows by applying $[\nabla_\mu,\nabla_\nu]$ to the scalar $\omega_\rho V^\rho$, which must give zero, and using Leibniz. For a general $(p,q)$-tensor there is one $+R$ term per upper index and one $-R$ term per lower index; problem P7 asks for the $(0,2)$ case, which is what proves the first symmetry in section 2.

Naming. "Ricci identity" is standard and slightly unfair: the identity is Riemann's, and what Ricci supplied was the absolute differential calculus in which it can be written down. The trace named after him is on a later slide.

33 · example

Worked example 1 — the curvature of a sphere

every step

On $S^2_a$, unit 32's Christoffel symbols are $\Gamma^\theta_{\phi\phi} = -\sin\theta\cos\theta$ and $\Gamma^\phi_{\theta\phi} = \Gamma^\phi_{\phi\theta} = \cot\theta$; every other one vanishes. Compute $R^\theta{}_{\phi\theta\phi}$, taking $\rho = \theta$, $\sigma = \phi$, $\mu = \theta$, $\nu = \phi$:

$$R^\theta{}_{\phi\theta\phi} = \underbrace{\partial_\theta\Gamma^\theta_{\phi\phi}}_{\text{(i)}} - \underbrace{\partial_\phi\Gamma^\theta_{\theta\phi}}_{\text{(ii)}} + \underbrace{\Gamma^\theta_{\theta\lambda}\Gamma^\lambda_{\phi\phi}}_{\text{(iii)}} - \underbrace{\Gamma^\theta_{\phi\lambda}\Gamma^\lambda_{\theta\phi}}_{\text{(iv)}}$$

(i) $\partial_\theta(-\sin\theta\cos\theta) = \sin^2\theta - \cos^2\theta$. (ii) $\Gamma^\theta_{\theta\phi} = 0$, so this is $0$. (iii) $\Gamma^\theta_{\theta\lambda} = 0$ for both $\lambda$, so this is $0$. (iv) the only surviving $\lambda$ is $\phi$: $\Gamma^\theta_{\phi\phi}\Gamma^\phi_{\theta\phi} = (-\sin\theta\cos\theta)(\cot\theta) = -\cos^2\theta$.

$$R^\theta{}_{\phi\theta\phi} = \sin^2\theta - \cos^2\theta - 0 + 0 + \cos^2\theta = \sin^2\theta$$

Lower the first index: $R_{\theta\phi\theta\phi} = g_{\theta\theta}R^\theta{}_{\phi\theta\phi} = a^2\sin^2\theta$.

checks

Dimensions. $[R_{\theta\phi\theta\phi}] = \mathrm L^2$ because $\theta,\phi$ are angles; the invariant $R_{\theta\phi\theta\phi}/(g_{\theta\theta}g_{\phi\phi}) = 1/a^2$ has $\mathrm L^{-2}$ ✓ — curvature is an inverse area.

Limit $a\to\infty$. $1/a^2 \to 0$: a sphere of huge radius is locally flat ✓. Limit $\theta\to0$. $R_{\theta\phi\theta\phi}\to0$, but so does $g_{\phi\phi}$; the ratio stays $1/a^2$ ✓ — the pole is a coordinate singularity, not a curvature one, exactly as unit 31 warned.

one component, not sixteen

$n = 2$ gives $2^4 = 16$ components, and we computed one. Section 2 proves that in two dimensions all sixteen are determined by that one — so this is the curvature of the sphere, finished.

Where the Christoffel symbols came from (unit 32's geodesic-Lagrangian trick, recomputed here so the example is self-contained). $L = \tfrac12 a^2(\dot\theta^2 + \sin^2\theta\,\dot\phi^2)$. The $\theta$ equation: $a^2\ddot\theta - a^2\sin\theta\cos\theta\,\dot\phi^2 = 0$, so $\ddot\theta - \sin\theta\cos\theta\,\dot\phi^2 = 0$, giving $\Gamma^\theta_{\phi\phi} = -\sin\theta\cos\theta$. The $\phi$ equation: $\tfrac{\mathrm d}{\mathrm ds}(a^2\sin^2\theta\,\dot\phi) = 0$, i.e. $\ddot\phi + 2\cot\theta\,\dot\theta\dot\phi = 0$, and since the geodesic equation reads $\ddot x^\rho + \Gamma^\rho_{\mu\nu}\dot x^\mu\dot x^\nu = 0$ with the symmetric sum counting the cross term twice, $\Gamma^\phi_{\theta\phi} = \Gamma^\phi_{\phi\theta} = \cot\theta$.

The other components. $R^\phi{}_{\theta\phi\theta} = 1$ by the same computation with $\theta$ and $\phi$ exchanged; lowering gives $R_{\phi\theta\phi\theta} = g_{\phi\phi}\cdot1 = a^2\sin^2\theta$, equal to $R_{\theta\phi\theta\phi}$, as the pair symmetry of section 2 demands. Everything with a repeated index in either antisymmetric pair vanishes.

A cross-check without any Christoffel symbols is on the conformal-curvature slide in section 3: writing the same sphere in unit 31's stereographic chart, where $g = 4a^2(\mathrm du^2+\mathrm dv^2)/(1+u^2+v^2)^2$, the formula $K = -e^{-2\varphi}\Delta\varphi$ returns $1/a^2$ in four lines. Two charts, two methods, one number — which is what it means for curvature to be a property of the sphere rather than of the description.

33 · transport

Curvature is the holonomy of a small loop

theorem · infinitesimal holonomy

Parallel-transport $V$ round the coordinate parallelogram with edges $\delta a^\mu$ and $\delta b^\nu$ — along $\delta a$, then $\delta b$, then back. The vector returns changed by

$$\Delta V^\rho = -R^\rho{}_{\sigma\mu\nu}\,V^\sigma\,\delta a^\mu\,\delta b^\nu + O(\delta^3) .$$

So curvature is not merely like path-dependence of transport: it is its leading coefficient, per unit coordinate area. A vanishing $R$ makes transport path-independent on a simply connected region (cited in the notes), and a vector field defined by transporting from one point is then well defined — which is what "flat" means.

two indices are a plane, not a direction

$\delta a^\mu\delta b^\nu$ enters only through its antisymmetric part, $R$ being antisymmetric in $\mu\nu$. Curvature is indexed by planes: it answers "go round this two-dimensional loop", never "go in this direction".

Out along $\delta a$, along $\delta b$, back and back, carrying $V$ the whole way (the turn is exaggerated). The mismatch $\Delta V$ on return is proportional to the enclosed area $\delta a\wedge\delta b$: halve both edges and it drops by four.

Reading the figure. The parallelogram is drawn in the coordinate plane of $x^\mu$ and $x^\nu$; on the manifold its sides are short curves. $V$ is carried round it by parallel transport, which has no notion of "keeping $V$ pointing the same way" except the one $\nabla$ supplies, and on return it is turned. The sign and size depend on the plane — the warning box.

Full proof

Parallel transport along a curve is $\mathrm dV^\rho/\mathrm d\lambda = -\Gamma^\rho_{\mu\sigma} \dot x^\mu V^\sigma$ (unit 32). Integrating once round the closed loop, $\Delta V^\rho = -\oint \Gamma^\rho_{\mu\sigma}V^\sigma\,\mathrm dx^\mu$, where $V$ inside the integral is itself the transported vector, so the equation is implicit and must be iterated.

To second order in the edge lengths, expand about the starting corner $p$: $\Gamma^\rho_{\mu\sigma}(x) = \Gamma^\rho_{\mu\sigma} + (\partial_\nu\Gamma^\rho_{\mu\sigma})\,\delta x^\nu$ and $V^\sigma(x) = V^\sigma - \Gamma^\sigma_{\nu\lambda}V^\lambda\,\delta x^\nu$, both evaluated at $p$. Then $\Delta V^\rho = -\big(\partial_\nu\Gamma^\rho_{\mu\sigma} - \Gamma^\rho_{\mu\lambda}\Gamma^\lambda_{\nu\sigma}\big)V^\sigma\oint \delta x^\nu\,\mathrm dx^\mu$. For the parallelogram, $\oint \delta x^\nu\,\mathrm dx^\mu = \delta a^\nu\delta b^\mu - \delta a^\mu\delta b^\nu$ — the signed coordinate area, antisymmetric. Contracting the antisymmetric part of the bracket with it gives exactly $-R^\rho{}_{\sigma\mu\nu}V^\sigma\delta a^\mu\delta b^\nu$. $\square$ (Problem P12 walks through this with every index in place.)

The converse direction. That $R = 0$ on a simply connected open set implies transport is path-independent, and hence that there is a chart with $g_{\mu\nu}$ constant, is the flatness theorem. It is a Frobenius-integrability argument, unit 31's theorem applied to the distribution of parallel frames; we take it as cited — Lee, Introduction to Riemannian Manifolds (2nd ed., 2018), Thm 7.10. Simple connectivity is not decoration: the cone in the notes to worked example 3, and of problem P1, has $R = 0$ everywhere off its apex and a nonzero holonomy round it.

33 · transport

In two dimensions: the angle is the integral of $K$

theorem · transport round a region

On an oriented surface, parallel transport once round the positively oriented boundary of a region $S$ that is a topological disc rotates every tangent vector by the same angle

$$\Delta\alpha = \int_S K\,\mathrm dA ,$$

where $K := R_{1212}/\det g$ is the Gaussian curvature. Worked example 1 found it on the sphere: $R_{\theta\phi\theta\phi}/(g_{\theta\theta}g_{\phi\phi}) = 1/a^2$. Section 3 proves it is chart-independent and is the surface's whole curvature.

Additivity does the rest: a big region is a union of small ones, each turning $V$ by $K\,\delta A$ (the previous slide), and the internal edges are traversed twice in opposite senses and cancel.

Cut $S$ into small cells and go round each one anticlockwise. Every inner edge is walked once each way, so its contributions cancel; what survives is one trip round the boundary $\partial S$, with $S$ on the left.

From the infinitesimal statement to the integral one. In two dimensions the holonomy slide's $\Delta V^\rho = -R^\rho{}_{\sigma\mu\nu}V^\sigma\delta a^\mu\delta b^\nu$ has only one independent component of $R$ to offer. Writing $R_{\rho\sigma\mu\nu} = K(g_{\rho\mu}g_{\sigma\nu} - g_{\rho\nu}g_{\sigma\mu})$ — the form proved in section 3, on In two dimensions there is only one plane, and used here ahead of its proof — and contracting with an oriented area element $\delta A^{\mu\nu} = \delta a^\mu\delta b^\nu - \delta a^\nu\delta b^\mu$ turns $\Delta V$ into $K\,\delta A$ times the rotation generator $\epsilon^\rho{}_\sigma$. That is a rotation by $K\,\delta A$. Summing over a triangulation of $S$ and letting the mesh go to zero gives the integral, the interior edges cancelling in pairs because each is traversed once in each direction. That the small rotations may simply be added uses one more fact: each small loop's holonomy is carried to the base point by transport along a path, which conjugates it by a rotation, and in $\mathrm{SO}(2)$, which is abelian, conjugation changes no angle. Composing the small loops in order therefore adds their angles. In higher dimensions the holonomy group is not abelian and this step is exactly what fails.

Why $K$ is defined as a ratio. $R_{1212}$ alone changes when the chart does — on the sphere it is $a^2\sin^2\theta$, which vanishes at the pole for no geometric reason. Dividing by $\det g$, which changes by the same factor, removes the chart; section 3 makes this precise.

33 · transport

Why one angle describes it, and its sign

why a rotation

Metric compatibility (unit 32) says transport preserves $g$, so the holonomy of a loop is an isometry of the tangent plane fixing the origin: an element of $\mathrm O(2)$. Transport along the loop up to time $t$ has a determinant, measured in oriented bases, that is continuous in $t$, equal to $\pm1$, and $1$ at $t=0$; so it stays $1$, and the holonomy lies in $\mathrm{SO}(2)$ — a rotation, by an angle.

orientation is not optional

Reverse the loop and $\Delta\alpha$ changes sign. "Positively oriented boundary" means the region is on the left. Get this wrong and a sphere appears to have negative curvature.

So all a surface's holonomy can say about a loop is one signed angle, the same for every vector carried round it.

Why "every tangent vector by the same angle". A rotation of the plane turns every vector through the same angle; that is what distinguishes it from a general linear map. So once the holonomy is known to lie in $\mathrm{SO}(2)$, one vector carried round the loop tells you what happens to all of them — which is why worked example 2 carries only one.

This is local Gauss–Bonnet. The familiar global statement, $\int_M K\,\mathrm dA = 2\pi\chi(M)$ for a closed surface, and the version with geodesic curvature along the boundary, are cited: do Carmo, Differential Geometry of Curves and Surfaces, §4-5. What we prove here — holonomy equals enclosed curvature — is the piece the physics needs, and it is the piece that generalises: in $n$ dimensions "rotation by an angle" becomes "an element of the holonomy group", and the Ambrose–Singer theorem says its Lie algebra is generated by the curvature. Also cited.

The historical order is the reverse of this one. Gauss proved the theorem about triangles; Levi-Civita invented parallel transport in 1917, ninety years later, and only then did it become obvious why the theorem was true.

33 · example

Worked example 2 — round the octant, by hand

transport, leg by leg

Take $a = 1$ and embed to keep the bookkeeping honest: $A = (1,0,0)$, $B = (0,1,0)$ on the equator, $N = (0,0,1)$. Traverse the octant $A\to B\to N\to A$, which is positively oriented. Start with $V = \hat y$ at $A$ — due east. On a great circle the travel direction is itself parallel-transported, and transport preserves the angle to it; that is all we need.

Leg $A\to B$ (equator, a geodesic). $V = \hat y$ is the travel direction, so it stays tangent: at $B$ the travel direction is $-\hat x$, so $V = -\hat x$.

Leg $B\to N$ (meridian $\phi = \tfrac\pi2$, in the $yz$-plane). Travel direction at $B$ is $\hat z$; $V = -\hat x$ is perpendicular to it and perpendicular to the plane of the great circle, so it is unmoved along the whole leg. At $N$, $V = -\hat x$.

Leg $N\to A$ (meridian $\phi = 0$, in the $xz$-plane). Travel direction at $N$ is $+\hat x$, so $V = -\hat x$ is anti-parallel to it and stays so. At $A$ the travel direction is $-\hat z$, hence $V = +\hat z$ — due north.

Schematic. Flat paper cannot hold a triangle with three right angles unless its sides bend — here by $30^\circ$ each. Along each side $V$ keeps its angle to the side, exactly the rule used on the sphere, so it turns with the sides: $3\times30^\circ = 90^\circ$, east to north.

Why "perpendicular to the plane of the great circle" settles leg 2. A great circle through $B$ and $N$ lies in the $yz$-plane. Along it, the surface normal is the position vector, which also lies in that plane, and so does the travel direction. Hence $\pm\hat x$ spans the tangent direction orthogonal to the travel direction at every point of the leg. A transported vector keeps its length and its angle to the (transported) travel direction, so a vector starting at exactly $-\hat x$ stays at exactly $-\hat x$ throughout.

Reading the figure. On the sphere the three sides are geodesics and do not bend at all; the bending is what a flat drawing must add to fit three right angles in. A flat triangle's tangent turns through $2\pi$ in all, so with three exterior angles of $\pi/2$ each the sides must make up the remaining $\pi/2$ between them. That missing $\pi/2$ is the angle excess, and since $V$ keeps its angle to each side, it is also the rotation of $V$.

33 · example

The octant, three ways to one number

both sides of the theorem

Holonomy. East $\to$ north is a rotation by $+\pi/2$ in the outward-normal orientation ✓.

Integral. $\displaystyle\int_\triangle K\,\mathrm dA = \frac{1}{a^2}\cdot\frac{4\pi a^2}{8} = \frac{\pi}{2}$ ✓. Independent of $a$, as the angle must be.

Angle excess. Three right angles: $\tfrac{3\pi}{2}-\pi = \tfrac\pi2$ ✓, the same number a third way.

and a real one

Gauss's own survey triangle, Hoher Hagen–Brocken–Inselsberg, sides $69$, $85$, $107$ km, area $2929\ \mathrm{km^2}$: excess $A/R_\oplus^2 = 7.22\times10^{-5}$ rad $= 14.9''$. Small, and larger than his instruments' error.

the excess is the curvature, not the size

The octant's excess is $\pi/2$ on a sphere of any radius; Gauss's triangle has $14.9''$ only because it covers $2929/(4\pi R_\oplus^2)$ of the Earth, about $5.7\times10^{-6}$ of it. Excess is $K$ times area.

Reading the final rotation. At $A = (1,0,0)$ the outward normal is $\hat x$, and the positively oriented tangent basis is $(\hat y,\hat z)$, since $\hat y\times\hat z = \hat x$. A positive rotation carries $\hat y\to\hat z$, i.e. east $\to$ north. We started at $\hat y$ and finished at $\hat z$: $+\pi/2$ exactly.

Traverse it the other way ($A\to N\to B\to A$) and the same leg-by-leg reasoning carries $V = \hat y$ to $\hat y$ at $N$, to $-\hat z$ at $B$, and to $-\hat z$ back at $A$: east has become south, a rotation by $-\pi/2$. The theorem's orientation clause is doing real work, not tidying.

The number 14.9 arcseconds. Heron on the plane triangle with sides 69, 85, 107 km gives $2929\ \mathrm{km^2}$; dividing by $R_\oplus^2 = (6371\ \mathrm{km})^2$ gives $7.217\times10^{-5}$ rad, and $\times(180/\pi)\times3600 = 14.89''$. Gauss's triangulation of Hanover (1821–25) is where he made the measurements that preceded the Disquisitiones circa superficies curvas of 1827. The once-common claim that he was testing whether physical space is Euclidean is not supported by his papers; the excess is a geodetic correction he needed, and it is the one he computed.

section 2

How many numbers is that?

Four symmetries, one count, and the two identities that everything later leans on.

33 · symmetries

Four symmetries

Work with all indices down, $R_{\rho\sigma\mu\nu} = g_{\rho\lambda}R^\lambda{}_{\sigma\mu\nu}$.

the symmetries

S1 $\;R_{\rho\sigma\mu\nu} = -R_{\rho\sigma\nu\mu}$
S2 $\;R_{\rho\sigma\mu\nu} = -R_{\sigma\rho\mu\nu}$
S3 $\;R_{\rho[\sigma\mu\nu]} = 0$, the first Bianchi identity
S4 $\;R_{\rho\sigma\mu\nu} = R_{\mu\nu\rho\sigma}$, pair symmetry

S1 is the definition ($R(X,Y) = -R(Y,X)$). S4 is not independent: it follows from S1–S3.

proof of S2

Apply the Ricci identity to the metric itself. For a $(0,2)$-tensor, $[\nabla_\mu,\nabla_\nu]T_{\rho\sigma} = -R^\lambda{}_{\rho\mu\nu}T_{\lambda\sigma} - R^\lambda{}_{\sigma\mu\nu}T_{\rho\lambda}$. Put $T = g$. Metric compatibility gives $\nabla g = 0$, so the left side is zero: $$0 = -R_{\sigma\rho\mu\nu} - R_{\rho\sigma\mu\nu} . \;\square$$

S2 is about the metric, S1 is not

S1 and S3 hold for any torsion-free connection. S2 needs $\nabla g = 0$ — so a connection that is not metric has a curvature with more independent components than the counting theorem allows.

Notation. Square brackets are antisymmetrisation with the $1/k!$: $T_{[\mu\nu]} = \tfrac12(T_{\mu\nu}-T_{\nu\mu})$ and $R_{\rho[\sigma\mu\nu]} = \tfrac16$(six signed terms), so S3 is the cyclic statement $R_{\rho\sigma\mu\nu}+R_{\rho\mu\nu\sigma}+R_{\rho\nu\sigma\mu}=0$ once S1 is used to reduce six terms to three.

Full proof of the $(0,2)$ Ricci identity used above

Let $\omega$ and $\chi$ be covector fields and consider $T = \omega\otimes\chi$. The scalar identity $[\nabla_\mu,\nabla_\nu]f = 0$ and the vector identity give, by Leibniz, $[\nabla_\mu,\nabla_\nu](\omega_\rho\chi_\sigma) = ([\nabla_\mu,\nabla_\nu]\omega_\rho)\chi_\sigma + \omega_\rho([\nabla_\mu,\nabla_\nu]\chi_\sigma)$ — the cross terms $(\nabla_\mu\omega_\rho)(\nabla_\nu\chi_\sigma)$ appear twice with opposite signs and cancel. Substituting the covector form $[\nabla_\mu,\nabla_\nu]\omega_\rho = -R^\lambda{}_{\rho\mu\nu}\omega_\lambda$ gives the stated result on decomposable tensors, and every $(0,2)$-tensor is a sum of those pointwise. $\square$

How much S2 buys. S1 alone leaves $n^2\cdot\tfrac{n(n-1)}2$ components — $96$ in four dimensions. S1 and S2 together leave $\big(\tfrac{n(n-1)}2\big)^2 = 36$. It is S4 and then S3 that bring it down to $20$.

33 · symmetries

The first Bianchi identity, proved in three lines

proof of S3

Both sides of S3 are tensors, so it is enough to prove it at one arbitrary point $p$ in one chart. Choose unit 32's normal coordinates at $p$, where $\Gamma^\rho_{\mu\nu}(p) = 0$ but $\partial\Gamma(p)$ need not vanish. There the $\Gamma\Gamma$ terms drop and

$$R^\rho{}_{\sigma\mu\nu}\big|_p = \partial_\mu\Gamma^\rho_{\nu\sigma} - \partial_\nu\Gamma^\rho_{\mu\sigma}.$$

Sum cyclically over $(\sigma,\mu,\nu)$:

$$\begin{aligned}&\big(\partial_\mu\Gamma^\rho_{\nu\sigma} - \partial_\nu\Gamma^\rho_{\mu\sigma}\big) + \big(\partial_\nu\Gamma^\rho_{\sigma\mu} - \partial_\sigma\Gamma^\rho_{\nu\mu}\big)\\ &\quad + \big(\partial_\sigma\Gamma^\rho_{\mu\nu} - \partial_\mu\Gamma^\rho_{\sigma\nu}\big) = 0,\end{aligned}$$

every term cancelling its partner because $\Gamma^\rho_{\mu\nu} = \Gamma^\rho_{\nu\mu}$ — torsion-freeness, and nothing else. $\square$

the technique, worth naming

"Tensor equation, so prove it in normal coordinates at a point" is the standard move and it is valid: if two tensors agree at $p$ in one basis they agree at $p$ in every basis, and $p$ was arbitrary. It will prove the second Bianchi identity in the same three lines.

what it may not be used for

Only for equations between tensors. Setting $\Gamma(p) = 0$ inside an expression that is not tensorial — $\partial_\mu\Gamma$ on its own, say — proves nothing, because the conclusion would not transport to another chart.

Why S3 is called an identity and not a symmetry. S1 and S2 are statements about which index slots are antisymmetric; S3 is a linear relation among components with distinct index patterns. In the algebraic language of section 2's counting slide, S1, S2 and S4 say that $R$ is a symmetric bilinear form on the space of bivectors, and S3 says that this form kills the totally antisymmetric part — one extra linear condition for each choice of four distinct indices.

An alternative proof, without coordinates. Sum $R(X,Y)Z + R(Y,Z)X + R(Z,X)Y$ for commuting fields and use torsion-freeness in the form $\nabla_XY - \nabla_YX = [X,Y]$; the nine terms cancel in three groups of three. Working it through is a good exercise; the coordinate proof is shorter and is the one to remember.

Where it is used. S3 is what reduces $36$ to $20$ in four dimensions, it is what makes S4 a theorem rather than an axiom, and it is the reason the Einstein tensor is symmetric. It is also the algebraic identity that, in unit 34's language, is $\mathrm d$ applied to the connection one-form, and there it will become a single line.

33 · symmetries

Pair symmetry is a consequence, not an axiom

S4 from S1, S2, S3

Write the cyclic identity S3 four times, once with each of $\rho,\sigma,\mu,\nu$ held in the first slot; add two of them and subtract the other two. Terms cancel in pairs by S1 and S2, and what survives, after relabelling, is

$$2R_{\rho\sigma\mu\nu} - 2R_{\mu\nu\rho\sigma} = 0 .$$

full proof in the notes — the bookkeeping is the whole content, and it is worth doing once.

what S4 means

Think of $R$ as a map $\Lambda^2 \times \Lambda^2 \to \mathbb R$, eating two planes: $R(u\wedge v,\,x\wedge y) := R_{\rho\sigma\mu\nu}u^\rho v^\sigma x^\mu y^\nu$. S1 and S2 say it is well defined on planes; S4 says it is symmetric in them. Curvature is a symmetric bilinear form on the space of planes — which is the sentence the counting theorem turns into a number.

$\Lambda^2$ here just means antisymmetric pairs, written $u\wedge v$; unit 34 gives the wedge its algebra. Nothing on this slide needs more than the index pattern.

Full proof of S4

Write the cyclic identity (S3, in its three-term form) four times:

(a) $R_{\rho\sigma\mu\nu}+R_{\rho\mu\nu\sigma}+R_{\rho\nu\sigma\mu}=0$
(b) $R_{\sigma\rho\mu\nu}+R_{\sigma\mu\nu\rho}+R_{\sigma\nu\rho\mu}=0$
(c) $R_{\mu\nu\rho\sigma}+R_{\mu\rho\sigma\nu}+R_{\mu\sigma\nu\rho}=0$
(d) $R_{\nu\mu\rho\sigma}+R_{\nu\rho\sigma\mu}+R_{\nu\sigma\mu\rho}=0$

Form (a) + (b) − (c) − (d). By S2, $R_{\rho\sigma\mu\nu}+R_{\sigma\rho\mu\nu}=0$, so the first terms of (a) and (b) cancel each other; likewise the first terms of (c) and (d). Of the eight remaining terms, two pairs cancel: $R_{\mu\rho\sigma\nu} = R_{\rho\mu\nu\sigma}$ by S1 and S2 together, so $R_{\rho\mu\nu\sigma} - R_{\mu\rho\sigma\nu} = 0$, and in the same way $R_{\sigma\nu\rho\mu} - R_{\nu\sigma\mu\rho} = 0$. The other two pairs double, by S2 alone: $R_{\rho\nu\sigma\mu} - R_{\nu\rho\sigma\mu} = 2R_{\rho\nu\sigma\mu}$ and $R_{\sigma\mu\nu\rho} - R_{\mu\sigma\nu\rho} = 2R_{\sigma\mu\nu\rho}$. Collecting, $2R_{\rho\nu\sigma\mu} + 2R_{\sigma\mu\nu\rho} = 0$, and S1 on the second term turns this into $R_{\rho\nu\sigma\mu} = R_{\sigma\mu\rho\nu}$. Relabelling $(\rho,\nu,\sigma,\mu)\to(\rho,\sigma,\mu,\nu)$ gives S4. $\square$

This is the one proof in the unit whose content is entirely bookkeeping. It is here because S4 is used constantly — the Ricci tensor's symmetry is a two-line consequence of it — and because a reader who has not seen that S4 is derived tends to assume the four symmetries are independent, and then miscounts.

The name. Luigi Bianchi published the differential identity in 1902; the algebraic one (S3) and the differential one were both known to Ricci, and the differential one had been found earlier by Aurel Voss (1880) and by Padova. "First" and "second" are the usual labels and are not chronological.

33 · counting

How many independent components

counting theorem

In $n$ dimensions, a tensor with S1–S4 has exactly

$$\frac{n^2(n^2-1)}{12}$$

independent components.

proof

Let $N = \binom n2 = \tfrac{n(n-1)}2$ be the number of index pairs $[\rho\sigma]$. By S1, S2 and S4, $R$ is a symmetric bilinear form on an $N$-dimensional space: $\tfrac{N(N+1)}2$ numbers. Given those, S3 says exactly that the totally antisymmetric part $R_{[\rho\sigma\mu\nu]}$ vanishes, and every totally antisymmetric tensor occurs as such a part, so S3 removes exactly $\binom n4$:

$$\frac{N(N+1)}{2} - \binom{n}{4} = \frac{n^2(n^2-1)}{12}. \;\square$$

The algebra, and why S3 is exactly $\binom n4$ conditions, are in the notes.

$n = 4$, so $N = 6$ index pairs. S1 and S2 make $R$ a $6\times6$ matrix indexed by pairs, S4 makes it symmetric: $21$ entries. S3 is then a single linear relation, tying the three entries whose pairs share no index.

The algebra, in full. $\tfrac{N(N+1)}2$ with $N = \tfrac{n(n-1)}2$ is $\tfrac{n(n-1)}{4}\big(\tfrac{n(n-1)}2+1\big) = \tfrac{n(n-1)(n^2-n+2)}{8}$. Subtracting $\binom n4 = \tfrac{n(n-1)(n-2)(n-3)}{24}$ and taking out $\tfrac{n(n-1)}{24}$ leaves $3(n^2-n+2) - (n-2)(n-3) = 3n^2-3n+6-n^2+5n-6 = 2n^2+2n = 2n(n+1)$, so the total is $\tfrac{n(n-1)\cdot2n(n+1)}{24} = \tfrac{n^2(n^2-1)}{12}$. Check: $n=2$ gives $1$, $n=3$ gives $6$, $n=4$ gives $20$, $n=5$ gives $50$.

Reading the figure. The rows and columns are the pairs $\rho\sigma$ with $\rho \lt \sigma$; the entry in row $\rho\sigma$, column $\mu\nu$ is $R_{\rho\sigma\mu\nu}$. The cells below the diagonal repeat the cells above (S4). S3 with all four indices distinct is the only cyclic sum that is not already zero by S1, S2 and S4, and in four dimensions there is one such choice, $\{1,2,3,4\}$ — so one relation, $\binom44 = 1$.

Full proof that S3 removes exactly $\binom n4$

Let $W$ be the space of $(0,4)$-tensors obeying S1, S2 and S4, of dimension $\tfrac{N(N+1)}2$. For $R \in W$ put $Z_{\rho\sigma\mu\nu} = R_{\rho\sigma\mu\nu}+R_{\rho\mu\nu\sigma}+R_{\rho\nu\sigma\mu}$, the cyclic sum S3 asks to vanish. (i) $Z$ is totally antisymmetric. It is invariant under cyclic permutations of $(\sigma\mu\nu)$ by construction, and S1 applied to each term shows $Z_{\rho\mu\sigma\nu} = -Z_{\rho\sigma\mu\nu}$; so $Z$ is antisymmetric in its last three slots. For the first two, $Z_{\sigma\rho\mu\nu} = R_{\sigma\rho\mu\nu} + R_{\sigma\mu\nu\rho} + R_{\sigma\nu\rho\mu}$, and S2, then S4 with S1 and S2, turn the three terms into $-R_{\rho\sigma\mu\nu}$, $-R_{\rho\nu\sigma\mu}$ and $-R_{\rho\mu\nu\sigma}$: $Z_{\sigma\rho\mu\nu} = -Z_{\rho\sigma\mu\nu}$. A transposition of the first two slots and antisymmetry in the last three generate every permutation. (ii) The map $R \mapsto Z$ from $W$ onto $\Lambda^4$ is surjective. A totally antisymmetric $A$ lies in $W$ (S4 is an even permutation) and its cyclic sum is $3A$. So by rank–nullity the kernel — the tensors obeying S1–S4 — has dimension $\tfrac{N(N+1)}2 - \binom n4$. $\square$ Problem P6 asks for this argument with every index written out.

33 · counting

The count, dimension by dimension

$n$	$\tfrac{N(N+1)}2$	Riemann	Ricci	Weyl
2	1	1	—	—
3	6	6	6	0
4	21	20	10	10
5	55	50	15	35

The last two columns look ahead to section 5. Ricci is the one trace of Riemann, a symmetric tensor with $\tfrac{n(n+1)}2$ components; Weyl is the part of Riemann the trace does not see, so its count is Riemann minus Ricci.

the twenty, again

Twenty in four dimensions — the same twenty the chart-counting slide could not remove. Two arguments with nothing in common, one answer.

the growth is the point

$n^2(n^2-1)/12$ grows like $n^4$: one number at $n = 2$, where the sphere has lived so far, and twenty at $n = 4$. Pictures from surfaces carry over plane by plane — section 3 — but never all at once.

33 · counting

The second Bianchi identity

the second Bianchi identity $$\nabla_\lambda R^\rho{}_{\sigma\mu\nu} + \nabla_\mu R^\rho{}_{\sigma\nu\lambda} + \nabla_\nu R^\rho{}_{\sigma\lambda\mu} = 0$$

Lowered and with S4, equivalently $\nabla_{[\lambda}R_{\rho\sigma]\mu\nu} = 0$.

proof — the same three lines

Both sides are tensors; work at $p$ in normal coordinates, where $\Gamma(p) = 0$, so $\nabla_\lambda$ acting on $R$ at $p$ is $\partial_\lambda$, and the $\Gamma\Gamma$ terms in $R$ contribute nothing to a single derivative at $p$ (each is a product of two factors, one of which vanishes). Hence at $p$

$$\nabla_\lambda R^\rho{}_{\sigma\mu\nu} = \partial_\lambda\partial_\mu\Gamma^\rho_{\nu\sigma} - \partial_\lambda\partial_\nu\Gamma^\rho_{\mu\sigma} .$$

Sum cyclically over $(\lambda,\mu,\nu)$: six terms, which cancel in pairs because partial derivatives commute (Clairaut, unit 03). $\square$

why this one matters most

S1–S4 are algebra at a point; the second Bianchi identity is the only differential constraint, and it is not optional or approximate — it holds for every metric whatsoever. Contract it twice and it says $\nabla^\mu G_{\mu\nu} = 0$ identically, which is section 5, and which is why unit 35's field equation can be consistent with $\nabla^\mu T_{\mu\nu} = 0$.

which indices cycle

The derivative index cycles with one antisymmetric pair — either one, by S4. Cycling it with one index from each pair gives a false statement.

Care with "the $\Gamma\Gamma$ terms contribute nothing". At $p$, $\partial_\lambda(\Gamma\Gamma) = (\partial_\lambda\Gamma)\Gamma + \Gamma(\partial_\lambda\Gamma)$, and both terms carry an undifferentiated $\Gamma$, which is zero at $p$. It is only the first derivative of $R$ that is safe this way — a second derivative of $R$ would pick up $(\partial\Gamma)(\partial\Gamma)$ and the trick fails.

The cyclic sum, written out. The six terms are $\partial_\lambda\partial_\mu\Gamma^\rho_{\nu\sigma} - \partial_\lambda\partial_\nu\Gamma^\rho_{\mu\sigma} + \partial_\mu\partial_\nu\Gamma^\rho_{\lambda\sigma} - \partial_\mu\partial_\lambda\Gamma^\rho_{\nu\sigma} + \partial_\nu\partial_\lambda\Gamma^\rho_{\mu\sigma} - \partial_\nu\partial_\mu\Gamma^\rho_{\lambda\sigma}$. The first and fourth cancel, the second and fifth cancel, the third and sixth cancel, each by Clairaut alone. $\square$

The lowered form. Lower $\rho$ with $g$, which passes through $\nabla$ because $\nabla g = 0$: $\nabla_{[\lambda}R_{|\rho\sigma|\mu\nu]} = 0$, the derivative cycling with the second pair. S4 moves that pair to the front, which gives $\nabla_{[\lambda}R_{\rho\sigma]\mu\nu} = 0$, the form contracted in section 5.

The deeper reason both Bianchi identities hold. In unit 34's language the curvature is $F = \mathrm dA + A\wedge A$ for the connection one-form $A$, and the second Bianchi identity is $\mathrm dF + A\wedge F - F\wedge A = 0$, which is $\mathrm d^2 = 0$ in disguise. Every gauge theory has it, for the same reason, and in electromagnetism it is the pair of Maxwell equations with no source, $\partial_{[\lambda}F_{\mu\nu]} = 0$ — for a reader who has taken unit 28, a branch this unit does not need. That two apparently unrelated facts are one fact is the argument for unit 34 existing at all.

section 3

Curvature belongs to the space

Sectional curvature, Gauss's theorem, and why no flat map of the Earth is possible.

33 · sectional

Sectional curvature: one number per plane

sectional curvature

For a plane $\Pi \subset T_pM$ spanned by independent $X, Y$,

$$K(\Pi) = \frac{R_{\rho\sigma\mu\nu}X^\rho Y^\sigma X^\mu Y^\nu}{\;|X|^2|Y|^2 - \langle X,Y\rangle^2\;} .$$

The denominator is the squared area of the parallelogram on $X,Y$ (Gram determinant), positive whenever $X,Y$ are independent and $\Pi$ is spacelike. Under a change of basis of $\Pi$ both numerator and denominator are multiplied by the square of its determinant, so $K$ depends on $\Pi$ alone.

this is the whole tensor

The sectional curvatures of all planes determine $R$ completely — a polarisation argument, like recovering a symmetric bilinear form from its quadratic form. So "curvature" can be thought of as a function on planes, which is how Riemann thought of it and how a physicist should picture it.

The next slide: what happens when every plane at every point gives the same number.

Check on the sphere. $X = \partial_\theta$, $Y = \partial_\phi$; numerator $R_{\theta\phi\theta\phi} = a^2\sin^2\theta$; denominator $g_{\theta\theta}g_{\phi\phi} - 0 = a^2\cdot a^2\sin^2\theta$. So $K = 1/a^2$, independent of $\theta$ ✓ — the sphere has constant curvature, which is the invariant version of "every point looks like every other".

Lorentzian caution. In signature $(-,+,+,+)$ the Gram determinant can be zero or negative, and $K(\Pi)$ is undefined for a null plane and sign-flipped for a timelike one. Sectional curvature is a Riemannian idea used carefully; in relativity one quotes components in an orthonormal frame instead — which is exactly what the geodesic-deviation section will do with $R^i{}_{0j0}$.

Polarisation, cited. That the sectional curvatures determine $R$ is do Carmo, Riemannian Geometry, ch. 4, Lemma 3.3; the proof is the same trick as recovering a symmetric bilinear form $B$ from $Q(x) = B(x,x)$, run twice, and it uses S3 at the last step.

33 · sectional

Constant curvature: the three model spaces

constant curvature

If $K(\Pi) = K$ for all $\Pi$ at all points, then $R_{\rho\sigma\mu\nu} = K\,(g_{\rho\mu}g_{\sigma\nu} - g_{\rho\nu}g_{\sigma\mu})$, the only tensor with S1–S4 built from $g$ alone. The sphere, hyperbolic space and flat space are the model cases; unit 39 needs this for cosmology.

Schematic. One geodesic triangle on each model surface. Its angle excess is $\int K\,\mathrm dA$ — the transport theorem of section 1 — so its sign is the sign of $K$; and the same sign decides whether geodesics that start parallel close up or spread, as section 4 will show.

Check that the constant-curvature form has the right sectional curvature. Substitute: numerator $= K\big((X\!\cdot\!X)(Y\!\cdot\!Y) - (X\!\cdot\!Y)^2\big)$, which is $K$ times the denominator. ✓ And it satisfies S1–S4 by inspection: antisymmetric in each pair by construction, symmetric under pair exchange, and the cyclic sum telescopes to zero.

Reading the figure. The sphere's triangle is the octant of worked example 2, with angle sum $3\pi/2$. The saddle is only a picture of a piece of the hyperbolic plane: a whole surface of constant negative curvature cannot sit smoothly in $\mathbb R^3$ (Hilbert's theorem, cited: do Carmo, Curves and Surfaces, §5-11), which is one more reason to want curvature defined from inside. Problem P2 computes $K = -1/a^2$ for the upper half-plane model.

Why "parallel geodesics" on a curved surface. There are no parallel lines in the Euclidean sense; "start parallel" means two geodesics that leave a common geodesic at right angles, the way two meridians leave the equator. Worked example 4 makes it quantitative.

33 · sectional

In two dimensions there is only one plane

A surface has a single tangent plane at each point, so a single sectional curvature — a function, the Gaussian curvature $K(p)$. Consistently with the counting theorem's answer of $1$:

the whole curvature of a surface $$R_{\rho\sigma\mu\nu} = K\,(g_{\rho\mu}g_{\sigma\nu} - g_{\rho\nu}g_{\sigma\mu}), \qquad K = \frac{R_{1212}}{\det g}.$$

full proof in the notes; problem P2 does it again with the indices out.

the sphere, again

$R_{\theta\phi\theta\phi} = a^2\sin^2\theta$, $\det g = a^4\sin^2\theta$, so $K = 1/a^2$ ✓ — and $K$ is the same at every point, including where $\sin\theta$ is nearly zero.

a formula worth memorising

For a conformally flat surface metric $g = e^{2\varphi}(\mathrm dx^2+\mathrm dy^2)$,

$$K = -e^{-2\varphi}\,\Delta\varphi,$$

with $\Delta = \partial_x^2+\partial_y^2$ the flat Laplacian.

Every surface metric is locally of this form (isothermal coordinates, cited), so this computes any $K$ with no Christoffel symbols at all.

the sphere in unit 31's chart

Stereographic: $e^{2\varphi} = 4a^2/(1+r^2)^2$, $r^2 = u^2+v^2$, so $\varphi = \ln(2a) - \ln(1+r^2)$ and $\Delta\varphi = -\tfrac1r\partial_r\!\big(\tfrac{2r^2}{1+r^2}\big) = -\tfrac{4}{(1+r^2)^2}$. Hence $K = \tfrac{(1+r^2)^2}{4a^2}\cdot\tfrac{4}{(1+r^2)^2} = \tfrac{1}{a^2}$ ✓.

Full proof that one function is the whole curvature

At $n = 2$, S1 forces $R_{\rho\sigma\mu\nu} = 0$ unless $\mu\nu$ is $12$ or $21$, and S2 the same for $\rho\sigma$; so every component is $0$ or $\pm R_{1212}$. The tensor $T_{\rho\sigma\mu\nu} = g_{\rho\mu}g_{\sigma\nu} - g_{\rho\nu}g_{\sigma\mu}$ has the same symmetries, so the same is true of it, and $T_{1212} = g_{11}g_{22} - g_{12}^2 = \det g \ne 0$. So at each point the tensors with S1 and S2 form a one-dimensional space spanned by $T$, and $R = f\,T$ with $f = R_{1212}/\det g$. Since $R$ and $T$ are both tensors and $T \ne 0$, $f$ is a function on $M$, the same in every chart — the sectional curvature of the only plane. $\square$

Full proof of the conformal formula

With $g_{11}=g_{22}=e^{2\varphi}$, $g_{12}=0$, the Christoffel symbols are $\Gamma^1_{11} = \varphi_x$, $\Gamma^1_{12} = \varphi_y$, $\Gamma^1_{22} = -\varphi_x$, $\Gamma^2_{22} = \varphi_y$, $\Gamma^2_{12} = \varphi_x$, $\Gamma^2_{11} = -\varphi_y$ (subscripts on $\varphi$ are partial derivatives). Substituting into the component formula and keeping only $R^1{}_{212}$: $R^1{}_{212} = \partial_1\Gamma^1_{22} - \partial_2\Gamma^1_{12} + \Gamma^1_{1\lambda}\Gamma^\lambda_{22} - \Gamma^1_{2\lambda}\Gamma^\lambda_{12}$. Term by term: $-\varphi_{xx} - \varphi_{yy} + (\varphi_x(-\varphi_x) + \varphi_y\varphi_y) - (\varphi_y\varphi_y + (-\varphi_x)\varphi_x) = -\Delta\varphi$. Then $R_{1212} = g_{11}R^1{}_{212} = -e^{2\varphi}\Delta\varphi$ and $K = R_{1212}/\det g = -e^{2\varphi}\Delta\varphi/e^{4\varphi} = -e^{-2\varphi}\Delta\varphi$. $\square$

Check on a case where the answer is known: the flat plane, $\varphi = 0$, gives $K = 0$ ✓. The hyperbolic upper half-plane $g = a^2(\mathrm dx^2+\mathrm dy^2)/y^2$ has $\varphi = \ln a - \ln y$, $\Delta\varphi = 1/y^2$, and $K = -(y^2/a^2)(1/y^2) = -1/a^2$ — constant negative curvature, which is problem P2.

Isothermal coordinates — that every smooth surface metric can locally be written $e^{2\varphi}(\mathrm dx^2+\mathrm dy^2)$ — is cited: S.-S. Chern, "An elementary proof of the existence of isothermal parameters on a surface", Proc. AMS 6 (1955). It is a genuinely two-dimensional fact, resting on the Beltrami equation, and has no analogue in higher dimensions.

33 · intrinsic

Theorema Egregium

Gauss, 1827

For a surface $\Sigma \subset \mathbb R^3$, the product of the principal curvatures $\kappa_1\kappa_2$ — defined by how $\Sigma$ bends in space — equals the Gaussian curvature $K$ computed from the induced metric alone. Hence $K$ is intrinsic: it is unchanged by any bending that does not stretch.

Cited: the equality $\det(\mathrm{II}) = R_{1212}$ is Gauss's equation, proved in do Carmo, Curves and Surfaces, §4-3. It needs the second fundamental form, which this course does not build — it has no use for a surrounding space.

the corollary we do prove

If $F : (M,g) \to (M',g')$ is an isometry, then $K' \circ F = K$. Immediate: $K$ was defined from $R$, $R$ from $\nabla$, $\nabla$ from $g$ — and $F^*g' = g$ carries each construction to the next. $\square$

the consequence everyone has met

No region of the sphere is isometric to any region of the plane, because $1/a^2 \ne 0$. Every flat map of the Earth distorts lengths — not through the cartographer's incompetence but because $K$ is an invariant. Mercator preserves angles and ruins areas; Peters preserves areas and ruins shapes; nothing preserves both.

33 · example

Worked example 3 — the cylinder is flat

every step

The cylinder of radius $a$, with coordinates $z$ along the axis and angle $\phi$ round it, has $g = \mathrm dz^2 + a^2\mathrm d\phi^2$.

Every component is constant, so every $\partial g = 0$, every $\Gamma$ vanishes, and $R = 0$: $K = 0$ everywhere.

Yet the cylinder is visibly bent in $\mathbb R^3$: round it the principal curvature is $\kappa_1 = 1/a$, along it $\kappa_2 = 0$. Product zero ✓, as Gauss says it must be.

A sheet of paper rolls into a cylinder without tearing; it will not wrap a ball.

Rolling the sheet bends it in space but stretches nothing, so every length and angle drawn on it survives: the triangle still has angle sum $\pi$. Bending in one direction only — one principal curvature zero — is what makes that possible.

The coordinates are the flat ones in disguise. Put $x = a\phi$: then $g = \mathrm dz^2 + \mathrm dx^2$, the flat plane, with $x$ running over $[0, 2\pi a)$ and then repeating. Locally the cylinder is the plane, which is the corollary on the previous slide in action; globally it is the plane rolled up, and a geodesic is a straight line on the unrolled sheet — a helix, a circle or an axial line once rolled.

The cone, and where its curvature went. $g = \mathrm dr^2 + r^2\sin^2\alpha\,\mathrm d\phi^2$ with $\phi \in [0,2\pi)$ is the cone of half-angle $\alpha$. Set $\tilde\phi = \phi\sin\alpha$ and it becomes the flat plane in polar coordinates — but with $\tilde\phi$ running only over $[0, 2\pi\sin\alpha)$. So $K = 0$ everywhere off the apex, and yet transport round a loop enclosing the apex rotates a vector by the deficit angle $2\pi(1-\sin\alpha)$: for $\alpha = 30^\circ$, exactly $\pi$, a half-turn. All the curvature is concentrated at the apex as a delta function. This is the standard counterexample to "zero curvature means trivial holonomy", and the reason the flatness theorem carries the hypothesis "simply connected". Problem P1 computes it.

section 4

Geodesic deviation

Two freely falling particles, and the equation that says curvature is the tidal force.

33 · deviation

A family of geodesics, and the field between them

Let $\gamma(\tau, s)$ be a smooth one-parameter family of geodesics: for each fixed $s$, $\tau \mapsto \gamma(\tau,s)$ is a geodesic with affine parameter $\tau$. Put

$$u = \frac{\partial\gamma}{\partial\tau}, \qquad \xi = \frac{\partial\gamma}{\partial s} .$$

the deviation vector

$\xi$ points from one geodesic to its neighbour. $u$ and $\xi$ are coordinate fields of the chart $(\tau,s)$ on the surface swept out, so $[u,\xi] = 0$ — unit 31's statement that coordinate fields commute. Torsion-freeness then gives $\nabla_u\xi = \nabla_\xi u$.

Both facts are used once each, in the proof of the geodesic deviation equation, and nothing else is needed.

The family as a chart on the surface it sweeps: geodesics are the lines of constant $s$, and $u$ and $\xi$ are the two coordinate directions. Each small cell of the $(\tau, s)$ grid closes up — the shaded one, for instance — and that is $[u,\xi] = 0$.

Why coordinate fields commute. Unit 31 proved $[\partial_\mu,\partial_\nu] = 0$ from Clairaut. Here $(\tau,s)$ are coordinates on the two-dimensional surface swept out by the family — a map $\gamma : \mathbb R^2 \to M$ — and $u$, $\xi$ are the images of $\partial_\tau$, $\partial_s$ under its differential. The bracket is natural under smooth maps, so it vanishes. (If the family degenerates, $\gamma$ fails to be an immersion at isolated points; the identity $\nabla_u\xi = \nabla_\xi u$ survives because it is a statement about the map, not the image.)

Torsion-freeness in this form. Unit 32 defined torsion as $T(X,Y) = \nabla_XY - \nabla_YX - [X,Y]$ and proved the Levi-Civita connection has $T = 0$. With $[u,\xi] = 0$ that is exactly $\nabla_u\xi = \nabla_\xi u$: "the rate at which the neighbour drifts along my worldline equals the rate at which my velocity changes as I step sideways". It is the only place torsion-freeness is used in this section, and it is load-bearing — in a theory with torsion the deviation equation acquires extra terms.

33 · deviation

What the deviation vector is, and is not

$\xi$ is not "the distance between two particles"

There is no such thing without a convention: the two worldlines are not at the same time in any absolute sense. $\xi$ is defined by the family, and it is a genuine vector field along each geodesic. What is physical is $\xi$ measured in the freely falling observer's own orthonormal frame — which is what the Newtonian limit will make explicit.

The freedom to reparametrise $s$ rescales $\xi$; the freedom to shift $\tau$ by a function of $s$ adds a multiple of $u$ to $\xi$. Neither changes anything that follows, because $R$ is antisymmetric in the relevant slots.

Shifting where each neighbour's clock starts joins my point to a different point of the same neighbour's worldline: $\xi$ becomes $\xi + f\,u$. The curvature term cannot tell the difference, since $R(u,u) = 0$.

33 · deviation

The geodesic deviation equation

the geodesic deviation equation $$\frac{\mathrm D^2\xi^\rho}{\mathrm d\tau^2} = -R^\rho{}_{\sigma\mu\nu}\,u^\sigma\,\xi^\mu\,u^\nu$$

where $\mathrm D/\mathrm d\tau = \nabla_u$ is unit 32's covariant derivative along the curve.

proof — four steps, each one line

$\nabla_u\nabla_u\xi = \nabla_u\nabla_\xi u$ (torsion-free, with $[u,\xi] = 0$)
$\phantom{\nabla_u\nabla_u\xi} = \nabla_\xi\nabla_u u + R(u,\xi)u$ (definition of $R$, with $[u,\xi]=0$)
$\phantom{\nabla_u\nabla_u\xi} = R(u,\xi)u$ (each curve is a geodesic: $\nabla_uu = 0$)
and $R(u,\xi)u$ has components $R^\rho{}_{\sigma\mu\nu}u^\sigma u^\mu\xi^\nu = -R^\rho{}_{\sigma\mu\nu}u^\sigma\xi^\mu u^\nu$ by S1. $\square$

read what is and is not there

No first derivative of the metric, no $\Gamma$, no acceleration, no force. Both particles are in free fall and feel nothing; the only thing that appears is curvature. If $R = 0$ the separation is linear in $\tau$ — Euclid's parallel postulate, as a differential equation.

checks

Dimensions. $[\mathrm D^2\xi/\mathrm d\tau^2] = \mathrm{L\,T^{-2}}$; $[R]=\mathrm L^{-2}$, $[u] = \mathrm{L\,T^{-1}}$, so the right side is $\mathrm{L^{-2}}\cdot\mathrm L\cdot\mathrm{L^2T^{-2}} = \mathrm{L\,T^{-2}}$ ✓.

Special case $\xi \parallel u$: right side vanishes by S1 ✓ — no relative acceleration along the direction of motion, as it must be.

The index gymnastics of the last step, in full. The curvature operator is defined by $R(X,Y)Z = R^\rho{}_{\sigma\mu\nu}Z^\sigma X^\mu Y^\nu\partial_\rho$, matching $R(\partial_\mu,\partial_\nu)\partial_\sigma = R^\rho{}_{\sigma\mu\nu}\partial_\rho$. So $R(u,\xi)u$ has $Z = u$, $X = u$, $Y = \xi$: components $R^\rho{}_{\sigma\mu\nu}u^\sigma u^\mu\xi^\nu$. Swapping the last two indices with S1 and renaming gives the boxed form. Convention-checking of exactly this kind is where sign errors in general relativity live; the antidote is the sphere, on the next slide, where the answer is known in advance.

Jacobi fields. A solution $\xi$ of this equation along a fixed geodesic is a Jacobi field. It is a second-order linear ODE system, so by unit 01's existence and uniqueness theorem the solution space has dimension $2n$: a Jacobi field is fixed by $\xi(0)$ and $(\mathrm D\xi/\mathrm d\tau)(0)$. Two points where a nontrivial Jacobi field vanishing at the first also vanishes at the second are conjugate. On the sphere, every pair of antipodes is conjugate, which is why a great circle stops being the shortest path once it passes the antipode — unit 11's conjugate points of the second variation, now on a manifold and with curvature as the cause.

The name. The equation goes back to Jacobi's theory of the second variation (1837); Levi-Civita gave it the curvature form in 1926. In relativity it is due to Synge (1926), and the detector of unit 38 is a device for measuring $\mathrm D^2\xi/\mathrm d\tau^2$.

33 · example

Worked example 4 — two meridians fall together

from the general equation to a number

On $S^2_a$, the meridians $\gamma(s,\phi_0)$ run north from the equator, $s$ arc length, $\phi_0$ the family parameter: $u = \partial_s$, $\xi = \Delta\phi\,\partial_{\phi_0}$ for two of them $\Delta\phi$ apart.

In two dimensions $R_{\rho\sigma\mu\nu} = K(g_{\rho\mu}g_{\sigma\nu} - g_{\rho\nu}g_{\sigma\mu})$. Take $\xi \perp u$ and write $\xi$ for the length of $\xi$. Then

$$-R^\rho{}_{\sigma\mu\nu}u^\sigma\xi^\mu u^\nu = -K\big(\xi^\rho (u\!\cdot\!u) - u^\rho(u\!\cdot\!\xi)\big) = -K\,\xi^\rho$$

using $u\cdot u = 1$ (arc length) and $u\cdot\xi = 0$. Hence

$$\xi'' = -K\xi = -\frac{\xi}{a^2}, \qquad \xi(s) = a\,\Delta\phi\,\cos(s/a),$$

taking $\xi(0) = a\,\Delta\phi$ at the equator and $\xi'(0) = 0$ (the meridians leave parallel).

Two meridians $\Delta\phi$ apart (exaggerated), drawn from the equator. They leave it parallel, never turn, and the gap between them shrinks as $\cos(s/a)$ until they meet at the pole.

33 · example

Two meridians: what the solution says

what the solution says

$\xi$ vanishes at $s = \pi a/2$: a quarter of a great circle, the pole. Two paths that start parallel, never turn, and meet. That is a tidal force, with no force in it.

Direct check: at colatitude $\theta$ the separation is $a\sin\theta\,\Delta\phi$, and $s = a(\tfrac\pi2-\theta)$, so $a\sin\theta = a\cos(s/a)$ ✓ — the ODE reproduces plane trigonometry on the sphere.

checks, and a real number

Limit $a\to\infty$: $\xi \to a\Delta\phi$, constant — parallel lines stay parallel ✓. Dimensions: $[K] = \mathrm L^{-2}$, $[\xi''] = \mathrm{L^{-1}}$ ✓.

On the Earth, meridians $1$ km apart at the equator are $1000\cos 45^\circ = 707$ m apart at latitude $45^\circ$. A flat map says $1000$ m.

The same equation $\xi'' = -K\xi$ for three signs of $K$, all starting parallel. Positive $K$ closes the gap ($0.707$ at latitude $45^\circ$, zero at the pole); zero keeps it; negative $K$ opens it exponentially.

Reading the plot. The horizontal axis is distance travelled in units of the radius of curvature, so one picture serves every sphere: the Earth, a football, a sphere of radius one light-year. Only $s/a$ appears, which is the dimensional statement that $K = 1/a^2$ is the only scale in the problem.

Conjugate points, concretely. $\xi$ vanishes at $s = \pi a/2$, the north pole, and next at $s = 3\pi a/2$, the south pole: this Jacobi field vanishes at both poles, so the poles are conjugate along every meridian. A meridian leaving the north pole is a shortest path all the way to the south pole — there no longer the unique one, since every meridian is — and past the south pole it is a geodesic and not a minimiser, which is exactly what the vanishing Jacobi field detects.

Negative curvature, for contrast. On the hyperbolic plane of curvature $-1/a^2$ the same derivation gives $\xi'' = +\xi/a^2$, so $\xi = a\Delta\phi\cosh(s/a)$: geodesics that start parallel diverge exponentially, at the rate $\sqrt{|K|}$ per unit length. That is the geometry behind chaos in the strict sense: geodesic flow on a negatively curved surface was the first system shown to have this sensitivity everywhere (Hadamard, 1898), and unit 17, a branch this unit does not need, calls $\sqrt{|K|}$ a Lyapunov exponent.

The Earth numbers. $\cos(45^\circ) = 0.7071$ so $707$ m; $\cos(60^\circ) = 0.5$ so $500$ m. Arc length from the equator to latitude $45^\circ$ is $R_\oplus\pi/4 = 5.004\times10^6$ m, and $\cos(s/R_\oplus) = \cos(\pi/4)$ ✓ — the two descriptions are the same statement.

33 · Newton

The Newtonian limit, set up

Unit 32 fixed the weak, static, slow-motion regime: with $x^0 = ct$,

$$g_{00} = -\Big(1 + \frac{2\Phi}{c^2}\Big), \qquad g_{ij} = \delta_{ij}, \qquad g_{0i} = 0,$$

with $|\Phi|/c^2 \ll 1$ and $\partial_t \Phi= 0$, and showed the geodesic equation becomes $\ddot{\vec x} = -\nabla\Phi$. The only Christoffel symbol that survives at this order is

$$\Gamma^i_{00} = -\tfrac12 g^{ij}\partial_j g_{00} = \frac{1}{c^2}\,\partial_i\Phi .$$

A freely falling observer has $u^\mu \approx (c, \vec v)$ with $|\vec v| \ll c$, so $u^0 \approx c$ dominates, and $\mathrm d\tau \approx \mathrm dt$.

which index is which

$\partial_i\Phi$ means $\partial\Phi/\partial x^i$ with $x^i$ a length, while $\partial_0 = c^{-1}\partial_t$. Keeping $c$ explicit in every step is tedious and is the only way the factors come out right; the identification on the next slide is worthless with a stray $c^2$.

what we expect to find

Unit 05 derived, from Newton alone, $\ddot\xi_i = -(\partial_i\partial_j\Phi)\,\xi_j$ — the tidal deviation equation. The geodesic deviation equation must reduce to exactly that, or the whole construction is wrong.

Deriving $\Gamma^i_{00}$. $\Gamma^i_{00} = \tfrac12 g^{i\lambda}(2\partial_0 g_{\lambda 0} - \partial_\lambda g_{00})$. Static kills the first piece; $g^{ij} = \delta^{ij}$ at this order leaves $-\tfrac12\partial_i g_{00} = \tfrac12\partial_i(1 + 2\Phi/c^2) = \partial_i\Phi/c^2$. Dimensions: $[\Phi] = \mathrm{L^2T^{-2}}$, so $\partial_i\Phi/c^2$ is $\mathrm{L^{-1}}$ ✓ — a Christoffel symbol in coordinates that are lengths is an inverse length.

Why $g_{ij} = \delta_{ij}$ is allowed here. The spatial metric perturbation is of the same order as $g_{00}$'s, but it enters the geodesic equation multiplied by $v^2/c^2$ and enters $R^i{}_{0j0}$ only through time derivatives, which vanish for a static field. It matters for light bending — unit 37's factor of two — and not here. Unit 32 made the same restriction for the same reason.

The "factor of two everyone gets wrong once" is the 2 in $g_{00} = -(1+2\Phi/c^2)$. It is not a choice: it is fixed by requiring the geodesic equation to give $\ddot{\vec x} = -\nabla\Phi$ rather than $-\tfrac12\nabla\Phi$, which unit 32 did. Everything on the next slide inherits it.

33 · example

Worked example 5 — Riemann is the tidal tensor

the identification, term by term

Step 1 — the component. To first order in $\Phi/c^2$ the $\Gamma\Gamma$ terms are second order and drop:

$$R^i{}_{0j0} = \partial_j\Gamma^i_{00} - \partial_0\Gamma^i_{j0} = \partial_j\Big(\frac{\partial_i\Phi}{c^2}\Big) - 0 = \frac{\partial_i\partial_j\Phi}{c^2}.$$

($\Gamma^i_{j0}$ vanishes for a static field, and $\partial_0$ of anything static is zero.)

Step 2 — the deviation equation. With $u^\mu \approx (c,\vec 0)$ and $\xi^\mu \approx (0,\vec\xi)$, only $\sigma = \nu = 0$ and $\mu = j$ survive:

$$\begin{aligned}\frac{\mathrm D^2\xi^i}{\mathrm d\tau^2} &= -R^i{}_{0j0}\,u^0\xi^j u^0 = -c^2\cdot\frac{\partial_i\partial_j\Phi}{c^2}\,\xi^j\\ &= -\big(\partial_i\partial_j\Phi\big)\xi^j .\end{aligned}$$

Step 3 — compare. The left side is $\ddot\xi_i$ at this order. This is unit 05's equation, character for character.

checks

Dimensions: $[\partial_i\partial_j\Phi] = \mathrm{T^{-2}}$, over $c^2$ gives $\mathrm L^{-2}$ ✓ — curvature is an inverse area, and the conversion factor is $c^2$.

Trace: $R^i{}_{0i0} = \nabla^2\Phi/c^2 = 4\pi G\rho/c^2$, zero in vacuum ✓ — the trace-free tidal tensor of unit 05, and the seed of unit 35's field equation.

Limit $\Phi \to$ const: no tidal field, $R^i{}_{0j0} = 0$ ✓ — a uniform gravitational field is no curvature at all, which is unit 30's equivalence principle.

Step 2, without the approximation $\vec v = 0$. Keeping $u^\mu = (c,\vec v)$ adds terms of relative order $v/c$ carrying $R^i{}_{0jk}$ and $R^i{}_{kj0}$. For a static weak field those components involve $\partial_t\Phi$ or are second order, so they vanish at this order. The approximation is therefore not a truncation but an exact statement about the regime.

Why $\mathrm D^2\xi^i/\mathrm d\tau^2 \to \ddot\xi^i$. $\mathrm D/\mathrm d\tau = \mathrm d/\mathrm d\tau + \Gamma^i_{\mu\nu}u^\mu(\cdot)^\nu$, and the correction terms carry one $\Gamma \sim \partial\Phi/c^2$ against the leading $\ddot\xi \sim \partial\partial\Phi\,\xi$; their ratio is $\Phi/c^2$, the same small parameter. And $\mathrm d\tau = \mathrm dt(1 + O(\Phi/c^2, v^2/c^2))$.

This slide is the point of the unit, and it is worth being explicit about what it does and does not establish. It shows that the geometry contains Newtonian gravity: the object the geometry calls curvature acts, in the weak static limit, exactly as the object Newton called the second derivative of the potential. It does not yet say which metrics occur — that is a field equation, and it is unit 35. What it does supply is the boundary condition that fixes the constant in that equation.

33 · example

The tidal tensor, drawn

the dictionary

$$\boxed{\;R^i{}_{0j0} = \frac{\partial_i\partial_j\Phi}{c^2}\;}$$

For a point mass, $\partial_i\partial_j\Phi = \tfrac{GM}{r^3}(\delta_{ij}-3n_in_j)$: eigenvalues $(-2,1,1)\,GM/(c^2r^3)$, stretch along $\hat n$, squeeze across. The two tidal bulges of unit 05 are two components of the Riemann tensor.

Relative acceleration is $\ddot\xi_i = -c^2R^i{}_{0j0}\,\xi^j$: along $\hat n$, $+2GM/r^3$ times the offset, outward; across, $GM/r^3$ times it, inward.

A ring of particles falling freely with an observer, a mass $M$ off to the left. Each arrow is $-c^2R^i{}_{0j0}\xi^j$ at that particle: twice as long outward along $\hat n$ as inward across it. Left alone, the ring becomes the ellipse (exaggerated) — the two tidal bulges.

Reading the figure. The arrows are computed from the point-mass tidal tensor: at a particle offset by $\xi = \rho(\cos\vartheta, \sin\vartheta)$ from the observer, with $\hat n$ along the first axis, the relative acceleration is $(GM/r^3)\,\rho\,(2\cos\vartheta, -\sin\vartheta)$. The ring is a slice through a sphere of particles; the third direction, out of the page, is squeezed like the vertical one, so the sphere becomes a prolate ellipsoid pointing at the mass, with its volume unchanged to first order because the eigenvalues sum to zero.

Scale of the numbers. $R^i{}_{0j0}$ has units $\mathrm m^{-2}$, and the radius of curvature is its inverse square root. The Moon's field at the Earth: $GM_{\mathrm m}/d^3 = 8.632\times10^{-14}\ \mathrm{s^{-2}}$, so $9.604\times10^{-31}\ \mathrm{m^{-2}}$, radius $1.020\times10^{15}$ m $= 0.108$ light-year. The Sun at the Earth: $4.411\times10^{-31}\ \mathrm{m^{-2}}$, radius $1.506\times10^{15}$ m. The Earth at its own surface: $GM_\oplus/R_\oplus^3 = 1.536\times10^{-6}\ \mathrm{s^{-2}}$, so $1.709\times10^{-23}\ \mathrm{m^{-2}}$, radius $2.42\times10^{11}$ m — about 1.6 astronomical units. Spacetime here is very nearly flat, and the whole of terrestrial gravity is a curvature radius larger than the Earth's orbit.

33 · Newton

What the identification settles

unit 05, Newton	unit 33, geometry
$\ddot\xi_i = -(\partial_i\partial_j\Phi)\xi_j$	$\mathrm D^2\xi^\rho/\mathrm d\tau^2 = -R^\rho{}_{\sigma\mu\nu}u^\sigma\xi^\mu u^\nu$
$T_{ij} = \partial_i\partial_j\Phi$	$c^2R^i{}_{0j0}$
$\operatorname{tr}T = 4\pi G\rho$	$c^2R_{00}$, and hence unit 35
trace-free in vacuum	Weyl carries vacuum tides
eigenvalues $(-2,1,1)\tfrac{GM}{r^3}$	the same, over $c^2$

why "gravity is curvature" is not a metaphor

Both equations are second-order linear ODEs for the same physical separation, with the same solutions and the same eigenvalues. They are not analogous: they are the same computation, written in two notations, one of which happens to survive a Lorentz boost.

what is not claimed

Not that the metric is determined — that needs a field equation. Not that $R^i{}_{0j0}$ exhausts $R$ — it is $6$ of the $20$, the "electric" part; the rest shows up in unit 38's waves. Not that $\Phi$ exists in general — it does not; only the weak static limit has one.

The six "electric" components. $R^i{}_{0j0}$ is symmetric in $ij$ by S4 and so has six independent components in four dimensions, one combination of which, the trace, is fixed by the local density. It is called the electric part of the curvature — in vacuum, where the trace vanishes, the electric part of the Weyl tensor — by analogy with the split of the electromagnetic field tensor into $\vec E$ and $\vec B$ relative to an observer (unit 28, for a reader who has taken it); the magnetic part, $R^i{}_{0jk}$, has no Newtonian counterpart at all and is the piece that makes gravitomagnetism and frame dragging (unit 40) possible.

Where Newton's theory actually fails is not here. A Newtonian theory reformulated geometrically — Newton–Cartan, mentioned in unit 30 — reproduces everything on the left of the table with a connection that is not metric. What it cannot do is be Lorentz-invariant, and that is the argument for the right-hand column, not this correspondence.

33 · Newton

A tidal force you could measure

a number that can be measured

Two balls dropped $1$ m apart horizontally, falling $100$ m: they converge by $\tfrac12(GM_\oplus/R_\oplus^3)\,\xi\,t^2$ with $t = 4.52$ s, that is $15.7\ \mu$m. Small, real, and entirely a statement about the curvature of spacetime.

check, by similar triangles

Each ball falls straight at the Earth's centre, so the two lines of fall meet $R_\oplus$ below. Over a fall of $h$ they close by $\xi h/R_\oplus = (1\ \mathrm m)(100)/(6.378\times10^6) = 15.7\ \mu$m ✓ — the same number with no curvature in sight, as it must be: the squeeze eigenvalue $GM/r^3$ is exactly what converging radial lines give.

Not to scale: the lines of fall really converge $6378$ km below. Horizontal neighbours squeeze together — the $+1$ eigenvalue — just as the two meridians did on the sphere.

The 15.7 micrometres, computed. $GM_\oplus/R_\oplus^3 = 3.986004418\times10^{14}/ (6.378137\times10^6)^3 = 1.5362\times10^{-6}\ \mathrm{s^{-2}}$. Horizontal separation has the squeeze eigenvalue $+GM/r^3$, so $\ddot\xi = -1.5362\times10^{-6}\xi$. Fall time from $100$ m is $t = \sqrt{2h/g} = 4.5160$ s with $g = 9.80665\ \mathrm{m\,s^{-2}}$, and over that time $\Delta\xi \approx \tfrac12(1.5362\times10^{-6})(1\ \mathrm m)(4.5160)^2 = 1.567\times10^{-5}$ m. Galileo's leaning-tower experiment, done to $10\ \mu$m, measures the Riemann tensor.

Why the two routes agree exactly. With $g = GM/R^2$ and $t^2 = 2h/g$, $\tfrac12(GM/R^3)\,\xi\,t^2 = \tfrac12(GM/R^3)\,\xi\,(2hR^2/GM) = \xi h/R$. The tidal formula is the geometry of converging radii, to first order in $h/R$. (The small difference between $1.567$ and $1.568\times10^{-5}$ m is only because $g = 9.80665$ is not exactly $GM_\oplus/R_\oplus^2 = 9.798$.)

section 5

The traces, and the one that is conserved

Ricci, scalar, Weyl, Einstein — and the identity that makes a field equation possible.

33 · traces

Ricci and the scalar curvature

the Ricci tensor and scalar curvature $$R_{\sigma\nu} := R^\rho{}_{\sigma\rho\nu}, \qquad R := g^{\sigma\nu}R_{\sigma\nu} .$$

this is the only trace

Contract $\rho$ with $\mu$: $R_{\sigma\nu}$. With $\nu$: $-R_{\sigma\mu}$, by S1. With $\sigma$: $R^\rho{}_{\rho\mu\nu} = g^{\rho\lambda}R_{\lambda\rho\mu\nu} = 0$, by S2. A trace of two lower indices uses $g^{-1}$: over $\mu\nu$ it is zero by S1, and over $\sigma\mu$ or $\sigma\nu$ it is $\pm$ the Ricci trace again by S4. So up to sign there is exactly one. $\square$

$R_{\sigma\nu}$ is symmetric: $R_{\sigma\nu} = g^{\rho\lambda}R_{\lambda\sigma\rho\nu} = g^{\rho\lambda}R_{\rho\nu\lambda\sigma}$ (S4) $= R_{\nu\sigma}$.

what each one means

Ricci: a small ball of dust released at rest has $\ddot V/V = -R_{\mu\nu}u^\mu u^\nu$ — the trace of the deviation equation, so the trace of the tidal tensor, which unit 05 said is $4\pi G\rho$. Scalar: $R$ measures how the volume of a small geodesic ball falls short of the Euclidean one (cited, in the notes).

the sphere

$R_{\mu\nu} = \tfrac{1}{a^2}g_{\mu\nu}$ and $R = \tfrac{2}{a^2}$. Check: $R = g^{\mu\nu}(g_{\mu\nu}/a^2) = 2/a^2$ ✓, and in two dimensions $R = 2K$ always. Positive, as a sphere must be with these conventions.

Computing the sphere's Ricci tensor. $R_{\theta\theta} = R^\rho{}_{\theta\rho\theta} = R^\phi{}_{\theta\phi\theta}$ (the $\rho=\theta$ term vanishes by S1). Lowering, $R^\phi{}_{\theta\phi\theta} = g^{\phi\phi}R_{\phi\theta\phi\theta} = (a^2\sin^2\theta)^{-1}(a^2\sin^2\theta) = 1$. And $g_{\theta\theta}/a^2 = 1$ ✓. Similarly $R_{\phi\phi} = R^\theta{}_{\phi\theta\phi} = \sin^2\theta = g_{\phi\phi}/a^2$ ✓.

The sign of the Ricci trace and the sign of the course. With the conventions on the title slide, a sphere has $R > 0$, ordinary matter gives $R_{\mu\nu}u^\mu u^\nu > 0$, and a ball of dust contracts. Every one of those three statements flips together under a convention change, which is why they are a usable self-test: if your sphere comes out with $R < 0$, a sign is wrong somewhere and the physics will be wrong too.

Full proof of the Ricci statement

In the freely falling observer's orthonormal frame, $u = c\,e_0$, and the deviation equation for a spatial $\xi$ reads $\ddot\xi^i = -c^2R^i{}_{0j0}\xi^j =: -M^i{}_j\xi^j$. Particles released at rest from the points $\xi(0)$ of a small ball are at $\xi(t) = (1 - \tfrac12 M t^2)\xi(0) + O(t^3)$, a linear map whose determinant is $1 - \tfrac12(\operatorname{tr}M)t^2 + O(t^3)$. So $\ddot V/V = -\operatorname{tr}M = -c^2R^i{}_{0i0}$ at $t = 0$. And $R^i{}_{0i0} = R^\rho{}_{0\rho0} = R_{00}$, since the $\rho = 0$ term is $R^0{}_{000} = 0$ by S1; with $u^0 = c$ that is $R_{\mu\nu}u^\mu u^\nu/c^2$. $\square$ In the Newtonian limit of section 4 this is $-\nabla^2\Phi = -4\pi G\rho$: matter makes a ball of dust contract.

The volume formula, stated precisely (cited: A. Gray, Tubes, 2nd ed., ch. 9): for a geodesic ball of radius $r$ in an $n$-manifold, $\mathrm{vol} = \omega_n r^n\big(1 - \tfrac{R}{6(n+2)}r^2 + O(r^4)\big)$. It is the cleanest answer to "what does scalar curvature measure", and one way of seeing why $R$ is the natural scalar for the Einstein–Hilbert action of unit 35.

33 · traces

Three dimensions: Ricci is everything

Counting already told us: at $n=3$, Riemann has $6$ components and the symmetric Ricci tensor has $\tfrac{3\cdot4}{2} = 6$. Equal counts do not by themselves prove determination — but here the map is explicit.

$n = 3$ $$\begin{aligned}R_{\rho\sigma\mu\nu} = {}&g_{\rho\mu}R_{\sigma\nu} + g_{\sigma\nu}R_{\rho\mu} - g_{\rho\nu}R_{\sigma\mu} - g_{\sigma\mu}R_{\rho\nu}\\ &- \tfrac{R}{2}\big(g_{\rho\mu}g_{\sigma\nu} - g_{\rho\nu}g_{\sigma\mu}\big)\end{aligned}$$

Check by tracing. Contract with $g^{\rho\mu}$, using $g^{\rho\mu}g_{\rho\mu} = 3$: the five terms give $3R_{\sigma\nu} + g_{\sigma\nu}R - R_{\sigma\nu} - R_{\sigma\nu} - \tfrac R2(3-1)g_{\sigma\nu} = R_{\sigma\nu}$ ✓.

the physical consequence

In three dimensions, $R_{\mu\nu} = 0$ forces $R_{\rho\sigma\mu\nu} = 0$: vacuum is flat. A universe with two space dimensions and one of time has no gravitational waves, no Schwarzschild solution, no orbits — gravity exists only where matter is. In four dimensions $20 - 10 = 10$ components are left free by the trace, and those ten are everything this course has left to study.

"flat" does not mean "dull"

Three-dimensional gravity is locally trivial, not globally: a point mass produces a cone, flat everywhere but with a deficit angle — exactly the cone of section 3. Curvature concentrated where the matter is, holonomy round it, and no field outside.

Full proof

Call the right-hand side $F(S)$, with the Ricci tensor replaced by any symmetric $S_{\mu\nu}$ and $R$ by its trace $s$. (i) $F(S)$ obeys S1–S4: each group of terms has the pattern $h_{\rho\mu}k_{\sigma\nu} + h_{\sigma\nu}k_{\rho\mu} - h_{\rho\nu}k_{\sigma\mu} - h_{\sigma\mu}k_{\rho\nu}$ with $h, k$ symmetric, antisymmetric in each pair and symmetric under exchanging them by inspection, and its cyclic sum over $\sigma\mu\nu$ has twelve terms cancelling in pairs by the symmetry of $h$ and $k$. (ii) Its Ricci trace is $S$ — the check on the slide. So the linear map $\mathrm{Ric}$ from the space of tensors with S1–S4 to the symmetric tensors has a right inverse $F$, and is onto. (iii) At $n = 3$ both spaces have dimension $6$, by the counting theorem and by $\tfrac{3\cdot4}{2}$. An onto linear map between spaces of equal finite dimension is bijective, so $\mathrm{Ric}$ is injective and $F$ is its inverse: $R = F(\mathrm{Ric})$ for every curvature tensor. $\square$ Problem P6 does the same with the steps as exercises. The next slide reads the same fact as "the Weyl tensor vanishes identically at $n = 3$".

2+1 gravity is a real research subject for exactly this reason: with no local degrees of freedom the theory is exactly soluble, and it became the laboratory in which quantum gravity questions could be asked without the four-dimensional difficulties (Deser–Jackiw–'t Hooft 1984; Witten 1988). The BTZ black hole of 1992 exists only because a negative cosmological constant puts curvature back in the vacuum.

And at $n = 2$? Riemann has one component, Ricci has three but only one independent, since $R_{\mu\nu} = \tfrac12 Rg_{\mu\nu}$ identically. So Ricci determines Riemann there too, even more trivially — and the Einstein tensor is identically zero, which the last slide of this section makes the punchline.

33 · traces

The Weyl tensor: what the traces leave behind

the Weyl tensor, $n \ge 3$ $$\begin{aligned}C_{\rho\sigma\mu\nu} = {}&R_{\rho\sigma\mu\nu} - \tfrac{2}{n-2}\big(g_{\rho[\mu}R_{\nu]\sigma} - g_{\sigma[\mu}R_{\nu]\rho}\big)\\ &+ \tfrac{2R}{(n-1)(n-2)}\,g_{\rho[\mu}g_{\nu]\sigma}\end{aligned}$$

the unique combination with S1–S4 and every trace zero: $g^{\rho\mu}C_{\rho\sigma\mu\nu} = 0$.

So Riemann splits: $\;20 = 10 \;(\text{Ricci}) + 10\;(\text{Weyl})$ in four dimensions, $6 = 6 + 0$ in three, and at $n=2$ the split does not exist.

two facts, cited

$C$ is conformally invariant: rescaling $g \to e^{2\omega}g$ leaves $C^\rho{}_{\sigma\mu\nu}$ unchanged (Wald, General Relativity, App. D). And for $n \ge 4$, $C = 0$ on an open set iff the metric is conformally flat there (Weyl 1918, Schouten 1921).

The component count, as a formula. Weyl has $\tfrac{n^2(n^2-1)}{12} - \tfrac{n(n+1)}{2} = \tfrac{n(n+1)(n+2)(n-3)}{12}$ components: zero at $n = 3$, ten at $n = 4$, thirty-five at $n = 5$. The vanishing at $n=3$ is the previous slide's theorem, obtained here by arithmetic. Problem P9 asks for the algebra.

Checking the definition is trace-free is a short computation worth doing once: contract the boxed line with $g^{\rho\mu}$, use $g^{\rho\mu}g_{\rho\mu} = n$ and the antisymmetrisation brackets, and the three terms give $R_{\sigma\nu} - R_{\sigma\nu} = 0$. The coefficients $2/(n-2)$ and $2/((n-1)(n-2))$ are not aesthetic choices; they are the unique solution of that requirement, which is why they blow up at $n = 2$.

Petrov classification, for orientation only: the ten components of Weyl in four dimensions are classified algebraically into types I, II, D, III, N, O by the multiplicity of their principal null directions. Schwarzschild and Kerr are type D, a plane gravitational wave is type N. It is cited — Stephani et al., Exact Solutions of Einstein's Field Equations (2nd ed.), ch. 4 — and not used here, but it is the vocabulary unit 38 and unit 40 borrow.

33 · traces

Ricci shrinks a ball, Weyl shears it

what Weyl does physically

Unit 35's field equation will fix Ricci algebraically from the matter at each point. Weyl is what is left over, and it is what propagates: in vacuum $R_{\mu\nu} = 0$ and $C = R$, so the tides you feel in empty space are Weyl. The Moon's pull on the oceans, light bending near the Sun, and the gravitational waves of unit 38 are all Weyl curvature.

Schematic, exaggerated: a ball of dust released at rest. The trace of the tidal tensor — Ricci, $\ddot V/V = -R_{\mu\nu}u^\mu u^\nu$ — changes its volume and only where there is matter. The trace-free part changes its shape at fixed volume; in vacuum that part is all there is, and it is Weyl.

Reading the figure. The left ball is the case of pure trace: tidal tensor proportional to the identity, every particle accelerating inward at the same rate per unit offset, as inside a uniform dust cloud. The right ball is the vacuum tide of the previous section's ring, eigenvalues $(-2,1,1)$ summing to zero. A general tidal tensor is a sum of the two, which is the split of unit 05 done covariantly.

Why "the tides in empty space are Weyl" is the same sentence as unit 05's. Unit 05 split the tidal tensor as $\tfrac13(\operatorname{tr}T)\delta + \text{trace-free}$ and observed that the trace is fixed by the local density while the trace-free part is not. That split, done covariantly on the Riemann tensor, is exactly Ricci plus Weyl. The unit 05 notes said so in advance; this is the slide that cashes it.

33 · Bianchi

The Einstein tensor is divergence-free, and nobody chose that

contract the second Bianchi identity twice

Start from $\nabla_\lambda R_{\rho\sigma\mu\nu} + \nabla_\rho R_{\sigma\lambda\mu\nu} + \nabla_\sigma R_{\lambda\rho\mu\nu} = 0$ and contract with $g^{\lambda\mu}$. Using $\nabla g = 0$ to move the inverse metric through, and $R_{\sigma\nu} = g^{\rho\lambda}R_{\lambda\sigma\rho\nu}$:

$$\nabla^\mu R_{\rho\sigma\mu\nu} - \nabla_\rho R_{\sigma\nu} + \nabla_\sigma R_{\rho\nu} = 0 .$$

Now contract with $g^{\sigma\nu}$. The first term becomes $\nabla^\mu R_{\rho\mu}$, the second $-\nabla_\rho R$, the third $\nabla^\nu R_{\rho\nu}$:

$$2\nabla^\mu R_{\rho\mu} - \nabla_\rho R = 0 . \;\square$$

the Einstein tensor $$\begin{gathered}G_{\mu\nu} := R_{\mu\nu} - \tfrac12 R\,g_{\mu\nu},\\ \nabla^\mu G_{\mu\nu} \equiv 0 .\end{gathered}$$

The last line rearranges the proof: $\nabla^\mu(R_{\rho\mu} - \tfrac12 g_{\rho\mu}R) = \tfrac12\nabla_\rho R - \tfrac12\nabla_\rho R = 0$.

why this is the unit's second punchline

Local conservation of energy and momentum is $\nabla^\mu T_{\mu\nu} = 0$ for any matter (unit 35 sets up $T_{\mu\nu}$). A field equation $(\text{geometry})_{\mu\nu} = \kappa\,T_{\mu\nu}$ is therefore inconsistent unless the left side is identically divergence-free. $R_{\mu\nu} = \kappa T_{\mu\nu}$ fails. $G_{\mu\nu} = \kappa T_{\mu\nu}$ cannot fail. Unit 35 has almost no freedom left.

Full proof, with every contraction spelled out

First contraction. $g^{\lambda\mu}\nabla_\lambda R_{\rho\sigma\mu\nu} = \nabla^\mu R_{\rho\sigma\mu\nu}$. For the second term, $g^{\lambda\mu}R_{\sigma\lambda\mu\nu} = -g^{\lambda\mu}R_{\lambda\sigma\mu\nu} = -R_{\sigma\nu}$, using S2 and then the definition of Ricci. For the third, $g^{\lambda\mu}R_{\lambda\rho\mu\nu} = R_{\rho\nu}$ directly. Since $\nabla g^{-1} = 0$, the inverse metric passes through every $\nabla$.

Second contraction. $g^{\sigma\nu}R_{\rho\sigma\mu\nu}$: apply S1 and S2 together ($R_{\rho\sigma\mu\nu} = R_{\sigma\rho\nu\mu}$) to get $g^{\sigma\nu}R_{\sigma\rho\nu\mu} = R_{\rho\mu}$. So the first term is $\nabla^\mu R_{\rho\mu}$. The second is $-g^{\sigma\nu}\nabla_\rho R_{\sigma\nu} = -\nabla_\rho R$. The third is $g^{\sigma\nu}\nabla_\sigma R_{\rho\nu} = \nabla^\nu R_{\rho\nu}$, the same object as the first. Total $2\nabla^\mu R_{\rho\mu} - \nabla_\rho R = 0$. $\square$

The word "identically" is the whole point. $\nabla^\mu G_{\mu\nu} = 0$ holds for every metric on every manifold, solution or not. It is not a conservation law that holds on shell; it is an algebraic-differential identity, in the same family as $\nabla\cdot\nabla\times = 0$ from unit 04 (and, for a reader who has taken unit 28, $\partial_\mu\partial_\nu F^{\mu\nu} = 0$ in electromagnetism). Noether's second theorem associates such identities with a gauge symmetry — here diffeomorphism invariance (unit 31) — and it is always an identity of this kind that makes a gauge field equation consistent with its source's conservation. Cited: Kosmann-Schwarzbach, The Noether Theorems (2011), which also translates Noether's paper; unit 35 sees the diffeomorphism argument in the variational route.

Einstein's own path. In his Zurich notebook of 1912–13 he considered field equations built on the Ricci tensor and set them aside; the "Entwurf" theory of 1913, which was not generally covariant, and the hole argument followed. He came back to the Ricci tensor only in November 1915, first with $R_{\mu\nu} = \kappa T_{\mu\nu}$ — which, by the identity above, forces the trace of $T$ to be constant (problem P8) — and then, on 25 November, with the trace term that makes $G_{\mu\nu}$. The identity had been in print since Bianchi's paper of 1902; Einstein reached $G_{\mu\nu}$ without it, and Hilbert reached it the same month from the action of unit 35. That it took the century's most determined physicist three years to find what one identity hands over in four lines is the argument for learning the identity.

33 · closing the example

The sphere, all the way down

object	on $S^2_a$
$R_{\theta\phi\theta\phi}$	$a^2\sin^2\theta$
$K$	$1/a^2$
$R_{\mu\nu}$	$g_{\mu\nu}/a^2$
$R$	$2/a^2$
$G_{\mu\nu}$	$g_{\mu\nu}/a^2 - \tfrac12(2/a^2)g_{\mu\nu} = 0$
holonomy of $\triangle$	$\pi/2 = \int_\triangle K\,\mathrm dA$
deviation of two meridians	$\xi = a\Delta\phi\cos(s/a)$

the last row of the table is a theorem

$G_{\mu\nu} \equiv 0$ on every two-dimensional manifold, because $R_{\mu\nu} = \tfrac12 Rg_{\mu\nu}$ identically there. So the Einstein equation in two dimensions reads $0 = \kappa T_{\mu\nu}$: it permits no matter at all. Gravity, as a field theory, does not exist below three dimensions — and, by the $n=3$ slide, has no vacuum degrees of freedom below four.

four dimensions is the first interesting case

$n=2$: $G \equiv 0$, no theory. $n=3$: $G$ determines $R$ entirely, no waves, no exterior field. $n=4$: ten components of $G$ fixed by matter, ten of Weyl left free to propagate. The next five units are about those ten.

Why $R_{\mu\nu} = \tfrac12 Rg_{\mu\nu}$ in two dimensions. From the constant-curvature form of section 3, valid at each point with $K = K(p)$: $R_{\rho\sigma\mu\nu} = K(g_{\rho\mu}g_{\sigma\nu}-g_{\rho\nu}g_{\sigma\mu})$. Contract: $R_{\sigma\nu} = g^{\rho\mu}R_{\rho\sigma\mu\nu} = K(2g_{\sigma\nu} - g_{\sigma\nu}) = Kg_{\sigma\nu}$, and $R = 2K$. So $R_{\mu\nu} - \tfrac12 Rg_{\mu\nu} = Kg_{\mu\nu} - Kg_{\mu\nu} = 0$. ✓

There is a two-dimensional gravity worth having, but it is not this one: Jackiw–Teitelboim gravity adds a scalar field whose equation of motion supplies what $G$ cannot. It is outside this course, and it is named here only so that "gravity does not exist in two dimensions" is not remembered as more sweeping than it is.

The sphere earned its keep. It supplied a curvature to compute, a holonomy to check against an integral, a pair of geodesics that meet, a case where Theorema Egregium bites, and a dimension in which the Einstein tensor is empty. Every one of those was a different face of the same single number $1/a^2$, which is what "curvature is one function on a surface" means in practice.

33 · checkpoint

Checkpoint

you can now do all four

Compute the Riemann tensor of a 2-surface from its metric and verify it has exactly one independent component, the Gaussian curvature.
Prove the contracted Bianchi identity $\nabla^\mu G_{\mu\nu} = 0$ from the second Bianchi identity, every contraction justified.
Reduce the geodesic deviation equation to unit 05's tidal equation, with $R^i{}_{0j0} = \partial_i\partial_j\Phi/c^2$ and the $c^2$ in its place.
Count the components of $R$ in $n$ dimensions and say what $20 = 10 + 10$ means in four.

the unit in one line

Covariant derivatives fail to commute by a tensor; no chart removes it, it is the tidal tensor, and one identity makes its trace-reversed trace conserved for free.

Unit 34 rewrites all of it with forms, and unit 35 writes down the field equation that the last slide of section 5 has already almost forced.

Twelve problems, about seven hours: ./mr problems 33. When the four items above are true: ./mr done 33.

The falsifiable form: take the hyperbolic upper half-plane $g = a^2(\mathrm dx^2+\mathrm dy^2)/y^2$ (problem P2) and the weak-field metric of the Newtonian-limit slide. If you cannot produce $K = -1/a^2$ with a justification that exactly one component was needed, cannot get from $\nabla_{[\lambda}R_{\rho\sigma]\mu\nu} = 0$ to $\nabla^\mu G_{\mu\nu} = 0$ without looking, and cannot recover $\ddot\xi_i = -(\partial_i\partial_j\Phi)\xi_j$ with the $c^2$ in the right place, then the checkpoint is false and unit 35 will be a list of symbols.

Two habits for what follows. First: every time a curvature component is computed, check it against a limit where the answer is known — flat space, large radius, or a symmetry that forces a component to vanish. Unit 36's Schwarzschild computation has dozens of components and no way to check the answer except this. Second: when a new tensor appears, ask what its traces are and whether the remainder is conformally invariant; that question is what organises everything from here to unit 40.