General Relativity — worked problems

(a) Angle preservation

For tangent vectors $u,v\in T_p M$, the angle $\theta$ is defined by $$\cos\theta = \frac{g(u,v)}{\sqrt{g(u,u)\cdot g(v,v)}}.$$ Under conformal rescaling $\tilde g = fg$ with $f>0$:

$$\cos\tilde\theta = \frac{\tilde g(u,v)}{\sqrt{\tilde g(u,u)\tilde g(v,v)}} = \frac{f\,g(u,v)}{\sqrt{f\,g(u,u)\cdot f\,g(v,v)}} = \frac{f\,g(u,v)}{f\sqrt{g(u,u)g(v,v)}} = \cos\theta.$$

The conformal factor cancels exactly. $$\boxed{\;\cos\tilde\theta = \cos\theta\;\text{(angles preserved).}\;}$$

(b) Null-curve preservation

A curve $\gamma$ is null if $g(\dot\gamma,\dot\gamma) = 0$ everywhere. Under $\tilde g = fg$:

$$\tilde g(\dot\gamma,\dot\gamma) = f\cdot g(\dot\gamma,\dot\gamma) = f\cdot 0 = 0.$$

$$\boxed{\;\text{Null curves remain null under conformal transformation.}\;}$$

Why this matters

Together: angles and light cones depend only on the conformal class $[g]$, not on the specific representative.

Penrose diagrams. Conformal rescaling can bring asymptotic infinity to a finite "edge" while preserving causal structure exactly. The Penrose diagram of Minkowski spacetime (or Schwarzschild, Kerr) is a conformally compactified picture of the original geometry.

AdS/CFT. Anti-de Sitter space has a conformal boundary at $r\to\infty$. The boundary CFT (Conformal Field Theory) lives on the conformal class of this boundary, not on a specific metric. This is why the CFT side has conformal symmetry — it's reading off the angles and causal structure of the bulk AdS at infinity.

Photon kinematics. The vacuum Maxwell equations are conformally invariant in 4D: rescaling $g\to fg$ leaves photon trajectories unchanged. The 4D coincidence is the structural reason for the masslessness of the photon — consistent with conformal invariance.

Induced metric

From $x = R\cos\phi\sin\theta$, $y = R\sin\phi\sin\theta$, $z = R\alpha\cos\theta$ compute differentials:

$dx = -R\sin\phi\sin\theta\,d\phi + R\cos\phi\cos\theta\,d\theta$,
$dy = R\cos\phi\sin\theta\,d\phi + R\sin\phi\cos\theta\,d\theta$,
$dz = -R\alpha\sin\theta\,d\theta$.

Squaring and summing with cross terms canceling by $\sin^2\phi+\cos^2\phi = 1$:

$dx^2 + dy^2 = R^2\cos^2\theta\,d\theta^2 + R^2\sin^2\theta\,d\phi^2$.
$dz^2 = R^2\alpha^2\sin^2\theta\,d\theta^2$.

Summing:

$$\boxed{\;ds^2 = R^2(\cos^2\theta + \alpha^2\sin^2\theta)\,d\theta^2 + R^2\sin^2\theta\,d\phi^2.\;}$$

Metric and inverse

$g_{\theta\theta} = R^2(\cos^2\theta + \alpha^2\sin^2\theta)$, $g_{\phi\phi} = R^2\sin^2\theta$. Inverse: $g^{\theta\theta} = 1/g_{\theta\theta}$, $g^{\phi\phi} = 1/g_{\phi\phi}$.

Metric derivatives

$\partial_\theta g_{\theta\theta} = R^2(-2\cos\theta\sin\theta + 2\alpha^2\sin\theta\cos\theta) = R^2(\alpha^2-1)\sin(2\theta)$.

$\partial_\theta g_{\phi\phi} = 2R^2\sin\theta\cos\theta = R^2\sin(2\theta)$.

Christoffels

$$\Gamma^\theta_{\theta\theta} = \tfrac12 g^{\theta\theta}\partial_\theta g_{\theta\theta} = \frac{(\alpha^2-1)\sin(2\theta)}{2(\cos^2\theta + \alpha^2\sin^2\theta)}.$$

$$\Gamma^\theta_{\phi\phi} = -\tfrac12 g^{\theta\theta}\partial_\theta g_{\phi\phi} = -\frac{\sin(2\theta)}{2(\cos^2\theta + \alpha^2\sin^2\theta)}.$$

$$\Gamma^\phi_{\theta\phi} = \tfrac12 g^{\phi\phi}\partial_\theta g_{\phi\phi} = \frac{R^2\sin(2\theta)}{2 R^2\sin^2\theta} = \cot\theta.$$

$$\boxed{\;\Gamma^\theta_{\theta\theta} = \tfrac{(\alpha^2-1)\sin(2\theta)}{2(\cos^2\theta+\alpha^2\sin^2\theta)},\;\;\Gamma^\theta_{\phi\phi} = -\tfrac{\sin(2\theta)}{2(\cos^2\theta+\alpha^2\sin^2\theta)},\;\;\Gamma^\phi_{\theta\phi} = \cot\theta.\;}$$

Cross-check: round-sphere limit

Setting $\alpha = 1$ gives $\cos^2\theta + \sin^2\theta = 1$, so $\Gamma^\theta_{\theta\theta}\to 0$, $\Gamma^\theta_{\phi\phi}\to -\sin(2\theta)/2 = -\tfrac12\sin(2\theta)$, $\Gamma^\phi_{\theta\phi}\to\cot\theta$ — matching Problem 2.7. ✓

Physical interpretation

For $\alpha < 1$: oblate spheroid (flattened at poles, like Earth). For $\alpha > 1$: prolate (elongated, like an American football). The non-zero $\Gamma^\theta_{\theta\theta}$ encodes the latitude-dependent stretching. Used in modelling Earth's geoid, rotating fluid drops, and gravitating stars in slow rotation.

(a) Contracted Christoffel identity

Apply $\Gamma^a_{bc} = \tfrac12 g^{ad}(\partial_b g_{cd}+\partial_c g_{bd}-\partial_d g_{bc})$ with $a = \mu$, $b = \mu$, $c = \nu$:

$$\Gamma^\mu_{\mu\nu} = \tfrac12 g^{\mu\lambda}(\partial_\mu g_{\nu\lambda} + \partial_\nu g_{\mu\lambda} - \partial_\lambda g_{\mu\nu}).$$

The first and third terms cancel by symmetry of $g^{\mu\lambda}$: relabel $\lambda\leftrightarrow\mu$ in the third term gives $g^{\lambda\mu}\partial_\mu g_{\lambda\nu} = g^{\mu\lambda}\partial_\mu g_{\nu\lambda}$, equal to the first. So $$\Gamma^\mu_{\mu\nu} = \tfrac12 g^{\mu\lambda}\partial_\nu g_{\mu\lambda}.$$

Apply Jacobi's formula for derivative of a determinant: $\partial_\nu\bar g = \bar g\cdot\mathrm{tr}(g^{-1}\partial_\nu g) = \bar g\cdot g^{\mu\lambda}\partial_\nu g_{\mu\lambda}$. Therefore $$\boxed{\;\Gamma^\mu_{\mu\nu} = \frac{1}{2\bar g}\partial_\nu\bar g = \tfrac12\partial_\nu\ln|\bar g| = \partial_\nu\ln\sqrt{|\bar g|}.\;}$$

(b) Transformation rule

From the demand that $\nabla_\mu V^\nu = \partial_\mu V^\nu + \Gamma^\nu_{\mu\rho}V^\rho$ transform as a $(1,1)$-tensor, one derives (cf. Problem 2.5):

$$\boxed{\;\Gamma'^b_{ac} = \frac{\partial x'^b}{\partial x^i}\frac{\partial x^j}{\partial x'^a}\frac{\partial x^k}{\partial x'^c}\Gamma^i_{jk} + \frac{\partial x'^b}{\partial x^i}\frac{\partial^2 x^i}{\partial x'^a\partial x'^c}.\;}$$

The first term is the homogeneous tensor part. The second term — the inhomogeneous second-derivative piece — is what makes $\Gamma$ a connection, not a tensor.

(c) Parallel transport preserves length

$|V|^2 = g_{\mu\nu}V^\mu V^\nu$. Differentiate along the curve $x(\lambda)$ with parallel-transported $V$ (so $\dot V^\mu + \Gamma^\mu_{\rho\sigma}\dot x^\rho V^\sigma = 0$, i.e. $\dot V^\mu = -\Gamma^\mu_{\rho\sigma}\dot x^\rho V^\sigma$):

$$\frac{d|V|^2}{d\lambda} = \dot x^\rho\partial_\rho g_{\mu\nu}V^\mu V^\nu + 2g_{\mu\nu}\dot V^\mu V^\nu = \dot x^\rho V^\mu V^\nu\bigl[\partial_\rho g_{\mu\nu} - 2g_{\lambda\nu}\Gamma^\lambda_{\rho\mu}\bigr].$$

The bracket is $\partial_\rho g_{\mu\nu} - \Gamma_{\rho\mu\nu} - \Gamma_{\rho\nu\mu}$ (using $\Gamma_{\mu\nu\rho} = \Gamma_{\nu\mu\rho}$ symmetry of last two indices in lowered form), which equals $-D_\rho g_{\mu\nu}$. By metric compatibility $D_\rho g_{\mu\nu} = 0$:

$$\boxed{\;\frac{d|V|^2}{d\lambda} = 0\;\text{(parallel transport preserves length).}\;}$$

This is the structural reason the Levi-Civita connection is the unique torsion-free metric-compatible connection (Fundamental Theorem, Problem 2.25): it's the unique one such that parallel transport preserves the geometric structure of vectors.

Step-by-step

By metric compatibility $\nabla_\mu g_{\nu\sigma} = 0$ (Problem 2.33c). Apply the Leibniz rule to $V_\nu = g_{\nu\sigma}V^\sigma$:

$$\nabla_\mu V_\nu = \nabla_\mu(g_{\nu\sigma}V^\sigma) = \underbrace{(\nabla_\mu g_{\nu\sigma})}_{=\,0}V^\sigma + g_{\nu\sigma}\nabla_\mu V^\sigma = g_{\nu\sigma}\nabla_\mu V^\sigma.$$

Multiply both sides by $U^\mu$ and use linearity of contraction:

$$U^\mu\nabla_\mu V_\nu = g_{\nu\sigma}\underbrace{(U^\mu\nabla_\mu V^\sigma)}_{= W^\sigma} = g_{\nu\sigma}W^\sigma = W_\nu.$$

$$\boxed{\;W_\nu = U^\mu\nabla_\mu V_\nu\;\;\text{(index lowering commutes with covariant differentiation).}\;}$$

Significance

This is the operational content of metric compatibility: raising/lowering indices commutes with $\nabla$. In tensor calculus on a Riemannian manifold one routinely raises and lowers indices and applies $\nabla$, and the order doesn't matter — just like ordinary calculus where commuting $\partial$ with constants works.

Were metric compatibility violated ($\nabla g \neq 0$), tensor calculus would become considerably more complicated: $\nabla_\mu V_\nu \neq g_{\nu\sigma}\nabla_\mu V^\sigma$ in general, requiring careful tracking of which "version" of the vector one differentiates. This is why all of physics is formulated on a metric-compatible manifold (Levi-Civita connection).

Orthonormal frame

$e_1 = \partial_\theta$ (unit since $g_{\theta\theta} = 1$), $e_2 = (1/\sin\theta)\partial_\phi$ (unit since $g(\partial_\phi,\partial_\phi) = \sin^2\theta$). Check: $\langle e_2,e_2\rangle = (1/\sin^2\theta)\sin^2\theta = 1$ ✓.

Frame Christoffels via $\nabla_{e_a}e_b$

Using coordinate Christoffels from Problem 2.7 ($\Gamma^\theta_{\phi\phi} = -\tfrac12\sin(2\theta)$, $\Gamma^\phi_{\theta\phi} = \cot\theta$):

$\nabla_{e_1}e_1 = \nabla_{\partial_\theta}\partial_\theta = \Gamma^a_{\theta\theta}\partial_a = 0$ (no $\Gamma$ has lower indices $(\theta,\theta)$).

$\nabla_{e_1}e_2$: $\nabla_{\partial_\theta}[(1/\sin\theta)\partial_\phi] = \partial_\theta(1/\sin\theta)\partial_\phi + (1/\sin\theta)\nabla_{\partial_\theta}\partial_\phi$ $= -(\cos\theta/\sin^2\theta)\partial_\phi + (1/\sin\theta)\cot\theta\,\partial_\phi$ $= -(\cos\theta/\sin^2\theta)\partial_\phi + (\cos\theta/\sin^2\theta)\partial_\phi = 0$.

$\nabla_{e_2}e_1$: $\nabla_{(1/\sin\theta)\partial_\phi}\partial_\theta = (1/\sin\theta)\Gamma^\phi_{\phi\theta}\partial_\phi = (\cot\theta/\sin\theta)\partial_\phi = \cot\theta\cdot e_2$.

$\nabla_{e_2}e_2$: $(1/\sin\theta)\nabla_{\partial_\phi}[(1/\sin\theta)\partial_\phi] = (1/\sin^2\theta)\Gamma^\theta_{\phi\phi}\partial_\theta = (1/\sin^2\theta)(-\sin\theta\cos\theta)\partial_\theta = -\cot\theta\cdot e_1$.

$$\boxed{\;\nabla_{e_1}e_1 = \nabla_{e_1}e_2 = 0,\quad \nabla_{e_2}e_1 = \cot\theta\,e_2,\quad \nabla_{e_2}e_2 = -\cot\theta\,e_1.\;}$$

Spin connection 1-form

Define $\omega^{ab}{}_c$ by $\nabla_{e_c}e_b = \omega^a{}_{bc}e_a$. From above: $\omega^2{}_{12} = \cot\theta$, $\omega^1{}_{22} = -\cot\theta$. The 1-form $\omega^{12} \equiv \omega^1{}_{2c}dx^c$. Using $dx^c = (d\theta,d\phi)$ and $e_2 = (1/\sin\theta)\partial_\phi$ so $d\phi = \sin\theta\,e^2$:

$$\omega^{12} = -\cos\theta\,d\phi.$$

Holonomy by Green's theorem

Around a closed loop $\gamma$ bounding a region $D$ on $S^2$:

$$\Omega = \oint_\gamma\omega^{12} = -\oint\cos\theta\,d\phi.$$

By Green's theorem (Stokes in 2D): $-\oint\cos\theta\,d\phi = -\iint_D d(\cos\theta\wedge d\phi) = -\iint_D(-\sin\theta\,d\theta\wedge d\phi) = \iint_D\sin\theta\,d\theta\,d\phi$.

But $\sin\theta\,d\theta\,d\phi$ is exactly the area element of the unit $S^2$. Hence:

$$\boxed{\;\Omega = \iint_D \sin\theta\,d\theta\,d\phi = \mathrm{Area}(D).\;}$$

Significance

This is the analytic counterpart of Problem 2.12's Gauss–Bonnet derivation. The "spin connection" $\omega^{12}$ is the curvature 1-form of the orthonormal frame bundle, and its integral around a loop equals the holonomy. Same machinery underlies the Foucault pendulum precession, Berry phases in QM, and the Atiyah–Singer index theorem.

Setup

$g_{rr} = 1$, $g_{\theta\theta} = r^2$, $g_{\phi\phi} = r^2\sin^2\theta$, off-diagonals zero. Inverse: $g^{rr} = 1$, $g^{\theta\theta} = 1/r^2$, $g^{\phi\phi} = 1/(r^2\sin^2\theta)$.

Non-zero metric derivatives

$\partial_r g_{\theta\theta} = 2r$, $\partial_r g_{\phi\phi} = 2r\sin^2\theta$, $\partial_\theta g_{\phi\phi} = 2r^2\sin\theta\cos\theta = r^2\sin(2\theta)$.

Computing Christoffels

$\Gamma^r_{\theta\theta} = -\tfrac12 g^{rr}\partial_r g_{\theta\theta} = -\tfrac12\cdot 1\cdot 2r = -r$.

$\Gamma^r_{\phi\phi} = -\tfrac12 g^{rr}\partial_r g_{\phi\phi} = -r\sin^2\theta$.

$\Gamma^\theta_{r\theta} = \tfrac12 g^{\theta\theta}\partial_r g_{\theta\theta} = \tfrac12\cdot(1/r^2)\cdot 2r = 1/r$.

$\Gamma^\theta_{\phi\phi} = -\tfrac12 g^{\theta\theta}\partial_\theta g_{\phi\phi} = -\tfrac12\cdot(1/r^2)\cdot r^2\sin(2\theta) = -\tfrac12\sin(2\theta) = -\sin\theta\cos\theta$.

$\Gamma^\phi_{r\phi} = \tfrac12 g^{\phi\phi}\partial_r g_{\phi\phi} = \tfrac12\cdot[1/(r^2\sin^2\theta)]\cdot 2r\sin^2\theta = 1/r$.

$\Gamma^\phi_{\theta\phi} = \tfrac12 g^{\phi\phi}\partial_\theta g_{\phi\phi} = \tfrac12\cdot[1/(r^2\sin^2\theta)]\cdot 2r^2\sin\theta\cos\theta = \cot\theta$.

$$\boxed{\;\Gamma^r_{\theta\theta} = -r,\;\;\Gamma^r_{\phi\phi} = -r\sin^2\theta,\;\;\Gamma^\theta_{r\theta} = \tfrac{1}{r},\;\;\Gamma^\theta_{\phi\phi} = -\sin\theta\cos\theta,\;\;\Gamma^\phi_{r\phi} = \tfrac{1}{r},\;\;\Gamma^\phi_{\theta\phi} = \cot\theta.\;}$$

Lesson

Non-zero Christoffels in flat space: $R^a{}_{bcd} = 0$ identically with these $\Gamma$. The Christoffels reflect coordinate curvilinearity, not curvature. They show up in the textbook polar Laplacian:

$$\Box\phi = \frac{1}{r^2}\partial_r(r^2\partial_r\phi) + \frac{1}{r^2\sin\theta}\partial_\theta(\sin\theta\,\partial_\theta\phi) + \frac{1}{r^2\sin^2\theta}\partial_\phi^2\phi,$$

and in the polar divergence familiar from EM. The deep insight: curvilinear charts on flat space have non-zero $\Gamma$, but Riemann vanishes.

Setup

Substitute $\rho \equiv 2R\tan(\theta/2)$ so $x = \rho\cos\phi$, $y = \rho\sin\phi$, $\rho^2 = x^2+y^2$. Useful identities: $\sin\theta = 4R\rho/(\rho^2+4R^2)$ (using $\sin\theta = 2\sin(\theta/2)\cos(\theta/2)$ and $\tan(\theta/2) = \rho/2R$).

Differentials

$d\rho = \frac{R}{\cos^2(\theta/2)}d\theta = \frac{R(\rho^2+4R^2)}{4R^2}d\theta$, so $d\theta = \frac{4R\,d\rho}{\rho^2+4R^2}$.

Substitute into $ds^2 = R^2(d\theta^2 + \sin^2\theta\,d\phi^2)$

$R^2 d\theta^2 = R^2\cdot \frac{16R^2 d\rho^2}{(\rho^2+4R^2)^2} = \frac{16R^4 d\rho^2}{(\rho^2+4R^2)^2}$.

$R^2\sin^2\theta\,d\phi^2 = R^2\cdot\frac{16R^2\rho^2}{(\rho^2+4R^2)^2}d\phi^2 = \frac{16R^4\rho^2 d\phi^2}{(\rho^2+4R^2)^2}$.

Sum: $\frac{16R^4(d\rho^2 + \rho^2 d\phi^2)}{(\rho^2+4R^2)^2}$. Using $d\rho^2 + \rho^2 d\phi^2 = dx^2 + dy^2$:

$$\boxed{\;ds^2 = \frac{16R^4(dx^2+dy^2)}{(x^2+y^2+4R^2)^2}.\;}$$

Conformally flat structure

$g_{ij} = \Omega^2\delta_{ij}$ with conformal factor $$\Omega(x,y) = \frac{4R^2}{x^2+y^2+4R^2}.$$ Note $\Omega\to 1$ at the origin (south pole) and $\Omega\to 0$ as $r^2\to\infty$ (the north pole at infinity in the plane).

Christoffels of a conformally flat metric

Standard formula: $$\Gamma^k_{ij} = \delta^k_i\partial_j\ln\Omega + \delta^k_j\partial_i\ln\Omega - \delta_{ij}\delta^{kl}\partial_l\ln\Omega.$$

With $\ln\Omega = \ln(4R^2) - \ln(x^2+y^2+4R^2)$, $\partial_x\ln\Omega = -2x/(x^2+y^2+4R^2)$, $\partial_y\ln\Omega = -2y/(x^2+y^2+4R^2)$. Substituting yields explicit non-zero Christoffels everywhere smooth.

Geometric significance

This is the Riemann sphere $\mathbb{CP}^1 \cong S^2$. Foundation of complex analysis: holomorphic maps on $\mathbb{CP}^1$ are exactly Möbius transformations $z\mapsto(az+b)/(cz+d)$. Penrose's twistor program builds on the conformal structure here. Stereographic projection preserves angles (it's conformal) but distorts areas; meridians map to rays from origin, latitudes to concentric circles. Used in cartography (e.g. Mercator-related projections preserving angle/heading).

(a) Parametrisation

With $z = 0$, the constraint $x^2+y^2-t^2 = 1$ describes a 2D surface in 3D Minkowski $\mathbb R^{2,1}$ (signature $-++$). At each $t$, $(x,y)$ lie on a circle of radius $\sqrt{1+t^2}$. Use parameters $(\chi,\phi)$:

$$\boxed{\;t = \sinh\chi,\;\;x = \cosh\chi\cos\phi,\;\;y = \cosh\chi\sin\phi,\;\;\chi\in\mathbb R,\;\phi\in[0,2\pi).\;}$$

Check: $x^2+y^2-t^2 = \cosh^2\chi - \sinh^2\chi = 1$ ✓.

(b) Induced metric

Ambient $ds^2 = dt^2 - dx^2 - dy^2$ (in $(+--)$ convention). Differentials:

$dt = \cosh\chi\,d\chi$,
$dx = \sinh\chi\cos\phi\,d\chi - \cosh\chi\sin\phi\,d\phi$,
$dy = \sinh\chi\sin\phi\,d\chi + \cosh\chi\cos\phi\,d\phi$.

$dt^2 = \cosh^2\chi\,d\chi^2$.

$dx^2 + dy^2 = (\sinh^2\chi)(\cos^2\phi+\sin^2\phi)d\chi^2 + \cosh^2\chi(\sin^2\phi+\cos^2\phi)d\phi^2 + (\text{cross terms cancel})$
$= \sinh^2\chi\,d\chi^2 + \cosh^2\chi\,d\phi^2$.

So $ds^2 = (\cosh^2\chi - \sinh^2\chi)d\chi^2 - \cosh^2\chi\,d\phi^2 = d\chi^2 - \cosh^2\chi\,d\phi^2$:

$$\boxed{\;ds^2 = d\chi^2 - \cosh^2\chi\,d\phi^2.\;}$$

Signature $(+,-)$ — Lorentzian. $\chi$ is timelike, $\phi$ is spacelike. Global slicing of 2D de Sitter: the spatial circle has circumference $2\pi\cosh\chi$, contracting from infinity at $\chi\to -\infty$ to minimum $2\pi$ at $\chi = 0$, then re-expanding.

(c) Christoffel symbols

$g_{\chi\chi} = 1$, $g_{\phi\phi} = -\cosh^2\chi$. Only non-zero derivative: $\partial_\chi g_{\phi\phi} = -2\cosh\chi\sinh\chi = -\sinh(2\chi)$.

$\Gamma^\chi_{\phi\phi} = -\tfrac12 g^{\chi\chi}\partial_\chi g_{\phi\phi} = -\tfrac12\cdot 1\cdot(-\sinh(2\chi)) = \tfrac12\sinh(2\chi) = \cosh\chi\sinh\chi$.

$\Gamma^\phi_{\chi\phi} = \tfrac12 g^{\phi\phi}\partial_\chi g_{\phi\phi} = \tfrac12\cdot(-1/\cosh^2\chi)\cdot(-\sinh(2\chi)) = \tanh\chi$.

$$\boxed{\;\Gamma^\chi_{\phi\phi} = \cosh\chi\sinh\chi,\;\;\Gamma^\phi_{\chi\phi} = \Gamma^\phi_{\phi\chi} = \tanh\chi.\;}$$

These are FRW Christoffels with scale factor $a(\chi) = \cosh\chi$: $\Gamma^\chi_{\phi\phi} = a\dot a$, $\Gamma^\phi_{\chi\phi} = \dot a/a$. Ricci scalar $R = -2$ (in $(+,-)$ signature), i.e. $\text{dS}_2$ with curvature radius $\ell = 1$.

(a) Riemann components in coordinates

Take basis $X = \partial_\nu$, $Y = \partial_\lambda$, $Z = \partial_\mu$ (commute), so $[X,Y] = 0$ and

$$R(\partial_\nu,\partial_\lambda)\partial_\mu = \nabla_\nu\nabla_\lambda\partial_\mu - \nabla_\lambda\nabla_\nu\partial_\mu.$$

Compute $\nabla_\lambda\partial_\mu = \Gamma^\sigma_{\lambda\mu}\partial_\sigma$, then apply $\nabla_\nu$ via Leibniz:

$$\nabla_\nu\nabla_\lambda\partial_\mu = (\partial_\nu\Gamma^\omega_{\lambda\mu})\partial_\omega + \Gamma^\sigma_{\lambda\mu}\nabla_\nu\partial_\sigma = (\partial_\nu\Gamma^\omega_{\lambda\mu} + \Gamma^\omega_{\nu\sigma}\Gamma^\sigma_{\lambda\mu})\partial_\omega.$$

Subtracting the $(\nu\leftrightarrow\lambda)$ version:

$$\boxed{\;R^\omega{}_{\mu\nu\lambda} = \partial_\nu\Gamma^\omega_{\lambda\mu} - \partial_\lambda\Gamma^\omega_{\nu\mu} + \Gamma^\omega_{\nu\sigma}\Gamma^\sigma_{\lambda\mu} - \Gamma^\omega_{\lambda\sigma}\Gamma^\sigma_{\nu\mu}.\;}$$

First Bianchi (cyclic in $\mu,\nu,\lambda$)

Cycle $(\mu,\nu,\lambda)\to(\nu,\lambda,\mu)\to(\lambda,\mu,\nu)$ and sum. The six $\partial\Gamma$ terms pair by symmetry of the lower Christoffel indices ($\Gamma^\sigma_{\mu\nu} = \Gamma^\sigma_{\nu\mu}$, torsion-free): each pair has form $\partial_\nu\Gamma^\omega_{\lambda\mu} - \partial_\nu\Gamma^\omega_{\mu\lambda} = 0$. The six $\Gamma\Gamma$ terms group as $\Gamma^\omega_{\sigma}(\Gamma^\sigma_{\lambda\mu}-\Gamma^\sigma_{\mu\lambda})$, all vanishing. Thus:

$$\boxed{\;R^\omega{}_{\mu\nu\lambda} + R^\omega{}_{\nu\lambda\mu} + R^\omega{}_{\lambda\mu\nu} = 0\quad (\text{first Bianchi, torsion-free}).\;}$$

(b) Second Bianchi identity

Work in Riemann normal coordinates at point $p$, where $\Gamma^\rho_{\mu\nu}(p) = 0$ (always possible; first derivatives generally non-zero). At $p$:

$$R_{\alpha\beta\mu\nu}\big|_p = g_{\alpha\rho}(\partial_\mu\Gamma^\rho_{\nu\beta} - \partial_\nu\Gamma^\rho_{\mu\beta})\big|_p.$$

Differentiate with $\nabla_\lambda$; at $p$, $\nabla$ reduces to $\partial$:

$$R_{\alpha\beta\mu\nu;\lambda}\big|_p = g_{\alpha\rho}(\partial_\lambda\partial_\mu\Gamma^\rho_{\nu\beta} - \partial_\lambda\partial_\nu\Gamma^\rho_{\mu\beta}).$$

Cyclic permute $(\lambda,\mu,\nu)$ and sum. The six second-derivative terms pair as $(\partial_\lambda\partial_\mu - \partial_\mu\partial_\lambda)\Gamma^\rho_{\nu\beta}$, all of which vanish (partials commute). Result holds at $p$ in normal coords; being a tensor equation, it holds everywhere:

$$\boxed{\;R_{\alpha\beta\mu\nu;\lambda} + R_{\alpha\beta\nu\lambda;\mu} + R_{\alpha\beta\lambda\mu;\nu} = 0\quad(\text{second Bianchi}).\;}$$

Vanishing of $\nabla_\mu T^{\mu\nu}$

Contract the second Bianchi with $g^{\alpha\mu}g^{\beta\nu}$. Using contractions and antisymmetries:

$g^{\alpha\mu}g^{\beta\nu}R_{\alpha\beta\mu\nu} = R$ (scalar curvature),
$g^{\beta\nu}R_{\alpha\beta\nu\lambda} = -R_{\alpha\lambda}$,
$g^{\alpha\mu}R_{\alpha\beta\lambda\mu} = -R_{\beta\lambda}$.

The contracted second Bianchi becomes: $\nabla_\lambda R - \nabla_\mu R^\mu{}_\lambda - \nabla_\mu R^\mu{}_\lambda = 0$, i.e. $\nabla_\mu R^\mu{}_\lambda = \tfrac12\nabla_\lambda R$. Equivalently:

$$\boxed{\;\nabla_\mu G^{\mu\nu} = \nabla_\mu(R^{\mu\nu} - \tfrac12 g^{\mu\nu}R) = 0\quad(\text{identically true}).\;}$$

Combined with Einstein's equation $G^{\mu\nu} = 8\pi G T^{\mu\nu}/c^4$:

$$\boxed{\;\nabla_\mu T^{\mu\nu} = 0.\;}$$

Physical content

In flat space, this reduces to $\partial_\mu T^{\mu\nu} = 0$ — the local energy-momentum conservation law. In curved space, it generalises but does NOT yield globally-conserved integrated quantities unless Killing vectors exist. The structure $\nabla G = 0$ is an identity (Bianchi), not an equation; Einstein derived the form of the field equations from this constraint.

Historical note

Einstein insisted from the start that matter must satisfy a local conservation law $\nabla T = 0$. He searched for a geometric tensor with identically-vanishing divergence to match. The Bianchi identities supply exactly one: $G^{\mu\nu}$ (up to a metric multiple $\Lambda g^{\mu\nu}$ — the cosmological constant term). So the field equations $$G^{\mu\nu} + \Lambda g^{\mu\nu} = \kappa T^{\mu\nu}$$ are forced uniquely by Bianchi identities + equivalence principle + diffeomorphism invariance.

Setup

Vector $V^\rho$ at base point $p$. Infinitesimal parallelogram with sides $\epsilon A^\mu$ and $\delta B^\nu$ ($\epsilon,\delta\to 0$). Traverse in four legs:

1. $p\to p_1 = p + \epsilon A$ along $A$.
2. $p_1\to p_2 = p_1 + \delta B$ along $B$ (where $B$ is evaluated at $p_1$).
3. $p_2\to p_3 = p_2 - \epsilon A$ along $-A$.
4. $p_3\to p$ back along $-B$.

Leg-by-leg parallel transport

Parallel transport along $\epsilon A$: $V^\rho(p_1) = V^\rho(p) - \epsilon A^\mu\Gamma^\rho_{\mu\sigma}V^\sigma + O(\epsilon^2)$.

Along $\delta B$ from $p_1$: $V^\rho(p_2) = V^\rho(p_1) - \delta B^\nu\Gamma^\rho_{\nu\sigma}(p_1)V^\sigma(p_1) + O(\delta^2)$. Expand $\Gamma$ at $p_1 = p + \epsilon A$: $\Gamma^\rho_{\nu\sigma}(p_1) = \Gamma^\rho_{\nu\sigma}(p) + \epsilon A^\mu\partial_\mu\Gamma^\rho_{\nu\sigma}(p) + O(\epsilon^2)$. Multiply through and keep terms to order $\epsilon\delta$.

Net change at order $\epsilon\delta$

The first-order corrections cancel (closed loop), and the second-order $\epsilon\delta$ term gives:

$$\Delta V^\rho = -\epsilon\delta(\partial_\mu\Gamma^\rho_{\nu\sigma} - \partial_\nu\Gamma^\rho_{\mu\sigma} + \Gamma^\rho_{\mu\lambda}\Gamma^\lambda_{\nu\sigma} - \Gamma^\rho_{\nu\lambda}\Gamma^\lambda_{\mu\sigma})A^\mu B^\nu V^\sigma.$$

The bracket is exactly $R^\rho{}_{\sigma\mu\nu}$ (cf. Problem 2.39a):

$$\boxed{\;\Delta V^\rho = -R^\rho{}_{\sigma\mu\nu}V^\sigma\,(\epsilon A^\mu)(\delta B^\nu).\;}$$

Commutator form

Equivalently, applied to general vector fields:

$$\boxed{\;[\nabla_\mu,\nabla_\nu]V^\rho = R^\rho{}_{\sigma\mu\nu}V^\sigma.\;}$$

The Riemann tensor is the obstruction to commutativity of covariant derivatives, equivalently the obstruction to path-independence of parallel transport.

Geometric significance

On a flat manifold ($R = 0$), parallel transport around any closed loop returns $V$ unchanged — consistent with the existence of global Cartesian-like coordinates. On a curved manifold, the rotation around an infinitesimal loop equals $R$ times the enclosed area. Around finite loops, the total rotation is $\oint = \int_D R\,dA$ (Stokes' theorem). This is the prototype for every curvature 2-form in gauge theory: Yang–Mills, Berry phases, characteristic classes (Chern, Pontryagin), and the Atiyah–Singer index theorem. In every case, "curvature is the obstruction to path-independence" is the underlying geometric statement.