The Riemann Curvature Tensor

Give a manifold a rule for differentiating vectors — a connection — and one sharp question decides everything about its shape: do the two covariant derivatives $\nabla_\mu$ and $\nabla_\nu$ commute? In flat space with Cartesian axes they do, because partial derivatives commute and the correction terms vanish. On a curved manifold they need not, because a connection is a prescription for carrying tangent vectors from point to neighbouring point, and that prescription can be path-dependent. The exact obstruction is a single object, the Riemann curvature tensor $R^{\rho}{}_{\sigma\mu\nu}$ , and its vanishing is the precise local test for flatness. The tensor wears two faces that turn out to be one. Algebraically it is the commutator: compute $[\nabla_\mu,\nabla_\nu]V^\rho$ and every derivative of $V$ cancels, leaving a pointwise linear map on $V$ — a $(1,3)$ tensor built purely from the connection and its first derivatives. Geometrically it is holonomy: parallel-transport a vector around a small closed loop and it comes back rotated, the rotation per unit enclosed area being exactly the curvature. These are the same statement, because parallel transport is just the integral of the covariant derivative around the loop. Because $R^{\rho}{}_{\sigma\mu\nu}$ is a tensor, its vanishing in one chart forces its vanishing in all of them — that is what makes curvature the honest, intrinsic measure of shape, blind to how a surface sits in a higher-dimensional embedding. Its symmetries slash its component count to $\tfrac{1}{12}n^2(n^2-1)$ , and its contractions build the Ricci, scalar, Weyl, and Einstein tensors. Physically it is the seat of gravity: it drives geodesic deviation — the tidal stretching of freely falling bodies — and its differential Bianchi identity is exactly what makes the Einstein equations conserve energy and momentum.

Key takeaways

The Riemann tensor $R^{\rho}{}_{\sigma\mu\nu}$ measures the failure of two covariant derivatives to commute — equivalently, the rotation a vector picks up when parallel-transported around an infinitesimal loop.
It is built from the connection alone: $R^{\rho}{}_{\sigma\mu\nu}=\partial_\mu\Gamma^{\rho}{}_{\nu\sigma}-\partial_\nu\Gamma^{\rho}{}_{\mu\sigma}+\Gamma^{\rho}{}_{\mu\lambda}\Gamma^{\lambda}{}_{\nu\sigma}-\Gamma^{\rho}{}_{\nu\lambda}\Gamma^{\lambda}{}_{\mu\sigma}$ , with no derivatives of the test vector surviving.
Flatness is exactly $R^{\rho}{}_{\sigma\mu\nu}=0$ : a manifold is locally isometric to Euclidean or Minkowski space if and only if the Riemann tensor vanishes identically.
Its symmetries cut the independent components to $\tfrac{1}{12}n^2(n^2-1)$ — that is $1,\,6,\,20$ in dimensions $2,\,3,\,4$ — and split the $n=4$ twenty into $10$ Ricci and $10$ Weyl.
Contractions give the Ricci tensor $R_{\mu\nu}=R^{\lambda}{}_{\mu\lambda\nu}$ , the scalar $R=g^{\mu\nu}R_{\mu\nu}$ , and, through the twice-contracted second Bianchi identity, the divergence-free Einstein tensor $\nabla^\mu G_{\mu\nu}=0$ .
Geodesic deviation gives curvature its physical teeth: tidal forces are a direct readout of $R^{\rho}{}_{\sigma\mu\nu}$ .
Key idea: curvature is the commutator of covariant derivatives, $[\nabla_\mu,\nabla_\nu]V^\rho=R^{\rho}{}_{\sigma\mu\nu}V^\sigma$ . Holonomy, geodesic deviation, the Bianchi identities, and the Ricci, Weyl, and Einstein tensors all unfold from this one defining equation.

The core insight

Curvature is the commutator of covariant derivatives: $[\nabla_\mu,\nabla_\nu]V^\rho=R^{\rho}{}_{\sigma\mu\nu}V^\sigma$ . Everything else — holonomy, geodesic deviation, the Bianchi identities, and the Ricci, Weyl, and Einstein tensors — unfolds from this single defining equation.

Where this lives today

The Riemann tensor is the daily working object of gravitational physics. Numerical-relativity codes for merging black holes evolve the full curvature and monitor coordinate-independent scalars built from it: the Kretschmann invariant $\mathcal{K}=R_{\rho\sigma\mu\nu}R^{\rho\sigma\mu\nu}$ , which for Schwarzschild equals $48G^2M^2/(c^4r^6)$ , diverges at the singularity $r=0$ but stays finite at the horizon — proof that the horizon is a coordinate artefact, not a physical edge. The same codes produce the waveform templates that LIGO and Virgo match against detected gravitational waves, ripples that are at heart propagating Weyl curvature. Satellite gravity-gradiometry missions such as GOCE measured the curvature components $\partial_i\partial_j\Phi$ of the Earth's field directly, mapping the geoid to centimetre precision. And the structural template of curvature — a connection whose non-commuting derivatives define a field strength — reappears verbatim in Yang-Mills gauge theory, where $F_{\mu\nu}=\partial_\mu A_\nu-\partial_\nu A_\mu+[A_\mu,A_\nu]$ is the Riemann formula in disguise, and integrals of its curvature two-forms yield topological invariants (Euler, Chern, Pontryagin numbers).

The physical picture

Carry an arrow across a sphere, keeping it as parallel as the surface allows. Start on the equator pointing east, walk north to the pole, come down a different meridian, and return along the equator. The arrow arrives home rotated. That rotation, per unit area enclosed by the loop, is the curvature — on a flat plane the same round trip returns the arrow untouched, so curvature is precisely the loop's memory of the shape it circled. The first figure makes this holonomy quantitative: for a spherical triangle the rotation angle equals the enclosed solid angle exactly.

Figure

Parallel transport of a vector around a spherical triangle with three right angles — one octant of a sphere of radius a. The vector returns rotated by exactly 90° = π/2, equal to the enclosed solid angle Ω = area/a² = (πa²/2)/a². On a flat plane the same closed loop would return the vector unchanged, so this 90° is pure curvature.

Think of $\nabla_\mu$ as "differentiate, but correct for the twisting of the coordinate frame as you go." Apply two such corrections in opposite orders and they need not agree, because the frame can twist differently along the $\mu$ and $\nu$ directions. The mismatch $[\nabla_\mu,\nabla_\nu]$ is not a differential operator at all — every derivative of the test vector cancels, leaving pure multiplication by the curvature, which is why $R^{\rho}{}_{\sigma\mu\nu}$ is a genuine tensor and not an artefact of the chart. It is also intrinsic: roll a flat sheet into a cylinder and it looks bent, yet its Riemann tensor is zero because you can unroll it without stretching. A sphere cannot be flattened without tearing, and that obstruction is exactly its curvature — Gauss's Theorema Egregium is the two-dimensional shadow of this fact. The second figure shows the payoff: curvature deforms a ring of freely falling particles into an ellipse, the tidal signature of gravity.

Figure

Geodesic deviation as the physical face of curvature: a ring of freely falling test particles around a mass is stretched along the radial direction and squeezed transversely, following D²ξ/dτ² = −Rᵖ_σμν uˢ ξᵘ uᵛ. In vacuum the Ricci trace vanishes, so stretch and squeeze balance and the enclosed volume is preserved. Near a 10-solar-mass black-hole horizon the radial tidal gradient reaches 2GM/r³ ≈ 1.0×10⁸ s⁻² per metre.

The core equations

The definition is the commutator. For an arbitrary connection there is one extra piece — the torsion — but for the Levi-Civita connection of general relativity torsion vanishes and the curvature term stands alone. Read it as: the failure of $\nabla_\mu$ and $\nabla_\nu$ to commute, acting on any vector, is a pure pointwise rotation of that vector, generated by $R^{\rho}{}_{\sigma\mu\nu}$ .

[\nabla_\mu,\nabla_\nu]V^\rho = R^{\rho}{}_{\sigma\mu\nu}\,V^\sigma - T^{\lambda}{}_{\mu\nu}\,\nabla_\lambda V^\rho \qquad (1)

where: $\nabla_\mu$ — covariant derivative of the connection; $R^{\rho}{}_{\sigma\mu\nu}$ — Riemann curvature tensor, of type $(1,3)$ ; $V^\sigma$ — components of any vector field; $T^{\lambda}{}_{\mu\nu}=\Gamma^{\lambda}{}_{\mu\nu}-\Gamma^{\lambda}{}_{\nu\mu}$ — torsion tensor, which vanishes for Levi-Civita. Expanding the commutator in Christoffel symbols and collecting the coefficient of $V^\sigma$ gives the explicit component formula. It has a memorable shape — a derivative of the connection minus its index-swap, plus a product of connections minus its swap — manifestly antisymmetric in the last pair $\mu\nu$ .

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} \qquad (2)

where: $\Gamma^{\rho}{}_{\mu\nu}$ — connection (Christoffel) coefficients; $\partial_\mu$ — ordinary partial derivative; $\lambda$ — summed dummy index. When the connection is metric-compatible ( $\nabla_\lambda g_{\mu\nu}=0$ ) and torsion-free, the fully-lowered tensor $R_{\rho\sigma\mu\nu}=g_{\rho\lambda}R^{\lambda}{}_{\sigma\mu\nu}$ gains three extra symmetries beyond the automatic antisymmetry in $\mu\nu$ — the source of the drastic reduction in component count.

R_{\rho\sigma\mu\nu}=-R_{\sigma\rho\mu\nu}=-R_{\rho\sigma\nu\mu}=R_{\mu\nu\rho\sigma} \qquad (3)

where: $R_{\rho\sigma\mu\nu}$ — fully covariant Riemann tensor. Read left to right: antisymmetry in the first pair, antisymmetry in the second pair, symmetry under exchange of the pairs. First-pair antisymmetry is the fingerprint of metric compatibility; pair-exchange follows from the two antisymmetries plus the first Bianchi identity. The cyclic sum of the curvature over its last three indices vanishes — the first, purely algebraic, Bianchi identity. It is a direct consequence of the connection being symmetric in its lower indices (zero torsion).

R^{\rho}{}_{\sigma\mu\nu}+R^{\rho}{}_{\mu\nu\sigma}+R^{\rho}{}_{\nu\sigma\mu}=0 \qquad (4)

where: the three terms cycle $\sigma\to\mu\to\nu\to\sigma$ in the lower slots. Differentiating the curvature and cycling the derivative index against the last pair gives the second, differential, Bianchi identity — the deepest of the lot, since two contractions of it deliver the conservation law that makes general relativity consistent.

\nabla_\lambda R^{\rho}{}_{\sigma\mu\nu}+\nabla_\mu R^{\rho}{}_{\sigma\nu\lambda}+\nabla_\nu R^{\rho}{}_{\sigma\lambda\mu}=0 \qquad (5)

where: $\nabla_\lambda$ — covariant derivative acting on the curvature tensor; the sum cycles $\lambda\to\mu\to\nu$ . Contracting the Riemann tensor once gives the Ricci tensor, again the scalar curvature, and a trace-adjusted combination the Einstein tensor — the objects that appear in the field equations of gravity.

R_{\mu\nu}=R^{\lambda}{}_{\mu\lambda\nu},\qquad R=g^{\mu\nu}R_{\mu\nu},\qquad G_{\mu\nu}=R_{\mu\nu}-\tfrac12 R\,g_{\mu\nu} \qquad (6)

where: $R_{\mu\nu}$ — Ricci tensor, the unique non-trivial contraction, symmetric for Levi-Civita; $R$ — Ricci scalar; $G_{\mu\nu}$ — Einstein tensor, automatically divergence-free; $g^{\mu\nu}$ — inverse metric. Finally, the equation that gives curvature its observable meaning: nearby geodesics do not stay parallel, and their relative acceleration is set entirely by the Riemann tensor. This is the tidal force, stripped of coordinates.

\frac{D^2 \xi^\rho}{\mathrm{d}\tau^2}=-R^{\rho}{}_{\sigma\mu\nu}\,u^\sigma\xi^\mu u^\nu \qquad (7)

where: $\xi^\rho$ — separation vector between two neighbouring geodesics; $u^\sigma=\mathrm{d}x^\sigma/\mathrm{d}\tau$ — tangent four-velocity; $D^2/\mathrm{d}\tau^2$ — second covariant derivative along the worldline.

Derivation ACurvature as the commutator of covariant derivatives

The goal is to prove that the commutator $[\nabla_\mu,\nabla_\nu]V^\rho$ is not a differential operator but a pointwise linear map on $V$ , and to extract the explicit component formula for $R^{\rho}{}_{\sigma\mu\nu}$ in the torsion-free case. We differentiate twice — carefully treating the first result as a rank-two tensor whose every index carries its own connection correction — then antisymmetrize in $\mu,\nu$ and watch every term containing a derivative of $V$ cancel, leaving a coefficient that must be a tensor because the left side is.

Step 1. Take the inner derivative and name the result: $\nabla_\nu V^\rho=\partial_\nu V^\rho+\Gamma^{\rho}{}_{\nu\sigma}V^\sigma\equiv W^{\rho}{}_\nu$ . Why: the second derivative acts on a $(1,1)$ tensor, so it must correct the lower index too.

Step 2. Take the outer derivative of $W^{\rho}{}_\nu$ , giving each index its own correction: $\nabla_\mu W^{\rho}{}_\nu=\partial_\mu W^{\rho}{}_\nu+\Gamma^{\rho}{}_{\mu\lambda}W^{\lambda}{}_\nu-\Gamma^{\lambda}{}_{\mu\nu}W^{\rho}{}_\lambda$ . Why: a covariant derivative adds $+\Gamma$ for each upper slot and $-\Gamma$ for each lower slot.

Step 3. Substitute $W^{\rho}{}_\nu=\partial_\nu V^\rho+\Gamma^{\rho}{}_{\nu\sigma}V^\sigma$ and expand fully to obtain $\nabla_\mu\nabla_\nu V^\rho=\partial_\mu\partial_\nu V^\rho+(\partial_\mu\Gamma^{\rho}{}_{\nu\sigma})V^\sigma+\Gamma^{\rho}{}_{\nu\sigma}\partial_\mu V^\sigma+\Gamma^{\rho}{}_{\mu\lambda}\partial_\nu V^\lambda+\Gamma^{\rho}{}_{\mu\lambda}\Gamma^{\lambda}{}_{\nu\sigma}V^\sigma-\Gamma^{\lambda}{}_{\mu\nu}\nabla_\lambda V^\rho$ . Why: writing every term out lets the next step's cancellations be seen, not asserted.

Step 4. Antisymmetrize in $\mu\leftrightarrow\nu$ . The second-derivative term $\partial_\mu\partial_\nu V^\rho$ is symmetric and drops; the two single-derivative terms $\Gamma^{\rho}{}_{\nu\sigma}\partial_\mu V^\sigma$ and $\Gamma^{\rho}{}_{\mu\lambda}\partial_\nu V^\lambda$ cancel against their swaps; and for zero torsion $\Gamma^{\lambda}{}_{\mu\nu}=\Gamma^{\lambda}{}_{\nu\mu}$ , so the last term cancels too. Why: every surviving term is now algebraic in $V$ — the derivative content has annihilated itself.

Step 5. Read off the coefficient of $V^\sigma$ : $[\nabla_\mu,\nabla_\nu]V^\rho=\big(\partial_\mu\Gamma^{\rho}{}_{\nu\sigma}-\partial_\nu\Gamma^{\rho}{}_{\mu\sigma}+\Gamma^{\rho}{}_{\mu\lambda}\Gamma^{\lambda}{}_{\nu\sigma}-\Gamma^{\rho}{}_{\nu\lambda}\Gamma^{\lambda}{}_{\mu\sigma}\big)V^\sigma=R^{\rho}{}_{\sigma\mu\nu}V^\sigma$ . Why: the left side is a tensor and $V^\sigma$ arbitrary, so the coefficient must be a tensor — the non-tensorial pieces of the $\Gamma$ symbols have cancelled in exactly the right combination.

Go deeper · Derivation B · 1 minCounting the independent components and the contracted Bianchi identity

Derivation BCounting the independent components and the contracted Bianchi identity

The goal is twofold: to count how many numbers the Riemann tensor carries once all its symmetries are imposed, and to show that contracting the second Bianchi identity twice forces the Einstein tensor to be divergence-free — the fact that lets curvature be equated to a conserved source. The counting is a symmetric-matrix problem minus the first Bianchi constraint; the conservation follows from a double metric trace, legal because $\nabla g=0$ .

Step 1. Treat the antisymmetric index pairs $[\rho\sigma]$ and $[\mu\nu]$ as single composite indices, each taking $M=\tfrac{n(n-1)}{2}$ values. Pair-exchange symmetry makes $R$ a symmetric $M\times M$ matrix, with $\tfrac{M(M+1)}{2}$ entries, and the first Bianchi identity removes a further $\binom{n}{4}$ , leaving $N(n)=\tfrac{1}{12}n^2(n^2-1)$ . Why: this gives $N(2)=1$ , $N(3)=6$ , $N(4)=20$ — one curvature number for a surface, twenty for spacetime.

Step 2. Lower the first index of the second Bianchi identity (5) with the metric, which is allowed because $\nabla_\lambda g_{\mu\nu}=0$ , giving $\nabla_\lambda R_{\rho\sigma\mu\nu}+\nabla_\mu R_{\rho\sigma\nu\lambda}+\nabla_\nu R_{\rho\sigma\lambda\mu}=0$ . Why: all four indices downstairs lets us contract cleanly with the inverse metric.

Step 3. Contract with $g^{\rho\mu}$ . Using $g^{\rho\mu}R_{\rho\sigma\mu\nu}=R_{\sigma\nu}$ and the antisymmetries, the three terms collapse to $\nabla_\lambda R_{\sigma\nu}-\nabla^\rho R_{\rho\sigma\nu\lambda}-\nabla_\nu R_{\sigma\lambda}=0$ . Why: the first contraction reduces the four-index identity to Ricci terms plus one Riemann term.

Step 4. Contract again with $g^{\sigma\nu}$ . This yields $\nabla_\lambda R-2\nabla^\mu R_{\mu\lambda}=0$ , i.e. the contracted Bianchi identity $\nabla^\mu R_{\mu\lambda}=\tfrac12\nabla_\lambda R$ . Why: the second trace expresses the Ricci divergence purely through the scalar gradient.

Step 5. Assemble the Einstein tensor and take its divergence: $\nabla^\mu\big(R_{\mu\lambda}-\tfrac12 g_{\mu\lambda}R\big)=\tfrac12\nabla_\lambda R-\tfrac12\nabla_\lambda R=0$ . Why: the two half-gradients of $R$ cancel identically, so $\nabla^\mu G_{\mu\lambda}=0$ is a geometric identity — geometry automatically respects the conservation $\nabla^\mu T_{\mu\lambda}=0$ that the matter source must obey.

Go deeper · Derivation C · 1 minCurvature of the round two-sphere from its metric

Derivation CCurvature of the round two-sphere from its metric

The goal is to run the whole machine end to end on the simplest curved space — a sphere of radius $a$ — starting from nothing but its metric, computing the Christoffel symbols, the one independent Riemann component, the Ricci tensor and scalar, and confirming the elementary Gaussian curvature $K=1/a^2$ . This is the calculation that shows the abstract commutator really reproduces a shape you can see, and it pins down the sign convention concretely.

Step 1. From the round metric $\mathrm{d}s^2=a^2(\mathrm{d}\theta^2+\sin^2\theta\,\mathrm{d}\phi^2)$ , so $g_{\theta\theta}=a^2$ and $g_{\phi\phi}=a^2\sin^2\theta$ , the non-vanishing Christoffels are $\Gamma^{\theta}{}_{\phi\phi}=-\sin\theta\cos\theta$ and $\Gamma^{\phi}{}_{\theta\phi}=\Gamma^{\phi}{}_{\phi\theta}=\cot\theta$ . Why: only $\partial_\theta g_{\phi\phi}$ is non-zero, so these three symbols are the whole input.

Step 2. Evaluate the one independent component $R^{\theta}{}_{\phi\theta\phi}=\partial_\theta\Gamma^{\theta}{}_{\phi\phi}-\partial_\phi\Gamma^{\theta}{}_{\theta\phi}+\Gamma^{\theta}{}_{\theta\lambda}\Gamma^{\lambda}{}_{\phi\phi}-\Gamma^{\theta}{}_{\phi\lambda}\Gamma^{\lambda}{}_{\theta\phi}$ . The first term is $\partial_\theta(-\sin\theta\cos\theta)=\sin^2\theta-\cos^2\theta$ , the second and third vanish, and the fourth is $-\Gamma^{\theta}{}_{\phi\phi}\Gamma^{\phi}{}_{\theta\phi}=+\cos^2\theta$ . Why: the $\Gamma\Gamma$ product supplies the $\cos^2\theta$ that cleans up the bare derivative.

Step 3. Sum: $R^{\theta}{}_{\phi\theta\phi}=(\sin^2\theta-\cos^2\theta)+\cos^2\theta=\sin^2\theta$ , and lowering, $R_{\theta\phi\theta\phi}=g_{\theta\theta}R^{\theta}{}_{\phi\theta\phi}=a^2\sin^2\theta$ . Why: this one positive component fixes the entire curvature, since $n=2$ has a single independent number.

Step 4. Contract to the Ricci tensor and scalar: $R_{\theta\theta}=R^{\phi}{}_{\theta\phi\theta}=1$ , $R_{\phi\phi}=R^{\theta}{}_{\phi\theta\phi}=\sin^2\theta$ , so $R=g^{\theta\theta}R_{\theta\theta}+g^{\phi\phi}R_{\phi\phi}=\tfrac{1}{a^2}+\tfrac{1}{a^2}=\tfrac{2}{a^2}$ . Why: positive, constant scalar curvature is the signature of a maximally symmetric space.

Step 5. Read off the Gaussian curvature $K=\tfrac{R}{2}=\tfrac{1}{a^2}$ , and check that $R_{\mu\nu\rho\sigma}=\tfrac{R}{2}(g_{\mu\rho}g_{\nu\sigma}-g_{\mu\sigma}g_{\nu\rho})$ holds in $n=2$ . Why: recovering the elementary $K=1/a^2$ from the full machinery is the sanity check that it is calibrated correctly.

Limiting cases

The Riemann tensor simplifies dramatically in several important regimes, each with its own geometric meaning:

Case	Result	Meaning
Flat space	$R^{\rho}{}_{\sigma\mu\nu}=0$	Necessary and sufficient (with zero torsion) for coordinates in which $\Gamma^{\rho}{}_{\mu\nu}=0$ everywhere and the metric is constant; covariant derivatives commute and parallel transport is path-independent.
Two dimensions	$R_{\mu\nu\rho\sigma}=\tfrac{R}{2}(g_{\mu\rho}g_{\nu\sigma}-g_{\mu\sigma}g_{\nu\rho})$	A single independent component; the whole curvature is the scalar $R=2K$ with $K$ the Gaussian curvature. The Einstein tensor vanishes identically.
Maximally symmetric	$R_{\mu\nu\rho\sigma}=\tfrac{R}{n(n-1)}(g_{\mu\rho}g_{\nu\sigma}-g_{\mu\sigma}g_{\nu\rho})$	Constant-curvature spaces — sphere, hyperbolic space, de Sitter, anti-de Sitter — with curvature built from the metric alone and $R$ everywhere constant.
Vacuum, $n\ge4$	$R_{\mu\nu}=0,\ C_{\rho\sigma\mu\nu}\ne0$	The Ricci part vanishes but the trace-free Weyl tensor survives, carrying tidal and gravitational-wave information — the Schwarzschild exterior is the canonical example.
Weak field	$R_{0i0j}\approx\partial_i\partial_j\Phi$	Newtonian limit: the relevant curvature components reduce to the Hessian of the gravitational potential, and geodesic deviation becomes the classical tidal-force equation.

Common trapRicci-flat does not mean flat, and sign conventions bite

Two traps snare almost everyone. First, $R_{\mu\nu}=0$ does not imply $R^{\rho}{}_{\sigma\mu\nu}=0$ in four or more dimensions: the Ricci tensor is only the trace part of Riemann, and the trace-free Weyl tensor can be — and around any isolated mass, is — non-zero in vacuum. The Schwarzschild exterior is Ricci-flat yet violently curved, which is why an astronaut is torn apart by tides where there is no local matter. Second, sign and index conventions are not standardized: $R^{\rho}{}_{\sigma\mu\nu}$ is always antisymmetric in $\mu\nu$ , but the placement of $\sigma$ and the overall sign differ between texts (MTW and Weinberg differ by a sign). State your convention $[\nabla_\mu,\nabla_\nu]V^\rho=R^{\rho}{}_{\sigma\mu\nu}V^\sigma$ before computing, and recall that torsion adds $-T^{\lambda}{}_{\mu\nu}\nabla_\lambda V^\rho$ to the commutator.

Worked examples

1 · Curvature of the two-sphere, from symbol to number (fully worked)

Take a sphere of radius $a$ with the round metric $\mathrm{d}s^2=a^2(\mathrm{d}\theta^2+\sin^2\theta\,\mathrm{d}\phi^2)$ . Find the Ricci scalar and Gaussian curvature, then evaluate both for the Earth, $a=6.371\times10^6$ m.

Step 1. Derivation 3 carries the metric through to the single independent component $R^{\theta}{}_{\phi\theta\phi}=\sin^2\theta$ , giving Ricci $R_{\theta\theta}=1$ , $R_{\phi\phi}=\sin^2\theta$ and scalar $R=g^{\theta\theta}R_{\theta\theta}+g^{\phi\phi}R_{\phi\phi}=\dfrac{2}{a^2}$ , so the Gaussian curvature is $K=R/2=\dfrac{1}{a^2}$ .

Step 2. Plug in the Earth's radius. With $a=6.371\times10^6$ m, $a^2=4.059\times10^{13}\ \mathrm{m}^2$ , so $K=1/a^2=\mathbf{2.46\times10^{-14}\ m^{-2}}$ and the Ricci scalar $R=2/a^2=\mathbf{4.93\times10^{-14}\ m^{-2}}$ . A sphere is maximally symmetric with a single constant curvature $K=1/a^2$ ; for the Earth that is $2.46\times10^{-14}\ \mathrm{m^{-2}}$ .

2 · Tidal stretching at a black-hole horizon

A radially infalling body of height $\ell=1.8$ m approaches a Schwarzschild black hole. Using the Newtonian-limit tidal gradient $2GM/r^3$ (the leading curvature component), estimate the head-to-foot relative acceleration at the horizon $r_s=2GM/c^2$ for a $10\,M_\odot$ hole, and contrast with the $4\times10^6\,M_\odot$ hole at the Galactic centre.

Step 1. For $M=10\,M_\odot=1.989\times10^{31}$ kg, the horizon radius is $r_s=2GM/c^2=2(6.674\times10^{-11})(1.989\times10^{31})/(2.998\times10^8)^2=\mathbf{2.95\times10^4\ m}$ , about 30 km.

Step 2. The tidal gradient there is $2GM/r_s^3=\mathbf{1.03\times10^{8}\ s^{-2}}$ per metre. Over $\ell=1.8$ m the relative acceleration is $\approx1.9\times10^8\ \mathrm{m\,s^{-2}}$ , roughly $\mathbf{1.9\times10^{7}\ g}$ — lethal spaghettification long before the horizon.

Step 3. The horizon tide scales as $2GM/r_s^3\propto 1/M^2$ . The Galactic-centre hole, $M=4\times10^6\,M_\odot$ , is $4\times10^5$ times more massive, so its horizon tide is $1.6\times10^{11}$ times weaker: $\approx6.4\times10^{-4}\ \mathrm{s^{-2}}$ per metre, only $\sim1\times10^{-4}\,g$ over a person. Big black holes have gentle horizons: you would cross Sgr A*'s unharmed while a stellar-mass hole shreds you outside it — a direct consequence of curvature going as $1/M^2$ at $r_s$ .

Physics of the portrait

The curvature tensor is the bequest of a man who lived to thirty-nine and reshaped geometry in a single lecture. Bernhard Riemann took Gauss's intrinsic curvature of surfaces and lifted it to any dimension, giving physics the language in which, sixty years later, Einstein would write gravity.

Origins · 1 minBernhard Riemann and how this was discovered

Bernhard Riemann

1826–1866 · founder of higher-dimensional differential geometry and the curvature tensor that bears his name

Georg Friedrich Bernhard Riemann was a shy, sickly pastor's son from Hanover whose brief career recast three fields. In 1854, for his Habilitation at Göttingen, he had to lecture before the faculty; Gauss, his advisor, chose from Riemann's list the topic Riemann had prepared least — the foundations of geometry. The result, "Über die Hypothesen, welche der Geometrie zu Grunde liegen," was a nearly formula-free work that reportedly moved the aged Gauss to rare public praise. In it Riemann proposed that space need not be flat or three-dimensional, that geometry is fixed by a locally-varying metric, and that the metric's departure from flatness is captured by a curvature. He introduced the multi-index object we now call the Riemann tensor in a Paris Academy prize essay around 1861, though the full tensor calculus was systematized later by Christoffel, Ricci-Curbastro, and Levi-Civita. He died of tuberculosis at Lake Maggiore in 1866, aged thirty-nine, never knowing his curved spaces would become physical spacetime — when Einstein needed a mathematics for gravity in 1915, it was Riemann's, reaching him through Ricci and Levi-Civita through the tutoring of his friend Marcel Grossmann.

History

1827

— Gauss proves the Theorema Egregium: the curvature of a surface is intrinsic

1854

— Riemann's Habilitation lecture founds higher-dimensional geometry

1861

— Riemann introduces the curvature tensor in a Paris Academy prize essay

1900

— Ricci-Curbastro and Levi-Civita codify the absolute differential calculus

1915

— Einstein makes Riemann curvature the field of gravitation

Going further

Texts. Misner, Thorne & Wheeler, Gravitation, chs. 11–13 (curvature, geodesic deviation, Bianchi identities) · Wald, General Relativity, ch. 3 · Nakahara, Geometry, Topology and Physics, ch. 7 · Stone & Goldbart, Mathematics for Physics, ch. on differential geometry · Arfken, Weber & Harris, Mathematical Methods for Physicists, 7th ed., tensor-analysis chapters · Hassani, Mathematical Physics, differential-geometry chapters. All formulas use the convention $[\nabla_\mu,\nabla_\nu]V^\rho=R^{\rho}{}_{\sigma\mu\nu}V^\sigma$ (Wald/MTW sign). Key anchor values: component counts $1,6,20$ from $N=\tfrac{1}{12}n^2(n^2-1)$ ; two-sphere $R=2/a^2$ , $K=1/a^2$ , giving $K=2.46\times10^{-14}$ m $^{-2}$ and $R=4.93\times10^{-14}\ \mathrm{m^{-2}}$ for Earth; contracted-Bianchi identity $\nabla^\mu G_{\mu\nu}=0$ ; Schwarzschild horizon $r_s=2.95\times10^4$ m and tidal gradient $1.03\times10^8\ \mathrm{s^{-2}}$ for a $10\,M_\odot$ hole; Kretschmann scalar $\mathcal{K}=48G^2M^2/(c^4r^6)$ .

Prerequisites. The Covariant Derivative · Christoffel Symbols and Affine Connections · Parallel Transport and Geodesics · The Metric Tensor · Tensor Contraction and the Trace. Leads to. Einstein Field Equations · Geodesic Deviation · Hodge Star and the Codifferential · Exterior Calculus and the Exterior Derivative.

Ask the tutor: Ask me to derive the Riemann tensor from the commutator, count its components in any dimension, or work the two-sphere or Schwarzschild curvature step by step.

Is the Riemann tensor the same thing as the Ricci tensor or the scalar curvature?

No — they are a hierarchy of decreasing information. The Riemann tensor $R^{\rho}{}_{\sigma\mu\nu}$ is the complete curvature, with twenty independent components in four dimensions. The Ricci tensor $R_{\mu\nu}=R^{\lambda}{}_{\mu\lambda\nu}$ is a contraction that keeps only ten of those, and the scalar $R=g^{\mu\nu}R_{\mu\nu}$ keeps just one. So a vacuum region can have $R_{\mu\nu}=0$ and $R=0$ yet a thoroughly non-zero Riemann tensor — exactly the situation outside any star or black hole.

What does it physically mean that curvature is the commutator of covariant derivatives?

It means curvature is path-dependence made local. A covariant derivative compares a vector with one at an infinitesimally close point by parallel transport. Doing $\nabla_\mu$ then $\nabla_\nu$ transports along one pair of edges of a tiny parallelogram; the opposite order transports along the other pair. If the answers differ, then transporting a vector all the way around the loop does not return it unchanged — and the mismatch, per unit area, is the Riemann tensor.

Why does the Riemann tensor need the connection but the metric only enters through it?

Curvature is fundamentally a property of the connection — the rule for parallel transport — not of the metric. You can define $R^{\rho}{}_{\sigma\mu\nu}$ for any affine connection, metric or not, straight from equation (2). The metric enters two ways. First, general relativity uses the Levi-Civita connection, the unique torsion-free connection compatible with the metric, so the Christoffels and hence the curvature become functions of the metric. Second, the metric lets you lower the upper index to form $R_{\rho\sigma\mu\nu}$ , unlocking the extra symmetries (first-pair antisymmetry, pair exchange) a bare connection lacks.

How is the Riemann tensor related to tidal forces I can actually feel?

Directly, through geodesic deviation. Two nearby free-falling particles each follow a geodesic, and the Riemann tensor sets their relative acceleration: $D^2\xi^\rho/\mathrm{d}\tau^2=-R^{\rho}{}_{\sigma\mu\nu}u^\sigma\xi^\mu u^\nu$ . In the weak-field limit this reduces to the Newtonian tidal equation $\ddot\xi^i=-\partial_i\partial_j\Phi\,\xi^j$ , the second derivatives of the potential. So the stretching you feel falling toward a black hole, the ocean tides raised by the Moon, and the deformation of an orbiting satellite are all readouts of curvature components. Curvature is not bolted onto gravity; it is the tidal field.

What is the difference between intrinsic and extrinsic curvature?

Intrinsic curvature is what the Riemann tensor measures: it depends only on distances within the surface itself, with no reference to any embedding. A cylinder has zero intrinsic curvature — a flat ant on it finds every triangle's angles summing to exactly 180°, and you can unroll the cylinder onto a plane without stretching. Extrinsic curvature describes how a surface bends within a higher-dimensional space it sits in, and the cylinder has plenty of that. The sphere is different: no flattening is possible without distortion, so its intrinsic curvature is genuinely non-zero. Gauss's Theorema Egregium says the Gaussian curvature of a surface is intrinsic, and the Riemann tensor generalizes it to all dimensions.

Why do sign conventions for the Riemann tensor differ between textbooks, and does it matter?

It matters enough to derail a calculation, so fix a convention and stick to it. Two independent choices float around: the ordering of indices, and the overall sign — MTW, Wald, and Nakahara use $[\nabla_\mu,\nabla_\nu]V^\rho=+R^{\rho}{}_{\sigma\mu\nu}V^\sigma$ , while Weinberg differs by an overall sign, and some texts permute the lower indices. These propagate into the sign of the Ricci tensor and the Einstein equations, though physical predictions are convention-independent once you are consistent. The safe practice, used throughout this article, is to write your defining commutator at the top of any computation and check the sign against the two-sphere, whose scalar curvature $R=+2/a^2$ must come out positive.