Gauss-Codazzi-Mainardi Equations

A hypersurface is a slice — a three-dimensional "now" cut through spacetime, or classically an ordinary surface in space. It carries two notions of curvature. The intrinsic curvature is everything an inhabitant confined to the slice could measure: triangle angle sums, the failure of parallel transport, all encoded in the induced metric γij\gamma_{ij}. The extrinsic curvature KijK_{ij} is invisible from inside — it records how the slice bends within the surrounding space, how its unit normal nμn^\mu tilts as you slide along it. Paper rolled into a cylinder is extrinsically curved but intrinsically flat; a sphere is curved both ways. The Gauss-Codazzi-Mainardi equations are the exact laws relating these two curvatures to the curvature of the ambient space. They come from one operation: take the ambient Riemann tensor and project each of its four indices either tangent to the slice or along the normal. All four tangential gives the Gauss equation — ambient curvature equals intrinsic curvature plus a quadratic in KijK_{ij}, the descendant of Gauss's Theorema Egregium. One index normal gives the Codazzi-Mainardi equation, first-order in derivatives of KijK_{ij}. Two indices normal gives the Ricci equation, governing how KijK_{ij} evolves. Feed Einstein's equations into the first two and they become the Hamiltonian and momentum constraints — the four conditions any valid initial data (γij,Kij)(\gamma_{ij},K_{ij}) must satisfy before a numerical-relativity simulation can take a single step.

Key takeaways
  • A hypersurface Σ\Sigma carries two curvatures — intrinsic (from its induced metric γij\gamma_{ij}) and extrinsic (KijK_{ij}, how it bends in the ambient space). The equations tie both to the ambient Riemann tensor by projecting its four indices onto Σ\Sigma and along the normal nμn^\mu.
  • All-tangential is the Gauss equation ((3)Rabcd^{(3)}R_{abcd} plus a quadratic in KK); one-normal is Codazzi-Mainardi (derivatives of KK); two-normal is the Ricci equation (evolution of KK).
  • With Einstein's equations, twice-contracted Gauss becomes the Hamiltonian constraint (3)R+K2−KijKij=16πρ^{(3)}R + K^2 - K_{ij}K^{ij} = 16\pi\rho and once-contracted Codazzi becomes the momentum constraint Dj(Kij−γijK)=8πjiD_j(K^{ij}-\gamma^{ij}K)=8\pi j^i.
  • A flat FLRW slice collapses the Hamiltonian constraint to the Friedmann equation H2=8π3ρH^2 = \tfrac{8\pi}{3}\rho; a surface in flat space collapses Gauss to (2)R=2κ1κ2^{(2)}R = 2\kappa_1\kappa_2.
  • These are pure geometric identities for any codimension-one slice; only the normal's norm ϵ=nμnμ=±1\epsilon = n_\mu n^\mu=\pm1 flips signs between a spacelike slice and a timelike wall.
  • Key idea: project the ambient Riemann tensor through the tangential projector and the normal — all-tangential gives Gauss (intrinsic curvature plus K2K^2), one-normal gives Codazzi (derivatives of KK), two-normal gives evolution; feeding Einstein's equations into the first two yields the Hamiltonian and momentum constraints of the 3+1 initial-value problem.
The core insight

Project the ambient Riemann tensor through the tangential projector and the normal — all-tangential gives Gauss (intrinsic curvature plus a quadratic in KijK_{ij}), one-normal gives Codazzi-Mainardi (derivatives of KijK_{ij}), two-normal gives the Ricci evolution equation — and contracting the first two against Einstein's equations yields the Hamiltonian and momentum constraints on initial data.

Where this lives today

The Gauss-Codazzi-Mainardi equations are the daily tools of numerical relativity, and every gravitational-wave detection rests on them. Before LIGO-Virgo-KAGRA can predict a binary black-hole waveform, someone must build initial data — a spatial metric γij\gamma_{ij} and extrinsic curvature KijK_{ij} satisfying the Hamiltonian and momentum constraints to machine precision. This is the conformal method: solving the elliptic constraint equations (contracted Gauss and Codazzi) for a conformal factor and a longitudinal vector, using Bowen-York or conformal-thin-sandwich data. The result is handed to an evolution code (BSSN or generalized-harmonic), whose equations come from the Ricci projection. In the continuum the constraints are preserved by the Bianchi identity ∇μGμν=0\nabla_\mu G^{\mu\nu}=0; in practice discretization seeds violations that BSSN and constraint-damping suppress, so constraint monitors on equations (6) and (7) are a running simulation's standard health check. The same machinery drives the Israel junction conditions for thin shells, apparent-horizon finders solving Θ(ℓ)=0\Theta_{(\ell)}=0, and — in mathematical relativity — the positive-mass theorem and the Penrose inequality.

The physical picture

Imagine an ant on an eggshell. It can never step off, yet it can still discover its world is curved: it walks a triangle and finds the angles sum to more than 180 degrees, or carries an arrow around a loop and finds it rotated on return. That is intrinsic curvature. A geometer floating above sees what the ant cannot: the shell bends away into the third dimension, and the direction "straight out" — the unit normal — swings around as you move from the fat end to the point. The rate of that swing is the extrinsic curvature KijK_{ij}. The first figure sets the two side by side and pins the sphere to a number.

Figure
A curved surface Σ carries intrinsic curvature (a geodesic triangle closes with angle sum over 180 degrees) and extrinsic curvature K (how its unit normal n tilts along Σ); for a sphere of radius r both principal curvatures equal 1/r, so the Gaussian curvature is 1/r² and the intrinsic Ricci scalar is 2/r²
Two curvatures on one surface: intrinsic (what the ant measures) and extrinsic (how Σ tilts its normal) ambient space M hypersurface Σ normal nᵘ tilts along Σ ⇒ Kᵢⱼ geodesic triangle: angle sum over 180° ⇒ ³R Sphere radius r: principal curvatures κ₁ = κ₂ = 1/r → Gaussian curvature κ₁κ₂ = 1/r² → intrinsic Ricci scalar ³R = 2/r²

The two curvatures are not independent — that is the whole subject. For a surface in flat space the ambient curvature is zero, so the intrinsic curvature is forced to equal a fixed combination of the extrinsic curvature: bend a flat sheet without stretching and its intrinsic geometry never changes, because rolling alters KijK_{ij} but leaves K2−KijKijK^2 - K_{ij}K^{ij} alone. In general relativity the ambient space is not flat, and the same bookkeeping becomes physics: spacetime curvature, sourced by matter, ties the geometry of a spatial slice to how that slice sits in time. The second figure turns the three projections into the machinery of the initial-value problem.

Figure
Spacetime foliated into spatial slices Σ threaded by the unit normal n with lapse N and shift β; projecting the ambient Riemann tensor gives three relations — all-tangential is Gauss (Hamiltonian constraint ³R + K² − KᵢⱼKⁱʲ = 16πρ), one-normal is Codazzi (momentum constraint), two-normal is the evolution of K — and a flat FLRW slice makes the Hamiltonian constraint the Friedmann law H² = 8πρ/3
The 3+1 split: three projections of the Riemann tensor across a spatial slice Σₜ Σₜ₊dt Σₜ₊₂dt nᵘ lapse N shift βⁱ 4 tangential → Gauss ³R + K² − KᵢⱼKⁱʲ = 16πρ Hamiltonian constraint 1 normal → Codazzi Dⱼ(Kⁱʲ − γⁱʲK) = 8πjⁱ momentum constraint 2 normal → Ricci ∂ₜKᵢⱼ = … (evolution) not a constraint Flat FLRW slice: K = −3H, K² − KᵢⱼKⁱʲ = 9H² − 3H² = 6H² ⇒ Hamiltonian constraint gives H² = 8πρ/3 (Friedmann)

Why should only some projections be constraints? The answer is causal. A constraint is imposed at a single instant, before evolution begins, so it must contain no second time derivatives. The Gauss and Codazzi projections involve γij\gamma_{ij}, KijK_{ij}, and their spatial derivatives only; since KijK_{ij} is essentially the metric's first time derivative, these are at most first-order in time and checkable on the slice itself. The doubly-normal Ricci projection contains LnKij\mathcal{L}_n K_{ij}, a second time derivative of γij\gamma_{ij} — it does not constrain the data, it evolves it. That distinction is why relativity has a well-posed Cauchy problem: four of the ten Einstein equations are constraints from Gauss and Codazzi, the remaining six push the data forward.

The core equations

Everything begins with splitting the ambient metric into a tangent piece and a normal piece. Let Σ\Sigma be a spacelike hypersurface in (M,g)(\mathcal{M},g) of signature (−,+,+,+)(-,+,+,+), with future-directed unit normal nμn^\mu, nμnμ=−1n_\mu n^\mu = -1. The induced metric doubles as the orthogonal projector onto Σ\Sigma.

γμν=gμν+nμnν,γμν=δμν+nμnν(1)\gamma_{\mu\nu}=g_{\mu\nu}+n_\mu n_\nu,\qquad \gamma^\mu{}_\nu=\delta^\mu{}_\nu+n^\mu n_\nu \qquad (1)

where: gμνg_{\mu\nu} — ambient spacetime metric; nμn^\mu — unit normal, nμnμ=ϵ=−1n_\mu n^\mu=\epsilon=-1 for a spacelike slice (+1+1 for a timelike wall); γμν\gamma_{\mu\nu} — induced (spatial) metric on Σ\Sigma; γμν\gamma^\mu{}_\nu — projector, γμνnν=0\gamma^\mu{}_\nu n^\nu=0, γμαγαν=γμν\gamma^\mu{}_\alpha\gamma^\alpha{}_\nu=\gamma^\mu{}_\nu. Reading: any tensor splits into parts along nn and parts in Σ\Sigma by hitting each index with γ\gamma or nn. The extrinsic curvature measures how the normal fails to be parallel-transported within Σ\Sigma — equivalently minus half the Lie derivative of the spatial metric along the normal, which is why it acts as the metric's "velocity."

Kμν=−γμαγνβ∇αnβ=−12 Lnγμν(2)K_{\mu\nu}=-\gamma_\mu{}^{\alpha}\gamma_\nu{}^{\beta}\nabla_\alpha n_\beta=-\tfrac12\,\mathcal{L}_{n}\gamma_{\mu\nu} \qquad (2)

where: KμνK_{\mu\nu} — extrinsic curvature, symmetric and purely tangential (Kμνnν=0K_{\mu\nu}n^\nu=0); ∇α\nabla_\alpha — ambient Levi-Civita connection; Ln\mathcal{L}_n — Lie derivative along nn; K≡γμνKμνK\equiv\gamma^{\mu\nu}K_{\mu\nu} — mean curvature. Reading: KμνK_{\mu\nu} lives on Σ\Sigma yet remembers how Σ\Sigma sits in M\mathcal{M}; it is symmetric because the normal is hypersurface-orthogonal. The Gauss equation is the fully tangential projection: the intrinsic Riemann tensor of Σ\Sigma differs from the projected ambient one by a quadratic in KK.

γαμγβνγγργδσ (4)Rμνρσ=(3)Rαβγδ+KαγKβδ−KαδKβγ(3)\gamma_\alpha{}^{\mu}\gamma_\beta{}^{\nu}\gamma_\gamma{}^{\rho}\gamma_\delta{}^{\sigma}\,{}^{(4)}R_{\mu\nu\rho\sigma}={}^{(3)}R_{\alpha\beta\gamma\delta}+K_{\alpha\gamma}K_{\beta\delta}-K_{\alpha\delta}K_{\beta\gamma} \qquad (3)

where: (4)R{}^{(4)}R — ambient Riemann tensor; (3)R{}^{(3)}R — intrinsic Riemann tensor of Σ\Sigma (from the connection DD compatible with γ\gamma); the KKKK terms are the "bending" correction. Reading: in general the quadratic term carries −ϵ-\epsilon, so a spacelike slice (ϵ=−1\epsilon=-1) shows the ++ sign here while a Euclidean surface (ϵ=+1\epsilon=+1) flips it — the single sign separating the two settings. The Codazzi-Mainardi equation is the one-index-normal projection, equating a mixed component of the ambient Riemann tensor to the antisymmetrized covariant derivative of KK.

γαμγβνγγρ nσ (4)Rμνρσ=DβKαγ−DαKβγ(4)\gamma_\alpha{}^{\mu}\gamma_\beta{}^{\nu}\gamma_\gamma{}^{\rho}\,n^{\sigma}\,{}^{(4)}R_{\mu\nu\rho\sigma}=D_\beta K_{\alpha\gamma}-D_\alpha K_{\beta\gamma} \qquad (4)

where: nσn^\sigma — the single normal-projected index; DαD_\alpha — the intrinsic (Levi-Civita) covariant derivative of γ\gamma; DβKαγD_\beta K_{\alpha\gamma} — spatial gradient of KK. Reading: first order in KK and free of time derivatives of KK, so it is a genuine constraint whose contraction is a momentum-balance law. The Ricci equation is the doubly-normal projection; it contains a time derivative of KK and so is dynamical.

γαμnνγγρnσ (4)Rμνρσ=LnKαγ+1NDαDγN+KαμKμγ(5)\gamma_\alpha{}^{\mu}n^{\nu}\gamma_\gamma{}^{\rho}n^{\sigma}\,{}^{(4)}R_{\mu\nu\rho\sigma}=\mathcal{L}_n K_{\alpha\gamma}+\tfrac{1}{N}D_\alpha D_\gamma N+K_{\alpha\mu}K^{\mu}{}_{\gamma} \qquad (5)

where: NN — lapse function (proper time per unit coordinate time along nn); LnKαγ\mathcal{L}_n K_{\alpha\gamma} — the normal Lie derivative carrying ∂tKij\partial_t K_{ij}; 1NDαDγN\tfrac{1}{N}D_\alpha D_\gamma N — acceleration term, aμ=Dμln⁡Na_\mu=D_\mu\ln N. Reading: rearranged, this is the ADM evolution equation for KijK_{ij} — the six equations pushing data forward once the constraints hold. Contracting Gauss twice and inserting Einstein's equations yields the Hamiltonian constraint.

(3)R+K2−KijKij=16πρ(6){}^{(3)}R+K^2-K_{ij}K^{ij}=16\pi\rho \qquad (6)

where: (3)R{}^{(3)}R — intrinsic Ricci scalar of the slice; KK — mean curvature; KijKijK_{ij}K^{ij} — squared extrinsic curvature; ρ≡Tμνnμnν\rho\equiv T_{\mu\nu}n^\mu n^\nu — Eulerian energy density. Reading: this ties the intrinsic curvature of "space now" to the matter energy density and how space embeds in time; in vacuum the right side vanishes. Contracting Codazzi once and inserting Einstein's equations yields the momentum constraint.

Dj(Kij−γijK)=8π ji(7)D_j\left(K^{ij}-\gamma^{ij}K\right)=8\pi\,j^i \qquad (7)

where: ji≡−γiμnνTμνj^i\equiv -\gamma^{i}{}_\mu n_\nu T^{\mu\nu} — Eulerian momentum density; DjD_j — covariant derivative of γij\gamma_{ij}; Kij−γijKK^{ij}-\gamma^{ij}K — trace-reversed extrinsic curvature. Reading: the divergence of trace-reversed KK balances the momentum flux — the structure York's conformal method exploits.

Derivation AThe Gauss equation from a double covariant derivative

The goal is to derive equation (3) by relating the intrinsic covariant derivative DD on Σ\Sigma to the ambient derivative ∇\nabla and computing a commutator. The intrinsic connection is the ambient one with every index projected back onto Σ\Sigma; iterating that projected derivative on a tangent vector and antisymmetrizing gives the intrinsic Riemann tensor on one side, and the projected ambient Riemann tensor plus projector derivatives on the other. Those projector derivatives are exactly where KijK_{ij} enters.

Step 1. For any vector vγv^\gamma tangent to Σ\Sigma, define the intrinsic derivative as the fully projected ambient one, Dαvγ=γαμγγν∇μvνD_\alpha v^\gamma = \gamma_\alpha{}^\mu\gamma^\gamma{}_\nu\nabla_\mu v^\nu. Why: this is the unique torsion-free derivative compatible with γij\gamma_{ij}, so it is the connection an inhabitant of Σ\Sigma would build; writing it through ∇\nabla is what lets us import the ambient curvature.

Step 2. Apply DD a second time, DαDβvγ=γαμγβνγγρ∇μ ⁣(γνσγρλ∇σvλ)D_\alpha D_\beta v^\gamma = \gamma_\alpha{}^\mu\gamma_\beta{}^\nu\gamma^\gamma{}_\rho\nabla_\mu\!\left(\gamma_\nu{}^\sigma\gamma^\rho{}_\lambda\nabla_\sigma v^\lambda\right), and let ∇μ\nabla_\mu act on the projectors inside the bracket. Why: the derivative of the projector is not zero — from (1), ∇μγνσ\nabla_\mu\gamma_\nu{}^\sigma contains ∇μ(nνnσ)\nabla_\mu(n_\nu n^\sigma), and the tangential part of ∇n\nabla n is exactly −K-K, so each differentiated projector deposits a factor of KK.

Step 3. Antisymmetrize in α\alpha and β\beta. The left side becomes intrinsic curvature by definition, (DαDβ−DβDα)vγ=(3)Rγδαβvδ(D_\alpha D_\beta - D_\beta D_\alpha)v^\gamma = {}^{(3)}R^\gamma{}_{\delta\alpha\beta}v^\delta; both derivatives hitting vv give the projected ambient commutator (4)R{}^{(4)}R; the cross terms (one derivative on a projector) give KαγKβδ−KβγKαδK_{\alpha}{}^\gamma K_{\beta\delta}-K_{\beta}{}^\gamma K_{\alpha\delta}. Why: antisymmetrizing kills the symmetric second-derivative pieces, cleanly separating the ambient contribution from the embedding one.

Step 4. Lower the free index and rearrange, giving equation (3). Why: no field equations were used, so this is an identity holding for any hypersurface in any (pseudo-)Riemannian manifold, with the KKKK sign set by ϵ=n⋅n\epsilon = n\cdot n. *For a spacelike slice the quadratic sign is ++; setting (4)R=0{}^{(4)}R=0 with ϵ=+1\epsilon=+1 reproduces the classical Euclidean surface relation.*

Go deeper · Derivation B · 1 minThe Hamiltonian constraint from the twice-contracted Gauss equation
Derivation BThe Hamiltonian constraint from the twice-contracted Gauss equation

The goal is to reduce the tensor identity (3) to the scalar equation (6) by tracing it twice with the inverse spatial metric and inserting Einstein's equations. Contracting the projected Riemann on both index pairs turns the left side into a combination of the ambient Ricci scalar and the doubly-normal Ricci component that reorganizes exactly into the Einstein tensor contracted twice with the normal — the Eulerian energy density — while the right side gives the intrinsic scalar plus the two curvature invariants.

Step 1. Contract (3) with γαγγβδ\gamma^{\alpha\gamma}\gamma^{\beta\delta}. The right side gives (3)R+K2−KijKij{}^{(3)}R + K^2 - K_{ij}K^{ij}, since γαγγβδKαγKβδ=K2\gamma^{\alpha\gamma}\gamma^{\beta\delta}K_{\alpha\gamma}K_{\beta\delta}=K^2 and γαγγβδKαδKβγ=KijKij\gamma^{\alpha\gamma}\gamma^{\beta\delta}K_{\alpha\delta}K_{\beta\gamma}=K_{ij}K^{ij}. Why: the double trace of the intrinsic Riemann tensor is the intrinsic Ricci scalar, and the two contractions of the quadratic term give the trace-squared and norm-squared of KK.

Step 2. On the left, write each projector as γμρ=gμρ+nμnρ\gamma^{\mu\rho}=g^{\mu\rho}+n^\mu n^\rho: the pure-gg piece gives (4)R{}^{(4)}R, the two cross terms each give +(4)Rμνnμnν+{}^{(4)}R_{\mu\nu}n^\mu n^\nu, and the double-nn piece vanishes by Riemann antisymmetry. Why: the projected double trace equals (4)R+2 (4)Rμνnμnν=2 Gμνnμnν{}^{(4)}R + 2\,{}^{(4)}R_{\mu\nu}n^\mu n^\nu = 2\,G_{\mu\nu}n^\mu n^\nu, using Gnn=Rnn+12(4)RG_{nn}=R_{nn}+\tfrac12{}^{(4)}R for a unit timelike normal.

Step 3. Assemble to (3)R+K2−KijKij=2 Gμνnμnν{}^{(3)}R + K^2 - K_{ij}K^{ij} = 2\,G_{\mu\nu}n^\mu n^\nu. Why: this is the purely geometric contracted Gauss equation, true before any physics — spatial curvature plus embedding invariants equals the doubly-normal Einstein tensor.

Step 4. Insert Gμν=8πTμνG_{\mu\nu}=8\pi T_{\mu\nu} and Tμνnμnν=ρT_{\mu\nu}n^\mu n^\nu=\rho, giving (3)R+K2−KijKij=16πρ{}^{(3)}R + K^2 - K_{ij}K^{ij}=16\pi\rho. Why: the Hamiltonian constraint — no time derivatives, imposable on one slice; in vacuum (3)R=KijKij−K2{}^{(3)}R = K_{ij}K^{ij}-K^2. *The FLRW specialization below reproduces the Friedmann equation exactly.*

Go deeper · Derivation C · 1 minThe momentum constraint from the contracted Codazzi equation
Derivation CThe momentum constraint from the contracted Codazzi equation

The goal is to reduce the Codazzi-Mainardi identity (4) to the vector equation (7) by tracing over one index pair and inserting Einstein's equations. Contracting the once-normal projection produces a mixed normal-tangential component of the ambient Ricci tensor which — because the metric trace drops under a tangential-normal projection — equals the same component of the Einstein tensor, the Eulerian momentum density; the right side of (4) becomes the divergence of the trace-reversed extrinsic curvature.

Step 1. Contract (4) with γαγ\gamma^{\alpha\gamma}. The right side gives DβK−DαKαβD_\beta K - D_\alpha K^\alpha{}_\beta. Why: after raising indices this is exactly Dj(Kij−γijK)D_j(K^{ij}-\gamma^{ij}K), the divergence form of a symmetric tensor with its trace removed.

Step 2. On the left, γαγγαμγγρ=γμρ\gamma^{\alpha\gamma}\gamma_\alpha{}^\mu\gamma_\gamma{}^\rho=\gamma^{\mu\rho} turns the projected Riemann into γβνnσ(4)Rνσ\gamma_\beta{}^\nu n^\sigma {}^{(4)}R_{\nu\sigma}, the once-normal projection of the ambient Ricci tensor. Why: the double-nn contribution vanishes by Riemann antisymmetry, so only the Ricci piece survives.

Step 3. Replace Ricci by Einstein: γβνnσ(4)Rνσ=γβνnσGνσ\gamma_\beta{}^\nu n^\sigma {}^{(4)}R_{\nu\sigma}=\gamma_\beta{}^\nu n^\sigma G_{\nu\sigma}, since the trace term is killed by γβνnσgνσ=0\gamma_\beta{}^\nu n^\sigma g_{\nu\sigma}=0. Why: this is why the momentum constraint carries no explicit (4)R{}^{(4)}R — the metric trace cannot survive a mixed projection.

Step 4. Insert Gμν=8πTμνG_{\mu\nu}=8\pi T_{\mu\nu} and γβνnσTνσ=−jβ\gamma_\beta{}^\nu n^\sigma T_{\nu\sigma}=-j_\beta, giving Dj(Kij−γijK)=8πjiD_j(K^{ij}-\gamma^{ij}K)=8\pi j^i. Why: the momentum constraint; in vacuum the trace-reversed KK is divergence-free, the condition Bowen-York and conformal-TT data satisfy. *York's conformal decomposition solves precisely this divergence equation for the longitudinal part of KijK^{ij}.*

Limiting cases

Each special case recovers a landmark result.

CaseResultMeaning
Flat ambient space (ϵ=+1\epsilon=+1)(2)R=K2−KijKij=2κ1κ2{}^{(2)}R = K^2 - K_{ij}K^{ij}=2\kappa_1\kappa_2;   DiKjk=DjKik\;D_iK_{jk}=D_jK_{ik}Classical surface theory: Gauss's Theorema Egregium plus the Codazzi integrability conditions for embedding a surface into R3\mathbb{R}^3.
Vacuum spacetime(3)R+K2−KijKij=0{}^{(3)}R + K^2 - K_{ij}K^{ij}=0;   Dj(Kij−γijK)=0\;D_j(K^{ij}-\gamma^{ij}K)=0Constraints with Tμν=0T_{\mu\nu}=0; vacuum data such as Brill waves and black-hole punctures.
Time-symmetric slice (Kij=0K_{ij}=0)(3)R=16πρ{}^{(3)}R = 16\pi\rho (vacuum: (3)R=0{}^{(3)}R=0)Momentum constraint automatic; Hamiltonian becomes one elliptic equation. Schwarzschild t=0t=0 is the archetype.
Maximal slice (K=0K=0)(3)R−KijKij=16πρ{}^{(3)}R - K_{ij}K^{ij}=16\pi\rhoTrace-free KijK_{ij}; the maximal-slicing gauge used to avoid singularities.
Spacelike vs. timelikeϵ=nμnμ=−1\epsilon=n_\mu n^\mu=-1 (spacelike), +1+1 (timelike)ϵ\epsilon multiplies the quadratic KK term in Gauss; the Cauchy problem uses ϵ=−1\epsilon=-1, timelike walls use +1+1.

Worked examples

1 · The Friedmann equation as a Hamiltonian constraint (fully worked)

Slice a flat Friedmann-Lemaître-Robertson-Walker universe along constant cosmic time and show the Hamiltonian constraint (6) is exactly the Friedmann equation.

Step 1. The flat FLRW slice has γij=a(t)2δij\gamma_{ij}=a(t)^2\delta_{ij}, lapse N=1N=1, zero shift. It is flat space rescaled, so (3)R=0{}^{(3)}R=0 — the slices are intrinsically flat even though spacetime is curved; all the curvature is in how they stack in time.

Step 2. Compute the extrinsic curvature from (2): with N=1N=1 and βi=0\beta^i=0, Kij=−12∂tγij=−12∂t(a2)δij=−aa˙ δijK_{ij}=-\tfrac12\partial_t\gamma_{ij}=-\tfrac12\partial_t(a^2)\delta_{ij}=-a\dot a\,\delta_{ij}. Its trace is K=γijKij=a−2δij(−aa˙ δij)=−3a˙a=−3HK=\gamma^{ij}K_{ij}=a^{-2}\delta^{ij}(-a\dot a\,\delta_{ij})=-3\dfrac{\dot a}{a}=\mathbf{-3H}, where H=a˙/aH=\dot a/a is the Hubble rate.

Step 3. Form the two invariants. K2=9H2K^2=9H^2. For the norm, KijKij=γikγjlKijKkl=a−4(aa˙)2 δikδjlδijδkl=3 a˙2a2=3H2K_{ij}K^{ij}=\gamma^{ik}\gamma^{jl}K_{ij}K_{kl}=a^{-4}(a\dot a)^2\,\delta^{ik}\delta^{jl}\delta_{ij}\delta_{kl}=3\,\dfrac{\dot a^2}{a^2}=3H^2. Hence K2−KijKij=9H2−3H2=6H2K^2-K_{ij}K^{ij}=9H^2-3H^2=\mathbf{6H^2}.

Step 4. Insert into the Hamiltonian constraint (6): 0+6H2=16πρ0+6H^2=16\pi\rho, i.e. H2=8π3ρH^2=\dfrac{8\pi}{3}\rho — the Friedmann equation (G=c=1G=c=1). The central equation of cosmology is nothing but the twice-contracted Gauss equation of a comoving slice; the curvature term k/a2k/a^2 reappears once the slices carry (3)R=6k/a2{}^{(3)}R=6k/a^2.

2 · The Schwarzschild time-symmetric slice (fully worked)

Show that the t=0t=0 slice of Schwarzschild in isotropic coordinates satisfies both constraints, so it is valid vacuum initial data.

Step 1. In isotropic coordinates the slice is conformally flat, γij=ψ4δij\gamma_{ij}=\psi^4\delta_{ij} with ψ=1+M2r\psi=1+\dfrac{M}{2r}. It is a moment of time symmetry, so Kij=0K_{ij}=0 — the geometry looks the same forward or backward in time, so its "velocity" Kij∝∂tγijK_{ij}\propto\partial_t\gamma_{ij} vanishes.

Step 2. With Kij=0K_{ij}=0 the momentum constraint (7) reads Dj(0)=8πjiD_j(0)=8\pi j^i, and in vacuum ji=0j^i=0, so it is satisfied identically — every time-symmetric slice solves the momentum constraint.

Step 3. The Hamiltonian constraint (6) with Kij=0K_{ij}=0 and ρ=0\rho=0 collapses to (3)R=0{}^{(3)}R=0. For γij=ψ4δij\gamma_{ij}=\psi^4\delta_{ij} the intrinsic Ricci scalar is (3)R=−8ψ−5∇^2ψ{}^{(3)}R=-8\psi^{-5}\hat\nabla^2\psi with ∇^2\hat\nabla^2 the flat Laplacian, so (3)R=0{}^{(3)}R=0 iff ψ\psi is flat-harmonic.

Step 4. Check that ψ=1+M2r\psi=1+\dfrac{M}{2r} is flat-harmonic for r>0r>0: ∇^2(1+M2r)=M2∇^2 ⁣(1r)=0\hat\nabla^2\left(1+\dfrac{M}{2r}\right)=\dfrac{M}{2}\hat\nabla^2\!\left(\dfrac1r\right)=0, since 1/r1/r is the Green's function of the 3D Laplacian. Both constraints hold, so the slice is legitimate vacuum initial data. This is the seed of the puncture method: a black hole is a harmonic conformal factor with a 1/r1/r pole, and superposing poles gives constraint-satisfying multi-black-hole data.

3 · Gaussian curvature of a sphere (Theorema Egregium)

Recover the intrinsic curvature of a round 2-sphere of radius rr in flat space and check it against Gauss-Bonnet.

Step 1. For a sphere in Euclidean R3\mathbb{R}^3 the ambient Riemann tensor vanishes and ϵ=+1\epsilon=+1, so the Gauss equation reads (2)Rabcd=KacKbd−KadKbc{}^{(2)}R_{abcd}=K_{ac}K_{bd}-K_{ad}K_{bc} with equal principal curvatures κ1=κ2=1/r\kappa_1=\kappa_2=1/r. Tracing gives (2)R=K2−KijKij=2κ1κ2=2/r2{}^{(2)}R=K^2-K_{ij}K^{ij}=2\kappa_1\kappa_2=\mathbf{2/r^2} and Gaussian curvature K=1/r2\mathcal{K}=1/r^2 — Gauss's remarkable result that the product of principal curvatures is intrinsic.

Step 2. Cross-check with Gauss-Bonnet: ∫ΣK dA=1r2⋅4πr2=4π=2πχ\int_\Sigma \mathcal{K}\,dA=\dfrac{1}{r^2}\cdot 4\pi r^2=\mathbf{4\pi}=2\pi\chi, giving Euler characteristic χ=2\chi=2 — correct for a sphere. The local Gauss equation and the global topological invariant agree.

Physics of the portrait

The Gauss-Codazzi-Mainardi equations were essentially complete in classical surface theory a century before physics needed them — then turned out to be exactly the language relativity required to make sense of time.

Origins · 1 minGauss-Codazzi-Mainardi equations and how this was discovered
GE
Gauss-Codazzi-Mainardi equations
projecting the ambient Riemann tensor onto a hypersurface Σ: intrinsic curvature plus a quadratic in the extrinsic curvature K · the 3+1 initial-value problem of general relativity

The story begins with Carl Friedrich Gauss and his 1827 Disquisitiones generales circa superficies curvas, where he proved the Theorema Egregium: the Gaussian curvature of a surface — the product of its principal curvatures — is intrinsic, computable from distances within the surface alone. He called it remarkable because the principal curvatures individually are manifestly extrinsic. The accompanying integrability conditions were found independently by Karl Peterson (1853), Gaspare Mainardi (1856), and Delfino Codazzi (1868) — the constraints an extrinsic curvature field must satisfy to arise from a real embedding. Ricci-Curbastro and Levi-Civita's tensor calculus (around 1900) recast it all in the form Einstein borrowed in 1915. The relativistic payoff came in stages: Yvonne Choquet-Bruhat (1952) used the constraint structure to prove the Einstein equations have a well-posed Cauchy problem; Arnowitt, Deser, and Misner (1959) built the canonical formulation in which Gauss becomes the Hamiltonian constraint and Codazzi the momentum constraint; and Lichnerowicz and York turned the constraints into a solvable elliptic system — the conformal method still used to build every set of initial data fed to a black-hole simulation. A theorem about eggshells became the reason we can compute colliding black holes.

History
1827
— Gauss proves the Theorema Egregium for surfaces in space
1856
— Mainardi writes the integrability relations for an embedded surface
1868
— Codazzi independently derives the same normal-projection equations
1959
— Arnowitt, Deser and Misner cast the constraints in canonical 3+1 form
1972
— York's conformal method parametrizes constraint-satisfying initial data

Going further

Texts. MTW, Gravitation (1973), §21.5, Box 21.2 · Wald, General Relativity (1984), Ch. 10 and App. E · Carroll, Spacetime and Geometry (2004), App. D · Gourgoulhon, 3+1 Formalism in General Relativity (2012), Ch. 2-4 · Baumgarte & Shapiro, Numerical Relativity (2010), Ch. 2-3 · do Carmo, Differential Geometry of Curves and Surfaces (1976), Ch. 4. Key checks: (1) the Gauss equation follows from the antisymmetrized projected double derivative, with a ++ sign for a spacelike slice; (2) the contracted Gauss relation gives (3)R+K2−KijKij=2Gμνnμnν=16πρ{}^{(3)}R + K^2 - K_{ij}K^{ij}=2G_{\mu\nu}n^\mu n^\nu=16\pi\rho; (3) the FLRW reduction yields K=−3HK=-3H, K2−KijKij=6H2⇒H2=8π3ρK^2-K_{ij}K^{ij}=6H^2\Rightarrow H^2=\tfrac{8\pi}{3}\rho; (4) ψ=1+M/2r\psi=1+M/2r is flat-harmonic; (5) the sphere gives (2)R=2/r2{}^{(2)}R=2/r^2 with Gauss-Bonnet ∫K dA=4π\int\mathcal{K}\,dA=4\pi. No constants beyond 8π8\pi/16π16\pi appear.

Prerequisites. ADM 3+1 Decomposition · Riemann curvature and covariant derivatives · Hypersurfaces and the induced metric. Leads to. Hamiltonian and Momentum Constraints · Initial-Value (Cauchy) Problem · Conformal Method for Constraint Solving · BSSN Formulation · Numerical Relativity and Binary Black-Hole Mergers.

Ask the tutor: ask me to walk through any one projection index-by-index, or to build fresh constraint-satisfying initial data and watch the Hamiltonian constraint fall out.

What is the difference between intrinsic and extrinsic curvature, in one sentence each?

Intrinsic curvature is everything a two-dimensional inhabitant could measure without leaving the surface — triangle angle sums, holonomy of parallel transport, encoded in γij\gamma_{ij} and its Riemann tensor (3)R{}^{(3)}R. Extrinsic curvature KijK_{ij} is what only an outside observer sees — how the surface bends into the surrounding space, the rate the unit normal tilts. A cylinder is the cleanest case: nonzero extrinsic curvature (it bends in space) but zero intrinsic curvature (unroll it flat), which is why a flat map wraps perfectly onto a can but never onto a globe.

Why are there exactly four constraint equations in general relativity?

Because the two constraint projections supply four scalar conditions: twice-contracted Gauss is one scalar (Hamiltonian) and once-contracted Codazzi is a three-component vector (momentum) — four of Einstein's ten equations. The remaining six, from the Ricci projection, push γij\gamma_{ij} and KijK_{ij} forward in time. The count matches the gauge freedom: four constraints for the four coordinate choices (one lapse, three shift), leaving two propagating degrees of freedom — the two gravitational-wave polarizations.

How can the Friedmann equation be a special case of the Gauss equation?

Slice an FLRW universe along constant cosmic time. Each spatially flat slice has zero intrinsic Ricci scalar and extrinsic curvature Kij=−aa˙ δijK_{ij}=-a\dot a\,\delta_{ij}, trace K=−3HK=-3H. The Hamiltonian constraint gives 9H2−3H2=16πρ9H^2 - 3H^2 = 16\pi\rho, i.e. H2=8π3ρH^2=\tfrac{8\pi}{3}\rho — the Friedmann equation. The law of cosmic expansion is literally the twice-contracted Gauss equation of a comoving slice; the curvature term k/a2k/a^2 returns once the slices carry nonzero intrinsic curvature.

Does the sign of the extrinsic curvature convention matter for the physics?

For the constraints, no; for the evolution equations, yes. Textbooks split over the sign on γγ∇n\gamma\gamma\nabla n in KμνK_{\mu\nu}, flipping KK everywhere. But the Hamiltonian constraint contains only the even invariants K2K^2 and KijKijK_{ij}K^{ij}, so it is convention-independent; the momentum and evolution equations are linear in KijK_{ij}, so their written form flips while the physics does not. The rule: fix one convention and check a known case — if Schwarzschild is valid data and FLRW reproduces Friedmann, your signs are consistent.

What exactly does the momentum constraint mean physically?

It is a statement of momentum balance on the slice: Dj(Kij−γijK)=8πjiD_j(K^{ij}-\gamma^{ij}K)=8\pi j^i says the divergence of the trace-reversed extrinsic curvature equals the momentum density jij^i seen by the Eulerian observer (moving along the normal). In vacuum the right side vanishes, so trace-reversed KijK^{ij} is transverse — exactly the condition York's conformal method builds into its transverse-traceless decomposition. Physically, you cannot specify an arbitrary velocity field for the geometry; the way space contracts or shears must be consistent with the momentum flowing through it, just as a fluid's velocity must respect its momentum density.

Why do numerical-relativity codes still monitor the constraints if they are preserved by evolution?

In the continuum, the contracted Bianchi identity ∇μGμν=0\nabla_\mu G^{\mu\nu}=0 guarantees that constraints holding on the initial slice hold on every later slice. But a computer evolves a finite-difference or spectral approximation, and discretization error seeds small violations of equations (6) and (7) at every step. In a badly formulated scheme these grow exponentially — which is exactly why BSSN and generalized-harmonic reformulations were developed, adding terms proportional to the constraints so violations decay rather than blow up. Tracking the constraint residual over time is the standard diagnostic that a simulation is still solving Einstein's equations and not drifting into unphysical geometry.

Before this

Where this leads