Curvature in Two and Three Dimensions

The starkest modern use of low-dimensional curvature is 2+1 quantum gravity. Because 3D vacuum gravity is locally flat, it has no propagating gravitons and reduces to a topological field theory of the Chern-Simons type — an exactly solvable model where the only observables are holonomies and the topology of spacetime. That exact solvability is why it remains a favourite laboratory for testing ideas about quantum spacetime, from spin foams to holography. Adding a negative cosmological constant escapes triviality without adding local dynamics: the field equation becomes $R_{ab}=2\Lambda g_{ab}$, giving constant-curvature AdS$_3$, and global identifications of that space produce the BTZ black hole — a genuine 2+1 black hole with horizon, temperature, and entropy, and the cleanest testing ground for the AdS/CFT correspondence. Meanwhile the 2D end of the story lives in every Regge-calculus and lattice-gravity simulation, where curvature is stored as angle deficits at vertices, exactly the discrete Gauss-Bonnet picture, and in the optical geometry of light bundles on the celestial 2-sphere. The open frontier is whether the topological cleanliness of 2+1 gravity survives, in any form, as a guide to the far harder 3+1 quantum theory.

Picture the Riemann tensor not as a forest of indices but as a single symmetric matrix. Its two antisymmetric index pairs $[ab]$ and $[cd]$ each label an oriented area element — a bivector — and there are exactly $N=n(n-1)/2$ of them. The pair-exchange symmetry $R_{abcd}=R_{cdab}$ then makes curvature a symmetric $N\times N$ operator on these area elements, and the first Bianchi identity trims a handful more entries. In two dimensions there is only one area element (the surface is its own single plane), so curvature is one number. In three dimensions there are three area elements, a symmetric $3\times3$ matrix, six numbers — precisely the six of a symmetric Ricci tensor. The curvature has no room to be anything more than its own traces.

The Weyl tensor is the part of curvature that angle-preserving (conformal) rescalings can never remove — geometrically, the shape-distorting tidal field that survives even in vacuum. Below four dimensions it has nowhere to live: every 2D metric is locally conformally flat, and in 3D the obstruction to conformal flatness is carried by the Cotton tensor, not Weyl, which is zero by construction. The consequence is vivid. A curved surface stays genuinely curved — a sphere is not flat — but all of that curvature is one scalar. And a 2+1 point mass warps space only into a cone: flat everywhere it can be, with the entire curvature crushed into a conical deficit at the source.

The whole subject follows from one counting statement. Impose the antisymmetries, the pair-exchange symmetry, and the first Bianchi identity on a rank-four tensor and only a fixed number of components survive at each point. That number is the gatekeeper for everything below.

where: $n$ — the dimension of the space or spacetime; $R_{abcd}$ — the Riemann curvature tensor. Reading it physically: the count is $1,6,20,50$ for $n=2,3,4,5$, so a surface carries one curvature number and a three-space carries six, exactly the sizes of a scalar and of a symmetric $3\times3$ matrix. In two dimensions there is only one tensor, up to scale, that has all of Riemann's index symmetries: the antisymmetrized product of two metrics. So Riemann must be proportional to it, and matching the trace fixes the coefficient to $R/2$.

where: $R$ — the Ricci scalar; $g_{ab}$ — the metric tensor. This says the surface has a single curvature knob: specify $R$ at a point and every component of the curvature there is determined. Contract that 2D form once and the Ricci tensor comes out proportional to the metric. Trace again and the Einstein tensor collapses to zero — a fact special to two dimensions and to no other.

where: $R_{ab}$ — the Ricci tensor; $G_{ab}$ — the Einstein tensor. The vanishing of $G_{ab}$ means the Einstein equations carry no local content in 2D: there is no field equation to solve, only topology. The one scalar of a surface is exactly the Gaussian curvature that Gauss defined intrinsically a century before Riemann. Matching the single component $R_{1212}$ to the sectional curvature of the only 2-plane gives the identification.

where: $K$ — the Gaussian curvature of the surface; $R$ — the Ricci scalar. For the unit 2-sphere $K=1$, so $R=2$ and $R_{ab}=g_{ab}$: a curved space whose entire curvature is one positive number. Three dimensions has six curvature components and the symmetric Ricci tensor has six too, so they must be in one-to-one correspondence. The explicit reconstruction expresses every Riemann component in terms of Ricci and the scalar.

where: $R_{bd}$ — the Ricci tensor, six independent components in 3D; $R$ — the Ricci scalar; $g_{ab}$ — the metric. Contracting the right side reproduces $R_{ab}$ exactly, confirming there is no curvature information beyond the Ricci data. The Weyl tensor is defined as whatever is left of Riemann after its trace parts are subtracted off — the trace-free, conformally invariant remainder. The definition works for any $n\ge 3$, but its content depends dramatically on dimension.

where: $C_{abcd}$ — the Weyl conformal tensor; $n$ — the dimension. Substituting the 3D reconstruction (5) makes every term cancel: $C_{abcd}\equiv 0$ for $n\le 3$. Weyl is not small below four dimensions — it is identically nothing. Finally, the physical payoff. In 3D the Ricci tensor is all of curvature, so a vacuum region where $R_{ab}=0$ has zero Riemann tensor — genuinely flat, not merely Ricci-flat.

where: $R_{ab}=0$ — the vacuum Einstein equation with no cosmological constant. Reading it: there is no curved vacuum exterior and no propagating wave in 2+1 gravity, because there is simply no curvature that empty space can hold.

1 · The unit 2-sphere — a curved space with one curvature number

Take the unit 2-sphere with Gaussian curvature $K=1$. Find its Ricci scalar, Ricci tensor, the single Riemann component, and confirm the Gauss-Bonnet integral.

Step 1. From $K=R/2$ with $K=1$, the Ricci scalar is $R=2K=\mathbf{2}$. The surface is manifestly curved, yet this one number is the whole story. The Gaussian curvature and the general-relativistic scalar differ by exactly the factor of two in equation (4).

Step 2. The Ricci tensor follows from the 2D identity $R_{ab}=\tfrac{R}{2}g_{ab}=\tfrac{2}{2}g_{ab}=\mathbf{g_{ab}}$. So Ricci equals the metric — the sphere is an Einstein space — and the Einstein tensor is $G_{ab}=R_{ab}-\tfrac12 R g_{ab}=g_{ab}-g_{ab}=\mathbf{0}$, as it must be in 2D.

Step 3. The lone Riemann component is $R_{1212}=\tfrac{R}{2}(g_{11}g_{22}-g_{12}^2)=1\cdot g=K\,g$, with $g=\det g_{ab}$. Every other component is fixed by antisymmetry, so specifying $R=2$ fixes the entire curvature tensor.

Step 4. Gauss-Bonnet: $\int_{S^2}K\,\mathrm{d}A = 1\cdot(4\pi)=\mathbf{4\pi}$, and independently $2\pi\chi=2\pi\cdot 2=4\pi$ since $\chi(S^2)=2$. The two agree exactly, and the integrated curvature is a topological number blind to any smooth deformation of the sphere. A genuinely curved surface, and its curvature is the single scalar $R=2$.

2 · The 3-sphere — Ricci rebuilds all of Riemann

Take the round 3-sphere $S^3$ of radius $a$, a maximally symmetric curved 3-space with $R_{ab}=\tfrac{2}{a^2}g_{ab}$. Reconstruct the full Riemann tensor from the 3D formula and show the curvature is constant and Weyl-free.

Step 1. Get the scalar: $R=g^{ab}R_{ab}=\tfrac{2}{a^2}\,g^{ab}g_{ab}=\tfrac{2}{a^2}\cdot 3=\mathbf{6/a^2}$. All curvature data is now the single Ricci tensor $R_{ab}=(2/a^2)g_{ab}$ plus this scalar.

Step 2. Feed these into the 3D reconstruction (5). The four Ricci terms give $\tfrac{2}{a^2}\big[g_{ac}g_{bd}+g_{bd}g_{ac}-g_{ad}g_{bc}-g_{bc}g_{ad}\big]=\tfrac{2}{a^2}\big[2g_{ac}g_{bd}-2g_{ad}g_{bc}\big]=\tfrac{4}{a^2}(g_{ac}g_{bd}-g_{ad}g_{bc})$.

Step 3. The scalar term contributes $-\tfrac{R}{2}(g_{ac}g_{bd}-g_{ad}g_{bc})=-\tfrac{3}{a^2}(g_{ac}g_{bd}-g_{ad}g_{bc})$. Adding, $R_{abcd}=\big(\tfrac{4}{a^2}-\tfrac{3}{a^2}\big)(g_{ac}g_{bd}-g_{ad}g_{bc})=\mathbf{\tfrac{1}{a^2}(g_{ac}g_{bd}-g_{ad}g_{bc})}$.

Step 4. This is exactly the maximally symmetric form with constant sectional curvature $1/a^2$ — the direct $S^3$ answer — recovered from the Ricci tensor alone. Because it already equals its own trace reconstruction, the Weyl tensor (6) evaluates to $\mathbf{C_{abcd}=0}$. For the unit 3-sphere ($a=1$): $R=6$, $R_{ab}=2g_{ab}$, sectional curvature $1$, Weyl zero. Six Ricci numbers rebuild all six Riemann components, with nothing left over.

The low-dimensional curvature story begins not with Einstein but with a surveyor measuring the Kingdom of Hanover, who realized that a two-dimensional creature could detect the bending of its world without ever leaving it.

Texts. Misner, Thorne and Wheeler, Gravitation (1973), §13.2 and §21.5 · Wald, General Relativity (1984), §3.2 with eq. 3.2.28 and Appendix C · Carroll, Spacetime and Geometry (2004), §3.7-3.8 · Schutz, A First Course in General Relativity, 2nd ed., §6.7 · Hartle, Gravity (2003), Ch. 21 · Weinberg, Gravitation and Cosmology (1972), §6.10. The component formula $n^2(n^2-1)/12$ yields the values $1,6,20,50$ for $n=2,3,4,5$. Contracting the 3D reconstruction formula recovers $R_{ab}$ exactly, and substituting it into the Weyl definition gives $C_{abcd}=0$. The 2D identities $K=R/2$, $G_{ab}=0$, and Gauss-Bonnet $\int K\,\mathrm{d}A=2\pi\chi$ hold in full generality. The local triviality of 2+1 gravity was established by Deser, Jackiw and 't Hooft (1984) and Witten (1988); the BTZ black hole by Bañados, Teitelboim and Zanelli (1992).

Prerequisites. Riemann Curvature Tensor · Algebraic Symmetries of the Riemann Tensor · Independent Components of the Riemann Tensor · Ricci Tensor · Ricci Scalar · Weyl Conformal Tensor. Leads to. Weyl Conformal Tensor · Einstein Tensor · Curvature as the Commutator of Covariant Derivatives.

Ask the tutor: Ask me to walk through why the ten Weyl components that switch on at four dimensions are exactly the two polarizations of a gravitational wave.