Ricci, Scalar, and Einstein Tensors

Ricci curvature, scalar curvature, and the Einstein tensor are canonical contractions of Riemann curvature. With the contraction $R_{\sigma\nu}=R^\rho{}_{\sigma\rho\nu}$ fixed explicitly, the differential Bianchi identity selects $G_{\mu\nu}=R_{\mu\nu}-\tfrac12Rg_{\mu\nu}$ as a divergence-free geometric tensor. This conclusion is kinematical; no field equation or matter interpretation is assumed here.

Required background. Affine Connections and Covariant Derivatives supplies the general tensor-derivative framework; The Levi-Civita Connection supplies the metric-compatible torsion-free derivative used here.

Helpful background. Calculus and Linear Algebra for Relativity supplies linear maps and index notation used in tensor traces; Exterior Algebra and Differential Forms supplies alternating forms and wedge products, while the contracted Bianchi calculation on this page is tensorial and self-contained.

Curvature contractions

For the adopted curvature convention,

$$R_{\sigma\nu}=R^\rho{}_{\sigma\rho\nu}, \qquad R=g^{\sigma\nu}R_{\sigma\nu}.$$

The placement matters: contracting the first and fourth positions instead would differ by a sign because Riemann is antisymmetric in its final pair. Levi–Civita curvature symmetries make $R_{\mu\nu}$ symmetric. The scalar $R$ is its metric trace, and the trace-free Ricci tensor is

$$S_{\mu\nu}=R_{\mu\nu}-\frac{R}{n}g_{\mu\nu}.$$

Ricci is not “the curvature” in full. It records a particular trace of the curvature endomorphism, whereas Weyl curvature is invisible to every single contraction. In dimensions two and three the algebraic situation is special; in four and higher dimensions, Ricci-flat metrics can still possess nonzero Riemann curvature.

The contraction definitions and their relation to the Riemann tensor are treated with the same sign convention in Tong 2019, §§3.2 and 3.4, pp. 100–117 and 130–132.

From the differential Bianchi identity to a conserved tensor

For a Levi–Civita connection, the differential Bianchi identity is

$$\nabla_{[\lambda}R_{\mu\nu]\rho\sigma}=0.$$

Expand the cyclic sum and contract with $g^{\lambda\rho}g^{\nu\sigma}$. The three terms are respectively $\nabla^\rho R_{\mu\rho}$, $-\nabla_\mu R$, and $\nabla^\sigma R_{\mu\sigma}$. Their sum vanishes, so Ricci symmetry gives

$$\nabla^\mu R_{\mu\nu}=\frac12\nabla_\nu R.$$

Metric compatibility gives $\nabla^\mu g_{\mu\nu}=0$. Therefore

$$G_{\mu\nu} \equiv R_{\mu\nu}-\frac12Rg_{\mu\nu}, \qquad \nabla^\mu G_{\mu\nu}=0.$$

Why the factor $1/2$? For a combination $R_{\mu\nu}-cRg_{\mu\nu}$, the divergence is $(1/2-c)\nabla_\nu R$; it vanishes identically for arbitrary geometry only when $c=1/2$. One may add a constant multiple $\Lambda g_{\mu\nu}$ without changing the divergence, but that introduces an additional constant rather than another curvature contraction.

This geometric identity is one reason the Einstein tensor is suitable for a gravitational field equation, but suitability is not the field equation itself. The General Relativity volume supplies the physical postulates, stress–energy, and dynamics.

Constant-curvature normalization check

In $n$ dimensions, a constant-sectional-curvature metric obeys

$$R_{abcd}=K(g_{ac}g_{bd}-g_{ad}g_{bc}).$$

Contracting the first and third indices gives

$$R_{bd}=(n-1)Kg_{bd}, \qquad R=n(n-1)K.$$

Hence

$$G_{ab} =-\frac{(n-1)(n-2)}{2}K g_{ab}.$$

The trace provides an independent check: $g^{ab}G_{ab}=(1-n/2)R$, which equals $-n(n-1)(n-2)K/2$. In two dimensions $G_{ab}$ vanishes identically for every metric, not merely for constant curvature; the Einstein tensor therefore carries no local curvature information there. On a round two-sphere, $R_{ab}=Kg_{ab}$ and $R=2K$, confirming the cancellation.

Meaning and limitations

Different contractions answer different geometric questions. $R$ is one scalar trace; $R_{\mu\nu}v^\mu v^\nu$ traces the directional curvature operator; $G_{\mu\nu}$ is the divergence-free trace reversal of Ricci. None of these tensors by itself is a complete local classifier in generic four-dimensional Lorentzian geometry. In particular, a vanishing scalar $R$ neither implies Ricci flatness nor full flatness, and $G_{\mu\nu}=0$ still permits Weyl curvature.

The derivation also depends on Levi–Civita assumptions. Torsion or nonmetricity modifies the Bianchi identities and the divergence calculation. Within its hypotheses, however, the result is exact and dimension independent. A complementary tensor-decomposition derivation appears in Calegari 2022, §7.2, pp. 54–55.

Ricci curvature also has a direct geometric averaging interpretation in Riemannian signature. For a unit vector $v$ and an orthonormal basis $v,e_2,\ldots,e_n$,

$$R_{ab}v^av^b=\sum_{i=2}^{n}K(v,e_i).$$

Thus Ricci sums the sectional curvatures of planes containing $v$. In Lorentzian signature the analogous trace carries signs from the orthonormal basis, and null directions require a screen-space formulation rather than the undefined sectional curvature of a degenerate plane. The contraction remains well defined even where the simple Riemannian average picture does not.

Trace reversal is dimension sensitive. The trace of the Einstein tensor is

$$G^a{}_a=\left(1-\frac n2\right)R.$$

For $n\ne2$ this relation can be inverted to recover $R$ from $G$, after which $R_{ab}=G_{ab}-G^c{}_c g_{ab}/(n-2)$. In two dimensions the trace vanishes and $G_{ab}\equiv0$ identically, reflecting the fact that the Riemann tensor has only its scalar part. This provides a second, dimension-based test of the normalization.

When computing from a metric, it is safer to form $R_{\mu\nu}$ from the explicitly declared Riemann convention, check symmetry, and only then take $R$ and $G_{\mu\nu}$. Importing a Ricci formula from a source with the opposite Riemann sign reverses every curvature contraction. A divergence check alone may not reveal that error because both $G$ and $-G$ are divergence free; the round-sphere sign check does.

Exercises

Compute the Einstein tensor of a four-dimensional constant-curvature metric.

Solution

Set $n=4$ in the checked formulas: $R_{ab}=3Kg_{ab}$ and $R=12K$. Therefore

$$G_{ab}=3Kg_{ab}-\frac12(12K)g_{ab}=-3Kg_{ab}.$$

Its divergence vanishes because $K$ is constant and $\nabla g=0$.

References

• Danny Calegari, Introduction to Differential Geometry, preliminary version, May 26, 2022, University of Chicago. PDF.
• David Tong, General Relativity, University of Cambridge Part III Mathematical Tripos lecture notes, 2019. PDF.