Contracted Bianchi Identity and the Einstein Tensor

Derivation

Contracted Bianchi Identity and the Einstein Tensor

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Statement

Starting from the second (differential) Bianchi identity for the Riemann tensor, \(\nabla_{[\lambda}R_{\rho\sigma]\mu\nu}=0\) written cyclically as \(\nabla_\lambda R_{\rho\sigma\mu\nu}+\nabla_\mu R_{\rho\sigma\nu\lambda}+\nabla_\nu R_{\rho\sigma\lambda\mu}=0\), contracting twice with the inverse metric yields the contracted Bianchi identity \(\nabla^\mu R_{\mu\nu}=\tfrac12\nabla_\nu R\), equivalently \(\nabla^\mu G_{\mu\nu}=0\) with the Einstein tensor \(G_{\mu\nu}=R_{\mu\nu}-\tfrac12 g_{\mu\nu}R\); moreover \(G_{\mu\nu}+\Lambda g_{\mu\nu}\) is the unique symmetric two-tensor built algebraically from \(R_{\mu\nu}\), \(g_{\mu\nu}\), and \(R\) that is identically divergence-free.

Why it matters

The contracted Bianchi identity is the geometric backbone of general relativity. Because \(\nabla^\mu G_{\mu\nu}=0\) holds identically for every metric, setting \(G_{\mu\nu}\) proportional to the stress–energy tensor \(T_{\mu\nu}\) automatically enforces local energy–momentum conservation \(\nabla^\mu T_{\mu\nu}=0\). Einstein was led to the correct field equations precisely because the naive candidate \(R_{\mu\nu}\propto T_{\mu\nu}\) is inconsistent: \(\nabla^\mu R_{\mu\nu}=\tfrac12\nabla_\nu R\neq0\) in general.

The uniqueness statement (a low-order case of Lovelock's theorem) explains why there is essentially no freedom in the left-hand side of the field equations in four dimensions, apart from an overall constant and the cosmological term \(\Lambda g_{\mu\nu}\). This rigidity ties the symmetry thread (diffeomorphism invariance and the Bianchi identities as its Noether shadow) to the energy thread (conservation of \(T_{\mu\nu}\)).

Assumptions

The connection is the Levi-Civita connection (metric-compatible, torsion-free).If \(\nabla_\alpha g_{\mu\nu}\neq0\), the metric cannot be moved through covariant derivatives during contraction and the clean identity \(\nabla^\mu G_{\mu\nu}=0\) fails; a torsion or non-metricity term survives.

The Riemann tensor is defined from the commutator of covariant derivatives (prior result), so it satisfies the algebraic symmetries and the second Bianchi identity.Without \(\nabla_{[\lambda}R_{\rho\sigma]\mu\nu}=0\) there is nothing to contract; the whole derivation has no premise.

Spacetime is a smooth (\(C^3\)) pseudo-Riemannian manifold so third derivatives of the metric exist and mixed partials commute.At curvature singularities or shock fronts the interchange \(\partial_\lambda\partial_\mu=\partial_\mu\partial_\lambda\) used to prove the Bianchi identity in normal coordinates is not licensed, and the identity holds only distributionally, if at all.

Derivation

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

Second Bianchi identity. In Riemann normal coordinates at a point \(p\), \(\Gamma^\alpha_{\beta\gamma}(p)=0\), so \(R_{\rho\sigma\mu\nu}=g_{\rho\alpha}(\partial_\mu\Gamma^\alpha_{\nu\sigma}-\partial_\nu\Gamma^\alpha_{\mu\sigma})\) and \(\nabla_\lambda R=\partial_\lambda R\) at \(p\). Cyclically antisymmetrizing \(\partial_\lambda\partial_\mu\Gamma^\alpha_{\nu\sigma}\) over \([\lambda\mu\nu]\) vanishes because partials commute; being a tensor equation true at arbitrary \(p\), it holds everywhere. C

\[ R_{\mu\nu}\equiv g^{\lambda\rho}R_{\rho\mu\lambda\nu}=R^{\lambda}{}_{\mu\lambda\nu},\qquad R\equiv g^{\mu\nu}R_{\mu\nu} \]

Fix conventions: Ricci is the (1,3)-contraction of Riemann, scalar curvature its trace. We use only the symmetries \(R_{\rho\sigma\mu\nu}=-R_{\sigma\rho\mu\nu}=-R_{\rho\sigma\nu\mu}=R_{\mu\nu\rho\sigma}\), inherited from the commutator definition of \(R\). A

\[ g^{\rho\mu}\big(\nabla_\lambda R_{\rho\sigma\mu\nu}+\nabla_\mu R_{\rho\sigma\nu\lambda}+\nabla_\nu R_{\rho\sigma\lambda\mu}\big)=0 \]

Contract Step 1 with \(g^{\rho\mu}\). Metric compatibility \(\nabla_\alpha g^{\rho\mu}=0\) lets \(g^{\rho\mu}\) pass through every \(\nabla\) as a constant, so contraction and differentiation commute. B

\[ \nabla_\lambda R_{\sigma\nu}-\nabla_\nu R_{\sigma\lambda}+\nabla^{\rho}R_{\rho\sigma\nu\lambda}=0 \]

Evaluate the three traces. Term 1: \(g^{\rho\mu}R_{\rho\sigma\mu\nu}=R_{\sigma\nu}\) (definition). Term 3: \(g^{\rho\mu}R_{\rho\sigma\lambda\mu}=-g^{\rho\mu}R_{\rho\sigma\mu\lambda}=-R_{\sigma\lambda}\) using antisymmetry in the last pair. Term 2 keeps a raised index, \(g^{\rho\mu}\nabla_\mu=\nabla^\rho\). This is the once-contracted Bianchi identity. B

\[ g^{\sigma\nu}\big(\nabla_\lambda R_{\sigma\nu}-\nabla_\nu R_{\sigma\lambda}+\nabla^{\rho}R_{\rho\sigma\nu\lambda}\big)=0 \]

Contract Step 4 a second time with \(g^{\sigma\nu}\), again moving the metric through the derivatives by compatibility. B

\[ \nabla_\lambda R-\nabla^{\sigma}R_{\sigma\lambda}-\nabla^{\rho}R_{\rho\lambda}=0 \]

Evaluate the traces. Term 1: \(g^{\sigma\nu}R_{\sigma\nu}=R\). Term 2: \(g^{\sigma\nu}\nabla_\nu R_{\sigma\lambda}=\nabla^\sigma R_{\sigma\lambda}\). Term 3: \(g^{\sigma\nu}R_{\rho\sigma\nu\lambda}=-g^{\sigma\nu}R_{\sigma\rho\nu\lambda}=-R_{\rho\lambda}\) (antisymmetry in the first pair, then the definition). B

\[ \nabla_\lambda R-2\,\nabla^{\mu}R_{\mu\lambda}=0\ \Longrightarrow\ \nabla^{\mu}R_{\mu\lambda}=\tfrac12\,\nabla_\lambda R \]

The two divergence terms in Step 6 are equal (Ricci is symmetric and the dummy index is renamed \(\sigma,\rho\to\mu\)); collect and solve. This is the contracted Bianchi identity. A

\[ \nabla^{\mu}R_{\mu\lambda}-\tfrac12\,\nabla^{\mu}\!\big(g_{\mu\lambda}R\big)=0\ \Longrightarrow\ \nabla^{\mu}\!\Big(R_{\mu\lambda}-\tfrac12 g_{\mu\lambda}R\Big)=0 \]

Rewrite \(\tfrac12\nabla_\lambda R=\tfrac12\nabla^\mu(g_{\mu\lambda}R)\), valid because \(\nabla^\mu g_{\mu\lambda}=0\) so the metric factors out of the derivative. Factor the common divergence. Define \(G_{\mu\nu}\equiv R_{\mu\nu}-\tfrac12 g_{\mu\nu}R\). A

\[ H_{\mu\nu}=a\,R_{\mu\nu}+b\,g_{\mu\nu}R+\Lambda\,g_{\mu\nu},\qquad \nabla^\mu H_{\mu\nu}=\Big(\tfrac{a}{2}+b\Big)\nabla_\nu R \]

Uniqueness. The most general symmetric two-tensor algebraic in \(R_{\mu\nu}\), \(g_{\mu\nu}\), \(R\) is \(H_{\mu\nu}\). Its divergence uses Step 7 for \(\nabla^\mu R_{\mu\nu}=\tfrac12\nabla_\nu R\) and \(\nabla^\mu(g_{\mu\nu}R)=\nabla_\nu R\); the \(\Lambda g_{\mu\nu}\) term is divergence-free automatically. Requiring \(\nabla^\mu H_{\mu\nu}=0\) for a generic metric (where \(\nabla_\nu R\neq0\)) forces \(b=-a/2\). Up to the overall scale \(a\), \(H_{\mu\nu}=a\,G_{\mu\nu}+\Lambda g_{\mu\nu}\). C

Result

\[ \nabla^{\mu}R_{\mu\nu}=\tfrac12\,\nabla_\nu R \qquad\Longleftrightarrow\qquad \nabla^{\mu}G_{\mu\nu}=0,\quad G_{\mu\nu}=R_{\mu\nu}-\tfrac12\,g_{\mu\nu}R \]

Reading. The divergence of the Ricci tensor is not zero but is locked to half the gradient of the scalar curvature. The particular combination that subtracts \(\tfrac12 g_{\mu\nu}R\) exactly cancels this "leftover," producing the unique (up to scale and a cosmological term) symmetric tensor whose covariant divergence vanishes identically on every metric. This is a geometric identity, true independently of any field equation.

Units check. With lengths as the only scale, curvature carries dimension \([R_{\mu\nu}]=[g_{\mu\nu}R]=L^{-2}\), so \(G_{\mu\nu}\) is dimensionally homogeneous. A covariant derivative adds \(L^{-1}\), giving \([\nabla^\mu G_{\mu\nu}]=L^{-3}\); both sides of \(\nabla^\mu G_{\mu\nu}=0\) match, and in Einstein's equation \(G_{\mu\nu}=\tfrac{8\pi G}{c^4}T_{\mu\nu}\) the prefactor \(\tfrac{8\pi G}{c^4}\approx2.1\times10^{-43}\ \mathrm{s^2\,kg^{-1}\,m^{-1}}\) converts \([T_{\mu\nu}]=\mathrm{J\,m^{-3}}\) to \(L^{-2}\).

Limiting cases

Flat spacetime: \(R_{\mu\nu\rho\sigma}=0\Rightarrow R_{\mu\nu}=R=0\Rightarrow G_{\mu\nu}=0\); the identity holds trivially, \(0=0\).
Constant scalar curvature (\(\nabla_\nu R=0\)): the contracted identity collapses to \(\nabla^\mu R_{\mu\nu}=0\); Ricci is itself conserved, as for maximally symmetric spaces.
Vacuum with cosmological constant: \(R_{\mu\nu}=\Lambda g_{\mu\nu}\Rightarrow R=n\Lambda\) constant, so \(G_{\mu\nu}=(1-\tfrac{n}{2})\Lambda g_{\mu\nu}\) is trivially divergence-free.
Two dimensions (\(n=2\)): \(G_{\mu\nu}\equiv0\) identically (the Einstein tensor is the trivial zero tensor), so the identity carries no dynamical content there.
Weak field, linearized: to first order \(\nabla\to\partial\) and the identity becomes \(\partial^\mu G^{(1)}_{\mu\nu}=0\), the linearized conservation law guaranteeing gauge consistency of \(h_{\mu\nu}\).

Breaks when

Nonzero torsion. In Einstein–Cartan or metric-affine gravity the connection is not symmetric; the second Bianchi identity acquires torsion source terms and \(\nabla^\mu G_{\mu\nu}\neq0\) in general. The clean divergence-free property is a property of the Levi-Civita connection specifically.
Non-metric connection. If \(\nabla_\alpha g_{\mu\nu}\neq0\), the metric cannot be pulled through the covariant derivatives in Steps 3–8; extra non-metricity terms survive and the Einstein tensor is no longer conserved.
Higher-curvature terms in the action. The identity \(\nabla^\mu G_{\mu\nu}=0\) is exact, but the field-equation tensor for \(f(R)\) or Gauss–Bonnet gravity is a different object; only in \(n>4\) does the Gauss–Bonnet (Lovelock) combination give a new independent divergence-free tensor, so uniqueness of \(G_{\mu\nu}\) is dimension-dependent.
Distributional curvature. At a curvature singularity, thin shell, or shock the third derivatives of the metric do not exist classically; mixed partials need not commute and the Bianchi identity holds only in a distributional sense (Israel junction conditions replace it).

Failure modes

Forgetting metric compatibility. Contracting with \(g^{\rho\mu}\) but leaving it outside the derivative as if \(g^{\rho\mu}\nabla_\lambda R=\nabla_\lambda(g^{\rho\mu}R)\) were an assumption to be checked — it is guaranteed only because \(\nabla g=0\); students who drop this get spurious \(\nabla g\) terms.
Sign error from antisymmetry. Writing \(g^{\rho\mu}R_{\rho\sigma\lambda\mu}=+R_{\sigma\lambda}\) instead of \(-R_{\sigma\lambda}\), forgetting the last-pair antisymmetry, flips a term and destroys the factor of \(\tfrac12\).
Wrong Ricci convention. Contracting the wrong pair of Riemann indices (or using \(R_{\mu\nu}=R^\lambda{}_{\mu\nu\lambda}\), which differs by a sign) scrambles the bookkeeping; the two internal divergence terms then fail to combine.
Claiming \(\nabla^\mu R_{\mu\nu}=0\). Asserting Ricci is conserved by analogy with \(G_{\mu\nu}\); it is not, except when \(R\) is constant. This is exactly the error the derivation corrects.
Dropping the cosmological term in uniqueness. Stating \(G_{\mu\nu}\) is the only divergence-free combination, ignoring that \(\Lambda g_{\mu\nu}\) is also identically conserved.
Using coordinate divergence \(\partial_\mu G^{\mu\nu}\). The ordinary partial divergence is not zero and not a tensor; only the covariant \(\nabla_\mu G^{\mu\nu}\) vanishes.

Discussion

The contracted Bianchi identity is best understood as the differential-geometric shadow of diffeomorphism invariance. By Noether's second theorem, a Lagrangian invariant under the infinite-dimensional group of coordinate changes yields not conservation laws for specific solutions but identities holding for all field configurations. Varying the Einstein–Hilbert action \(S=\tfrac{1}{2\kappa}\int R\sqrt{-g}\,d^4x\) produces \(G_{\mu\nu}\) as the metric equation of motion, and diffeomorphism invariance forces \(\nabla^\mu G_{\mu\nu}=0\) off-shell — precisely the identity derived here purely from curvature symmetries.

The identity is what makes Einstein's equations self-consistent. With \(G_{\mu\nu}=\kappa T_{\mu\nu}\), taking the covariant divergence gives \(\nabla^\mu T_{\mu\nu}=0\) as a mathematical necessity, not an extra postulate. This entangles geometry and matter: one cannot specify an arbitrary non-conserved source. It also reduces the ten field equations to six independent ones, the four Bianchi constraints reflecting the four-parameter coordinate freedom — the correct count for a theory whose dynamical content is the geometry modulo diffeomorphisms.

The uniqueness result generalizes to Lovelock's theorem: in \(n\) dimensions, the most general symmetric divergence-free tensor built from the metric and its first two derivatives, and linear in the second derivatives, is \(a\,G_{\mu\nu}+\Lambda g_{\mu\nu}\). Relaxing linearity admits the Gauss–Bonnet and higher Lovelock tensors, but these are topological (total derivatives) in four dimensions and contribute nothing to the four-dimensional field equations. This is the deep reason general relativity is essentially forced once one demands a metric theory with second-order, conserved field equations.

A sharper statement: the second Bianchi identity is the \(d\) of the curvature two-form, \(DR=0\), the Riemannian analogue of the Yang–Mills identity \(D F=0\) for the field-strength two-form. Contracting twice with the metric is the Hodge-dual operation that converts this exterior identity into the vector statement \(\nabla^\mu G_{\mu\nu}=0\). Equivalently, \(G_{\mu\nu}\) is (up to normalization) the double dual of the Riemann tensor's trace, and its conservation is the covariant exactness \(\nabla^\mu({*}R{*})_{\mu\nu}=0\). This viewpoint makes the parallel with gauge theory, where \(\nabla^\mu(\text{current})=0\) follows identically from the Bianchi identity of the connection, manifest.

Common misconceptions. The contracted Bianchi identity is not a conservation law that holds "because energy is conserved" — the logic runs the other way: it is a pure geometric identity that enforces conservation once the field equations are imposed. Nor does \(\nabla^\mu G_{\mu\nu}=0\) mean there is a globally conserved energy in curved spacetime; there is generally no such quantity, because \(\nabla^\mu T_{\mu\nu}=0\) is a local (covariant) statement, not an integral one, absent a timelike Killing vector.

Worked examples

Example 1 — Cosmological conservation (FLRW continuity equation).

\[ G_{\mu\nu}=\frac{8\pi G}{c^4}T_{\mu\nu}\ \Longrightarrow\ \nabla^\mu G_{\mu\nu}=0\ \Longrightarrow\ \nabla^\mu T_{\mu\nu}=0 \]

The contracted Bianchi identity applied to Einstein's equation forces the source to be conserved. Symbols first: take the covariant divergence of both sides. A

\[ T^{\mu}{}_{\nu}=\mathrm{diag}\!\big(-\rho c^2,\,p,\,p,\,p\big),\qquad \nabla_\mu T^{\mu}{}_{0}=0\ \Rightarrow\ \dot\rho+3\frac{\dot a}{a}\Big(\rho+\frac{p}{c^2}\Big)=0 \]

Insert a perfect fluid into the FLRW metric \(ds^2=-c^2dt^2+a(t)^2 d\vec x^{\,2}\); the time component of \(\nabla_\mu T^{\mu}{}_\nu=0\) is the continuity equation, with \(H\equiv\dot a/a\). B

\[ p=0\ \Rightarrow\ \dot\rho=-3H\rho;\qquad H_0=2.27\times10^{-18}\,\mathrm{s^{-1}},\ \rho_{m,0}=2.7\times10^{-27}\,\mathrm{kg\,m^{-3}} \]

Pressureless matter (dust). Substitute present-day numbers (\(H_0=70\ \mathrm{km\,s^{-1}\,Mpc^{-1}}\)). A

\[ \dot\rho=-3(2.27\times10^{-18})(2.7\times10^{-27})\ \mathrm{kg\,m^{-3}\,s^{-1}} \]

Numbers in. A

\[ \dot\rho_{m,0}\approx-1.84\times10^{-44}\ \mathrm{kg\,m^{-3}\,s^{-1}},\qquad \rho_m\propto a^{-3} \]

Reading. The Bianchi identity, via Einstein's equation, dictates that cosmic matter density dilutes as the inverse cube of the scale factor — the familiar \(\rho\propto a^{-3}\) — with the present decline rate shown. No independent conservation postulate is needed.

Example 2 — Schur's theorem: an Einstein space has constant curvature.

\[ R_{\mu\nu}=\frac{R}{n}\,g_{\mu\nu}\quad(\text{Einstein space, }n\ \text{dimensions}) \]

Assume Ricci is pointwise proportional to the metric with (possibly position-dependent) factor \(R/n\); tracing with \(g^{\mu\nu}\) fixes the coefficient as \(R/n\). B

\[ \nabla^\mu R_{\mu\nu}=\nabla^\mu\!\Big(\frac{R}{n}g_{\mu\nu}\Big)=\frac{1}{n}\nabla_\nu R \]

Take the divergence, moving the metric through \(\nabla\) by compatibility. A

\[ \frac{1}{n}\nabla_\nu R=\frac12\nabla_\nu R\ \Longrightarrow\ \Big(\frac1n-\frac12\Big)\nabla_\nu R=0 \]

Equate with the contracted Bianchi identity \(\nabla^\mu R_{\mu\nu}=\tfrac12\nabla_\nu R\). For \(n\neq2\) the prefactor is nonzero. A

\[ n=4,\ \Lambda=1.1\times10^{-52}\,\mathrm{m^{-2}}:\quad R_{\mu\nu}=\Lambda g_{\mu\nu}\ \Rightarrow\ R=4\Lambda=4.4\times10^{-52}\,\mathrm{m^{-2}} \]

De Sitter space is the concrete \(n=4\) realization: a genuine Einstein space, and Schur's theorem is satisfied because \(\Lambda\) (hence \(R\)) is indeed constant, consistent with the observed cosmological constant scale. A

\[ n\ge3:\quad R_{\mu\nu}=\frac{R}{n}g_{\mu\nu}\ \Longrightarrow\ \nabla_\nu R=0\quad(R=\text{const}) \]

Reading. The contracted Bianchi identity alone forces any three-or-higher-dimensional Einstein space to have constant scalar curvature — pointwise isotropy of Ricci curvature upgrades to global constancy. This is Schur's rigidity theorem, and it is why de Sitter (\(R=4\Lambda\)) is self-consistent while a spatially varying \(\Lambda(x)\) would violate the identity.

Problems

Starting from \(\nabla^\mu R_{\mu\nu}=\tfrac12\nabla_\nu R\), show that the Einstein tensor with a cosmological constant, \(G_{\mu\nu}+\Lambda g_{\mu\nu}\), is also divergence-free, and explain why \(\Lambda\) must be a spacetime constant.
Solution
\(\nabla^\mu(G_{\mu\nu}+\Lambda g_{\mu\nu})=\nabla^\mu G_{\mu\nu}+g_{\mu\nu}\nabla^\mu\Lambda+\Lambda\nabla^\mu g_{\mu\nu}\). The first term is zero by the contracted Bianchi identity; the last term is zero by metric compatibility. Hence \(\nabla^\mu(G_{\mu\nu}+\Lambda g_{\mu\nu})=\nabla_\nu\Lambda\). For the combination to be identically conserved for all metrics we need \(\nabla_\nu\Lambda=0\), i.e. \(\Lambda\) constant. (If \(\Lambda\) varied it would behave as a dynamical scalar field, not a constant of nature.)
In \(n\) dimensions the Einstein tensor is \(G_{\mu\nu}=R_{\mu\nu}-\tfrac12 g_{\mu\nu}R\). Compute its trace \(g^{\mu\nu}G_{\mu\nu}\) and show \(G_{\mu\nu}\equiv0\) when \(n=2\).
Solution
\(g^{\mu\nu}G_{\mu\nu}=g^{\mu\nu}R_{\mu\nu}-\tfrac12 g^{\mu\nu}g_{\mu\nu}R=R-\tfrac12\,n\,R=\big(1-\tfrac{n}{2}\big)R\). For \(n=2\) the trace is \(0\). More strongly, in two dimensions Riemann has a single independent component, \(R_{\mu\nu}=\tfrac12 g_{\mu\nu}R\); substituting gives \(G_{\mu\nu}=\tfrac12 g_{\mu\nu}R-\tfrac12 g_{\mu\nu}R=0\) identically. Two-dimensional gravity therefore has no local Einstein dynamics.
Take the trace-reversed form of Einstein's equation \(R_{\mu\nu}=\tfrac{8\pi G}{c^4}\big(T_{\mu\nu}-\tfrac12 g_{\mu\nu}T\big)\) in \(n=4\). Using the contracted Bianchi identity, show it is equivalent to \(G_{\mu\nu}=\tfrac{8\pi G}{c^4}T_{\mu\nu}\).
Solution
Trace \(R_{\mu\nu}=\kappa(T_{\mu\nu}-\tfrac12 g_{\mu\nu}T)\) with \(g^{\mu\nu}\) (\(n=4\), \(\kappa=8\pi G/c^4\)): \(R=\kappa(T-\tfrac12\cdot4\cdot T)=\kappa(T-2T)=-\kappa T\), so \(T=-R/\kappa\). Substitute back: \(R_{\mu\nu}=\kappa T_{\mu\nu}-\tfrac12\kappa g_{\mu\nu}T=\kappa T_{\mu\nu}+\tfrac12 g_{\mu\nu}R\). Rearrange: \(R_{\mu\nu}-\tfrac12 g_{\mu\nu}R=\kappa T_{\mu\nu}\), i.e. \(G_{\mu\nu}=\kappa T_{\mu\nu}\). The two forms are algebraically identical; the Bianchi identity guarantees both imply \(\nabla^\mu T_{\mu\nu}=0\).
Consider the candidate field equation \(a R_{\mu\nu}+b g_{\mu\nu}R=\kappa T_{\mu\nu}\). Given that \(T_{\mu\nu}\) is conserved and generic (so \(\nabla_\nu R\) need not vanish), determine the ratio \(b/a\).
Solution
Take the divergence: \(\nabla^\mu(aR_{\mu\nu}+bg_{\mu\nu}R)=a\cdot\tfrac12\nabla_\nu R+b\nabla_\nu R=\big(\tfrac{a}{2}+b\big)\nabla_\nu R\) (using the contracted Bianchi identity and \(\nabla^\mu(g_{\mu\nu}R)=\nabla_\nu R\)). The right side gives \(\kappa\nabla^\mu T_{\mu\nu}=0\). Since \(\nabla_\nu R\neq0\) in general, \(\tfrac{a}{2}+b=0\), so \(b/a=-\tfrac12\) — the Einstein tensor ratio, recovered without ever assuming it.
For a maximally symmetric space, \(R_{\mu\nu\rho\sigma}=K(g_{\mu\rho}g_{\nu\sigma}-g_{\mu\sigma}g_{\nu\rho})\) with \(K\) constant in \(n\) dimensions. Compute \(R_{\mu\nu}\), \(R\), and \(G_{\mu\nu}\), and verify \(\nabla^\mu G_{\mu\nu}=0\) directly.
Solution
Contract with \(g^{\mu\rho}\): \(R_{\nu\sigma}=K(g^{\mu\rho}g_{\mu\rho}g_{\nu\sigma}-g^{\mu\rho}g_{\mu\sigma}g_{\nu\rho})=K(n g_{\nu\sigma}-g_{\nu\sigma})=K(n-1)g_{\nu\sigma}\). Trace: \(R=K(n-1)g^{\nu\sigma}g_{\nu\sigma}=Kn(n-1)\). Then \(G_{\mu\nu}=K(n-1)g_{\mu\nu}-\tfrac12 g_{\mu\nu}Kn(n-1)=K(n-1)\big(1-\tfrac{n}{2}\big)g_{\mu\nu}\). Since \(K\) is constant and \(\nabla^\mu g_{\mu\nu}=0\), \(\nabla^\mu G_{\mu\nu}=K(n-1)(1-\tfrac{n}{2})\nabla^\mu g_{\mu\nu}=0\). The identity holds trivially because scalar curvature is constant, consistent with the constant-\(R\) limiting case.