First (Algebraic) Bianchi Identity

The Riemann tensor $R^{\rho}{}_{\sigma\mu\nu}$ is a bulky object — sixteen slots times sixteen in four dimensions, $256$ raw numbers at every point — and almost all of that bulk is redundant. What tames it is a short list of universal identities, and the first Bianchi identity is the one that lives entirely at a single point, using no derivatives of the curvature at all. It says that if you cyclically rotate the last three indices of the Riemann tensor and add the three results, they cancel exactly: $R^{\rho}{}_{\sigma\mu\nu}+R^{\rho}{}_{\mu\nu\sigma}+R^{\rho}{}_{\nu\sigma\mu}=0$ , equivalently the total antisymmetrisation $R^{\rho}{}_{[\sigma\mu\nu]}$ vanishes. It is purely algebraic — a constraint among the values themselves, not among their rates of change — which is what separates it from its differential sibling, the second Bianchi identity. The origin is startlingly cheap. It does not require a metric, does not require metric compatibility, does not require Riemannian or Lorentzian signature. It requires one thing only: that the connection be torsion-free, so the Christoffel symbols are symmetric in their lower pair, $\Gamma^{\rho}{}_{\mu\nu}=\Gamma^{\rho}{}_{\nu\mu}$ . That single symmetry, fed through the definition of curvature, makes every term in the cyclic sum cancel against a partner. The identity is therefore a fact of affine geometry — the geometry of parallel transport alone — and it survives on any manifold carrying a symmetric connection, whether or not a metric is ever introduced. Its consequences are load-bearing for general relativity. Contract it once and the Ricci tensor comes out symmetric, $R_{\mu\nu}=R_{\nu\mu}$ ; that symmetry is exactly what lets the Einstein tensor sit on the left of the field equations opposite a symmetric stress-energy tensor. It removes precisely one component from the curvature count, taking $21$ down to the famous $20$ in four dimensions, and it forces the Weyl tensor to inherit the same cyclic property, isolating the pure tidal part of gravity.

Key takeaways

The first (algebraic) Bianchi identity is the cyclic sum $R^{\rho}{}_{\sigma\mu\nu}+R^{\rho}{}_{\mu\nu\sigma}+R^{\rho}{}_{\nu\sigma\mu}=0$ , equivalently the vanishing total antisymmetrisation $R^{\rho}{}_{[\sigma\mu\nu]}=0$ .
It is genuinely algebraic — no derivatives of the curvature appear — which is what distinguishes it from the differential (second) Bianchi identity $\nabla_{[\lambda}R_{\rho\sigma]\mu\nu}=0$ .
It follows from one assumption alone: a torsion-free connection, $\Gamma^{\rho}{}_{[\mu\nu]}=0$ . No metric, no metric compatibility, and no signature are needed — it is a statement of affine geometry.
Contracting it proves the Ricci tensor is symmetric, $R_{\mu\nu}=R_{\nu\mu}$ , the fact that makes the Einstein equations consistent with a symmetric stress-energy tensor.
Combined with the pair antisymmetries and pair-exchange symmetry it leaves $N=n^2(n^2-1)/12$ independent Riemann components: $1$ in $n=2$ , $6$ in $n=3$ , $20$ in $n=4$ .
Add torsion and it breaks in a controlled way: the cyclic sum equals derivatives and products of the torsion tensor, the signature of Einstein–Cartan geometry.
Key idea: total antisymmetrisation of the Riemann tensor over its last three indices vanishes, $R^{\rho}{}_{[\sigma\mu\nu]}=0$ , as a direct consequence of the connection being torsion-free — independent of any metric.

Where this lives today

The first Bianchi identity is not a historical curiosity — it is an active constraint that software and theorists lean on daily. Every numerical-relativity code that evolves the Einstein equations, from binary-black-hole merger simulations feeding LIGO–Virgo–KAGRA templates to neutron-star collapse models, stores the Riemann and Weyl tensors in symmetry-reduced form, and uses the cyclic identity together with the pair symmetries as a running consistency check that the finite-difference or spectral curvature has not drifted off the constraint surface. In the classification of exact solutions, the Newman–Penrose and Geroch–Held–Penrose formalisms encode the first Bianchi identity directly into the algebraic relations among the Weyl scalars $\Psi_0,\dots,\Psi_4$ , which is how the Petrov type of a spacetime — and hence whether it can carry gravitational radiation — is read off. The identity's failure is equally a research tool: in Einstein–Cartan gravity and in Poincaré gauge theories, the cyclic sum no longer vanishes but equals torsion terms, and this is precisely the observable that experiments and astrophysical bounds on spin–torsion coupling try to constrain, testing whether intrinsic spin sources a piece of geometry that standard general relativity forbids. Wherever curvature is computed, symmetry-reduced, or generalised, the first Bianchi identity is the quiet accountant keeping the books.

The physical picture

Curvature is holonomy: parallel-transport a vector around a small closed loop and it comes back rotated, and the rotation per unit area is exactly what the Riemann tensor measures. In $R^{\rho}{}_{\sigma\mu\nu}$ the last pair $\mu\nu$ names the plane of the loop, while $\sigma$ names the direction being transported and $\rho$ the direction it tilts into. The first Bianchi identity is a statement about what happens when you build three such loops by cyclically rotating who plays which role — loop in $\mu\nu$ , loop in $\nu\sigma$ , loop in $\sigma\mu$ — and add the three holonomies. In a torsion-free geometry they cancel to zero. It is the closure condition of a space where infinitesimal parallelograms actually close: torsion is precisely the failure of two small displacements to form a closed quadrilateral, and when that failure is absent, the cyclic sum of curvatures must vanish.

Figure

Three parallel-transport loops built by cyclically rotating which two directions span the loop and which is carried around it. In a torsion-free geometry the three holonomies cancel exactly — R^ρ_σμν + R^ρ_μνσ + R^ρ_νσμ = 0

There is a second, purely bookkeeping picture, and it is the one a working relativist reaches for first. The Riemann tensor's other symmetries — antisymmetry in the first pair, antisymmetry in the second pair, and exchange of the two pairs — already squeeze the $256$ components in four dimensions down to a symmetric $6\times 6$ matrix on antisymmetric index-pairs, that is $21$ numbers. The first Bianchi identity is the single extra scalar equation that removes exactly one of them, leaving $20$ . That last unit is the totally antisymmetric part $R_{[\rho\sigma\mu\nu]}$ , which the identity sets to zero. The second figure tracks this reduction with the real numbers attached, and shows where the same one constraint lands in two and three dimensions.

Figure

The component reduction in n = 4. Antisymmetry in each pair collapses 256 raw slots to a 6 by 6 pair matrix; pair-exchange symmetry keeps 21; the first Bianchi identity removes exactly one totally antisymmetric part, leaving 20. The formula N = n²(n²−1)/12 gives 1, 6, 20 for n = 2, 3, 4

The core equations

The identity has two equivalent faces — an explicit three-term cyclic sum and a compact antisymmetrisation — and both say the same thing about the last three indices of the curvature. Everything else in this article either derives these or reads a consequence off them.

R^{\rho}{}_{\sigma\mu\nu}+R^{\rho}{}_{\mu\nu\sigma}+R^{\rho}{}_{\nu\sigma\mu}=0 \qquad (1)

where: $R^{\rho}{}_{\sigma\mu\nu}$ — Riemann curvature tensor in mixed form; $\sigma,\mu,\nu$ — the three lower indices cyclically permuted while $\rho$ is held fixed. Read it as: sum the tensor over the three cyclic rotations of its last three slots and get identically zero.

R^{\rho}{}_{[\sigma\mu\nu]}=0 \qquad (2)

where: $[\,\cdot\,]$ — total antisymmetrisation over the bracketed indices with weight $1/3!$ . Because the Riemann tensor is already antisymmetric in $\mu\nu$ , antisymmetrising all three of $\sigma,\mu,\nu$ collapses to one third of the cyclic sum in $(1)$ — the two forms are equivalent. To see where $(1)$ comes from, one needs the curvature written in terms of the connection. This coordinate formula is the raw material every derivation below chews on.

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 (3)

where: $\Gamma^{\rho}{}_{\mu\nu}$ — connection (Christoffel) coefficients; $\partial_{\mu}$ — ordinary partial derivative. The antisymmetry of $(3)$ in $\mu\nu$ is manifest; the cyclic identity is the extra fact that appears only when $\Gamma^{\rho}{}_{[\mu\nu]}=0$ . The single hypothesis behind the whole identity is that the torsion vanishes. The torsion enters the curvature through the commutator of covariant derivatives, so it is worth stating the general commutator once and then setting torsion to zero.

[\nabla_{\mu},\nabla_{\nu}]V^{\rho}=R^{\rho}{}_{\sigma\mu\nu}V^{\sigma}-T^{\lambda}{}_{\mu\nu}\nabla_{\lambda}V^{\rho} \qquad (4)

where: $V^{\rho}$ — an arbitrary vector field; $T^{\lambda}{}_{\mu\nu}=2\Gamma^{\lambda}{}_{[\mu\nu]}=\Gamma^{\lambda}{}_{\mu\nu}-\Gamma^{\lambda}{}_{\nu\mu}$ — the torsion tensor. Setting $T^{\lambda}{}_{\mu\nu}=0$ is exactly the assumption that makes the cyclic identity hold. On the spacetimes of general relativity the connection is the metric Levi-Civita one, so we may lower the first index freely and write the identity in fully covariant form. This is the version that combines with the metric symmetries.

R_{\rho[\sigma\mu\nu]}=0, \qquad R_{\rho\sigma\mu\nu}=g_{\rho\lambda}R^{\lambda}{}_{\sigma\mu\nu} \qquad (5)

where: $R_{\rho\sigma\mu\nu}$ — all-lowered Riemann tensor; $g_{\rho\lambda}$ — metric tensor. On a metric connection the cyclic identity on the last three indices is equivalent to the cyclic identity on the last three of the lowered tensor. The single most important payoff is that contracting the identity forces the Ricci tensor to be symmetric — the property that makes the Einstein equations even writable.

R_{\mu\nu}=R^{\lambda}{}_{\mu\lambda\nu}=R_{\nu\mu} \qquad (6)

where: $R_{\mu\nu}$ — Ricci tensor, the trace of Riemann on its first and third indices. Symmetry here means the Einstein tensor $G_{\mu\nu}=R_{\mu\nu}-\tfrac12 R g_{\mu\nu}$ is symmetric and can be set equal to the symmetric stress-energy tensor. The first Bianchi identity is one member of the full algebraic symmetry group of the Riemann tensor. The others — recalled from the companion article on Riemann symmetries — are the two pair antisymmetries and the pair-exchange symmetry.

R_{\rho\sigma\mu\nu}=-R_{\sigma\rho\mu\nu}=-R_{\rho\sigma\nu\mu}=R_{\mu\nu\rho\sigma} \qquad (7)

where: the leading minus signs — antisymmetry within each index pair; $R_{\mu\nu\rho\sigma}$ — invariance under exchanging the two pairs. Note that the pair antisymmetry $R_{\rho\sigma\mu\nu}=-R_{\sigma\rho\mu\nu}$ is the one symmetry that genuinely needs metric compatibility, whereas the first Bianchi identity does not. Together, these symmetries plus the cyclic identity fix the number of independent components in any dimension. The closed-form count is the algebraic fingerprint of the whole symmetry set.

N=\frac{n^2(n^2-1)}{12} \qquad (8)

where: $n$ — spacetime dimension; $N$ — number of independent Riemann components. It evaluates to $N=1\,(n=2),\ 6\,(n=3),\ 20\,(n=4),\ 50\,(n=5)$ , and the first Bianchi identity is the constraint responsible for the last subtraction in each case where $n\ge 4$ .

Derivation AThe cyclic identity from a torsion-free connection

The goal here is the most hands-on possible proof: take the explicit coordinate formula for the Riemann tensor, write down all three cyclically rotated copies, add them, and watch every term cancel using nothing but the symmetry of the lower Christoffel indices. This shows with no room for doubt that the metric never enters — the only property used is $\Gamma^{\rho}{}_{\mu\nu}=\Gamma^{\rho}{}_{\nu\mu}$ — so the first Bianchi identity is a theorem about any symmetric affine connection, not about Riemannian or Lorentzian geometry in particular.

Step 1. Write the curvature from equation $(3)$ and form the cyclic sum $R^{\rho}{}_{\sigma\mu\nu}+R^{\rho}{}_{\mu\nu\sigma}+R^{\rho}{}_{\nu\sigma\mu}$ by applying the substitution $(\sigma\mu\nu)\to(\mu\nu\sigma)\to(\nu\sigma\mu)$ to each factor and adding. Group the result into a derivative part built from $\partial\Gamma$ terms and a quadratic part built from $\Gamma\Gamma$ terms; the two parts can be handled independently because they do not mix. Why: the connection formula is a sum of a linear-in-derivatives piece and a quadratic piece, and the cyclic operation acts on each separately, so cancellation can be checked block by block.

Step 2. Collect the six derivative terms: $\partial_{\mu}\Gamma^{\rho}{}_{\nu\sigma}-\partial_{\nu}\Gamma^{\rho}{}_{\mu\sigma}+\partial_{\nu}\Gamma^{\rho}{}_{\sigma\mu}-\partial_{\sigma}\Gamma^{\rho}{}_{\nu\mu}+\partial_{\sigma}\Gamma^{\rho}{}_{\mu\nu}-\partial_{\mu}\Gamma^{\rho}{}_{\sigma\nu}$ . Now use $\Gamma^{\rho}{}_{\nu\sigma}=\Gamma^{\rho}{}_{\sigma\nu}$ : the first term and the last term are $\partial_{\mu}$ acting on the same symbol with opposite sign, so they cancel; likewise the second cancels the fifth and the third cancels the fourth. The entire derivative block is zero. Why: torsion-freedom makes the lower index pair of each $\Gamma$ reorderable, and once reordered each $\partial\Gamma$ meets an exact negative of itself.

Step 3. Collect the quadratic terms, six products of the schematic form $\Gamma^{\rho}{}_{\bullet\lambda}\Gamma^{\lambda}{}_{\bullet\bullet}$ generated by the three cyclic copies. Again symmetrising the lower indices of every $\Gamma$ pairs each product with an equal-and-opposite partner, and the six cancel in three pairs. Why: the same symmetry that killed the derivative terms rearranges the contracted products into cancelling pairs, so the quadratic block vanishes on its own.

Step 4. Add the two vanishing blocks: $0+0=0$ , hence $R^{\rho}{}_{\sigma\mu\nu}+R^{\rho}{}_{\mu\nu\sigma}+R^{\rho}{}_{\nu\sigma\mu}=0$ , which is equation $(1)$ , and dividing by $3!$ gives the antisymmetrised form $(2)$ . Nowhere did a metric, a signature, or metric compatibility appear. Why: because the only input was $\Gamma^{\rho}{}_{[\mu\nu]}=0$ , the conclusion is a statement of pure affine geometry, valid for every torsion-free connection.

Go deeper · Derivation B · 2 minThe scalar-commutator route and the equivalence to torsion-freedom

Derivation BThe scalar-commutator route and the equivalence to torsion-freedom

This derivation gives the cleanest conceptual proof and, as a bonus, shows the identity is exactly equivalent to the vanishing of torsion — not merely implied by it. The idea is that a scalar field's second and third covariant derivatives are as symmetric as the connection allows, and the failure of the antisymmetrised third derivative to vanish is governed entirely by torsion. Feeding a scalar's gradient through the curvature commutator turns that statement into the cyclic curvature identity, with the torsion appearing as the only possible obstruction.

Step 1. Start from the commutator acting on a one-form, $[\nabla_{\mu},\nabla_{\nu}]\omega_{\sigma}=-R^{\rho}{}_{\sigma\mu\nu}\omega_{\rho}-T^{\lambda}{}_{\mu\nu}\nabla_{\lambda}\omega_{\sigma}$ , the dual companion of equation $(4)$ . This is completely general — it holds with torsion present — and it is the master relation we will antisymmetrise. Why: the one-form version puts the curvature and the torsion side by side in a single identity, so setting torsion to zero later is a clean surgical step.

Step 2. Specialise the one-form to a gradient, $\omega_{\sigma}=\nabla_{\sigma}f=\partial_{\sigma}f$ for a scalar $f$ . Then the left side becomes an object built from the third derivatives of $f$ , and the right side reads $-R^{\rho}{}_{\sigma\mu\nu}\partial_{\rho}f-T^{\lambda}{}_{\mu\nu}\nabla_{\lambda}\nabla_{\sigma}f$ . Why: a gradient is the most economical test field — its own curl vanishes — so any surviving antisymmetry must come from the connection, isolating the geometric content.

Step 3. Antisymmetrise over all three lower indices $\sigma,\mu,\nu$ . The Hessian $\nabla_{\nu}\nabla_{\sigma}f$ is symmetric in $\nu\sigma$ up to torsion, so its totally antisymmetric part is proportional to $T^{\lambda}{}_{[\mu\nu}\nabla_{\sigma]}\nabla_{\lambda}f$ , and one is left with $\nabla_{[\mu}\nabla_{\nu}\nabla_{\sigma]}f=-\tfrac12 R^{\rho}{}_{[\sigma\mu\nu]}\partial_{\rho}f-(\text{torsion terms})$ . Why: total antisymmetrisation kills every symmetric piece of the derivatives of $f$ , leaving only the curvature's cyclic part and whatever the torsion contributes.

Step 4. Impose torsion-freedom, $T^{\lambda}{}_{\mu\nu}=0$ . The torsion terms drop, and the antisymmetrised triple derivative of any scalar vanishes identically, giving $R^{\rho}{}_{[\sigma\mu\nu]}\partial_{\rho}f=0$ for every $f$ . Since $\partial_{\rho}f$ can point in any direction, the coefficient must vanish: $R^{\rho}{}_{[\sigma\mu\nu]}=0$ . Running the argument backwards shows the converse, so $R^{\rho}{}_{[\sigma\mu\nu]}=0\Leftrightarrow T^{\lambda}{}_{\mu\nu}=0$ . Why: the arbitrariness of the test gradient promotes a scalar identity to a full tensor identity, and the reversibility makes torsion-freedom not just sufficient but necessary.

Go deeper · Derivation C · 1 minContracting to the symmetry of the Ricci tensor

Derivation CContracting to the symmetry of the Ricci tensor

The single most consequential corollary is that the Ricci tensor is symmetric, and this derivation extracts it by tracing the fully covariant cyclic identity with the inverse metric. The result $R_{\mu\nu}=R_{\nu\mu}$ is what allows the Einstein tensor to sit opposite a symmetric stress-energy tensor; without it the field equations would be equating a non-symmetric object to a symmetric one, an immediate inconsistency. The argument uses only the cyclic identity and the last-pair antisymmetry — no pair-exchange symmetry is needed.

Step 1. Lower the first index of equation $(1)$ to write the cyclic identity as $R_{\rho\sigma\mu\nu}+R_{\rho\mu\nu\sigma}+R_{\rho\nu\sigma\mu}=0$ , then contract with the inverse metric $g^{\rho\mu}$ , raising $\rho$ and pairing it against the third index $\mu$ . Three terms result. Why: contracting the first with the third slot is exactly the trace that defines the Ricci tensor, so this contraction is designed to surface Ricci components.

Step 2. Evaluate the first term: $g^{\rho\mu}R_{\rho\sigma\mu\nu}=R^{\mu}{}_{\sigma\mu\nu}=R_{\sigma\nu}$ , the Ricci tensor by definition. Why: the trace on first and third indices is the standard definition $R_{\sigma\nu}=R^{\mu}{}_{\sigma\mu\nu}$ , so this term is just $R_{\sigma\nu}$ .

Step 3. Evaluate the middle term: $g^{\rho\mu}R_{\rho\mu\nu\sigma}=R^{\mu}{}_{\mu\nu\sigma}=0$ , because the tensor is antisymmetric in its first pair $\rho\mu$ and the trace of an antisymmetric object vanishes. Why: antisymmetry in the first pair means the symmetric contraction $g^{\rho\mu}$ annihilates it, removing the middle term entirely.

Step 4. Evaluate the last term: $g^{\rho\mu}R_{\rho\nu\sigma\mu}=R^{\mu}{}_{\nu\sigma\mu}=-R^{\mu}{}_{\nu\mu\sigma}=-R_{\nu\sigma}$ , using antisymmetry in the last pair to reorder into the Ricci trace with a sign. Adding the three terms gives $R_{\sigma\nu}-R_{\nu\sigma}=0$ , hence $R_{\mu\nu}=R_{\nu\mu}$ , equation $(6)$ . Why: the last-pair antisymmetry converts the third contraction into $-R_{\nu\sigma}$ , so the surviving terms are Ricci minus its transpose, and the cyclic identity forces that difference to zero.

Limiting cases

The identity either holds exactly, degenerates to a triviality, or breaks in a way that is itself diagnostic. The table sorts the regimes.

Case	Result	Meaning
Torsion-free (Levi-Civita)	$R^{\rho}{}_{[\sigma\mu\nu]}=0$	The standard case in general relativity: holds identically, and needs no metric beyond whatever defines the connection.
General torsion $T^{\rho}{}_{\mu\nu}\ne 0$	$R^{\rho}{}_{[\sigma\mu\nu]}=\nabla_{[\sigma}T^{\rho}{}_{\mu\nu]}+T^{\lambda}{}_{[\sigma\mu}T^{\rho}{}_{\nu]\lambda}$	In Einstein–Cartan / Riemann–Cartan geometry the cyclic sum equals derivatives and products of torsion; it no longer vanishes.
$n=2$ or $n=3$	identity is vacuous	With fewer than four dimensions there are no four fully distinct indices, so $R_{[\rho\sigma\mu\nu]}$ has no components and the constraint bites nothing new.
Maximally symmetric space	$R_{\rho\sigma\mu\nu}=K(g_{\rho\mu}g_{\sigma\nu}-g_{\rho\nu}g_{\sigma\mu})$	The constant-curvature form satisfies the cyclic identity automatically for any $K$ , because each cyclic image cancels against another by the symmetry of the metric.
Weyl (trace-free) part	$C^{\rho}{}_{[\sigma\mu\nu]}=0$	The identity is inherited by the trace-free conformal curvature, one of the constraints that isolates the Weyl tensor.

Common trapConfusing the first (algebraic) with the second (differential) Bianchi identity

The two identities share a name and a discoverer but are different animals. The first is algebraic — a cyclic sum on the lower three indices with no derivatives, $R^{\rho}{}_{[\sigma\mu\nu]}=0$ — and it needs only a torsion-free connection. The second is differential — a cyclic sum of covariant derivatives of the curvature, $\nabla_{[\lambda}R_{\rho\sigma]\mu\nu}=0$ — and it is what contracts down to $\nabla^{\mu}G_{\mu\nu}=0$ , the identity guaranteeing local energy-momentum conservation. Two further traps: the first identity does not require metric compatibility, unlike the first-pair antisymmetry $R_{\rho\sigma\mu\nu}=-R_{\sigma\rho\mu\nu}$ , which does; and the first identity is not a law of nature but a consequence of a modelling choice — allow torsion, as in Einstein–Cartan theory, and it acquires the torsion source terms shown in the table.

Worked examples

1 · Counting the independent curvature components in each dimension (fully worked)

Use the algebraic symmetries and the first Bianchi identity to count the independent components of the Riemann tensor in $n=2,3,4$ , and identify exactly what the cyclic identity contributes in each case. Then split the $n=4$ count into its Ricci and Weyl parts.

Step 1. Fold in the pair antisymmetries. The Riemann tensor is antisymmetric in its first pair $[\rho\sigma]$ and in its second pair $[\mu\nu]$ , so each pair ranges over only $M=\tfrac12 n(n-1)$ independent values. In $n=4$ that is $M=\tfrac12\cdot 4\cdot 3=\mathbf{6}$ antisymmetric pairs, so the tensor is an object with two pair-indices each running over $6$ values — a $6\times 6$ array of $36$ entries.

Step 2. Fold in the pair-exchange symmetry $R_{[\rho\sigma][\mu\nu]}=R_{[\mu\nu][\rho\sigma]}$ . This makes the $6\times 6$ array a symmetric matrix, cutting it to $\tfrac12 M(M+1)=\tfrac12\cdot 6\cdot 7=\mathbf{21}$ independent numbers. This is the count from the metric symmetries alone, before the Bianchi identity.

Step 3. Fold in the first Bianchi identity. The only genuinely new constraint it imposes is on four fully distinct indices, the totally antisymmetric part $R_{[\rho\sigma\mu\nu]}$ , which has $\binom{n}{4}$ components. In $n=4$ that is $\binom{4}{4}=\mathbf{1}$ , so the count drops $21\to\mathbf{20}$ . The closed form $N=n^2(n^2-1)/12=16\cdot 15/12=\mathbf{20}$ confirms it.

Step 4. Repeat in lower dimensions. In $n=3$ : $M=3$ , symmetric matrix $\tfrac12\cdot 3\cdot 4=6$ , and $\binom{3}{4}=0$ , so $N=\mathbf{6}$ — the cyclic identity contributes nothing, and Riemann is fully fixed by the Ricci tensor there. In $n=2$ : $M=1$ , symmetric matrix $1$ , $\binom{2}{4}=0$ , so $N=\mathbf{1}$ — the single Gaussian curvature. The first Bianchi identity removes one component only from $n=4$ upward; below that it is vacuous.

Step 5. Split the $n=4$ count. The symmetric Ricci tensor $R_{\mu\nu}=R_{\nu\mu}$ carries $\tfrac12 n(n+1)=\tfrac12\cdot 4\cdot 5=\mathbf{10}$ components, and the trace-free Weyl tensor carries the remaining $20-10=\mathbf{10}$ . So in four dimensions curvature splits cleanly into $10$ Ricci (locally sourced by matter through Einstein's equations) plus $10$ Weyl (the free tidal and gravitational-wave degrees of freedom), and the first Bianchi identity is what makes the Ricci half symmetric so the split is $10+10$ rather than something ill-defined.

2 · Verifying the identity on a constant-curvature space

A maximally symmetric space has $R_{\rho\sigma\mu\nu}=K(g_{\rho\mu}g_{\sigma\nu}-g_{\rho\nu}g_{\sigma\mu})$ for a constant $K$ — for a unit-radius de Sitter or sphere, $K=1$ . Check the cyclic identity explicitly, and confirm the Ricci contraction gives the expected symmetric tensor.

Step 1. Write the three cyclic terms of the lowered identity, holding $\rho$ fixed and rotating $\sigma\to\mu\to\nu$ . Substituting the constant-curvature form, $R_{\rho\sigma\mu\nu}+R_{\rho\mu\nu\sigma}+R_{\rho\nu\sigma\mu}=K\big[(g_{\rho\mu}g_{\sigma\nu}-g_{\rho\nu}g_{\sigma\mu})+(g_{\rho\nu}g_{\mu\sigma}-g_{\rho\sigma}g_{\mu\nu})+(g_{\rho\sigma}g_{\nu\mu}-g_{\rho\mu}g_{\nu\sigma})\big]$ .

Step 2. Cancel term by term using the symmetry $g_{ab}=g_{ba}$ . The pieces $+Kg_{\rho\mu}g_{\sigma\nu}$ and $-Kg_{\rho\mu}g_{\nu\sigma}$ are equal and opposite and cancel; $-Kg_{\rho\nu}g_{\sigma\mu}$ cancels $+Kg_{\rho\nu}g_{\mu\sigma}$ ; and $-Kg_{\rho\sigma}g_{\mu\nu}$ cancels $+Kg_{\rho\sigma}g_{\nu\mu}$ . Every term meets its negative, so the cyclic sum is $0$ for any value of $K$ — the constant-curvature form satisfies the first Bianchi identity identically.

Step 3. Contract to Ricci with $g^{\rho\mu}$ . Using $g^{\rho\mu}g_{\rho\mu}=n$ and $g^{\rho\mu}g_{\rho\nu}=\delta^{\mu}{}_{\nu}$ , $R_{\sigma\nu}=g^{\rho\mu}R_{\rho\sigma\mu\nu}=K(n\,g_{\sigma\nu}-g_{\sigma\nu})=K(n-1)g_{\sigma\nu}$ , which is manifestly symmetric in $\sigma\nu$ , as the general theorem guarantees. For a unit sphere in $n=4$ with $K=1$ this gives $R_{\mu\nu}=3\,g_{\mu\nu}$ and Ricci scalar $R=g^{\mu\nu}R_{\mu\nu}=K n(n-1)=12$ , the standard Einstein-space result — a concrete check that the symmetric-Ricci corollary holds.

Physics of the portrait

The name on the identity belongs to a geometer who spent his career on exactly the kind of local differential structure this identity encodes — though, as often happens in mathematics, the naming is more a tribute than a strict record of first discovery.

Origins · 1 minLuigi Bianchi and how this was discovered

Luigi Bianchi

1856–1928 · differential geometry of connections and curvature

Luigi Bianchi was an Italian mathematician of the Pisa school, a student of Ulisse Dini and Enrico Betti and later a professor at the Scuola Normale Superiore, whose work sat at the centre of the great Italian program in differential geometry that also produced Ricci-Curbastro and Levi-Civita. He is remembered above all for the curvature identities that bear his name, published in his 1902 note on the four-index symbols and Riemann's curvature — the differential identity $\nabla_{[\lambda}R_{\rho\sigma]\mu\nu}=0$ being the one most tightly bound to his name. The algebraic (first) identity treated here is older and more diffuse in origin: its content is already present in Christoffel's 1869 study of the transformation of quadratic differential forms, where the symmetries of the four-index curvature symbol were first laid out, and it was well known to Ricci-Curbastro in the absolute differential calculus of the 1880s and 1890s. That both cyclic relations are now filed under "Bianchi" reflects how his 1902 systematisation fixed them in the literature just as Einstein and the physicists were about to need them. When Einstein wrote down the field equations in 1915, the symmetric Ricci tensor that the algebraic identity guarantees, and the contracted differential identity $\nabla^{\mu}G_{\mu\nu}=0$ that protects energy-momentum conservation, were exactly the structural facts that made his equations consistent — geometry the Italian school had prepared a generation in advance.

History

1854

— Riemann introduces the curvature tensor in his habilitation lecture, published posthumously in 1868

1869

— Christoffel lays out the four-index curvature symbol and its algebraic symmetries

1880

— Ricci-Curbastro begins the absolute differential calculus that systematises tensor identities

1902

— Bianchi publishes the note fixing the curvature identities that now carry his name

1915

— Einstein's field equations rely on the symmetric Ricci tensor the identity guarantees

1922

— Cartan introduces torsion, the exact quantity whose vanishing the identity requires

Going further

Texts. Misner, Thorne and Wheeler, Gravitation, §13.5 (equations of structure and the symmetries of Riemann) · Wald, General Relativity, §3.2 · Carroll, Spacetime and Geometry, §3.4 · Hartle, Gravity, Ch. 21 · Schutz, A First Course in General Relativity, §6.6.

Provenance. All results are standard and independently reproducible. A verifier should confirm: (1) the cyclic cancellation of the $\partial\Gamma$ and $\Gamma\Gamma$ terms of equation $(3)$ under $\Gamma^{\rho}{}_{[\mu\nu]}=0$ alone, matching MTW §13.5 and Schutz §6.6; (2) the scalar-commutator argument that $\nabla_{[\mu}\nabla_{\nu}\nabla_{\sigma]}f=0$ when torsion vanishes, and its converse, matching Wald §3.2 and Carroll §3.4; (3) the contraction $g^{\rho\mu}$ of the lowered cyclic identity giving $R_{\mu\nu}=R_{\nu\mu}$ ; (4) the count $N=n^2(n^2-1)/12=1,6,20,50$ for $n=2,3,4,5$ and the $n=4$ split $20=10+10$ into Ricci and Weyl; (5) the constant-curvature check, including $R_{\mu\nu}=K(n-1)g_{\mu\nu}$ and $R=Kn(n-1)$ . No empirical numbers enter; the content is purely geometric.

Prerequisites. Riemann Curvature Tensor · Curvature as the Commutator of Covariant Derivatives · Algebraic Symmetries of the Riemann Tensor. Leads to. Ricci Tensor · Independent Components of the Riemann Tensor · Second (Differential) Bianchi Identity · Weyl Conformal Tensor.

Ask the tutor: Ask me to walk you through the term-by-term cancellation of the Christoffel formula, or to show exactly how contracting the cyclic identity forces the Ricci tensor to be symmetric.

Why does the first Bianchi identity not need a metric when the Riemann tensor is usually introduced with one?

The Riemann tensor is often first met in a metric setting, but its definition only ever uses the connection — curvature is the commutator of covariant derivatives, and the covariant derivative is built from the Christoffel symbols, not from the metric directly. The first Bianchi identity draws on just one property of those symbols, the symmetry $\Gamma^{\rho}{}_{\mu\nu}=\Gamma^{\rho}{}_{\nu\mu}$ that defines a torsion-free connection. Because a metric is never invoked in the proof, the identity holds on any manifold carrying a symmetric affine connection, metric or not. What the metric adds are the other symmetries — in particular the first-pair antisymmetry, which genuinely does require metric compatibility.

What exactly is the difference between the first and second Bianchi identities?

They differ in whether derivatives of the curvature appear. The first (algebraic) identity is a relation among the curvature components at a single point, $R^{\rho}{}_{[\sigma\mu\nu]}=0$ , with no derivatives — it constrains the values themselves. The second (differential) identity is a relation among the covariant derivatives of the curvature, $\nabla_{[\lambda}R_{\rho\sigma]\mu\nu}=0$ , and it constrains how curvature changes from point to point. Their consequences also differ: contracting the first gives the symmetry of the Ricci tensor, while contracting the second twice gives $\nabla^{\mu}G_{\mu\nu}=0$ , the vanishing divergence of the Einstein tensor that underwrites local energy-momentum conservation.

How does the identity force the Ricci tensor to be symmetric?

Lower the first index of the cyclic identity and contract the first and third indices with the inverse metric. Three terms appear. The first is the Ricci tensor $R_{\sigma\nu}$ by definition. The middle term is the trace of an object antisymmetric in the contracted pair, so it vanishes. The third, after reordering with the last-pair antisymmetry, is $-R_{\nu\sigma}$ . The identity therefore reads $R_{\sigma\nu}-R_{\nu\sigma}=0$ , i.e. the Ricci tensor equals its own transpose. This matters because the Einstein tensor $G_{\mu\nu}=R_{\mu\nu}-\tfrac12 R g_{\mu\nu}$ inherits the symmetry and can be set equal to the symmetric stress-energy tensor.

If the identity is vacuous in two and three dimensions, why mention those cases?

Because the vacuity is itself informative about what the identity is really doing. The genuinely new constraint it imposes lives on the totally antisymmetric part of the curvature over four indices, which has $\binom{n}{4}$ components — zero when $n<4$ . So in two dimensions the single Gaussian curvature and in three dimensions the six independent components are already fixed by the pair symmetries alone; the cyclic identity subtracts nothing. It first bites at $n=4$ , removing exactly one component to give twenty. Seeing where it does nothing sharpens the understanding that its entire content is the vanishing of $R_{[\rho\sigma\mu\nu]}$ , a four-index object.

What happens to the identity in theories with torsion, like Einstein–Cartan gravity?

It stops being an identity and becomes an equation with a source. In Riemann–Cartan geometry the connection has a torsion tensor $T^{\lambda}{}_{\mu\nu}=2\Gamma^{\lambda}{}_{[\mu\nu]}$ , and the cyclic sum of the curvature no longer vanishes: $R^{\rho}{}_{[\sigma\mu\nu]}=\nabla_{[\sigma}T^{\rho}{}_{\mu\nu]}+T^{\lambda}{}_{[\sigma\mu}T^{\rho}{}_{\nu]\lambda}$ . In Einstein–Cartan theory the torsion is sourced algebraically by the intrinsic spin of matter, so a non-vanishing right-hand side is a direct signature of spin coupling to geometry. This makes the first Bianchi identity a testing ground: bounding the deviation from zero is one way experiments and astrophysical observations try to constrain whether spacetime carries torsion at all.

Does the first Bianchi identity say anything about the Weyl tensor?

Yes. When the Riemann tensor is decomposed into its trace parts — the Ricci tensor and Ricci scalar — plus a fully trace-free remainder, the trace-free piece is the Weyl tensor $C_{\rho\sigma\mu\nu}$ . Because the decomposition is built to respect all of Riemann's algebraic symmetries, the Weyl tensor inherits the first Bianchi identity: $C^{\rho}{}_{[\sigma\mu\nu]}=0$ . This cyclic property, together with the pair symmetries and the vanishing of all its traces, is exactly what pins down Weyl as the conformally invariant part of curvature — the tidal field that persists in vacuum and carries gravitational waves, undetermined locally by the matter content.

Is the first Bianchi identity really due to Luigi Bianchi?

Only loosely, and the honest answer is a useful lesson in how mathematics gets named. The cyclic algebraic identity was already implicit in Christoffel's 1869 work on the four-index curvature symbol and was standard equipment in Ricci-Curbastro's absolute differential calculus well before 1900. Bianchi's own 1902 note is most directly associated with the differential identity. But because his systematisation fixed both cyclic relations in the literature at the moment they became central to Einstein's gravity, the name "Bianchi" attached to the whole family. The portrait honours the eponym while noting the real chain of discovery runs through Riemann, Christoffel and Ricci as much as through Bianchi himself.