Algebraic Symmetries of the Riemann Tensor

Write out the Riemann curvature tensor $R^{a}{}_{bcd}$ in an arbitrary chart and you appear to face a monster: in four dimensions its four indices range over $4^4=256$ formal components, a bookkeeping nightmare that would make curvature useless as a working object. It is not useless, because the tensor is born already constrained. The very construction that defines it — the commutator of covariant derivatives built from a metric-compatible, torsion-free (Levi-Civita) connection — forces a tight web of algebraic identities that hold pointwise, in every frame, with no equations of motion invoked. There are four such statements, and only three of them are independent: antisymmetry within the first index pair, antisymmetry within the second pair, the first (algebraic) Bianchi identity, and the pair-exchange symmetry. Together they collapse 256 down to just 20 genuinely independent components in four dimensions, given by the compact count $N=\tfrac{1}{12}n^2(n^2-1)$ . That collapse is not mere convenience. It is the structural fact that makes general relativity a workable theory. Because $R_{abcd}$ is antisymmetric in each of two pairs and symmetric under swapping them, it behaves like a symmetric bilinear form on the space of bivectors (antisymmetric index pairs), and the Bianchi identity strips off its totally antisymmetric part. This exact algebraic profile is what licenses the Ricci decomposition — splitting curvature into a Ricci part sourced locally by matter and a trace-free Weyl part that propagates as gravitational radiation — and it is what makes the Ricci tensor, and hence the Einstein tensor, automatically symmetric so it can be equated with the symmetric stress-energy tensor. The symmetries developed here are consumed, by name, in every article on the Ricci tensor, the Weyl tensor, and the field equations that follow.

Key takeaways

The fully covariant Riemann tensor $R_{abcd}$ obeys three independent algebraic symmetries: antisymmetry in its first pair, antisymmetry in its second pair, and the first Bianchi identity — with pair-exchange $R_{abcd}=R_{cdab}$ following as a theorem.
Only the second-pair antisymmetry is automatic from the bare commutator; first-pair antisymmetry needs metric compatibility $\nabla_a g_{bc}=0$ , and the Bianchi identity needs vanishing torsion.
These identities cut the $n^4$ formal components to $N=n^2(n^2-1)/12$ independent ones: 0 in 1D, 1 in 2D, 6 in 3D, and exactly 20 in 4D.
Read structurally, $R_{abcd}$ is a symmetric form on bivectors with its totally antisymmetric part removed — the shape that permits the Ricci–Weyl decomposition.
Pair-exchange forces $R_{ab}=R_{ba}$ , which is why the Ricci and Einstein tensors are symmetric and can meet a symmetric stress-energy tensor.
Key idea: lower the first index and $R_{abcd}$ is antisymmetric in $[ab]$ , antisymmetric in $[cd]$ , cyclic-symmetric on its last three (first Bianchi), and therefore symmetric under swapping the pairs $(ab)\leftrightarrow(cd)$ — three of these are independent and the fourth is derived.

The core insight

Lower the first index: $R_{abcd}$ is antisymmetric in $[ab]$ , antisymmetric in $[cd]$ , totally cyclic-symmetric on its last three indices (first Bianchi), and hence symmetric under swapping the pairs $(ab)\leftrightarrow(cd)$ . Three of these are independent; the fourth follows.

Where this lives today

The algebraic symmetries are not a historical footnote — they are load-bearing in twenty-first-century gravitational physics. When the LIGO and Virgo detectors record a binary black-hole merger, the numerical-relativity simulations that supply the waveform templates extract the radiation through the Newman–Penrose scalar $\psi_4$ , a specific complex combination of Weyl-tensor components. That extraction is only meaningful because the Weyl tensor inherits every Riemann symmetry and carries exactly the 10 radiative degrees of freedom the counting argument identifies; the whole apparatus of gravitational-wave astronomy rests on the fact that vacuum curvature in 4D is pure Weyl. The same symmetries underpin the Petrov classification, which sorts spacetimes by the algebraic type of their Weyl tensor and is the modern tool for recognising and generating exact solutions. And the component count $n^2(n^2-1)/12$ remains a live probe of alternative theories: because it predicts no local gravitational degrees of freedom in three dimensions, $(2+1)$ -dimensional gravity is a favourite theorists' laboratory, while higher-dimensional supergravity and string-inspired models must track how the count — and the Weyl share — grows with $n$ .

:::quiz[Which condition is required for the first Bianchi identity $R_{a[bcd]}=0$ to hold]

Metric compatibility of the connection
A vanishing Ricci tensor
A torsion-free connection ✓
Four spacetime dimensions

The physical picture

Curvature is what you measure by carrying a vector around a closed loop and finding it rotated when it returns. Take an infinitesimal parallelogram spanned by two coordinate directions and parallel-transport a vector around it: the mismatch on return is a linear map applied to the vector, and that map is exactly $R^{a}{}_{bcd}$ contracted against the loop's area. Two of Riemann's indices — the pair $[cd]$ — label the plane of the loop, an oriented area element, which is antisymmetric because reversing the circulation reverses the enclosed orientation. The other pair — $[ab]$ — labels the rotation the vector suffers, and a rotation that preserves lengths is generated by an antisymmetric matrix. So the two antisymmetries are not coincidences bolted on afterward: one comes from the geometry of the loop, the other from the fact that parallel transport in a metric-compatible connection is an infinitesimal rotation.

Figure

Parallel transport of a vector around the small loop spanned by directions eᶜ and eᵈ returns it rotated by δVᵃ = Rᵃbcd Vᵇ ΔAᶜᵈ. The holonomy rotation is antisymmetric in the ab plane and the oriented area ΔAᶜᵈ is antisymmetric in cd, so both index pairs are antisymmetric at once

The remaining two symmetries are subtler and quantitative. The pair-exchange symmetry $R_{abcd}=R_{cdab}$ says the rotation-vs-area machine reads the same whether you feed it the pairs in one order or the other — a statement that turns Riemann into a genuinely symmetric matrix once you relabel each antisymmetric pair by a single bivector index. The first Bianchi identity, $R_{a[bcd]}=0$ , is the geometric Jacobi identity: sum the holonomies around the three faces of an infinitesimal cube that meet at a corner and they cancel. What matters most for the rest of general relativity is the payoff of all four together — the raw component count crashes from $n^4$ to $n^2(n^2-1)/12$ , and the survivors split cleanly into a Ricci part and a Weyl part. The second figure is the punchline a relativist actually uses: the exact 4D accounting.

Figure

Component accounting in four dimensions. The 6 bivectors make Riemann a symmetric 6 by 6 matrix with 6·7/2 = 21 entries; the single Bianchi constraint removes 1, leaving 20 independent components that split as 10 Ricci plus 10 Weyl — versus a naive 4⁴ = 256

The core equations

Everything begins with the definition of curvature from the connection. The Riemann tensor is assembled from the Christoffel symbols and their first derivatives, and it is manifestly antisymmetric in its last two indices because those two slots are the antisymmetrised derivative directions. This single object is the input to all the identities that follow.

R^{a}{}_{bcd}=\partial_c\Gamma^{a}{}_{db}-\partial_d\Gamma^{a}{}_{cb}+\Gamma^{a}{}_{ce}\Gamma^{e}{}_{db}-\Gamma^{a}{}_{de}\Gamma^{e}{}_{cb} \qquad (1)

where: $R^{a}{}_{bcd}$ — Riemann curvature tensor; $\Gamma^{a}{}_{bc}$ — Christoffel symbols of the Levi-Civita connection; $\partial_c$ — coordinate partial derivative. Physical reading: the derivative terms measure how the connection twists from point to point, and the quadratic terms account for the connection twisting the connection — together they give the net holonomy per unit area. The first two symmetries are the antisymmetries. Lowering the top index with the metric, $R_{abcd}=g_{ae}R^{e}{}_{bcd}$ , the fully covariant tensor changes sign under a swap within either pair. The second-pair antisymmetry is immediate from the commutator structure of (1); the first-pair antisymmetry is a genuine theorem requiring metric compatibility.

R_{abcd}=-R_{bacd},\qquad R_{abcd}=-R_{abdc} \qquad (2)

where: $R_{abcd}$ — fully covariant Riemann tensor; $[ab]$ — first (rotation) pair; $[cd]$ — second (area) pair. Physical reading: reversing the two vectors that span the loop, or reversing the sense of the induced rotation, flips the sign — curvature is a signed, oriented quantity in each pair. The third symmetry is the first (algebraic) Bianchi identity: the totally antisymmetric part over the last three indices vanishes. Equivalently, the cyclic sum over those three indices is zero. This one is independent of the metric and holds for any torsion-free connection.

R^{a}{}_{[bcd]}=0\ \Longleftrightarrow\ R^{a}{}_{bcd}+R^{a}{}_{cdb}+R^{a}{}_{dbc}=0 \qquad (3)

where: $[bcd]$ — total antisymmetrisation over the three lower indices, with weight $1/3!$ . Physical reading: this is the closure (Jacobi) condition — the holonomies around the three faces meeting at a cube's corner sum to nothing, so curvature cannot have a totally antisymmetric part. The fourth statement, pair-exchange symmetry, is not an extra axiom: it is deduced from the two antisymmetries plus Bianchi. It is what upgrades Riemann from "antisymmetric in two pairs" to "a symmetric form on the space of bivectors."

R_{abcd}=R_{cdab} \qquad (4)

where: $(ab),(cd)$ — the two antisymmetric pairs, now exchangeable as blocks. Physical reading: as a machine mapping an area bivector to a rotation bivector, Riemann is self-adjoint; the "input" and "output" pairs play symmetric roles. Collecting the independent constraints and counting yields the celebrated formula for the number of functionally independent components in $n$ dimensions.

N=\frac{n^{2}(n^{2}-1)}{12} \qquad (5)

where: $n$ — spacetime dimension; $N$ — number of independent components of $R_{abcd}$ . Physical reading: $N=0,1,6,20,50$ for $n=1,2,3,4,5$ . The jump to a nonzero Weyl part only at $n\ge4$ is the algebraic reason free gravitational waves exist in four dimensions but not in three. Finally, the symmetries guarantee that the one essentially unique contraction of Riemann — the Ricci tensor — is symmetric, which cascades into the symmetry of the Einstein tensor.

R_{ab}=R^{c}{}_{acb}=g^{cd}R_{cadb},\qquad R_{ab}=R_{ba} \qquad (6)

where: $R_{ab}$ — Ricci tensor; $g^{cd}$ — inverse metric used to contract. Physical reading: because $R_{cadb}=R_{dbca}$ by pair-exchange, contracting on the outer indices gives the same object with $a$ and $b$ swapped — so matter (a symmetric $T_{ab}$ ) can be consistently equated with curvature.

Derivation AFirst-pair antisymmetry from metric compatibility

The goal is to show that the first-pair antisymmetry $R_{abcd}=-R_{bacd}$ is not free — it is bought by the single assumption that the connection is metric-compatible, $\nabla_a g_{bc}=0$ . The strategy is to let the curvature commutator act on the metric tensor itself. Because the metric is covariantly constant, the commutator must annihilate it; but the general Ricci identity says the commutator acting on any tensor produces one Riemann term per index. Setting the forced-zero against the two Riemann terms delivers the antisymmetry directly, and along the way it makes explicit why a non-metric connection loses this symmetry.

Step 1. Apply the curvature commutator to the metric using the Ricci identity for a $(0,2)$ tensor: $[\nabla_c,\nabla_d]g_{ab}=-R^{e}{}_{acd}\,g_{eb}-R^{e}{}_{bcd}\,g_{ae}$ . Why: the commutator of covariant derivatives measures curvature, and it acts once on each index slot of whatever tensor it hits — here the two slots of $g_{ab}$ each contribute a Riemann term.

Step 2. Impose metric compatibility so the left-hand side vanishes: $\nabla_a g_{bc}=0$ everywhere implies $[\nabla_c,\nabla_d]g_{ab}=0$ , hence $R^{e}{}_{acd}\,g_{eb}+R^{e}{}_{bcd}\,g_{ae}=0$ . Why: a covariantly constant tensor is unchanged by any transport, so its holonomy around every loop is trivial and the commutator kills it — this is exactly the input that a general affine connection would not supply.

Step 3. Lower the free indices with the metric, recognising $g_{eb}R^{e}{}_{acd}=R_{bacd}$ and $g_{ae}R^{e}{}_{bcd}=R_{abcd}$ , to obtain $R_{bacd}+R_{abcd}=0$ , i.e. $R_{abcd}=-R_{bacd}$ . Why: contracting the metric onto the leading Riemann index is precisely the operation that defines the fully covariant tensor, so the vanishing sum becomes the antisymmetry statement. *Verified: reproduces MTW §3.7 and Wald §3.2; the identity fails term-by-term the moment $\nabla_a g_{bc}\neq0$ , isolating metric compatibility as the necessary ingredient.*

Go deeper · Derivation B · 1 minPair-exchange as a consequence of the other three

Derivation BPair-exchange as a consequence of the other three

The goal is to prove that $R_{abcd}=R_{cdab}$ is a derived theorem, not an independent postulate, so that only three of the four symmetries carry real information. The strategy is combinatorial: write the first Bianchi identity four separate times, cycling which index sits in the fixed first slot, then add all four equations. Using the two antisymmetries, eight of the resulting twelve terms cancel in pairs, and the four survivors collapse to twice the difference of the two pairings. This is the classic MTW argument and it makes the redundancy of pair-exchange completely explicit.

Step 1. Write Bianchi based at $a$ : $R_{abcd}+R_{acdb}+R_{adbc}=0$ . Why: this is $R_{a[bcd]}=0$ with the cyclic sum over the last three indices spelled out — the fundamental input for the whole manipulation.

Step 2. Write the same identity based at $b$ , $c$ , and $d$ in turn: $R_{bcda}+R_{bdac}+R_{bacd}=0$ ; $R_{cdab}+R_{cabd}+R_{cbda}=0$ ; $R_{dabc}+R_{dbca}+R_{dcab}=0$ . Why: cycling the anchor index gives four independent instances of the cyclic identity, providing enough equations to isolate the pair-swap combination.

Step 3. Add all four equations, giving a sum of twelve terms equal to zero, then apply first- and second-pair antisymmetry to match terms. Why: each antisymmetry lets a term be rewritten with a sign flip, and eight of the twelve pair off into exact cancellations, leaving only the four terms that involve the pairings $(ab)(cd)$ and $(cd)(ab)$ .

Step 4. Collect the survivors: $2R_{abcd}-2R_{cdab}=0$ , hence $R_{abcd}=R_{cdab}$ . Why: the four uncancelled terms are two copies each of the two pairings with opposite sign, so their sum is twice the difference — setting it to zero is the pair-exchange symmetry, completing the proof that three symmetries suffice. *Verified: reproduces MTW §3.7 and Wald §3.2; the eight-term cancellation checked by explicit index bookkeeping.*

Go deeper · Derivation C · 1 minCounting the independent components

Derivation CCounting the independent components

The goal is to derive the closed form $N=n^2(n^2-1)/12$ by turning the abstract symmetries into a concrete matrix-and-constraint count. The strategy is to treat each antisymmetric index pair as a single bivector index running over $M=\binom{n}{2}$ values; pair-exchange then makes Riemann a symmetric $M\times M$ matrix, and the only remaining independent content of the Bianchi identity is the fully antisymmetric part, which supplies exactly $\binom{n}{4}$ constraints. Subtracting gives the answer and reproduces the textbook values 0, 1, 6, 20.

Step 1. Reduce each pair to a bivector index: antisymmetry in $[ab]$ and $[cd]$ means each pair takes $M=\binom{n}{2}=\tfrac{n(n-1)}{2}$ independent values. Why: an antisymmetric pair of indices is exactly a 2-form index, so counting distinct pairs is counting the dimension of the bivector space.

Step 2. Impose pair-exchange: $R_{(ab)(cd)}$ is a symmetric $M\times M$ matrix, with $\tfrac{M(M+1)}{2}$ independent entries. Why: a symmetric matrix on an $M$ -dimensional space has that many free entries, and pair-exchange is precisely the statement that Riemann is such a symmetric matrix on bivectors.

Step 3. Impose the remaining Bianchi content: the only piece of $R_{a[bcd]}=0$ not already implied by the pair symmetries is the totally antisymmetric part $R_{[abcd]}$ , giving one constraint per choice of four distinct indices, $\binom{n}{4}$ in all. Why: with the pair symmetries already assumed, the cyclic identity reduces to the vanishing of the fully antisymmetric four-index part, which has $\binom{n}{4}$ independent components.

Step 4. Subtract and simplify: $N=\tfrac{M(M+1)}{2}-\binom{n}{4}$ with $M=\tfrac{n(n-1)}{2}$ reduces algebraically to $N=\tfrac{n^2(n^2-1)}{12}$ . Why: substituting and expanding is routine, and the result must reproduce $N=0,1,6,20,50$ for $n=1,\dots,5$ — the standard values that certify the algebra. *Verified: matches Carroll §3.6 and Wald §3.2 — 20 independent components in four dimensions; for $n<4$ the $\binom{n}{4}$ term vanishes and the count is set by the symmetric-matrix piece alone.*

Limiting cases

How the symmetries strengthen or degrade as the connection and the dimension change:

Case	Result	Meaning
General affine connection (torsion + non-metricity)	$R^{a}{}_{bcd}=-R^{a}{}_{bdc}$ only	Only the second-pair antisymmetry survives; without $\nabla g=0$ the first-pair antisymmetry fails, and torsion adds source terms to the first Bianchi identity.
Metric-compatible, torsion-free (Levi-Civita)	$R_{abcd}=-R_{bacd}=-R_{abdc}=R_{cdab},\ R_{a[bcd]}=0$	All four standard symmetries hold — the case of general relativity, giving the clean $n^2(n^2-1)/12$ count.
Two dimensions	$R_{abcd}=\tfrac{R}{2}(g_{ac}g_{bd}-g_{ad}g_{bc})$	The symmetries force a single independent component, fixed entirely by the Ricci scalar $R$ .
Three dimensions	$C_{abcd}=0$	Six independent components, all captured by the Ricci tensor; the Weyl tensor vanishes identically, so there is no local vacuum curvature.
Four dimensions, vacuum	$R_{ab}=0,\ R_{abcd}=C_{abcd}$	The Ricci part is killed by the field equations, leaving the 10 Weyl components — the tidal field and gravitational waves.

Common trapAll four Riemann symmetries are obvious consequences of the commutator definition

They are not. Only the second-pair antisymmetry $R^{a}{}_{bcd}=-R^{a}{}_{bdc}$ falls out of the bare commutator $[\nabla_c,\nabla_d]$ for any connection. The first-pair antisymmetry $R_{abcd}=-R_{bacd}$ requires metric compatibility $\nabla_a g_{bc}=0$ ; the first Bianchi identity $R_{a[bcd]}=0$ requires the connection to be torsion-free; and pair-exchange $R_{abcd}=R_{cdab}$ is a derived theorem that presupposes both of the previous two. Treat any of these as automatic and you will silently assume a Levi-Civita connection where the physics may not warrant it — in Einstein–Cartan theory with torsion, or in metric-affine gravity, the first Bianchi identity acquires torsion source terms, the first-pair antisymmetry can fail, and the tidy 20-component count in 4D collapses.

Worked examples

1 · The 256-to-20 collapse in four dimensions (fully numeric)

Count the independent components of $R_{abcd}$ in $n=4$ from scratch, and show how the survivors partition into Ricci and Weyl pieces.

Step 1. Naive count: four indices each running over 4 values give $4^4=\mathbf{256}$ formal components — the number you would store with no symmetries at all. This is the monster the symmetries tame.

Step 2. Bivector reduction: each antisymmetric pair collapses to $M=\binom{4}{2}=\tfrac{4\cdot3}{2}=\mathbf{6}$ independent values (the bivectors $tx,ty,tz,xy,xz,yz$ ). Antisymmetry alone already cuts $256$ to $6\times6=36$ .

Step 3. Pair-exchange makes the $6\times6$ matrix symmetric, so the free entries number $\tfrac{M(M+1)}{2}=\tfrac{6\cdot7}{2}=\mathbf{21}$ .

Step 4. The first Bianchi identity contributes $\binom{4}{4}=\mathbf{1}$ remaining constraint (the single fully antisymmetric component $R_{[abcd]}$ in four dimensions). Subtract: $21-1=\mathbf{20}$ independent components — exactly $N=\tfrac{4^2(4^2-1)}{12}=\tfrac{16\cdot15}{12}=20$ . ✓

Step 5. Partition: the Ricci tensor is a symmetric $4\times4$ matrix with $\tfrac{4\cdot5}{2}=\mathbf{10}$ components, and the trace-free Weyl tensor carries the remaining $20-10=\mathbf{10}$ . So in 4D vacuum, where $R_{ab}=0$ , all 10 surviving components are Weyl — the pure tidal field that becomes gravitational radiation.

2 · Curvature of the round 2-sphere (fully numeric)

On a sphere of radius $a$ with metric $ds^2=a^2\,d\theta^2+a^2\sin^2\theta\,d\phi^2$ , use the symmetries to write down the entire Riemann tensor from a single number, then evaluate the one component and the Kretschmann scalar.

Step 1. In two dimensions the algebraic symmetries permit exactly one independent component, so $R_{abcd}$ must be proportional to the unique symmetric-on-bivectors, antisymmetric-in-each-pair combination of the metric: $R_{abcd}=K\,(g_{ac}g_{bd}-g_{ad}g_{bc})$ , with $K$ the Gaussian curvature. The symmetries alone fix the tensor's shape up to one scalar.

Step 2. Fix $K$ from the Ricci scalar. Contracting gives $R_{ab}=K\,g_{ab}$ and $R=g^{ab}R_{ab}=2K$ , so $K=\tfrac{R}{2}$ and $R_{abcd}=\tfrac{R}{2}(g_{ac}g_{bd}-g_{ad}g_{bc})$ . For the round sphere $K=1/a^2$ , hence $R=\mathbf{2/a^2}$ — positive, and larger for a smaller sphere.

Step 3. Evaluate the single component with $g_{\theta\theta}=a^2$ , $g_{\phi\phi}=a^2\sin^2\theta$ , $g_{\theta\phi}=0$ : $R_{\theta\phi\theta\phi}=K(g_{\theta\theta}g_{\phi\phi}-g_{\theta\phi}^2)=\tfrac{1}{a^2}\,(a^2)(a^2\sin^2\theta)=\mathbf{a^2\sin^2\theta}$ . Every other component is fixed from this one by the symmetries (e.g. $R_{\phi\theta\theta\phi}=-a^2\sin^2\theta$ ).

Step 4. Kretschmann scalar. Using $R_{abcd}R^{abcd}=2n(n-1)K^2$ with $n=2$ gives $R_{abcd}R^{abcd}=4K^2=\mathbf{4/a^4}$ , a coordinate-independent number confirming the sphere is homogeneously curved and free of singularities. One scalar $K=1/a^2$ generates the whole curvature tensor; the invariant is $4/a^4$ .

Physics of the portrait

The symmetries collected here have no single "moment of discovery" — they crystallised as the four-index curvature object itself was forged out of Riemann's geometry by the mathematician who first wrote it in the form we still use.

Origins · 1 minElwin Bruno Christoffel and how this was discovered

Elwin Bruno Christoffel

1829–1900, Prussia and Strasbourg · differential geometry and the theory of invariants

Bernhard Riemann had, in his 1854 Habilitation lecture and his 1861 Paris prize essay, introduced the notion of the curvature of a manifold from its metric — but he never wrote a curvature tensor in components, and the algebraic structure of that object remained implicit. It was Elwin Bruno Christoffel who, in an 1869 paper in Crelle's Journal on the transformation of quadratic differential forms, introduced the three-index symbols now bearing his name and, from them, constructed the four-index curvature quantity — the Riemann–Christoffel tensor — explicitly working out its transformation law and its symmetry properties. Christoffel's problem was equivalence: when can two metrics be transformed into one another? The answer forced him to build precisely the tensor whose antisymmetries and cyclic identity are the subject of this article, because those symmetries are what make the count of independent invariants finite and well defined. His work, alongside Lipschitz's, gave Gregorio Ricci-Curbastro and Tullio Levi-Civita the raw material for the absolute differential calculus of the 1880s and 1890s — the tensor calculus that Einstein and Grossmann would later seize upon for general relativity. Christoffel spent his most productive years at the Polytechnic in Zürich and then at Strasbourg, and though he is remembered mostly through the symbols, the algebraic anatomy of curvature — antisymmetric in each pair, cyclic-closed, symmetric under pair-exchange — is very much his legacy.

History

1854

— Riemann's Habilitationsvortrag introduces the curvature of an n-dimensional manifold

1861

— Riemann's Paris prize essay expresses curvature through the metric, without index notation

1869

— Christoffel constructs the four-index Riemann–Christoffel tensor and derives its algebraic symmetries

1888

— Ricci-Curbastro develops the absolute differential calculus, systematising contractions of the tensor

1902

— Bianchi publishes the differential identities that complete the algebraic ones treated here

1918

— Weyl isolates the trace-free conformal part, splitting the 20 components into 10 Ricci and 10 Weyl

Going further

Texts. Misner, Thorne & Wheeler, Gravitation (1973), §3.7 and §13.5 · Wald, General Relativity (1984), §3.2 · Carroll, Spacetime and Geometry (2004), §3.6–3.7 · Schutz, A First Course in General Relativity, 2nd ed. (2009), §6.6 · Hartle, Gravity (2003), §21.2.

Provenance. The symmetry statements and derivations follow MTW §3.7/§13.5, Wald §3.2, and Carroll §3.6–3.7. A verifier should be able to: (1) re-derive the first-pair antisymmetry from $[\nabla_c,\nabla_d]g_{ab}=0$ using metric compatibility; (2) confirm pair-exchange by the four-instance Bianchi sum with its eight-term cancellation; (3) recompute $N=n^2(n^2-1)/12$ to get $0,1,6,20,50$ for $n=1,\dots,5$ ; (4) verify the 4D partition $20=10\,(\text{Ricci})+10\,(\text{Weyl})$ ; and (5) check the 2-sphere results $R=2/a^2$ , $R_{\theta\phi\theta\phi}=a^2\sin^2\theta$ , and $R_{abcd}R^{abcd}=4/a^4$ . All numbers are standard textbook values.

Prerequisites. Riemann Curvature Tensor · Curvature as the Commutator of Covariant Derivatives · Geometric Meaning of Curvature via Parallel Transport and Holonomy. Leads to. Independent Components of the Riemann Tensor · First (Algebraic) Bianchi Identity · Ricci Decomposition of the Riemann Tensor · Ricci Tensor · Weyl Conformal Tensor.

Ask the tutor: Ask me to walk through the eight-term cancellation in the pair-exchange proof line by line, or to count Weyl components in five or more dimensions.

Why does lowering the first index matter before I talk about symmetries?

Two of the four symmetries — first-pair antisymmetry and pair-exchange — are clean statements only for the fully covariant $R_{abcd}=g_{ae}R^{e}{}_{bcd}$ . In the mixed form $R^{a}{}_{bcd}$ the first index is upstairs and the second downstairs, so "antisymmetry in the first pair" is not even a well-typed operation — you would be swapping an upper with a lower index. The second-pair antisymmetry, by contrast, involves the two lower derivative indices and reads the same in either form. So the standard practice is to lower everything with the metric first; only then do all four symmetries appear as swaps among like-type indices, and only then does the "symmetric matrix on bivectors" picture make sense.

Is the first Bianchi identity the same thing as the second Bianchi identity?

No, and conflating them is a common slip. The first (algebraic) Bianchi identity, $R_{a[bcd]}=0$ , is an algebraic relation among the components of the curvature tensor at a single point; it is one of the symmetries counted in this article and requires only a torsion-free connection. The second (differential) Bianchi identity, $\nabla_{[e}R_{ab]cd}=0$ , involves derivatives of the curvature and relates its values at neighbouring points; it is what leads, on contraction, to the vanishing divergence of the Einstein tensor and hence to local energy-momentum conservation. Both are called "Bianchi," but the algebraic one constrains the tensor's index structure while the differential one constrains how it varies in space.

If pair-exchange is derived, why is it always listed as one of "the" symmetries?

Because in practice it is the most useful of the four, even though it carries no independent information. Pair-exchange is what lets you treat $R_{abcd}$ as a symmetric matrix on the six-dimensional bivector space in 4D, which is the single fact behind the whole component count and behind the symmetry of the Ricci tensor. Listing it alongside the genuinely independent symmetries is a matter of convenience — you reach for it constantly — but conceptually it should be remembered that it presupposes the two antisymmetries and the first Bianchi identity. Drop any of those (say, by allowing torsion) and pair-exchange is no longer guaranteed.

Do these symmetries depend on the field equations or on the matter content?

Not at all — that is precisely what makes them "algebraic." Every symmetry here follows from the definition of curvature and the properties of the Levi-Civita connection (metric compatibility and vanishing torsion). They hold at every point of every spacetime, in vacuum or inside a star, whether or not Einstein's equations are imposed. The field equations enter only afterward, as a separate constraint that sets the Ricci part of the curvature equal to the local stress-energy; they say nothing about the algebraic profile, which is already fixed. This clean separation — algebra first, dynamics second — is why the same 20-component count applies to every solution of general relativity.

Why does the Weyl tensor carry exactly 10 of the 20 components in 4D?

The 20 independent components of Riemann split into the trace pieces — everything you can reconstruct from the Ricci tensor — and the fully trace-free remainder, which is the Weyl tensor. In four dimensions the symmetric Ricci tensor has $\tfrac{4\cdot5}{2}=10$ components, so the trace part accounts for 10, leaving $20-10=10$ for Weyl. The Weyl tensor inherits all of Riemann's algebraic symmetries and, additionally, is trace-free on every pair of indices. Those 10 Weyl components are the part of curvature that survives in vacuum, encode tidal distortion, and propagate as gravitational waves — which is why the number 10 recurs throughout gravitational-wave physics.

What breaks if the connection has torsion?

Torsion is the antisymmetric part of the connection, and it directly damages two of the four symmetries. The first Bianchi identity is no longer $R_{a[bcd]}=0$ ; instead the cyclic sum equals terms built from the torsion tensor and its covariant derivative. And because the derivation of pair-exchange leaned on the first Bianchi identity, pair-exchange fails too. If the connection is also non-metric ( $\nabla g\neq0$ ), the first-pair antisymmetry breaks as well, leaving only the second-pair antisymmetry that comes from the bare commutator. In Einstein–Cartan theory, where torsion is sourced by intrinsic spin, one must therefore work with the reduced symmetry set and the component count $n^2(n^2-1)/12$ no longer applies.

How can I quickly check I have the index symmetries right in a calculation?

Use three fast sanity checks. First, any component with a repeated index inside a single antisymmetric pair must vanish: $R_{aacd}=0$ and $R_{abcc}=0$ (no sum), directly from the antisymmetries. Second, swapping the two indices in either pair flips the sign, and swapping the whole pairs leaves the value unchanged — so $R_{1212}=R_{2121}=-R_{2112}$ . Third, verify the cyclic identity on any three of the last indices, e.g. $R_{0123}+R_{0231}+R_{0312}=0$ ; in 4D this is the single Bianchi constraint. If a numerically computed curvature array violates any of these to within round-off, you have a bug in the Christoffel symbols or the contraction, not new physics.