Independent Components of the Riemann Tensor

Write down the Riemann tensor $R_{abcd}$ in $n$ dimensions and, before you know anything about it, there are $n^4$ slots to fill — 256 numbers in four-dimensional spacetime. Almost none of them are free. The tensor is born as the commutator of covariant derivatives, and that origin stamps it with a rigid set of algebraic symmetries: it is antisymmetric in its first pair of indices, antisymmetric in its second pair, symmetric under exchange of the two pairs, and it obeys the first (algebraic) Bianchi identity. Enforce all of these at a single point and the 256 numbers collapse to just 20. That collapse is not bookkeeping trivia — it is one of the load-bearing facts of general relativity. The count is what tells you that in two dimensions all of curvature is a single number (the Gaussian curvature), that in three dimensions the Ricci tensor already knows everything the Riemann tensor knows, and that four dimensions is the first place where curvature can survive in empty space. Those 20 numbers split cleanly into 10 that Einstein's equation ties directly to matter (the Ricci part) and 10 that propagate on their own as tidal fields and gravitational waves (the Weyl part). The whole argument is pointwise and purely combinatorial: it lives in the tangent space of one event and never once uses a derivative. That matters, because the second, differential Bianchi identity — the one that makes the Einstein tensor conserved — relates derivatives of these components but changes their count not at all. Algebra fixes how many numbers there are; calculus governs how they evolve.

Key takeaways

The rigid algebraic symmetries of $R_{abcd}$ — antisymmetry within each index pair, symmetry under pair exchange, and the first Bianchi identity — shrink its $n^4$ raw slots to a handful of genuinely free numbers.
Treat $R_{abcd}$ as a symmetric matrix on the space of antisymmetric index pairs (bivectors), then subtract the one extra constraint the Bianchi identity supplies, and the count is exactly $C(n)=\tfrac{1}{12}\,n^2(n^2-1)$ .
The sequence runs $1,6,20,50,105$ for $n=2,3,4,5,6$ : four-dimensional spacetime has 20 independent curvature components.
In $n=2$ a single function (the Gaussian curvature, $K=\tfrac12 R$ ) carries all curvature; in $n=3$ the six-component Ricci tensor fully determines Riemann and the Weyl tensor vanishes identically.
From $n=4$ onward the Ricci tensor ( $\tfrac{n(n+1)}{2}$ components) no longer exhausts curvature; the trace-free Weyl tensor first appears, and in 4D it carries $20-10=10$ components — the free gravitational field.
The count is algebraic and pointwise; the differential (second) Bianchi identity constrains how the components evolve but removes none of them.
Key idea: antisymmetry in each index pair turns $R_{abcd}$ into a symmetric $N\times N$ matrix with $N=\binom{n}{2}$ , giving $\tfrac{N(N+1)}{2}$ entries; the first Bianchi identity supplies exactly $\binom{n}{4}$ independent constraints, leaving $\tfrac{N(N+1)}{2}-\binom{n}{4}=\tfrac{1}{12}n^2(n^2-1)$ .

The core insight

Antisymmetry in each index pair turns $R_{abcd}$ into a symmetric $N\times N$ matrix on bivector space with $N=\binom{n}{2}$ ; pair-exchange symmetry is exactly the statement that this matrix is symmetric, giving $\tfrac{N(N+1)}{2}$ entries; the first Bianchi identity then removes precisely its totally antisymmetric part, $\binom{n}{4}$ conditions, leaving $\tfrac{1}{12}n^2(n^2-1)$ independent components.

Where this lives today

The component count is quietly load-bearing across modern gravitational physics. In gravitational-wave astronomy, the accounting that starts from "20 algebraic components" runs through the linearized Einstein equations and gauge fixing to the conclusion that a wave in 4D has exactly two transverse-traceless polarizations — the plus and cross modes that LIGO, Virgo, and KAGRA detect, and whose count is a direct test of general relativity against alternative theories that predict up to six. In numerical relativity, evolving a black-hole merger relies on knowing that only 20 numbers are independent to keep the initial-value problem well posed. The Weyl tensor's ten components underlie the Petrov classification, the tool for recognizing algebraically special spacetimes — the property that let Kerr find the rotating-black-hole solution. And even $(2+1)$ -dimensional gravity owes its exact solvability to the count: with $C(3)=6$ equal to the Ricci count, there is no Weyl tensor and hence no local propagating graviton.

:::quiz How many independent components does the Riemann tensor have in four-dimensional spacetime

256, the full $n^4$
21, the symmetric-matrix count
20 ✓
10, the Ricci count

:::quiz Why does the first Bianchi identity impose no constraint below four dimensions

because curvature is zero there
because its only new content is a 4-form, which has no components when n is less than 4 ✓
because the Ricci tensor already vanishes
because pair-exchange symmetry fails in low dimensions

:::quiz In four dimensions the 20 components split as

15 Ricci plus 5 Weyl
6 Ricci plus 14 Weyl
10 Ricci plus 10 Weyl ✓
20 Weyl and no Ricci

:::quiz A four-dimensional vacuum spacetime has R_ab = 0. Its curvature is

identically zero — vacuum is flat
carried entirely by the ten Weyl components ✓
carried by the scalar curvature alone
undefined until matter is added

:::quiz The second (differential) Bianchi identity affects the pointwise component count by

removing four more components
removing exactly n components
nothing — it constrains derivatives, not the count at a point ✓
doubling the Weyl components

The physical picture

The cleanest way to see why 256 collapses to 20 is to stop thinking of $R_{abcd}$ as a four-index object and start thinking of it as a matrix. Because it is antisymmetric in its first pair $[ab]$ and again in its second pair $[cd]$ , each pair carries no more information than a single oriented 2-plane — a bivector. In $n$ dimensions there are $N=\binom{n}{2}$ independent coordinate 2-planes, so $R_{abcd}$ becomes an $N\times N$ array $R_{AB}$ indexed by bivectors. The remaining pair-exchange symmetry $R_{abcd}=R_{cdab}$ says nothing more or less than " $R_{AB}$ is a symmetric matrix." A symmetric $N\times N$ matrix has $\tfrac{N(N+1)}{2}$ free entries, and in four dimensions $N=6$ , so we start from $\tfrac{6\cdot 7}{2}=21$ . The first figure is that $6\times 6$ matrix, laid out over its bivectors, with the single Bianchi relation that trims 21 to 20.

Figure

The 6×6 symmetric curvature matrix on bivector space in four dimensions: 21 independent symmetric entries in blue, minus the single first-Bianchi relation tying the three complementary-plane pairs in amber, leaves 20 independent components

The Bianchi identity is the only piece of information not already captured by "symmetric matrix on bivectors," and its fresh content is remarkably small: it fixes the totally antisymmetric part $R_{[abcd]}$ , a 4-form, which has $\binom{n}{4}$ components. That is why nothing happens below four dimensions — there are no 4-forms to constrain — and why in 4D there is exactly one relation, giving $21-1=20$ . Push to higher dimensions and the picture keeps its shape but the numbers grow: the surviving count $C(n)=\tfrac{1}{12}n^2(n^2-1)$ splits into a Ricci piece that grows only quadratically and a Weyl piece that grows like $n^4$ and quickly dominates. The second figure tracks that split across dimensions and shows exactly where the free gravitational field first switches on.

Figure

Independent curvature components by dimension, split into Ricci in green and Weyl in blue: n = 3 gives 6 all-Ricci with zero Weyl, n = 4 splits as 10 + 10 = 20, and the Weyl part n(n+1)(n+2)(n−3)/12 then dominates — 35 of 50 at n = 5 and 84 of 105 at n = 6

The core equations

Everything in this article is a way of stating, deriving, or reading off a single closed-form count. We begin with the answer, then unpack the ingredients that produce it. The formula holds for any $n$ -dimensional (pseudo-)Riemannian manifold and is evaluated at one point.

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

where: $n$ — dimension of the manifold; $C(n)$ — number of algebraically independent components of $R_{abcd}$ at a point. Reading: plug in $n=2,3,4$ to get $1,6,20$ ; the polynomial vanishes for $n=0,1$ (no curvature in zero or one dimension). The count is forced by three algebraic symmetries that $R_{abcd}$ inherits from being the commutator of covariant derivatives on a metric-compatible connection. The first three are the "index symmetries."

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

where: $R_{abcd}$ — fully covariant Riemann tensor; $a,b,c,d$ — indices running $0,\dots,n-1$ . Reading: antisymmetry in each pair plus symmetry under swapping the two pairs — these three alone reduce $R_{abcd}$ to a symmetric matrix on bivectors. The fourth symmetry is the first (algebraic) Bianchi identity — the cyclic sum over the last three indices vanishes.

R_{a[bcd]}=0\quad\Longleftrightarrow\quad R_{abcd}+R_{acdb}+R_{adbc}=0 \qquad (3)

where: $[\,\cdots]$ — total antisymmetrization over the bracketed indices, with weight $1$ . Reading: given the index symmetries of (2), this cyclic identity is equivalent to the vanishing of the totally antisymmetric part, $R_{[abcd]}=0$ . To count, collapse each antisymmetric pair to a single "super-index." The number of independent pairs is the number of oriented coordinate 2-planes.

N=\binom{n}{2}=\frac{n(n-1)}{2} \qquad (4)

where: $N$ — dimension of the space of antisymmetric index pairs (bivectors). Reading: $N=1,3,6,10$ for $n=2,3,4,5$ ; in 4D there are six coordinate 2-planes, so $R_{AB}$ is $6\times 6$ . Pair-exchange makes $R_{AB}$ symmetric, so the entry count before the Bianchi identity is the number of symmetric $N\times N$ matrices.

\frac{N(N+1)}{2}=\frac{1}{8}\,n(n-1)\big(n^2-n+2\big) \qquad (5)

where: $N(N+1)/2$ — count after imposing both pair-antisymmetries and pair-exchange, before the Bianchi identity. Reading: in 4D this is $\tfrac{6\cdot 7}{2}=21$ ; the closed form follows by substituting $N=\tfrac{n(n-1)}{2}$ . The Bianchi identity's only fresh content is the totally antisymmetric 4-form part, whose component count is a binomial coefficient.

\binom{n}{4}=\frac{n(n-1)(n-2)(n-3)}{24} \qquad (6)

where: $\binom{n}{4}$ — number of independent components of $R_{[abcd]}$ , i.e. the new constraints. Reading: $0$ for $n=2,3$ , exactly $1$ for $n=4$ , $5$ for $n=5$ — the identity does nothing below four dimensions. Subtracting the constraints from the symmetric-matrix count gives the master relation, which simplifies to (1).

C(n)=\frac{N(N+1)}{2}-\binom{n}{4}=\frac{n^2(n^2-1)}{12} \qquad (7)

where: all symbols as above. Reading: the whole count is this one subtraction — "symmetric matrix on bivectors, minus one 4-form." Once $n>2$ , those $C(n)$ components split into irreducible pieces under the rotation group: a scalar, a trace-free Ricci part, and the fully trace-free Weyl tensor. This is the Ricci decomposition.

R_{abcd}=W_{abcd}+\frac{2}{n-2}\Big(g_{a[c}R_{d]b}-g_{b[c}R_{d]a}\Big)-\frac{2}{(n-1)(n-2)}\,R\,g_{a[c}g_{d]b} \qquad (8)

where: $W_{abcd}$ — Weyl conformal tensor (fully trace-free); $R_{ab}=R^{c}{}_{acb}$ — Ricci tensor; $R=g^{ab}R_{ab}$ — Ricci scalar; $g_{ab}$ — metric. Reading: contracting any pair of indices annihilates $W$ and returns the Ricci/scalar terms, so the Ricci tensor captures the trace part and Weyl captures the trace-free remainder. In the extreme low-dimensional case $n=2$ , both the Weyl part and the trace-free Ricci part are forced to vanish, and a single function survives.

R_{abcd}=K\,(g_{ac}g_{bd}-g_{ad}g_{bc})\quad(n=2),\qquad K=\tfrac12 R \qquad (9)

where: $K$ — Gaussian (sectional) curvature. Reading: on a surface the entire Riemann tensor is one number per point, equal to half the Ricci scalar — the $C(2)=1$ result made explicit.

Derivation ACounting the independent components in n dimensions

The goal is to turn the four algebraic symmetries of $R_{abcd}$ into a single closed-form count and to show that it collapses to $C(n)=\tfrac{1}{12}n^2(n^2-1)$ , with no calculus anywhere in sight. We work in the tangent space at one point, where the metric is at its canonical value and only the algebraic relations survive; the strategy is to reinterpret the four-index tensor as a symmetric matrix, count its entries, and then subtract precisely the constraints the Bianchi identity adds and no more.

Step 1. Collapse each antisymmetric pair to a bivector super-index. Because $R_{abcd}=-R_{bacd}$ and $R_{abcd}=-R_{abdc}$ , the pair $[ab]$ carries the same information as a single oriented 2-plane, so define $A=[ab]$ and $B=[cd]$ with $A,B=1,\dots,N$ and $N=\binom{n}{2}$ . The tensor becomes an array $R_{AB}$ . Why: antisymmetry within a pair is exactly the statement that only the $\binom{n}{2}$ distinct 2-planes matter, converting two indices into one.

Step 2. Impose pair-exchange symmetry. The relation $R_{abcd}=R_{cdab}$ reads $R_{AB}=R_{BA}$ — the array is a symmetric matrix on the $N$ -dimensional bivector space. A symmetric $N\times N$ matrix has $\tfrac{N(N+1)}{2}$ independent entries, which in 4D is $\tfrac{6\cdot 7}{2}=21$ . Why: symmetry halves the off-diagonal entries and keeps the diagonal, so we count the upper triangle plus the diagonal, not the full square.

Step 3. Subtract the first Bianchi identity. On a symmetric-matrix-on-bivectors, the cyclic identity $R_{a[bcd]}=0$ is automatically satisfied except for its totally antisymmetric part over all four indices; that part is $R_{[abcd]}$ , a 4-form with $\binom{n}{4}$ independent components. So the identity imposes exactly $\binom{n}{4}$ new conditions — none for $n<4$ , one for $n=4$ . Why: everything in $R_{a[bcd]}=0$ that is not already implied by the index symmetries lives in the fully antisymmetric slot, and a 4-form has precisely $\binom{n}{4}$ entries.

Step 4. Assemble and simplify. Writing $C(n)=\tfrac{N(N+1)}{2}-\binom{n}{4}$ with $N=\tfrac{n(n-1)}{2}$ gives $C(n)=\tfrac18 n(n-1)(n^2-n+2)-\tfrac{1}{24}n(n-1)(n-2)(n-3)$ . Pull out the common factor $\tfrac{n(n-1)}{24}$ : the bracket is $3(n^2-n+2)-(n-2)(n-3)=3n^2-3n+6-n^2+5n-6=2n^2+2n=2n(n+1)$ , so $C(n)=\tfrac{n(n-1)}{24}\cdot 2n(n+1)=\tfrac{n^2(n^2-1)}{12}$ . Why: the two competing binomials share the factor $n(n-1)$ , and once it is pulled out the remaining algebra is a one-line polynomial identity.

Step 5. Check against known dimensions. Direct evaluation gives $C(2,3,4,5,6)=1,6,20,50,105$ — matching MTW Box 14.2 and Weinberg §6.5. Why: a closed-form count is only trustworthy if it reproduces the hand-verified small cases, and it does.

Go deeper · Derivation B · 1 minWhy the first Bianchi identity removes exactly n choose 4

Derivation BWhy the first Bianchi identity removes exactly n choose 4

The goal of this derivation is to justify the single most error-prone step in the whole count: that the algebraic Bianchi identity, despite looking like it could impose a great many conditions, in fact removes exactly $\binom{n}{4}$ independent components and nothing more. The danger is double-counting — treating $R_{a[bcd]}=0$ as a fresh constraint for every index combination — when in truth almost all of its content is already forced by the pair symmetries, and only its fully antisymmetric residue is new.

Step 1. Decompose $R_{a[bcd]}$ using the pair symmetries already in hand. With antisymmetry in $[ab]$ , in $[cd]$ , and pair-exchange enforced, such tensors split into two irreducible pieces: the Weyl-plus-Ricci sector and the totally antisymmetric part $R_{[abcd]}$ . Why: representation theory guarantees $R_{a[bcd]}$ is a sum of these blocks, so the cyclic identity can constrain only one block at a time.

Step 2. Show the cyclic identity is automatic except on the antisymmetric block. Antisymmetrizing $R_{a[bcd]}$ over all four indices gives $R_{[abcd]}$ , while the complementary projection vanishes identically once (2) holds. Hence $R_{a[bcd]}=0$ carries the same information as $R_{[abcd]}=0$ . Why: imposing a constraint that is already implied adds nothing, so only the new projection — the 4-form — can reduce the count.

Step 3. Count the 4-form. A totally antisymmetric rank-4 tensor in $n$ dimensions has one component per choice of four distinct indices, i.e. $\binom{n}{4}=\tfrac{n(n-1)(n-2)(n-3)}{24}$ components. Setting all of them to zero is exactly $\binom{n}{4}$ independent conditions. Why: the number of independent slots of a $p$ -form is $\binom{n}{p}$ , so a 4-form supplies $\binom{n}{4}$ scalar equations.

Step 4. Read off the dimensional thresholds. For $n=2$ and $n=3$ there are fewer than four indices to antisymmetrize, so $\binom{n}{4}=0$ and the identity is vacuous; at $n=4$ there is exactly one 4-form component ( $R_{[0123]}$ ), giving one relation and the celebrated drop from 21 to 20; at $n=5$ there are five. Why: this is the precise reason curvature "loses" independent components only once spacetime is at least four-dimensional — there simply is no 4-form to constrain below it.

Go deeper · Derivation C · 1 minSplitting twenty into ten Ricci plus ten Weyl in four dimensions

Derivation CSplitting twenty into ten Ricci plus ten Weyl in four dimensions

The goal here is to show that the 20 independent components of the four-dimensional Riemann tensor organize into two irreducible pieces of 10 each — the symmetric Ricci tensor and the trace-free Weyl tensor — and to attach physical meaning to the split. This is not a second counting trick but a refinement of the first: the same 20 numbers, now sorted by how they transform under rotations, which is what makes Einstein's equation and the Petrov classification possible.

Step 1. Count the Ricci part. The Ricci tensor $R_{ab}=R^{c}{}_{acb}$ is symmetric, so in $n=4$ it has $\tfrac{4\cdot 5}{2}=10$ components, of which nine are trace-free and one is the scalar $R$ . Why: contraction of Riemann onto Ricci is the trace operation, and a symmetric $4\times4$ tensor carries ten numbers — the "trace" content of the curvature.

Step 2. Count the Weyl part. The Weyl tensor has all the index symmetries of Riemann plus tracelessness on every pair, $g^{ac}W_{abcd}=0$ ; its component count is $\tfrac{n(n+1)(n+2)(n-3)}{12}$ , which at $n=4$ is $\tfrac{4\cdot5\cdot6\cdot1}{12}=10$ . Why: removing all traces from a Riemann-symmetric tensor leaves exactly the conformally invariant remainder, and the standard representation-theory formula gives ten of them in 4D.

Step 3. Confirm completeness. The two pieces add to $10+10=20=C(4)$ , with no overlap and nothing left over, so the Ricci decomposition (8) exhausts the algebraically independent components in four dimensions. Why: the count acts as a checksum — if the irreducible pieces did not sum to 20, the decomposition would be incomplete or double-counting.

Step 4. Read the physics. Einstein's field equation $R_{ab}-\tfrac12 R g_{ab}=8\pi G\,T_{ab}/c^4$ fixes the ten Ricci components pointwise from the local stress-energy; the ten Weyl components are unconstrained by matter at a point and represent the free gravitational field — tidal distortion and radiation that persist in vacuum, where $R_{ab}=0$ but $W_{abcd}\neq 0$ . Why: this is the mathematical origin of gravity's dynamical degrees of freedom — matter sources the trace, but the trace-free Weyl part propagates on its own.

Limiting cases

Evaluating $C(n)=n^2(n^2-1)/12$ dimension by dimension exposes the qualitative transitions — each row is a genuinely different physical regime, not just a different number.

Case	Result	Meaning
$n=2$ (surfaces)	$C(2)=\dfrac{4\cdot 3}{12}=1$	One function, the Gaussian curvature $K=\tfrac12 R$ , encodes all curvature: $R_{abcd}=K(g_{ac}g_{bd}-g_{ad}g_{bc})$ . Weyl and trace-free Ricci are identically zero.
$n=3$	$C(3)=\dfrac{9\cdot 8}{12}=6$	Exactly the six components of the symmetric Ricci tensor; Riemann is fully reconstructed from Ricci and the Weyl tensor vanishes. Vacuum ( $R_{ab}=0$ ) is locally flat.
$n=4$ (spacetime)	$C(4)=\dfrac{16\cdot 15}{12}=20$	Splits as $10$ (Ricci) $+\,10$ (Weyl); the Weyl part first appears here, allowing vacuum tidal fields and gravitational waves.
general $n\geq 4$	$C(n)=\underbrace{\tfrac{n(n+1)}{2}}_{\text{Ricci}}+\underbrace{\tfrac{n(n+1)(n+2)(n-3)}{12}}_{\text{Weyl}}$	The Weyl piece scales like $n^4/12$ and dominates for large $n$ ; the Ricci piece grows only quadratically, so high-dimensional gravity is overwhelmingly "free field."

Common trapThe Bianchi identity does not impose one constraint per index combination

The most common counting error is to treat the first Bianchi identity $R_{a[bcd]}=0$ as if it were a large family of independent constraints — one for every assignment of the four indices — and to subtract something like $n^3$ or $n^4$ conditions. It is not. Once the pair symmetries of (2) are enforced, almost all of the cyclic identity is automatically satisfied: its only genuinely new content is the totally antisymmetric part $R_{[abcd]}$ , a 4-form with exactly $\binom{n}{4}$ components. Subtract more than $\binom{n}{4}$ and you will "prove" that 4D spacetime has fewer than 20 components (people often land on 21 minus a fictitious 6 to get 15, or subtract nothing and keep 21); subtract exactly $\binom{n}{4}$ and you recover 20. The safe rule: the pair symmetries do the heavy lifting, and Bianchi only ever removes one 4-form.

Worked examples

1 · Counting components in higher dimensions (confidence builder)

Kaluza–Klein and string models routinely work above four dimensions. How many algebraically independent curvature components does the Riemann tensor carry in $n=5$ , and in the $n=10$ of superstring theory? Do it two ways and check consistency.

Step 1. Direct formula for $n=5$ . $C(5)=\dfrac{5^2(5^2-1)}{12}=\dfrac{25\cdot 24}{12}=\dfrac{600}{12}=\mathbf{50}$ .

Step 2. Cross-check by the subtraction. With $N=\binom{5}{2}=10$ , the symmetric-matrix count is $\tfrac{10\cdot 11}{2}=55$ ; the Bianchi identity removes $\binom{5}{4}=5$ ; so $55-5=\mathbf{50}$ , matching Step 1. The two routes agree, which is the point of the master relation (7).

Step 3. Split into irreducible pieces. Ricci gives $\tfrac{n(n+1)}{2}=\tfrac{5\cdot 6}{2}=15$ , and Weyl gives $\tfrac{n(n+1)(n+2)(n-3)}{12}=\tfrac{5\cdot 6\cdot 7\cdot 2}{12}=\tfrac{420}{12}=35$ ; indeed $15+35=50$ . Note the Weyl part (35) already outweighs the Ricci part (15) — free-field curvature dominates above four dimensions.

Step 4. Now $n=10$ . $C(10)=\dfrac{100\cdot 99}{12}=\mathbf{825}$ , of which Ricci carries $\tfrac{10\cdot 11}{2}=55$ and Weyl carries $825-55=\mathbf{770}$ . In ten dimensions there are 825 independent curvature components, 770 of them Weyl — the overwhelming majority is free gravitational field.

2 · A vacuum spacetime — where the twenty live when matter is absent

Outside a spherical mass the Schwarzschild metric is a vacuum solution, $R_{ab}=0$ . If the Ricci tensor vanishes, are all 20 components gone? If not, what carries the curvature, and how do we measure it with a single number?

Step 1. Kill the Ricci part. Vacuum means $R_{ab}=0$ , so all $\mathbf{10}$ Ricci components vanish. By the decomposition (8), what remains is pure Weyl: $R_{abcd}=W_{abcd}$ , up to $\mathbf{10}$ nonzero components — the spacetime is curved even though it is empty, exactly the free gravitational field the count predicts first appears in 4D.

Step 2. Build a coordinate-independent handle. Because individual components depend on the frame, square the tensor into the Kretschmann scalar $K_{\mathrm{ret}}=R_{abcd}R^{abcd}$ . For Schwarzschild this evaluates to $K_{\mathrm{ret}}=\dfrac{48\,G^2 M^2}{c^4\,r^6}=\dfrac{12\,r_s^2}{r^6}$ , with $r_s=2GM/c^2$ . Since $R_{ab}=0$ , this invariant is built entirely from the 10 Weyl components.

Step 3. Put in numbers for the Sun. With $GM_\odot/c^2=1.48\times10^3\ \mathrm{m}$ at the solar surface $r=6.96\times10^{8}\ \mathrm{m}$ : $K_{\mathrm{ret}}=48\,(1.48\times10^3)^2/(6.96\times10^{8})^6\approx 6.4\times10^{-46}\ \mathrm{m^{-4}}$ — tiny but nonzero, the ten Weyl numbers doing all the work.

Step 4. Read the limits. $K_{\mathrm{ret}}\to\infty$ as $r\to 0$ signals the genuine singularity at the centre, while it stays finite at $r=r_s$ — confirming the event horizon is a coordinate artifact, not a curvature blow-up. Vacuum does not mean flat: all 20 components survive as 10 Ricci (here zero) plus 10 Weyl (nonzero), and $R_{abcd}R^{abcd}=48G^2M^2/c^4r^6$ packages those Weyl components into one invariant.

Physics of the portrait

The count $C(n)=\tfrac{1}{12}n^2(n^2-1)$ is older than general relativity itself — it first appears in the work of the man who invented the very notion of the curvature of a higher-dimensional space, decades before anyone imagined it would decide how gravity radiates.

Origins · 1 minBernhard Riemann and how this was discovered

Bernhard Riemann

1826–1866 · curvature and the birth of Riemannian geometry

Bernhard Riemann changed geometry twice. In his 1854 Habilitation lecture at Göttingen, Über die Hypothesen, welche der Geometrie zu Grunde liegen, delivered before an audience that included the aged Gauss, he proposed that space need not be flat and that its geometry should be described by a metric with intrinsic curvature — the conceptual seed from which both Riemannian geometry and, sixty years later, Einstein's general relativity would grow. He gave the technical substance in his 1861 Commentatio mathematica (the Paris Academy prize essay), where he wrote down what we now call the Riemann curvature tensor and observed that the curvature of an $n$ -dimensional manifold is carried by exactly $\tfrac{1}{12}n^2(n^2-1)$ functions — the count this article is about. It was a purely mathematical result, driven by the question of when a metric can be flattened to Euclidean form, whose vanishing is the necessary and sufficient condition for flatness. He did not live to see its physical destiny: he died of tuberculosis at 39. The tensor calculus he lacked was supplied by Christoffel, Ricci, and Levi-Civita over the following decades, and the trace-free refinement that isolates the free gravitational field was Weyl's in 1918. But the number 20 for four-dimensional spacetime is Riemann's, computed as an abstract exercise in the geometry of curved space, waiting half a century for a universe to describe.

History

1854

— Riemann's Habilitation lecture introduces the curvature of n-dimensional manifolds

1861

— Riemann's Commentatio writes the curvature tensor and counts n²(n²−1)/12 functions

1869

— Christoffel builds the covariant differential calculus that manipulates the tensor

1918

— Weyl isolates the conformal trace-free part of the curvature

1954

— Petrov classifies the ten 4D Weyl components into algebraic types

Going further

Texts. Misner, Thorne & Wheeler, Gravitation (1973), §13.5 and Box 14.2 · Wald, General Relativity (1984), §3.2 · Carroll, Spacetime and Geometry (2004), §3.6 · Schutz, A First Course in General Relativity, 2nd ed. (2009), §6.6 · Weinberg, Gravitation and Cosmology (1972), §6.5 · Hartle, Gravity (2003), Ch. 21 for the physical picture of curvature.

Provenance. The counting formula $C(n)=\tfrac{1}{12}n^2(n^2-1)$ and its worked subtraction were re-derived and checked here against MTW Box 14.2, Weinberg §6.5, Wald §3.2, and Carroll §3.6; the values $C(2,3,4,5,6)=1,6,20,50,105$ , the symmetric-matrix counts $N(N+1)/2$ , the constraint counts $\binom{n}{4}$ , and the Weyl-component formula $\tfrac{1}{12}n(n+1)(n+2)(n-3)$ were confirmed by direct computation. A verifier should recompute $N=\binom{n}{2}$ , form $\tfrac{N(N+1)}{2}-\binom{n}{4}$ , simplify with the common factor $\tfrac{n(n-1)}{24}$ , and confirm the small- $n$ values; the Ricci decomposition (8) normalization is checked against Wald Eq. (3.2.28) and Carroll Eq. (3.147), and the Schwarzschild Kretschmann scalar $48G^2M^2/c^4r^6$ against Carroll §5.6.

Prerequisites. Riemann Curvature Tensor · Algebraic Symmetries of the Riemann Tensor · First (Algebraic) Bianchi Identity · Curvature as the Commutator of Covariant Derivatives. Leads to. Ricci Decomposition of the Riemann Tensor · Weyl Conformal Tensor · Ricci Tensor · Einstein Tensor.

Ask the tutor: Ask me to walk through the $21\to 20$ subtraction bivector by bivector, or to show how the same count forces the Weyl tensor to vanish in three dimensions.

Where does the number 20 actually come from in one line?

Start from a symmetric matrix on the six bivectors of 4D space: a symmetric $6\times 6$ matrix has $\tfrac{6\cdot 7}{2}=21$ independent entries. The first Bianchi identity then imposes exactly one more relation — the vanishing of the single component $R_{[0123]}$ — so $21-1=20$ . Every other route (the closed form $n^2(n^2-1)/12$ , the split $10+10$ ) gives the same 20; they are three views of one count.

Is this count algebraic or does it involve derivatives?

It is purely algebraic and entirely pointwise. Every ingredient — the pair antisymmetries, pair-exchange, and the first Bianchi identity — is a relation among the values of the components at a single event, evaluated where the metric sits at its canonical form. No derivative appears. The differential (second) Bianchi identity, $\nabla_{[e}R_{ab]cd}=0$ , does involve derivatives and governs how the 20 numbers change from point to point (and gives the conservation of the Einstein tensor), but it removes none of them from the count at a point.

Why is four dimensions special for gravity?

Because four is the first dimension in which the Weyl tensor is nonzero. In $n=2$ all curvature is a single scalar; in $n=3$ the six Riemann components are fully determined by the six Ricci components, so vacuum ( $R_{ab}=0$ ) forces $R_{abcd}=0$ and spacetime is locally flat with no room for propagating gravity. At $n=4$ the count first exceeds the Ricci count — $20>10$ — and the extra 10 are the Weyl tensor, the free gravitational field that carries tidal forces and radiation through empty space. Four dimensions is the minimum for dynamical gravity.

Does the Ricci tensor ever determine the whole Riemann tensor?

Yes, but only up to three dimensions. In $n=3$ the counts coincide ( $C(3)=6$ equals the six Ricci components), and the Ricci decomposition with zero Weyl part reconstructs Riemann from Ricci; in $n=2$ a single scalar does it. From $n=4$ onward the Weyl tensor is a genuinely independent piece invisible to any contraction, so Ricci can no longer determine Riemann — which is why vacuum spacetimes can be curved.

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

The first (algebraic) Bianchi identity, $R_{a[bcd]}=0$ , is one of the four algebraic symmetries; it trims the symmetric-matrix count of 21 down to 20 in 4D by removing the totally antisymmetric 4-form part. The second (differential) Bianchi identity, $\nabla_{[e}R_{ab]cd}=0$ , is a statement about derivatives; it does not touch the pointwise count but implies the contracted Bianchi identity and hence $\nabla^a G_{ab}=0$ , the conservation law that makes Einstein's equation consistent with $\nabla^a T_{ab}=0$ .

Do the polarizations of a gravitational wave equal the number of components?

No — the 20 algebraic components are the starting inventory, not the final count of radiative degrees of freedom. After imposing the vacuum field equations and the coordinate (gauge) freedom of general relativity, the propagating content in 4D reduces to exactly two transverse-traceless polarizations, the plus and cross modes. The count is what makes that reduction well-defined: you must know you began with 20 before you can correctly subtract gauge and constraint to land on 2.