Ricci Decomposition of the Riemann Tensor

Hand someone the Riemann tensor $R_{abcd}$ and they see one intimidating object with twenty independent numbers in four dimensions. The Ricci decomposition is the statement that those twenty numbers are not an unstructured heap: they sort themselves, cleanly and uniquely, into three piles that never mix. Contract $R_{abcd}$ once and you get the Ricci tensor; contract again and you get the Ricci scalar. Those two traces carry ten of the twenty components — and Einstein's equation nails them, pointwise, to the matter sitting at that event. Subtract every possible trace and what is left over is the Weyl tensor, ten components that no contraction can touch: the free gravitational field, the part that survives in empty space as tidal stretching and as radiation streaming out to infinity. The reason the piles never mix is representation theory. As a tensor with Riemann symmetries, $R_{abcd}$ lives in a reducible representation of the orthogonal group, and that representation shatters into three mutually orthogonal irreducible subspaces — a one-dimensional scalar piece, a traceless-Ricci piece, and the totally traceless Weyl piece. Taking traces is the projection onto the lower subspaces; whatever is annihilated by every trace is the genuinely irreducible core. The algebraic glue that reassembles the trace pieces into a curvature-shaped object is the Kulkarni–Nomizu product ⊚, which feeds two symmetric tensors in and returns something with exactly the antisymmetries and Bianchi symmetry of Riemann. Package the trace data as the Schouten tensor $P_{ab}$ and the whole decomposition collapses to a single line, $R_{abcd}=C_{abcd}+(P⊚g)_{abcd}$ — Weyl plus one Kulkarni–Nomizu product. Everything downstream, from why gravity propagates through vacuum to the conformal invariance that classifies spacetimes, is written in this split.

Key takeaways

The Riemann tensor of an $n$ -dimensional manifold splits into three mutually orthogonal, irreducible pieces: a scalar part built from the Ricci scalar $R$ , a part built from the traceless Ricci tensor, and the totally traceless Weyl tensor $C_{abcd}$ .
Taking traces is the projection: the Ricci tensor and scalar are the trace content, and whatever survives every contraction is the Weyl tensor — the part invisible to any trace.
The two trace pieces are packaged by the Kulkarni–Nomizu product of the metric with the Schouten tensor, collapsing the split to one line, $R_{abcd}=C_{abcd}+(P⊚g)_{abcd}$ .
The three subspaces are orthogonal, so the squared norm splits with no cross terms: the Kretschmann scalar separates cleanly into a Weyl invariant plus Ricci and scalar terms.
All three pieces are present only for $n\ge 4$ ; in $n=3$ the Weyl tensor vanishes identically and Ricci determines Riemann, and in $n=2$ only the scalar piece survives.
Physically the split divides curvature sourced locally by matter (Ricci, fixed by Einstein's equation) from the free tidal and radiative field (Weyl, constrained only by the Bianchi identities).
Key idea: the Riemann tensor is the orthogonal sum of three irreducible pieces under the orthogonal group — a scalar (Ricci-scalar) part, a traceless-Ricci part, and the totally traceless Weyl tensor — and the first two are packed by the Kulkarni–Nomizu product of the Schouten tensor with the metric, $R_{abcd}=C_{abcd}+(P⊚g)_{abcd}$ .

The core insight

The Riemann tensor is the orthogonal sum of three irreducible pieces under the orthogonal group: a scalar (Ricci-scalar) part, a traceless-Ricci part, and the totally traceless Weyl tensor; the first two are packaged by the Kulkarni–Nomizu product of the Schouten tensor with the metric, so that $R_{abcd}=C_{abcd}+(P⊚g)_{abcd}$ .

Where this lives today

The decomposition is quietly load-bearing across gravitational physics. In gravitational-wave astronomy the radiation reaching LIGO, Virgo, and KAGRA is Weyl curvature: outside the source $R_{ab}=0$ , so the strain a detector measures is a projection of $C_{abcd}$ transported along null rays, and the Newman–Penrose scalar $\Psi_4$ that numerical-relativity codes extract from a black-hole merger is a specific Weyl component. The Petrov classification — the sorting of the ten Weyl components into algebraic types I through N — is how relativists recognize algebraically special spacetimes, and it was the structural clue that led Kerr to the rotating-black-hole metric (type D). In cosmology the conformal invariance of the Weyl tensor makes it the natural measure of inhomogeneity: exact Friedmann–Lemaître–Robertson–Walker models are conformally flat with $C_{abcd}=0$ , so the growth of the Weyl tensor tracks structure formation, and Penrose's Weyl curvature hypothesis proposes that it was near zero at the Big Bang, encoding the arrow of time. The orthogonal norm split feeds quantum field theory in curved spacetime, where the conformal anomaly is written in terms of $C_{abcd}C^{abcd}$ and the Gauss–Bonnet density $R_{abcd}R^{abcd}-4R_{ab}R^{ab}+R^2$ ; and conformal (Weyl) gravity, a live alternative theory, takes $C_{abcd}C^{abcd}$ as its entire action.

The physical picture

Picture the space of all tensors that share the Riemann symmetries — antisymmetric in the first pair of indices, antisymmetric in the second, symmetric under swapping the pairs, and obeying the first Bianchi identity. This is a vector space, and in four dimensions it is twenty-dimensional. The orthogonal group acts on it, and under that action the space does not stay whole: it breaks into three invariant subspaces that no rotation can mix. One is a single dimension, a pure multiple of a constant-curvature template built from the metric alone. The next carries the traceless part of the Ricci tensor, nine dimensions in 4D. The last is the Weyl tensor, ten dimensions, defined by the property that every trace of it is zero. The first figure lays out this shattering with the actual component counts.

Figure

The 20 independent Riemann components in four dimensions split into three orthogonal irreducible pieces — 1 scalar (Ricci scalar R), 9 traceless-Ricci (the tensor S_ab), and 10 Weyl (C_abcd) — adding to 1 + 9 + 10 = 20, with taking traces acting as the projection onto the lower two boxes

Why should matter care about this particular split? Because Einstein's field equation is an algebraic statement about the trace pieces alone: $G_{ab}=8\pi T_{ab}$ ties the Ricci tensor, and hence the scalar and traceless-Ricci pieces, directly to the local stress-energy. The Weyl piece is left untouched at a point — it is constrained only by the differential Bianchi identities, which is exactly why it can carry information away from its source. In vacuum the trace pieces switch off entirely and the curvature is pure Weyl; in a maximally symmetric universe the Weyl piece switches off and the curvature is pure scalar. The second figure shows these two extremes side by side with their actual curvature invariants, the Kretschmann scalar $R_{abcd}R^{abcd}$ split cleanly into its irreducible contributions.

Figure

The orthogonal norm split of the Kretschmann scalar for two exact spacetimes: Schwarzschild vacuum is pure Weyl with R_abcd R^abcd = 9.2×10⁻⁴⁶ m⁻⁴ at the solar surface, while de Sitter space is pure scalar with R_abcd R^abcd = 3.2×10⁻¹⁰⁴ m⁻⁴ — in each case only one of the three orthogonal boxes is nonzero

The core equations

The whole decomposition is one master identity, and everything else in this section either defines a symbol in it or reads off a consequence. The master form writes the Riemann tensor as its irreducible core plus a single Kulkarni–Nomizu product carrying all the trace information. It holds on any $n$ -dimensional pseudo-Riemannian manifold with $n\ge 3$ .

R_{abcd}=C_{abcd}+(P\owedge g)_{abcd} \qquad (1)

where: $R_{abcd}$ — Riemann curvature tensor; $C_{abcd}$ — Weyl conformal tensor, traceless on every pair; $P_{ab}$ — Schouten tensor; $g_{ab}$ — metric; $\owedge$ — Kulkarni–Nomizu product. Reading: two objects on the right, the traceless core $C$ and one product $(P\owedge g)$ that repackages the Ricci and scalar data — nothing else is needed. The product $\owedge$ is the machine that turns two symmetric $(0,2)$ -tensors into a tensor with Riemann symmetries. It is defined by four terms, arranged so that antisymmetry in each pair and pair-exchange symmetry come out automatically.

(A\owedge B)_{abcd}=A_{ac}B_{bd}+A_{bd}B_{ac}-A_{ad}B_{bc}-A_{bc}B_{ad} \qquad (2)

where: $A_{ab},\,B_{ab}$ — any symmetric $(0,2)$ -tensors. Reading: the four terms are the two "diagonal" products minus the two "off-diagonal" products; setting $A=B=g$ gives $(g\owedge g)_{abcd}=2(g_{ac}g_{bd}-g_{ad}g_{bc})$ , the constant-curvature template. The Schouten tensor is the specific combination of the Ricci tensor and Ricci scalar whose Kulkarni–Nomizu product with the metric reproduces the trace pieces exactly. It is the natural variable of conformal geometry.

P_{ab}=\frac{1}{n-2}\left(R_{ab}-\frac{R}{2(n-1)}\,g_{ab}\right) \qquad (3)

where: $R_{ab}$ — Ricci tensor; $R$ — Ricci scalar; $n$ — dimension, $n\ge 3$ . Reading: its trace is $g^{ab}P_{ab}=R/[2(n-1)]$ , and both the $1/(n-2)$ and $1/[2(n-1)]$ factors are fixed uniquely by demanding that $C$ be totally traceless. Solving (1) for the core gives the Weyl tensor written out explicitly in terms of Riemann, Ricci, and the scalar — the form you actually compute with.

C_{abcd}=R_{abcd}-\frac{1}{n-2}\big(g_{ac}R_{bd}-g_{ad}R_{bc}-g_{bc}R_{ad}+g_{bd}R_{ac}\big)+\frac{R}{(n-1)(n-2)}\big(g_{ac}g_{bd}-g_{ad}g_{bc}\big) \qquad (4)

where: all symbols as above. Reading: the middle bracket is $(\mathrm{Ric}\owedge g)_{abcd}$ and the last is $\tfrac12(g\owedge g)_{abcd}$ , so $C=R-\tfrac{1}{n-2}(\mathrm{Ric}\owedge g)+\tfrac{R}{2(n-1)(n-2)}(g\owedge g)$ — the same content as (1) with $P$ unpacked. The defining property of $C$ is that it is annihilated by every contraction: no trace of the Weyl tensor is anything but zero. This is what makes it the irreducible remainder.

g^{ac}C_{abcd}=0,\qquad C^{a}{}_{bad}=0 \qquad (5)

where: $g^{ac}$ — inverse metric used to contract. Reading: because all traces vanish, the Ricci tensor and scalar are carried entirely by the other two pieces — the Weyl tensor is completely invisible to $R_{ab}$ and $R$ . The trace pieces themselves are worth naming separately. The scalar piece is a pure multiple of the constant-curvature template and is the entire curvature of a maximally symmetric space.

R^{\text{(scalar)}}_{abcd}=\frac{R}{2n(n-1)}\,(g\owedge g)_{abcd}=\frac{R}{n(n-1)}\big(g_{ac}g_{bd}-g_{ad}g_{bc}\big) \qquad (6)

where: $R$ — Ricci scalar. Reading: comparing with $R_{abcd}=K(g_{ac}g_{bd}-g_{ad}g_{bc})$ identifies the sectional curvature $K=R/[n(n-1)]$ of a space of constant curvature. The middle piece is the traceless Ricci tensor $S_{ab}$ glued to the metric. It vanishes precisely for Einstein spaces, where $R_{ab}\propto g_{ab}$ .

R^{\text{(Ricci)}}_{abcd}=\frac{1}{n-2}\,\big(S\owedge g\big)_{abcd},\qquad S_{ab}=R_{ab}-\frac{R}{n}\,g_{ab} \qquad (7)

where: $S_{ab}$ — traceless Ricci tensor, $g^{ab}S_{ab}=0$ . Reading: $R_{abcd}=R^{\text{(scalar)}}_{abcd}+R^{\text{(Ricci)}}_{abcd}+C_{abcd}$ , the three-way form of the one-line master identity. Because the three pieces sit in orthogonal subspaces, the squared norm of the curvature splits with no cross terms — the Kretschmann scalar decomposes into a Weyl invariant plus Ricci and scalar contributions.

R_{abcd}R^{abcd}=C_{abcd}C^{abcd}+\frac{4}{n-2}\,R_{ab}R^{ab}-\frac{2}{(n-1)(n-2)}\,R^2 \qquad (8)

where: $R_{abcd}R^{abcd}$ — Kretschmann scalar; $C_{abcd}C^{abcd}$ — Weyl (Petrov) invariant. Reading: in $n=4$ the coefficients are $2$ and $\tfrac13$ , giving $R_{abcd}R^{abcd}=C_{abcd}C^{abcd}+2R_{ab}R^{ab}-\tfrac13 R^2$ — the identity used to isolate the Weyl invariant from tabulated curvature scalars.

Derivation AForcing all traces to vanish to isolate the Weyl tensor

The goal is to build, from scratch, the unique tensor that has all the Riemann symmetries and is annihilated by every possible contraction, starting only from $R_{abcd}$ itself. The strategy is to subtract from Riemann the most general trace-carrying correction that the Kulkarni–Nomizu product allows — one term linear in the Ricci tensor and one term linear in the scalar — leaving two undetermined coefficients, and then to fix those coefficients by the single demand that the result be traceless. That demand turns out to determine both coefficients uniquely, which is why the Weyl tensor exists and is unique.

Step 1. Write the ansatz $C_{abcd}=R_{abcd}+\alpha\,(\mathrm{Ric}\owedge g)_{abcd}+\beta\,R\,(g\owedge g)_{abcd}$ , using the only two Riemann-symmetric tensors that are linear in $R_{ab}$ and $g$ , or in $R$ and $g$ . Why: the Kulkarni–Nomizu product guarantees each correction term already lives in the Riemann symmetry class, so $C$ automatically inherits all four algebraic symmetries no matter what $\alpha,\beta$ are.

Step 2. Compute the trace of the Ricci correction. Contracting with $g^{ac}$ and using $g^{ac}g_{ac}=n$ , $g^{ac}R_{ac}=R$ , and $g^{ac}g_{ad}=\delta^c{}_d$ , the four terms of $(\mathrm{Ric}\owedge g)_{abcd}$ give $nR_{bd}+R\,g_{bd}-R_{bd}-R_{bd}=(n-2)R_{bd}+R\,g_{bd}$ . Why: this is the only place the dimension $n$ enters, and it is what will force the $1/(n-2)$ factor into the answer.

Step 3. Compute the trace of the scalar correction. Since $(g\owedge g)_{abcd}=2(g_{ac}g_{bd}-g_{ad}g_{bc})$ , contracting gives $g^{ac}(g\owedge g)_{abcd}=2(n-1)g_{bd}$ . Why: the constant-curvature template traces to a pure multiple of the metric, so it can only cancel the $g_{bd}$ part of the Ricci trace, not the $R_{bd}$ part.

Step 4. Trace the whole ansatz using $g^{ac}R_{abcd}=R_{bd}$ (the definition of the Ricci tensor). The result is $g^{ac}C_{abcd}=R_{bd}+\alpha[(n-2)R_{bd}+R\,g_{bd}]+2\beta(n-1)R\,g_{bd}$ . Why: setting this to zero is the tracelessness condition, and it separates into an $R_{bd}$ part and a $g_{bd}$ part that must each vanish independently.

Step 5. Solve the two conditions. The $R_{bd}$ coefficient gives $1+\alpha(n-2)=0$ , so $\alpha=-\tfrac{1}{n-2}$ ; the $g_{bd}$ coefficient gives $\alpha+2\beta(n-1)=0$ , so $\beta=\tfrac{1}{2(n-1)(n-2)}$ . Why: two linear equations for two unknowns have a unique solution, which is exactly why the traceless piece is unique — there is no freedom left.

Step 6. Substitute back to get $C_{abcd}=R_{abcd}-\tfrac{1}{n-2}(\mathrm{Ric}\owedge g)_{abcd}+\tfrac{R}{2(n-1)(n-2)}(g\owedge g)_{abcd}$ , which is equation (4) once the products are expanded. Why: the coefficients reproduce the textbook Weyl tensor of Wald and Carroll, confirming the construction landed on the standard object.

Go deeper · Derivation B · 1 minPacking the trace pieces into the Schouten tensor

Derivation BPacking the trace pieces into the Schouten tensor

The goal is to show that the two separate trace corrections of the Weyl definition are not really two objects but one — a single symmetric tensor Kulkarni–Nomizu-multiplied into the metric — and to identify that tensor as the Schouten tensor. This is what turns the three-term Weyl formula into the compact master form $R_{abcd}=C_{abcd}+(P\owedge g)_{abcd}$ , and it is the step that makes conformal geometry tractable, because the Schouten tensor is the object with the simplest conformal transformation law.

Step 1. Move the trace terms of (4) to the other side to read Riemann as $R_{abcd}=C_{abcd}+\tfrac{1}{n-2}(\mathrm{Ric}\owedge g)_{abcd}-\tfrac{R}{2(n-1)(n-2)}(g\owedge g)_{abcd}$ . Why: the master form is just the Weyl definition solved for Riemann, so the trace terms return with flipped signs.

Step 2. Use bilinearity of the Kulkarni–Nomizu product, $A\owedge g+\lambda\,g\owedge g=(A+\lambda g)\owedge g$ , to factor the two trace terms into one product with the metric. Why: both trace terms already carry $g$ as one factor, so the distributive law of $\owedge$ collapses them into a single product against a combined symmetric tensor.

Step 3. Read off the combined tensor as $P_{ab}=\tfrac{1}{n-2}R_{ab}-\tfrac{R}{2(n-1)(n-2)}g_{ab}=\tfrac{1}{n-2}\big(R_{ab}-\tfrac{R}{2(n-1)}g_{ab}\big)$ , the Schouten tensor of equation (3). Why: this is precisely the coefficient of $g$ in the factored product, and it matches the standard Schouten definition term for term.

Step 4. Check the trace. Contracting gives $g^{ab}P_{ab}=\tfrac{1}{n-2}\big(R-\tfrac{nR}{2(n-1)}\big)=\tfrac{R}{2(n-1)}$ , so tracing the master identity twice returns $R_{ab}$ and $R$ identically — the decomposition loses no information. Why: a valid decomposition must reconstruct every trace exactly, and this check confirms that Weyl plus Schouten package recovers the full Ricci content.

Go deeper · Derivation C · 1 minWhy the squared norm splits with no cross terms

Derivation CWhy the squared norm splits with no cross terms

The goal is to prove that the Kretschmann scalar $R_{abcd}R^{abcd}$ decomposes into a sum of squares of the three pieces with no mixing, and to extract equation (8). The engine is orthogonality: the inner product of any two distinct irreducible pieces vanishes because contracting a traceless tensor against a trace-built tensor always produces a trace of the traceless one, which is zero. This is the concrete, index-level meaning of the abstract claim that the three subspaces are orthogonal.

Step 1. Write $R_{abcd}=R^{\text{(scalar)}}_{abcd}+R^{\text{(Ricci)}}_{abcd}+C_{abcd}$ and expand the full contraction $R_{abcd}R^{abcd}$ into six terms — three squares and three cross terms. Why: the squared norm of a sum is the sum of squares plus twice the cross terms, so the whole question reduces to whether those cross terms survive.

Step 2. Kill the Weyl cross terms. Any contraction of $C_{abcd}$ against either trace piece reduces, after using that $(g\owedge g)$ and $(S\owedge g)$ are built from metrics, to a trace of $C$ — and every trace of $C$ is zero by (5). Why: the traceless core is orthogonal to everything trace-built, so both $C\!\cdot\!R^{\text{(scalar)}}$ and $C\!\cdot\!R^{\text{(Ricci)}}$ vanish outright.

Step 3. Kill the scalar–Ricci cross term. Contracting $R^{\text{(Ricci)}}_{abcd}\propto(S\owedge g)$ against $R^{\text{(scalar)}}_{abcd}\propto(g\owedge g)$ produces the trace $g^{ab}S_{ab}$ , which is zero because $S_{ab}$ is the traceless Ricci tensor. Why: the scalar template is a multiple of the metric on each pair, so pairing it with $S$ always squeezes out the trace of $S$ , and that trace was removed by construction.

Step 4. Assemble the surviving squares. Using $R_{ab}R^{ab}=S_{ab}S^{ab}+R^2/n$ to re-express the traceless-Ricci square in terms of the full Ricci tensor, and evaluating the scalar square, the three self-contractions combine into $C_{abcd}C^{abcd}+\tfrac{4}{n-2}R_{ab}R^{ab}-\tfrac{2}{(n-1)(n-2)}R^2$ , which is equation (8). Why: only the diagonal terms survive, so the Kretschmann scalar is genuinely a sum of independent invariants — the algebraic signature of an orthogonal decomposition.

Limiting cases

The character of the decomposition changes sharply with dimension, and the two physical extremes — vacuum and maximal symmetry — each switch off a different pair of pieces.

Case	Result	Meaning
$n=2$	$R_{abcd}=\tfrac{R}{2}(g_{ac}g_{bd}-g_{ad}g_{bc})$	Only the scalar piece exists; all curvature is the single number $R$ . No traceless-Ricci or Weyl part.
$n=3$	$C_{abcd}\equiv 0,\ \ R_{abcd}=(P\owedge g)_{abcd}$	Weyl vanishes identically; Riemann is fully fixed by the Ricci tensor. The Cotton tensor takes over as the conformal-flatness obstruction.
$n=4$	$\dim:\ 1+9+10=20$	All three pieces nontrivial: 1 scalar, 9 traceless-Ricci, 10 Weyl — the physical case for general relativity.
Vacuum, $n\ge 4$	$R_{ab}=0\Rightarrow R_{abcd}=C_{abcd}$	Both trace pieces vanish; Riemann equals Weyl. All curvature is tidal and radiative.
Maximally symmetric	$S_{ab}=0,\ C_{abcd}=0$	Only the scalar piece; $R_{abcd}=K(g_{ac}g_{bd}-g_{ad}g_{bc})$ with $K=R/[n(n-1)]$ .

Common trapVacuum does not mean flat, and Weyl is not zero in empty space

The most common trap is to reason "no matter, so $R_{ab}=0$ , so no curvature." Setting $R_{ab}=0$ kills only the two trace pieces; it says nothing about the Weyl tensor, which is exactly the piece no contraction can see. Schwarzschild and Kerr are vacuum solutions with $R_{ab}=0$ everywhere outside the source, yet their Weyl tensors are large — that Weyl curvature is the tidal field that spaghettifies an infalling astronaut and the radiation that a merging binary sends to LIGO. A second, quieter trap is convention drift: textbooks differ on whether the Kulkarni–Nomizu product carries a factor of 2, on the sign of the Riemann tensor, and on the exact $1/(n-2)$ and $1/[2(n-1)]$ coefficients. Always pin down the convention before quoting a Weyl or Schouten formula, or the coefficients in (3), (4), and (8) will silently disagree with your source.

Worked examples

1 · Schwarzschild vacuum is pure Weyl (fully worked)

Outside a spherical mass the Schwarzschild metric is a vacuum solution, $R_{ab}=0$ . Use the decomposition to say which of the three pieces carry the curvature, then evaluate the surviving invariant at the surface of the Sun.

Step 1. Switch off the trace pieces. Vacuum means $R_{ab}=0$ , hence $R=0$ and $S_{ab}=R_{ab}-\tfrac{R}{n}g_{ab}=0$ . By (6) and (7) both the scalar and traceless-Ricci pieces vanish identically, so the master identity (1) collapses to $R_{abcd}=C_{abcd}$ . All twenty components (ten independent, in 4D) are pure Weyl.

Step 2. Reduce the invariant. With $R_{ab}=0$ and $R=0$ , the orthogonal split (8) loses its last two terms, leaving $R_{abcd}R^{abcd}=C_{abcd}C^{abcd}$ . The Kretschmann scalar and the Weyl invariant are numerically equal in any vacuum spacetime — a direct, checkable consequence of the decomposition.

Step 3. Put in the Schwarzschild value. For Schwarzschild the Kretschmann scalar is $R_{abcd}R^{abcd}=\dfrac{48\,G^2M^2}{c^4\,r^6}=\dfrac{12\,r_s^2}{r^6}$ , with $r_s=2GM/c^2$ . Using $GM_\odot/c^2=1.476\times10^3$ m (so $r_s=2.95\times10^3$ m) at the solar surface $r=6.96\times10^8$ m gives $R_{abcd}R^{abcd}=48\,(1.476\times10^3)^2/(6.96\times10^8)^6\approx\mathbf{9.2\times10^{-46}\ m^{-4}}$ .

Step 4. Read the physics. This number is small but nonzero, and by Step 2 it is carried entirely by the ten Weyl components — the Sun curves empty space through its Weyl tensor alone. It diverges as $r\to0$ (the true singularity) but stays finite at $r=r_s$ , confirming the horizon is a coordinate artifact. In Schwarzschild vacuum $R_{abcd}=C_{abcd}$ and $R_{abcd}R^{abcd}=C_{abcd}C^{abcd}=9.2\times10^{-46}$ m⁻⁴ at the solar surface — pure free field.

2 · de Sitter space is pure scalar

A universe with only a positive cosmological constant $\Lambda$ is de Sitter space, maximally symmetric with $R_{ab}=\Lambda g_{ab}$ in four dimensions. Which pieces survive, and what is its Kretschmann scalar? Check the split both ways with the observed $\Lambda$ .

Step 1. Switch off Weyl and traceless-Ricci. Maximal symmetry forces $C_{abcd}=0$ , and $R_{ab}=\Lambda g_{ab}$ gives $R=4\Lambda$ in $n=4$ , so $S_{ab}=R_{ab}-\tfrac{R}{4}g_{ab}=\Lambda g_{ab}-\Lambda g_{ab}=0$ . Only the scalar piece survives: $R_{abcd}=\tfrac{R}{12}(g_{ac}g_{bd}-g_{ad}g_{bc})=\tfrac{\Lambda}{3}(g_{ac}g_{bd}-g_{ad}g_{bc})$ .

Step 2. Compute the invariant directly. For a constant-curvature space $R_{abcd}R^{abcd}=\tfrac{2R^2}{n(n-1)}=\tfrac{R^2}{6}$ in 4D. With $R=4\Lambda$ this is $\tfrac{16\Lambda^2}{6}=\tfrac{8\Lambda^2}{3}$ .

Step 3. Cross-check with the split (8). Here $C=0$ and $R_{ab}R^{ab}=\Lambda^2 g_{ab}g^{ab}=4\Lambda^2=R^2/4$ , so the right side is $2R_{ab}R^{ab}-\tfrac13 R^2=2\cdot\tfrac{R^2}{4}-\tfrac{R^2}{3}=\tfrac{R^2}{2}-\tfrac{R^2}{3}=\tfrac{R^2}{6}$ , matching Step 2 exactly. The orthogonal decomposition is internally consistent.

Step 4. Put in numbers. The observed cosmological constant is $\Lambda\approx1.1\times10^{-52}$ m⁻², so $R=4\Lambda\approx4.4\times10^{-52}$ m⁻² and $R_{abcd}R^{abcd}=\tfrac{8\Lambda^2}{3}\approx\mathbf{3.2\times10^{-104}\ m^{-4}}$ . de Sitter is the mirror image of Schwarzschild: pure scalar rather than pure Weyl, with every bit of curvature carried by the single Ricci scalar $R$ .

Physics of the portrait

The decomposition is a nineteenth- and twentieth-century collaboration: Riemann built the tensor, Ricci and Levi-Civita built the calculus that traces it, and one man isolated the piece that no trace can reach — the conformally invariant core that turned out to be the free gravitational field.

Origins · 1 minHermann Weyl and how this was discovered

Hermann Weyl

1885–1955 · conformal curvature and the geometry of gravitation

Hermann Weyl was the rare mathematician who reshaped physics from the inside. A student of Hilbert at Göttingen, he moved to the ETH in Zürich in 1913, where he shared a corridor with Einstein during the years general relativity was being digested by mathematicians. In the 1918 fourth edition of his masterwork Raum, Zeit, Materie ("Space, Time, Matter") and the companion paper Reine Infinitesimalgeometrie, Weyl introduced the tensor now bearing his name: the totally traceless part of the Riemann tensor, obtained by stripping away every contraction. His deeper motivation was conformal geometry — he proved that this tensor is unchanged when the metric is rescaled by any positive function, $g_{ab}\to\Omega^2 g_{ab}$ , and that its vanishing (for $n\ge 4$ ) is the exact condition for a metric to be conformally flat. The same 1918 work launched gauge theory: Weyl's attempt to unify gravity and electromagnetism by making length itself position-dependent failed as physics but seeded the gauge principle that now underlies the entire Standard Model. His reach was extraordinary — the Peter–Weyl theorem in harmonic analysis, the Weyl group and character formula in representation theory, foundational work on Riemann surfaces, and the Weyl equation for massless spin-half particles. When the Nazis came to power in 1933 he left Germany for the Institute for Advanced Study in Princeton, joining Einstein and Gödel. He did not set out to describe gravitational radiation, but the tensor he defined to answer a question about conformal rescaling turned out to be exactly the part of curvature that propagates through empty space — the carrier of tidal forces and gravitational waves, and the object the Petrov classification would later use to sort every vacuum spacetime.

History

1854

— Riemann conceives the intrinsic curvature of n-dimensional space

1900

— Ricci-Curbastro and Levi-Civita codify the absolute differential calculus and the Ricci trace

1918

— Weyl isolates the conformally invariant traceless part of the curvature

1921

— Schouten introduces the tensor packaging the trace parts of the curvature

1954

— Petrov classifies the algebraic types of the Weyl tensor

Going further

Texts. Misner, Thorne & Wheeler, Gravitation (1973), §13.5 and Box 13.1 · Wald, General Relativity (1984), §3.2, eq. (3.2.28) · Carroll, Spacetime and Geometry (2004), §3.7, eqs. (3.147)–(3.151) · Hawking & Ellis, The Large Scale Structure of Space-Time (1973), §2.6 · Besse, Einstein Manifolds (1987), Ch. 1 (Kulkarni–Nomizu product and curvature decomposition) · Schutz, A First Course in General Relativity, 2nd ed. (2009), §6.7.

Provenance. The Weyl coefficients $\alpha=-1/(n-2)$ and $\beta=1/[2(n-1)(n-2)]$ were re-derived here by imposing tracelessness and checked against Wald §3.2 and Carroll §3.7; the master form $R_{abcd}=C_{abcd}+(P\owedge g)_{abcd}$ and the Schouten tensor match Besse Ch. 1 after convention alignment. The component split $1+9+10=20$ at $n=4$ , the Kretschmann-split coefficients $4/(n-2)$ and $2/[(n-1)(n-2)]$ , and the two worked numbers — Schwarzschild $R_{abcd}R^{abcd}=48G^2M^2/c^4r^6=9.2\times10^{-46}$ m⁻⁴ at the solar surface, and de Sitter $R_{abcd}R^{abcd}=8\Lambda^2/3=3.2\times10^{-104}$ m⁻⁴ with $\Lambda=1.1\times10^{-52}$ m⁻² — were verified numerically, including the two independent routes to the de Sitter invariant. Conventions: signature and Riemann sign as in Wald; the $(A\owedge B)$ definition carries no extra factor of 2.

Prerequisites. Riemann Curvature Tensor · Algebraic Symmetries of the Riemann Tensor · Ricci Tensor · Ricci Scalar · Independent Components of the Riemann Tensor. Leads to. Weyl Conformal Tensor · Einstein Tensor · Contracted Bianchi Identities · Second (Differential) Bianchi Identity.

Ask the tutor: Ask me to derive the Weyl coefficients by hand from the tracelessness condition, or to show why the Weyl tensor vanishes in three dimensions but the Cotton tensor does not.

What is the Ricci decomposition in one sentence?

It is the unique splitting of the Riemann tensor into three mutually orthogonal, irreducible pieces — a scalar part built from the Ricci scalar, a part built from the traceless Ricci tensor, and the totally traceless Weyl tensor — sorted by how each transforms under rotations. Taking traces projects onto the first two pieces; the third, the Weyl tensor, is defined by being annihilated by every trace. In four dimensions the three pieces carry 1, 9, and 10 of the twenty independent curvature components. The whole thing compresses to $R_{abcd}=C_{abcd}+(P\owedge g)_{abcd}$ , Weyl plus one Kulkarni–Nomizu product of the Schouten tensor with the metric.

Why is the Weyl tensor called the "free" gravitational field?

Because it is the part of the curvature that matter does not fix at a point. Einstein's equation $G_{ab}=8\pi T_{ab}$ is an algebraic relation involving only the Ricci and scalar pieces, so wherever you specify the stress-energy you have determined those two pieces completely and pointwise. The Weyl tensor appears nowhere in that equation; it is constrained only by the differential (Bianchi) identities, which relate its derivatives to derivatives of the Ricci part. That is what lets Weyl curvature exist and propagate in vacuum, where $T_{ab}=0$ and $R_{ab}=0$ but $C_{abcd}\ne 0$ : tidal fields between separated bodies and gravitational waves are pure Weyl, the gravitational field's own degrees of freedom rather than a local response to matter.

What is the Kulkarni–Nomizu product and why use it?

It is a product that takes two symmetric $(0,2)$ -tensors $A$ and $B$ and returns a $(0,4)$ -tensor $(A\owedge B)_{abcd}=A_{ac}B_{bd}+A_{bd}B_{ac}-A_{ad}B_{bc}-A_{bc}B_{ad}$ that automatically has all the algebraic symmetries of the Riemann tensor — antisymmetry in each pair, pair-exchange symmetry, and the first Bianchi identity. Its usefulness is that it lets you build genuine curvature-type tensors out of lower-rank data: the trace pieces of Riemann are just $(g\owedge g)$ and $(\mathrm{Ric}\owedge g)$ with numerical coefficients. Because the product is bilinear, two separate trace terms fuse into a single $(P\owedge g)$ , which is what makes the master decomposition a one-line identity rather than a three-term formula.

How does the Schouten tensor relate to the Ricci and Einstein tensors?

All three are trace-adjustments of the Ricci tensor differing only in their coefficient of $R\,g_{ab}$ . The Einstein tensor is $G_{ab}=R_{ab}-\tfrac12 R g_{ab}$ , tuned so that its divergence vanishes. The Schouten tensor is $P_{ab}=\tfrac{1}{n-2}\big(R_{ab}-\tfrac{R}{2(n-1)}g_{ab}\big)$ , tuned instead so that $(P\owedge g)$ reproduces exactly the trace pieces of Riemann and leaves the Weyl tensor traceless. The Schouten tensor is the natural variable of conformal geometry because it has the simplest transformation law under a conformal rescaling of the metric, whereas the Einstein tensor is the natural variable of dynamics because of its conservation. They encode the same Ricci data in coordinates suited to different questions.

Why does the Weyl tensor vanish in three dimensions?

It is forced by counting. In $n$ dimensions the Riemann tensor has $\tfrac{1}{12}n^2(n^2-1)$ independent components, which is 6 for $n=3$ , exactly matching the six components of the symmetric Ricci tensor. So in three dimensions the Ricci tensor already carries all of the curvature information, leaving nothing for a traceless remainder — the Weyl subspace has dimension zero. Concretely, the traceless-Weyl piece has dimension $\tfrac{1}{12}n(n+1)(n+2)(n-3)$ , which is zero at $n=3$ and first becomes positive (equal to 10) at $n=4$ . In three dimensions the role of conformal-flatness obstruction is instead played by the Cotton tensor, a derivative of the Schouten tensor.

Is the decomposition unique, or are there other ways to split Riemann?

It is unique as an orthogonal decomposition into pieces irreducible under the orthogonal group. The reason is representation theory: the space of Riemann-symmetric tensors carries a representation that splits into exactly these three irreducible summands (for $n\ge 4$ ), and irreducible pieces of a representation are unique up to the obvious labeling. Any other "split" you might write down is either a relabeling of these three pieces or is not orthogonal and not irreducible. This uniqueness is what gives the decomposition physical force — the Weyl tensor is not one arbitrary choice of traceless part but the traceless part, which is why its vanishing is a coordinate-independent statement about conformal flatness and why its algebraic type is an invariant of the spacetime.