03.02.16 · differential-geometry / manifolds

Weyl tensor and conformally flat metrics

shipped3 tiersLean: none

Anchor (Master): Weyl 1918 Math. Z. 2; Eisenhart Riemannian Geometry 1926 Ch. III; Sternberg Curvature in Mathematics and Physics Ch. 9

Intuition Beginner

Imagine stretching a rubber map of the world. You can shrink one region and blow up another, and the angles between roads stay the same even though distances change wildly. A transformation that rescales distances but preserves angles is called conformal. Two metrics that differ only by such a rescaling describe the same "shape of angles" even if they disagree about size.

The Riemann curvature tensor records every way a space bends. Some of that bending is just an artifact of how you chose to measure size locally, and a conformal rescaling can wash it away. But some bending is stubborn: no rescaling removes it. The Weyl tensor is the part of the curvature that survives every conformal rescaling. It is the curvature you cannot massage away by changing the local scale.

When the Weyl tensor is zero, the space is conformally flat: you can rescale it, region by region, until it looks like ordinary flat space as far as angles are concerned. So the Weyl tensor answers a sharp question. Is this curved space secretly flat once you stop caring about size?

In relativity this part has a name beyond mathematics. It is the free gravitational field, the tidal stretching that travels as gravitational waves through empty space where no matter sits.

Visual Beginner

Picture three panels. On the left, a round sphere: its curvature comes entirely from the averaged pieces (Ricci and scalar), and after a conformal rescaling it can be flattened onto a plane, so its Weyl part is zero. In the middle, flat space drawn with a distorted grid that has been conformally stretched; the angles between grid lines are preserved, and the Weyl part is still zero. On the right, the region outside a star: empty, so the averaged curvature vanishes, yet space is genuinely curved. All of that curvature lives in the Weyl tensor.

The Weyl tensor is the curvature that no rescaling of local size can erase. It is invisible to the volume-measuring parts of curvature and carries the tidal, wave-like information.

Worked example Beginner

Any 3-dimensional space has zero Weyl tensor. We check this on a concrete case, the round 3-sphere, by counting.

Step 1. Count the independent pieces of curvature in dimension 3. The full Riemann tensor in dimension $n$ has $n^{2} (n^{2} - 1) /12$ independent components. For $n = 3$ this is $9 \cdot 8/12 = 6$ .

Step 2. Count the averaged piece, the Ricci tensor. It is a symmetric $3 \times 3$ table, which has $3 \cdot 4/2 = 6$ independent entries.

Step 3. Compare. Both counts are $6$ . The Ricci tensor already holds as much information as the full curvature. There is nothing left over for the Weyl tensor to record, so it must be zero.

Step 4. Read off the meaning for the 3-sphere. Its curvature is fully captured by the averaged Ricci part, and indeed the round 3-sphere is conformally flat: a rescaling maps it to flat space minus a point.

What this tells us: in dimension 3 the averaged curvature already uses up every degree of freedom, so the Weyl tensor carries no information at all. The Weyl tensor only becomes a genuinely new object starting in dimension 4.

Check your understanding Beginner

Exercise (easy, multiple choice).

What does it mean for a space to be conformally flat?

A. The space is already flat with zero curvature B. After a local rescaling that preserves angles, the space looks flat C. The space has constant positive curvature D. The volume of every region equals the flat-space volume

Hint

Conformal means angle-preserving rescaling. Conformally flat means you can reach flat space by such a rescaling.

Answer

B. Feedback-correct: a conformally flat space becomes flat after multiplying the metric by a positive function, which rescales distances while keeping angles fixed. Feedback-wrong: it need not be flat to begin with (a sphere is conformally flat but curved); constant positive curvature is a stronger condition; volumes are generally not preserved by a conformal rescaling.

Formal definition Intermediate+

Let $(M, g)$ be a Riemannian (or pseudo-Riemannian) manifold of dimension $n \geq 3$ with Riemann curvature tensor written in fully covariant form $R_{ab c d} = g (R (e_{a}, e_{b}) e_{d}, e_{c})$ , Ricci tensor $R_{ab} = g^{c d} R_{c ab d}$ , and scalar curvature $R = g^{ab} R_{ab}$ . We use square brackets for antisymmetrisation, $T_{[ab]} = \frac{1}{2} (T_{ab} - T_{ba})$ .

Definition (Schouten tensor). For $n \geq 3$ the Schouten tensor is

$P_{ab} = \frac{1}{n - 2} (R_{ab} - \frac{R}{2 ( n - 1 )} g_{ab}) .$

Its trace is $g^{ab} P_{ab} = R / (2 (n - 1))$ . The Schouten tensor repackages the Ricci tensor and the scalar curvature into the single object that transforms simply under conformal change.

Definition (Kulkarni-Nomizu product). For symmetric 2-tensors $h, k$ , the Kulkarni-Nomizu product $h \owedge k$ is the algebraic curvature tensor

$(h \owedge k)_{ab c d} = h_{a c} k_{b d} + h_{b d} k_{a c} - h_{a d} k_{b c} - h_{b c} k_{a d} .$

It has all the algebraic symmetries of the Riemann tensor.

Definition (Weyl conformal curvature tensor). The Weyl tensor is the totally trace-free part of the Riemann tensor, defined by the decomposition

$R_{ab c d} = C_{ab c d} + (P \owedge g)_{ab c d}, n \geq 3.$

Equivalently, writing the Ricci and scalar pieces out,

$C_{ab c d} = R_{ab c d} - \frac{1}{n - 2} (g_{a c} R_{b d} - g_{a d} R_{b c} - g_{b c} R_{a d} + g_{b d} R_{a c}) + \frac{R}{( n - 1 ) ( n - 2 )} (g_{a c} g_{b d} - g_{a d} g_{b c}) .$

The Weyl tensor inherits every algebraic symmetry of the Riemann tensor: it is antisymmetric in $ab$ and in $c d$ , symmetric under the pair swap $ab \leftrightarrow c d$ , and satisfies the first Bianchi identity $C_{a [b c d]} = 0$ . Beyond these it is totally trace-free: every contraction with the inverse metric vanishes,

$g^{a c} C_{ab c d} = 0.$

Counterexamples to common slips

The Weyl tensor is not the traceless Ricci tensor. The traceless Ricci tensor $R_{ab} - (R / n) g_{ab}$ is a 2-tensor living in the Ricci part of the decomposition. The Weyl tensor is a 4-tensor that is what remains after both the Ricci and scalar parts are removed.
Ricci-flat does not mean Weyl-flat. If $R_{ab} = 0$ then $C_{ab c d} = R_{ab c d}$ , so a Ricci-flat space carries all of its curvature in the Weyl tensor. The Schwarzschild metric is the standard witness.
Conformal flatness is local. $C = 0$ guarantees a conformal factor exists in a neighbourhood of each point, not a single global rescaling to flat space. The round sphere is conformally flat yet not globally conformal to all of flat space.

Key theorem with proof Intermediate+

Theorem (conformal invariance of the Weyl tensor). Let $n \geq 3$ and let $\tilde{g} = Ω^{2} g$ for a positive smooth function $Ω$ . Then the Weyl tensor in mixed $(1, 3)$ form is unchanged: $\tilde{C}^{a}_{b c d} = C^{a}_{b c d}$ .

Proof. Write $Ω = e^{φ}$ and set $Υ_{a} = \nabla_{a} φ$ . The Levi-Civita connection of $\tilde{g}$ differs from that of $g$ by the tensor

$\tilde{Γ}_{ab}^{c} = Γ_{ab}^{c} + δ_{a}^{c} Υ_{b} + δ_{b}^{c} Υ_{a} - g_{ab} Υ^{c},$

where $Υ^{c} = g^{c d} Υ_{d}$ . Substituting this into the curvature formula $R^{a}_{b c d} = \partial_{c} Γ_{d b}^{a} - \partial_{d} Γ_{c b}^{a} + Γ_{ce}^{a} Γ_{d b}^{e} - Γ_{d e}^{a} Γ_{c b}^{e}$ and collecting terms gives the transformation of the $(1, 3)$ Riemann tensor,

$\tilde{R}^{a}_{b c d} = R^{a}_{b c d} - δ_{c}^{a} T_{b d} + δ_{d}^{a} T_{b c} - g_{b d} T^{a}_{c} + g_{b c} T^{a}_{d},$

where the symmetric tensor $T_{ab} = \nabla_{a} Υ_{b} - Υ_{a} Υ_{b} + \frac{1}{2} ∣Υ ∣^{2} g_{ab}$ collects all first and second derivatives of $φ$ , and $T^{a}_{c} = g^{ab} T_{b c}$ . The point is that the inhomogeneous correction has exactly the index structure of a Kulkarni-Nomizu product $(T \owedge g)$ , raised on one slot.

Now contract to obtain the Ricci and scalar transformations. Tracing $a$ with $c$ in the displayed formula,

$\tilde{R}_{b d} = R_{b d} - (n - 2) T_{b d} - (g^{a c} T_{a c}) g_{b d} .$

A further trace with $\tilde{g}^{b d} = Ω^{- 2} g^{b d}$ gives the scalar curvature. Combining these two results, the Schouten tensor of $\tilde{g}$ is found to transform as

$\tilde{P}_{ab} = P_{ab} - T_{ab},$

so that $T_{ab} = P_{ab} - \tilde{P}_{ab}$ is precisely the difference of Schouten tensors. Substituting back, the inhomogeneous correction to $\tilde{R}^{a}_{b c d}$ equals the difference of the two $(P \owedge g)$ terms. Therefore the combination

$\tilde{R}^{a}_{b c d} - (\tilde{P} \owedge g)^{a}_{b c d} = R^{a}_{b c d} - (P \owedge g)^{a}_{b c d}$

is conformally invariant. By the definition of the Weyl tensor, the left side is $\tilde{C}^{a}_{b c d}$ and the right side is $C^{a}_{b c d}$ , so $\tilde{C}^{a}_{b c d} = C^{a}_{b c d}$ . $□$

Bridge. This invariance builds toward the Weyl-Schouten characterisation of conformal flatness in the Master section, and appears again in the conformal compactification techniques of general relativity and in the construction of conformally invariant operators. The structural reason the argument closes is that the Schouten tensor absorbs every inhomogeneous derivative term in the conformal change, leaving the Weyl tensor as the conformally rigid residue. The same mechanism explains why the Weyl tensor descends to an invariant of the conformal class $[g]$ rather than of the metric $g$ itself, and it connects to the representation theory of the orthogonal group, where the Weyl tensor spans the unique irreducible summand of the curvature module annihilated by every trace.

Exercises Intermediate+

Exercise 4 (medium, multiple choice).

Why is the Weyl tensor used in the mixed $(1, 3)$ form $C^{a}_{b c d}$ when discussing conformal invariance, rather than the fully covariant $C_{ab c d}$ ?

A. The covariant form does not satisfy the Bianchi identity B. Only the $(1, 3)$ form is exactly invariant; the covariant form scales by $Ω^{2}$ C. The covariant form is not trace-free D. The two forms differ by the Schouten tensor

Hint

Lowering an index uses the metric, and the metric itself scales by $Ω^{2}$ .

Answer

B. The $(1, 3)$ tensor $C^{a}_{b c d}$ is invariant. Lowering its first index multiplies by $g_{a e}$ , and since $\tilde{g}_{a e} = Ω^{2} g_{a e}$ , the covariant form transforms as $\tilde{C}_{ab c d} = Ω^{2} C_{ab c d}$ . Both forms satisfy the Bianchi identity and are trace-free; the distinction is purely the conformal weight introduced by lowering an index.

Exercise 5 (medium, symbolic).

Show that in dimension $n = 2$ the Weyl tensor as defined would vanish identically, so the construction starts at $n = 3$ .

Hint

In dimension 2 the Riemann tensor is $R_{ab c d} = \frac{R}{2} (g_{a c} g_{b d} - g_{a d} g_{b c})$ . Substitute and also note $R_{ab} = \frac{R}{2} g_{ab}$ .

Answer

In dimension 2, $R_{ab c d} = \frac{R}{2} (g_{a c} g_{b d} - g_{a d} g_{b c})$ and $R_{ab} = \frac{R}{2} g_{ab}$ . The defining formula has $n - 2 = 0$ in two denominators, so the Schouten construction is undefined at $n = 2$ ; the genuine statement is that the orthogonal-group curvature module has no trace-free 4-tensor summand below dimension 4, and every algebraic curvature tensor in dimension 2 (and 3) is determined by its Ricci contraction. Hence no independent Weyl part exists, which is why the definition is stated for $n \geq 3$ and the tensor carries independent information only for $n \geq 4$ .

Exercise 6 (hard, symbolic).

Prove that the Weyl tensor vanishes identically in dimension $n = 3$ .

Hint

Count components: the trace-free conditions overdetermine the symmetries. Alternatively, show $C_{ab c d}$ , being trace-free with Riemann symmetries, must be zero in dimension 3 by exhibiting the curvature in terms of Ricci.

Answer

In dimension 3 the space of algebraic curvature tensors has dimension $n^{2} (n^{2} - 1) /12 = 6$ , equal to the dimension of symmetric $3 \times 3$ tensors. The map sending a curvature tensor to its Ricci contraction is therefore a linear isomorphism, so every curvature tensor is recovered from its Ricci part as $R_{ab c d} = (P \owedge g)_{ab c d}$ with the dimension-3 Schouten tensor. By definition $C_{ab c d} = R_{ab c d} - (P \owedge g)_{ab c d} = 0$ . Concretely, $R_{ab c d} = g_{a c} R_{b d} - g_{a d} R_{b c} - g_{b c} R_{a d} + g_{b d} R_{a c} - \frac{R}{2} (g_{a c} g_{b d} - g_{a d} g_{b c})$ in $n = 3$ , which is exactly $(P \owedge g)_{ab c d}$ .

Exercise 7 (hard, symbolic).

For a manifold of constant sectional curvature $κ$ , the Riemann tensor is $R_{ab c d} = κ (g_{a c} g_{b d} - g_{a d} g_{b c})$ . Show that the Weyl tensor vanishes, so every constant-curvature space is conformally flat.

Hint

Compute $R_{ab}$ and $R$ for this curvature, then substitute everything into $C_{ab c d}$ .

Answer

From $R_{ab c d} = κ (g_{a c} g_{b d} - g_{a d} g_{b c})$ , contraction gives $R_{ab} = (n - 1) κ g_{ab}$ and $R = n (n - 1) κ$ . Substituting into the definition: the Ricci block contributes $- \frac{1}{n - 2} (n - 1) κ (2 g_{a c} g_{b d} - 2 g_{a d} g_{b c}) \cdot$ (collecting the four terms) $= - 2 (n - 1) κ / (n - 2) (g_{a c} g_{b d} - g_{a d} g_{b c})$ , and the scalar block contributes $+ \frac{n ( n - 1 ) κ}{( n - 1 ) ( n - 2 )} (g_{a c} g_{b d} - g_{a d} g_{b c}) = \frac{nκ}{n - 2} (g_{a c} g_{b d} - g_{a d} g_{b c})$ . Adding the Riemann term $κ (g_{a c} g_{b d} - g_{a d} g_{b c})$ and combining coefficients: $κ - 2 (n - 1) κ / (n - 2) + nκ / (n - 2) = κ \cdot \frac{( n - 2 ) - 2 ( n - 1 ) + n}{n - 2} = κ \cdot \frac{0}{n - 2} = 0$ . Hence $C_{ab c d} = 0$ .

Exercise 8 (hard, multiple choice).

In dimension $n = 3$ the Weyl tensor vanishes, so it cannot detect conformal flatness. Which tensor replaces it as the obstruction?

A. The traceless Ricci tensor B. The Cotton tensor $C_{ab c} = \nabla_{c} P_{ab} - \nabla_{b} P_{a c}$ C. The Bach tensor D. The second fundamental form

Hint

The obstruction in dimension 3 is a third-order quantity built from derivatives of the Schouten tensor.

Answer

B. In dimension 3 a metric is conformally flat if and only if the Cotton tensor $C_{ab c} = \nabla_{c} P_{ab} - \nabla_{b} P_{a c}$ vanishes. For $n \geq 4$ the divergence of the Weyl tensor equals (up to a factor) the Cotton tensor, so vanishing Weyl already forces vanishing Cotton; in dimension 3 the Weyl tensor is identically zero and the Cotton tensor becomes the genuine obstruction. The Bach tensor is the relevant conformal invariant in dimension 4 variational problems, not the flatness obstruction here.

Advanced results Master

Theorem 1 (Ricci decomposition). For $n \geq 4$ the space of algebraic curvature tensors over an $n$ -dimensional inner product space decomposes under the orthogonal group $O (n)$ into three irreducible summands,

$R_{ab c d} = scalar \frac{R}{2 n ( n - 1 )} (g \owedge g)_{ab c d} + traceless Ricci \frac{1}{n - 2} (\overset{˚}{R} \owedge g)_{ab c d} + Weyl C_{ab c d},$

where $\overset{˚}{R}_{ab} = R_{ab} - (R / n) g_{ab}$ is the traceless Ricci tensor. The three pieces are mutually orthogonal in the natural inner product on curvature tensors, and the Weyl summand is the unique component on which every trace vanishes.

Theorem 2 (Weyl-Schouten). Let $(M, g)$ be a (pseudo-)Riemannian manifold of dimension $n \geq 4$ . Then $(M, g)$ is locally conformally flat, meaning every point has a neighbourhood on which $g = Ω^{2} δ$ for the flat metric $δ$ and some positive $Ω$ , if and only if the Weyl tensor vanishes identically, $C_{ab c d} = 0$ . In dimension $n = 3$ the Weyl tensor is identically zero and the analogous criterion is the vanishing of the Cotton tensor $C_{ab c} = \nabla_{c} P_{ab} - \nabla_{b} P_{a c}$ .

Theorem 3 (divergence and the Cotton tensor). In every dimension $n \geq 4$ the divergence of the Weyl tensor recovers the Cotton tensor,

$\nabla^{d} C_{ab c d} = (n - 3) C_{c ab},$

so for $n \geq 4$ a vanishing Weyl tensor forces a vanishing Cotton tensor, while in $n = 3$ the Cotton tensor stands alone. This is the structural reason the obstruction to conformal flatness shifts from a fourth-order algebraic tensor to a third-order differential tensor as the dimension drops from $4$ to $3$ .

Theorem 4 (Petrov classification, Lorentzian $n = 4$ ). On a four-dimensional Lorentzian manifold the Weyl tensor, viewed as a complex-linear self-dual operator on 2-forms, has a characteristic-polynomial structure that sorts spacetimes into the Petrov types I, II, III, D, N, and O according to the multiplicities of its principal null directions. Type O is exactly $C = 0$ , the conformally flat case; type D contains the Schwarzschild and Kerr geometries; type N is the algebraic signature of pure radiation, the form taken by gravitational waves far from their source.

Synthesis. The Weyl tensor is the conformally invariant residue of curvature, and the decomposition that isolates it is the foundational fact that organises conformal differential geometry. The central insight is that the orthogonal group splits the curvature module into trace-carrying and trace-free parts, and the trace-free part is exactly what conformal rescaling cannot touch, which is the bridge from representation theory to the Weyl-Schouten theorem. Putting these together with the divergence identity, the Cotton tensor emerges as the derivative shadow of the Weyl tensor and takes over as the conformal-flatness obstruction in dimension 3, where the Weyl tensor has run out of components. In Lorentzian signature the same algebraic object refines into the Petrov classification, so the abstract decomposition becomes a physical taxonomy of gravitational fields. Across all of these the pattern is one structure seen from four angles: an irreducible summand of curvature, a conformal invariant of the metric class, an obstruction to flatness, and the free gravitational field of general relativity.

Full proof set Master

Proposition 1 (the Weyl tensor is totally trace-free). For $n \geq 3$ , the tensor $C_{ab c d}$ defined by $R_{ab c d} = C_{ab c d} + (P \owedge g)_{ab c d}$ satisfies $g^{a c} C_{ab c d} = 0$ , and by its symmetries every other single contraction vanishes as well.

Proof. Contract the decomposition with $g^{a c}$ . On the left, $g^{a c} R_{ab c d} = R_{b d}$ . On the right we need $g^{a c} (P \owedge g)_{ab c d}$ . From the Kulkarni-Nomizu definition,

$(P \owedge g)_{ab c d} = P_{a c} g_{b d} + P_{b d} g_{a c} - P_{a d} g_{b c} - P_{b c} g_{a d} .$

Applying $g^{a c}$ term by term, using $g^{a c} g_{a c} = n$ , $g^{a c} P_{a c} = tr P = R / (2 (n - 1))$ , and $g^{a c} g_{a d} = δ_{d}^{c}$ ,

$g^{a c} (P \owedge g)_{ab c d} = P_{b d} + n P_{b d} - P_{b d} - (tr P) g_{b d} = (n - 2) P_{b d} + (tr P) g_{b d} .$

Now substitute the Schouten definition $P_{b d} = \frac{1}{n - 2} (R_{b d} - \frac{R}{2 ( n - 1 )} g_{b d})$ , giving $(n - 2) P_{b d} = R_{b d} - \frac{R}{2 ( n - 1 )} g_{b d}$ and $(tr P) g_{b d} = \frac{R}{2 ( n - 1 )} g_{b d}$ . These sum to exactly $R_{b d}$ . Therefore $g^{a c} C_{ab c d} = g^{a c} R_{ab c d} - g^{a c} (P \owedge g)_{ab c d} = R_{b d} - R_{b d} = 0$ . The antisymmetries $C_{ab c d} = - C_{ba c d} = - C_{ab d c}$ and the pair symmetry $C_{ab c d} = C_{c d ab}$ convert this single vanishing trace into the vanishing of all four single contractions, and the totally antisymmetric trace vanishes by the first Bianchi identity. $□$

Proposition 2 (conformal weight of the covariant Weyl tensor). Under $\tilde{g} = Ω^{2} g$ the fully covariant Weyl tensor scales as $\tilde{C}_{ab c d} = Ω^{2} C_{ab c d}$ .

Proof. The mixed Weyl tensor is conformally invariant, $\tilde{C}^{a}_{b c d} = C^{a}_{b c d}$ , by the Key theorem. Lowering the first index uses the metric: $\tilde{C}_{ab c d} = \tilde{g}_{a e} \tilde{C}^{e}_{b c d} = Ω^{2} g_{a e} C^{e}_{b c d} = Ω^{2} C_{ab c d}$ . $□$

Proposition 3 (Schwarzschild has nonzero Weyl tensor). The Schwarzschild metric is Ricci-flat but not flat; its Weyl tensor is nonzero, and a representative orthonormal-frame component scales as $m / r^{3}$ .

Proof. In Schwarzschild coordinates the exterior metric is $g = - (1 - 2 m / r) d t^{2} + (1 - 2 m / r)^{- 1} d r^{2} + r^{2} d Ω^{2}$ . A direct computation of the Einstein equations in vacuum gives $R_{ab} = 0$ , so $R = 0$ and consequently $C_{ab c d} = R_{ab c d}$ by Proposition 1's decomposition with vanishing Schouten tensor. The Riemann tensor is not identically zero: in the orthonormal frame $e_{\hat{0}} = (1 - 2 m / r)^{- 1/2} \partial_{t}$ , $e_{\hat{1}} = (1 - 2 m / r)^{1/2} \partial_{r}$ , the radial tidal component is

$R_{\hat{0} \hat{1} \hat{0} \hat{1}} = - \frac{2 m}{r ^{3}} .$

This is nonzero for $m \neq = 0$ , hence $C_{\hat{0} \hat{1} \hat{0} \hat{1}} = - 2 m / r^{3} \neq = 0$ . Therefore the Weyl tensor is nonzero and carries the entire curvature of the vacuum region. The $1/ r^{3}$ falloff is the tidal-force signature: it is exactly the Newtonian tidal tensor $\partial_{i} \partial_{j} (m / r)$ in the weak-field limit. $□$

Proposition 4 (FRW and maximally symmetric spaces are conformally flat). Every Friedmann-Robertson-Walker metric, and more generally every spatially homogeneous and isotropic spacetime, has vanishing Weyl tensor.

Proof. The spatial slices of an FRW metric have constant curvature, and the full metric $g = - d t^{2} + a (t)^{2} γ$ , with $γ$ a constant-curvature 3-metric, can be written after a conformal-time substitution $d η = d t / a$ as $g = a (η)^{2} (- d η^{2} + γ)$ . The bracketed metric is a static product of a line with a constant-curvature 3-space; such a metric has vanishing Weyl tensor because each constant-curvature factor contributes only Ricci and scalar curvature (Exercise 7) and the product structure introduces no Weyl component in this isotropic case. Since the Weyl tensor is conformally invariant and the conformal factor $a (η)^{2}$ does not change it, the original FRW metric has $C = 0$ . Maximally symmetric spacetimes (de Sitter, anti-de Sitter, Minkowski) are special cases of constant curvature and are conformally flat directly by Exercise 7. $□$

Connections Master

Sectional, Ricci, and scalar curvature 03.02.05. The Weyl tensor is the third and final irreducible piece of the curvature decomposition begun there: that unit isolates the Ricci and scalar traces, and the Weyl tensor is precisely what remains after both traces are removed. The decomposition $R =$ scalar $+$ traceless-Ricci $+$ Weyl is the completion of the curvature anatomy.
Constant curvature and the Killing-Hopf theorem 03.02.06. Every constant-curvature model space has vanishing Weyl tensor and is therefore conformally flat; the Weyl tensor refines the constant-curvature classification by measuring how far a general metric departs from conformal flatness, a strictly weaker condition than constant curvature.
Bochner technique and curvature vanishing 03.02.15. The orthogonal-group decomposition that isolates the Weyl tensor is the same representation-theoretic splitting used to define the curvature operator in Bochner-type vanishing theorems; positivity of the Weyl-free part of the curvature operator drives several refined vanishing results.

Historical & philosophical context Master

Hermann Weyl introduced the conformal curvature tensor in his 1918 paper "Reine Infinitesimalgeometrie" in Mathematische Zeitschrift ^{[Weyl 1918]}, as part of a programme to build a "pure infinitesimal geometry" in which only the comparison of nearby lengths, not lengths at a distance, was meaningful. The conformal tensor was the invariant that survived his gauge freedom in local scale. Jan Schouten gave the tensor decomposition its modern algebraic form in 1921, introducing the trace-adjusted tensor that now bears his name. Émile Cotton had earlier, in 1899, found the third-order tensor that controls conformal flatness in three dimensions ^{[Cotton 1899]}, the case where Weyl's tensor degenerates to zero.

Luther Eisenhart codified the conformal theory in his 1926 treatise Riemannian Geometry ^{[Eisenhart 1926]}, establishing the Weyl-Schouten theorem in the form used today. The physical reading of the tensor as the free gravitational field, the part of spacetime curvature not fixed pointwise by the matter content through Einstein's equations, was sharpened by Aleksei Petrov, whose 1954 algebraic classification of the Weyl tensor ^{[Petrov 1954]} organised exact solutions of general relativity by the degeneracy structure of their conformal curvature and gave gravitational radiation its invariant algebraic signature.

Bibliography Master

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}

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