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The exchange symmetries of a curvature-like tensor do not imply the first Bianchi identity

Statement refuted

Statement refuted: "Let $M$ be a smooth four-manifold with a global chart and let $T$ be a smooth $(0, 4)$ -tensor field on $M$ whose components satisfy, at every point, $T_{a b c d} = - T_{b a c d}, T_{a b c d} = - T_{a b d c}, T_{a b c d} = T_{c d a b}$ for all $a, b, c, d$ — the skew-symmetries and the exchange of the two index pairs of The algebraic symmetries of the curvature tensor and the symmetry of the Ricci and Einstein tensors clauses (a)-(b). Then $T$ also satisfies the first Bianchi identity $T_{a b c d} + T_{a c d b} + T_{a d b c} = 0$ of clause (c)."

Facts & Assumptions

Given: The smooth four-manifold $M = R^{4}$ with the identity chart $x = (x^{0}, x^{1}, x^{2}, x^{3})$ and the flat Lorentzian metric $g = diag (- 1, 1, 1, 1)$ with time orientation $N = \partial_{0}$ (Smooth time-oriented Lorentzian four-manifolds: the geometric objects of the model).

[L1]

For the curvature tensor of a metric-selected connection the identities $R_{a b c d} = - R_{b a c d}$ , $R_{a b c d} = - R_{a b d c}$ , $R_{a b c d} = R_{c d a b}$ and $R_{a b c d} + R_{a c d b} + R_{a d b c} = 0$ all hold; the displayed implicational statement above asserts only the first three for an arbitrary tensor field (The algebraic symmetries of the curvature tensor and the symmetry of the Ricci and Einstein tensors clauses (a)-(c)).

[L2]

A smooth tensor field of type $(0, 4)$ on $M$ is determined by its component functions in the maximal atlas, and a family of component functions that is $C^{\infty}$ in every chart — in particular a family that is constant in the global identity chart — defines a smooth $(0, 4)$ -tensor field (Tensor fields of type (r,s) on a Lorentzian four-manifold and their components clauses (b)-(c)).

[L3]

An algebraic curvature tensor at a point is defined to satisfy the four identities of [L1] together, the Bianchi identity included (A curvature tensor vanishes if the tidal operator vanishes for every timelike direction).

Counterexample

In the identity chart define the constant component families $ω_{01} = ω_{23} : = 1, ω_{10} = ω_{32} : = - 1, ω_{a b} : = 0 otherwise,$ and $T_{a b c d} : = ω_{a b} ω_{c d} .$ By [L2] these constants define a smooth $(0, 2)$ -tensor field $ω$ and a smooth $(0, 4)$ -tensor field $T$ on $M$ .

1.1L2constructalgebra

(The skew-symmetries.) Since $ω_{a b} = - ω_{b a}$ for all $a, b$ , $T_{a b c d} = ω_{a b} ω_{c d} = - ω_{b a} ω_{c d} = - T_{b a c d}$ and $T_{a b c d} = - ω_{a b} ω_{d c} = - T_{a b d c}$ : the first two identities of the statement hold everywhere.

1.2L2algebra

(Exchange of the pairs.) Since multiplication of real numbers is commutative, $T_{c d a b} = ω_{c d} ω_{a b} = ω_{a b} ω_{c d} = T_{a b c d}$ : the third identity holds everywhere.

2.1L2step 1.2algebra

(The Bianchi sum does not vanish.) For $(a, b, c, d) = (0, 1, 2, 3)$ the sum of the statement is $T_{0123} + T_{0231} + T_{0312}$ . The components are $T_{0123} = ω_{01} ω_{23} = 1 \cdot 1 = 1$ , $T_{0231} = ω_{02} ω_{31} = 0 \cdot 0 = 0$ and $T_{0312} = ω_{03} ω_{12} = 0 \cdot 0 = 0$ . Hence $T_{0123} + T_{0231} + T_{0312} = 1 \neq 0$ at every point of $M$ : $T$ violates the first Bianchi identity.

3.1step 1.1step 1.2step 2.1L1L3algebra∎

(Conclusion.) Steps 1.1 and 1.2 show that $T$ satisfies the two skew-symmetries and the pair exchange, and step 2.1 shows that it does not satisfy the Bianchi identity. Therefore the displayed statement is refuted: the exchange symmetries alone do not imply the Bianchi identity, which is an independent condition and is imposed separately in the definition of an algebraic curvature tensor [L3]. In particular, by clause (c) of [L1] no metric-selected connection on $M$ has $T$ as its curvature tensor.

Remarks

Why the example works. The tensor is defined directly by $T_{a b c d} = ω_{a b} ω_{c d}$ for the two-form $ω = d x^{0} \land d x^{1} + d x^{2} \land d x^{3}$ ; no Kulkarni-Nomizu construction is involved. The three exchange identities are immediate from the skew-symmetry of $ω$ , while the Bianchi sum for $(0, 1, 2, 3)$ receives the single term $ω_{01} ω_{23}$ that the other two orderings of the indices miss. The example is the promised companion of the independence remark in The algebraic symmetries of the curvature tensor and the symmetry of the Ricci and Einstein tensors.
What is not claimed. The counterexample is algebraic: it refutes the implication, not the Bianchi identity for curvature tensors; that identity is a theorem of this page (The algebraic symmetries of the curvature tensor and the symmetry of the Ricci and Einstein tensors clause (c)). No claim is made that $T$ arises geometrically, and no count of independent components is asserted.

The exchange symmetries of a curvature-like tensor do not imply the first Bianchi identity

Statement refuted

Facts & Assumptions

Counterexample

Remarks

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