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A nonzero algebraic curvature tensor whose tidal operator vanishes for one timelike direction

Statement refuted

Statement refuted: "Let $R_{p}$ be an algebraic curvature tensor at a point $p$ of a time-oriented smooth Lorentzian four-manifold, with the four identities of A curvature tensor vanishes if the tidal operator vanishes for every timelike direction, and suppose there is one timelike vector $T \in T_{p} M$ such that the tidal operator $J \mapsto R (T, J) T$ vanishes for every $J \in T_{p} M$ . Then $R_{p} = 0$ ."

Facts & Assumptions

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

[L1]

A vector $R (T, J) T$ associated with a $(0, 4)$ -tensor $R$ is defined by the pairing $g (R (T, J) T, W) = R_{a b c d} W^{a} T^{c} J^{d} T^{b}$ for all $W$ , the convention fixed in The Riemann curvature operator and the derived Ricci, scalar and Einstein tensors clause (b) and A curvature tensor vanishes if the tidal operator vanishes for every timelike direction; $g$ is nondegenerate, so such a vector is zero if and only if the pairing vanishes for all $W$ .

[L2]

An algebraic curvature tensor at a point is a multilinear map whose components in a basis satisfy $R_{a b c d} = - R_{b a c d} = - R_{a b d c} = R_{c d a b}$ and $R_{a b c d} + R_{a c d b} + R_{a d b c} = 0$ (A curvature tensor vanishes if the tidal operator vanishes for every timelike direction).

[L3]

A family of component functions that is constant in the global identity chart defines a smooth tensor field (Tensor fields of type (r,s) on a Lorentzian four-manifold and their components clauses (b)-(c)).

Counterexample

Let $h$ be the smooth $(0, 2)$ -tensor field on $M$ with constant components in the identity chart $h_{00} = h_{0 i} = h_{i 0} : = 0, h_{i j} : = δ_{i j} (i, j \in {1, 2, 3}),$ the Euclidean metric on the spacelike coordinate three-plane extended by zero, and let $R$ be the smooth $(0, 4)$ -tensor field with constant components $R_{a b c d} : = h_{a c} h_{b d} - h_{a d} h_{b c} .$ Both exist by [L3]. For a future timelike vector $T$ write $R (T, J) T$ for the vector defined in [L1].

1.1L2L3algebra

(The four algebraic identities.) Since $h_{a b} = h_{b a}$ , one has $R_{b a c d} = h_{b c} h_{a d} - h_{b d} h_{a c} = - R_{a b c d}$ and $R_{a b d c} = h_{a d} h_{b c} - h_{a c} h_{b d} = - R_{a b c d}$ ; and $R_{c d a b} = h_{c a} h_{d b} - h_{c b} h_{d a} = h_{a c} h_{b d} - h_{a d} h_{b c} = R_{a b c d}$ . Moreover $R_{a b c d} + R_{a c d b} + R_{a d b c} = (h_{a c} h_{b d} - h_{a d} h_{b c}) + (h_{a d} h_{c b} - h_{a b} h_{c d}) + (h_{a b} h_{d c} - h_{a c} h_{d b}) = 0$ , each of the three pairs cancelling by $h_{x y} = h_{y x}$ . Hence $R$ satisfies the four identities of [L2] at every point.

1.2L3algebra

(Nonvanishing.) $R_{1212} = h_{11} h_{22} - h_{12} h_{21} = 1 \cdot 1 - 0 = 1 \neq 0$ , so $R \neq 0$ .

1.3L1L3algebra

(The tidal operator vanishes for $T = \partial_{0}$ .) Let $T : = \partial_{0}$ , a future timelike vector with $g (T, T) = - 1$ . For all $J, W$ , by [L1] and the definition of $R$ , $g (R (T, J) T, W) = R_{a b c d} W^{a} T^{c} J^{d} T^{b} = R_{a 00 d} W^{a} J^{d}$ and $R_{a 00 d} = h_{a 0} h_{0 d} - h_{a d} h_{00} = 0 - 0 = 0$ , because all components of $h$ with an index $0$ vanish. Hence the pairing vanishes for all $J, W$ and, by nondegeneracy [L1], $R (T, J) T = 0$ for every $J \in T_{p} M$ .

2.1L1step 1.3algebra

(A tilted timelike direction detects the tensor.) Let $T^{'} : = \partial_{0} + \frac{1}{2} \partial_{1}$ ; then $g (T^{'}, T^{'}) = - 1 + \frac{1}{4} = - \frac{3}{4} < 0$ , so $T^{'}$ is timelike. Expanding in the constant components, $g (R (T^{'}, \partial_{3}) T^{'}, W) = R_{a b c d} W^{a} T^{' c} (\partial_{3})^{d} T^{' b} = \frac{1}{4} R_{a 113} W^{a}$ , with the remaining three terms $R_{a 003}$ , $R_{a 013}$ and $R_{a 103}$ vanishing: each of their products contains a factor $h_{0 b}$ or $h_{b 0}$ for some $b$ , and all such components of $h$ are zero. Now $R_{a 113} = h_{a 1} h_{13} - h_{a 3} h_{11} = - h_{a 3}$ , so $g (R (T^{'}, \partial_{3}) T^{'}, W) = - \frac{1}{4} W^{3}$ and, by [L1], $R (T^{'}, \partial_{3}) T^{'} = - \frac{1}{4} \partial_{3} \neq 0$ : the tidal operator of $R$ does not vanish for every timelike direction.

3.1step 1.1step 1.2step 1.3step 2.1L2algebra∎

(Conclusion.) By steps 1.1 and 1.2, $R$ is a nonzero algebraic curvature tensor; by step 1.3 there is a timelike vector, namely $\partial_{0}$ , for which the tidal operator vanishes identically; by step 2.1 the operator is nonzero for the timelike direction $T^{'}$ . Hence the displayed statement is refuted: the vanishing of the tidal operator for a single timelike direction does not force $R = 0$ , and the hypothesis "for every timelike $T$ " in A curvature tensor vanishes if the tidal operator vanishes for every timelike direction cannot be weakened to a single observer.

Remarks

Where the single oblique direction matters. For $T = \partial_{0}$ every slot of $R$ built from $T$ contains a factor $h_{\cdot 0} = 0$ , so the whole tidal pairing collapses; the tilt $T^{'} = \partial_{0} + \frac{1}{2} \partial_{1}$ gives the slot $h_{11} = 1$ a nonzero weight and produces the component $R (T^{'}, \partial_{3}) T^{'} = - \frac{1}{4} \partial_{3}$ . The example thus separates the hypothesis "one timelike direction" from the hypothesis "all timelike directions" used by the criterion.
Scope. The counterexample is algebraic and concerns the stated implication only; no realization of $R$ as the curvature of a metric-selected connection is claimed, and no empirical statement is made.

A nonzero algebraic curvature tensor whose tidal operator vanishes for one timelike direction

Statement refuted

Facts & Assumptions

Counterexample

Remarks

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