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The half-trace subtraction with the rank-three spatial projector is not trace-free

Statement refuted

Statement refuted: "Let $n$ be an observer and $s$ a propagation direction of the flat model, let $h^{a}_{b} = δ^{a}_{b} + n^{a} n_{b}$ be the spatial projector of clause (a) of The wave screen of an observer and propagation direction, and the transverse-traceless projection of a symmetric two-tensor, let $h_{a b} = η_{a c} h^{c}_{b}$ , and let $A$ be any symmetric covariant $2$ -tensor with $A (n, \cdot) = 0$ . Then the tensor

$A^{T T} : = h_{a}^{c} h_{b}^{d} A_{c d} - \frac{1}{2} h_{a b} h^{c d} A_{c d}$

is transverse-traceless: $A^{T T} (n, \cdot) = 0$ , $A^{T T} (s, \cdot) = 0$ and $η^{a b} A_{a b}^{T T} = 0$ ."

Facts & Assumptions

Given: The flat model with inertial chart, the observer $n = \partial_{0}$ and propagation direction $s = \partial_{3}$ , the spatial projector $h^{a}_{b} = δ^{a}_{b} + n^{a} n_{b}$ and the symmetric spatial tensor $A = diag (0, 1, 1, 0)$ .

[L1]

In the basis $(e_{0}, e_{1}, e_{2}, e_{3})$ with $e_{0} = n$ , $e_{3} = s$ , the spatial projector of $n$ has components $h^{a}_{b} = diag (0, 1, 1, 1)$ and $h_{a b} = η_{a c} h^{c}_{b} = diag (0, 1, 1, 1)$ ; it fixes every spatial tensor, $h_{a}^{c} h_{b}^{d} A_{c d} = A_{a b}$ whenever $A (n, \cdot) = 0$ (The wave screen of an observer and propagation direction, and the transverse-traceless projection of a symmetric two-tensor clause (a), The screen and spatial projectors are orthogonal projections of rank two and three, and the TT projection is the orthogonal projection onto the two-dimensional transverse-traceless space clause (a), Observers, rest spaces and the spatial projection at an event, The inverse metric, musical isomorphisms and index raising).

[L2]

For the rank-two screen projector $p = diag (0, 1, 1, 0)$ of the same pair, $p_{a}^{c} p_{b}^{d} A_{c d} = A$ for the tensor $A$ above, $p^{c d} A_{c d} = A_{11} + A_{22} = 2$ and the screen TT projection is $Π_{n, s} (A) = A - \frac{1}{2} \cdot 2 p = diag (0, 1, 1, 0) - diag (0, 1, 1, 0) = 0$ (The screen and spatial projectors are orthogonal projections of rank two and three, and the TT projection is the orthogonal projection onto the two-dimensional transverse-traceless space clauses (b)-(d), The wave screen of an observer and propagation direction, and the transverse-traceless projection of a symmetric two-tensor clause (b)).

Counterexample

Work in the inertial chart of the flat model with $n = \partial_{0}$ , $s = \partial_{3}$ , and take the symmetric spatial tensor

$(A_{a b}) = diag (0, 1, 1, 0)$

(the screen projector of the pair $(n, s)$ ; in particular $A (n, \cdot) = 0$ ). Then

$A^{T T} = diag (0, 0, 0, - 1),$

which is spatial, $A^{T T} (n, \cdot) = 0$ , but is not transverse to $s$ , since $A^{T T} (s, s) = - 1$ , and has trace $η^{a b} A_{a b}^{T T} = η^{33} (- 1) = - 1 \neq 0$ ; so it is not trace-free.

1.1L1algebra

(The "half-trace" tensor.) By [L1], $h_{a}^{c} h_{b}^{d} A_{c d} = A$ because $A$ is spatial, and $h^{c d} A_{c d} = A_{11} + A_{22} + A_{33} = 1 + 1 + 0 = 2$ ; hence $A^{T T} = A - \frac{1}{2} \cdot 2 h = diag (0, 1, 1, 0) - diag (0, 1, 1, 1) = diag (0, 0, 0, - 1) .$

2.1step 1.1L2algebra∎

(Failure of trace-freeness and the rank-two contrast.) The tensor $A^{T T} = diag (0, 0, 0, - 1)$ kills $n = \partial_{0}$ , but $A^{T T} (s, s) = - 1$ for $s = \partial_{3}$ , so it is not propagation-transverse; its trace is $η^{a b} A_{a b}^{T T} = η^{33} (- 1) = - 1 \neq 0$ , so the asserted conclusion fails. By contrast, [L2] computes that the same half-trace formula with the rank-two screen projector gives $Π_{n, s} (A) = 0$ , which is trace-free: the failure is exactly the rank-three projector weight. This refutes the statement.

Remarks

The rank-two repair. The contrast computed in step 2.1 shows which weight is correct: the same half-trace formula is trace-free for every input with the rank-two screen projector $p$ , while the rank-three spatial projector fails it for the displayed input. The general identity is $tr (q_{a}^{c} q_{b}^{d} A_{c d} - \frac{1}{2} q_{a b} q^{c d} A_{c d}) = (1 - \frac{r}{2}) q^{c d} A_{c d}$ for an $η$ -self-adjoint idempotent $q$ of rank $r$ . This trace vanishes when $r = 2$ or $q^{c d} A_{c d} = 0$ ; for nonzero input trace it vanishes exactly when $r = 2$ ; this is why the half weight belongs to the two-dimensional screen and not to the three-dimensional rest space (The screen and spatial projectors are orthogonal projections of rank two and three, and the TT projection is the orthogonal projection onto the two-dimensional transverse-traceless space remark).
Scope. The counterexample concerns the algebraic projector formula in the flat model; it makes no statement about metric perturbations, field equations or observations.

The half-trace subtraction with the rank-three spatial projector is not trace-free

Statement refuted

Facts & Assumptions

Counterexample

Remarks

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