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Counterexample: the reversed Cauchy-Schwarz inequality fails outside same-orientation timelike pairs

Statement refuted

The following universal claim is false. Let $(A, V, η, C^{+})$ be the oriented time-oriented affine Minkowski space.

(U1) The reversed Cauchy-Schwarz inequality $- η (v, w) \geq \sqrt{- η (v, v)} \sqrt{- η (w, w)}$ holds for every pair of vectors $v, w \in V$ that are individually timelike, regardless of their time orientations.

(U2) The analogous reversed form $∣ η (v, w) ∣ \geq \sqrt{η (v, v)} \sqrt{η (w, w)}$ holds for every pair of spacelike vectors $v, w \in V$ .

First witness, refuting (U1): in the inertial chart with diagonal metric $diag (- 1, 1, 1, 1)$ take $v : = (5, 3, 0, 0)$ , which is future timelike, and $w : = (- 5, 3, 0, 0)$ , which is past timelike. Then $η (v, w) = 34 > 0$ , so $- η (v, w) = - 34$ , while $\sqrt{- η (v, v)} \sqrt{- η (w, w)} = 4 \cdot 4 = 16$ , and the asserted inequality $- 34 \geq 16$ is false; the sum $v + w = (0, 6, 0, 0)$ is spacelike. Second witness, refuting (U2): $v : = (0, 1, 0, 0)$ and $w : = (0, 1, 1, 0)$ are spacelike with $η (v, v) = 1$ , $η (w, w) = 2$ , $η (v, w) = 1$ , so $∣ η (v, w) ∣ = 1 < 1 \cdot \sqrt{2} = \sqrt{η (v, v)} \sqrt{η (w, w)}$ . Hence the same-orientation hypothesis of The reversed Cauchy-Schwarz inequality for future timelike vectors cannot be dropped, and the spacelike reversed inequality is not universal.

Facts & Assumptions

Given: The oriented time-oriented affine Minkowski space $(A, V, η, C^{+})$ and an inertial frame $(o; e_{0}, e_{1}, e_{2}, e_{3})$ with chart $x$ and coordinates $(u^{0}, u^{1}, u^{2}, u^{3})$ of a vector $u = u^{0} e_{0} + u^{1} e_{1} + u^{2} e_{2} + u^{3} e_{3}$ (Inertial frames and inertial charts of the affine Minkowski model).

[L1]

Coordinates and pairing: in the frame's ordered basis $η (e_{a}, e_{b}) = diag (- 1, 1, 1, 1)_{a b}$ , and by bilinearity $η (u, u^{'}) = - u^{0} u^{' 0} + u^{1} u^{' 1} + u^{2} u^{' 2} + u^{3} u^{' 3}$ for all $u, u^{'} \in V$ (Inertial frames and inertial charts of the affine Minkowski model, Lorentz forms of inertia (3,1,0) on a 4-dimensional real vector space).

[L2]

Cone coordinates: $T_{+} = {u : u^{0} > ∣ u_{s} ∣}$ and $T_{-} = {u : u^{0} < - ∣ u_{s} ∣}$ are exactly the two components of the timelike set, and in a frame with $e_{0} \in C^{+}$ one has $C^{+} = T_{+}$ , $C^{-} = T_{-}$ ; hence $u$ is future timelike exactly when $u^{0} > ∣ u_{s} ∣$ and past timelike exactly when $u^{0} < - ∣ u_{s} ∣$ , where $∣ u_{s} ∣ = \sqrt{(u^{1})^{2} + (u^{2})^{2} + (u^{3})^{2}}$ (The timelike set has exactly two convex components, described in orthonormal coordinates, Time orientation, future and past timelike/null/causal vectors, and the closed cones).

[L3]

Causal classification: $u$ is timelike when $η (u, u) < 0$ , spacelike when $η (u, u) > 0$ (Timelike, null, spacelike and causal vectors of a Lorentz vector space).

[L4]

The reversed Cauchy-Schwarz inequality and its equality case: for future timelike $v, w$ with $a : = \sqrt{- η (v, v)}$ , $b : = \sqrt{- η (w, w)}$ one has $- η (v, w) \geq a b$ , with equality exactly for positive proportional pairs; the same inequality holds for past timelike pairs (The reversed Cauchy-Schwarz inequality for future timelike vectors).

[L5]

Square roots and monotonicity of squaring: every real $c \geq 0$ has a unique $\sqrt{c} \geq 0$ with $(\sqrt{c})^{2} = c$ , in particular $\sqrt{1} = 1$ ; for $x, y \geq 0$ one has $x < y ⟺ x^{2} < y^{2}$ (Square roots exist: a unique $\sqrt{a} \geq 0$ with $(\sqrt{a})^{2} = a$ ; the positives are ${x^{2} : x \neq 0}$ , Squaring is monotone on the nonnegatives).

[L6]

Order of $R$ : the order is transitive, so $x < y < z$ implies $x < z$ (Ordered field).

Counterexample

technique · direct

1.1givenL1algebra

In the given inertial frame evaluate the four vectors of the two witnesses: $η (v, v) = - 25 + 9 = - 16 < 0$ and $η (w, w) = - 25 + 9 = - 16 < 0$ for $v = (5, 3, 0, 0)$ and $w = (- 5, 3, 0, 0)$ ; and $η (v, w) = 25 + 9 = 34$ for this pair; for the second pair $v^{'} = (0, 1, 0, 0)$ , $w^{'} = (0, 1, 1, 0)$ one has $η (v^{'}, v^{'}) = 1 > 0$ , $η (w^{'}, w^{'}) = 1 + 1 = 2 > 0$ and $η (v^{'}, w^{'}) = 1$ .

2.1step 1.1L2L3

Both $v$ and $w$ are timelike by [L3]; since $v^{0} = 5 > 3 = ∣ v_{s} ∣$ and $w^{0} = - 5 < - 3 = - ∣ w_{s} ∣$ , [L2] identifies $v$ as future timelike and $w$ as past timelike, so the pair $(v, w)$ is individually timelike with opposite time orientations.

2.2step 1.1L3L5algebra

For the second pair, [L3] makes $v^{'}$ and $w^{'}$ spacelike, and $\sqrt{η (v^{'}, v^{'})} \sqrt{η (w^{'}, w^{'})} = \sqrt{1} \cdot \sqrt{2} = 1 \cdot \sqrt{2} = \sqrt{2}$ by [L5]; since $1 < 2$ and $1, \sqrt{2} \geq 0$ , strict monotonicity of squaring gives $\sqrt{2} > 1$ ; hence $∣ η (v^{'}, w^{'}) ∣ = 1 < \sqrt{2} = \sqrt{η (v^{'}, v^{'})} \sqrt{η (w^{'}, w^{'})}$ , so (U2) is refuted.

2.3step 1.1L1algebra

The span of the second pair is ${(0, s, t, 0) : s, t \in R}$ , on which $η ((0, s, t, 0), (0, s, t, 0)) = s^{2} + t^{2}$ vanishes only for $s = t = 0$ : the subspace spanned by the two spacelike witnesses contains no null direction, so the failure of (U2) is not an artifact of a null direction in the span.

3.1step 1.1step 2.1L1L2L3algebra

The sum $v + w = (0, 6, 0, 0)$ has $η (v + w, v + w) = 36 > 0$ , so it is spacelike by [L3]. This is an additional property of this witness, not a criterion for failure of (U1): the pairing in (U1) and the causal character of the sum are distinct claims. Closure of the future timelike cone under addition does not apply here because $w$ is past timelike.

3.2step 1.1step 2.1L5L6algebra

For this pair $\sqrt{- η (v, v)} = \sqrt{16} = 4 = \sqrt{- η (w, w)}$ , so the right-hand side of (U1) is $16$ , while the left-hand side is $- η (v, w) = - 34 < 0$ ; since $0 < 16$ , transitivity gives $- 34 < 16$ , so $- 34 \geq 16$ is false and (U1) is refuted.

4.1step 3.1step 2.2step 2.3step 3.2L4∎

The two witnesses therefore refute the claimed universality: the reversed Cauchy-Schwarz inequality requires the same (future) time orientation of the pair, and the spacelike reversed form fails even on a spacelike plane with no null direction; neither witness contradicts the positive results [L4], which are stated for same-orientation timelike pairs.

Counterexample: the reversed Cauchy-Schwarz inequality fails outside same-orientation timelike pairs

Statement refuted

Facts & Assumptions

Counterexample

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