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The parity and time-reversal maps are Lorentz maps that are not proper, and det = 1 does not imply orthochrony

Statement refuted

The statements "every Lorentz map is proper and orthochronous" and, separately, "every Lorentz map with $\det L = 1$ is orthochronous" are false.

Facts & Assumptions

Given: An inertial frame $(o; e_{0}, e_{1}, e_{2}, e_{3})$ with $η_{0} = diag (- 1, 1, 1, 1)$ , and the coordinate columns $x = (x^{0}, x^{1}, x^{2}, x^{3})$ of a vector in that frame.

[L1]

A linear map $L$ is a Lorentz map exactly when its matrix $M$ satisfies $M^{T} η_{0} M = η_{0}$ in the frame's ordered basis; $L$ is proper when $\det L = 1$ and orthochronous when $L (C^{+}) = C^{+}$ (Linear and affine Lorentz maps, and the Lorentz group).

[L2]

A Lorentz map satisfies $\det L = \pm 1$ ; $L (C^{+})$ is one of the two components $C^{+}, C^{-}$ ; and $L (C^{+}) = C^{+}$ if and only if $L (C^{+}) \subseteq C^{+}$ (Lorentz maps are isomorphisms of determinant ±1 that preserve causal character and permute the cones; the Lorentz sets are groups).

[L3]

Since $e_{0} \in C^{+}$ in an inertial frame, the components are described in the frame's coordinates by $C^{+} = {x : x^{0} > ∣ x_{v e c} ∣}$ and $C^{-} = {x : x^{0} < - ∣ x_{v e c} ∣}$ (Inertial frames and inertial charts of the affine Minkowski model, Time orientation, future and past timelike/null/causal vectors, and the closed cones).

[L4]

A vector is timelike when $η (x, x) < 0$ , and $η (x, x) = - (x^{0})^{2} + ∣ x_{v e c} ∣^{2}$ in an $η$ -orthonormal basis (Timelike, null, spacelike and causal vectors of a Lorentz vector space).

[L5]

The determinant of a diagonal matrix is the product of its diagonal entries (The determinant of a triangular matrix is the product of its diagonal entries), and the matrix product of two diagonal matrices is the diagonal matrix of the products of the corresponding entries (Rectangular matrix multiplication and the identity matrix $I_{n}$ , including zero-sized shapes).

Counterexample

1.1L1L5algebra

Define $Π$ and $T$ by their matrices in the frame basis, $Π = diag (1, - 1, - 1, - 1)$ and $T = diag (- 1, 1, 1, 1)$ . For any diagonal $M = diag (m_{0}, m_{1}, m_{2}, m_{3})$ , $M^{T} η_{0} M = diag (- m_{0}^{2}, m_{1}^{2}, m_{2}^{2}, m_{3}^{2})$ by [L5]; for $Π$ and for $T$ each $m_{a}^{2} = 1$ , so both matrices equal $η_{0} = diag (- 1, 1, 1, 1)$ . Hence $Π$ and $T$ are Lorentz maps by the matrix criterion [L1].

1.2L1L5algebra

By the diagonal product rule [L5], $\det Π = 1 \cdot (- 1)^{3} = - 1$ and $\det T = - 1$ , so by [L1] neither $Π$ nor $T$ is proper: they are Lorentz maps failing properness.

1.3L2L3algebra

For every $x$ , $(Π x)^{0} = x^{0}$ and $(Π x)_{v e c} = - x_{v e c}$ , so $∣ (Π x)_{v e c} ∣ = ∣ x_{v e c} ∣$ ; if $x \in C^{+}$ then $x^{0} > ∣ x_{v e c} ∣$ by [L3], hence $(Π x)^{0} = x^{0} > ∣ (Π x)_{v e c} ∣$ and $Π x \in C^{+}$ by [L3] again. Thus $Π (C^{+}) \subseteq C^{+}$ , and [L2] gives $Π (C^{+}) = C^{+}$ : the improper map $Π$ is orthochronous, so $Π$ refutes "orthochronous implies proper".

1.4L2L3L4algebra

For every future timelike $x$ , $(T x)^{0} = - x^{0} < 0$ and $∣ (T x)_{v e c} ∣ = ∣ x_{v e c} ∣ < x^{0} = - (T x)^{0}$ , so $T x \in C^{-}$ by [L3]; since $T$ is a bijection with inverse $T$ , $T (C^{+}) = C^{-}$ and $T$ is not orthochronous.

1.5L1L2L3L5algebra

By the diagonal product rule [L5], $Π T = diag (- 1, - 1, - 1, - 1) = - I_{4}$ , whose determinant is $(- 1)^{4} = + 1$ ; and for every $x \in C^{+}$ one has $x^{0} > ∣ x_{v e c} ∣ \geq 0$ , so $(Π T) x = - x$ has $((Π T) x)^{0} = - x^{0} < 0$ and $∣ ((Π T) x)_{v e c} ∣ = ∣ x_{v e c} ∣ < x^{0} = - ((Π T) x)^{0}$ , whence $(Π T) x \in C^{-}$ by [L3] and $(Π T) (C^{+}) \subseteq C^{-}$ ; since $Π T$ is a bijection, $Π T (C^{+}) = C^{-}$ by [L2]. Hence $\det (Π T) = + 1$ and $Π T$ is not orthochronous: $\det L = 1$ does not imply orthochrony, and this proper non-orthochronous map refutes "proper implies orthochronous".

2.1step 1.1step 1.2step 1.3step 1.4step 1.5L1L2∎

Steps 1.1 and 1.2 exhibit Lorentz maps that are not proper ( $Π$ and $T$ ), step 1.3 exhibits a Lorentz map that is not proper yet is orthochronous, and steps 1.4 and 1.5 exhibit Lorentz maps that are not orthochronous, including one, $Π T$ , with $\det = + 1$ . The identity supplies the proper orthochronous sign class, so together with $Π, T, Π T$ all four classes are realized. Therefore $O (V, η)$ strictly contains $S O^{+} (V, η)$ , the four combinations of the sign of the determinant and orthochrony are genuinely distinct, and the convention of Linear and affine Lorentz maps, and the Lorentz group(e), which reserves "Lorentz transformation" for the proper orthochronous maps, is not automatic. Both refuted statements are false.

The parity and time-reversal maps are Lorentz maps that are not proper, and det = 1 does not imply orthochrony

Statement refuted

Facts & Assumptions

Counterexample

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