Orthochronous Lorentz group

The orthochronous Lorentz group $𝐿 ↑ = O ↑ (3, 1)$ is the group of all Lorentz transformations preserving the direction of time, #m/def/group/lorentz i.e.

O ↑ (3, 1) = {Λ \in O (3, 1) : Λ 0 0 \geq 1}

since $∣ Λ 0 0 ∣ < 1$ is forbidden (see discussion below), this consists of all Lorentz transformations with positive $Λ 0 0$ .

Discussion

First note that by definition for all $(Λ 𝜇 𝜈) \in 𝐿$ ,¹

Λ 𝜇 𝜅 𝜂 𝜇 𝜈 Λ 𝜈 𝜆 = 𝜂 𝜅 𝜆

and taking adjoints

Λ 𝜅 𝜇 𝜂 𝜇 𝜈 Λ 𝜆 𝜈 = 𝜂 𝜅 𝜆

the $00$ component of which yield

- (Λ 0 0) 2 + Λ 𝑖 0 Λ 𝑖 0 = Λ 𝜇 0 𝜂 𝜇 𝜈 Λ 𝜈 0 = 𝜂 00 = - 1 - (Λ 0 0) 2 + Λ 0 𝑖 Λ 0 𝑖 = Λ 0 𝜇 𝜂 𝜇 𝜈 Λ 0 𝜈 = 𝜂 00 = - 1

implying the properties

(Λ 0 0) 2 = 1 + Λ 𝑖 0 𝛿 𝑖 𝑗 Λ 𝑗 0 \geq 1 = 1 + Λ 0 𝑖 𝛿 𝑖 𝑗 Λ 0 𝑗 \geq 1

the former yields equality iff $Λ 𝑖 0 = 0$ for $𝑖 = 1, 2, 3$ the latter yields equality iff $Λ 0 𝑖 = 0$ for each $𝑖$ . Thus in particular $Λ 0 0 = 1$ iff $Λ 0 𝑖 = Λ 𝑖 0 = 0$ for each $𝑖$ , in which case $(Λ 𝑖 𝑗) \in O (3)$ . Now since $(Λ 0 0) 2 \geq 1$ , $Λ 0 0 \in ℝ ∖ (- 1, 1)$ which splits $𝐿$ into disconnected components based on the sign of $Λ 0 0$ .

Proof of defining property

Let $Λ \in O ↑ (3, 1)$ . Consider a timelike vector $𝑣 \in ℝ 4$ , i.e. $𝜂 𝜇 𝜈 𝑣 𝜇 𝑣 𝜈 < 0$ implying $𝑣 0 𝑣 0 > 𝑣 𝑖 𝑣 𝑖$ . Now if $𝑣 0 > 0$ , then

Λ 0 𝜇 𝑣 𝜇 = Λ 0 0 𝑣 0 + Λ 0 𝑖 𝑣 𝑖 \geq Λ 0 0 𝑣 0 - \sqrt Λ 0 𝑖 Λ 0 𝑖 \sqrt 𝑣 𝑖 𝑣 𝑖 = Λ 0 0 𝑣 0 - \sqrt (Λ 0 0) 2 + 1 \sqrt 𝑣 𝑖 𝑣 𝑖 > Λ 0 0 𝑣 0 - Λ 0 0 𝑣 0 = 0

Now let $Λ' \in O ↓ (3, 1)$ , and $𝑣$ be timelike with $𝑣 0 < 0$ . It follows $- 𝑣$ is timelike with $𝑣 0 > 0$ and $- Λ' \in O ↑ (3, 1)$ , thus

Λ 0 𝜇 (- 1) 𝑣' 𝜇 > 0 ⟹ Λ 0 𝜇 𝑣' 𝜇 < 0 (- 1) Λ' 0 𝜇 𝑣 𝜇 > 0 ⟹ Λ' 0 𝜇 𝑣 𝜇 < 0 (- 1) Λ' 0 𝜇 (- 1) 𝑣' 𝜇 > 0 ⟹ Λ' 0 𝜇 𝑣' 𝜇 > 0

Hence an arbitrary $Λ \in O (3, 1)$ preserves the direction of time iff $Λ 0 0 \geq 1$ and reverses the direction of time iff $Λ 0 0 \leq - 1$ .

From this it immediately follows that $O ↑ (3, 1)$ is a group, since if $Λ$ and $Λ'$ preserve the direction of time so too must $Λ Λ'$ and $Λ - 1$ . Furthermore it follows that $O ↑ (3, 1)$ is a Normal subgroup since for any $Λ \in O ↑ (3, 1)$ and $Λ' \in O ↓ (3, 1)$ we the result of $Λ' - 1 Λ Λ'$ will preserve the direction of time.

Alternate proof of normal subgroup

Define $𝜑 (Λ) = s g n (Λ 0 0)$ . Let $Λ, Λ' \in O (3, 1)$ . If $𝜑 (Λ) 𝜑 (Λ') = 1$ then applying the inequalities in ^eq1 and the Cauchy-Schwarz inequality

(Λ Λ') 0 0 = Λ 0 𝜇 Λ' 𝜇 0 = Λ 0 0 Λ' 0 0 + Λ 0 𝑖 Λ' 𝑖 0 \geq Λ 0 0 Λ' 0 0 - ∣ Λ 0 𝑖 Λ 𝑖 0 ∣ = ∣ Λ 0 0 ∣ ∣ Λ' 0 0 ∣ - ∣ Λ 0 𝑖 Λ 𝑖 0 ∣ > \sqrt Λ 0 𝑖 Λ 0 𝑖 \sqrt Λ 𝑖 0 Λ 𝑖 0 - ∣ Λ 0 𝑖 Λ 𝑖 0 ∣ \geq 0

and indeed $(Λ Λ') 0 0 > 0$ since otherwise it would have null determinant. Similarly if $𝜎 = - 1$

(Λ Λ') 0 0 = Λ 0 𝜇 Λ' 𝜇 0 = Λ 0 0 Λ' 0 0 + Λ 0 𝑖 Λ' 𝑖 0 \leq Λ 0 0 Λ' 0 0 + ∣ Λ 0 𝑖 Λ 𝑖 0 ∣ = - ∣ Λ 0 0 ∣ ∣ Λ' 0 0 ∣ + ∣ Λ 0 𝑖 Λ 𝑖 0 ∣ < \sqrt Λ 0 𝑖 Λ 0 𝑖 \sqrt Λ 𝑖 0 Λ 𝑖 0 - ∣ Λ 0 𝑖 Λ 𝑖 0 ∣ \leq 0

and by the same $(Λ Λ') 0 0 < 0$ since otherwise it would have null determinant. Hence $𝜑 : O (3, 1) \to ℤ 2$ is a group homomorphism and its kernel $k e r 𝜑 = O ↑ (3, 1)$ is a normal subgroup.

#state/tidy | #lang/en | #SemBr

2018. From the Lorentz Group to the Celestial Sphere, §1.3, p. 9 ↩

Orthochronous Lorentz group

Footnotes