Manuscript I-B-G@2 · I-B-G-v2.tex · section · 3b4d700f

Angular lifts, geometric terms, and apex length

The prescribed host and its causal geometry › Angular lifts, geometric terms, and apex length · lines 378–420

I-B-G@2#eq:lengths, I-B-G@2#eq:ordered

Choose lifts of $φ, φ^{'}$ to $R$ , and set

ϑ = α (φ - φ^{'}) .

Unlike $ψ$ , $ϑ$ is not reduced modulo $2 π α$ . Put

K (ϑ) R_{k} (ϑ)^{2} ρ (u)^{2} ρ_{0} = {k \in Z : ∣ ϑ - 2 π α k ∣ < π}, = r^{2} + r^{'2} - 2 r r^{'} cos (ϑ - 2 π α k) + Δ z^{2}, = r^{2} + r^{'2} + 2 r r^{'} cosh u + Δ z^{2}, u \geq 0, = ρ (0) = (r + r^{'})^{2} + Δ z^{2} .

The pair is called generic if

∣ ϑ - 2 π α k ∣ \neq = π for every k \in Z .

These excluded loci are the geometric image boundaries.

If $ϑ = ψ$ , then $0 \in K (ψ)$ , and

d (x, x^{'}) = R_{0} \leq R_{k} < ρ_{0} (k \in K (ψ)) .

At the angular cut $∣ ψ ∣ = π α$ , two shortest geometric terms can have equal length. Thus it is not correct globally to assert $R_{k} > R_{0}$ for every $k \neq = 0$ . The strict inequality does hold for these additional terms in an angularly small chart defined in Section sec:transfer.

Displayed formulas in this passage

I-B-G@2#eq:lengths · line 386
$K (ϑ) R_{k} (ϑ)^{2} ρ (u)^{2} ρ_{0} = {k \in Z : ∣ ϑ - 2 π α k ∣ < π}, = r^{2} + r^{'2} - 2 r r^{'} cos (ϑ - 2 π α k) + Δ z^{2}, = r^{2} + r^{'2} + 2 r r^{'} cosh u + Δ z^{2}, u \geq 0, = ρ (0) = (r + r^{'})^{2} + Δ z^{2} .$
I-B-G@2#eq:ordered · line 409
$d (x, x^{'}) = R_{0} \leq R_{k} < ρ_{0} (k \in K (ψ)) .$
unlabeled · line 381
$ϑ = α (φ - φ^{'}) .$
unlabeled · line 402
$∣ ϑ - 2 π α k ∣ \neq = π for every k \in Z .$