LemmaStatement: Literature-sourcedProof: AI-generatedSession-authored (Fable 5 assisted)precheck passjudge pass (gpt-5.6-terra)audited 2026-08-27

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Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
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Hyperbolic spaces have thin geodesic quadrilaterals

Statement

Let $X$ be a geodesic $δ$ -hyperbolic space. Then every geodesic quadrilateral in $X$ is $2 δ$ -thin: each side lies in the closed $2 δ$ -neighborhood of the union of the other three sides.

Facts & Assumptions

Given: A geodesic $δ$ -hyperbolic space $X$ and a geodesic quadrilateral with vertices $a, b, c, d$ .

[L1]

Every geodesic triangle in $X$ is $δ$ -slim (Delta-slim triangles and hyperbolic spaces).

[A1]

The diagonal $[a, c]$ cuts the quadrilateral into the geodesic triangles $a b c$ and $a c d$ .

Proof

technique · direct

1.1

By [A1], each point of the side $[a, b]$ lies either within distance $δ$ of $[a, c]$ or within distance $δ$ of $[b, c]$ by applying [L1] to triangle $a b c$ .

A1L1

2.1

If such a point lies near $[a, c]$ , then applying [L1] to triangle $a c d$ shows that the nearby point on $[a, c]$ lies within distance $δ$ of $[a, d] \cup [d, c]$ . Hence every point of $[a, b]$ lies within distance $2 δ$ of $[b, c] \cup [c, d] \cup [d, a]$ . Cyclic symmetry gives the same bound for each side, so the quadrilateral is $2 δ$ -thin.

L1step 1.1algebra∎

Depends on

Delta-slim triangles and hyperbolic spaces

Used by

Free abelian groups of rank at least two are not hyperbolic Proposition
Morse stability of quasi-geodesics Theorem

Dependency tree · two levels

2 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Clara Löh, Geometric Group Theory, Section 6.2.3 (standard reference, not scraped)