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✓ 7 results · all verified · 4 also independently AI-judged

Every result on this page is machine-checked by a proof checker and read in full and owner-audited; the judge is an additional, independent cross-model AI review of the proofs. The 3 not AI-judged were verified by owner audit (typically over a confirmed judge false positive), not failures.

Hyperbolic Spaces and Hyperbolic Groups — Examples

1 · Prerequisites

2 · Summary

These examples and counterexamples anchor the page’s main geometry: trees and the hyperbolic plane as model spaces, free and surface groups as standard hyperbolic groups, a concrete small-cancellation witness, and the square grid as the canonical obstruction to hyperbolicity.

3 · Logical flowchart

4 · Definitions, theorems and proofs

None yet.

5 · Examples, counterexamples and false statements

ExampleConstruction: Literature-sourcedVerification: AI-generatedprecheck passaudited 2026-08-27Open item page →

Every tree is 0-hyperbolic

Example

Every tree is $0$ -hyperbolic.

Facts & Assumptions

Given: A tree $T$ .

[L1]

Cayley trees are $0$ -hyperbolic (Cayley trees are 0-hyperbolic).

Verification

technique · direct

1.1

The proposition [L1] states exactly that every tree is $0$ -hyperbolic.

2.1

Therefore $T$ is a model example of a hyperbolic space with the best possible constant $δ = 0$ .

step 1.1∎

ExampleConstruction: Literature-sourcedVerification: AI-adaptedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-27Open item page →

The hyperbolic plane is hyperbolic

Example

The hyperbolic plane $H^{2}$ is a hyperbolic geodesic metric space.

Facts & Assumptions

Given: The standard geodesic metric on $H^{2}$ .

[A1]

Classical hyperbolic geometry gives a uniform slimness constant for geodesic triangles in $H^{2}$ .

[L1]

A geodesic metric space is hyperbolic exactly when all geodesic triangles are $δ$ -slim for some $δ \geq 0$ (Delta-slim triangles and hyperbolic spaces).

Verification

technique · direct

1.1

By [A1], geodesic triangles in $H^{2}$ are uniformly slim.

givenA1

2.1

Therefore [L1] shows that $H^{2}$ is hyperbolic.

L1step 1.1∎

ExampleConstruction: Literature-sourcedVerification: AI-adaptedprecheck passaudited 2026-08-27Open item page →

Free groups have Cantor-set boundaries

Example

If $F_{r}$ is a free group of rank $r \geq 2$ , then its Gromov boundary is a Cantor set.

Facts & Assumptions

Given: A free group $F_{r}$ of rank $r \geq 2$ .

[L1]

Free groups are hyperbolic (Finite groups and free groups are hyperbolic).

[A1]

The boundary of a regular tree of valence at least $3$ is homeomorphic to a Cantor set.

Verification

technique · direct

1.1

By [L1], the Cayley graph of $F_{r}$ with respect to a free basis is a hyperbolic tree.

2.1

Because $r \geq 2$ , that tree has valence at least $3$ , so [A1] identifies its boundary with a Cantor set. Hence the boundary of $F_{r}$ is a Cantor set.

A1step 1.1∎

ExampleConstruction: Literature-sourcedVerification: AI-generatedprecheck passaudited 2026-08-27Open item page →

Closed surface groups are hyperbolic

Example

The fundamental group of a closed hyperbolic surface is a hyperbolic group.

Facts & Assumptions

Given: A closed hyperbolic surface $Σ$ and its fundamental group $π_{1} (Σ)$ .

[L1]

The hyperbolic plane is hyperbolic (The hyperbolic plane is hyperbolic).

[L2]

The Švarc-Milnor lemma transfers geometric actions on proper geodesic spaces to quasi-isometries with finitely generated groups (The Svarc-Milnor lemma).

[L3]

Hyperbolicity is invariant under quasi-isometry (Hyperbolicity is a quasi-isometry invariant of geodesic spaces).

Verification

technique · direct

1.1

The group $π_{1} (Σ)$ acts properly discontinuously and cocompactly by deck transformations on the universal cover $H^{2}$ of $Σ$ .

given

2.1

By [L2], $π_{1} (Σ)$ is quasi-isometric to $H^{2}$ , and [L1] shows that $H^{2}$ is hyperbolic. Therefore [L3] makes $π_{1} (Σ)$ hyperbolic.

L1L2L3step 1.1∎

ExampleConstruction: Literature-sourcedVerification: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-27Open item page →

A small-cancellation presentation gives a hyperbolic group

Example

Consider the one-relator presentation

G = ⟨ x_{1}, x_{2}, x_{3}, x_{4}, x_{5}, x_{6}, x_{7} ∣ x_{1} x_{2} x_{3} x_{4} x_{5} x_{6} x_{7} ⟩ .

This is a finite $C^{'} (1 / 6)$ presentation and hence defines a hyperbolic group.

Facts & Assumptions

Given: The displayed presentation of $G$ .

[A1]

In the single relator $x_{1} x_{2} x_{3} x_{4} x_{5} x_{6} x_{7}$ , no nonempty subword occurs as an initial segment of two distinct cyclic conjugates or inverse cyclic conjugates, so the symmetrized presentation has no nontrivial pieces and therefore satisfies $C^{'} (1 / 6)$ vacuously.

[L1]

Finite $C^{'} (1 / 6)$ presentations define hyperbolic groups (Finite C'(1/6) presentations define hyperbolic groups).

Verification

technique · direct

1.1

By [A1], the displayed finite presentation satisfies the $C^{'} (1 / 6)$ condition.

givenA1

2.1

Therefore [L1] shows that the presented group $G$ is hyperbolic.

L1step 1.1∎

CounterexampleConstruction: Literature-sourcedVerification: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-27Open item page →

Z^2 is not hyperbolic

Statement refuted

The page proves that not every finitely generated group is hyperbolic; the standard witness is $Z^{2}$ .

Facts & Assumptions

Given: The free abelian group $Z^{2}$ .

[L1]

Free abelian groups of rank at least two are not hyperbolic (Free abelian groups of rank at least two are not hyperbolic).

Counterexample

technique · direct

1.1

The group $Z^{2}$ is free abelian of rank $2$ .

given

2.1

Therefore [L1] shows that $Z^{2}$ is not hyperbolic, providing the required counterexample.

L1step 1.1∎

CounterexampleConstruction: Literature-sourcedVerification: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-27Open item page →

A product of two infinite groups need not be hyperbolic

Statement refuted

The product of two infinite groups need not be hyperbolic.

Facts & Assumptions

Given: The direct product $Z \times Z$ .

[L1]

Free abelian groups of rank at least two are not hyperbolic (Free abelian groups of rank at least two are not hyperbolic).

Counterexample

technique · direct

1.1

The group $Z \times Z$ is a direct product of two infinite groups and is free abelian of rank $2$ .

given

2.1

Therefore [L1] shows that $Z \times Z$ is not hyperbolic. So the general claim fails.

L1step 1.1∎

Hyperbolic Spaces and Hyperbolic Groups — Examples

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, theorems and proofs

5 · Examples, counterexamples and false statements

Every tree is 0-hyperbolic

Example

Facts & Assumptions

Verification

The hyperbolic plane is hyperbolic

Example

Facts & Assumptions

Verification

Free groups have Cantor-set boundaries

Example

Facts & Assumptions

Verification

Closed surface groups are hyperbolic

Example

Facts & Assumptions

Verification

A small-cancellation presentation gives a hyperbolic group

Example

Facts & Assumptions

Verification

Z^2 is not hyperbolic

Statement refuted

Facts & Assumptions

Counterexample

A product of two infinite groups need not be hyperbolic

Statement refuted

Facts & Assumptions

Counterexample

Sources