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Math-1791 · mathematics · Group theory

Questions in Geometric Group Theory — Q 1.15

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Statement

Definitions and mathematical statement

Choose a finite symmetric generating set $S$ for a group $G$. Give its Cayley graph unit-length edges and the induced path metric. The group is word-hyperbolic if there is a number $\delta\geq0$ such that each side of every geodesic triangle lies within distance $\delta$ of the other two sides. Changing the finite generating set does not change whether this property holds.

A group is residually finite if each nonidentity element survives in some finite quotient: \[ \forall g\in G\setminus\{1\}\ \exists Q\text{ finite}\ \exists\phi:G\to Q\text{ a homomorphism},\quad\phi(g)\ne1. \] The target may be replaced by the image of $\phi$, making the homomorphism surjective. Equivalently, the intersection of all finite-index normal subgroups of $G$ is trivial.

Is every word-hyperbolic group residually finite? The finite quotient and its cardinality may depend on the element being separated; the question requests neither a uniform bound nor an efficient procedure for finding such a quotient. It includes groups with torsion as well as torsion-free groups. Hyperbolicity alone is the hypothesis: linearity, a cubical action, or membership in a particular manifold class is not assumed. A negative answer requires a hyperbolic group containing a nonidentity element killed by every homomorphism to every finite group, rather than merely a group with few small quotients.

Short English statement

Can every nonidentity element of every hyperbolic group be detected in a finite quotient?

Sources

Catalogue entry

Statement licence: CC-BY-4.0 · Source file

Contributions 1

Math-1791-A1 · Literature report · LLM-generated · unknown ·

OpenAI claim summary for Math-1791

The torsion-free hyperbolic example has one fixed nonidentity element killed by every finite quotient, exactly refuting residual finiteness. Within these assumptions, this is a claimed negative answer to the stated question.

Questions in Geometric Group Theory — Q 1.15

OpenAI's claim

OpenAI presents a counterexample in the selected manuscript. A torsion-free hyperbolic group is claimed not residually finite and consequently not linear over any field.

Mathematical statement of the claimed result

Result 1.
There is claimed to be a torsion-free word-hyperbolic group $G$ whose finite residual is nontrivial: \[ R_f(G)=\bigcap_{\substack{Q\text{ finite}\\\phi:G\to Q}}\ker\phi\ne\{1\}. \] For this finitely generated example, the linear-obstruction corollary states \[ \forall K\text{ commutative field},\ \forall n\ge1,\ \forall\rho:G\to\operatorname{GL}_n(K),\qquad R_f(G)\subseteq\ker\rho. \] Thus a single nonidentity element is annihilated by every such finite-dimensional representation.

Outline of the argument

Punctured-line blocks with few short cycles give triangular relations meeting a geometric filling criterion. Real and prime-field rank comparisons produce a persistent finite-quotient obstruction, while local factorization certificates and planar metric bounds establish hyperbolicity and exclude torsion.

Formal verification

For family 252, the selected example is torsion-free hyperbolic and fails residual finiteness. Nonlinearity is a further group-theoretic consequence, not a separate selected conclusion. The listed Comparator files are TorsionFreeHyperbolic.lean. Named declarations include OAI.Release075.main. This site has not executed the Lean code. The scope descriptions do not certify the remaining manuscript arguments.

Status caveat

OpenAI attributes these manuscripts to work by an internal model. Its release cautions that arguments without formalization can contain mistakes. This catalogue has not independently checked the proofs; the labels report the mathematical claims attributed to OpenAI.

Sources

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