A torsion-free hyperbolic group that is neither residually finite nor linear over any field

The explanation

Take the integers and the element 5. Reduce modulo 7, a finite group of seven clock positions. The number 5 lands on position 5, not on 0. So 5 survives in a finite image. Every nonzero integer n survives modulo |n| + 1, so the integers are residually finite. A group is residually finite when each nonidentity element has a nonidentity image in some finite groupSource 1. Such groups can be studied through finite snapshots.

A word-hyperbolic group is a finitely generated group whose distance graph has thin triangles. Every point on one side of a geodesic triangle lies near the other two sidesSource 2. Free groups are the standard example, and they are residually finite. Since Gromov's work in 1987, a central question asked whether every hyperbolic group is residually finiteSource 2.

Several facts made this hard to test. Kapovich and Wise showed it is equivalent to every nontrivial hyperbolic group having a nontrivial finite quotientSource 3. Agol showed hyperbolic groups acting nicely on cube complexes are virtually special, hence residually finiteSource 4. So a counterexample cannot have that cubical structure. Wise had groups that are not residually finite with nonpositive curvature, but they contain flat planes and are not hyperbolicSource 5.

The paper proves there is a torsion-free word-hyperbolic group that is not residually finiteSource 6. Torsion-free means no nonidentity element has finite order. The group comes from a finite complex of flat triangles glued along edgesSource 7. Hyperbolicity comes from a test on angles around each vertex. Consider equilateral triangles with 60° corners. Six around a point give 6 × 60° = 360°, which is flat. Seven give 420°, more than a full turn, which bends like a saddle. The criterion asks that every closed walk around a vertex have angle at least 2π + η, for some η > 0Source 8. Then the group is torsion-free and hyperbolicSource 8.

The obstacle is to kill an element in all finite quotients at once. The paper uses a fixed finite family of four-edge "rectangular words." If each survived somewhere, a product of finite quotients would keep them allSource 9. Then a rank count in matrices gives two incompatible boundsSource 9,Source 10. So one fixed member vanishes in every finite quotient, though the proof does not say whichSource 11. The paper also claims that every finite matrix representation, over any commutative field, kills that elementSource 12.

How sure is this? A computer proof checker called Lean has a formal proof of a torsion-free hyperbolic group that is not residually finite. The Lean note says nonlinearity is not a separate conclusion there. No specialist review of the paper has been published.

What it doesn't show. This does not name which element always vanishes, and it does not address skew fields or infinite-dimensional representations.