A counterexample to Wall's finite D(2) problem

The explanation

Write down five letters, x₁, a₁, x₂, a₂ and s. Add four rules, such as x₁a₁x₁⁻¹ = a₁⁴ and x₁ = a₂¹³. Letters and rules like these define a group. They also define a space, the presentation complex B. It has one point, five loops, and four discs glued along the rules. Its Euler characteristic is 1 − 5 + 4 = 0.

Now give each letter a complex number. Send x₁ and a₂ to −1. Send a₁ to ζ, a cube root of 1, x₂ to ζ², and s to 1. Each rule still holds. For example, ζ⁴ = ζ and (−1)¹³ = −1. This choice is a character, a way to read loops as numbers. It defines twisted homology, a count of holes where going around a loop multiplies by its number. For B, the key boundary map is a 4-by-5 matrix of complex numbers. Computed from the four rules, its rank is 4. So B has no twisted second homology.

Why does this matter? A CW complex is a space glued from points, edges, discs and higher cells. Wall's D(2) condition is an algebraic test. It says the complex looks two-dimensional to homology and cohomology. The finite D(2) problem asks whether every finite complex that passes the test is homotopy equivalent to a finite complex of dimension at most twoSource 1. Homotopy equivalent means the same shape up to continuous deformation. The problem reappeared as Problem D3 in Wall's 1979 listSource 2.

Partial answers existed. Adding enough two-spheres always gives a two-dimensional complexSource 3. But such stabilization does not answer the original questionSource 4. Hambleton proved the D(2) property for finite subgroups of SO(3), the rotation group of spaceSource 5. Hofmann and Nicholson proved it for quaternion groups of orders 24 and 32Source 6. Bridson and Tweedale had candidate counterexamples. Their failure was conditional on an unproved relation-gap assertionSource 7.

The paper claims a negative answer to the problemSource 8. Its example has infinite fundamental groupSource 9. The construction starts with Quillen's plus construction. It kills a perfect subgroup, one equal to its own commutators, while keeping homology. Mannan showed that this turns a finite two-dimensional complex into a finite D(2) complexSource 10. In the example, the subgroup is made perfect by coprime numbers: 4² − 1 = 15 and 3³ − 1 = 26 share no factor. The quotient group has an element of order two that the character seesSource 11. Here x₁ squares to the identity, but the character sends it to −1. A short boundary calculation shows the twisted second homology is zeroSource 12. That is the rank computation above.

The obstacle is to show no two-dimensional model exists at all. The paper proves a rule for finite two-dimensional complexes. If twisted second homology vanishes, the character must send every finite-order element to 1. The proof follows the cyclic-tower method of HowieSource 13. It climbs a tower of infinite cyclic covers. Each lifted image contains strictly more simplices, yet all come from one fixed finite complexSource 14. That is a contradiction. Our example breaks the rule, since x₁ has order two and reads as −1.

How sure is this? No Lean check covers this paper. No specialist review has been published. The key matrix calculation is small and easy to repeat. The tower argument carries the main weight, so treat the result as a claim.

What it doesn't show. This does not settle the problem for finite fundamental groups, where the known positive cases live.