Counterexamples to finite generation at chromatic height two

The explanation

Fix the prime p = 11. A standard spectrum called BP⟨2⟩ has homotopy built from two variables, v₁ and v₂. Their degrees are 2(p − 1) and 2(p² − 1). For p = 11 that gives degrees 20 and 240. So the coefficients form a polynomial ring in two variables, sitting in degrees 20 and 240. This spectrum is the basic building block in this story.

Some background first. A spectrum is the stable version of a space, the object studied in stable homotopy theory. A thick subcategory is everything you can build from a starting object. You may take finite sums, shifts, cones of maps and retracts, a finite number of times. The question is which infinite spectra you can build from BP⟨n⟩ this way.

The Hahn–Wilson conjecture gives a cohomological answer. It says finite presentation over the Steenrod algebra, plus a height bound, should pick out exactly these spectraSource 1. The Steenrod algebra is a standard ring of operations on mod-p cohomology. Finitely presented means described by finitely many generators and relations. Such a spectrum is called fp. Mahowald and Rezk proved an equivalent test. Smashing with some finite spectrum must give only finitely many nonzero homotopy groups, each finiteSource 2. The height is a level in chromatic homotopy theory. Lee and Pstrągowski wrote the conjecture down and report learning it from Dylan Wilson in 2021Source 3.

Some cases were known. Lee and Pstrągowski proved the conjecture at height oneSource 4. A related claim, the telescope conjecture, compares two ways to localize a spectrum. Burklund, Hahn, Levy and Schlank disproved it at every prime and every height of at least twoSource 5. But those examples fail the fp finiteness test. So they do not give an fp counterexampleSource 6.

The paper claims to disprove the Hahn–Wilson conjecture at height twoSource 7. For each large enough prime, it builds a spectrum X of exact fp-type two. Yet X cannot be built from BP⟨2⟩ in finitely many steps. The result holds for sufficiently large primesSource 8. The spectrum X also passes both localization comparisons. Lee and Pstrągowski had suggested such a hypothesis might be enough. The paper says this extra hypothesis does not suffice at height twoSource 9.

Why is X hard to catch? Its local pieces, seen one height at a time, keep strong finiteness properties. The obstruction lies in how those local pieces are joinedSource 10. So any local test will pass. Only the global gluing shows that X is not finitely built.

The construction uses chromatic algebraicity. At large primes, spectra can be modelled by sheaves on a stack of formal group laws. The paper uses Pstrągowski's comparison, available at height two for p > 8Source 11. Its own operator needs larger primes stillSource 12. It builds the example by gluing a complete part to an open part, through a carefully chosen map on the overlap.

How sure is this? No Lean check covers this paper. No specialist review has been published. The proof relies on deep comparison theorems, so the counterexample remains a claim.

What it doesn't show. This does not cover small primes, and it does not address the conjecture at height three or above.