Hahn–Wilson finite generation: a gluing obstruction

Source and verification

This article explains family 319 of the October 6, 2026 OpenAI collection. Research conclusions are attributed to its manuscripts. The review covers statements, introductions, and selected proof mechanisms; it is not an independent verification of the complete proofs.

Finite presentation is an algebraic finiteness condition; building a spectrum from a fixed generator by finitely many attachments is a geometric one. The Hahn–Wilson conjecture proposes that these conditions agree at each chromatic height. This manuscript claims counterexamples at height two whose localizations still have the expected good behavior. The obstruction is carried by the way the local pieces are glued. (OpenAI, 2026)

🏷️ Preliminaries: three different finiteness conditions

A spectrum $X$ is connective if ${\pi }_{i}X=0$ for $i<0$. It is $p$-complete if it is local for mod-$p$ homology equivalences; derived completion retains the inverse-limit information implicit in this condition. The mod-$p$ Steenrod algebra $\mathcal{A}$ consists of stable operations on mod-$p$ cohomology.

A bounded-below $p$-complete spectrum is fp when $H^{\ast }(X;{\mathbb{F}}_p)$ has a presentation with finitely many generators and relations over $\mathcal{A}$. The Mahowald–Rezk criterion used in the source says this is equivalent to the existence of a nonzero finite $p$-local spectrum $V$ for which ${\pi }_{\ast }(V\wedge X)$ is finite in total: finitely many nonzero groups, each finite.

Morava $K(j)$-theory detects chromatic height $j$, with $K(0)$ rational homology. A nonzero finite spectrum has type $j$ when $j$ is the least height at which its Morava homology is nonzero. An fp-spectrum has fp-type at most $n$ if the total-finiteness test holds for a finite spectrum of type $n+1$; the criterion makes the choice of such a test immaterial. Exact fp-type two means at most two and not at most one.

For a spectrum $B$, its ordinary thick closure $\mathrm{Thick}(B)$ allows finite sums, integer suspensions, cofiber sequences and retracts. Here a retract means maps $X\to Y\to X$ whose composite is the identity. Arbitrary tensor factors and infinite constructions are not among these operations.

The standard completed truncated Brown–Peterson spectrum has coefficients

$$\begin{array}{c}{\pi }_{\ast }\mathrm{BP}⟨2{⟩}_p^{\wedge }={\mathbb{Z}}_p[v_1,v_2],\\ |v_i|=2(p^i-1).\end{array}$$

These graded pieces are finitely generated over ${\mathbb{Z}}_p$, generally not finite groups. Even degreewise finite groups would not imply finite homotopy in total. This distinction will matter both in constructing the example and in proving that it is fp.

🏷️ The claimed counterexample and its range

Let $L_2$ denote Johnson–Wilson $E(2)$-localization, which retains heights zero through two. Let $L_2^f$ be finite localization away from finite spectra of type greater than two. A telescope $T(2)$ is the homotopy colimit of iterates of a periodic self-map of a finite type-two spectrum; $L_{T(2)}$ and $L_{K(2)}$ are the corresponding localizations.

Manuscript's height-two counterexample

For every sufficiently large prime $p$, there is a connective $p$-complete spectrum $X$ of exact fp-type two such that

$$X\notin \mathrm{Thick}(\mathrm{BP}⟨2{⟩}_p^{\wedge }).$$

The same $X$ satisfies

$$\begin{array}{rl}{L}_2^{f}X& \simeq L_{2}X,\\ L_{T(2)}X& \simeq L_{K(2)}X.\end{array}$$

The generator is the specified standard regular Brown–Peterson quotient used in the source. (OpenAI, 2026)

The source invokes odd-prime uniqueness results to relate standard quotient choices. It does not prove this theorem for every prime. In particular, $p>8$ is only the range for its chosen algebraic comparison; the operator construction imposes larger bounds. The two telescope comparisons are properties of this example, not a resolution of the restricted telescope conjecture for all fp-spectra.

🏷️ An algebraic gluing problem

At sufficiently large primes, the proof uses a homology-compatible triangulated comparison between $E(2)$-local spectra and a periodic derived category. “Triangulated” means that it preserves suspensions and cofiber triangles, so it transports finite constructions and retracts. The source supplies additional Ext and convergence arguments for its chosen version of this comparison. All inverse-limit and gluing constructions are performed on the algebraic side before this comparison is applied.

The algebraic setting has an open height-one region and a completed height-two region, separated by a parameter $t$ corresponding to $v_1$. Write $L$ for inversion of $t$ and $\hat{U}$ for derived $t$-completion, formed as the inverse limit of the cofibers of powers of $t$. A generator $T^{\prime }$ represents the relevant coefficient chart after finite-field extension, reduction modulo $p$ and removal of central automorphisms. These changes and their descent data are substantial geometric inputs; we use only their stated interfaces here.

The required object $U$ must satisfy

$$\begin{array}{c}{\pi }_{i}U\text{ finite for every }i,\\ \hat{U}\in \mathrm{Thick}({\hat{T}}^{\prime }),\\ U\notin \mathrm{Thick}(T^{\prime }).\end{array}$$

Here ${\pi }_i$ denotes derived global sections. The completed object will be a finite sum of shifts of the cone of a single operation $a:{\hat{T}}^{\prime }\to {\hat{T}}^{\prime }$. The open object will be chosen by a lattice in its overlap with the completed object. (OpenAI, 2026)

🏷️ An operator with persistent kernel and cokernel

On the overlap, operations act on a Laurent-series field over $k={\mathbb{F}}_{p^2}$. In a coordinate $w$ its compact open subspace is $k[[w^{-1}]]$. A Laurent series has a greatest exponent of $w$; this is its leading position.

The construction chooses an operator $a$ with closed image and two persistent families: kernel vectors with distinct nonpositive leading positions, and positive monomials whose cokernel classes remain independent even after quotienting by the compact open image. For constants $c>0$ and $\beta >1/3$, there are at least $c{N}^{\beta }$ of each kind within precision $N$, for every sufficiently large $N$.

A triangular matrix matching constructs these families. At each depth, finite rank tests decide whether a row and column are paired. The decision reduces to the exceptional corner value of a block matrix: after eliminating its old rows and columns, one remaining scalar determines whether the rank increases. Interpolation over ${\mathbb{F}}_{p^2}$ permits selected base-$p$ digits to avoid pivots while forcing pivots at the other eligible digits. The source preserves $B_0$ central choices per new digit, giving the exponent $\beta ={\log }_p{B}_0>1/3$.

The construction also ensures that only finitely many matched pairs cross any fixed Laurent cutoff. This controls the topology of the infinite matrix and gives closed image. A count along a sparse subsequence, without this cutoff control and the uniform count at all large scales, would not supply the gluing argument. (OpenAI, 2026)

🏷️ Pairing defects to obtain finite homotopy

Let $Y$ be the finite sum of shifts of the cone of $a$ through one periodicity interval. Its overlap $LY$ has kernel and cokernel components. Pair a distinguished polar cokernel vector $q_j$ with a compact kernel vector $k_l$, ordering the lists by increasing $j$ and $|l|$. The first $\min \{n_{+}(N),n_{-}(N)\}$ pairs lie within both cutoffs, so at least $c{N}^{\beta }$ simultaneous pairs remain available.

The vectors $(q_j,k_l)$ are independent modulo a compact open subspace. Extend their quotient classes to a basis, lift that basis, and take its algebraic span. This gives a discrete cocompact lattice: it is discrete, and the quotient of the overlap by it is compact. Because the open category is a split periodic category of $k$-vector spaces, this lattice is realized by an open object $V$ and a map $V\to LY$.

Define the homotopy pullback

$$U=Y{\times }_{LY}V.$$

The map ${\pi }_{i}Y\to {\pi }_{i}LY$ has finite kernel and compact open image. That image is commensurable with the compact open subspace complemented by the lattice, meaning their intersection has finite index in both. Consequently the joint map from ${\pi }_{i}Y\oplus {\pi }_{i}V$ has finite kernel and cokernel. The pullback long exact sequence expresses ${\pi }_{i}U$ as an extension of that kernel by the next-degree cokernel, so ${\pi }_{i}U$ is finite.

Completion kills the two $t$-invertible objects $V$ and $LY$, giving $\hat{U}\simeq Y$. Inverting $t$ instead gives $LU\simeq V$. This proves degreewise finiteness and completed thick membership; it has not yet excluded global finite constructions. (OpenAI, 2026)

🏷️ Why no finite tower can produce the glued object

The obstruction uses both homology and dual homology. After a flat coefficient extension, an exact functor $\Psi$ sends $T^{\prime }$ to a periodic free module and sends $U$ to a perfect module with homology torsion in every degree and nonzero in at least one degree. Here “perfect” means a retract of a finite complex of finite free modules; the coefficient domain has principal left and right ideals. More precisely, the periodic coefficient ring is $\mathcal{B}=S_0[t^{\pm 1}]$, with $t$ central of positive even degree and $S_0$ a principal left and right domain. The module dual is

$$M^{\ast }=R\mathrm{Hom}(M,\mathcal{B}).$$

A homology ghost is a map inducing zero on homology in every degree. The special gluing enforces the following implication for every global map $g:U\to P$ with $P\in \mathrm{Thick}(T^{\prime })$:

$$\begin{array}{c}H(\Psi g)=0\\ ⟹H((\Psi g{)}^{\ast })=0.\end{array}$$

This is not a general property of maps of perfect modules. Its restriction to maps coming from the original global category is the essential extra input.

Why does it hold here? The source bounds joint kernels of two integral operators satisfying the source’s initial-symbol conditions (one symbol generates a distinguished principal ideal and the other lies outside it), at Laurent precision $N$ with a fixed lower Laurent-order bound, by $O(N^{1/3})$. A global map must send the paired defect vectors into a discrete target lattice. If it becomes a homology ghost after coefficient extension, an integral second annihilator gives a rank bound on the images of the polar components. Compactness bounds the images of the kernel components. Together these leave a killed subspace of dimension at least

$$c{N}^{\beta }-O(N^{1/3}).$$

Composing with maps to free shifts turns this into a large joint kernel of operations. Unless those composites lie in the ideal making the cone map vanish, the independent $O(N^{1/3})$ bound is contradicted because $\beta >1/3$. The resulting vanishing of composites is exactly the required dual-homology vanishing. The full integral operator, flatness and adjoint calculations are inputs to this explanation. (OpenAI, 2026)

The categorical conclusion can now be proved explicitly. Every object of the thick closure is a retract of a finite tower whose successive layers are finite sums of shifts of $T^{\prime }$. Suppose $U$ were such a retract, and follow its insertion into each tail of that tower. Starting at the zero tail, prove backward that these maps vanish on both kinds of homology.

At one step, split the tested next tail into free and torsion parts; the principal-domain hypothesis permits this periodic derived splitting. Dual-homology vanishing makes the composite into the free part zero as a derived map. Remove that part by taking a fiber. The remaining triangle has a free layer and a torsion tail. Ordinary homology vanishing forces the induced homology map to factor through the homology of the free layer, but a map from torsion homology to a free module over a domain is zero. The special ghost implication then recovers dual-homology vanishing for the preceding stage.

At the first stage, the retract insertion is therefore zero on homology. It also has a left inverse and hence is split injective on homology, contradicting the nonzero homology of $\Psi U$. This excludes towers of arbitrary finite length and their retracts; excluding only a retract of a finite sum would not suffice. (OpenAI, 2026)

🏷️ From degreewise finiteness to an fp-spectrum

The comparison produces an $E(2)$-local spectrum $Z$ with finite $p$-group homotopy in every degree. Set $X={\tau }_{\ge 0}Z$. Its finite Postnikov stages are $p$-complete, and their inverse limit proves completeness of $X$.

Let $N={\tau }_{<0}Z$. Its finite Postnikov ranges are built from shifts of $H{\mathbb{F}}_p$, hence are $E(2)$-acyclic. Thus $L_{2}X\simeq Z$. For a finite type-three spectrum $V$, finite duality and locality give $V\wedge Z=0$, so

$$V\wedge X\simeq {\Sigma }^{-1}(V\wedge N).$$

The left side is bounded below and degreewise finite; the right side is bounded above. Therefore its homotopy is finite in total. This proves fp-ness and fp-type at most two by the criterion in the preliminaries.

A hypothetical global construction of $X$ from the Brown–Peterson generator would localize and transport to a finite construction of the algebraic object. The source then reverses the coefficient changes using nilpotent boundary maps in comparison triangles. These triangles need not split. That descent would contradict the obstruction for $U$.

Exact type two follows using the known height-one theorem: type at most one would put $X$ in the thick closure of $\mathrm{BP}⟨1{⟩}_p^{\wedge }$. This generator is itself obtained by the $v_2$ cofiber from the chosen height-two generator, again contradicting nonmembership. (OpenAI, 2026)

🏷️ Good localizations do not repair the global gluing

The negative tail $N$ is also telescope-acyclic: positive periodic self-maps telescope to zero on $H{\mathbb{F}}_p$, and rational homology vanishes there as well. This gives $L_2^{f}X\simeq Z\simeq L_{2}X$. For the height-two comparison, the fiber of $Z\to L_{K(2)}Z$ is $E(2)$-local and $K(2)$-acyclic. After smashing with $T(2)$, it has zero homology at all heights zero through two while remaining $E(2)$-local, so it vanishes. Hence $L_{T(2)}X\simeq L_{K(2)}X$.

The positive local classification then places $L_{K(2)}X$ in the local thick closure of Morava $E_2$. Its test asks for finiteness in each degree, not finite total homotopy. Neither that local theorem nor the telescope comparisons construct the missing global gluing maps. This explains how the claimed counterexample can retain good local finiteness while escaping the ordinary global thick closure. (OpenAI, 2026)

The review underlying this exposition covers the statement and selected operator, gluing, finite-tower and transfer mechanisms. The full chart geometry, completed operations and algebraic comparison are substantial remaining inputs, not independently certified here.

🏷️ Related notes

Chromatic splitting distinguishes local filtrations from wedge decompositions and canonical retractions. Here the obstruction concerns ordinary global thick generation even when both telescope comparisons hold.

🏷️ Key steps and proof outlines

Why torsion alone cannot obstruct a finite construction

The cone of multiplication by $2$ on $\mathbb{Z}$ is a finite two-cell construction with torsion homology. Thus finding torsion after a change of coefficients does not prove failure of thick generation. The manuscript adds a condition on which maps come from the original gluing problem.

Its abstract criterion uses an exact functor $\Psi$ sending the proposed generator $T_0$ to the free periodic coefficient object ${\mathcal{B}}_0$. Write $Y=\Psi (U_0)$ for the image of the proposed counterexample and $M^{\ast }=R\mathrm{Hom}(M,{\mathcal{B}}_0)$ for the dual. The coefficient ring has the principal-ideal and splitting properties established earlier in the source. The two required tests are:

1. $H(Y)$ is torsion in every degree and nonzero in some degree.
2. For every original-category map $g:U_0\to P$ with $P\in \mathrm{Thick}(T_0)$,

$$H(\Psi (g))=0⟹H((\Psi (g){)}^{\ast })=0.$$

The second statement concerns the dual map in the opposite direction. It is not asserted for every map between arbitrary coefficient objects. The gluing lattice restricts maps that descend from the original category, and that restriction supplies the implication. (OpenAI, 2026)

Backward induction excludes every finite tower

Suppose, toward a contradiction, that $U_0$ lies in the thick closure. It is then a retract of an object $P_0$ with a finite tower

$$F_j⟶P_j⟶P_{j+1},P_s=0,$$

where each $F_j$ is a finite sum of shifts of $T_0$. This normal form includes arbitrary retracts: after adding complements, extend a map between retracts by zero; its cone contains the desired cone as a retract. Consequently retracts can be moved to the end of the construction.

Let $g_j:U_0\to P_j$ be the composite of the retract insertion with the tower maps. Start with $g_s=0$ and induct backward, maintaining vanishing on both ordinary and dual homology. For one induction step, the periodic splitting lemma separates the next tail into a free part and a part with torsion homology.

Dual-homology vanishing makes the composite to the free part zero as a derived map: maps to free shifts are precisely the corresponding dual-homology classes. The current map therefore lifts to the fiber obtained by removing that free quotient. This fiber sits in a triangle whose first term is the free layer and whose last term has torsion homology.

In its long exact homology sequence, the incoming map from torsion to free homology vanishes, so the free layer injects. Ordinary-homology vanishing at the next tail places the current map’s image inside this free submodule. A map from the torsion module $H(Y)$ to a free module over a domain is zero. This proves ordinary-homology vanishing at the current stage; the original-category implication restores dual-homology vanishing for the next backward step.

At the end, $H(\Psi (g_0))=0$. But $g_0$ is a retract insertion, so its image under any functor has a left inverse. Its homology map must be a split injection. Since $H(Y)\ne 0$, this is impossible. The argument excludes towers of every finite length without needing a uniform bound on their lengths.

The exponent gap enforces the global-map condition

The analytic input behind the second test is a dimension bound for simultaneous operator kernels. Fix a lower Laurent-order bound, so the vectors lie in $x^{-L_0}R$ with $L_0$ fixed. At precision $N$, two fixed sufficiently independent operator constraints have joint kernel dimension at most

$$C(1+N{)}^{1/3}.$$

Here $C$ may depend on $L_0$ and the two fixed operators. The proof follows orbit functions of putative independent kernel vectors. A divided-difference interpolation estimate bounds their coefficient valuations by $O(N)$. This permits reduction to functions on a finite congruence quotient; the two initial-symbol relations then bound the dimension of that finite function space. The quotient indices grow cubically in the relevant scale, producing the exponent $1/3$.

The constructed lattice has at least $c{N}^{\beta }$ paired defects at every sufficiently large precision, with $\beta >1/3$. If a global map violated the required dual test, the lattice compatibility equations would force too many of those defect vectors into such a joint kernel. The inequality $c{N}^{\beta }>C(1+N{)}^{1/3}$ eventually contradicts the bound. This is why the operator construction needs a strict exponent advantage and counts at all sufficiently large cutoffs. These operator and descent estimates remain substantive manuscript inputs; the backward-tower argument explains how their output yields the stated obstruction.

🏷️ Manuscripts and verification scope

• Counterexamples to the Hahn-Wilson conjecture at height two — pinned manuscript and companion source.

No family scope page was present at lean/docs/319.md in the pinned snapshot. This does not establish that all related formalization code is absent. No Lean build, axiom audit, or manuscript-to-formal-statement equivalence check was performed for these notes.

OpenAI collection provides prerequisites and the reading sequence; coverage register records completed and pending families.

🐻  OpenAI 2026. Counterexamples to the Hahn–Wilson conjecture at height two.