The stable Hurewicz image: last suspensions and cone detection

Source and verification

This article explains family 316 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.

The homology of the stable sphere’s infinite loop space is explicitly describable and enormous. Very little of it is represented by maps from spheres. Family 316 claims Curtis’s conjecture: in positive degrees, these spherical homology classes are accounted for by the Hopf-invariant-one classes and the Kervaire-invariant-one classes that exist. The proof uses the last nonzero homology suspension, an actual mapping cone, and a weight obstruction. (OpenAI, 2026)

🏷️ Preliminaries: spherical classes and operations

Write $Q{S}^n={\mathrm{colim}}_r{\Omega }^r{S}^{n+r}$. This infinite loop space packages stable maps into the sphere of dimension $n$. The space $Q{S}^0$ has components indexed by integers; $Q_0{S}^0$ is its basepoint component. For $d>0$, the stable stem ${\pi }_d^S$ identifies with ${\pi }_d(Q_0{S}^0)$.

The mod-two Hurewicz map sends a sphere map to the image of its fundamental class:

$$h_d:{\pi }_d^S⟶H_d(Q_0{S}^0;{\mathbb{F}}_2).$$

A class in its image is spherical. It is primitive for the space diagonal, meaning its coproduct is $z\otimes 1+1\otimes z$, and every positive dual Steenrod operation ${\mathrm{Sq}}_{\ast }^i$ annihilates it. These conditions are necessary, not a realization theorem.

The homology product is the additive Pontryagin product induced by infinite loop addition. Dyer–Lashof operations generate the homology algebra from particle classes. Homology suspension ${\sigma }_{\ast }$ goes from one infinite loop level to the next and kills products of positive-dimensional classes. It is different from stable homology evaluation $H_{\ast }({\Omega }^{\infty }E)\to H_{\ast }(E)$.

The Hopf classes $\eta ,\nu ,\sigma$ lie in stable degrees $1,3,7$. Their invariant can be detected by a Steenrod square between the two cells of a stable mapping cone. A Kervaire-invariant-one class ${\theta }_j$ has degree $2^{j+1}-2$; under the framed-manifold interpretation its invariant is the Arf invariant of a quadratic form. This notation asserts no new existence result.

Manuscript's stable Hurewicz classification

The positive-degree mod-two Hurewicz image is spanned over ${\mathbb{F}}_2$ by the images of $\eta ,\nu ,\sigma$ and the Kervaire-invariant-one classes that exist. (OpenAI, 2026)

Combining this claim with the stated Kervaire nonexistence input gives the support bound

$$d\in \{1,2,3,6,7,14,30,62,126\}$$

for any positive degree with nonzero image. This is an upper bound, not a claim that every listed degree has nonzero image or an assertion of its exact dimension.

🏷️ Weights and the special role of level zero

The extended power $D_{r}T=(T^{\wedge r}{)}_{h{\Sigma }_r}$ is the homotopy quotient of an $r$-fold smash product by permutation of inputs. Its weight is $r$. The free commutative ring spectrum is $\mathbb{P}(T)={\bigvee }_{r\ge 0}{D}_{r}T$. Multiplication adds weights; a Dyer–Lashof operation doubles them.

At positive sphere levels the splitting gives direct weight summands. Primitive homology occurs at powers-of-two weights. The dual cohomology modules are

$$\begin{array}{c}{M}_0(n)={\Sigma }^n{\mathbb{F}}_2,\\ M_k(n)={\Sigma }^n{d}_0^n{D}_k(k\ge 1),\end{array}$$

where $D_k={\mathbb{F}}_2[d_0,\dots ,d_{k-1}]$ is the binary Dickson algebra, with $|d_j|=2^k-2^j$. It can be defined as the polynomials in $k$ degree-one variables invariant under invertible binary changes of variables. The displayed suspension shifts total degree; it does not change the polynomial exponents. Dual homology suspension is inclusion of the next ideal $d_0^{n+1}{D}_k\subset d_0^n{D}_k$.

At level zero, particle group completion is essential. If $t$ denotes a particle and $u$ the reduced generator, then $t=1+u$. The source constructs

$${\Sigma }_{+}^{\infty }Q{S}^0\simeq \mathbb{P}(S\{u\})[(1+u{)}^{-1}].$$

Compatible projections to finite weight truncations jointly detect homology. Thus a nonzero class has a smallest nonzero, or leading, weight. This is a filtration statement, not an unqualified direct splitting of the actual level-zero diagonal. Its leading term has the appropriate linear primitive structure. For an odd-degree primitive annihilated class, that leading term survives homology suspension. (OpenAI, 2026)

🏷️ The last nonzero suspension is a square

Start with a stable class $\alpha$ of degree $d>0$ and let $z_j={\sigma }_{\ast }^j{h}_d(\alpha )$. For $j>d$, its degree $d+j$ lies strictly between $j$ and $2j$, where $H_{\ast }Q{S}^j$ has no classes. Hence a nonzero Hurewicz image has a last nonzero suspension $z_e$.

The positive-degree kernel of homology suspension consists exactly of decomposables. In the polynomial Hopf algebras involved, a decomposable primitive is a square with a unique primitive root. The Cartan formula and injectivity of squaring make the root Steenrod-annihilated as well. Consequently

$$\begin{array}{c}{z}_e=w^2,|w|=m-1,\\ d+e=2m-2,m\ge 2.\end{array}$$

The root must still suspend nontrivially. At $e\ge 1$, the preceding spherical class supplies a functional on the preceding weight module. The first allowed polynomial exponent, together with the parity relations in its Steenrod coinvariants, forces $|w|$ odd. An odd-degree root cannot itself be a square, so it is outside the suspension kernel.

At $e=0$, there is no preceding level. A separate argument excludes nonzero spherical fourth powers. This distinction prevents the proof from silently assuming the information that is available only after a positive suspension.

🏷️ Excluding fourth powers without confusing two products

The adjoints of weight projections are generally not additive infinite loop maps. Applying one to a Pontryagin square introduces a correction involving the block pairing $\circ$ between extended powers. The source distinguishes this pairing from the additive product $\ast$ throughout.

Suppose a spherical image were $z^{\ast 4}\ne 0$. Let $r$ be the leading weight of $z$. The leading weight-$2r$ term $p$ of $z^{\ast 2}$ is nonzero, but its weight suspension vanishes. In the Dickson description, its functional kills the ideal generated by $d_0$. A detecting class can therefore be chosen as a polynomial in even-index Stiefel–Whitney classes of a symmetric-group permutation representation.

The block correction disappears by a separate lemma: a positive primitive annihilated class $B$ with zero stable homology evaluation satisfies $B^{\circ 3}=0$. The relevant correction is a fourth block power of a projection below the leading weight, where stable evaluation is zero. Only after removing this correction does the projected spherical class become an actual additive square suitable for a mapping-cone argument.

That cone forces a nonzero top square ${\mathrm{Sq}}^{b+1}\tilde{u}$, with $b$ even. On the other hand, the square of the characteristic detector lifts to an integral polynomial in Pontryagin classes, using ${\rho }_2(p_i)=w_{2i}^2$. Positive-degree integral cohomology of the finite symmetric group is torsion. Its pullback to the attaching sphere is zero because the sphere’s relevant integral cohomology is torsion-free, so the integral class extends over the cone.

Restriction from degree $2b+1$ cohomology of the cone to that of the target is injective: the preceding relative term in the attachment sequence vanishes. The integral reduction and ${\mathrm{Sq}}^b\tilde{u}$ have the same restriction, so they are equal. Hence ${\mathrm{Sq}}^b\tilde{u}$ is killed by ${\mathrm{Sq}}^1$. The Adem relation for even $b$ now gives

$${\mathrm{Sq}}^{b+1}\tilde{u}={\mathrm{Sq}}^1{\mathrm{Sq}}^b\tilde{u}=0,$$

contradicting the cone detector. Thus spherical fourth powers, and therefore higher iterated squares, are excluded. (OpenAI, 2026)

🏷️ The mapping cone retains realization information

Put $n=e+1$. The next adjoint is a map $g:S^{2m-1}\to Q{S}^n$ with zero Hurewicz image, while its loop adjoint has image $w^2$. The general cone-square lemma requires $u^2=0$. For $Q{S}^n$ with $n>0$, the positive cohomology algebra is exterior, so every positive-degree class has square zero. For a degree-$m$ class $u$ in this target, its unique extension to the mapping cone satisfies

$${\tilde{u}}^2=⟨u,{\sigma }_{\ast }w⟩a,$$

where $a$ is the nonzero top relative class. The cone-square lemma is proved for connected targets and includes $m=2$; simple connectivity of the original target is not assumed.

The resulting Steenrod-module extension must retain $a$. Taking ordinary algebra indecomposables of the cone could kill it, since $a$ may itself be a square. Instead, the source removes only products of distinct lifted exterior generators. Cartan’s formula and annihilation of the detecting functional show that this span is Steenrod-stable. The top class survives in an extension of the target’s indecomposable cohomology.

Let $A$ be the Steenrod algebra and $I$ its ideal of positive operations. On a detected degree-$m$ monomial $v\in M_k(n)$, this actual extension supplies

$${\mathrm{Sq}}^m\otimes v\ne 0\text{in }I{\otimes }_A{M}_k(n).$$

Balancing means $(ab)\otimes v=a\otimes bv$. A section of the extension turns its failure to commute with operations into a functional on this balanced tensor product, taking the displayed tensor to one. This is the realization input absent from merely primitive, annihilated homology classes.

🏷️ The weight obstruction and its two exponent thresholds

The decisive algebra lemma requires both that tensor nonvanishing and a detecting functional lifted from the preceding module $M_k(n-1)$. Under these hypotheses, for a nonbottom weight $k\ge 1$, it forces

$$n=k=1,m\text{ a power of two}.$$

The proof first takes coinvariants for Milnor primitives and applies a Cartier operation that halves exponents and degrees. A monomial with $d_0$-exponent $c$, detected by the lifted functional, satisfies

$$\begin{array}{c}b=2^{{\nu }_2(m)},n\equiv k(\mathrm{mod}b),\\ c\ge n+b-2\end{array}$$

when $m$ is even. If $m$ is a power of two, repeated halving ends in degree zero and gives $n+(m-1)k=m$, forcing $n=k=1$.

If $m$ is even but not a power of two, an injection on even-degree Tor cycles (relations in the kernel of the operation-action map) allows halving the tensor without losing its nonvanishing. The functional starts with exponent threshold $n-1$; the tensor starts with threshold $n$. Keeping these distinct is necessary when passing to successive ideals. At the final odd parameter, balancing and explicit near-top Steenrod identities make the tensor zero. Odd $m\ge 3$ has a separate tensor vanishing argument. These are the source’s algebraic obstruction mechanisms; a full reconstruction of their coinvariant and Tor calculations is outside this exposition.

🏷️ Hopf detection and Kervaire detection

If the terminal level is positive, then $n=e+1\ge 2$. The obstruction excludes every nonbottom weight, leaving the bottom class. Its degree gives $m=n$ and $d=e=n-1$. The stable counit carries the cone detector to the actual two-cell stable cofiber of $\alpha$. Adams’s Hopf invariant theorem then restricts $d$ to $1,3,7$.

If the terminal level is zero, the argument first forces $m$ even. The odd-degree root has a leading component that survives suspension. The obstruction now gives root weight two and $m=2^j$, so $z_0=w^2$ has leading weight four and $d=2^{j+1}-2$.

Leading weight four supplies an Adams-filtration bound, not an equality between weight and filtration. An imported lifting theorem sends a class of Adams filtration at least three into the derived eighth power of the augmentation ideal. Project to weights at most seven. In a positive-weight free resolution, every term of the derived eighth power has at least eight positive-weight inputs. Its canonical map into that truncation is zero. The nonzero weight-four detector therefore rules out filtration at least three.

Positive even stems have no filtration-zero or filtration-one sphere class. In the relevant filtration-two bidegree, the only possible class is $h_j^2$: the degree equation $2^a+2^b=2^{j+1}$ forces $a=b=j$. Browder’s elementwise detection theorem identifies the invariant of the particular stable class as Kervaire invariant one. Merely observing a permanent cycle somewhere in this bidegree would not suffice.

Finally choose a Hopf or Kervaire representative with the same invariant. The detecting invariant is additive, so the difference has invariant zero. If the difference had nonzero Hurewicz image, the proved necessity would force its invariant to be one. Its Hurewicz image is therefore zero, proving that the named images span the entire image.

For a positive sphere $S^n$, homology suspension kills the Kervaire square images. The manuscript thus deduces the sphere case of Eccles’s conjecture: spherical classes in $H_{\ast }Q{S}^n$ are the bottom class or nonzero suspensions of Hopf images. This assertion concerns positive-dimensional spheres, not arbitrary spaces.

The selected-source audit follows the cone construction, terminal-stage deductions, fourth-power detector, and stated algebra interfaces. It does not independently verify every weight-model construction, pairing identity, Tor-injection calculation, or imported detection and lifting theorem.

🏷️ Related posts

• The prime-three Kervaire family concerns a different prime and detector; the mod-two framed invariant and Curtis image studied here should not be identified with it.
• Smith–Toda complexes explains another use of filtered stable-homotopy detection, where permanence must lead to an actual map before taking a cofiber.

🏷️ Key steps and proof outlines

A mapping cone detects information absent from homology alone

The central topological lemma starts with an actual map $g:S^{2m-1}\to T$, for a connected target $T$ and $m\ge 2$. Assume $h(g)=0$ and that the loop adjoint has Hurewicz image $x^2$, where $|x|=m-1$. If $u\in H^m(T;{\mathbb{F}}_2)$ has $u^2=0$, then its extension to the mapping cone satisfies

$${\tilde{u}}^2=⟨u,{\sigma }_{\ast }x⟩a,$$

with $a$ the top relative class. The attachment exact sequence gives uniqueness of $\tilde{u}$, because $m<2m-1$, and injectivity of the relative top class, because $h(g)=0$. The remaining issue is the coefficient of that class.

To compute it, represent a class together with a nullhomotopy of its square by the homotopy fiber

$$F=\mathrm{hofib}({\mathrm{Sq}}^m:K({\mathbb{F}}_2,m)\to K({\mathbb{F}}_2,2m)).$$

The map representing $u$ lifts to $F$. Looping gives a fibration with fiber $K({\mathbb{F}}_2,2m-2)$ over $K({\mathbb{F}}_2,m-1)$. It splits as a space over its base by instability, but that splitting need not preserve loop multiplication.

Let $c$ be the bottom homology class of $\Omega F$. Its square maps to zero in the base: positive-degree homology classes of an additive Eilenberg–Mac Lane space have square zero over ${\mathbb{F}}_2$. At chain level, the shuffle terms pair by exchanging their two blocks, giving cancellation in characteristic two. Thus $c^2$ lies in the one-dimensional fiber contribution.

Why is $c^2$ nonzero? In the cobar spectral sequence, it is represented by $[c_0\mid c_0]$, with each $c_0$ in degree $m$. Its first differential vanishes because the reduced coproduct of $c_0$ is zero. Vanishing of the bottom cohomology square in $F$ separately shows that this tensor is not a boundary for that differential. No higher incoming differential has an allowed source. A higher outgoing differential would require $2+r$ positive homology factors in total degree $2m+r-1$, but

$$(2+r)m>2m+r-1.$$

Connectivity therefore excludes that target. The tensor-length filtration is finite in each total degree, so survival detects actual loop homology. Consequently $c^2$ is the nonzero fiber class. Mapping $x$ into this universal example identifies its coefficient as $⟨u,{\sigma }_{\ast }x⟩$, proving the cone formula. (OpenAI, 2026)

Quotienting by all products would erase the detector

For $T=Q{S}^n$ with $n>0$, the argument passes from cone cohomology to an extension of indecomposable cohomology modules. There is a trap: the top class $a$ may itself equal the square of a lifted generator. Quotienting cone cohomology by all products would kill the information just obtained.

Choose exterior generators downstairs so that at most one degree-$m$ generator has nonzero pairing with ${\sigma }_{\ast }x$. Upstairs, remove only square-free products of at least two distinct lifted generators. Their span is stable under Steenrod operations by the Cartan formula. A repeated distinguished generator could produce $a$, but it cannot arise from a positive operation on one factor because the detecting functional annihilates positive Steenrod images; if another positive factor remains, its product with $a$ is zero anyway.

This restricted quotient keeps $a$ and produces the module extension needed by the manuscript’s separate algebraic obstruction. It explains why realization by a sphere map is stronger than being primitive and Steenrod-annihilated in the homology algebra.

Leading weight bounds filtration in only one direction

At terminal suspension stage zero, the source’s obstruction isolates a nonzero leading weight-four component. Kuhn’s lifting theorem says that Adams filtration at least three would lift the Hurewicz image through the derived eighth power of the augmentation ideal.

Project to the algebra retaining only weights one through seven. In a free resolution of its positive-weight ideal, every term of the derived eighth power has at least eight inputs, each of weight at least one. Hence its canonical map to the truncated algebra is zero by weight. Naturality would force the original Hurewicz image to vanish in every weight below eight, contradicting its nonzero weight-four component.

Thus the given element has Adams filtration at most two. This is a lower-weight obstruction to high filtration, not an equality between weight and filtration. In the positive even stem $2^{j+1}-2$, filtration zero and one contribute nothing, and the filtration-two degree equation

$$2^a+2^b=2^{j+1}$$

forces $a=b=j$. The only possible detector is therefore $h_j^2$, leading to the imported Kervaire detection theorem. The separate positive-terminal-stage argument instead yields a two-cell stable Hopf detector. Neither branch constructs new Kervaire classes; both classify an element assumed to have nonzero Hurewicz image.

🏷️ Manuscripts and verification scope

• The Stable Hurewicz Image of the Sphere at Two — pinned manuscript and companion source.

No family scope page was present at lean/docs/316.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. The Stable Hurewicz Image of the Sphere at Two.