The Stable Hurewicz Image of the Sphere at Two

Proof. For a spherical representative f:S^{n+d}\to QS^n with d>0, adjunction gives f_0:S^d\to Q_0S^0, and naturality gives h(f)=\sigma_*^n h_d([f_0]). By Theorem 1.1, the level-zero image is spanned by the Hopf and Kervaire images. The latter are squares killed by the first homology suspension, leaving the stated Hopf possibilities; this is the Curtis-to-Eccles sphere implication recorded in [33, Chapter 5, pp. 81–82]. In degree n, \pi_n(QS^n)=\pi_0^S=\Z gives the bottom image represented by the unit S^n\to QS^n; connectivity excludes lower degrees.□

Proof. Hill, Hopkins, and Ravenel prove that \theta_j can exist only for j\le6 [14, Theorem 1.1]. By Theorem 1.1, the positive image is therefore spanned by the named images in Hopf degrees 1,3,7 and Kervaire degrees 2,6,14,30,62,126. Finitely many generators over \F_2 give a finite image and the stated support.□

Proof. Let F denote the free unbased commutative monoid in spaces. The reduced free monoid is the homotopy pushout

Indeed, mapping this pushout into a commutative monoid Y gives the homotopy fibre of \operatorname{Map}(X,Y)\to\operatorname{Map}(*,Y)=Y over the unit. This is exactly the mapping space of based maps from X to Y, based at its unit. The pushout therefore has the required reduced free universal property, including the basepoint homotopy.

The functor \Sigma^\infty_+ is a strong symmetric monoidal left adjoint from cartesian spaces to spectra. On commutative algebra objects it is again a left adjoint, and hence preserves this pushout. Write T=\Sigma^\infty X. The basepoint section and augmentation give a natural stable splitting

Under it, the spectrum map induced by the basepoint is the inclusion of the first summand. Since the free commutative algebra functor sends a direct sum to a coproduct of algebras, the pushout after suspension is

Here \mathbb P(S)\to S sends its generator to the ring unit. The augmentation of the resulting algebra sends T to zero. Thus the positive-weight summands are precisely the augmentation fibre, or equivalently the reduced suspension spectrum at the unit. In this description the generating spectrum map is the difference between the inclusion of X and its basepoint section.

We identify the weight-one adjoint using the algebra projection

where the target is the split square-zero algebra. For any spectrum M, the fibre over the unit point of

identifies with \Omega^\infty M, via y\mapsto1+y, as a grouplike commutative monoid. To see the structure as well as the underlying space, expand a multiproduct using distributivity in (S\oplus M)^{\wedge t}. The term with no M factors gives the base unit. The terms with one M factor give the sum of the inputs. Every term with two or more M factors vanishes by the square-zero structure. These formulas respect permutations and iterated products: the structure maps on the fibre are the additive sum-over-fibres maps. They therefore give the stated identification with the additive infinite loop space of M.

The adjunction from the suspension ring now gives an E_\infty map C(X)\to\Omega^\infty T. On the generating based space it is the usual map X\to\Omega^\infty\Sigma^\infty X. Reduced freeness identifies it with the canonical stabilization map. Equivalently, maps from its group completion into a grouplike commutative monoid \Omega^\infty E are based maps X\to\Omega^\infty E, or spectrum maps \Sigma^\infty X\to E. This identifies the group completion with QX. For connected X, the reduced free monoid has trivial \pi_0, so it is already grouplike and no group completion is needed. Under C(X)\simeq QX, the adjoint of the weight-one projection is then the identity of QX. By the suspension-spectrum adjunction this says exactly that the projection is the stable counit.□

Proof. For v\in S^1, the based map x\mapsto v\wedge x induces C(X)\to C(\Sigma X). These maps form a pointed family, since the basepoint of S^1 induces the constant map to the monoid unit. Its adjoint is

Since \Sigma X is connected, the target is Q\Sigma X. On the generating space X, adjunction identifies this family with the usual stabilization into \Omega Q\Sigma X\simeq QX. Reduced freeness and group completion therefore identify its adjoint with the usual structure map of the infinite loop spaces.

The positive-weight equivalence in Proposition 2.1 is a natural equivalence between the reduced functors

Both send the one-point based space to zero. Their action on mapping spaces therefore respects the basepoint given by factorization through that space, including its nullhomotopy. Apply this naturality to the pointed mapping-space family v\mapsto(v\wedge-), and then take the stable adjunction. On the rth extended power the same v is used in every tensor factor. This gives (1) with no mixing of weights. For a sphere, D_r(\Sigma^nS) is the Thom spectrum of n\rho_r over B\Sigma_r, where \rho_r is the permutation representation. The map in question is the Thom map associated to

using the diagonal line \mathbb R\subset\rho_r.

The commutation rule with upper-index Dyer–Lashof operations is the usual suspension rule for free infinite loop space homology [25, §20, p. 481]. The product rules can also be read directly from the adjunction of loop multiplication: the suspension of loop concatenation factors through the pinch map and gives the sum of the two suspensions. Thus a cross product of two positive-dimensional classes has zero image. For a product with a component point, the surviving term is the suspension of the positive-dimensional factor, multiplied by the augmentation of the point class, which is one.□

Proof. In terms of the reduced generator, the diagonal is the sum of its two linear inclusions and the term induced by the reduced diagonal X\to X\wedge X. The latter is null for a suspension. Because the source ring is free, a homotopy on the generating spectrum gives the required homotopy of commutative ring maps.

Apply D_r to the linear map T\to T\vee T. The terms choosing the first summand i times are indexed by the transitive \Sigma_r-set \Sigma_r/(\Sigma_i\times\Sigma_{r-i}). Equivariantly, the map into these terms is the diagonal into a finite sum indexed by that set. On homotopy orbits it is the stable transfer to the indicated subgroup. This describes every component, including the two end components.□

Proof. The product over a finite binary vector space is additive in U. Indeed, if W'=W\oplus\langle a\rangle and its root polynomial for W is P_W, then

Induction from P_{\{0\}}(U)=U proves additivity and the displayed powers of U. Homogeneity gives the degrees of the coefficients. They are invariant because an invertible binary linear change of variables permutes the roots.

Put B=\F_2[d_0,\ldots,d_{k-1}]\subset P_k and L=\F_2(x_1,\ldots,x_k). Each x_i satisfies the monic root polynomial, so P_k is integral over B. Therefore B has transcendence degree k; its k indicated generators are algebraically independent and B is integrally closed. The derivative of the root polynomial is the nonzero product d_0 of all nonzero binary linear forms. Its splitting field over \operatorname{Frac}B is L. Every automorphism of this splitting field acts invertibly and linearly on the root space; conversely every such linear transformation fixes the coefficients. Consequently L^{\GL_k(\F_2)}=\operatorname{Frac}B. An invariant polynomial lies in this fraction field and is integral over B, hence belongs to B. This proves the assertion.□

Proof. We first determine dimensions and the weights in which primitives can occur. The free infinite loop space homology theorem gives, for n>0,

including the empty word g_n, where Q_jz=Q^{j+|z|}z and |g_n|=n [9]. The generator g_n is primitive. Naturality of the operations for the coproduct, together with Cartan and the vanishing of positive operations on the unit, shows that every displayed polynomial generator is primitive.

For n=0, the same polynomial description holds for H_*\mathbb P(S) on the free degree-zero generator, which we denote here by u. One may see this from the free unbased monoid \coprod_rB\Sigma_r, whose suspension ring is free on a degree-zero spectrum generator. The Dyer–Lashof operations for commutative ring spectra agree with the space operations on such suspension rings [6]. Naturality for the linear ring coproduct again makes the polynomial generators primitive. This argument does not identify that coproduct with the actual space diagonal at level zero.

In both calculations, the bottom generator has weight one, a Dyer–Lashof operation doubles weight, and multiplication adds weights. In a polynomial Hopf algebra on primitive generators, the primitives have basis the iterated 2-powers of those generators. For example, this follows by writing the primitive identity as f(\mathbf x+\mathbf y)=f(\mathbf x)+f(\mathbf y): a monomial involving distinct variables produces a nonzero cross term, and for a single variable all interior binomial coefficients vanish exactly when the exponent is a power of two. Every polynomial involves only finitely many variables, so the argument also applies to the infinite polynomial algebra.

It follows that a primitive basis is given by the words

with u in place of g_n at n=0. Initial zero indices record iterated squares. A word of length k has weight 2^k and degree

Thus primitives vanish at the other positive weights. Setting a_1=j_1 and a_i=j_i-j_{i-1} for i>1 rewrites this degree as

These are exactly the monomial degrees in \Sigma^nd_0^nD_k. At n=0 we perform this count at fixed weight, allowing degree zero; it remains finite in each degree at that weight.

We next identify the Steenrod module, rather than just its graded dimension. Fix r=2^k with k\geq1 and let V=(\Z/2)^k\subset\Sigma_r be the regular elementary abelian subgroup. The extended power D_r(\Sigma^nS) is the Thom spectrum of n\rho_r over B\Sigma_r. Restrict its cohomology to BV. Every proper two-block transfer restricts to zero. Indeed, the double coset formula expresses its restriction as transfers from subgroups

Each is proper, because V acts transitively on the particles whereas a two-block subgroup preserves a nonempty proper subset. If W<V, restriction H^*BV\to H^*BW is surjective. The projection formula then expresses transfer of any class as a class from H^*BV times the transfer of 1, which is the even index [V:W], hence zero. Naturality of the mod-2 Thom class gives the same assertion with the representation twist.

The normalizer of V acts through \GL_k(\F_2), so the restriction image is contained in D_k times the Thom class. It contains all of that module: the Stiefel–Whitney classes of \rho_r restrict to the Dickson generators in (2). More explicitly, \rho_r|_V is the sum of all the real one-dimensional characters, and its total Stiefel–Whitney class is the product of 1+a over all a\in V^*. Thus w_{2^k-2^j}(\rho_r)|_{BV}=d_j, and polynomials in these classes, multiplied by the Thom class, give the claimed image. Restriction therefore gives a surjection from the transfer quotient onto D_k times the Thom class. The dimension computation (6) makes this surjection an isomorphism in every degree.

Finally, split the regular representation over BV as \mathbb R\oplus\bar\rho_r. Pulling the Thom class back along the trivial n-plane inclusion

multiplies it by the Euler class of n\bar\rho_r, namely d_0^n, and leaves the trivial Thom shift \Sigma^n. Multiplication by d_0^n is injective in the polynomial algebra. All maps used here are stable cohomology maps; hence the resulting identification is the Steenrod module \Sigma^nd_0^nD_k with its inherited action. In particular this also proves stability of that ideal under A. Weight one is immediate from D_1(\Sigma^nS)=\Sigma^nS.

For compatibility with suspension, the Thom map in Lemma 2.3 has complement \bar\rho_r. Its pullback sends a Thom coefficient f to d_0f. The level-n identification then sends this coefficient to d_0^n(d_0f)=d_0^{n+1}f, which is its image under the level-(n+1) identification. Thus, after the common suspension shift, the map is

given by inclusion. This proves (4), with all degree shifts as asserted.□

Proof. Under Proposition 2.1, the particle is the ring element t=1+u. Group completion of C(S^0) inverts this particle component. For a commutative ring spectrum A', a map from its suspension ring is a map from the additive E_\infty particle monoid to the multiplicative E_\infty space of A'. Such a map extends over group completion precisely when the particle maps to a unit. Equivalently, the resulting ring map inverts 1+u. This is the universal property of the localization in (7), including its mapping spaces. The augmentation sends u to zero and t to one, so the equivalence is augmented. The underlying module of localization at a degree-zero element is its multiplication telescope. Therefore ordinary homology satisfies

In \mathbb P_{\leq N}(S\{u\}), the element u is nilpotent in \pi_0. Thus 1+u is a unit, with inverse \sum_{j=0}^N(-u)^j in \pi_0, and the truncation extends across localization to give (8). Their compatibility follows from the localization universal property.

Every class of H_*R can be multiplied by a sufficiently large power of t to become a class in the free homology algebra H_*\mathbb P(S\{u\}). This algebra is polynomial, as used in Lemma 2.6, and is a direct sum of weights. Hence such a numerator has only finitely many nonzero weights. If its smallest nonzero component has weight r, multiplying by (1+u)^a leaves that component unchanged: the weight-zero term is one and every other term has positive weight. Conversely, in a finite truncation the inverse of 1+u has weight-zero term one and only positive weights otherwise. Thus a nonzero numerator remains nonzero in every truncation retaining its leading weight. This proves joint detection, independence of the leading component from the chosen numerator, and invariance under powers of t. The inclusion H_*Q_0S^0\to H_*QS^0 is injective, since it is the inclusion of one component, so these conclusions apply to the base component as well.

Before localization, the actual space diagonal on the reduced generator is

Consequently, a weight-r class maps only to terms of total weight at least r, and its terms of total weight r are given by the linear coproduct. This follows by applying extended powers to (9) and then multiplying. It remains true after localization: the diagonal of t is t\otimes t, whose inverse has constant term one and only positive total weights otherwise. These statements may all be made in finite truncations in the two target factors, so no convergence of an infinite sum is required.

The kernels of the projections through weight r-1 define a decreasing weight filtration. The preceding calculation identifies its associated graded coproduct with the linear coproduct. Since z is primitive, comparison in its leading total weight gives

The weight projections are maps of spectra, so they commute with the stable Steenrod action. Thus z_r is also A-annihilated. Lemma 2.6 makes r a power of two; weight one has no positive-dimensional homology, so r=2^k with k\geq1.

For the suspension assertion, choose a\geq0 such that t^az has a finite numerator in the particle monoid homology. The translation rule in Lemma 2.3 gives

The numerator has the same leading component z_r and no smaller weights. Apply the weight-preserving family of Lemma 2.3 before group completion. It follows that the weight-r component of the suspension is exactly the image of z_r under (1). By (4), the corresponding functional is the restriction from D_k to d_0D_k, with one suspension shift.

Finally, all the generators of D_k/(d_0) have even degrees: |d_j|=2^k-2^j is even for j\geq1 (and for k=1 the quotient is just \F_2). A nonzero odd-degree functional on D_k therefore cannot vanish on d_0D_k. This proves that the leading component survives suspension when d is odd.□

Proof. The product rule of Lemma 2.3 kills the decomposables. For the reverse inclusion, use the polynomial generators in (5). Suspension commutes with upper-index operations. When the bottom class changes from g_n to g_{n+1}, every intermediate word with those same upper indices increases in degree by one. Thus each lower index of a nonempty polynomial generator decreases by one. The original indices satisfy 0<j_1\leq\cdots\leq j_k, so the resulting indices satisfy

These are distinct nonzero primitive words in the target. The empty-word generator g_n suspends to g_{n+1}, also nonzero. Hence suspension is injective on the span of the polynomial generators modulo decomposables.

At level zero, localization at the particle gives

A word of length k has component degree 2^k. Replacing each such generator by its product with t^{-2^k} identifies this algebra with the Laurent polynomial algebra on t over a polynomial algebra concentrated in component zero. Thus H_*Q_0S^0 is polynomial, with generators the translates to component zero of the nonempty positive-excess particle generators Q_{j_1}\cdots Q_{j_k}t with 0<j_1\leq\cdots\leq j_k. A word of length k is translated by t^{-2^k}. Suspension ignores that translation and commutes with its upper-index operations, giving the same distinct nonzero words on g_1 with all lower indices decreased by one. One may equivalently start with the reduced generator u=t-1: positive operations on the additive monoid unit vanish, so the positive operations on u and t agree. This proves injectivity on indecomposables at level zero as well.

The injectivity here is into the full homology of the target. Some of the resulting words have initial lower index zero and are themselves squares; they must not be discarded by projecting the target to indecomposables. Their independence as primitive words is exactly what gives the asserted kernel description.□

Proof. Put QH=H^+/(H^+)^2, let \pi:H\to QH send 1 to zero, and define

The identity \pi(ab)=\epsilon(a)\pi(b)+\epsilon(b)\pi(a) shows that \partial is a derivation. If y_i are homogeneous polynomial generators, connectedness and grading give

For a polynomial f, the derivation rule yields

Only finitely many variables occur in f. If \partial f=0, inspect the coefficient of [y_i] for variables of largest degree occurring in f. There is no contribution from a higher-degree generator, so their ordinary polynomial partial derivatives vanish. Descending through generator degrees shows that every \partial f/\partial y_i vanishes. Over \F_2, this means that every exponent in every monomial of f is even; equivalently, f=z^2 for a unique polynomial z.

If f is a decomposable primitive, then \partial f=1\otimes\pi(f)=0, so this argument applies. Moreover,

The tensor product H\otimes H is again a polynomial algebra, so its Frobenius map is injective. Hence the displayed reduced coproduct is zero and z is primitive.

Finally, the homology Cartan formula in characteristic two gives

If z^2 is A-annihilated, injectivity of Frobenius forces \Sq_*^a z=0 for every a>0. Since these squares generate the positive Steenrod action, the root is A-annihilated. The polynomial homology descriptions above verify the algebra hypotheses for the infinite loop spaces under consideration.□

Proof. The cohomology sequence of the attachment shows that restriction in degree m is an isomorphism: m<2m-1. It also shows that the top relative class injects, since g^*:H^{2m-1}(T)\to H^{2m-1}(S^{2m-1}) is zero. Thus \widetilde u^2 is a scalar multiple of a.

To determine the scalar, we use the universal space for a cohomology class equipped with a nullhomotopy of its square. Write K(j)=K(\F_2,j) and set

A map representing u lifts to F because its square is zero. Looping the defining map represents \Sq^m on the fundamental class of degree m-1, which is zero by instability. Consequently the fibration

is trivial up to fibre homotopy. This is a statement about spaces over K(m-1); its splitting need not preserve loop products.

We need the additive square-zero property of Eilenberg–MacLane spaces: every positive-degree mod-two homology class of K(j), j\ge1, has zero square for its additive Pontryagin product. This follows from Serre’s cohomology calculation [30]; the following chain argument gives the precise property used here. Use the simplicial abelian group G_r=Z^j(\Delta[r];\F_2) of normalized simplicial j-cocycles, with faces given by restriction. Its Moore complex is \F_2 in degree j and zero elsewhere: below that degree there are no j-cochains; in higher degrees the vanishing faces and the cocycle equation force the remaining face to vanish. Thus its realization is an additive model for K(j). The normalized linearized complex N(\F_2[G]), rather than the Moore complex of G, computes its homology. Addition induces the chain product through the shuffle map and [a][b]=[a+b]. For a chain z of degree q>0, the terms in z*z pair under exchange of the two complementary (q,q)-shuffle blocks and of the two chosen simplices of z. This involution has no fixed point, and the paired summands agree because addition is commutative. They cancel over \F_2, even before normalization. Hence z*z=0, proving the property.

Let c\in H_{m-1}(\Omega F) be the bottom class. Its image in H_{m-1}K(m-1) has square zero for the additive multiplication. This is the additive square-zero property just proved. It follows that c^2 lies in the kernel of the projection to K(m-1). In degree 2m-2 this kernel is one-dimensional, generated by the bottom class of the fibre. The homotopy long exact sequence gives \pi_{2m-2}\Omega F=\F_2, and the space splitting shows that its Hurewicz map identifies its nonzero element with that fibre class.

It remains to show that c^2 is the nonzero fibre class. We do this by showing that the product of bottom classes survives the cobar spectral sequence. The space F is (m-1)-connected, so it is simply connected even when m=2. Adams’s cobar construction [1] has a multiplicative comparison with chains on its based loop space; we use the modern comparison in [28, Theorem 1]. Choose a 1-reduced Kan model for F; the comparison for its cobar algebra follows from [28, Corollary 8 and Theorem 9]. Each desuspended reduced chain has positive degree, so the tensor-length filtration is finite in each total degree. The resulting spectral sequence therefore converges to actual loop homology, without a completion ambiguity. Write its first page as

Let c_0\in H_m(F) be the bottom class. In length two and internal degree 2m, the term [c_0\mid c_0] is not a boundary for the coproduct differential: the relevant coefficient is dual to the square of the bottom cohomology class of F, and that square vanishes by the defining fibre construction. The term is a cycle since the reduced coproduct of c_0 is zero.

There is no higher incoming differential: its source would be (2-r,2m-r+1), hence have negative length or length zero and positive internal degree. A higher outgoing differential would land in length 2+r and internal degree 2m+r-1, where r\ge2. Such a term vanishes, since each reduced homology factor has degree at least m and

This argument includes m=2: the term is then in (s,t)=(2,4), and a higher outgoing target has internal degree r+3<2(r+2). The same connectivity bound shows that [c_0] is permanent and detects the unique bottom class c. Cobar concatenation represents the loop product on the associated graded. The nonzero surviving product [c_0\mid c_0] consequently implies c^2\ne0 in actual loop homology. It is therefore the nonzero fibre class.

Under any lift T\to F, the loop image of x is \langle u,\sigma_*x\rangle c, by homology suspension and the one-dimensional bottom homology. Thus the adjoint of the composite S^{2m-1}\to T\to F has Hurewicz image \langle u,\sigma_*x\rangle c^2. The injectivity just proved shows that this composite is null precisely when the coefficient is zero.

Finally, \widetilde u^2=0 is equivalent to the existence of a lift of its representing map over C_g to F. Such a lift restricts to a lift on T whose composite with g is null; conversely, any such lift extends over the attachment. The calculation applies to every choice of lift. One can also see the independence directly: changes of lift are measured by H^{2m-1}(T), whose pullback along g is zero. This proves (10).□

Proof. The hypothesis h(g)=0 gives an exact sequence of positive cohomology A-modules

The relative module structure implies that a times any positive-degree class is zero. Choose exterior algebra generators of B lifting a homogeneous basis of Q B, choosing the basis in degree m so that at most one generator has nonzero \lambda. Choose lifts of these generators to \widetilde H^*(C_g). Their squares are zero except possibly for that one distinguished degree-m generator, whose square is a by Lemma 3.1.

The nonempty square-free monomials in these lifts, together with a, form a graded vector-space basis of \widetilde H^*(C_g). Quotienting by all products could kill a, since it may itself be a square. Instead we remove only products of distinct generators. Let L be the span of the products of at least two distinct lifted generators, with no repetitions. Restriction maps L isomorphically onto B^2. We claim that L is an A-submodule. Apply Cartan to such a product and expand each factor in the lifted basis. A term involving a vanishes against any remaining positive factor. A repeated generator also gives zero, except that two occurrences of the distinguished generator give a; if another positive factor remains, this term again vanishes.

The only possible contribution outside L is therefore a product of exactly two linear occurrences of the distinguished generator. A positive Steenrod operation on a factor cannot supply such a linear coefficient, because \lambda annihilates positive Steenrod images. If both operations are zero, the original two distinct generators cannot both be the distinguished generator. This proves the claim.

Taking the quotient by L gives (11). In degree m, the square detector is unaffected, and its value is still \lambda.□

Proof. Pull back (11) to the summand M=M_k(n) and choose a graded additive section s. In total degree 2m, define the scalar \varphi(\alpha\otimes z) by

It respects balancing. Indeed, in comparing \alpha\beta\otimes z and \alpha\otimes\beta z, the intermediate degree |\beta|+|z| is strictly less than 2m because |\alpha|>0. The possible discrepancy in the section therefore vanishes, and associativity gives equality. Thus \varphi descends to (I\otimes_A M)_{2m}.

Since n>0, the module M_k(n) is a positive suspension of an unstable polynomial module; hence \Sq^m v=0. This is also true in the one-dimensional bottom summand. Proposition 3.2 therefore gives \varphi(\Sq^m\otimes v)=\lambda(v)=1, proving (12).□

Proof. Under the adjunction between suspension ring spectra and E_\infty spaces, \tau gives an E_\infty map from the additive structure of QS^0 to the multiplicative space of the truncated ring. The source is grouplike, so this map lands in units. Its weight-zero coordinate is constantly 1, since \tau is augmented; the positive coordinates at the source basepoint are zero. Multiplication in the truncated graded ring has weight-i coordinate equal to the right side of (13). This proves that formula, including its compatibility with all lower coordinates in the single chosen truncation. Group completion causes no further terms: 1+u is a unit in the target, so the map is precisely the extension of the free-monoid map through the localization (7).

To evaluate (13) on A_1\times A_2, distribute the two primitive diagonals across its additive summands. Either the two positive inputs enter the two linear terms, giving their Pontryagin product, or both enter one bilinear summand, giving its block pairing. Every other allocation pairs a positive class with the zero point and vanishes. This gives (14). Although j_i need not preserve addition, it is a based space map, so it preserves ordinary primitivity and commutes with the dual Steenrod operations. When A_1=A_2, the terms with indices t and i-t cancel by block commutativity; only the equal-block term can remain. This proves (15).□

Proof. Put X=\Omega^\infty E_1 and Y=\Omega^\infty E_2. The pairing is a coherent family of additive infinite loop maps in its first variable. Indeed its adjoint is a spectrum map E_2\to F(E_1,E_3); passing to zero spaces gives a map from Y to the space of spectrum maps E_1\to E_3, and hence to the space of additive infinite loop maps. Connective covers may be taken without changing the zero spaces. In particular, the naturality homotopies for the additive E_\infty structure vary coherently over Y.

For an unbased space U, write

where C_2 interchanges the factors. The naturality just described factors the pairing of an additive square operation with Y through

Let W be the free periodic \F_2[C_2]-resolution with generators e_j and differential d e_j=(1+\tau)e_{j-1}. The chain model

computes H_*\mathcal E_2U. Choosing cycles for a homology basis of U gives an equivariant quasi-isomorphism before taking homotopy orbits. It follows that a basis of the resulting homology consists of unordered unequal pairs in resolution degree zero and classes e_j\otimes z\otimes z, j\geq0, on equal basis elements.

We claim the universal formula

with terms of negative resolution degree omitted. This is the extended-square calculation underlying May’s formula [24, §2, formula (2.11), pp. 246–247]; we include its verification.

Transfer from \mathcal E_2U to U^2 sends an unequal pair to the sum of its two ordered versions and kills all equal-pair classes. Thus it detects the unequal-pair part injectively. Transfer is natural for \delta, since its pulled-back double cover is the double cover of \mathcal E_2X\times Y obtained from \mathcal E_2X. The input of (17) has transfer zero. The output therefore has no unequal-pair terms.

For \beta\in H^qU, let P(\beta)\in H^{2q}\mathcal E_2U be its external square, and let h\in H^1BC_2 be the characteristic class of the double cover. In the dual resolution model, a cocycle f representing \beta gives P(\beta) by f\otimes f in resolution column zero. This description is independent of f: if f'=f+dc, then P(f')-P(f) is the total coboundary of the cochain whose columns zero and one are, respectively,

Multiplication by h^l is the resolution shift, so h^lP(\beta) is represented by f\otimes f in column l. It evaluates on e_j\otimes z\otimes z as \delta_{lj}\langle\beta,z\rangle^2.

External squares respect cross products over the common base BC_2. Moreover, pulling P(\beta) back along the Borel diagonal BC_2\times U\to\mathcal E_2U gives

this is the total-square definition of the Steenrod operations. Consequently, for \alpha\in H^*X and \beta\in H^qY,

Its evaluation on the input in (17) can be nonzero only when

The coefficient is then \langle\alpha,u\rangle\langle\beta,\Sq_*^t y\rangle. Tensor cohomology classes detect all equal-pair coefficients; in each evaluation only finitely many homology basis elements are involved. Together with the transfer calculation, these evaluations prove (17).

Now take j=i-a and apply the label pairing, followed by the additive square operation. The term with index t becomes Q^{i+t}(u\circ\Sq_*^t y). When y is A-annihilated, only t=0 remains. Its resolution index is i-a-b, which proves both cases of (16).

Finally, for a primitive y in the basepoint component, \Delta y=y\otimes[0]+[0]\otimes y. Distributivity of a spectrum pairing over addition gives

because pairing a positive-dimensional class with the zero point is zero. If one first factor is a component point [c], its pairing with [0] is the additive unit [0] in the target; the same formula proves the translation assertion.□

Proof. The free infinite loop space homology theorem and group completion describe H_*Q_0X_v as a polynomial Pontryagin algebra [9, Part I]. Its generators are the positive-dimensional particle classes from a basis of H_*B\Sigma_v, and the admissible positive-excess Dyer–Lashof words of positive length on particle classes, including the degree-zero particle. Each is translated to the basepoint component. There is one degree-zero particle component to invert in group completion; these translations do not change stable homology evaluation in positive dimensions.

The counit takes the length-zero positive-dimensional generators independently to their corresponding basis classes in H_*E_v. Stable homology evaluation kills products of positive-dimensional classes. It also kills positive-length words: upper-index operations commute with homology suspension, and repeated suspension eventually makes their outer operation unstable. Therefore, modulo Pontryagin decomposables, the hypothesis \varepsilon_{v*}B=0 expresses B as a sum of positive-length words only.

Pair this expression with B. The preceding lemma discards the decomposables and translations in the first factor. A remaining term has the form Q^q c with q+|c|=i. Its pairing can be nonzero only if

so |c|=0. In this case

Thus B\circ B is Pontryagin decomposable, and another pairing with B is zero by Lemma 4.3.

We also record a property used below: the block pairing of two positive primitives in their basepoint components is primitive. Indeed, naturality of the ordinary diagonal leaves the two end terms; each crossed term pairs a positive-dimensional class with the zero point and vanishes.□

Proof of Proposition 4.1. It suffices to exclude fourth powers, since every higher iterated square is a fourth power. Suppose, therefore, that

Applying Lemma 2.9 twice shows that the homogeneous positive-dimensional root z is primitive and A-annihilated.

Choosing the detector. Let r=2^{k_0}\ge2 be the leading weight of this root, as in Lemma 2.7, and let z_r\in H_*D_rS be its leading component. Put

Here b is positive and even, \ell\ge2, and d=2b. In the polynomial associated weight algebra the leading component of x is

and is nonzero. This square is taken in the graded homology algebra H_*\mathbb P(S), rather than by the unstable pairing \circ.

The class p is primitive for the linear weight coproduct and is a product of positive-dimensional factors. It therefore dies under the weight suspension (1). One may see this before group completion: homology suspension from the free monoid kills positive-dimensional Pontryagin products, and Lemma 2.3 identifies its weight components. Thus, under the dual description in Lemma 2.6, the functional defined by p on the Dickson algebra D_\ell kills d_0D_\ell. It is nonzero on some monomial in d_1,\ldots,d_{\ell-1}. The restriction description of that lemma lifts d_j to w_{s-2^j}(\rho_s), where \rho_s is the permutation representation of \Sigma_s. Hence there is a class

which is a polynomial in the even-index Stiefel–Whitney classes of \rho_s.

Removing the block correction. Use one truncation through weight s=2r to define all the maps j_i. We claim

The first equality is (15). For the second, apply that formula to x=z^{*2}:

Each J_i z lies in the basepoint component and is primitive and A-annihilated, by Lemma 4.2. Thus (J_rz)^{*2} is a primitive Pontryagin decomposable, and its block square is zero by Lemma 4.3. Block commutativity over \F_2 cancels the mixed terms. This gives (J_rx)^{\circ2}=(J_{r/2}z)^{\circ4}. The indices are integral because r is a power of two at least two, and all block products have total weight at most s.

For the last equality in (19), observe that stable counit followed by an adjoint recovers the original spectrum map. Thus, on positive-dimensional homology,

In particular B=J_{r/2}z has zero stable homology evaluation, because r/2 is below the leading weight of z. It is a primitive, A-annihilated class in Q_0X_{r/2}. Lemma 4.4 gives B^{\circ3}=0, hence B^{\circ4}=0. This proves the last equality in (19). The same counit identity gives

Contradictory cone squares. Let

be the adjoint of j_s\circ f:S^{2b}\to Q_0X_s, using the equivalence QX_s\simeq\Omega Q\Sigma X_s. The space \Sigma X_s, and hence T, is connected. The loop product under this equivalence is the additive Pontryagin product. Equation (19) therefore says that the loop adjoint has Hurewicz class (J_sx)^{*2}, and gives h(g)=0 after homology suspension.

Let \varepsilon:\Sigma^\infty T\to\Sigma^\infty\Sigma X_s be the stable counit and define

Counit compatibility with homology suspension and (21) give

Stable Steenrod naturality gives

With m=b+1, the hypotheses of Lemma 3.1 now hold. The unique extension \widetilde u\in H^mC_g consequently satisfies

On the other hand, the restriction of \Sq^{m-1}\widetilde u to T is

This class has an integral torsion lift. Indeed, for real vector bundles the reduction of the integral Pontryagin class is

For completeness, this follows from p_i(\xi)=(-1)^ic_{2i}(\xi\otimes\C): the underlying real bundle of \xi\otimes\C is \xi\oplus\xi, and reduction of its Chern classes gives its even Stiefel–Whitney classes. Because v is a polynomial in even-index Stiefel–Whitney classes, its square is the reduction of an integral polynomial V\in H^{2b}(B\Sigma_s;\Z) in the Pontryagin classes of \rho_s. The degree is positive. Restriction to the trivial subgroup followed by transfer shows that positive-degree integral cohomology of the finite group \Sigma_s is annihilated by |\Sigma_s|; thus V is torsion. The integral class

is torsion as well.

Since H^{2b+1}(S^{2b+1};\Z)=\Z is torsion-free, g^*U=0. The integral cohomology sequence of the mapping cone supplies an extension \widetilde U\in H^{2b+1}(C_g;\Z). Restriction

is injective, since the preceding group in the cone sequence is \widetilde H^{2b}(S^{2b+1};\F_2)=0. Consequently

The mod-2 Bockstein \Sq^1 annihilates an integral reduction. Finally b is even, so the Adem relation gives

contradicting (22).□

Proof. Milnor’s polynomial dual and coproduct description [26, §5, Theorem 3, and Appendix 1] gives A_*=\F_2[\xi_1,\xi_2,\ldots]. It implies that restriction to E_* is the quotient by the squares of these generators: their images are the exterior generators dual to the Milnor primitives. Squaring is an injective Hopf algebra homomorphism on A_*; its degreewise dual is the indicated map V. The formulas on the squares follow by pairing with the powers of \xi_1.

For completeness, the kernel can be read directly from the Milnor coproduct. After restricting its first factor to E_*, that coproduct becomes

The other terms have an even positive exponent in the first factor and therefore vanish. A polynomial has trivial coaction under this map exactly when it is a square: the coefficients of the single generators \bar\xi_i are its ordinary partial derivatives, and their simultaneous vanishing in characteristic two is equivalent to every exponent being even. Conversely, a square has trivial coaction since \bar\xi_i^2=0. Thus the annihilator of I_EA is exactly the subalgebra of squares, which also annihilates \ker V. Degreewise duality proves I_EA=\ker V. Conjugation preserves E and the subalgebra of squares in A_*, so it also gives AI_E=\ker V. Finally KM=I_EM, using K=I_EA and AM=M, and hence the stated quotient module is well defined.□

Proof. The q_i are derivations and q_i x_s=x_s^{2^{i+1}}. The initial formulas on Dickson generators appear in [32, Corollary 2.3(b)] and are also recorded, with indexing shifted by one, in [27, Proposition 2.3]; we give the calculation. Apply q_i to P(x_s)=0. Only the linear power of x_s contributes a derivative, so

This identity holds on the whole binary span of the x_s, by additivity. A linearized polynomial of degree at most 2^{k-1} is uniquely determined by its values on that span: the difference of two such polynomials cannot have 2^k distinct roots unless it is zero. For i<k-1 this gives the partial-derivative formula. For i=k-1, substituting x_s^{2^k}=\sum_{j<k}d_jx_s^{2^j} gives the Euler formula.

For t\geq0, raising the same root equation to the power 2^{t+1} gives, as an equality of derivations on P_k,

Induction on t expresses every later primitive in the first k with square coefficients. These coefficients belong to D_k.

Since d_0 is the product of the nonzero binary linear forms and the total square of such a form a is a+a^2, the total square of d_0 is divisible by d_0. Its quotient is invariant and therefore belongs to D_k. The Cartan formula now proves the stability of every d_0^pD_k.

Each nonzero term in (26) raises the d_0-exponent by exactly one. Thus no monomial at exponent p is hit. Above this floor, a monomial with some even b_j is the q_{j-1}-image of the monomial obtained by decreasing c by one and increasing b_j by one. This preimage is still in d_0^pD_k. If all b_j are odd, the Euler primitive on the monomial at exponent c-1 has coefficient

It hits the given monomial exactly when c\not\equiv k\pmod2. The images of the first k primitives are spanned by these individual monomials. Later primitives add no new images: a square coefficient passes inside any of the derivations. This proves the basis assertion, including the case k=1, when the list of b_j is empty.□

Proof. On P_k define C_x to be zero on a monomial unless all its exponents are odd, and in the latter case set

This is the dual Kameko squaring map [19]; see [16, §2, p. 4069] for the dual formula. It is also the polynomial top-form version of Cartier’s differential-form construction [7]; see also [8, Chapter II, §6, pp. 200–203]. We include the polynomial identities and naturality needed for its restriction to the present ideals. The binomial formula for squares on a one-variable polynomial, together with the Cartan formula, gives

Indeed, on an odd exponent the parity of \binom{2a+1}{2i} is the parity of \binom ai; an odd square makes that exponent even. An even input exponent can never contribute an odd output exponent, since the corresponding binomial coefficient then vanishes. Applying these facts in all variables proves the identities.

The operator C_x commutes with \GL_k(\F_2). To see this, write a polynomial uniquely as a sum of square-free monomials with square coefficients. The coefficient of x_1\cdots x_k after a linear change of variables is multiplied by its determinant; no square-free monomial of smaller degree can contribute to that coefficient. Over \F_2 the determinant is one, and taking the square root of the coefficient commutes with the linear change of variables. Hence C_x preserves D_k.

We compute this restriction by an elementary form of Cartier naturality. If y_1,\ldots,y_k are algebraically independent polynomials and J=\det(\partial y_j/\partial x_i), then

Here C_y is the same odd-exponent extractor in formal variables y_j. For a proof, the polynomial differential-form complex in one variable has cohomology generated by 1 and [x\,dx] over \F_2[x^2]. Tensoring these complexes gives the corresponding assertion in k variables. Thus there is an isomorphism from the algebra of forms, regarded without differential, to its cohomology, given by

This isomorphism is natural under polynomial substitution. On differentials the required fact is that f\mapsto[f\,df] is additive and satisfies the derivation rule with coefficients squared: its additivity follows because the cross term is d(fg), and its product rule follows by expanding (fg)d(fg). In top degree the inverse isomorphism sends [h\,dx_1\cdots dx_k] to C_x(h)\,dx_1\cdots dx_k. Pulling back a top form in the y variables proves (30).

For Dickson coordinates y_j=d_{j-1} the Jacobian is

In fact, differentiating P(x_s)=0 with respect to each x_i gives the matrix identity whose two factors have entries x_s^{2^j} and \partial d_j/\partial x_i, and whose product is d_0 times the identity matrix. The determinant of the first factor is d_0: it is nonzero, is divisible by every nonzero binary linear form, and has degree 2^k-1. Taking determinants gives (31).

For c\geq k-1, apply (30) to f=d_0^{c-k+1}\prod_{j\geq1}d_j^{b_j}. It is zero unless c\equiv k\pmod2 and every b_j is odd; otherwise it is exactly (28). The same assertion holds for smaller nonnegative c. Multiply first by a sufficiently large even power d_0^{2N}, use C_x(d_0^{2N}f)=d_0^NC_x(f), and cancel d_0^N in the polynomial domain. The output exponents in the nonzero cases are nonnegative.

The description of the hits in Lemma 5.3 now shows that C_x descends to coinvariants and is a bijection on the stated above-floor monomials. Squares do not decrease the d_0-exponent: this follows from the divisibility of the total square of d_0 used above and the Cartan formula. Consequently the above-floor span is a submodule. Equation (29) gives the claimed action on its image. The least possible target exponent is the one displayed in the statement, and a direct degree calculation in (28) gives the change d\mapsto(d-k)/2.□

Proof. The monomial lies above the exponent floor e in M_k(e). Since the functional factors through exterior coinvariants, Lemma 5.3 forces c\equiv k\pmod2 and all its other exponents to be odd. Its total degree there is m-1, so (24) gives n\equiv k\pmod2.

Restrict the induced functional to the above-floor submodule and use Lemma 5.4. The first allowed exponent in this submodule is c=n, so its image has first exponent

Give that image the grading shift e', and put

The image is exactly M_k(e'), and its paired monomial has degree m'-1: if the old internal degree is d, then e'+(d-k)/2=(m-2)/2. The new functional is again annihilated, because the Cartier map intertwines the halved action. If m' is even, its paired monomial cannot be at the floor c'=e'; a monomial at that floor has even total degree, whereas m'-1 is odd. It therefore lies in the next ideal and satisfies the same hypothesis, allowing the step to repeat.

Iterate while the paired parameter m is even, for t=\nu_2(m) steps. With b=2^t, the resulting parameters obey

The integrality of n_t gives the congruence in (32), and the last two relations give its inequality. All shifts e_t=n_t-1 are nonnegative.

If m=b, the final paired monomial has degree zero in M_k(e_t). The smallest degree in that module is 2^ke_t, so e_t=0. Thus

Since n,k\geq1, this equality forces n=k=1.□

Proof. Put M=M_k(n) and B=A/K. The exact sequence of right A-modules K\to I\to I_B\to0, followed by tensoring with M, shows that the kernel of (34) is the image of K\otimes_A M. Here we have used

Multiplication gives a surjection I_E\otimes_EA\to K=I_EA; tensoring again gives a surjection I_E\otimes_EM\to K\otimes_AM. Thus every element in this kernel has a preimage in I_E\otimes_EM. The action maps from both tensor products to M commute with this map. A preimage of a cycle is therefore a cycle itself.

We prove that every even-total-degree cycle in I_E\otimes_EM is zero. Write a homogeneous such cycle as

Each z_i has odd total degree in M. By (24), its monomials have d_0-exponent c\geq n+1 and parity opposite to n, hence opposite to k. Lemma 5.3 gives z_i\in I_EM.

This last property allows us to replace later primitives by the first k inside the tensor product. If q_i=\sum_{j<k}a_j^2q_j on M, then

It suffices to check this for z_i=q_sy. Balancing, commutativity of the exterior generators, and their vanishing on squares give

Thus z has a cycle representative using only q_j, j<k, whose coefficients still have exponents at least n+1.

Let E_{[k]}=\Lambda(q_0,\ldots,q_{k-1}) and L=D_k[d_0^{-1}]. We first show that L is projective over E_{[k]}. For T\subseteq\{0,\ldots,k-1\} write q_T=\prod_{i\in T}q_i, and put

The first k-1 primitives take v_0 to d_0^{-1}, and the last, Euler primitive takes d_0^{-1} to 1. Hence q_{\{0,\ldots,k-1\}}v_0=1. The action map from the free module E_{[k]}\otimes_{\F_2}L to L has the following section:

It is E_{[k]}-linear: multiplying its first factor by q_i and reindexing the subsets gives s(q_ix), since terms with repeated q_i vanish. To compute the composite with the action map, expand each q_T(v_0q_{T^c}x) by the Cartan formula. For a fixed subset U, the term (q_Uv_0)(q_{U^c}x) occurs once for every T with U\subseteq T\subseteq\{0,\ldots,k-1\}. Its coefficient is 2^{k-|U|}, so only the full subset remains. The composite is therefore x. The formula preserves degrees, since v_0 has the negative of the degree of the product of all the q_i. This proves the asserted projectivity.

Consider now the tensor product of the usual periodic free resolutions for the one-generator exterior algebras. After tensoring with an E_{[k]}-module, its first differential is

The degree-two terms have one generator for each repeated pair i,i and each distinct pair i,j. Writing e_i for the degree-one generators, their images are

The cokernel of this differential is I_{E_{[k]}}\otimes_{E_{[k]}}M: the displayed relations become q_i^2\otimes y=0 and (q_iq_j+q_jq_i)\otimes y=0 under balancing. Projectivity of L implies that every degree-one cycle in this complex with coefficients in L has a degree-two filling.

Apply this to the cycle representative obtained above, ignoring its harmless shift by n. Choose a filling with Laurent-polynomial coefficients. In each of the first two differentials, every nonzero term raises the d_0-exponent of its coefficient by exactly one, by (26). Delete from the filling all monomials with exponent less than n. Their differential has exponents at most n, whereas the retained part has exponents at least n+1. Since the given cycle has only exponents at least n+1, the retained part is still a filling. Its coefficients belong to d_0^nD_k. All Laurent polynomials and sums used here are finite, so this deletion requires no completion.

We have proved that the representative is a boundary already over E_{[k]} with coefficients in M. Equivalently, it is zero in I_{E_{[k]}}\otimes_{E_{[k]}}M, and hence also in I_E\otimes_EM. Its image in I\otimes_AM is zero, which proves the lemma.□

Proof. The Adem relation \Sq^m=\Sq^1\Sq^{m-1} gives \Sq^m\otimes v=\Sq^1\otimes\Sq^{m-1}v. For n\geq2 the second operation exceeds the internal polynomial degree m-n, so this is zero. For n=1 it is the internal square v^2. By linearity we may take v=d_0^c\prod_{j\geq1}d_j^{b_j}, where c\geq1. For k\geq2, formula (26) gives

The preimage belongs to d_0D_k. For k=1 use instead \Sq^1(d_0^{2c-1})=d_0^{2c}. Balancing once more and using (\Sq^1)^2=0 proves the assertion.□

Proof. Apply the total square simultaneously to the variables x_s and U in (23). Since \Sq(U+a)=(U+a)(U+a+1) and P is additive,

Comparison of the coefficients of U^{2^j} gives \Sq(d_0)=P(1)d_0 and, for 1\leq j<k,

Summing these identities yields

which is (37). The term of degree 2|d_j|-1 in that formula is exactly d_0d_{j+1}. In the Cartan expansion of an operation one below the top on a product, exactly one factor receives its operation one below the top and every other factor receives its top operation. Equal choices cancel in pairs, leaving precisely (38).□

Proof. The tensor is a cycle, because \Sq^m exceeds the internal degree m-n of v. We induct on \nu_2(m), using Lemma 5.7 throughout; its parity assumption follows from (32). Here we halve coinvariants of the current ideal, with floor n. This differs from the floor n-1 used for the functional in Lemma 5.6.

Suppose first that b\geq4. The image of v in exterior coinvariants is either zero, in which case the injectivity lemma gives the result immediately, or is an above-floor survivor. Indeed c\geq n+b-2\geq n+2. In this case the first allowed exponent in the above-floor part of M_k(n) is n+2, and its Cartier image has first exponent

With this shift the image is M_k(n'), with total degrees halved. The image of our cycle in (34) consequently comes from \Sq^{m/2}\otimes v' in I\otimes_AM_k(n'), where v'=C(v). Put c'=(c+k-2)/2 and b'=b/2. Then

Thus the same bounds hold for m/2,n',c'. Since m/2 is still not a power of two, induction makes this latter tensor zero, and injectivity gives the required vanishing of the original cycle.

It remains to treat b=2. Write m=2L, with L\geq3 odd. All coinvariants of M_k(n) have even total degree: this is clear at the floor from (24), and above it from c\equiv k\equiv n\pmod2. Regrade them by half and call the resulting A-module C_n. The projected tensor is

The second factor is the class of the original operation \Sq^{m-2}v. It suffices to show that this class is zero or an image of \Sq^1 in C_n, since balancing then introduces (\Sq^1)^2=0.

If n\geq3, the operation already exceeds the internal degree of v. If n=2, then k is even and at least two, and the operation is the internal square v^2. This lies above the floor and has all nonzero-index exponents even, so it is zero in coinvariants by Lemma 5.3.

Suppose n=1, so k is odd. For k=1 the coinvariant basis, including the floor, consists of all odd monomials in d_0. The map C_x identifies C_1 with \Sigma D_1, and the image of v=d_0^{2L-1} is d_0^{L-1}. Its image under \Sq^{L-1} is

as required.

Finally, suppose n=1 and k\geq3 is odd. The internal degree of v is m-1, so Lemma 5.9 applies to \Sq^{m-2}v. In each summand of (38), at most one exponent of d_j with j\geq1 is odd. Since there are at least two such indices, every term above the floor dies in exterior coinvariants. The only possible floor term comes from the index j=0 when v=d_0w, where w involves only d_j, j\geq1. It is d_0d_1w^2.

This remaining term is an image for the halved first square. Formula (37) gives \Sq^2d_2=d_1 and \Sq^1d_0=\Sq^2d_0=0 for k\geq3. Together with (26) and the Cartan formula, it therefore gives the explicit identity

The extra term is an exterior hit in d_0D_k, with preimage

Hence the class of d_0d_1w^2 is the original \Sq^2-image of [d_0d_2w^2], or equivalently a \Sq^1-image after halving. This completes the final case, and the injectivity lemma finishes the proof.□

Proof of Lemma 5.1. The odd assertion is Lemma 5.8. For even m, the functional hypothesis gives (32) by Lemma 5.6. If m is not a power of two, Lemma 5.10 contradicts (25). If m is a power of two, the last assertion of Lemma 5.6 gives n=k=1.□

Proof. Since \sigma_*z_e=0, Lemmas 2.8 and 2.9 give (40) with a primitive A-annihilated root. Its degree is positive, so m\ge2.

If e=0 and \sigma_*w=0, the same lemmas make w a square. Then z_0 is a fourth power, contradicting Proposition 4.1. Thus the root suspends nontrivially in this case.

For e\ge1, we instead use the preceding spherical class z_{e-1} to prove that |w| is odd. First suppose that e\ge2, so both levels have the direct weight decomposition of Lemma 2.6. Choose a nonzero weight 2^l component of z_e. Here l\ge1, since the degree d+e is greater than the bottom degree e. Its functional on M_l(e) comes, after the grading shift, from the functional of z_{e-1} on M_l(e-1). Vanishing of z_{e+1} says that it annihilates d_0^{e+1}D_l, by (4). It must therefore pair nontrivially with some floor monomial

In M_l(e-1) this monomial is above the floor. Since the lifted functional is A-annihilated, Lemma 5.3 gives

The degree formula and (40) yield

For l\ge2, only the term with j=1 is odd. For l=1, the right side is e, which is odd by the congruence. Thus m-1 is odd.

If e=1, (40) shows that d is odd. Choose the leading weight 2^l of z_0. Lemma 2.7 says that it survives to z_1, with a functional lifted from the annihilated functional on M_l(0). Since z_2=0, the same floor argument applies to this component of z_1, and again (41) proves that m-1 is odd.

An odd-degree class cannot be a square. Therefore the suspension-kernel and primitive-root lemmas imply \sigma_*w\ne0 whenever e\ge1.□

Proof. Suppose that \sigma_*w has a nonzero component of weight 2^k with k\ge1. Its degree-m functional on M_k(n) is lifted from the A-annihilated functional of w on M_k(e). The nonzero tensor supplied by (42) therefore satisfies all the hypotheses of Lemma 5.1. That lemma forces n=1, contrary to n=e+1\ge2. Hence \lambda is nonzero on the bottom summand, giving

Lemma 3.1 now says that the extension of the weight-one bottom class in C_{f_n} has nonzero \Sq^n, equal to the top relative class.

Let \epsilon:\Sigma^\infty QS^n\to\Sigma^nS be the stable counit and define

This is the stable adjoint representing \alpha in stem d=n-1. The counit is the weight-one projection, by Proposition 2.1. The commutative square with identity on the attaching sphere induces a map

to the stable two-cell cofiber. Its pullback carries the bottom class to the weight-one bottom class and the top relative class to the top relative class. Both top relative classes inject in mod-2 cohomology. Naturality therefore makes \Sq^n nonzero in C_{\alpha_n}. This is nonzero mod-2 stable Hopf detection of \alpha.

Adams’s Hopf invariant one theorem, in its stable two-cell form [3, Theorem 1.1.1(d)], permits n=1,2,4,8. Since d=n-1>0, we obtain d=1,3,7.□

Proof. Leading weight. Here n=1, and we have proved that m is even. Thus |w|=m-1 is odd. Choose the leading weight 2^k of w in the level-zero expansion. It has k\ge1, because w has positive degree. By Lemma 2.7, this weight survives in \sigma_*w, with functional lifted from an annihilated functional on D_k=M_k(0). Combining Corollary 3.3 with Lemma 5.1 gives

Consequently z_0=w^2 has leading weight four: its leading term is the square of the nonzero weight-two term, and Frobenius is injective in the associated free polynomial homology algebra. Also d=2m-2=2^{j+1}-2.

Adams filtration. We bound the Adams filtration of the given element. Let \mathcal I be the augmentation ideal of \Sigma^\infty_+QS^0. Kuhn’s lifting theorem [21, Theorems 1.2–1.3], applied to X=S and the coefficient ring H\F_2, says that an element of mod-2 Adams filtration at least s_0 has Hurewicz image lifted from H\F_2\wedge\mathcal I^{2^{s_0}}, where \mathcal I^{2^{s_0}} is the corresponding derived augmentation-ideal power of \Sigma^\infty_+QS^0. Its hypotheses hold: both spectra are connective and \pi_0S\to\pi_0H\F_2 is onto. At the prime two the localization away from (p-1)! in that theorem is vacuous. The theorem does not require QS^0 to be connected.

To use this with the leading weight, take the augmented algebra map

from (8), and let J denote the positive ideal in its target. Regard J as a nonunital algebra in the full category of nonnegatively graded spectra, with ordinary Day convolution and with J_r=0 for r=0 and r\ge8. Using the reduced commutative operad \mathrm{Com}, whose arities are positive, its derived eighth power is the bar construction

see [21, Definition 2.7] and [22, Definition 3.1 and Example 3.2]. Compute this derived bar construction with a free resolution in strictly positive weights. Every simplicial summand then has at least eight inputs of weight at least one; no nullary operation can remove such an input. Its total weight is therefore at least eight. The canonical map J^8\to J preserves weight and is zero, since J has only weights one through seven. The direct-sum functor from the full graded category is strong symmetric monoidal and preserves realizations and homotopy orbits. It therefore carries the graded free resolution and bar construction to the corresponding ungraded derived construction, with the same zero canonical map. Extension of coefficients to H\F_2 also preserves this construction [22, Theorem 3.4(d)]; in particular, the canonical map remains zero after smashing with H\F_2.

By naturality, the Hurewicz image of an element of Adams filtration at least three would vanish under every positive-weight projection below eight. The weight-four image just found is nonzero. Thus \alpha has Adams filtration at most two. This uses only a lower bound on possible weights; it does not identify Adams filtration with an exact weight.

Kervaire detection. Positive stable stems are finite by Serre’s finiteness theorem [29, Chapter V, §3, Proposition 3]: choose an odd sphere dimension N>d+1, so that \pi_{N+d}(S^N) is both stable and finite. Their odd-primary parts have zero mod-2 Hurewicz image and zero mod-2 detecting invariants, so we may discard them and use the convergent mod-2 Adams spectral sequence for the remaining two-primary group. More precisely, Adams’s convergence theorem identifies its successive filtration quotients with E_\infty and its filtration intersection with the prime-to-two torsion, which has been removed [2, Theorem 2.1(i),(v),(vi)]. Write F^s\pi_d^S for the subgroup of Adams filtration at least s in this two-primary group. Filtration zero has no positive-stem sphere class. Filtration one has internal degrees 2^a, hence stems 2^a-1; none is the positive even stem in question. In filtration two, the classical calculation [3, Theorem 2.5.1(1)–(2)] gives basis elements h_bh_a, where b\ge a\ge0 and b\ne a+1. The identity 2^a+2^b=2^{j+1} forces a=b=j. Therefore

and the given element is detected by the surviving class h_j^2.

Browder’s detection theorem identifies the framed Kervaire invariant of an individual stable sphere element with its detection by h_j^2. The original framed Arf identification and secondary pairing occur in [5, Theorem 3.2, Proposition 7.6 and Theorems 7.7–7.8]; the precise elementwise statement is also recorded in [4, Theorem 1.1]. This statement includes the low framed cases j=1,2,3. It applies to the particular \alpha above, whose nonzero class in F^2\pi_d^S/F^3\pi_d^S is represented by the surviving h_j^2, and gives Kervaire invariant one.

In particular the detector is unchanged by adding an element of F^3\pi_d^S. Such an element cannot be detected by a filtration-two class, so the same elementwise theorem gives it Kervaire invariant zero; additivity of the framed invariant gives the assertion. This use of Browder’s theorem concerns the invariant of each element, in addition to the existence of a permanent cycle.□

Proof of Theorem 1.1. Every positive-degree stable class with nonzero Hurewicz image has a last nonzero suspension. Propositions 6.2 and 6.3 show that it has the stated nonzero Hopf or Kervaire invariant. These two degree cases are disjoint.

In a Hopf degree, choose the corresponding class among \eta,\nu,\sigma. In a Kervaire degree, choose any Kervaire-invariant-one representative \theta_j; its existence in this case is already supplied by the class under discussion. Each detecting invariant is additive with values in \F_2. For the stable Hopf invariant, attach two top cells to the same bottom sphere by representatives \alpha and \beta. If u is the bottom class and a_\alpha,a_\beta are the relative top classes, naturality on the two one-cell cofibers gives \Sq^{d+1}u=H(\alpha)a_\alpha+H(\beta)a_\beta. Pinching the attaching sphere induces a map from the cofiber of \alpha+\beta to this two-attachment cofiber; it pulls both relative classes to the single top class. Naturality therefore gives H(\alpha+\beta)=H(\alpha)+H(\beta). For the Kervaire invariant, disjoint union of framed manifolds takes their quadratic forms to orthogonal sums and adds their Arf invariants. Subtracting the chosen representative therefore gives an invariant-zero class. If this difference had nonzero Hurewicz image, the proved necessity would force its invariant to be one, a contradiction. Thus its Hurewicz image is zero.

Every nonzero image is consequently the image of the chosen Hopf or Kervaire representative in that degree. These representative images themselves belong to the image, so they span it. No existence assertion for an additional \theta_j is used or obtained from this last step. This proves Curtis’s conjecture in all positive degrees.□