The Kervaire invariant problem at the prime three

Proof. For n=3^r with r\geq1, the strong Kervaire condition in stem 4n-2 asks for an order-three representative detected by b_{r-1}; such a representative is Amelotte’s \theta_r. This is the index change in Lemma 6.2. Theorem 1.1 therefore gives this condition exactly for r=1,3,4, or n=3,27,81. Amelotte’s Theorem 1.1 [1] supplies a nontrivial H-space product at each of these indices, with Anick dimensions 2\cdot3^r+1 and 2\cdot3^{r+1}+1. At r=4 these dimensions are 163 and 487, and the same theorem gives the displayed BW_{81} equivalence. Looping that equivalence gives the assertion about W_{81}.

In the n>1 setting of Selick’s Theorem 3.3, his Corollary 3.4 [35] makes \Omega S^{2n+1}\{3\} mod-three atomic when the strong Kervaire condition in stem 4n-2 fails; this includes every n that is not a power of three. For the remaining powers, Theorem 1.1 excludes r=2 and all r\geq5. Atomicity excludes any product with two noncontractible factors, not only a product of the displayed Anick form; see [35, Definition 3.1 and its discussion]. This proves (ii). Amelotte’s Theorem 1.4 and its proof [1] identify, for every n>1, the existence of a homotopy-associative multiplication with the same order-three strong Kervaire condition in stem 4n-2; its necessity also forces n=3^r. This proves the classification in (iii).

Finally, the proof of Amelotte’s Corollary 3.3 [1] uses that the loop H-space \Omega S^{2n+1}\{3\} is homotopy associative and homotopy commutative and has null third-power map. These properties pass to H-space retracts. Repeating that argument for the r=4 product proves the stated commutativity and power-map assertions. The multiplication assertion is existential, and no conclusion about power maps or loop-space structures on other BW_n is asserted at the remaining indices.□

Proof. The class \beta_1\in\pi^S_{10} has order three and is detected by b_0. The classical stem-106 Kervaire element has an order-three representative detected by b_2; see [32, Theorem 3.4.4(c), discussion after Proposition 6.4.11, and Table A3.4] and [1, Theorem 1.1 and Corollary 1.2]. This is the surviving corrected beta combination, not an assertion of survival for \beta_{9/9} alone.

If b_1 survived, Lemma 2.1 would give a permanent Adams–Novikov two-line class in internal degree 36 with that Thom reduction. The Miller–Ravenel–Wilson two-line calculation [26, Theorem 2.6] identifies this group with the order-three group generated by \beta_{3/3}; see also [32, Lemma 6.4.12(i)]. The classical Toda differential

excludes such a permanent class; see [5, Lemma 2.5].□

Proof of Theorem 1.1. Lemma 1.3 settles the indices 0, 1, and 2. Theorem 5.1, proved in Section 5, excludes every index j\geq4. Theorem 6.1, proved in Section 6 using the finite algebra of Section 7, constructs an exact-order-three representative detected by b_3. These assertions exhaust all nonnegative indices. Since K_3\subseteq K by definition, both sets are as stated.□

Proof. The unit-preserving map \BP\to\HF induces a map between the unit fibres and their smash powers. Consequently the filtrations on sphere homotopy satisfy

In particular, a class of Adams filtration exactly two has Adams–Novikov filtration at most two. The connective Adams–Novikov spectral sequence converges to 3-local sphere homotopy [32, Theorems 2.2.3 and 2.2.13–14]; see also [26, p. 475, after Equation (1.10)]. These positive stable stems are finite, so their 3-local and 3-complete groups agree. Its E_2-term has internal degree divisible by 4. In the stem 4\cdot3^{j+1}-2, filtration zero or one is therefore impossible. Thus the filtration is exactly two. The induced map on E_2 is Thom reduction. There is no incoming Adams differential in the bidegree of b_j: the potential length-two source has filtration zero and positive odd internal degree, and longer sources have negative filtration. The associated-graded image of the chosen class consequently gives the asserted nonzero multiple of b_j.

For completeness, the comparison with group cochains can be constructed without requiring an equivariant chosen orientation. Use the semicosimplicial Amitsur object with level \MU_{(3)}^{\wedge(s+1)}. For K=G or H, apply the translated orientation to the \ell-th factor at g_\ell, and multiply in E:

The target is the spectrum of equivariant homogeneous group cochains. Deleting a group entry corresponds to inserting the unit in the indicated source factor. The orientation is a map under the unit; its translates form a coherent family under that unit. Their exterior powers, followed by the coherent commutative multiplication in E, thus make (2.6) compatible with all cofaces and with the augmentation from the sphere. Only unit insertions occur here, so no E_\infty refinement of the orientation is required.

Forgetting degeneracies gives the unnormalized rather than the normalized cochain complex; the degenerate subcomplex is contractible. It therefore leaves the comparison unchanged from E_2 onward. The finite-stage comparison also respects actual filtration-two lifts. The ordinary totalization tower maps under the augmentation to its fat tower. In the target, s^0:C^1\to C^0 has section d^0, and d^1-d^0 lands in N^1=\ker s^0. Hence

with an augmentation-compatible projection onto the first summand. Composing the fat group-cochain comparison with this projection therefore gives an actual ordinary filtration-two lift, whose H^2-symbol is the normalized comparison class. Quillen’s 3-typical splitting [29, Theorems 4–5] identifies the \MU_{(3)} and \BP descriptions of the ordinary Adams–Novikov spectral sequence. On the \BP cobar, the comparison is precisely substitution of the chosen coordinates and their strict coordinate changes. The augmentations ensure that it sends actual filtered sphere classes to the corresponding filtered homotopy fixed point classes.□

Proof. The two-line theorem of Miller–Ravenel–Wilson [26, Theorem 2.6], also stated in [32, Corollary 5.4.5(a)], describes a direct sum of cyclic groups with parameters m=L3^n, 3\nmid L, denominator v_1^d, and order 3^{i+1}. The internal degree is 4(4m-d). Its restrictions include

In particular d<2\cdot3^n. In the desired degree the equation is

If n\geq j+1, it gives d\geq3\cdot3^n, a contradiction. For n\leq j, division by 3^n shows that the positive integer 4L-3^{j+1-n} is less than 2, hence is 1. Thus d=3^n, L has the displayed value, and integrality forces n\equiv j\pmod2. For n=0, the bound i\le n gives i=0; for n\ge1, the inequality a_{n-1}<3^n again forces i=0. This bounds the dimension by the number of classes in (2.7), and shows that the group has exponent 3. The remaining index restrictions are satisfied by all these candidates: 3^i\mid d is automatic, the special bound d\leq3^n when L=1 is an equality, and maximality of i follows from a_{n-1}<3^n for n\geq1, while 3\nmid d when n=0.

We next prove independence of these uncorrected classes. Equation (2.4) and the Frobenius identity in characteristic 3 show that v_2^{Ld} is primitive modulo (3,v_1^d), so its inner connecting class

is defined. Its exact v_1-annihilator exponent is d. Indeed v_1^d c_n=0, whereas the connecting-sequence comparison gives

For the last assertion, the primitives of \BP_*/3 are \F_3[v_1]. One can see this directly: if a primitive polynomial depends on a highest variable v_s, s\geq2, set all positive t_i except t_{s-1} to zero. The lower v_i are fixed, while the right unit sends

Comparing the highest power of t_{s-1} in a polynomial depending nontrivially on v_s contradicts invariance. The image of the primitives in \BP_*/(3,v_1) is therefore constant, and does not contain v_2^{Ld}. This proves the nonzero connecting class. Since the d’s are distinct, multiplication by v_1^{d_{\max}-1} in a proposed relation isolates its largest-exponent summand. Hence the c_n’s are independent.

It remains to exclude relations created by \delta_3. Set N=3^{j+1}. The integral alpha-line calculation [26, Theorem 2.2] says that the image of integral \Ext^1 in this internal degree is spanned, after reduction modulo 3, by

Explicitly, an integral representative is the differential of v_1^N divided by 3^{1+\nu_3(N)}. Expanding (v_1+3t_1)^N-v_1^N, its linear term reduces to (2.8); its k-th term for k\geq2 has 3-adic valuation k-1-\nu_3(k)>0. The class (2.8) is not v_1-torsion. Evaluate it on the height-one Honda formal group in characteristic 3 at the strict automorphism 1+3. The coefficient v_1 is then invertible, the coefficient action is trivial, and the first strict coefficient is nonzero. Evaluation of a one-coboundary at that automorphism is zero, whereas every v_1-multiple of (2.8) evaluates nontrivially. The kernel of \delta_3 consequently meets the v_1-torsion span of the c_n’s trivially. Their images are independent and attain the previously established dimension bound.

To compute the Thom image, put I=(3,v_1,v_2,\ldots) in the left coefficient ring of the cobar complex. Modulo 3, write

Every term of this polynomial has at least (L-1)d positive coefficient factors. It therefore admits an integral lift in I^{(L-1)d}. The ideal I is invariant, so the cobar differential preserves its powers. Moreover, on each free left coefficient module,

this follows, for example, because the initial form of 3 is a non-zero-divisor in the I-graded polynomial ring. The integral differential of the lift is divisible by 3; after division it still lies in I^{(L-1)d-1}. When L>1, we have (L-1)d\geq2, so reduction of this representative under Thom reduction is zero.□

Proof. We use a fixed Lubin–Tate deformation to evaluate the comparison of Lemma 2.1. Let \zeta be a primitive ninth root of unity, and let \calO=W(k)[\zeta], allowing a residue extension if necessary. The Lubin–Tate construction for \Z_3[\zeta] [21, Theorem 1] gives a height-one formal \Z_3[\zeta]-module over \calO. Its underlying special fibre has height six. After residue extension it can be identified with the chosen Honda formal group. The embedding of \Q_3(\zeta) in its rational endomorphism algebra is conjugate to our chosen embedding by Skolem–Noether. The conjugator can be made a unit: multiply it by an element of the centralizing cyclotomic field whose S-valuation cancels its own, using v_S(\zeta-1)=1. This gives a framing identifying the ninth-root action with G.

We obtain a G-equivariant evaluation of graded coefficient rings

The group fixes \calO, and acts on w by \zeta or its inverse, according to the action convention. The elements v_1,v_2 have zero residue modulo \pi=\zeta-1. By (2.3), the first strict coefficient t_1(g) has nonzero residue. The latter statement concerns the strict coordinate change after grading: its residual automorphism has linear coefficient 1, and g-1 first occurs at the third power of the coordinate.

Put T=4\cdot3^{j+1}. The degree-T line has character \zeta^{\pm2\cdot3^{j+1}}=1, hence trivial G-action. After trivializing this line, the periodic cyclic cochain complex is

Its reduction modulo 3 has zero differentials. The connecting homomorphism for multiplication by 3 is therefore

This formula follows by lifting the one-cochain c, taking its differential 9c, and dividing by 3. The inner connecting homomorphism \delta_{v_1^d} is formed in the \BP cobar complex before evaluation. We evaluate its already formed polynomial one-cocycle; we do not form a connecting homomorphism for multiplication by the evaluated v_1^d on \calO/3. The comparison from the bar resolution to the cyclic resolution evaluates a one-cocycle at g, so it computes the image of the inner connecting class in (2.7).

Express x in the basis of Lemma 2.2. Every summand with L>1 has zero Thom image; therefore the coefficient of the n=j,L=1 summand is nonzero. The residue of the inner cocycle at g is

For L>1 it is zero, whereas for L=1 it is a unit. Thus the inner cocycle of the entire combination evaluates to an element c\in\calO/3 with unit residue. Equation (2.9) sends it to 3c\ne0 in \calO/9. The integral coefficient modules and their cochain complexes are 3-torsion-free, so evaluation commutes with the outer connecting homomorphism \delta_3. This naturality proves nonvanishing in H^2(G;E_T).□

Proof. If the image were zero, the class would lie in the image of \pi_{T-1}(Y/F^2), and Y/F^2\simeq\Tot_1C^\bullet. In the fibre sequence

the degree-(T-1) group of C^0 vanishes by evenness. Every such element therefore comes from \pi_T N^1C^\bullet. Its boundary in the second layer is the cochain differential from degree one to degree two. The given symbol would thus be a d_1-boundary, contrary to its nonzero class in H^2. Equivalently, after E_2 the only possible ordinary incoming differential has length two and starts in odd internal degree; larger incoming lengths start in negative filtration. The finite two-layer argument proves the assertion directly, without a convergence assertion for any localized tower.□

Proof. Choose the actual Adams–Novikov filtration-two lift of \alpha given by Lemma 2.1. The augmented comparison sends it to a class in \pi_{4\cdot3^{j+1}-2}F^2. Its symbol is nonzero by Lemma 2.4. Lemma 2.6 proves that its image in E^{hG} is nonzero. Finally, a positive stable sphere stem is finite, and a homomorphism preserves finite additive order.□

Proof. Put F=\operatorname{Frac}(A), let \alpha_1,\ldots,\alpha_m be the distinct automorphisms of F induced by J, and choose representatives \gamma_i. Write \gamma_i(e)=c_i e, where c_i\in A^\times, and let h be the order of the kernel. On Fe the trace is

Distinct field automorphisms are linearly independent over F. Indeed, from a nonzero relation with the fewest terms, evaluating at af and subtracting \alpha_m(a) times the relation at f removes its last term. Choosing a on which another \alpha_i differs from \alpha_m gives a shorter nonzero relation, a contradiction. Since hc_i\ne0, the displayed trace is nonzero.

Choose f\in F on which it is nonzero, and write f=a/b. The nonzero invariant \prod_{\gamma\in J}\gamma(b) clears the denominator of f. Multiplication by this invariant commutes with the trace, and therefore gives a nonzero trace of an element of Ae.□

Proof. Work in the category of spectra with coherent J-action. Induction and coinduction from the trivial subgroup agree: their underlying objects are the finite coproduct and finite product of |J| copies of a spectrum, which agree naturally in a stable category. A map f:S^d\to Z therefore gives a composite

The last map is adjoint to f; on the summand indexed by \gamma, it is \gamma f. The underlying composite is consequently \sum_{\gamma\in J}\gamma f. Maps from a trivial-action sphere to Z are maps from that sphere to Z^{hJ}, proving the assertion. For a finite group the continuous cochain construction gives the same homotopy limit.□

Proof. Each Ru^n is a semilinear R-line. We check the kernel hypothesis in Lemma 2.8. There is no nontrivial ninth root of unity in W(k). To see this first for a cube root, its residue is 1, so write it as 1+3c. The equation

forces c=0. Applying this twice handles ninth roots. If q^9=1 in the power-series domain R, its constant term is therefore 1, and

forces q=1, since the second factor has nonzero constant term 9.

Now suppose \gamma\in G acts trivially on R, and write \gamma(u)=q_\gamma u. The group relation gives q_\gamma^9=1, hence q_\gamma=1. It acts trivially on every Ru^n. Lemma 2.8 gives a nonzero integral trace in Ru^n, and Lemma 2.9 lifts it to \pi_{-2n}(E^{hG}). Its restriction has infinite order because R is a characteristic-zero domain. The lifted class has infinite order as well.□

Proof. For j\geq4,

A survivor would give a nonzero finite-order element of the group in this degree by Theorem 2.7. Periodicity would identify it with a nonzero torsion element in degree -2, a contradiction.□

Proof. Put C^q=F^q/F^{q+1}, and let i,j,k be the exact-couple maps obtained from the fibre sequences for these layers. The cumulative boundary group on E_r in filtration s is

This is the usual exact-couple formula B_r=j(\ker i^{r-1}): for r=1 it is zero, and deriving the exact couple once adds the images of elements killed by the next iterate of i. Induction gives the displayed expression. The symbol j_s\alpha has zero image under k, so it has no outgoing differential on any page where it is nonzero. The hypothesis puts it in B_{s+1}^s, proving the assertion. Equivalently, \alpha comes from \pi_{V+1}(F^0/F^s); this finite quotient involves exactly the intervening s layers.□

Proof. We give the construction for K=H,G simultaneously, retaining H=\langle\sigma\rangle and writing Q=K/H. Let

We use the derived genuine norm and its commutative-ring counit; their indexed multiplication and geometric fixed point formulas are recalled in [16, Section 2 and Appendix B]. Nonequivariantly, R_N is a smash of |K| copies of \MU, labelled by K. Let

be the coefficient of x^{27} in the strict change between the orientations labelled e and \sigma. Norming this class and then applying the counit gives

Geometric fixed points. We claim that \Phi^H z=0 as a genuine Q-class. The geometric fixed points of a free indexed norm descend to the norm indexed by the orbit set, with its residual group action; see [25, Corollary 2.11]. After applying \Phi^H, the factors in the domain of

are labelled by Q\times K, the target factors by Q, and the multiplication folds them by

This follows equally by viewing the counit on each free norm factor as right translation of its labels. The displayed map factors, equivariantly for Q, as

The first stage is the Q-norm of the multiplication \MU^{\wedge K}\to\MU^{\wedge Q} that identifies labels in each H-coset. Under that multiplication the e and \sigma orientations coincide. Their strict coordinate change is the identity, so c maps to zero before the remaining Q-norm. This proves the claim with the residual equivariance retained.

An Euler multiple vanishes in the norm spectrum. Every nontrivial subgroup of K contains H. The family of subgroups not containing H is therefore just the trivial subgroup. Isotropy separation gives

For a spectrum concentrated over H, geometric H-fixed points identify its maps from S^V with genuine Q-maps from S^{V^H}; see [12, Section 2.4]. Apply this also after smashing with S^{-52\rho_K}. The obstruction to lifting (2.11) through the first term is precisely its image

which is zero. Thus z has such a lift.

The group

vanishes if 52|K|-2d<0. Indeed, EK has free cells in nonnegative dimensions. A free q-cell contributes an induced spectrum, whose group in representation degree 52\rho_K-d\lambda is the underlying group of R_N in degree 52|K|-2d-q. It is zero because \operatorname{Res}_eR_N=\MU^{\wedge |K|} is connective. Induction over finite skeleta, followed by compactness of a representation sphere, proves the vanishing. Multiplication of the lifted class by a_\lambda^d therefore gives

The least such exponents are d=79 for H and d=235 for G.

The image of z is invertible. Let E^{\mathrm{Bor}} denote the cofree genuine K-spectrum associated to the given action. There is a map

obtained on underlying spectra with action by applying the translated orientation in each factor and multiplying:

The coherent action and the E_\infty-multiplication make this a map with coherent action; cofree adjunction gives the genuine map. Its underlying map is homotopy multiplicative, which is all that will be needed. In particular, an E_\infty refinement of the orientation has not been assumed.

Let c_E be the underlying image of c. On the special fibre, \sigma-1 has S-valuation 3, so its first nonzero nonlinear strict coefficient occurs at x^{27}. Hence c_E has unit residue and is a homogeneous unit in E_*. The underlying image of z is the product of its K-conjugates, and is again a unit. Multiplication by z_E is thus an underlying equivalence

Both spectra are cofree, since a finite representation sphere preserves cofreeness. An underlying equivalence between cofree spectra is a genuine equivalence. Transport (2.12) along the map of spectra and cancel multiplication by z_E. This proves the asserted vanishing of the actual Euler powers.

The finite-page bound. The Tate E_2-unit is nonzero: the equivariant residue map R\to k takes every additive trace to zero, since K acts trivially on k and |K|=0 there. Cyclic periodicity therefore makes every Euler-power symbol nonzero on the ordinary E_2-page. The skeletal filtration-two lift of a_\lambda has symbol (2.10). Multiplying these lifts with a cellular diagonal gives a lift of a_\lambda^d in filtration 2d, with the corresponding power as symbol. It maps to zero in filtration zero by the result just proved. Lemma 2.12 therefore makes the symbol an incoming boundary by length 2d. Its representation internal degree is d(2-\lambda), of underlying dimension zero. An incoming d_r has source internal dimension 1-r, so evenness forces r odd. The bounds improve to 2d-1, namely 157 and 469. After inverting the Euler symbol, a boundary on its power is a boundary on the unit. This proves the claim entirely on finite pages.□

Proof. For e=g,\sigma, comparison of the two 3-series under the strict coordinate change gives, through index six,

Here and below an expression involving a graded coefficient is read after multiplication by the appropriate power of the period. To justify the truncation, in f\circ[3]_{eF}=[3]_F\circ f all coefficients v_l,ev_l for l<6 belong to \m. Terms containing two such coefficients vanish modulo \m^2. Inside a term already multiplied by v_l one may work in characteristic three. The special-fibre Honda law is additive below total degree 3^6, and terms involving the index-six summand together with a positive nonlinear power exceed the degree under consideration. These observations leave precisely (3.1).

The S-valuations of g-1 and \sigma-1 are one and three. Consequently their first residual nonlinear strict coefficients have indices one and three, respectively. The identity (\sigma-1)^2=-3\sigma in the endomorphism algebra implies a\,a^{27}=-1 for the latter coefficient. Equation (3.1) gives the asserted descending terms. At index six the equation v_6=u^{-728}, together with -728\equiv1\pmod3, gives the period formulas.

For Teichmüller elements the lower strict coefficients vanish on the special fibre. It follows that P_i has weight 3^i-1. Projecting a period to weight one preserves its residual generator and the leading terms just calculated. In particular

The six displayed coordinates are independent: the three y_i have leading indices 3,4,5, and the three l_i have leading indices 0,1,2. Their asserted action follows either from this triangular calculation or from D_\sigma^2=0 on \m/\m^2.□

Proof. For one pair k[y,l] with \sigma y=y+l, the orbit norm is N=y^3-yl^2, and the ring is free over k[N,l] on 1,y,y^2. After removing powers of N, its homogeneous part of degree d\ge2 has basis

The action on this three-dimensional space is a single unipotent block of length three; it is a free k[H]-module. Degree zero is trivial, and degree one is the two-dimensional indecomposable module, a suspension of the trivial module in the stable category.

Tensor the three decompositions. Terms having a free factor have zero Tate cohomology. The stable suspension class of each degree-one factor is represented by its value y_i, and cup product with that class identifies the corresponding shifted Tate groups. This gives (3.2) as a module, and also identifies products on distinct factors. Odd anticommutation and the vanishing of odd squares give the displayed exterior algebra.

The elements N_i lift by taking multiplicative orbit norms of lifts of y_i; the norm of a period has initial term u^3. The periodicity operator b and its inverse act on the actual cyclic complex. These are therefore coefficient cycles.□

Proof. The definition of y_0 gives d_1(u)=uz_0. The linear part of \Tr_H(u)/u is P_0+l_0=0, so d_1(eu)=0. In the cyclic cup-product convention this is equivalently d_1(e)=ez_0.

We need the same trace assertion to one further order. The trace of the chosen period now lies in \m^2E_{-2}. It is invariant, so its leading term defines an even Tate class in filtration two. The only classes there with the required internal degree are the shifts of uz_i z_j, i<j, by powers of b. None has T-weight one: the added weights are 2,8,10. The leading trace class is therefore zero. Subtracting the trace of an element of \m^2E_{-2} corrects u so that

Since g commutes with H, the elements D_g u,D_g^2u,D_\sigma u satisfy the same trace condition. By Lemma 3.1 their first terms span uz_2,uz_1,uz_0, in triangular order, modulo l_i-terms. Those l_i-terms are boundaries in the odd cyclic group. Thus d_1(z_1u)=d_1(z_2u)=0. The equation d_1^2(u)=0 also gives d_1(z_0)=0. Leibniz yields the asserted formulas on e,z_1,z_2. Taking cohomology, a monomial survives precisely when its power of u is congruent modulo three to its number of factors among e,z_1,z_2; this gives (3.3).

The three generators eu,z_1u,z_2u all have period exponent one modulo three. An even target at that exponent must contain exactly one of these generators and also z_0. For a d_2 from z_1u or z_2u this is impossible in filtration three; for a d_2 from eu, the filtration-two candidates have the wrong weight. For z_0, a target has period exponent zero and filtration three. The norms N_1,N_2 have weights 6,24, whereas N_0 has weight zero. The remaining possible four-exterior-factor target, with its period corrected by u^{-3}, has weight 10. Thus the only possible differential is d_2z_0=c\,bN_0.

To determine c, let L=\Q_3(\zeta_3) and take a height-one Lubin–Tate formal module for the unramified cubic extension of L. Its underlying formal group has height six. After a residue-field extension, choose a framing identifying the action of \zeta_3 with the prescribed H-action. Skolem–Noether gives such a framing rationally; it can be made integral because the centralizer division algebra contains an element of S-valuation one. This construction uses formal-module deformation theory and the given stabilizer embedding, not the coefficient model being proved.

On this fixed lift the ungraded coefficients have trivial H-action and the Lie period has character \zeta_3, or its inverse if the opposite action convention is chosen. Put \pi=\zeta_3-1. Filter the target cyclic complex by powers of \pi. The class z_0 maps to a nonzero multiple of \pi e. Since 3 is a unit times \pi^2, the norm differential on this internal-degree-zero class sends \pi e to a nonzero multiple of \pi^3b. This is its d_2, and N_0 maps to \pi^3 up to a unit. Naturality proves c\ne0.

The resulting page is

Its three odd generators have no even target of period exponent one modulo three. Its even generators already lift to coefficient cycles. Hence all subsequent differentials vanish.

We finally justify passage to the complete complex. In each fixed internal and cohomological degree its terms are compact: their \m-adic quotients are finite because k is finite. Survival to every page means that every prescribed finite accuracy admits a lift with the given leading cohomology class. A boundary correction in the leading quotient permits us to fix a representative of that class. The sets of such lifts whose differentials vanish to accuracy N are nonempty nested closed subsets of a compact module. Their intersection supplies an exact cycle. Images of the differential, and images of deeper filtration pieces, are closed for the same reason. Thus the final page really is the associated graded of the coefficient Tate cohomology; no inverse-limit extension is being suppressed.

Apply this argument to eu. An odd cyclic cocycle is precisely an element in the kernel of \Tr_H, so it gives the claimed x_0. Projection to the weight-one character commutes with \Tr_H, preserves the residual period, and gives the last assertion.□

Proof. The actual trace equation of Proposition 3.3 and its first two g-translates give the three relations. All x_i/x_0 reduce to one, so the product being inverted is a unit.

Eliminate x_6,x_7,x_8, and use x_0 and the five ratios x_i/x_0, 1\le i\le5, as coordinates. The degree-zero part of the source is a power-series ring in five variables over W(k); the remaining variable is an invertible period. The tangent map is invertible. Indeed the first five positive differences D_g^j x_0/x_0 have leading terms in P_{6-j}, respectively, with nonzero coefficients. Any first-order correction from u to the weight-one x_0, after division by u, has weight zero and lies in the span of P_0,P_3. Its positive D_g-differences only change lower terms in this triangular calculation. Together with 3, the five displayed differences therefore give all six cotangent directions. The passage from these differences to the five ratio coordinates is an invertible triangular binomial change of variables. Formal inversion over W(k) proves that the degree-zero map is an isomorphism, and adjoining the period proves (3.5). Its equivariance holds by the definition x_i=g^i x_0.□

Proof. Lift the norm classes N_1,N_2 in (3.4), choosing weight 2,8 lifts before taking norms. Their norms have weights 6,24. The trace-zero x_i give odd cocycles, and division by q_i places them in internal degree four. The first two g-differences yield the z_2u,z_1u terms of (3.4); together with x_0, they give the three odd generators. One may first use a common denominator q to verify this associated-graded assertion.

Tate cohomology is killed by three, so it is a k-algebra. All the power series in the positive-filtration norm parameters converge in its complete filtration. The proposed generators account for the entire associated graded. Their only proposed relations are the exterior relations, which hold in cohomology by graded commutativity and invertibility of two. The complete associated-graded comparison therefore proves (3.6).

For the alternative parameters, the identity

holds in every commutative H-ring: the mixed terms in the triple product are the two complete cyclic orbits. Apply it to a=1, b=x_i/x_0-1. Modulo additive traces, q_i/q-1 is the norm of that ratio difference. The first two ratio differences span the y_1,y_2 directions modulo the y_0,l_i directions, by the triangular tangent calculation; the norms of the latter directions vanish on (3.4). Thus these two norm functions are another regular parameter system.

Finally q_{i+3}=q_i, and x_{i+3}-x_i=D_\sigma x_i is an odd cyclic boundary. Hence g(a_2)=a_0 in cohomology. Modulo \mathfrak n, semilinearity disappears and the cyclic permutation of three basis elements is a single Jordan block in characteristic three.□

Proof. The representation-graded multiplicative norm of x_0 is an actual class of degree minus twice the real regular representation of H. Its symbol is q u_\lambda^2. Division by the Euler square, whose symbol is b^2u_\lambda^2, gives A. Both numerator and Euler class have actual filtered lifts. Their quotient is a unit on each nonzero localized page and has no outgoing differential.

The ratios

are norms of elements of \pi_0E, so they have degree-zero homotopy lifts. The norm parameters used in Proposition 3.5 have the same property. Constants are cycles: a Teichmüller constant satisfies c^{729}=c, and Leibniz kills its differential on a characteristic-three page. For an arbitrary series use the finite decomposition

Leibniz kills each cube and each parameter. This proves the assertion without a continuity assumption on a spectral-sequence derivation.□

Proof. The formal-group difference of two homomorphisms with the same reduction is a homomorphism f whose coefficients lie in the maximal ideal \mathfrak b of B. Suppose inductively that they lie in \mathfrak b^i, i\ge1. In the equality

the left side is zero modulo \mathfrak b^{i+1}: its linear term is 3f, and every nonlinear term has at least two factors from \mathfrak b^i. On the right, only substitution by the special-fibre 3-series matters. That series has positive finite order, so substitution by it is injective on (\mathfrak b^i/\mathfrak b^{i+1})[[x]]. Thus f vanishes modulo \mathfrak b^{i+1}. Induction proves f=0.□

Proof. We first compute the fixed equations, without passing to reduced points. In the model of Theorem 3.4, put a_i=x_i/x_0, so a_0=1, and put r=a_3. Fixation of the ungraded base by \sigma says

It follows that a_4=ra_1, a_5=ra_2, and a_6=r^2. The first trace relation also gives a_6=-1-r. Thus r^2+r+1=0. Conversely, this equation and the displayed formulas impose all the base-fixation equations, including the remaining two trace relations. Writing r=1+s gives (3.7). Fixing its structural root r=\zeta gives precisely D_0. The action of \mathcal C preserves this locus because it centralizes \sigma.

On an Artinian test ring, a fixed deformation carries the lift of the framed automorphism \sigma. This lift is unique by Lemma 3.7. Its endomorphism 1+\sigma+\sigma^2 reduces to zero and is therefore zero. It consequently extends the \Z_3-action to an \calO-action. The eigenvalue condition imposes exactly the structural action on the Lie line. Conversely, such a strict formal-module deformation supplies this lifted automorphism and the eigenvalue condition. This proves the claimed identification of deformation functors on all Artinian tests. Since [\calO:\Z_3]=2, its underlying height six is relative height three.

On \bar D the structural uniformizer vanishes. The first nonlinear coefficient of the uniformizer endomorphism is t_1(\sigma), a section of w^{1-3}=w^{-2}. After this coefficient is zero, its next leading coefficient is t_2(\sigma), a section of w^{1-9}=w^{-8}. Subtracting the identity from \sigma gives these statements successively modulo the previous leading coefficients. Leading coefficients of an endomorphism transform in the indicated lines under coordinate change. They are therefore intrinsic \mathcal C-invariant sections, denoted h_1,h. Here is the parameter comparison explicitly. After the faithfully flat completed residue extension k\subseteq\bar k, the universal formal-module deformation admits parameters U_1,U_2 and a formal group coordinate z satisfying

this is [36, arXiv version, Section 1, Equation (1.1)], applied to the ramified local field \Q_3(\zeta). If z'=cz+O(z^2), with c a unit, conjugation sends a first nonzero coefficient \lambda z^N to c^{1-N}\lambda. Thus h_1w^2 is a unit times U_1, and, on its zero divisor, hw^8 is a unit times U_2. Nonvanishing of these successive cotangent classes descends along k\subseteq\bar k. The formal inverse function theorem therefore makes h_1w^2 a regular parameter of \bar D, and identifies its zero curve with k[[hw^8]]. This proves the disc assertion over the original finite residue field; no descent of the chosen normal-form coordinate is required.

At the fixed deformation, \Nm_H(x_i)=x_i^3, since 1+\zeta+\zeta^2=0 and \zeta^{0+1+2}=1. Hence

The parameter description of Proposition 3.5, and perfection of k, identify K with k[[r_1^3,r_2^3]]=\bar D^3. Using the common denominator q, the values of the three odd generators are w^{-2}, (1+r_1)w^{-2}, and (1+r_2)w^{-2}. The alternative denominators q_i multiply these by units from the cube subring, so give the same span.

The class of t_1 is the restriction of the integral Adams–Novikov one-cocycle and is fixed by the centralizer. Its value is h_1, up to the nonzero scalar determined by the cyclic generator convention. Its evaluated first jet is nonzero, so its class \theta is primitive in V. The residual weights of V, from (3.4) after division by the norm period, are -2,0,6. Character projection gives a basis v,\theta,y of these weights. The tangent chain in Lemma 3.1 puts \theta at the fixed end and gives (3.8), including nonvanishing of c_1,c_2.

Under the identification K=\bar D^3, F=q^2h_1^3 is (h_1w^2)^3; it is a parameter. This also gives S_0\cong(D')^3. The values and first derivatives of 1,r_1,r_2 span first jets. The class \theta is transverse to the curve, while v has nonzero constant value. The remaining class y has nonzero derivative along the curve. Equivalently its scalar first jet has weight 8, whereas the transverse parameter has weight 2. Thus, on the curve, the evaluated v,y have orders zero and one.

For completeness, these orders prove the asserted injectivity over the smaller ring. Write \tau=hw^8, so D'=k[[\tau]] and S_0=k[[\tau^3]]. A nonzero cube-series multiple of the value of v has order 0\pmod3, and one of the value of y has order 1\pmod3. They cannot cancel. Evaluation of W is therefore injective. All constructions used to define it are functorial for the centralizer action, proving equivariance.□

Proof. Suppose an open subgroup U fixed the curve. Its elements would lift to automorphisms of the formal \calO-module over D', with their given special fibres. The relative height at the generic point is two: the first height parameter is zero and the second is nonzero. After extending the fraction field to an algebraic closure, the centralizer of \Q_3(\zeta) in the rational endomorphism algebra therefore has dimension four over \Q_3(\zeta).

At the closed point the corresponding centralizer is a division algebra of degree three, hence dimension nine over \Q_3(\zeta). An open subgroup of its units spans this algebra: differences of elements in a sufficiently small neighborhood of one contain a nonzero multiple of its full integral lattice. Choose nine members of U linearly independent over \Q_3(\zeta). Their generic lifts would satisfy a nontrivial linear relation over that field of operators. Clear its denominators using the structural \calO-action. The resulting sum is an integral formal-module endomorphism; if it vanishes generically, all its power-series coefficients vanish over the domain D'. Specialization gives the same forbidden relation at the closed point. This proves that no open subgroup fixes the curve. A finite image would have an open kernel, since distinct automorphisms of a formal disc are distinguished on finite jets. Thus every open subgroup has infinite image.

If a Laurent function is nonconstant, subtract its constant term when necessary and invert it when it has a pole. Its leading order shows that the Laurent-series field of the disc is finite over the one-variable Laurent-series field it generates. A group of field automorphisms fixing that function is consequently finite. It cannot contain the image of an open subgroup of \mathcal C. The same argument and the same dimension contradiction apply after a finite extension of constants.□

Proof. Let z be such a section. Since h is invariant in w^{-8}, the product z^8h^i is an invariant Laurent function. It is a nonzero constant by Lemma 3.9. In a period trivialization, h has order one. Valuation therefore gives 8\ord(z)+i=0. Put d=\ord(z); then i=-8d, and z/h^d has eighth power constant. After a finite extension of constants it is itself constant. This proves the first assertion and its integral variant.

For a sign twist the same eighth-power calculation applies. The resulting constant z/h^d would have to transform by the nontrivial character \chi, although the group fixes constants. It must therefore vanish, a contradiction.□

Proof. Write A_0=\bar D, and let \mathcal X/A_0 be its universal strict formal \calO-module. Let \mathcal M denote the covariant de Rham Dieudonné module of its underlying 3-divisible group. Our convention is \mathcal M=\mathbb D(\mathcal X^D), where \mathbb D is the contravariant crystal and \mathcal X^D is the Cartier dual. The crystal and its duality are provided by [6, Section 2.2, Theorem 2, and Section 5.2, Theorem 5(i)]. It is free of rank six over A_0. The uniformizer endomorphism acts on \mathcal M by an operator \epsilon with \epsilon^2=-3(1+\epsilon)=0. We first show that

At the closed point, the integral Dieudonné lattice is torsion-free over W(k)\otimes_{\Z_3}\calO=W(k)[\zeta]. Indeed multiplication by three is injective, and the relation \pi^2=-3(1+\pi) implies that multiplication by \pi is injective as well. This coefficient ring is a discrete valuation ring, so the lattice is free of rank three over it. Reduction modulo three gives a rank-three free module over k[\epsilon]/(\epsilon^2). Lift such a basis to \mathcal M. The resulting B_0-linear map B_0^3\to\mathcal M is a map of free A_0-modules of rank six which is an isomorphism at the closed point. Its determinant is a unit, proving the assertion.

Set M_0=\mathcal M/\epsilon\mathcal M, and let L be the Lie line, denoted w. The Hodge quotient factors through M_0 because the structural uniformizer acts as zero on L. Write

Here M_0 has rank three and N has rank two.

We make the tangent identification on arbitrary Artinian tests. For a continuous map A_0\to R_0 to an Artinian local k-algebra and a finite R_0-module I, use the split square-zero extension R_0\oplus I. Give I the divided powers \gamma_1(x)=x and \gamma_j(x)=0 for j\ge2. Grothendieck–Messing applies to this nilpotent divided-power thickening. The splitting provides a distinguished pullback lift; the crystal evaluated on the thickening is the base change of the de Rham module of that lift. The theorem identifies deformations with lifts of the Hodge quotient, and its full-faithfulness assertion identifies lifted endomorphisms with operators preserving that quotient [8, Remark 2.3(iii), Theorem 2.4(ii),(iv)]. Thus the strict \calO-module lifts are exactly the quotient lifts on which \epsilon acts by zero, or equivalently lifts of the quotient M_0\to L in (3.11). Relative to the split lift these are parametrized by \Hom(N,L\otimes I). The representing deformation functor parametrizes them by continuous derivations of A_0 into I. Naturality on all these tests, and passage to the inverse limit, give equivariantly

In particular

This is the operator-compatible version of the usual Kodaira–Spencer calculation.

Let \phi be absolute Frobenius of A_0. Functoriality and crystalline base change give A_0-linear maps between \phi^*\mathcal M and \mathcal M induced by relative Frobenius and Verschiebung, with both composites multiplication by three and hence zero [6, Section 2.2, Equation (2.4), pp. 24–25]. At the closed point the integral covariant Lie quotient has dimension one, so the corresponding operator has elementary divisors five units and one factor three. The relation FV=VF=3 gives the complementary operator one unit and five factors three; see [8, Theorem 4.33(3) and Remark 4.34(2)]. Thus one of the reduced maps, say U, has rank five, and its complement has rank one. Their corresponding minors remain units. Over the fraction field of A_0 the ranks are therefore at least five and one, and their zero composition forces the first rank to be exactly five.

Both source and target of U are free rank-three modules over B_0. On the Frobenius twist the operator is the pullback of the endomorphism \epsilon_{\mathcal M}. Naturality of the crystalline maps makes U linear for these operators: in one direction the identity is U\phi^*(\epsilon_{\mathcal M})=\epsilon_{\mathcal M}U, and in the other direction it is the reversed identity. These identities follow on every Artinian test from functoriality for the endomorphism [\pi], and pass to the formal base by compatible inverse limits. This statement does not apply absolute Frobenius to the dual-number algebra, which would send its scalar \epsilon to zero.

In B_0-bases write

Its underlying A_0-linear matrix has block form

Consequently 0=\det_{A_0}(U)=a^2. The ring A_0 is reduced, so a=0. At the closed point, rank five forces elementary divisors 1,1,\epsilon over k[\epsilon]/(\epsilon^2). Thus b is a unit.

This division of the determinant is intrinsic. For a free rank-one B_0-module, multiplication by \epsilon identifies its reduction modulo \epsilon with its submodule multiplied by \epsilon. The determinant map just obtained has image in that submodule and kills \epsilon times its source. Dividing and reducing therefore gives an isomorphism of the reduced determinant lines. Reversing it if necessary, we obtain a \mathcal C-equivariant isomorphism

Choose a generator e, with image of 1\otimes e equal to a_0e, a_0\in A_0^\times. After a finite extension of k, Hensel’s lemma gives a unit b_0 with b_0^2=a_0^{-1}. Replacing e by b_0e makes it Frobenius-fixed. For c\in\mathcal C, write c(e)=\lambda_c e. Equivariance gives \lambda_c^3=\lambda_c, and a unit satisfying this identity in the local ring is 1 or -1. These scalars define a sign character \chi. The cotangent weights in (3.12) are 2,8 and the Lie weight is one. Hence

It remains to relate M_0 to V. Use the first principal-parts module of L^{-2}, with its exact sequence

The jet map is linear over cubes, since d(f^3)=0. Thus coefficient evaluation induces an A_0-linear, \mathcal C-equivariant map

Functoriality of principal parts includes simultaneous changes of the base and the line trivialization, so the asserted equivariance is independent of the chosen period. Using the basis whose scalar evaluations are 1,r_1,r_2, its matrix of values and derivatives is

It is therefore an isomorphism. Taking determinants and using (3.12) gives

Finally restrict to D'. The invariant element \theta is primitive and gives a trivial subline; in principal parts its transverse derivative is nonzero even though its value on the curve is zero. Taking the quotient by this subline gives D'\otimes_{S_0}W. Its determinant is unchanged, proving (3.10).□

Proof. For a rank-two module,

An equivariant homomorphism W\to P_n is an invariant vector in W^\vee P_n. Extend its determinant line to the curve and use Lemma 3.11, together with the injective evaluation of Theorem 3.8. The vector then evaluates injectively in

Tensoring the evaluation by a line over the smaller cube ring preserves this injectivity: after extending and trivializing the line, all values acquire the same nonzero factor. The argument does not assert injectivity after arbitrary extension of coefficients in the two-dimensional source. Lemma 3.10 rules out the resulting invariant section and proves the vanishing.

A vector in WP_n evaluates in w^{3n-2}. Lemma 3.10 gives exactly 8d=2-3n, and integrality gives d\ge0. If h occurs, the generator q^2h of the indicated submodule of W has scalar valuation one. It cannot be divisible by a nonunit of S_0=k[[\tau^3]], since such divisibility would raise both coefficients in the valuation-zero/valuation-one basis by at least three. Thus the line is primitive. The section h is invariant and P_2 extends to w^6. The line is consequently stable with extended character w^6; division of w^{-9}\chi by this character gives w^{-15}\chi for the quotient.

For the final assertion, multiply the vector by q^{2+8k} to regard it as an element of W. Its evaluation is

Here q^8h^3 is the scalar parameter \tau^3 of S_0. The evaluation has order 1+3k. In the basis of values of orders zero and one, distinct residue classes modulo three prevent leading-term cancellation. Both coefficients are therefore divisible by \tau^{3k}. Dividing within W gives the vector q^2h, or equivalently h\in WP_{-2}.□

Proof. Evaluate the integral Bockstein representatives on the fixed deformation over D_0=W(k)[\zeta_3][[r_1,r_2]]. The coefficient lines in the three even degrees under consideration have trivial H-action. In the cyclic resolution, the connecting map for reduction modulo three is therefore b times the value of the intermediate one-cocycle at \sigma. This is an integral connecting-map calculation before reduction by the uniformizer \zeta_3-1.

Modulo that uniformizer, the formal-module relation (\sigma-1)^2=-3\sigma gives

The intermediate v_1-Bocksteins for v_2, v_2^3, and v_2^4 therefore have values, respectively,

For example, the last value is the reduction of 4v_2^3t_1(\sigma)^3, and the middle value follows by cubing the right-unit difference before dividing by v_1^3. These calculations can use Hazewinkel generators; changing to the Araki parameters of the fixed-geometry calculation leaves these reductions unchanged. The scalar-cohomology evaluation in Theorem 3.8 is injective, giving the asserted classes in (4.1).

We use the classical Adams–Novikov differentials

and the permanence of \alpha_1 and \beta_1; see [5, Lemma 2.5 and Table 1]. The comparison of Section 2 carries these formulas to the present spectral sequence. Since \beta_{3/3}=b^{-2}\beta_1^3, the first formula gives d_5(b)=\theta b^3 after normalizing \theta. Cancellation by \beta_1^3 is valid on this page: its scalar factor is nonzero in the domain K[q^{\pm1},b^{\pm1}], and the exterior algebra is free over that domain. Permanence of \alpha_1 gives d_5(\theta)=0.□

Proof. The bidegree of \kappa is (1,-4). A monomial A^a b^i\Lambda^j\widehat V has bidegree (-4a+2i-j,-6a+4j). If \Delta a,\Delta i,\Delta j are the changes under \kappa, then

The first equation makes \Delta j odd; the second makes \Delta j\equiv-1\pmod3. Since 0\leq j\leq3, the only possibility is \Delta j=-1, and then \Delta a=\Delta i=0. Multiplication by the cycle A^2 preserves the derivation rule.

The scalar cohomology of d_5 is K[A^{\pm1},B^{\pm1}], where B=b^3, and it has no incoming d_5 boundaries. In particular,

is a non-zero-divisor on this scalar cohomology. The image of \alpha_1\beta_4 is, up to scalar, B^{-1}\Theta\beta_1^5. The second formula of (4.6), and the absence of a d_9 target for B, therefore give

Cancelling the regular scalar factor proves d_9(\Theta)=cA^{-2}F.□

Proof. Suppose first that \mu=0, and put z=\Theta\hat v\hat y in exponent sector zero. Every boundary in this sector lies in \mathfrak n^2. Since \Theta z=0, the derivation rule and (4.10) give

in cohomology. Choose a T-weight representative for \kappa(z). Its weight is 10: z has weight 4 and multiplication by A^2 adds weight 6. The exterior-two basis weights are

There is no constant term of weight 10, and the only possible linear term is a multiple of F\hat v\hat y, because the coefficient tangent weights are 6,-2. Equation (4.13) fixes its coefficient, so

Both gA/A and the determinant of g on \widehat V are 1 modulo \mathfrak n, and gF\equiv F\pmod{\mathfrak n^2}. Equivariance of d_9 therefore makes this representative g-fixed modulo \mathfrak n^2. On the other hand, the residual Jordan action makes g(\hat v\hat y)-\hat v\hat y have a nonzero \Theta\hat v-coefficient. Since F is a parameter, this is a contradiction. Thus \mu\ne0.

Choose a complement to K\Theta and write

on that complement, with M_0\equiv\mu I\pmod{\mathfrak n^2}. The exterior-three component of \Psi^2(z')=0 is

The matrix \operatorname{tr}M_0-M_0 has invertible reduction \mu I, so c_0=0.

If \mu\notin\F_3, the complementary exterior-one and exterior-two terms in every sector are paired with their \Theta-multiples by invertible matrices: their reductions are i+\mu and i+2\mu, for i\in\F_3. Only the scalar and \Theta terms in sector zero remain. The differential (4.10) then leaves only the even algebra S_0[A^{\pm1},B^{\pm1}]. It has no odd class capable of hitting the unit on a later page, contradicting the Euler bound. Hence \mu\in\F_3^{\times}.

It remains to remove possible higher coefficient corrections to M_0. In sector i=-\mu put N=M_0-\mu I. The complementary exterior-one and exterior-two cohomology groups are \ker N and \coker N, respectively; multiplication by \Theta is the natural map \ker N\to\coker N. For a class \alpha\in\coker N, one has \Theta\alpha=0. Applying \kappa shows that cF\alpha lies in the image of this natural map. In particular, the two standard cokernel generators have lifts k_1,k_2\in\ker N such that

Every entry of N lies in \mathfrak n^2, so the matrix with columns k_1,k_2 is cF I modulo \mathfrak n^2. Its determinant has nonzero initial form c^2F^2. The two columns thus have full rank over \operatorname{Frac}K, and N=0.

The cohomology of (4.9) now gives (4.11). The quotient notation V/K\theta is independent of the chosen complement: changing a lift by a multiple of \Theta changes the corresponding cycle by a boundary in its nonzero sector. The operation \kappa vanishes on the two complementary generators because it preserves their nonzero exponent sector, where no scalar cohomology occurs. It also vanishes on A and B. Finally, F is regular, so taking the Koszul cohomology of d_9(\Theta)=cA^{-2}F proves (4.12).□

Proof. In this case the first change has length 65 and k_0=1. The only later possible lengths up to the bound are 93,101,129,137. The coefficients in (4.21) are constants times Z. Hence every E_{66}(H) component is a direct sum of free S_0-modules and modules killed by Z, and the scalar components are free. Generically the plane component in a sector is the whole plane W, the line \ell, the quotient W/\ell, or zero.

At length 93, the plane-to-scalar map is zero. For the whole plane this follows from the no-homomorphism rule; for the quotient it follows from the sign obstruction. For the line, l=7 in (4.16), so a contraction would require a meromorphic invariant in

Its Lie exponent is -54, which is not divisible by eight. Thus the map is zero generically, and it is zero integrally because its scalar target is torsion-free.

Only a decreasing differential can hit the unit, so it would have to be d_{129}. Its source has exterior degree one and n=22. We may choose a T-invariant source representative, since |T|=26 is invertible in k. The two coefficient weights before quotienting are 3n-2 and 3n+6, while Z has weight -2. Consequently the orders of invariant coefficients at n=22 satisfy

They are respectively 6 and 10 modulo 13. Thus every such representative on E_{66}(H) is divisible by Z^6 within its component. This statement also holds for subquotient components: averaging lifts an invariant class to an invariant representative, and the displayed orders give the divisibility before passage to the quotient.

There is no outgoing d_{93} on a plane component. Therefore the E_{94}(H) plane is a quotient of the preceding plane, and every potential unit source still has a representative Z^6e. There may be incoming d_{93} maps from exterior degree two; their images do not affect this conclusion. The exterior-two group, in contrast, is a submodule of its E_{66}(H) predecessor, since no exterior-three group can map into it. Its torsion is consequently still killed by Z.

Suppose that Z^6e is a d_{101}-cycle. Then Z^6d_{101}(e)=0, and the preceding torsion statement gives Zd_{101}(e)=0. Hence Ze is also a cycle, and on E_{102}(H)

There is no change between this page and E_{129}(H). The scalar groups have received no d_{93} and have only been replaced by submodules under d_{101}; their unit is not divisible by Z. Since Z is a cycle, a Z-divisible source cannot map to that unit. Thus d_{129} cannot hit it. The increasing length 137 cannot do so either, contradicting the bound 157.□

Proof. All nonzero coefficients of d_{17} in (4.21) are units, so its cohomology is torsion-free over S_0. If \epsilon=0, then \eta\ne0 and the X,YX terms disappear, leaving the exterior algebra on Y over S_0[A^{\pm1},B^{\pm1}]. If \epsilon\ne0 but \eta/\epsilon\notin\F_3, the X,YX terms again disappear; the surviving algebra is exterior on B^2Y over S_0[A^{\pm1},B^{\pm3}].

In the remaining cases let p_x\in\{0,1,2\} represent -\eta/\epsilon\in\F_3 and set C=B^3. The cohomology is

Its scalar, line, quotient, and product supports in the p coordinate are respectively

If p_x=2, the two plane columns align and together carry the corresponding shifted W-action. If p_x\ne2, the second column carries the quotient W/\ell: the \ell component of a changed lift lies in a sector killed by the preceding differential.

We check all further operations before any other change. An increasing map involving the line is the invariant-vector problem for \ell P_{-2l}; after evaluation it has the same condition l\equiv1\pmod4 as (4.18). This also governs maps from the quotient to the determinant. Increasing maps to the quotient, or from the line to the determinant, have the nontrivial sign twist and vanish. The full-plane case has the original invariant-vector condition. None of the remaining increasing lengths 53,89,125 satisfies it.

For a decreasing map, contraction on \ell=hP_2 has coefficient line

The Hasse rule makes l\equiv2\pmod4; among the decreasing lengths in (4.17), only 33 meets this condition. The same condition governs contraction from the determinant to the quotient. The other two line/quotient possibilities have the sign obstruction, and the aligned full-plane possibilities vanish by the no-homomorphism rule. Thus the aligned case p_x=2 admits no further differential within the bound and is impossible.

The d_{33} must therefore occur. Its coefficient is a nonzero constant times h^3, by (4.25) with l=2. In the algebra (4.23), this says

for p_x=0 or 1. The map on the product follows by the derivation rule. The factors A^{-6} and C are surviving units. In the \epsilon=0 case the corresponding formula is d_{33}(Y)=aA^{-6}BZ; in the other one-generator case it is the first formula of (4.26). Either one-generator case would consequently leave only an even scalar algebra, incapable of killing the unit later. This excludes both of those cases and proves \epsilon\ne0 and p_x\in\{0,1\}.

Taking cohomology of (4.26) gives

The remaining odd coefficient has residual T-weight -2 in the W line. An increasing map now requires

and a decreasing map requires

Within the bound, these leave only lengths 53 and 105, respectively. Only the latter can hit the unit. At length 105 the shift is \Delta p=5, so a source in the odd support p_x can land in scalar support zero only if p_x=1. Thus \eta/\epsilon=2, and setting U=BX gives (4.22). Rescale h to absorb \epsilon in (4.20); then the two d_{17} coefficients become p and p+2=p-1.□

Proof. The possible lengths were determined in the proof of Lemma 4.5. Since

the increasing length 53, together with the derivation rule, has exactly the form (4.27). Its scalar is an element of k, and there is no exterior-two group to receive d_{53}(U).

If e_{53}=0, the algebra (4.22) persists until length 105. If e_{53}\ne0, then

The surviving scalar exponents are a\equiv0\pmod3, and the surviving odd exponents are a\equiv1\pmod3. Thus

In particular, the notation C^{-2}U still denotes a well-defined surviving class: in (4.29) it means (C^3)^{-1}(CU). Only the surviving Laurent unit C^3 is inverted in this expression.

The bidegree of A^{18}C^{-2}U is (-105,-104), exactly the bidegree of a possible d_{105} source for the unit. In either case its source group is one-dimensional. There is no other length after 33 and at most 157 that can hit the unit. The Euler bound therefore forces (4.28). Since every nonzero scalar is a unit, the derivation rule makes the cohomology after this differential zero. Equivalently, if e_{53}=0 it pairs all odd and even Laurent slots, while if e_{53}\ne0 it pairs the odd and even slots of (4.29).□

Proof. The preceding lemmas show that the exact normalized d_5 has \mu=1. Thus d_5(z)=2b^2\theta\wedge z for z\in V. Since A=qb^{-2} is a cycle, d_5(q)=2qb^2\theta. The derivation rule gives the coefficient m+2n+2j in part (1). In raw coordinates, the contraction d_9(\Theta)=cA^{-2}F is multiplication by J=q^{-2}F after restoring the corresponding b-shifts.

Parts (2) and (3) are Lemma 4.5, with the supports read from (4.24) at p_x=1. For the product-to-quotient map, apply the derivation rule to the nonzero line contraction; it becomes multiplication by the same scalar Hasse cube after using h\wedge(-) to identify the quotient with the determinant. Part (4) is Proposition 4.6.□

Proof. The multiplicative norm of the underlying class x_0 is an actual representation-graded homotopy class. In the real regular representation of G, three of the nontrivial line pairs are faithful, and the fourth is \lambda'. A power map of complex lines of degree prime to three gives an equivalence of their one-point compactifications after three-localization. Thus the three faithful line pairs may all be identified with \lambda, with unit changes of orientation. The norm symbol is consequently \Delta u_\lambda^6u'^2. Division by the sixth power of the faithful Euler symbol a_\lambda=bu_\lambda gives (5.2). Its norm coefficient is invertible. The derivation rule shows that the inverse of a cycle unit is again a cycle for as long as the page is nonzero.

All the Tate groups are annihilated by nine. A Teichmüller constant c satisfies c^{3^6}=c; applying any differential gives dc=3^6c^{3^6-1}dc=0. Every Witt constant modulo nine is a sum of a Teichmüller constant and three times one, proving the assertion about constants.

Finally, Witt Frobenius on the maximal order is inner, implemented by S, up to the chosen opposite-group convention. A uniformizer in the maximal subfield \Q_3(\zeta_9) has the same valuation. Multiplying S by a unit therefore makes its conjugation agree with conjugation by an element of that subfield. The resulting extended-stabilizer symmetry commutes with G. It induces Witt Frobenius on constants and preserves the ordinary cyclic and representation-orientation symbols b,u'.□

Proof. We first work in the polynomial coefficient model of Theorem 3.4, before localization and completion. Group monomials by their three degrees in the trace-zero blocks. A triplet whose entries are not all equal has a three-element orbit; the corresponding module is induced from H. The standard restriction and transfer for an induced module give precisely its invariant and coinvariant descriptions.

Consider an equal triplet (d,d,d). The module is tensor induction of \Sym^d L, where

To compute its stable type, use the exact multiplication sequence obtained by quotienting the polynomial permutation representation by its invariant linear sum. In degree d the polynomial permutation module has Tate cohomology only from the monomial with three equal exponents, and that occurs precisely when 3\mid d. The long exact Tate sequence therefore gives one k in even degrees, one k in odd degrees, or zero, according as d\equiv0,1,2\pmod3.

These Tate patterns determine the nonprojective lattice summands. For completeness, a torsion-free W(k)[H]-lattice is a sum of trivial, cyclotomic, and free lattices. Map it by (N_H,\sigma-1) into its invariant and trace-zero lattices. These are free over W(k) and W(k)[\zeta_3], respectively. Modulo their maximal ideals, the image glues two subspaces by an isomorphism. Choosing bases adapted to those subspaces splits the gluing into paired one-dimensional graphs and unpaired summands. The paired graph is W(k)[H], since

This gives the asserted three lattice types. Accordingly, \Sym^dL is stably W(k), L, or zero in the three residue classes of d.

Terms containing a free summand contribute nothing after tensor induction. For a nonconstant choice of factors this follows by induction from H. For repeated free factors, the diagonal H-action on a tensor permutation basis is free; its G-action is then free too, because every nontrivial subgroup of G contains H.

Tensor induction of L is stably the third suspension of the trivial G-lattice. To see this without losing the integral extension, tensor the three short exact quotient sequences defining L. Every intermediate term is projective by the preceding observation. The cyclic permutation of the three suspended factors has positive sign. The two surviving ends thus give a threefold stable suspension, whose odd Tate group is W(k)/9. Tensor induction of the trivial lattice gives the corresponding even W(k)/9. Restriction to H is reduction modulo three, and transfer is multiplication by three. In particular the exceptional transfer is nonzero; replacing these groups by k prematurely would lose the restriction kernel.

Here is the degree identification. At H, degree 3e+\epsilon in block i, with \epsilon=0,1, contributes q_i^ex_i^\epsilon; the odd case is interpreted as a cyclic one-cocycle value. After allowing \Delta^{-1}, use x_i/q_i for the odd generator. Then j counts odd factors and n=\sum_i(e_i+\epsilon_i). An equal triplet has 3\mid n and j=0 or 3, giving exactly the exceptional terms above. The odd equal-triplet generator supplies c_G. The remaining triplets give all the degree-one and degree-two invariants by transfer. The same monomial calculation, now on the three q_i, proves (5.3).

These calculations survive localization and completion. After inverting \Delta, the degree-zero coefficient ring is finite over its Noetherian invariant subring, and every graded piece is a finite module there. Completion on the invariant side at the image of the residue point is flat and gives the desired completion, since the primes in the fibre constitute one orbit. The exceptional terms are supported at that point: invariant ratios act on them by their values at x_i/x_0=1 in k. Flat completion of the periodic cyclic complexes therefore preserves the computed groups and maps. Finally, multiplication by c_G identifies their scalar and volume parts; on the exceptional kernels this follows from the nonzero transfer pairing, and on induced summands from restriction.□

Proof. Theorem 4.7 gives the H formula w_0\theta\wedge(-). On successive exterior degrees in a row of (5.4), its coefficient is r_0+j. Restriction determines the first arrow. Writing the cyclic basis modulo \theta identifies the middle arrow with D, up to an invertible translate and sign. Transfer determines the last arrow by the projection formula. Lemma 5.3 then gives (5.5): in each of the three residue cases one coefficient vanishes and the other two are units. Restriction of u' is the trivial orientation; its possible d_5 target has injective restriction. At length nine it has no target.□

Proof. In row zero of (5.5), the degree-one d_9 is injective by restriction. When its invariant source is a norm, the map is also specified by transfer. In row two both middle groups inject under restriction, so the degree-two map is injective. Its entire source is transferred from \theta-multiples, represented by DP_n; the degree-one outgoing is therefore zero by transfer.

For row one, choose L_1 with NL_1=J. Such an element exists by (5.3), since J has index -2. The top arrow sends the invariant restriction value f to [L_1f] in (P_{n-2})_Q, up to a common nonzero scalar. Restriction proves this except when the target index is divisible by three. In that case all the sources are transfers, and

in coinvariants gives the same formula. All remaining arrows vanish by degree or by restriction.

Take the quotients row by row. In degree zero, when n\equiv0, every incoming invariant multiplier lifts by a norm, so the transfer still embeds after passage to (S_n)_Q. When n\equiv1, restriction was injective on the whole transfer target; quotienting its invariant J-image leaves exactly NS_n. The n\equiv2 case has no additional Tate quotient and gives (S_n)_Q. In degree one, restriction was injective before d_9; the quotient is (W_n)_Q divided by the reduction of \widehat H^{-1}(Q;P_n\otimes\overline M). This removes the norm kernel exactly for n\equiv0. In degree two, the additional quotient is by L_1P_{n+2}^Q; when n\equiv1 it removes precisely the newly appearing norm kernel.

For these assertions, the long exact Tate sequence of multiplication by J gives

as statements about dimensions in every degree. The plane statement uses the stable suspension represented by \overline M. The same exact sequences identify the norm kernels just used. Finally, the endpoint kernels and cokernels left in (5.5) give the additional scalar and volume lines stated in the proposition.□

Proof. At length thirteen only the exterior-degree change 0\to3 can be nonzero. For the reverse change, the target index is n-4, where there is no restriction kernel. The main scalar-to-volume map is necessarily zero on transfers, by its H counterpart. Write its two unit coefficients as

For w_0=3p and 3\mid n, a scalar lift has symbol

and its logarithmic coefficient is p\varepsilon+(i-2n/3)\varepsilon'. If this vanished with p\not\equiv0\pmod3, the restriction would survive to d_{17}, where Theorem 4.7 gives a nonzero primitive h term. But the available restriction image in the G target is NW_{n-2}, which is zero modulo the curve parameter Z: the residual plane is a two-term Jordan block. Naturality gives a contradiction. Lemma 5.2 makes the ratio \varepsilon'/\varepsilon Frobenius-fixed when defined. If \varepsilon=0, take p=1 and i-2n/3=0; if \varepsilon'\ne0, choose i-2n/3 in \F_3 to cancel the coefficient with p=1. Both are impossible. Thus \varepsilon\ne0 and \varepsilon'=0.

On the two extra scalar families, the actual quotient Euler symbol 3bu' gives d_{13}(3b)=0. The product 3b^2T_0 is nonzero in its 3c_G target. Hence we must allow

At w_0\equiv1\pmod3, multiplication by T_0 vanishes, so the extra scalar remains. At w_0\equiv2\pmod3, the derivation rule for B makes the scalar remain exactly when w_0\equiv\rho=2-3\eta\pmod9. Taking the cokernels in the volume groups gives the three stated families. Finally the scalar lift coefficient is p\varepsilon, so lifts outside transfers remain exactly when w_0=3p\equiv0\pmod9.□

Proof. At H, the invariant line generated by h and its wedge quotient give

Apply coinvariants and their long exact Tate sequence. When n\equiv0, the last Tate group is zero, giving (5.8) directly. When n\equiv1, the middle Tate group is zero and exactness requires dividing the first coinvariant group by its norm kernel, exactly as in (5.6). When n\equiv2, the first Tate group is zero, and the corresponding norm kernels in the middle and last coinvariants are both divided out. These are all the cases. Equivalently the obstruction is

which has the dimensions already calculated. Transferring the H differentials now gives the two coefficients in (5.8); the shift between the two source coordinates changes p-1 in the H formula to p+1 in this anchored chain. The top outgoing is zero by transfer.

On scalar unit lifts the restriction of a possible extra map is zero. Its target is the single norm kernel in the first unscaled image of (5.8), occurring when p\equiv0 and 3\mid n. Derivation on C,u' and translation by \mathcal D give exactly (5.9). There is no other kernel target. The Frobenius assertion follows from Lemma 5.2.□

Proof. First consider the possible intervening lengths in (5.10). At length twenty-one, the degree-one sources are transfers and their H differential is zero; the other arrows are excluded by (5.11). At twenty-five the possible volume-to-scalar arrows have source residues

and target residues 7,\rho,\rho-2, respectively. The last target occurs only if \rho=2. Any degree-compatible scalar-to-volume alternative has transfer source and is zero. At twenty-nine only the extra scalar at residue 4 can hit a restriction kernel in degree one at residue 3: the other candidate has target with injective restriction and is zero. At thirty-three the H contractions give the main transfer maps, and there may additionally be a volume arrow from residue 6 to the restriction kernel in degree two at residue 0.

It remains to compute the kernels and cokernels of the main contractions. Set L=h^3, an invariant scalar section of index -8 and valuation one in the curve ring S_0=k[[Z]]. The Q-action on this disc is nontrivial: otherwise its action on the scalar lines would also be trivial, since a scalar of order three in the unit group of the characteristic-three disc is one. This contradicts their calculated Tate dimensions. Thus its invariant fraction field has degree three, and the complete disc extension is totally ramified. For example the orbit product of a parameter generates an invariant disc over which the original disc has rank three. Valuations of invariant functions are consequently multiples of three.

The section L shifts index by -8\equiv1\pmod3, while \Delta shifts it by three and is an invariant unit. These sections realize minimal invariant valuations 0,1,2 at indices n\equiv0,1,2, respectively. Any ratio of two invariant sections of one index is an invariant function, so there can be no smaller nonnegative valuation in that index. The Tate dimensions of Proposition 5.5 then show that the corresponding norm images begin at valuations 3,4,2.

It follows that multiplication by L on \mathcal T_n\to\mathcal T_{n-8} has the three effects

Indeed multiplication shifts each free invariant lattice by one valuation, and comparison with the minima 3,4,2 gives the stated free cokernels; the remaining kernel or cokernel is the explicit norm kernel in (5.6). Dividing a norm kernel out at length seventeen simply removes that kernel or cokernel from this list.

More precisely, the degree-one line at index n is identified by h with \mathcal T_{n+2}, and its contraction is

The degree-two group at index n contracts to the degree-one quotient identified by h\wedge, by

These formulas are transfers of the nonzero H length-thirty-three maps. Applying (5.12) to each index gives exactly the finite table. The extra volume arrow identified above is the only additional incoming on the residual degree-two kernel.□

Proof. By Skolem–Noether choose a stabilizer unit \delta conjugating g to g^{-1}; a conjugator is made a unit by adjusting its valuation in the ramified centralizing field. If d_0 is its residual Lie scalar, conjugation scales the first strict coefficient of g by d_0^{-2}, whereas inversion negates that coefficient. Thus d_0^2=-1.

A cyclic b-shift changes sign and contributes exponent two. A scalar q^n has residual Lie exponent 3n. Let \overline V=V/\mathfrak nV and let N_0=g-1 on \overline V. The quotient \overline V/N_0\overline V is represented by x_0/q_0. Its coefficient weight -2 and the cyclic one-cocycle inversion sign give total exponent zero. Since

\delta preserves the canonical Jordan flag and its successive diagonal characters, starting at this quotient, are 1,-1,1. The last line is k\theta=N_0^2\overline V. Thus on \overline W=\overline V/k\theta the quotient character is 1 and the invariant-line character is -1. The primitive line generated by q^2h reduces to this unique invariant line, so its extra exponent is two. The plane and volume determinants likewise have character -1=d_0^2.

Restriction and transfer preserve these characters on the finite quotients. In particular the constants representing the norm kernels \mathsf y,\mathsf M have the indicated determinant character. Their constant representatives inject into the residue at Z=0, so higher-order semilinear terms do not alter these characters. On \mathsf A only its two subquotient characters are needed; no splitting of their extension is asserted. Conjugation identifies the representation spheres with the same orientation factor on source and target at fixed i. Finally 2\Delta m=R-\Delta j by (5.1), giving (5.15).□

Proof. Apply (5.17)–(5.19) to the scalar entry \mathsf A. Its only incoming possibilities, modulo 216, are from \mathsf f at length 25,169,97 when \rho=2,5,8, respectively. Its outgoing possibilities are at 13,17,53, to \mathsf d,\mathsf y,\mathsf x, respectively. For clarity the complete incoming list within the Euler bound is

At length twenty-five this is a statement about the residual transfer kernel after the main length-thirty-three quotient, and may equivalently be read on the early page. If \rho=2, that arrow is zero. Indeed \mathsf f=\mathsf v\mathsf d up to shifts, \mathsf v has no outgoing there, and the possible \mathsf d differential lands in \mathsf r, whose product with \mathsf v is zero.

Now take an \mathsf A copy with e_{53}t=0. Its transfer-kernel generator is the transfer of an H scalar that survives d_{53} and is a d_{105} target, by Proposition 4.6. The corresponding H source has exterior degree one and index divisible by three. Its transfer is zero in Table 1. Naturality of d_{105} therefore forces the transferred scalar to vanish before the G page E_{105}: on that page its transfer would have to equal d_{105}(0)=0. It cannot have an outgoing earlier, since it is transferred from an H cycle. An incoming is thus necessary before length 105. The \rho=5 row of (5.21) has none, and the \rho=2 candidate is zero. Hence \rho=8 and the required incoming has length 97.

If e_{53}=0, this reasoning applies to every copy of the transfer kernel, and every \mathsf f is consumed at length 97. That differential cannot hit a restriction lift of the unit, since H restriction of the target is nonzero. No source then remains for the only later possible unit incoming, at 313. This contradicts the bound of Proposition 2.13. Therefore e_{53}\ne0.

An \mathsf A restriction lift with V(t,u)=0 survives to length fifty-three. If t\not\equiv0\pmod3, its restriction has nonzero H differential, while the only G target is \mathsf x, whose restriction is zero. Thus V(t,u)=0 implies t=0 modulo three. The Frobenius ratio condition in Lemma 5.7 makes this a linear equation over \F_3. Its kernel can be contained in t=0 only if \nu\ne0 and \nu'=0.□

Proof. At E_2 the comparison is the usual map from group cohomology to Tate cohomology: it is surjective in filtration zero and an isomorphism in positive filtration. Suppose the stated ranges hold on E_R. Lift a Tate cycle of filtration s\ge0 to an ordinary class. Its differential maps to zero, and the target has filtration s+R\ge R, where comparison is injective. Thus the lift is an ordinary cycle. For injectivity on the next page, an ordinary cycle in filtration s\ge R that is a Tate boundary has a Tate boundary source in filtration s-R\ge0. Lift that source by surjectivity and use injectivity on E_R at filtration s to identify its ordinary boundary with the given cycle. This proves the induction.

Apply it to a specified filtration-zero symbol. At each earlier length its ordinary differential maps to zero, and comparison is injective at that target; thus that very symbol survives. The finite tower description of cycles supplies its lift over the stated skeleton. No identification of an infinite Tate abutment enters this argument.□

Proof. Consider at H the ordinary orientation symbol

On Tate pages it is C times an inverse faithful Euler power. By Proposition 5.10 its first nonzero differential has length fifty-three. Lemma 5.11 supplies an actual H-equivariant representative on the 52-skeleton of the chosen free G-model for EG. This is a lift of the specified filtration-zero symbol: at each preceding length its ordinary differential vanishes because comparison is injective at that outgoing target. We use this same skeleton for the pointwise norm, restriction, and transfer.

Take its pointwise multiplicative norm on that same skeleton. Explicitly, tensor-induce its map, precompose with the diagonal of the skeleton, and use indexed multiplication into E. Include its virtual sphere factor before taking the norm. This gives induced degree

whose orientation symbol is

up to a constant unit. Here the three faithful lifts of \lambda|H have been identified by the prime-to-three power equivalences used in Lemma 5.2.

We justify the first obstruction of this norm directly. On the boundary of a free 53-cell the partial maps define a product over a 52-sphere. Stably, the disjoint-basepoint sphere splits into a constant summand and a reduced sphere. The reduced diagonal into two or more reduced factors maps S^{52} into S^{104} or a higher-dimensional sphere, and is zero. Hence only terms with one reduced factor contribute to the first obstruction. The other factors are their filtration-zero point values. Sum these single-factor terms over the three H-cells: the result is precisely the additive transfer cochain of the one-factor obstruction, multiplied by the other two translated symbols. All three partial factors extend over the preceding skeleton, so this formula is valid modulo exactly the previous obstruction indeterminacies. It is therefore a formula on E_{53}:

This local obstruction calculation is the filtration-zero instance of the general norm-transfer differential construction in [10, Theorem 4.6 and Proposition 5.4]. That construction is an antecedent for the argument; its prime-two periodicity applications are not inputs here.

The translates must in general be retained in (5.23). In this application Y has coefficient one and its H representation extends to G. Its cohomology symbol is thus fixed by Q, and its successive surviving page classes are fixed as well. Consequently the right side is \operatorname{tr}(Y^2d_{53}Y) here.

Multiply by the necessary faithful Euler power to remove the u_\lambda shift. The source becomes C^3u'^{18}, and the target becomes

Before reduction of indices, this target has n=-8 and w_0=39. It is the \mathsf x entry; that entry has no possible removal before length fifty-three and receives the H transfer isomorphically on this page. Explicitly its exterior degree is one, its twist is i=18, and (5.13) gives (t,u)=(4,24). The complete arrow rule gives no incoming to \mathsf x before 53; the Mackey calculation and the main d_{33} quotient identify the transferred H constant with its one-dimensional generator. This assertion is on E_{53}, before taking the cohomology of the differential now being computed. The target is therefore nonzero. On the source C^3u'^{18}, formula (5.22) reads \omega+18\omega'=\omega. This proves the lemma.□

Proof. Choose arbitrary copies z of \mathsf b and z' of \mathsf d. Their product is zero, so

Both arrows have shift (-1,4) by (5.20). The two terms, when nonzero, belong to the same \mathsf f copy, at (t_z+t_{z'},u_z+u_{z'}+4), and the products in (5.27) are nonzero. Thus one coefficient vanishes if and only if the other does. Because the two chosen copies were arbitrary, either both arrows vanish everywhere or both are nonzero everywhere. This proves uniformity even before any smaller unit lattice is used.

Suppose they vanish. Then all \mathsf v,\mathsf d copies persist to page 97: they have no intervening incoming, \mathsf v has no outgoing, and the only remaining possible outgoing of \mathsf d is the length-97 map to \mathsf p. Choose their product in a required \mathsf f_2 copy. On that page,

because its possible value is the product of two extra scalars. This contradicts the required nonzero differential on \mathsf f_2. Therefore both arrows are everywhere nonzero; their source and target groups are one-dimensional, so all four labels disappear.□

Proof. Suppose a copy of \mathsf a_2 survives forty-nine. The only possible subsequent incoming through ninety-seven is the length-eighty-five arrow from \mathsf q, with shift (3,8). But \mathsf q remains only at residue zero, so it can hit only \mathsf a_0. The potential length-ninety-seven outgoing has target \mathsf v, now absent. Thus \mathsf a_2 is a cycle on page 97. The Euler copy p^E is also still present: the length-fifty-seven arrow is zero, and the possible length-ninety-three incoming from \mathsf y_0 has target residue two, not zero. The product p^E\mathsf a_2 is consequently a required \mathsf f_2 that is a product of two cycles at ninety-seven, a contradiction.

Suppose instead that \mathsf a_0 survives forty-nine. The detector p^\dagger is present on page 85; its only earlier possible incoming was the zero length-fifty-seven map. If the hypothesized \mathsf a_0 were hit at eighty-five, multiplying by this detector would make its nonzero \mathsf f_2 product an earlier boundary, contrary to the requirement that every \mathsf f_2 remain until ninety-seven. Hence \mathsf a_0 remains to ninety-three. Nor can p^\dagger be hit there under this hypothesis: if d_{93}z=p^\dagger, then

would again make a required \mathsf f_2 an earlier boundary. Both factors therefore reach ninety-seven as cycles, and their product gives the same contradiction. Thus \mathsf a_0 and \mathsf a_2 cannot survive forty-nine. Their one-dimensional targets \mathsf R_0,\mathsf R_2 are removed with them. The common remaining flag at residue one follows from the same pairing and the unit symmetries through fifty-three.

Finally suppose Q_0=1. A remaining \mathsf q_0 stays to page 97: its potential eighty-five target \mathsf a_0 is absent. If C_1=1, an incoming on \mathsf c_1 after thirty-three and before ninety-seven, multiplied by \mathsf q_0, would make the nonzero product \mathsf q_0\mathsf c_1 a required \mathsf f_2 that is an earlier boundary. Thus \mathsf c_1 must itself remain to ninety-seven; then their product is a product of cycles there, also impossible. This proves (5.31).□

Proof. We first make the units giving the translations explicit. On E_{53} put

Equation (5.22) says that the logarithmic differentials of C^3 and u' are \Omega and \kappa\Omega. Their common target is a residue-field line, so 3\Omega=0. Hence the units

and their inverses are cycles. Choose any integer representative of \kappa; changing it by three changes U_1 by a power of U_0. Their shifts (9,0) and (-3\kappa,1) generate \mathcal L. Together with \mathcal D, they realize every required translation by actual multiplication on the page.

At the end of length ninety-seven, all transfer kernels of \mathsf A have disappeared. Only its restriction lifts at \mathcal L remain. These are products of surviving units, and their inverses are cycles by derivation. They have no outgoing between fifty-three and ninety-seven; the incoming at ninety-seven lands in their transfer kernel, not in their nonzero restriction quotient. In particular, passing to this quotient preserves the identities U_iU_i^{-1}=1: restriction sends these units to units on the still nonzero H page, so 1 is not a length-ninety-seven boundary. No splitting of the earlier \mathsf A extension is needed.

For a remaining \mathsf A the only possible outgoing before 313 are to \mathsf d,\mathsf y, \mathsf x at 229,233,269, respectively. The first target has been removed. The second has t\equiv1\pmod3, whereas \mathsf y occurs only at t\equiv0. The third has t\equiv2\pmod3, whereas \mathsf x occurs only at t\equiv1. Thus all those outgoing maps are excluded. Formula (5.21) has no remaining incoming before 313 and no other one through the bound 469. The bound in Proposition 2.13 forces a nonzero \mathsf f\to\mathsf A at 313 hitting the unit. Its shift, computed by (5.20), is (5,36), giving (5.33).

No other arrow in (5.32) has length 313. A differential hitting one contracts the differential graded algebra: if dz=1 and dy=0, then d(zy)=y up to the usual sign. Its cohomology is therefore zero. This proves the proposition, including the actual page range of the unit symmetries.□

Proof. There is no incoming to \mathsf p_1 in the remaining array. For explicit verification, the possible incoming lengths from (5.32) are 57,93,273,309: their sources have, respectively, residues \mathsf x_0,\mathsf y_2,\mathsf x_2, \mathsf y_1, all absent. The first arrow is also zero by transfer. The removed label \mathsf d would have supplied incoming at 97 or 313, and supplies none now. Its possible outgoing partners through 313 are

The potential removed-label target at 229 is \mathsf b, also absent. Thus every \mathsf p_1 must pair with \mathsf c_0 at 85 or \mathsf c_1 at 301.

Similarly the remaining possibilities for \mathsf M_2 are an outgoing at 149 to \mathsf f_2, already consumed at 97, and incoming at 177,233,249 from \mathsf a_2,\mathsf X_0,\mathsf c_0, respectively. Only the last is present. The earlier candidate at 105 came from the removed \mathsf b label. This list is exhaustive by the same congruences. Thus \mathsf M_2, if present, must be hit from \mathsf c_0 at 249.

If C_1=1, the length-thirty-three pairing left both \mathsf c_1 and \mathsf M_2. The unique partner of every \mathsf M_2 forces C_0=1 and reserves every \mathsf c_0 copy for length 249; the shift (5,28) is a bijection on the copies. Therefore all \mathsf p_1 must use \mathsf c_1 at 301. If C_1=0, neither \mathsf c_1 nor \mathsf M_2 occurs, and every \mathsf p_1 must use \mathsf c_0 at 85, again forcing C_0=1. This proves both alternatives. In particular \mathsf M_1 was present after length thirty-three in either case.□

Proof. Each row of Table 2 begins with six entries and then three optional pairs, so it has even cardinality. The row s_x has one additional entry, the singleton \mathsf x. By Proposition 5.15, every entry pairs within its row except the marked \mathsf f that leaves to \mathsf A at 313. Removing two endpoints preserves parity. Therefore the marked \mathsf f must belong to the same row as \mathsf x.

The marked source is (-5,-36)+\mathcal L by (5.33). It is an \mathsf f_1, whose offset is (16,86), so its row is

Equality with (5.35) says (-21,-124)\in\mathcal L, or

Since \kappa\in\F_3, this is equivalent to \kappa=2=-1. The conclusion follows.□

Proof of Theorem 5.1. By Proposition 5.17, copies are distinguished by the key t-3u modulo nine. The three rows have keys 0,3,6. The exceptional \mathsf x row s_x=(0,2) has key 3.

The detector p^\dagger for b_2 has coordinates (-4,12) and hence key 5. Its \mathsf p_2 offset (2,10) in Table 2 has key 8, so this detector lies in row 5-8=6. In that row the only available incoming on \mathsf p_2 is the length-ninety-three arrow from \mathsf y. The \mathsf x arrows belong to row 3, and all other incoming labels have been removed. Since b_2 is known to survive, the detector cannot leave by an outgoing differential while present. It must nevertheless disappear before 313 by Proposition 5.15. Consequently the \mathsf y of row 6 hits this \mathsf p_2 at length 93.

Consider now \mathsf p_0 in that same row. Its potential incoming from \mathsf x at 273 belongs to row 3 and cannot hit it. Its only other incoming is the length-309 map from the row’s \mathsf y, already consumed at 93. Thus this \mathsf p_0 has no incoming. It must disappear by an outgoing differential; the full table permits its outgoing at 85 to \mathsf c_2 or at 157 to \mathsf a_1. The argument does not require a choice between these two lengths. Figure 1 summarizes this constraint.

The \mathsf p_0 offset (6,34) has key 6-3\cdot34=3 modulo nine. In row 6 it therefore has total key zero. Every detector with \jmath\ge4 in (5.30) has

It is exactly a \mathsf p_0 copy of this key. The unit translations of Proposition 5.15 identify all such copies through the relevant pages. In particular none can be an earlier incoming Tate boundary, and each must have the forced outgoing just proved. This contradicts the cycle property of the image of any surviving sphere class. Hence no b_{\jmath} with \jmath\ge4 survives.

For comparison, the \jmath=3 detector has key 6, whereas the \jmath=2 detector has key 5. Thus the tail exclusion does not decide b_3 by the same argument. Its survival and order-three representative will be constructed separately in Section 6.□

Proof. The equivalence of the first two conditions is Selick’s strong odd-primary criterion, in its stable Moore-space formulation [35, Theorems 1.1 and 3.3]; see also the proof of [1, Theorem 1.1]. In the notation of that formulation the stem is 2m(p-1)-2, and the detecting Adams class is b_{k-1} when m=p^k. Thus at p=3 and k=j+1 the stem is 4m-2 and the class is precisely b_j.

For completeness, Spanier–Whitehead duality gives the last equivalence without changing the primary test. If g:\Sigma^{4m-2}M\to S, then DM\simeq\Sigma^{-1}M and

The cohomology degrees of C_g and C_a are, respectively, \{0,4m-1,4m\} and \{0,1,4m\}. Conjugation in the Steenrod algebra sends \mathcal P^m to -\mathcal P^m modulo products of positive reduced powers [28, Section 7, pp. 166–167]. Such products vanish on the two cones, because the necessary intermediate degrees are absent. Consequently duality preserves nonvanishing of the indicated operation. Finally, 3\vartheta=0 follows directly from (6.1); the detection and nonvanishing are part of the strong criterion. Thus the criterion concerns an actual order-three representative, rather than an order assertion about an associated-graded group.□

Proof. The first Adams obstruction measures whether the displayed comodule extension splits. Cone maps give its naturality. To add two maps, first take their direct sum and then use the diagonal on the source and the fold on the target; the corresponding pullback and pushout of extensions give their Baer sum. If u=vw and both v_* and w_* vanish, the cone comparison from w to vw identifies the extension of vw with the pushout of the extension of w along v_*=0. This pushout is split.

The maps between two successive stages of the mod-three Adams tower are homology-zero maps. Thus a factorization through its second stage has zero invariant. The comparison \BP\to\HF supplies this factorization for a map of Adams–Novikov filtration at least two.□

Proof. The class of a_0 has Adams–Novikov filtration one. Filtration zero is excluded by the support of \pi_*\BP M, while filtration at least two is excluded by its nonzero primary test and Lemma 6.3. Let

be its permanent leading class.

We use the notation of Section 7: v=v_1, a=[t_1], B is the reduced integral 3-Bockstein, and

By the unconditional basis calculation (7.5), write

After setting all positive coefficient generators v_i to zero, these representatives specialize to \lambda t_1^{27}. Every one-cobar boundary also specializes to zero, since the positive coefficient ideal is invariant. The Thom comparison sends t_1, up to a unit, to the first reduced-power dual generator; its conjugate Milnor convention changes only this unit. Hence \lambda, up to a unit, is the \mathcal P^{27} coefficient of C_{a_0}, and is nonzero. Rescale a_0 to make \lambda=1. Lemma 7.2 then gives

The boundary along q increases Adams–Novikov filtration by one and induces the integral 3-Bockstein; this is the geometric boundary theorem [32, Theorem 2.3.4]. Here it can be seen directly on two layers. Put X=F_2S, Y=F_1S, and Z=Y/X. A lift of a_0 to Y/3 and an integral lift to Z of its image in Z/3 give a class in

The integral lift exists because the homotopy of Z is torsion-free. Comparing fibres over Z yields the triangle

where \iota:X\to Y is the inclusion: the fibre of Y\to Q identifies with the fibre of X\to X/3, and its map into Y is 3\iota. Projection to Y/3 identifies the resulting boundary with a lift of x to F_2S. Projection to Z says that three times its layer symbol is the d_1 of the chosen integral lift. Division by three in the torsion-free layer gives exactly the stated Bockstein, with a harmless convention-dependent sign.

After reducing again by i, the filtration-two symbol is \pm B(z)=\pm B(y). Indeed the Bockstein is a derivation,

so B(v^{26}a)=0. Multiplicativity of the sphere tower and its pairing with the Moore tower puts ix^3 in filtration at least six, with filtration-six symbol represented by \pm B(y)^3. This symbol is zero by (6.4). Thus ix^3 has filtration greater than six. Its stem is 318\equiv2\bmod4, so (6.2) makes ten the next possible filtration.□

Proof. Since two is invertible, the spectral sequence for the homotopy orbits by C_2 takes invariants exactly. The cellular complex of L^{(2)} has dimension two, proving the assertion about local coefficients. With constant coefficients its possible reduced cohomology groups are \ker(3:A\to A) in degree one and A/3A in degree two. Inversion acts by minus one on both, so neither has nonzero C_2-invariants. In particular the augmentation \Sigma^\infty\mathcal B_+\to S is a 3-local equivalence: its cofibre is bounded below and has zero integral localized homology, hence is zero by the stable Hurewicz theorem.

The Thom spectrum of \rho over \mathcal B is D^\circ(S^1). Its orientation system is \mathrm{sgn}. On the lens skeleton, integral reduced homology is \Z/3 in degree one, and inversion acts by minus one. Twisting by \mathrm{sgn} therefore retains that group and removes the degree-zero group. The Thom homology is consequently just \Z/3 in degree four. To identify the spectrum, realize its lowest homotopy generator by S^4; it has order three by the lowest-degree Hurewicz isomorphism and extends to \Sigma^4M. The extension is an integral homology equivalence and therefore a 3-local equivalence of bounded-below spectra.

The difference bundle (n-1)\rho is oriented, since n-1=106 is even, and is stably spherically trivial after localization at three. Indeed its spherical classifying map lifts to the classifying space of degree-one stable sphere equivalences. This space is simply connected and its second homotopy group is \pi_1S_{(3)}=0. The subsequent Postnikov obstructions vanish by the cohomological dimension of \mathcal B. There is no inverse-limit ambiguity: for a fixed homotopy degree of the mapping space, the fibres of its Postnikov tower have zero homotopy in that degree once their Eilenberg–Mac Lane degree exceeds it by more than two. Thus the required compatible nullhomotopies exist. Trivializing this difference bundle identifies

Before taking Thom spectra, D^\circ(qa_0) is a map from the sphere bundle S^{n\rho} to S^\rho. After the difference bundle is trivialized, its fibrewise mapping spectrum is the constant spectrum S^{-3(n-1)}: endomorphisms of the remaining invertible sphere line identify canonically with scalars. Sections of this constant spectrum are determined, on homotopy classes, by evaluation at one point, because \Sigma^\infty\mathcal B_+\simeq S. At that point the map is the ordinary smash cube of qa_0, namely the suspension of x^3 in the indicated degrees. This proves (6.7), up to a 3-local unit which is absorbed in the source identification.□

Proof. Filter D_3M by the number of upper Moore cells in its three input factors. The four successive layers, before simplification, have stabilizers \Sigma_{3-r}\times\Sigma_r, 0\leq r\leq3. The bottom layer is \Sigma^\infty B\Sigma_{3+}. For one upper cell, the \Sigma_2-stabilizer acts trivially on its one-dimensional sphere; the layer is therefore \Sigma S after localization at three. For two upper cells, interchanging the two odd spheres is the sign action. Its \Sigma_2-homotopy orbits are zero at three, so this layer disappears. The top layer is D_3S^1.

The attachment of the one-cell layer to the bottom is degree three. One can compute it in the cellular complex of M^{\wedge3}: a chosen one-upper-cell generator has boundary three times the bottom generator. Passing to \Sigma_3-coinvariants identifies the three choices of this generator; it does not add their three boundaries. Thus the attachment has degree three, not nine. Since the reduced suspension spectrum of the connected space B\Sigma_3 has zero \pi_0, this attachment has no reduced component.

It follows that the part K below the top layer has a projection r:K\to M inducing an isomorphism on bottom homology; explicitly, K is the cofibre of the bottom degree-three map into \Sigma^\infty B\Sigma_{3+}, and the augmentation gives r. Pull back D_3M\to D_3S^1 along D^\circ S^1\to D_3S^1, obtaining U, and push out its fibre along r. These constructions give a diagram of triangles

Use (6.6) in the lower row and desuspend its boundary to define g. The natural comparison square for D^\circ q gives D^\circ M\to U\to T; composing it with D^\circ a_0 gives F. Its quotient is (6.7). The bottom-homology claims follow because the comparisons of base orbits preserve H_0, the projection r preserves the bottom generator, and the remaining quotient \Sigma^4M has no homology near degree zero.□

Proof. Throughout this proof homology has coefficients \F_3. We adapt Toda’s diagonal cone construction [37, Section 1, Theorem 2] to the truncated functor D^\circ. Applying the diagonal cone nullhomotopy gives a natural map

Compose it with the map D^\circ C_{a_0}\to D^\circ S^t induced by C_{a_0}\to S^t. This composite is zero on D^\circ M, and therefore factors through C_{D^\circ a_0}/D^\circ M\simeq\Sigma D^\circ S^n. The resulting map

comes from the diagonal trivial line in the permutation representation. More explicitly, use the same interval parameter in the three cone factors and then collapse their ends. Its normal representation is \rho-\mathbf1=\lambda. The resulting Thom map is therefore the Euler map for \lambda. Since N+2=3t=324, the map on the tested top homology is precisely the cap product (6.8); it is nonzero.

The spectrum D^\circ M has homology only in degrees at most five: its fibre M^{\wedge3} has homology in degrees zero through three and its base has homological dimension two. Consequently the suspended top class of D^\circ S^n\simeq\Sigma^NM gives a class c\in H_{3t}(C_{D^\circ a_0}). Let u\in H_t(C_{a_0}) be the top generator. In the homology spectral sequence over \mathcal B, the fibre of D^\circ C_{a_0} has tensor-product homology with degrees chosen from \{0,1,t\}. In total degree 3t, the sole possible term is

the action is trivial since t is even. There is no possible differential into or out of this term, nor another term in its total degree. Projection to D^\circ S^t identifies it with the fibre generator there. By (6.10), the image of c is a nonzero multiple of u^{\otimes3}.

The bottom fibre class survives as well. The Cartan formula for reduced powers [28, Section 2, Property 2.5] on the fibre gives

Indeed, to reach degree zero each tensor factor must be lowered from degree t=108 by \mathcal P_*^{27}; all other Cartan terms have zero contribution to the bottom. Naturality shows that \mathcal P_*^{81}c is nonzero in H_0(C_{D^\circ a_0}).

Finally, the map D^\circ M\to T and the identity on the source induce C_{D^\circ a_0}\to C_F. It is an isomorphism on the bottom and on the tested top homology. Applying naturality again proves the assertion.□

Proof. The Moore sequence gives \pi_1M=0 and \pi_0M=\F_3. Applying [-,M] to that sequence shows that restriction to the bottom identifies [M,M] with \pi_0M. In particular 3\,1_M=0. The cofibre sequence

then allows the identity of M to extend to a multiplication \mu with one unit law. The other unit composite has the same bottom restriction as the identity, so it is the identity by the same restriction calculation.

Applying F(-,M) to (6.1) identifies the fibre of \varepsilon:F(M,M)\to F(S,M)=M with F(\Sigma S,M)=\Sigma^{-1}M. Adjunction applied to \mu gives the section s. Its value on i is the identity by the unit law. Splitting this fibre sequence proves the assertions.□

Proof. The layer complex agrees with the full E_1-complex above filtration two. In filtration two it has the same cycles but omits incoming d_1-boundaries. This gives the stated comparison on E_2, hence on E_5 by sparseness.

For clarity, a full d_r-cycle lifts to a cycle whenever comparison is surjective in its slot and injective in its outgoing slot: the differential of any lift maps to zero and must therefore vanish. Comparison on the resulting homology is injective as well if it is injective in the current slot and surjective in the incoming slot, because a full boundary can then be lifted.

Apply this first to d_5. An outgoing slot from filtration f\geq2 has filtration at least seven, where c_5 is an isomorphism. Thus c_6 is surjective for every f\geq2. If f\geq7, the incoming slot has filtration f-5\geq2, so the same argument gives an isomorphism. Sparseness identifies this page with E_9. For d_9, all outgoing slots again compare isomorphically. When f\geq11, the incoming filtration f-9 is at least two, proving the last row. Everything is natural in maps of the underlying tower.□

Proof. Choose a lift \widehat c\in\pi_{318}F_{10}M of ix^3 using Proposition 6.4, and let c be its image in \pi_{318}F_2M. Smash the section in Lemma 6.8 with I^{\wedge2} and put

After forgetting filtration, w is exactly x^3 1_M, because s(i)=1_M.

Postcomposition with g:\Sigma^3M\to M defines

The obstruction to lifting w through the triangle (6.9) smashed with I^{\wedge2} is

We show separately that both components of this obstruction under (6.11) vanish.

The first component. Forgetting filtration makes the whole obstruction zero, since F is already an actual lift of x^3 1_M. Its first component lies in \pi_{321}F_2M, and the map

is injective. To see this, the quotient M/F_2M has its two layers L_0M and L_1M, both with zero \pi_{322}: their support degrees are, respectively, 0 and 3 modulo four, whereas 322\equiv2. Thus \pi_{322}(M/F_2M)=0, and the long exact sequence gives the injection. The first component is therefore zero already in the restricted tower.

The second component. The fixed composite

suspends to a map h:\Sigma^4M\to M. Under the splitting, the second obstruction is precisely h_*(c)\in\pi_{322}F_2M. The map h on \BP_*-homology is multiplication by a scalar multiple of v_1, since a \BP_*-linear degree-four map \BP_*M\to\BP_*M is determined by its value in \BP_4M=\F_3\{v_1\}. This is a comodule map: v_1 is primitive modulo three, by \eta_R(v_1)=v_1+3t_1. Functoriality of the normalized cobar complex therefore identifies E_2(h) with that literal coefficient multiplication. The map of Adams towers passes this identification to E_5 and E_6; the coefficient v_1 is permanent by the target-vanishing argument in Section 7. No multiplicativity assumption on h is used. Its action on the full E_6-page in the relevant slots is consequently a scalar multiple of

which is zero by Lemma 7.2. Lemma 6.9 is an isomorphism on both these E_6-slots. Consequently h induces zero on the filtration-ten associated-graded map of the restricted abutment. Since c\in\mathcal F^{10}\pi_{318}F_2M, this proves

The equality follows from sparseness: in stem 322 only filtrations congruent to two modulo four can occur.

For every such filtration f\geq14, Lemma 7.2, including its high-filtration estimate, gives

The comparison of Lemma 6.9 is an isomorphism there, so the restricted E_{10}-groups vanish as well. Every associated-graded quotient of \mathcal F^{14}\pi_{322}F_2M is therefore zero. Increasing connectivity makes this filtration finite, and hence \mathcal F^{14}\pi_{322}F_2M=0. This proves that the second component vanishes.

Notice why the two page arguments were separated: an incoming full d_9 at filtration ten may start in filtration one, which the restricted tower omits. No E_{10}-comparison in filtration ten was asserted or used. The zero map (6.12) removes that filtration on E_6 before the higher E_{10}-vanishing is applied.

We have now proved \gamma_*(w)=0. The triangle (6.9) smashed with I^{\wedge2} supplies the desired lift. More explicitly, M is dualizable, so

The exact sequence obtained by applying F(M,-) to that triangle therefore identifies the vanishing obstruction just calculated with the existence of F':\Sigma^NM\to F_2T. Its quotient represents w, whose underlying map is x^3 1_M=\pi_TF.□

Proof of Theorem 6.1. Let \overline F' be the composite of the map in Proposition 6.10 with F_2T\to T. Its quotient equals that of F, so the difference has a lift

The map \overline F' has Adams–Novikov filtration at least two. The maps F, \overline F', and F-\overline F' induce zero on mod-three homology for degree reasons, so Lemma 6.3 and additivity show that the primary invariant of F-\overline F' equals that of F. Since j_T is an isomorphism on bottom homology, the cone comparison from Z to j_TZ is an isomorphism on the bottom and on the tested top degrees. Proposition 6.7 thus gives a nonzero

Use the splitting of Lemma 6.8 to write the class of Z as the sum of its section component and its quotient-precomposition component. The section component is s_*(z_0) for some z_0\in\pi_NM=\pi_{322}M. This class has Adams–Novikov filtration at least two: the degree-322 homotopy of both L_0M and L_1M is zero, so it lifts successively through F_1M and F_2M. Applying the section to that lift shows that the corresponding map \Sigma^NM\to M also factors through F_2M. It can therefore be subtracted without altering the primary test.

The remaining component factors through the upper quotient of its source:

The cone comparison induced by \Sigma^Nq preserves the bottom class and the top class in degree N+2=324. Hence C_{a_1} has nonzero \mathcal P^{81} from its bottom to its top. Apply Lemma 6.2 with m=81=3^4: the map a_1:S^{323}\to M gives an element \theta\in\pi_{322}S detected by b_3. The actual Moore boundary satisfies

Nonzero detection implies \theta\ne0; therefore its additive order is exactly three.□

Proof. Let \mathcal P_m be the poset of nonempty subsets of [m]=\{0,\ldots,m\}. The functor

sends an inclusion to the induced order-preserving injection. We show that u is homotopy initial. For 0\leq k\leq m, the category (u\downarrow[k]) is the face poset of the simplicial complex K_{m,k} whose nonempty faces are weakly increasing partial maps [m]\dashrightarrow[k]. Indeed, a face consists of a subset S and an order-preserving map from its induced ordinal to [k].

The complex K_{m,k} is contractible when k\leq m. To see this, start with the subcomplex whose last argument is either undefined or has value k. This is a cone on K_{m-1,k}, with cone vertex (m,k). For each a<k, the remaining faces with last value a form the cone with vertex (m,a) on K_{m-1,a}. Its intersection with the previously constructed complex is exactly that base. Since a\leq m-1, the base is contractible by induction. Attaching a cone on a contractible subcomplex preserves contractibility. The initial case m=k=0 is a point. This proves the assertion, including the case k=m, where the first cone is contractible without any assumption on its base. The homotopy-initiality criterion therefore identifies \Tot_m X with the homotopy limit of the nonempty unit cube indexed by \mathcal P_m. Including its empty vertex gives the cube with initial value Z and one unit insertion in each of its m+1 directions. Its total fibre is I^{\wedge(m+1)}\wedge Z, because smash products of spectra preserve finite fibres in each variable. This construction is compatible with restriction of cubes and hence with the tower maps.

Choose the E_4 structure on \BP furnished by [3, Theorem 1.1]. In particular, its associative algebra structure in associative algebras supplies coherent pointwise pairings of the Amitsur construction. The external pairings, followed by \mu, give the pairing of totalization spectral sequences, and thus of the identified Adams towers. The boundary formula for a pairing of filtered objects is precisely the Leibniz rule; it uses the binary pairing and its compatibility with the tower, not associativity of \mu. The required unital \mu is provided by Lemma 6.8.

Finally, the sphere-input Amitsur levels have even, 3-torsion-free homotopy groups. Smashing with M reduces these groups modulo 3. Unitality makes the resulting coefficient pairing B\otimes_{\BP_*}B\longrightarrow B ordinary multiplication. Since 3 is invariant in the Hopf algebroid, the normalized cobar complex after this reduction computes D. Its induced cup product is the ordinary Ext product, proving the last assertion. In particular, associativity and graded commutativity hold on E_2 and pass to all subsequent pages by taking homology.□

Proof. Every A_i annihilates a vector in K^-. If A(u)x=0, then A_0x=-\sum_{i>0}u_iA_ix, proving the second kernel inclusion. If all A_i for i>0 annihilate x, then A(u)x=A_0x; this proves the first image inclusion. The last follows directly from the formula for A(u). No independence assumption on the parameters is used.□

Proof. Use the ordinary normalized cobar complex. Order its monomials first by increasing number of coefficient factors v_i, counted with multiplicity; next by decreasing total coefficient weight; and finally by the decreasing lexicographic order on the total t_i exponent tuple, with the largest index dominant. Filter by tails of this order. At a fixed internal weight this filtration is finite, including across all cobar lengths.

The ideal generated by the positive-degree coefficients is invariant under the right unit. Thus a right-unit correction cannot decrease the coefficient count. If it preserves that count and introduces a nonconstant t monomial, it decreases the coefficient weight. After setting the coefficients to zero, each nonadditive coproduct term of t_i uses lower-index t’s and hence decreases the final tuple. These assertions follow directly from the Hazewinkel recursion of Appendix A. They show that the associated graded cobar differential is the one for ordinary additive coordinates with primitive coefficients.

The cohomology of this associated graded complex is

where, for w_{i,j}=3^j(3^i-1)/2,

One obtains this formula by writing each additive coordinate in base-three digits. Its graded dual is a tensor product of truncated polynomial algebras \F_3[x]/(x^3), with x of weight w_{i,j}. The alternating free resolution with maps x and x^2 has one exterior Ext generator in degree (1,w_{i,j}) and one polynomial generator in degree (2,3w_{i,j}). Tensoring these resolutions gives (7.17); only finitely many factors occur in any fixed weight.

The leading class of b on this page is c_{1,0}. Indeed, h=t_1^3-v_1^2t_1, and its integral Bockstein has coefficient-free part

the standard polynomial generator for the weight-one digit. All the other terms have positive coefficient count.

For a monomial on (7.17) define its deficit to be \frac92f-d. Among exterior generators the only positive deficits occur at weights 1,3,4, and they are respectively 7/2,3/2,1/2. As these generators are exterior, their total contribution is at most 11/2. The polynomial generator b has deficit 6; every other polynomial generator and every coefficient factor has nonpositive deficit. Consequently a monomial not divisible by b^k has deficit at most

It follows that the quotient of (7.17) by b^k vanishes in bidegree (f,d) when d<S_k(f). Multiplication by b^k on that page is injective everywhere.

To transfer these assertions to D, take the filtered mapping cone Q of the cochain cup map

Shift the source filtration by the leading monomial of b^k. The first cohomology page of the cone is exactly the quotient just described, since its leading cup map is injective. The filtration is finite in fixed weight, so H^{f,d}(Q)=0 for d<S_k(f). The long exact sequence contains

Vanishing of the right term proves surjectivity. Vanishing of the left term uses d<S_k(f-1) and proves injectivity, including the one-filtration shift in (7.16).□

Proof. For (i), lift a cycle c\in D_t to x with qx=c. Then qLx=Lc=0, and injectivity into D_{t+J} gives Lx=0. For (ii), suppose x is a cycle and qx=Ly in D_t. Surjectivity into D_{t-J} gives y=qz. Hence q(x-Lz)=0; injectivity into D_t gives x=Lz.□

Proof. For a single digit, let e be the resolution degree and let j_0=0,1,2 be the operator digit. If e is even, the next boundary multiplies by P_m. On the two terms for which H is nonzero its coefficient is j_0!j_0=1 in \F_3. If e is odd, the only nonzero case has j_0=2; the next boundary is -P_{2m} and the combined coefficient is -2=1. On the other terms H\partial^0 supplies the identity, except at e=j_0=0, which is the augmentation summand.

In the tensor product all factors before the first active one have operator digit and resolution degree zero. Hence no additional tensor sign is required in (A.9). Terms differentiating a later factor cancel between \partial^0H and H\partial^0, because H changes the degree of the selected factor by one. Finally, the increase of C(g) on incrementing this factor is exactly h'm_\nu. This proves all the assertions.□

Proof. First consider a cycle e whose leading part is p. The leading part of its boundary is \partial^0p, so \partial^0p=0. In degree zero there is no P_0g_0 term, because \varepsilon(e)=0 and distinct coefficient monomials are independent. Lemma A.1 therefore gives \partial^0H(p)=p in every relevant degree. Subtracting \partial H(p) removes precisely the leading part and introduces only larger order values. It preserves the cycle condition. Since only finitely many order values occur below N, the procedure terminates and returns a preimage of e.

For the induction on s, \varepsilon\partial_1=0. The preceding argument with r=0 proves exactness at G_0. Assume the differentials through \partial_{s-1} have been constructed, their successive composites vanish, and cycles in G_{s-2} can be lifted. The error \partial_{s-1}p in (A.13) is such a cycle. Its additive leading part vanishes because (\partial^0)^2=0, so all corrections have order strictly larger than that of g. Equation (A.13) gives \partial_{s-1}\partial_sg=0 and preserves the required leading part. The first paragraph now lifts every cycle in G_{s-1}. This completes the induction and proves exactness at each finite cutoff.

We give the completion and comparison details to ensure that finite truncation has not changed the invariant being calculated. From (A.4), an output operator P_U in P_Sv^QP_T satisfies

To see this, take a contributing coefficient of \Delta(t^U). If its coefficient has weight c\geq0, homogeneity gives |U|=|S-I|+|T|+c. Thus an output expansion on a free generator may be truncated before acting on it from the left: discarded output terms cannot return below the cutoff. Input truncation is justified separately. Degree-zero differentials preserve a coefficient already on the far left, and their generator values have nonnegative new coefficient weight, so they do not decrease pure weight. The contraction preserves it. The constructions at successive cutoffs therefore agree under truncation. Taking their compatible limits gives exact lifting in the completed modules, rather than relying on an exactness theorem for an infinite dual.

For comparison use the normalized relative bar resolution of B over \mathcal U relative to B. It is left free on the words [P_{T_1}|\cdots|P_{T_s}], T_i\ne0. This description remains valid for noncentral B: move coefficient factors left by the balancing relation and (A.4). The differential multiplies successive entries with the alternating bar signs, including the first left multiplication and the last action on 1\in B; an entry in B at a normalized position is discarded. For example its degree-two boundary is

with the middle term expanded and normalized. A B-linear contracting homotopy inserts the left P_T as the first word entry when T\ne0 and is zero when T=0, while keeping its coefficient on the far left. The usual adjacent terms in the two composites cancel; the sole uncancelled term is the original word, with the augmentation term removed in degree zero. This verifies the contraction directly. It preserves pure weight, where a word generator has weight \sum|T_i|.

Construct a chain map from G to this bar resolution by sending g_0 to the empty word and, on each further generator, applying the bar contraction to the image of its boundary. Construct the reverse map using (A.12). Their composites are homotopic to the identity: after the homotopy has been defined in lower degrees, its next required value is a cycle, and the corresponding contraction or antidifferential supplies a preimage. This also proves independence of all choices of lifts. Every map and homotopy has the cutoff compatibility just proved.

Finally a cochain of weight d takes a generator of weight w(g) to a polynomial of weight d-w(g). Evaluation of v^QP_Tg uses P_T on that polynomial, which is zero if |T|+w(g)>d. Only finitely many terms of any comparison or homotopy can consequently be seen by such a cochain. Cochains on the relative bar identify with the ordinary cobar by coefficient extraction: multiplying adjacent operators is dual to inserting \Delta, and balancing coefficients is dual to applying \eta_R. Thus the comparison is a cochain homotopy equivalence in each weight d\leq N, as asserted.□

Proof. The difference of two lifts with the same augmentation is null homotopic: in each successive degree its uncancelled difference is a cycle, which is lifted by (A.12). The same argument applied to a coboundary representative proves independence of that representative. Composition of such lifts is the defining Yoneda composition.

One may also compare on the relative bar. An unsigned lift applies the cocycle to the last block of entries and retains the first block; move the resulting coefficient left by the balancing relation and discard scalar entries at normalized positions. The interior terms cancel in its chain identity, and the terms crossing the division between blocks cancel by the cocycle equation. Multiplying its component in degree s by (-1)^{js} gives (A.25). Evaluation by the other cocycle is the bar, and hence cobar, cup product, with the stated ordering sign.

To check the final assertion explicitly, suppose an odd-filtration differential D is a left derivation for a product juxtaposed as xy, and change to x\star y=(-1)^{fg}xy, where x,y have filtrations f,g. Then

The row rescaling L|_{D_{f,*}}=(-1)^fD therefore obeys the left Leibniz rule for \star, and L^2=-D^2=0. Row rescaling preserves every kernel, image, and line of possible cycles. A further common unit puts L(c')=ab^3 as in Section 7. The permitted unit in the d_9 formula absorbs the analogous convention change there.□

Proof. Each row produced by elimination is an \F_3-linear combination of the entries of (A.33). It therefore vanishes at each actual square-zero differential. Rows with no quadratic term are valid affine restrictions on the present parameter space. Substitution of an affine parametrization preserves all such points. Every nontrivial refinement decreases the finite affine dimension, which proves termination. Retaining only the linear consequences uses no claim about all polynomial consequences or about the full set of \F_3-points of the quadratic system.□

Proof. Every A_i kills K^-. If A(u)x=0, then A_0x=-\sum_{i>0}u_iA_ix lies in \sum_{i>0}\im A_i, which is the condition defining K^+. On the common kernel of the A_i with i>0, the map A(u) equals A_0, proving the lower image inclusion. The upper image inclusion follows by summing the images of its summands. Each argument applies to every single parameter value, so restricting the allowed parameter values cannot invalidate it.□

Finite calculation. The following intermediate outputs specify checks on the elimination at each stage, in addition to the final tables. They can be reconstructed without any external list of Ext groups.

First enumerate (A.7) and (A.15), construct (A.13), and form (A.16). For (f,d)=(1,27),(6,10),(6,81) the dimensions of the three successive cochain terms are respectively

Reducing the incoming and outgoing columns gives ranks

For the first slot append, in order, the cochains v^{26}a,y,t_7'; the rank becomes 14. For the second append vb^3; the rank becomes 2. For the third construct B(y) by (A.21), apply (A.19), and form its cube using (A.24)–(A.26). Appending this cubic leaves the incoming rank 5256. Keeping witnesses in column elimination expresses the cubic as a boundary. In these coordinates B(y) has eight coefficient terms on six generators, the cubic has 76 terms, and the resulting preimage can be taken to have 2642 terms. Substitution in (A.16) verifies its boundary directly. This proves the algebraic zero in the statement.

For weight 60, the grid contains 1213 slots and its complete pairs have 2779 scalar unknowns. Expanding the zero and Leibniz equations gives 15852 scalar equations. For each of the nine (k,e) choices, the coefficient rank is 2667. The augmented rank is 2667 for (1,0) and 2668 for each of the other eight choices. These values are obtained from (A.32), including its constant column; the pivot at the constant position is an explicit linear combination giving 1=0 in each inconsistent case.

At weight 92 there are 2269 slots, of total Ext dimension 6341, and 1820 differential pairs with 23384 entries. The Leibniz assembly uses 15453 factor multiplication matrices, including successive factors in (A.31). There are 137471 scalar zero and Leibniz equations, 110060 of them having nonzero augmented row. Coefficient and augmented ranks are both 23119. Thus the initial affine dimension is 265. The successive square eliminations have the following checkpoints:

In each row quadratic monomials precede linear ones, as specified after (A.33); after the final refinement the affine rank is 23158. Apply (A.34) and (A.35) to its projected matrices. They give the sixteen rows of Table 5. The two additional stacked ranks are 1 and 2. Equations (A.36)–(A.38), using (A.40) in the second case, give (4,5,5) and (6,5,5). Equation (A.39) gives the pairs (3,3),(2,2),(4,4).

Finally compute the five iterated b maps of Table 8 with cutoff 93. As a check independent of quotient-coordinate extraction, append the resulting product cochains to the incoming coboundary columns in their target slots. In the order of that table, incoming ranks and ranks after appending are

Their differences are the product ranks 8,6,6,4,4. All these operations have been reduced to the polynomial recursions, the terminating lifting recursion, and the explicitly stated elimination rules. The equalities used to infer the spectral-sequence conclusions are therefore finite algebra calculations; their use is proved in Section 7.□