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 54. Theorem 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 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]. 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] and [1]. This is the surviving corrected beta combination, not an assertion of survival for \(\beta_{9/9}\) alone.

If \(b_1\) survived, Lemma 4 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] identifies this group with the order-three group generated by \(\beta_{3/3}\); see also [32]. The classical Toda differential \[d_5(\beta_{3/3})=\pm\alpha_1\beta_1^3\neq0\] excludes such a permanent class; see [5]. ◻

Proof of Theorem 1. Lemma 3 settles the indices \(0\), \(1\), and \(2\). Theorem 36, proved in Section 5, excludes every index \(j\geq4\). Theorem 53, 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 \(\mathrm{BP}\to H\mathbf F_3\) induces a map between the unit fibres and their smash powers. Consequently the filtrations on sphere homotopy satisfy \[ F_{\mathrm{AN}}^s\pi_*S \subseteq F_{\mathrm{Adams}}^s\pi_*S . \tag{5}\] 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]; see also [26]. 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 \(\mathrm{MU}_{(3)}^{\wedge(s+1)}\). For \(K=G\) or \(H\), apply the translated orientation to the \(\ell\)-th factor at \(g_\ell\), and multiply in \(E\): \[ (a_0,\ldots,a_s)\longmapsto \bigl[(g_0,\ldots,g_s)\longmapsto \prod_{\ell=0}^s g_\ell f(a_\ell)\bigr]. \tag{6}\] 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 (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 \[\mathop{\mathrm{Tot}}^{\mathrm{fat}}_1 C^\bullet \simeq \mathop{\mathrm{Tot}}_1 C^\bullet\oplus\Omega C^0,\] 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] identifies the \(\mathrm{MU}_{(3)}\) and \(\mathrm{BP}\) descriptions of the ordinary Adams–Novikov spectral sequence. On the \(\mathrm{BP}\) 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], also stated in [32], 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 \[0\leq i\leq n,\qquad 1\leq d\leq a_{n-i},\qquad a_0=1,\qquad a_r=3^r+3^{r-1}-1\quad(r\geq1).\] In particular \(d<2\cdot3^n\). In the desired degree the equation is \[4L3^n-d=3^{j+1}.\] 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 (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 (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 \[c_n=\delta_{v_1^d}(v_2^{Ld}) \in\mathop{\mathrm{Ext}}_{\mathrm{BP}_*\mathrm{BP}}^{1,\,4\cdot3^{j+1}}(\mathrm{BP}_*,\mathrm{BP}_*/3)\] is defined. Its exact \(v_1\)-annihilator exponent is \(d\). Indeed \(v_1^d c_n=0\), whereas the connecting-sequence comparison gives \[v_1^{d-1}c_n=\delta_{v_1}(v_2^{Ld})\ne0.\] For the last assertion, the primitives of \(\mathrm{BP}_*/3\) are \(\mathbf F_3[v_1]\). 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 \[v_s\longmapsto v_s+v_1t_{s-1}^3-v_1^{3^{s-1}}t_{s-1}.\] Comparing the highest power of \(t_{s-1}\) in a polynomial depending nontrivially on \(v_s\) contradicts invariance. The image of the primitives in \(\mathrm{BP}_*/(3,v_1)\) 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] says that the image of integral \(\mathop{\mathrm{Ext}}^1\) in this internal degree is spanned, after reduction modulo \(3\), by \[ v_1^{N-1}t_1 . \tag{8}\] 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 (8); its \(k\)-th term for \(k\geq2\) has \(3\)-adic valuation \(k-1-\nu_3(k)>0\). The class (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 (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 \[A_1=t_1^3-v_1^2t_1,\qquad c_n=\frac{(v_2^d+v_1^d A_1^d)^L-v_2^{Ld}}{v_1^d}.\] 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, \[(I^q:3)=I^{q-1}\qquad(q\geq1);\] 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 4. Let \(\zeta\) be a primitive ninth root of unity, and let \(\mathcal O=W(k)[\zeta]\), allowing a residue extension if necessary. The Lubin–Tate construction for \(\mathbf Z_3[\zeta]\) [21] gives a height-one formal \(\mathbf Z_3[\zeta]\)-module over \(\mathcal O\). Its underlying special fibre has height six. After residue extension it can be identified with the chosen Honda formal group. The embedding of \(\mathbf Q_3(\zeta)\) 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 \[E_*\longrightarrow \mathcal O[w^{\pm1}],\qquad |w|=-2.\] The group fixes \(\mathcal O\), 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 (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 \[\mathcal O\xrightarrow{0}\mathcal O\xrightarrow{9}\mathcal O \xrightarrow{0}\mathcal O\xrightarrow{9}\cdots.\] Its reduction modulo \(3\) has zero differentials. The connecting homomorphism for multiplication by \(3\) is therefore \[ \delta_3:H^1(G;\mathcal O/3)\longrightarrow H^2(G;\mathcal O), \qquad \bar c\longmapsto3c\pmod9 . \tag{9}\] 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 \(\mathrm{BP}\) 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 \(\mathcal O/3\). 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 (7).

Express \(x\) in the basis of Lemma 5. 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 \[L\,v_2^{(L-1)d}t_1(g)^{3d}.\] 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 \(c\in\mathcal O/3\) with unit residue. Equation (9) sends it to \(3c\ne0\) in \(\mathcal O/9\). 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 \(Y/F^2\simeq\mathop{\mathrm{Tot}}_1C^\bullet\). In the fibre sequence \[\Omega N^1C^\bullet\longrightarrow \mathop{\mathrm{Tot}}_1C^\bullet\longrightarrow C^0,\] 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 4. The augmented comparison sends it to a class in \(\pi_{4\cdot3^{j+1}-2}F^2\). Its symbol is nonzero by Lemma 7. Lemma 9 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 \[fe\longmapsto h\sum_{i=1}^m c_i\alpha_i(f)e .\] 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 \[\operatorname{triv}(S^d) \xrightarrow{\mathrm{diag}} \operatorname{coInd}_e^J S^d \simeq \operatorname{Ind}_e^J S^d \longrightarrow Z.\] 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 11. 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 \[(1+3c)^3-1=9c(1+3c+3c^2)=0\] 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 \[(q-1)(1+q+\cdots+q^8)=0\] 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 11 gives a nonzero integral trace in \(Ru^n\), and Lemma 12 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 \(j\geq4\), \[4\cdot3^{j+1}-2=972\cdot3^{j-4}-2.\] A survivor would give a nonzero finite-order element of the group in this degree by Theorem 10. 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 \[B_r^s =j_s\ker\bigl(\pi_VF^s\longrightarrow \pi_VF^{s-r+1}\bigr), \qquad 1\leq r\leq s+1.\] 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 \[R_N=N_e^K\mathrm{MU},\qquad \rho_K=\text{the real regular representation of }K .\] We use the derived genuine norm and its commutative-ring counit; their indexed multiplication and geometric fixed point formulas are recalled in [16]. Nonequivariantly, \(R_N\) is a smash of \(|K|\) copies of \(\mathrm{MU}\), labelled by \(K\). Let \[c\in\pi_{52}\operatorname{Res}_eR_N\] 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 \[ z\in\pi_{52\rho_K}^K R_N. \tag{11}\]

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]. After applying \(\Phi^H\), the factors in the domain of \[N_e^K\operatorname{Res}_eR_N\longrightarrow R_N\] are labelled by \(Q\times K\), the target factors by \(Q\), and the multiplication folds them by \[Q\times K\longrightarrow Q,\qquad (aH,c)\longmapsto acH .\] 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 \[Q\times K\longrightarrow Q\times Q\longrightarrow Q.\] The first stage is the \(Q\)-norm of the multiplication \(\mathrm{MU}^{\wedge K}\to\mathrm{MU}^{\wedge Q}\) 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 \[EK_+\wedge R_N\longrightarrow R_N \longrightarrow \widetilde{EK}\wedge R_N .\] 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]. Apply this also after smashing with \(S^{-52\rho_K}\). The obstruction to lifting (11) through the first term is precisely its image \[\Phi^H z\in \pi_{(52\rho_K)^H}^{Q}\Phi^H R_N,\] which is zero. Thus \(z\) has such a lift.

The group \[\pi_{52\rho_K-d\lambda}^K(EK_+\wedge R_N)\] 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 \(\operatorname{Res}_eR_N=\mathrm{MU}^{\wedge |K|}\) 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 \[ a_\lambda^d z=0 \quad\text{when}\quad 52|K|-2d<0. \tag{12}\] 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 \[R_N\longrightarrow E^{\mathrm{Bor}}\] obtained on underlying spectra with action by applying the translated orientation in each factor and multiplying: \[(a_\gamma)_{\gamma\in K}\longmapsto \prod_{\gamma\in K}\gamma f(a_\gamma).\] 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 \[S^{52\rho_K}\wedge E^{\mathrm{Bor}} \longrightarrow E^{\mathrm{Bor}}.\] Both spectra are cofree, since a finite representation sphere preserves cofreeness. An underlying equivalence between cofree spectra is a genuine equivalence. Transport (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 (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 15 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, \[ ev_i\equiv v_i+\sum_{l<i}v_l\,t_{i-l}(e)^{3^l} \pmod{\mathfrak m^2}. \tag{13}\] 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 \(\mathfrak m\). Terms containing two such coefficients vanish modulo \(\mathfrak m^2\). 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 (13).

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 (13) 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 \[D_\sigma y_0=a^{27}aP_0=-P_0.\] 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 \(\mathfrak m/\mathfrak m^2\). ◻

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 \[l^d,\qquad yl^{d-1},\qquad y^2l^{d-2}.\] 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 (14) 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 \(\mathop{\mathrm{Tr}}_H(u)/u\) 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 \(\mathfrak m^2E_{-2}\). 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 \(\mathfrak m^2E_{-2}\) corrects \(u\) so that \[\mathop{\mathrm{Tr}}_H(u)\in\mathfrak m^3E_{-2}.\] Since \(g\) commutes with \(H\), the elements \(D_g u,D_g^2u,D_\sigma u\) satisfy the same trace condition. By Lemma 17 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 (15).

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 \(L=\mathbf Q_3(\zeta_3)\) 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 \[ k[b^{\pm1},u^{\pm3},N_1,N_2] \otimes\Lambda(eu,z_1u,z_2u). \tag{16}\] 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 \(\mathfrak m\)-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 \(\mathop{\mathrm{Tr}}_H\), so it gives the claimed \(x_0\). Projection to the weight-one character commutes with \(\mathop{\mathrm{Tr}}_H\), preserves the residual period, and gives the last assertion. ◻

Proof. The actual trace equation of Proposition 19 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 (17). Its equivariance holds by the definition \(x_i=g^i x_0\). ◻

Proof. Lift the norm classes \(N_1,N_2\) in (16), 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 (16); 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 (18).

For the alternative parameters, the identity \[\mathop{\mathrm{N}}_H(a+b)-\mathop{\mathrm{N}}_H(a)-\mathop{\mathrm{N}}_H(b)\in\mathop{\mathrm{im}}(\mathop{\mathrm{Tr}}_H)\] 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 (16). 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 \[q_i/q=\mathop{\mathrm{N}}_H(x_i/x_0)\] are norms of elements of \(\pi_0E\), so they have degree-zero homotopy lifts. The norm parameters used in Proposition 21 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 \[f=\sum_{0\le i,j<3}f_{ij}^{\,3}n_1^i n_2^j,\qquad f_{ij}\in K.\] 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 \[[3]_{F_2}\circ f=f\circ[3]_{F_1},\] 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 20, put \(a_i=x_i/x_0\), so \(a_0=1\), and put \(r=a_3\). Fixation of the ungraded base by \(\sigma\) says \[a_{i+3}=r a_i.\] 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 (19). 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 23. Its endomorphism \(1+\sigma+\sigma^2\) reduces to zero and is therefore zero. It consequently extends the \(\mathbf Z_3\)-action to an \(\mathcal O\)-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 \([\mathcal O:\mathbf Z_3]=2\), 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 \[ [\pi](z)\equiv U_i z^{3^i} \pmod{(\pi,U_1,\ldots,U_{i-1},z^{3^i+1})}, \qquad i=1,2; \tag{21}\] this is [36], applied to the ramified local field \(\mathbf Q_3(\zeta)\). 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, \(\mathop{\mathrm{N}}_H(x_i)=x_i^3\), since \(1+\zeta+\zeta^2=0\) and \(\zeta^{0+1+2}=1\). Hence \[q_i/q\longmapsto (x_i/x_0)^3=(1+r_i)^3.\] The parameter description of Proposition 21, 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 (16) 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 17 puts \(\theta\) at the fixed end and gives (20), 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 \(\mathcal O\)-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 \(\mathbf Q_3(\zeta)\) in the rational endomorphism algebra therefore has dimension four over \(\mathbf Q_3(\zeta)\).

At the closed point the corresponding centralizer is a division algebra of degree three, hence dimension nine over \(\mathbf Q_3(\zeta)\). 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 \(\mathbf Q_3(\zeta)\). Their generic lifts would satisfy a nontrivial linear relation over that field of operators. Clear its denominators using the structural \(\mathcal O\)-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 25. In a period trivialization, \(h\) has order one. Valuation therefore gives \(8\mathop{\mathrm{ord}}(z)+i=0\). Put \(d=\mathop{\mathrm{ord}}(z)\); 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 \(\mathcal O\)-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]. 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 \[\mathcal M\ \text{is free of rank three over }\ B_0=A_0[\epsilon]/(\epsilon^2).\]

At the closed point, the integral Dieudonné lattice is torsion-free over \(W(k)\otimes_{\mathbf Z_3}\mathcal O=W(k)[\zeta]\). 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 \[ 0\longrightarrow N\longrightarrow M_0\longrightarrow L \longrightarrow0. \tag{23}\] 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]. Thus the strict \(\mathcal O\)-module lifts are exactly the quotient lifts on which \(\epsilon\) acts by zero, or equivalently lifts of the quotient \(M_0\to L\) in (23). Relative to the split lift these are parametrized by \(\mathop{\mathrm{Hom}}(N,L\otimes I)\). 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 \[T_{A_0/k}\cong\mathop{\mathrm{Hom}}_{A_0}(N,L),\qquad \Omega^1_{A_0/k}\cong N\otimes L^{-1}.\] In particular \[ \det\Omega^1_{A_0/k}\cong(\det M_0)L^{-3}. \tag{24}\] 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]. 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]. 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 \[U=U_0+\epsilon U_1,\qquad \det_{B_0}(U)=a+\epsilon b.\] Its underlying \(A_0\)-linear matrix has block form \[\begin{pmatrix}U_0&0\\ U_1&U_0\end{pmatrix}.\] 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 \[\phi^*(\det M_0)\xrightarrow{\ \cong\ }\det M_0.\] 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 (24) are \(2,8\) and the Lie weight is one. Hence \[\operatorname{wt}_T(\det M_0)=2+8+3=13\pmod{26}.\]

It remains to relate \(M_0\) to \(V\). Use the first principal-parts module of \(L^{-2}\), with its exact sequence \[0\longrightarrow\Omega^1_{A_0/k}\otimes L^{-2} \longrightarrow P^1_{A_0/k}(L^{-2}) \longrightarrow L^{-2}\longrightarrow0.\] The jet map is linear over cubes, since \(d(f^3)=0\). Thus coefficient evaluation induces an \(A_0\)-linear, \(\mathcal C\)-equivariant map \[A_0\otimes_K V\longrightarrow P^1_{A_0/k}(L^{-2}).\] 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 \[\begin{pmatrix} 1&r_1&r_2\\ 0&1&0\\ 0&0&1 \end{pmatrix}.\] It is therefore an isomorphism. Taking determinants and using (24) gives \[\det(A_0\otimes_KV) \cong L^{-6}\det\Omega^1_{A_0/k} \cong L^{-9}\chi.\] 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 (22). ◻

Proof. For a rank-two module, \[W^\vee\cong W\otimes(\det W)^{-1}.\] 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 27, together with the injective evaluation of Theorem 24. The vector then evaluates injectively in \[w^{-2}w^9\chi\,w^{3n}=w^{7+3n}\chi.\] 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 26 rules out the resulting invariant section and proves the vanishing.

A vector in \(WP_n\) evaluates in \(w^{3n-2}\). Lemma 26 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 \[(q^2h)\,(q^8h^3)^k.\] 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 \[v_1=0,\qquad v_2=-t_1(\sigma)^4.\] The intermediate \(v_1\)-Bocksteins for \(v_2\), \(v_2^3\), and \(v_2^4\) therefore have values, respectively, \[t_1(\sigma)^3,\qquad t_1(\sigma)^9,\qquad \pm t_1(\sigma)^{15}.\] 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 24 is injective, giving the asserted classes in (25).

We use the classical Adams–Novikov differentials \[ d_5(\beta_{3/3})=\text{unit}\cdot\alpha_1\beta_1^3, \qquad d_9(\alpha_1\beta_4)=\text{unit}\cdot\beta_1^6, \tag{30}\] and the permanence of \(\alpha_1\) and \(\beta_1\); see [5]. 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 \[-4\Delta a+2\Delta i-\Delta j=1, \qquad -6\Delta a+4\Delta j=-4.\] 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, \[\beta_1=A^{-2}B^{-1}F\] 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 (30), and the absence of a \(d_9\) target for \(B\), therefore give \[B^{-1}d_9(\Theta)\beta_1^5 =\text{unit}\cdot\beta_1^6.\] 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 (34) give \[ \Theta\kappa(z)=cFz \tag{37}\] 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 \[\begin{array}{c|ccc} \text{basis vector}&\Theta\hat v&\Theta\hat y&\hat v\hat y\\ \text{weight}&-2&6&4. \end{array}\] 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 (37) fixes its coefficient, so \[\kappa(z)\equiv cF\hat v\hat y\pmod{\mathfrak n^2}.\] 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 \[\Psi(z')=\Theta M_0(z')+c_0(z')\hat v\hat y\] on that complement, with \(M_0\equiv\mu I\pmod{\mathfrak n^2}\). The exterior-three component of \(\Psi^2(z')=0\) is \[c_0\bigl((\operatorname{tr}M_0-M_0)z'\bigr)=0.\] The matrix \(\operatorname{tr}M_0-M_0\) has invertible reduction \(\mu I\), so \(c_0=0\).

If \(\mu\notin\mathbf F_3\), 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 \(i\in\mathbf F_3\). Only the scalar and \(\Theta\) terms in sector zero remain. The differential (34) 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 \(\mu\in\mathbf F_3^{\times}\).

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 \(\mathop{\mathrm{coker}}N\), respectively; multiplication by \(\Theta\) is the natural map \(\ker N\to\mathop{\mathrm{coker}}N\). For a class \(\alpha\in\mathop{\mathrm{coker}}N\), 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 \[k_j\equiv cF e_j\pmod{\mathop{\mathrm{im}}N}.\] 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 (33) now gives (35). 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 (36). ◻

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 (45) 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 (40), so a contraction would require a meromorphic invariant in \[\mathop{\mathrm{Hom}}(P_2,P_{-16})=P_{-18}.\] 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 \[66-2-2a\equiv0\pmod{26} \quad\hbox{or}\quad 66+6-2a\equiv0\pmod{26}.\] 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 \(E_{102}(H)\) \[[Z^6e]=Z^5[Ze].\] 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 (45) 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 \(\eta/\epsilon\notin\mathbf F_3\), 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 \(-\eta/\epsilon\in\mathbf F_3\) and set \(C=B^3\). The cohomology is \[ S_0[A^{\pm1},C^{\pm1}] \otimes\Lambda(B^2Y,B^{p_x}X). \tag{47}\] Its scalar, line, quotient, and product supports in the \(p\) coordinate are respectively \[ 0,\qquad2,\qquad p_x,\qquad p_x+2\pmod3. \tag{48}\] 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 (42). 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 \[ \mathop{\mathrm{Hom}}(P_2,P_{-2l-2})=P_{-2l-4}. \tag{49}\] The Hasse rule makes \(l\equiv2\pmod4\); among the decreasing lengths in (41), 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 (49) with \(l=2\). In the algebra (47), this says \[ d_{33}(B^2Y)=aA^{-6}CZ,\qquad d_{33}(B^{p_x}X)=0,\qquad a\in k^{\times}, \tag{50}\] 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 (50). 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 (50) gives \[k[A^{\pm1},C^{\pm1}]\otimes\Lambda(B^{p_x}X).\] The remaining odd coefficient has residual \(T\)-weight \(-2\) in the \(W\) line. An increasing map now requires \[3n\equiv3(n-2l)-2\pmod{26}, \quad\text{equivalently}\quad 6l+2\equiv0\pmod{26},\] and a decreasing map requires \[3n-2\equiv3(n-2l-2)\pmod{26}, \quad\text{equivalently}\quad 6l+4\equiv0\pmod{26}.\] 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 (46). Rescale \(h\) to absorb \(\epsilon\) in (44); then the two \(d_{17}\) coefficients become \(p\) and \(p+2=p-1\). ◻

Proof. The possible lengths were determined in the proof of Lemma 33. Since \[|C|=(18,0),\qquad |U|=(3,4),\qquad |A|=(-4,-6),\] the increasing length \(53\), together with the derivation rule, has exactly the form (51). 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 (46) persists until length \(105\). If \(e_{53}\ne0\), then \[d_{53}(C^a)=a e_{53}A^{-8}C^{a+1}U.\] The surviving scalar exponents are \(a\equiv0\pmod3\), and the surviving odd exponents are \(a\equiv1\pmod3\). Thus \[ E_{54}(H)=k[A^{\pm1},C^{\pm3}]\otimes\Lambda(CU) \qquad(e_{53}\ne0). \tag{53}\] In particular, the notation \(C^{-2}U\) still denotes a well-defined surviving class: in (53) 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 (52). 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 (53). ◻

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 33, with the supports read from (48) 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 34. ◻

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 (57). 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 \(\mathbf Q_3(\zeta_9)\) 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 20, 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 \(\mathop{\mathrm{Sym}}^d L\), where \[L=W(k)[H]/(N_H),\qquad N_H=1+\sigma+\sigma^2.\] 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 \[W(k)[H]=W(k)\mathbin{\times}_{k}W(k)[\zeta_3].\] This gives the asserted three lattice types. Accordingly, \(\mathop{\mathrm{Sym}}^dL\) 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 (58).

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 35 gives the \(H\) formula \(w_0\theta\wedge(-)\). On successive exterior degrees in a row of (59), 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 38 then gives (60): 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 (60), 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 (58), 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 \[[L_1Nf']=[Jf']\] 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 \[\widehat H^d(Q;S_n)\cong\widehat H^d(Q;W_n) \cong \begin{cases}k,&n\equiv0,1\pmod3,\\0,&n\equiv2\pmod3, \end{cases}\] 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 (60) 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 \[d_{13}B/B=\varepsilon T_0,\qquad d_{13}u'/u'=\varepsilon'T_0.\] For \(w_0=3p\) and \(3\mid n\), a scalar lift has symbol \[\mathcal D^{n/3}B^pu'^{\,i-2n/3},\] 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 35 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 37 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 \(\mathbf F_3\) 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 \[d_{13}(3b^2)=\eta\varepsilon\,3b^2T_0, \qquad \eta\in\mathbf F_3.\] 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 \[0\longrightarrow S_n\longrightarrow W_{n-2} \longrightarrow S_{n-4}\longrightarrow0.\] Apply coinvariants and their long exact Tate sequence. When \(n\equiv0\), the last Tate group is zero, giving (63) 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 (61). 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 \[\ker(N:M_Q\longrightarrow M) =\widehat H^{-1}(Q;M),\] which has the dimensions already calculated. Transferring the \(H\) differentials now gives the two coefficients in (63); 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 (63), occurring when \(p\equiv0\) and \(3\mid n\). Derivation on \(C,u'\) and translation by \(\mathcal D\) give exactly (64). There is no other kernel target. The Frobenius assertion follows from Lemma 37. ◻

Proof. First consider the possible intervening lengths in (65). At length twenty-one, the degree-one sources are transfers and their \(H\) differential is zero; the other arrows are excluded by (66). At twenty-five the possible volume-to-scalar arrows have source residues \[2,\quad\rho+4,\quad\rho+2\] 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 40 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 \[ \begin{array}{c|l} n\bmod3&\text{effect}\\\hline 0&\text{surjective, with the whole norm kernel as kernel;}\\ 1&\text{injective, with cokernel }k;\\ 2&\text{injective, with cokernel the target norm kernel }k. \end{array} \tag{67}\] 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 (61). 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 \[L:\mathcal T_{n+2}\longrightarrow\mathcal T_{n-6}.\] The degree-two group at index \(n\) contracts to the degree-one quotient identified by \(h\wedge\), by \[L:\mathcal T_n\longrightarrow\mathcal T_{n-8}.\] These formulas are transfers of the nonzero \(H\) length-thirty-three maps. Applying (67) 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 \[\delta N_0\delta^{-1}=-N_0+N_0^2,\] \(\delta\) 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 (56), giving (70). ◻

Proof. Apply (72)–(74) 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 \[ \begin{array}{c|l} \rho&\text{incoming lengths on }\mathsf A\text{ through }469\\\hline 2&25,241,457\\ 5&169,385\\ 8&97,313. \end{array} \tag{76}\] 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 34. 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 (76) 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 16. 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 42 makes this a linear equation over \(\mathbf F_3\). 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 \[Y=u_{\lambda|H}^{-9}.\] On Tate pages it is \(C\) times an inverse faithful Euler power. By Proposition 45 its first nonzero differential has length fifty-three. Lemma 46 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 \[9\operatorname{Ind}_H^G(\lambda|H) -18(1+\lambda'),\] whose orientation symbol is \[u_\lambda^{-27}u'^{18}\] 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 37.

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 \(E_{53}\): \[ d_{53}\operatorname{Nm}_H^G(Y) =\operatorname{tr}_H^G\bigl(d_{53}Y\cdot gY\cdot g^2Y\bigr). \tag{78}\] This local obstruction calculation is the filtration-zero instance of the general norm-transfer differential construction in [10]. 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 (78). 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 \[\operatorname{tr}\bigl(C^2d_{53}^H(C)\bigr).\] 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 (68) 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 (77) 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 \[0=d_{25}(zz')=d_{25}(z)z'-z\,d_{25}(z').\] Both arrows have shift \((-1,4)\) by (75). 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 (82) 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, \[d_{97}(\mathsf v\mathsf d) =\mathsf v\,d_{97}(\mathsf d)=0,\] 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 \[d_{93}(z\mathsf a_0)=p^\dagger\mathsf a_0\] 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 (86). ◻

Proof. We first make the units giving the translations explicit. On \(E_{53}\) put \[\Omega=(C^3)^{-1}d_{53}(C^3).\] Equation (77) 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 \[ U_0=C^9,\qquad U_1=C^{-3\kappa}u' \tag{89}\] 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 (76) has no remaining incoming before \(313\) and no other one through the bound \(469\). The bound in Proposition 16 forces a nonzero \(\mathsf f\to\mathsf A\) at \(313\) hitting the unit. Its shift, computed by (75), is \((5,36)\), giving (88).

No other arrow in (87) 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 (87) 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 \[\begin{array}{c|c} 85&\mathsf c_0\\ 101&\mathsf X_0\text{ (absent)}\\ 157&\mathsf a_2\text{ (absent)}\\ 301&\mathsf c_1. \end{array}\] 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 3 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 50, 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 (88). It is an \(\mathsf f_1\), whose offset is \((16,86)\), so its row is \[s_f=(-21,-122)+\mathcal L.\] Equality with (90) says \((-21,-124)\in\mathcal L\), or \[-21-372\kappa\equiv0\pmod9.\] Since \(\kappa\in\mathbf F_3\), this is equivalent to \(\kappa=2=-1\). The conclusion follows. ◻

Proof of Theorem 36. By Proposition 52, 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 3 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 50. 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 (85) has \[t\equiv0\pmod3,\qquad t-3u=-40\cdot3^{\jmath-2}\equiv0\pmod9.\] It is exactly a \(\mathsf p_0\) copy of this key. The unit translations of Proposition 50 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]; see also the proof of [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 \[a=\Sigma^{4m-1}Dg:S^{4m-1}\longrightarrow M,\qquad C_a\simeq\Sigma^{4m}DC_g.\] 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]. 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 (93); 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 \(\mathrm{BP}\to H\mathbf F_3\) 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 \(\pi_*\mathrm{BP}M\), while filtration at least two is excluded by its nonzero primary test and Lemma 55. Let \[z\in E_2^{1,108}(M)\] 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 \[\begin{align*} y&=\frac{\eta_R(v_2^9-v_1^8v_2^7)-(v_2^9-v_1^8v_2^7)}{v_1^9},\\ t_7'&=\frac{\eta_R(v_2^7)-v_2^7}{v_1} \qquad\text{in the mod-three cobar complex}. \end{align*}\] By the unconditional basis calculation (109), write \[z=\lambda y+\nu t'_7+\mu v^{26}a, \qquad \lambda,\nu,\mu\in\mathbf F_3.\] 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 65 then gives \[ z=y+\mu v^{26}a,\quad\mu\in\mathbf F_3,\qquad B(y)^3=0. \tag{96}\]

The boundary along \(q\) increases Adams–Novikov filtration by one and induces the integral 3-Bockstein; this is the geometric boundary theorem [32]. 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 \[Q=(Y/3)\mathbin{\times}_{Z/3}Z.\] The integral lift exists because the homotopy of \(Z\) is torsion-free. Comparing fibres over \(Z\) yields the triangle \[X\xrightarrow{\,3\iota\,}Y \longrightarrow Q\longrightarrow\Sigma X,\] 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, \[B(v)=\pm a,\qquad B(a)=0,\qquad a^2=0,\] 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 (96). Thus \(ix^3\) has filtration greater than six. Its stem is \(318\equiv2\bmod4\), so (94) 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 \(\mathbf Z/3\) 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 \(\mathbf Z/3\) 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 \[D^\circ(S^n)\simeq \Sigma^{3(n-1)}D^\circ(S^1) \simeq\Sigma^{3n+1}M=\Sigma^NM.\]

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 (99), 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 \[\begin{tikzcd}[column sep=large] K \arrow[r] \arrow[d,"r"'] & U \arrow[r] \arrow[d] & D^\circ S^1 \arrow[d,equal] \arrow[r] & \Sigma K \arrow[d,"\Sigma r"]\\ M \arrow[r,"j_T"'] & T \arrow[r,"\pi_T"'] & D^\circ S^1 \arrow[r] & \Sigma M . \end{tikzcd}\] Use (98) 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 (99). 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 \(\mathbf F_3\). We adapt Toda’s diagonal cone construction [37] to the truncated functor \(D^\circ\). Applying the diagonal cone nullhomotopy gives a natural map \[C_{D^\circ a_0}\longrightarrow D^\circ(C_{a_0}).\] 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 \[ \Sigma D^\circ S^n\longrightarrow D^\circ S^t \tag{102}\] 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 (100); 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 \[H_0\bigl(\mathcal B;\mathbf F_3\{u^{\otimes3}\}\bigr);\] 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 (102), 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] on the fibre gives \[\mathcal P_*^{81}(u^{\otimes3}) =(\mathcal P_*^{27}u)^{\otimes3}\ne0.\] 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 \(\pi_0M=\mathbf F_3\). 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 \[M\xrightarrow{3}M\longrightarrow M\wedge M\] 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 (93) 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 56, and let \(c\) be its image in \(\pi_{318}F_2M\). Smash the section in Lemma 60 with \(I^{\wedge2}\) and put \[w=s_*(c)\in\pi_{318}F_2\mathcal E.\] After forgetting filtration, \(w\) is exactly \(x^3 1_M\), because \(s(i)=1_M\).

Postcomposition with \(g:\Sigma^3M\to M\) defines \[\gamma:\Sigma^3\mathcal E\longrightarrow\mathcal E.\] The obstruction to lifting \(w\) through the triangle (101) smashed with \(I^{\wedge2}\) is \[\gamma_*(w)\in\pi_{321}F_2\mathcal E.\] We show separately that both components of this obstruction under (103) 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 \[\pi_{321}F_2M\longrightarrow\pi_{321}M\] 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 \[\Sigma^3M\xrightarrow{\Sigma^3s}\Sigma^3\mathcal E \xrightarrow{\gamma}\mathcal E \xrightarrow{\tau}\Sigma^{-1}M\] 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 \(\mathrm{BP}_*\)-homology is multiplication by a scalar multiple of \(v_1\), since a \(\mathrm{BP}_*\)-linear degree-four map \(\mathrm{BP}_*M\to\mathrm{BP}_*M\) is determined by its value in \(\mathrm{BP}_4M=\mathbf F_3\{v_1\}\). 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 \[ v_1:E_6^{10,328}(M)\longrightarrow E_6^{10,332}(M), \tag{104}\] which is zero by Lemma 65. Lemma 61 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 \[h_*(c)\in\mathcal F^{11}\pi_{322}F_2M =\mathcal F^{14}\pi_{322}F_2M.\] The equality follows from sparseness: in stem 322 only filtrations congruent to two modulo four can occur.

For every such filtration \(f\geq14\), Lemma 65, including its high-filtration estimate, gives \[E_{10}^{f,322+f}(M)=0.\] The comparison of Lemma 61 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 (104) removes that filtration on \(E_6\) before the higher \(E_{10}\)-vanishing is applied.

We have now proved \(\gamma_*(w)=0\). The triangle (101) smashed with \(I^{\wedge2}\) supplies the desired lift. More explicitly, \(M\) is dualizable, so \[F(M,I^{\wedge2}\wedge Z) \simeq I^{\wedge2}\wedge F(M,Z).\] 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 53. Let \(\overline F'\) be the composite of the map in Proposition 62 with \(F_2T\to T\). Its quotient equals that of \(F\), so the difference has a lift \[Z:\Sigma^NM\longrightarrow M,\qquad j_TZ=F-\overline F'.\] 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 55 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 59 thus gives a nonzero \[\mathcal P^{81}:H^0(C_Z;\mathbf F_3)\longrightarrow H^{324}(C_Z;\mathbf F_3).\]

Use the splitting of Lemma 60 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: \[\Sigma^NM\xrightarrow{\Sigma^Nq}S^{N+1} \xrightarrow{a_1}M.\] 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 54 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 \[qa_1=\Sigma\theta,\qquad 3\theta=0.\] 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 \[u\colon\mathcal P_m\longrightarrow\Delta_{\leq m},\qquad S\longmapsto[|S|-1]\] 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 \(\mathop{\mathrm{Tot}}_m X\) 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 \(\mathrm{BP}\) furnished by [3]. 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 60.

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 \(B\otimes_{\mathrm{BP}_*}B\longrightarrow B\) 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 8. 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 \[ \mathbf F_3[v_1,v_2,\ldots]\otimes \bigotimes_{i\geq1,\,j\geq0} \bigl(\Lambda(e_{i,j})\otimes\mathbf F_3[c_{i,j}]\bigr), \tag{121}\] where, for \(w_{i,j}=3^j(3^i-1)/2\), \[|e_{i,j}|=(1,w_{i,j}),\qquad |c_{i,j}|=(2,3w_{i,j}).\] One obtains this formula by writing each additive coordinate in base-three digits. Its graded dual is a tensor product of truncated polynomial algebras \(\mathbf F_3[x]/(x^3)\), 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 (121); 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 \[t_1^2\mid t_1+t_1\mid t_1^2,\] the standard polynomial generator for the weight-one digit. All the other terms have positive coefficient count.

For a monomial on (121) 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 \[\frac{11}{2}+6(k-1)=6k-\frac12.\] It follows that the quotient of (121) 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 \[b^k\colon C^{*-2k,d-3k}\longrightarrow C^{*,d}.\] 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 \[H^{f-1,d}(Q)\longrightarrow D_{f-2k,d-3k} \xrightarrow{\ b^k\ }D_{f,d}\longrightarrow H^{f,d}(Q).\] 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 (120). ◻

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 \(\mathbf F_3\). 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 (132). 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 70 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 (136) 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 (136) 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 (127), an output operator \(P_U\) in \(P_Sv^QP_T\) satisfies \[ |U|\geq |T|. \tag{137}\] 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 (127). 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 \[P_S[P_T]-[P_SP_T]+[P_S]P_T(1),\] 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 (135). 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 (135). 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 (148). 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 \[D(x\star y)=(-1)^gD(x)\star y+x\star D(y).\] 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 \(\mathbf F_3\)-linear combination of the entries of (156). 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 \(\mathbf F_3\)-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 \(\sum_{i>0}\mathop{\mathrm{im}}A_i\), 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 (130) and (138), construct (136), and form (139). For \((f,d)=(1,27),(6,10),(6,81)\) the dimensions of the three successive cochain terms are respectively \[(12,49,94),\qquad(3,2,1),\qquad(9301,11419,12864).\] Reducing the incoming and outgoing columns gives ranks \[(11,35),\qquad(1,0),\qquad(5256,6156).\] 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 (144), apply (142), and form its cube using (147)–(149). 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 (139) 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 (155), 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 (154). 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: \[\begin{array}{r|r|r|r} \text{parameters}&\text{scalar square equations}& \text{polynomial row rank}&\text{new affine restrictions}\\\hline 265&16412&72&38\\ 227&16412&20&1\\ 226&16412&19&0 \end{array}\] In each row quadratic monomials precede linear ones, as specified after (156); after the final refinement the affine rank is \(23158\). Apply (157) and (158) to its projected matrices. They give the sixteen rows of Table [finite:enclosures]. The two additional stacked ranks are \(1\) and \(2\). Equations (159)–(161), using (163) in the second case, give \((4,5,5)\) and \((6,5,5)\). Equation (162) gives the pairs \((3,3),(2,2),(4,4)\).

Finally compute the five iterated \(b\) maps of Table [finite:product-ranks] 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 \[(4491,4499),\quad(7540,7546),\quad(7589,7595),\quad (1868,1872),\quad(1000,1004).\] 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. ◻