==== TABLE OF CONTENTS ==== A-finite-Smith-Toda-complex-V4-at-the-prime-1009//00-introduction.tex A-finite-Smith-Toda-complex-V4-at-the-prime-1009//01-framework.tex A-finite-Smith-Toda-complex-V4-at-the-prime-1009//02-lower-complexes.tex A-finite-Smith-Toda-complex-V4-at-the-prime-1009//02a-cobar-preliminaries.tex A-finite-Smith-Toda-complex-V4-at-the-prime-1009//03-may-and-first-obstruction.tex A-finite-Smith-Toda-complex-V4-at-the-prime-1009//04-frobenius-elimination.tex A-finite-Smith-Toda-complex-V4-at-the-prime-1009//05-final-obstruction.tex A-finite-Smith-Toda-complex-V4-at-the-prime-1009//appendix-resolution.tex A-finite-Smith-Toda-complex-V4-at-the-prime-1009//main.tex A-finite-Smith-Toda-complex-V4-at-the-prime-1009//preamble.tex A-finite-Smith-Toda-complex-V4-at-the-prime-1009//references.bib =========================== ==== BEGIN A-finite-Smith-Toda-complex-V4-at-the-prime-1009//00-introduction.tex ==== \section{Introduction}\label{intro:section} The Smith--Toda realization problem asks how far one can kill the chromatic parameters in Brown--Peterson homology, one at a time, while retaining a finite spectrum. At a prime $p$, write \[ BP_* = \mathbb Z_{(p)}[v_1,v_2,\ldots], \qquad |v_i|=2(p^i-1), \] with Hazewinkel generators. A \emph{finite $p$-local spectrum} is a retract of the $p$-localization of a finite CW spectrum. A \emph{Smith--Toda complex $V(n)$} is such a spectrum with Brown--Peterson homology $BP_*/(p,v_1,\ldots,v_n)$, including the canonical $BP_*BP$-comodule structure and a generator in degree zero. In this quotient each listed generator is killed to its first power. The comodule records the Brown--Peterson cooperations, so its specification is part of the realization problem, in addition to the underlying graded module. The periodicity theorem gives generalized Moore spectra whose homology kills suitable powers of the initial generators \citep[Proposition~5.14]{HopkinsSmith1998}. Requiring every exponent to be one is a more restrictive realization problem. The mod-$p$ Moore spectrum gives $V(0)$ at every prime. The classical constructions of $V(1)$, $V(2)$, and $V(3)$ for $p>2$, $p>4$, and $p>6$ are due to Adams, Smith, and Toda \citep{Adams1966,Smith1970,Toda1971}; see also the historical account in \citet[p.~493]{Nave2010}. Successive cofibers turn a suitable first-power $v_i$ self-map into the next quotient. The difficulty is producing the self-map on the finite spectrum: an algebraically primitive coefficient need not survive the Adams--Novikov spectral sequence (ANSS) to an actual map. These constructions thus connect the algebra of complex cobordism with specific attaching maps in the stable homotopy category. The next case remained outside those constructions. The September 2023 version of \citet[Remark~3.29]{CulverZhang2024} records the existence of $V(4)$ as open at every prime. There are also height-dependent nonexistence results: \citet[Theorem~1.3]{Nave2010} proves that $V((p+1)/2)$ does not exist for $p\geq7$, and that $V((p-1)/2)$, if it exists, cannot be a ring spectrum. Those obstructions leave room for $V(4)$ at a sufficiently large prime. We establish the following specified-prime realization. \begin{theorem}\label{intro:main} At $p=1009$ there is a finite $p$-local spectrum $X$ and an isomorphism \[ BP_*X\ \cong\ BP_*/(p,v_1,v_2,v_3,v_4) \] of graded $BP_*BP$-comodules, with the canonical quotient comodule on the right and its generator in degree zero. \end{theorem} Theorem~\ref{intro:main} gives a positive answer to the Smith--Toda $V(4)$ existence problem at the stated prime. Its conclusion concerns the ordinary $p$-local stable category; finiteness is established before any chromatic localization. For comparison, \citet[Theorem~1.7]{KatoShimomuraShimomura2025} construct at this prime an $L_5$-local spectrum $W_5$ with \[ BP_*W_5=v_5^{-1}BP_*/(p,v_1,v_2,v_3,v_4), \] where $L_5$ denotes localization with respect to $v_5^{-1}BP$. That realization has $v_5$ inverted and does not supply the ordinary finite realization of the unlocalized quotient required here. The choice $1009$ makes the finite inequalities in our calculation generous. We do not require a product on the final spectrum $X$. \subsection{Methods and proof strategy} The algebraic setting comes from the spectrum of Brown and Peterson \citep{BrownPeterson1966}, Quillen's formal-group description of cobordism and its $p$-typical summand \citep{Quillen1969}, and Hazewinkel's explicit $p$-typical generators \citep{HazewinkelIII}. Our main tool is the sphere-source Adams--Novikov spectral sequence, whose cobordism-theoretic construction goes back to Novikov \citep{Novikov1967}. We use its ordinary $BP$ resolution, its cobar $E_2$ term, and its compatibility with actual binary maps; see \citet[Sections~2.2--2.3]{Ravenel2004}. The algebraic estimates use filtered Hopf-algebra cohomology as developed by \citet[Section~4, Theorem~4]{May1966}, with the odd-primary filtration of \citet[Theorem~3.2.5]{Ravenel2004} and an additional filtration adjustment constructed below. A closer predecessor is the work of \citet[author preprint, pp.~2--6]{LeeRavenel1994} on the nilpotence order of the first beta-family sphere class $\beta_1$. They use a modified May calculation on $V(3)$ and constrain Adams--Novikov differentials through multiplication by $\beta_1$ and the $V(2)$-module structure of $V(3)$. Our final obstruction calculation requires an annihilation relation for products of a beta power with specific filtration-five classes, together with a bound on the page at which those products vanish. Toda's work on sphere relations \citep[p.~839, Corollary]{Toda1967} and his subsequent cyclic extended-power constructions \citep[Sections~1--3, especially Theorem~3]{Toda1968} provide the topological antecedents. We give the particular coefficient-cohomology calculations and finite-cell arguments needed for this bound below. The construction has three stages. First we construct $Y=V(3)$ with a binary product and a two-sided unit. These unit identities are enough to turn a sphere map realizing $v_4$ into a self-map of $Y$. We obtain the lower complexes and their products by finite cofiber constructions and explicit vanishing of their attaching obstructions; we do not need an associative product on $Y$. Second, we prove that the filtration-zero class $v_4$ survives in the Adams--Novikov spectral sequence for $Y$. To describe the obstruction calculation, put $q=2p-2$ and $u_i=1+p+\cdots+p^{i-1}$ for $i\geq1$. Divide the internal degree by $q$ and call the resulting integer the \emph{weight}. A cobar class of cohomological degree $s$ and weight $w$ lies in stem $qw-s$. Every possible positive differential on $v_4$ therefore has target \begin{equation}\label{intro:targets} (s,w)=(qk+1,u_4+k),\qquad k\geq1. \end{equation} A finite May calculation leaves only $k=1,2,p-1,p$. For $k=1$, a relation supplied by the coboundary of $v_5$, together with an injective multiplication on the target, forces the differential to vanish. For $k=2,p-1$, we compute actual cohomology by taking the Frobenius kernel of the group of truncated additive coordinate changes. The quotient acts triangularly on kernel cohomology. Filtering this coefficient module by weight identifies the first possible differential and checks that both its source and its target are present. This step controls the polynomial generators as well as the exterior generators on the May page. The remaining case $k=p$ requires a topological argument. We write $\beta\in\pi_{pq-2}S^0_{(p)}$ for the first beta-family sphere class. The entire $E_2$ target consists of products of a large power of its detector with certain filtration-five classes of $Y$. We realize those low-filtration classes by maps $\gamma$ and construct an explicit relation \[ \beta^{\,p(p-2)+1}\gamma=0. \] The construction uses a cyclic extended power of a sphere and the averaging retract of a smash power of a two-cell spectrum. The corresponding Adams--Novikov product has filtration strictly smaller than the length of the prospective differential on $v_4$. Consequently the whole target has vanished before that differential could occur. The bound on the page is the point: eventual vanishing in homotopy, without such a bound, would not settle this obstruction. Finally the filtration-zero edge gives a map $S^{2(p^4-1)}\to Y$ with Brown--Peterson Hurewicz image $v_4$. Multiplication by the binary product on $Y$ extends it to $\Sigma^{2(p^4-1)}Y\to Y$. Its cofiber is finite, and the nonzerodivisor property of $v_4$ identifies its homology with the required quotient. The bottom sphere map identifies the quotient coaction and fixes the generator in degree zero. Section~\ref{fw:section} fixes the Brown--Peterson coordinates and the topological spectral-sequence conventions. Section~\ref{may:preliminaries} constructs the filtered cobar calculation, and Section~\ref{low:section} supplies the unital lower complexes. Section~\ref{may:section} lists the height-four targets and removes the first differential; Section~\ref{frob:section} proves the two Frobenius vanishings. Section~\ref{top:section} establishes the homotopy relation, its page bound, and the realizing cofiber. Appendix~\ref{app:resolution} provides the ordinary convergent resolution and the filtered binary pairing that connects the algebraic calculation to the actual homotopy product. ==== END A-finite-Smith-Toda-complex-V4-at-the-prime-1009//00-introduction.tex ==== ==== BEGIN A-finite-Smith-Toda-complex-V4-at-the-prime-1009//01-framework.tex ==== \section{Brown--Peterson coordinates and the Adams--Novikov resolution} \label{fw:section} Throughout the proof, \begin{equation} p=1009,\qquad q=2p-2=2016,\qquad u_i=\frac{p^i-1}{p-1}\quad(i\geq 0). \label{fw:parameters} \end{equation} The integer $1009$ is prime: its square root is less than $32$, and none of $2,3,5,7,11,13,17,19,23,29,31$ divides it. Put \[ R=BP_* =\mathbb Z_{(p)}[v_1,v_2,\ldots],\qquad \Gamma=BP_*BP=R[t_1,t_2,\ldots],\qquad I_m=(p,v_1,\ldots,v_{m-1}). \] Both $v_i$ and $t_i$ have internal degree $qu_i$. We call internal degree divided by $q$ the \emph{weight}. A cobar class of cohomological degree $s$ and weight $w$ has stem $qw-s$. We use the ordinary Brown--Peterson spectrum, its polynomial cooperation algebra, and a strictly unital associative model of its multiplication. For the identification of these polynomial rings with the homotopy and cooperations of $BP$, see \citet[Theorems~4.1.18--4.1.19]{Ravenel2004}. The Hazewinkel recursion and the integrality of its polynomial generators are given in \citet[Equation~A2.2.1 and Theorem~A2.2.3]{Ravenel2004}; the formal-sum coordinate convention is explained in the proof of Theorem~A2.1.27 there. The formal-group foundations are due to \citet[Section~5, Theorem~4]{Quillen1969}; the explicit generators and cooperations are developed in \citet[Sections~3.1 and~4.4]{HazewinkelIII}. A coherent associative model exists, in particular, by forgetting structure from \citet[Theorem~1.1]{BasterraMandell2013}. No commutative or $E_\infty$ model of $BP$ is required below. We give the particular coordinate, convergence, and pairing arguments needed for the construction. \subsection{Coordinates and additive ranges} Let $a_i$ be the coefficients in the formal-group sum expression \[ [p]_F(x)=\mathop{\sum}\nolimits_F a_i x^{p^i},\qquad a_0=p. \] The strict coordinate change is written in the same convention as $\sum_F t_i x^{p^i}$, with $t_0=1$. Reversing the convention for composition reverses the cobar conventions without changing any of the arguments. \begin{lemma}[Leading coefficients and low cooperations] \label{fw:coordinates} For each $m\geq 1$, $I_m$ is an invariant ideal and $v_m$ is primitive modulo $I_m$. Modulo $I_m$, the coproducts of $t_i$, $i\leq m$, are the additive-composition coproducts. At the specialization that kills $p$ and all $v_i$, the resulting Hopf algebra is \begin{equation} P=\mathbb F_p[t_1,t_2,\ldots],\qquad \Delta(t_i)=\sum_{\ell=0}^i t_\ell\otimes t_{i-\ell}^{p^\ell}. \label{fw:additive-hopf} \end{equation} \end{lemma} \begin{proof} Write the logarithm as $\log_F(x)=\sum_{i\geq0}l_i x^{p^i}$, where $l_0=1$. The Hazewinkel recursion is \[ pl_n=v_n+\sum_{i=1}^{n-1}l_i v_{n-i}^{p^i}. \] Applying the logarithm to the formal sum for $[p]_F$ and comparing the coefficient of $x^{p^n}$ shows that the coefficient of the indecomposable $v_n$ in $a_n$ is \begin{equation} c_n=1-p^{p^n-1}. \label{fw:coordinate-unit} \end{equation} Indeed the two terms involving $l_n$ are $pl_n$ and $p^{p^n}l_n$, and the coefficient of $v_n$ in $l_n$ is $1/p$. All other terms involve lower $v_i$. The coefficients $a_n$ are integral, so the grading and polynomial independence of the $v_i$ give $a_n-c_nv_n\in(v_1,\ldots,v_{n-1})R$. Each $c_n$ is a $p$-local unit. The invariance and primitivity follow inductively. The element $p$ is invariant. Once the lower coefficients vanish, the first surviving term of the $p$-series is $c_m v_m x^{p^m}$. A strict coordinate change has leading term $x$, so comparison of this first surviving term on the two sides of the coordinate-change identity fixes $c_m v_m$, and hence $v_m$. This proves both invariance of the next ideal and the asserted primitivity. After all base generators have been killed, the formal group is additive, and composition of additive $p$-polynomials gives Equation~\eqref{fw:additive-hopf}. Modulo $I_m$, a correction to this coproduct must have a positive base coefficient, of weight at least $u_m$. For $ik$ reduced factors. It therefore cannot occur. A bar containing $t_4$ would leave at most $k$ weight for the other $d-1\geq qk-1>k$ bars, and likewise cannot occur. Higher generators exceed the weight bound. The remaining coproducts are additive by Lemma~\ref{fw:coordinates}; equivalently, apply Lemma~\ref{fw:range-comparison}. This identifies both adjacent differentials and proves the cohomology assertion. \end{proof} ==== END A-finite-Smith-Toda-complex-V4-at-the-prime-1009//01-framework.tex ==== ==== BEGIN A-finite-Smith-Toda-complex-V4-at-the-prime-1009//02-lower-complexes.tex ==== \section{The unital complexes through height three} \label{low:section} We first construct the lower complexes, retaining a binary two-sided unit. The algebraic input is the May upper bound of Lemma~\ref{may:page}: a bidegree containing no initial-page monomial has zero cobar cohomology. The two vanishing families below serve different parts of the induction: the first makes each $v_n$ survive to a self-map, and the second permits the two unit maps to extend to a binary product over every pair of cells. \begin{lemma}[Lower obstruction groups] \label{low:algebraic-vanishing} The following cobar cohomology groups vanish. \begin{enumerate} \item For $1\leq n\leq3$ and $k\geq1$, the group with coefficients $R/I_n$ in bidegree \[ (s,w)=(qk+1,u_n+k). \] \item For $0\leq n\leq3$, choose $e_i\in\{0,1,2\}$ for $0\leq i\leq n$ and put \[ U=\sum_{i=0}^n e_i u_i,\qquad \rho=\sum_{i=0}^n e_i\geq2. \] For every $k\geq1$, the group with coefficients $R/I_{n+1}$ in bidegree \[ (s,w)=(qk+1-\rho,U+k) \] is zero. \end{enumerate} \end{lemma} \begin{proof} Include the first case in the notation of the second by setting $U=u_n$ and $\rho=0$, while retaining its indicated coefficient comodule. If the normalized cochain group in degree $(s,w)$ is nonzero, Lemma~\ref{fw:normalized-bound} gives \begin{equation} 2015k\leq U+\rho-1. \label{low:normalized-cutoff} \end{equation} The exact values \[ (u_0,u_1,u_2,u_3,u_4)=(0,1,1010,1{,}019{,}091,1{,}028{,}262{,}820) \] give Table~\ref{low:cutoffs}. Each upper bound is obtained by substituting the displayed maxima in Equation~\eqref{low:normalized-cutoff} and taking the integer part. A zero upper bound on $k$ excludes all $k\geq1$. \begin{table}[htbp] \centering \begin{tabular}{lrrrr} \toprule Case & Maximum $U$ & Maximum $\rho$ & Maximum $k$ & Maximum $w$\\ \midrule Product, $n=0$ & $0$ & $2$ & $0$ & ---\\ Product, $n=1$ & $2$ & $4$ & $0$ & ---\\ Product, $n=2$ & $2022$ & $6$ & $1$ & $2023$\\ Product, $n=3$ & $2{,}040{,}204$ & $8$ & $1012$ & $2{,}041{,}216$\\ $v_1$ target & $1$ & $0$ & $0$ & ---\\ $v_2$ target & $1010$ & $0$ & $0$ & ---\\ $v_3$ target & $1{,}019{,}091$ & $0$ & $505$ & $1{,}019{,}596$\\ \bottomrule \end{tabular} \caption{Preliminary bounds for all lower obstruction bidegrees.} \label{low:cutoffs} \end{table} These bounds also justify the additive comparison in the adjacent cochain degrees. For either remaining product case, $wk$ reduced bars when $d\in\{s-1,s,s+1\}$. Such a coefficient cannot occur. A bar containing $t_n$ or a higher coordinate leaves at most $k$ for the other $d-1\geq qk-1>k$ bars and cannot occur either. Hence the same three cochain degrees are additive. In all cases, both adjacent differentials are identified, so the desired cohomology is identified with that for $P$. Every remaining weight in Table~\ref{low:cutoffs} is at most $2{,}041{,}216$, whereas \[ p^3=1{,}027{,}243{,}729. \] A May letter covering position $3$ or above has weight at least $p^3$. Thus only positions $0,1,2$ can occur. We use the common exterior and polynomial alphabet of Equation~\eqref{may:low-alphabet}, with the interval coverage conventions of Section~\ref{may:preliminaries}. Allowing this entire alphabet only enlarges the upper bound if fewer coordinates were retained in the preceding comparison. Suppose a monomial of the desired bidegree exists. Let $m\leq6$ be its number of exterior letters and $L=(s-m)/2$ its total polynomial exponent. Every polynomial letter has weight at least $p$, so \[ U+k\geq pL=p(p-1)k+\frac{p(1-\rho-m)}2. \] For all profiles under consideration, $U\leq2p^2+4p+6$ and $\rho\leq8$. Consequently \begin{equation} (p^2-p-1)k \leq U+\frac{p(\rho+5)}2 \leq 2p^2+\frac{21}2p+6. \label{low:polynomial-cutoff} \end{equation} If $k\geq3$, the left side exceeds the last expression by at least \[ p^2-\frac{27}2p-9=\frac{2{,}008{,}901}{2}>0. \] It follows that $1\leq k\leq2$. Write $U=F_0+pF_1+p^2F_2$. In a product profile the digits are \[ F_0=e_1+e_2+e_3,\qquad F_1=e_2+e_3,\qquad F_2=e_3, \] with missing $e_i$ set to zero. For a $v_n$ target they are initial strings of ones. In every case, \begin{equation} 6\geq F_0\geq F_1\geq F_2\geq0, \qquad F_1\leq4,\quad F_2\leq2. \label{low:digits} \end{equation} Let $H_r$ count the exterior letters covering position $r$, and let $n_B,n_C$ be the polynomial exponents of $B,C$. Adding weights gives the exact identity \begin{equation} F_0+k+pF_1+p^2F_2 =H_0+p(H_1+L-n_C)+p^2(H_2+n_B+n_C). \label{low:weight-equation} \end{equation} We next extract its digits without an unproved no-carry assumption. First, $0\leq H_0\leq3$ and $1\leq F_0+k\leq80$. Hence $n_B+n_C\leq2$. Put $\delta=(m+\rho-1)/2$. It is an integer because $s-m$ is even. For a target with $\rho=0$, that parity forces $m$ odd, hence $m\geq1$; for a product profile $\rho\geq2$. Thus $\delta\geq0$. The bounds $m\leq6$, $\rho\leq8$ and the same parity give $\delta\leq6$. Now $L=(p-1)k-\delta$. Subtracting the position-zero equation from Equation~\eqref{low:weight-equation}, dividing by $p$, and rearranging gives \[ H_1-k-\delta-n_C-F_1 =p(F_2-k-H_2-n_B-n_C). \] Since $H_1\leq4$, the left side lies in $[-14,3]$. Its only possible multiple of $p=1009$ is zero. We have therefore proved the three exact relations \begin{equation} H_0=F_0+k,\qquad H_2+n_B+n_C=F_2-k,\qquad H_1=F_1+k+\delta+n_C. \label{low:coverage-relations} \end{equation} Among the exterior letters covering position zero, only $x,a$ fail to cover position two. Thus $H_2\geq H_0-2$. Equations~\eqref{low:digits} and~\eqref{low:coverage-relations} imply \[ F_0-F_2+2k\leq2. \] As $k\geq1$, equality is forced: $k=1$ and $F_0=F_1=F_2=f$. All the intervening inequalities are equalities, giving \[ n_B=n_C=0,\qquad H_0=f+1,\qquad H_2=f-1. \] The bounds on $H_0,H_2$ leave only $f=1,2$. If $f=1$, the exterior letters $x,a$ must occur and $c,b,z$ must be absent. Only $y$ is optional, so $H_1\leq2$ and $m\geq2$. On the other hand, the last relation in Equation~\eqref{low:coverage-relations} gives $H_1=2+\delta>2$, since $\delta=(m+\rho-1)/2>0$. If $f=2$, the letters $x,a,c$ must occur and $b,z$ must be absent. Again only $y$ is optional, so $H_1\leq3$ and $m\geq3$, whereas the same relation gives $H_1=3+\delta>3$. Both possibilities are impossible. There is no monomial on the May upper-bound page, and hence no cobar cohomology in any of the stated bidegrees. \end{proof} \begin{lemma}[Finite local cell attachments] \label{low:finite-localization} A spectrum obtained from finitely many $p$-local sphere cells is the $p$-localization of a finite CW spectrum. \end{lemma} \begin{proof} Proceed one attachment at a time. If the previously constructed spectrum is the localization of a finite spectrum $Z$, an attaching map belongs to $\pi_d(Z_{(p)})=\pi_d(Z)\otimes\mathbb Z_{(p)}$ and can be written $a/m$ with $p\nmid m$. After localization, multiplication by $m$ on its sphere source is an equivalence. The localized cone of the integral attaching map $a$ is therefore equivalent to the cone of $a/m$. Induction over the finitely many cells proves the assertion. \end{proof} \begin{proposition}[Construction through $V(3)$] \label{low:V3} For $0\leq n\leq3$ there is a finite $p$-local cell spectrum $V(n)$ with a bottom-cell unit $\eta_n:S_{(p)}\to V(n)$ and a binary multiplication \[ \mu_n:V(n)\wedge V(n)\longrightarrow V(n) \] whose two unit identities hold. Its cells are indexed by subsets of $\{0,\ldots,n\}$, in dimensions \begin{equation} \sum_{i\in S}(qu_i+1), \label{low:cell-dimensions} \end{equation} and its $BP$-homology is the canonical quotient comodule $R/I_{n+1}$, generated in degree zero by the bottom unit. In particular, we may fix \begin{equation} Y=V(3),\qquad BP_*Y=R/I_4, \label{low:Y} \end{equation} with such a binary two-sided unital multiplication. \end{proposition} \begin{proof} Start with $V(0)$, the cofiber of $p:S_{(p)}\to S_{(p)}$. Since multiplication by $p$ is injective on $R$, its $BP$-homology is $R/(p)$, with the canonical quotient coaction and bottom generator. Its cells are in dimensions zero and one, as in Equation~\eqref{low:cell-dimensions}. We give first the product-extension argument, which applies to this initial complex and to every later complex once its asserted homology and cells have been obtained. The subcomplex \[ U_n=(S_{(p)}\wedge V(n)) \mathop{\cup}_{S_{(p)}\wedge S_{(p)}} (V(n)\wedge S_{(p)}) \ \subseteq\ V(n)\wedge V(n) \] has a map to $V(n)$ given by the two identity maps, which agree on the common bottom sphere. Every remaining cell corresponds to a pair of nonempty subsets of $\{0,\ldots,n\}$. If $e_i$ counts occurrences of $i$ in that pair, its dimension is $qU+\rho$, with precisely the parameters in the second part of Lemma~\ref{low:algebraic-vanishing}. The obstruction to extending across the cell lies in $\pi_{qU+\rho-1}V(n)$. An Adams--Novikov term in that stem has $qw-s=qU+\rho-1$, or \[ w=U+k,\qquad s=qk+1-\rho. \] Since $2\leq\rho\leq81$, so they do not change $\pi_0$. Thus $p\eta_3=0$ in $\pi_0Y$. Applying the left unit identity of its binary multiplication gives \[ p\operatorname{id}_Y =\mu_3\bigl((p\eta_3)\wedge\operatorname{id}_Y\bigr)=0. \] \end{proof} ==== END A-finite-Smith-Toda-complex-V4-at-the-prime-1009//02-lower-complexes.tex ==== ==== BEGIN A-finite-Smith-Toda-complex-V4-at-the-prime-1009//02a-cobar-preliminaries.tex ==== \section{Filtered cobar preliminaries} \label{may:preliminaries} We use the filtered Hopf-algebra method of \citet{May1966}, with the odd-primary May-type filtration in the form of \citet[Theorem~3.2.5]{Ravenel2004}, and allow an additional digit-zero weight needed for the injectivity argument in Section~\ref{may:section}. For earlier applications of Ravenel's filtration to $V(3)$ and powers of $\beta_1$, see \citet{LeeRavenel1994}. We construct this modified filtration below, including its initial page and convergence. Throughout this section, cohomological degree and weight are denoted by $(s,w)$. Unless an Adams--Novikov differential is named explicitly, the spectral sequences in this section are algebraic spectral sequences of cobar complexes. Put \[ P=\mathbb F_p[t_1,t_2,\ldots],\qquad |t_i|_w=u_i, \qquad \overline\Delta(t_i)=\sum_{01. \] A nonprimitive coproduct summand of $t_i^{p^j}$ replaces it by $t_l^{p^j}\otimes t_{i-l}^{p^{j+l}}$ and decreases filtration by exactly one. These observations show that this is a multiplicative coalgebra filtration, and that its associated graded Hopf algebra is the tensor product of primitive truncated polynomial Hopf algebras \[ \bigotimes_{i,j}\mathbb F_p[z_{i,j}]/(z_{i,j}^p). \] For one such factor the dual algebra is again a truncated polynomial algebra: factorial rescaling identifies its basis in exponents less than $p$ with the ordinary polynomial basis. The free resolution alternating multiplication by its generator and its $(p-1)$st power computes its cohomology as one exterior generator in degree one and one polynomial generator in degree two. Their weights are respectively the weight of $z_{i,j}$ and $p$ times that weight. Tensoring these resolutions gives \eqref{may:initial-page}, including the stated auxiliary filtrations. The exterior square is zero since its cohomological degree is odd and $p$ is odd. The filtration-one part of the reduced coproduct gives the splitting formula for $h$. For the polynomial generator associated to a digit $z$, a filtration-one contribution would replace one of its $p$ copies by a split pair and leave $p-1$ unsplit copies of $z$. In the associated-graded cohomology of the $z$ factor, a nonzero multiplicity has the form $pa+\epsilon$, with $a\geq0$ and $\epsilon\in\{0,1\}$. The multiplicity $p-1$ has neither form, so that component has zero cohomology. Carries cannot contribute to this first differential by the strict inequality above. This proves $d_1b=0$. In a fixed positive internal weight only finitely many digit factors occur, every monomial exponent is bounded, and a normalized bar has positive weight. Its length is consequently bounded as well. Thus all the filtrations under discussion are finite in each weight, and their spectral sequences converge to the indicated cobar cohomology. The construction restricts to $P_r$ because its coproduct uses only the first $r$ coordinates. \end{proof} For later calculations we fix the following notation, always distinguishing the bare exterior letter $b$ from the indexed polynomial letters: \begin{equation} \begin{gathered} x=h_{1,0},\quad a=h_{2,0},\quad c=h_{3,0},\qquad y=h_{1,1},\quad b=h_{2,1},\quad z=h_{1,2},\\ A=b_{1,0},\qquad B=b_{2,0},\qquad C=b_{1,1}. \end{gathered} \label{may:low-alphabet} \end{equation} An exterior interval $h_{i,j}$ covers positions $j,\ldots,j+i-1$ in its weight, and a polynomial letter $b_{i,j}$ covers positions $j+1,\ldots,j+i$. Let $m$ be the number of exterior letters, let $H_r$ count those covering position $r$, and let $L=(s-m)/2$ be the number of polynomial letters, counted with multiplicity. If only positions $0,1,2$ occur, the alphabet is precisely \eqref{may:low-alphabet}. Writing $n_A,n_B,n_C$ for the polynomial multiplicities gives \begin{equation} w=H_0+p(H_1+L-n_C)+p^2(H_2+n_B+n_C), \qquad n_A=L-n_B-n_C. \label{may:three-position-weight} \end{equation} ==== END A-finite-Smith-Toda-complex-V4-at-the-prime-1009//02a-cobar-preliminaries.tex ==== ==== BEGIN A-finite-Smith-Toda-complex-V4-at-the-prime-1009//03-may-and-first-obstruction.tex ==== \section{The height-four obstruction calculation} \label{may:section} The algebraic spectral sequences below use the filtered cobar complexes of Section~\ref{may:preliminaries}. \subsection{Possible differential targets} We now use the finite spectrum $Y$ furnished by Proposition~\ref{low:V3}. Its homology is $R/I_4$, its bottom cell is a unit, and it has a binary two-sided unital multiplication. The possible positive Adams--Novikov differentials from $v_4$ have targets \begin{equation} s=qk+1,\qquad w=u_4+k,\qquad k\geq1. \label{may:obstruction-degrees} \end{equation} \begin{lemma}[Explicit cutoffs] \label{may:target-bound} The groups in \eqref{may:obstruction-degrees} are identified with $H^{s,w}(P)$. Their May page has only positions $0,1,2,3$, and it is zero unless $k\leq p+2$. For $k\geq2$, only positions $0,1,2$ can occur. \end{lemma} \begin{proof} If $wk$. Higher coefficient generators are excluded by $w0 \quad (p=1009). \] It follows that $k\leq p+2$. For $k\geq2$ we have $L\geq2p-6$. If any letter reaches position three, that letter has weight at least $p^3$, while all other polynomial letters have total weight at least $(L-1)p$. This is a valid lower bound whether the distinguished letter is exterior or polynomial. Thus \[ w\geq p^3+2p^2-7p >p^3+p^2+2p+3=u_4+p+2\geq w, \] since the strict difference is $p^2-9p-3>0$ at $p=1009$. This contradiction excludes position three in that range. \end{proof} \subsection{The first obstruction} The first possible differential is $d_{q+1}$, and its target is the actual cobar cohomology in degree $(q+1,u_4+1)$. We first explain why an injective multiplication on this group will rule out the differential. Lemma~\ref{fw:v5-relation} gives \[ v_4h_{1,4}+e v_4^p h_{1,0}=0 \] in the $E_2$ algebra for $Y$. By $q$-sparsity no earlier positive differential occurs, so Proposition~\ref{fw:anss-pairing} makes $d_{q+1}$ a derivation on this algebra. Let $\sigma$ denote the algebraic specialization to $P$-cohomology. Differentiate the relation and then apply $\sigma$. All terms containing $v_4$ specialize to zero, and $d_{q+1}(v_4^p)=p v_4^{p-1}d_{q+1}(v_4)=0$. Hence \begin{equation} \sigma\bigl(d_{q+1}(v_4)\bigr)h_{1,4}=0. \label{may:specialized-obstruction} \end{equation} Specialization is applied after the topological differential, to an identity of cobar-cohomology classes; no spectrum realizing $\sigma$ is required. It remains to prove that multiplication by $h_{1,4}$ is injective on $H^{q+1,u_4+1}(P)$. Its degree shift is $(1,p^4)$, so its product lies in degree $(q+2,u_5+1)$. An incoming May differential to that product must have source degree $(q+1,u_5+1)$. The next lemma lists the initial May monomials in the obstruction degree and in this incoming-source degree. We will use the digit-zero adjustment of Lemma~\ref{may:page} to separate the surviving source filtrations from the product filtration. In the alphabet of Equation~\eqref{may:low-alphabet}, the splitting differential is \begin{equation} da=xy,\qquad db=yz,\qquad dc=xb+az, \label{may:small-splitting} \end{equation} with the exterior order $x,a,c,y,b,z$ used for signs. \begin{lemma}[The first target and all incoming-source intervals] \label{may:first-list} In bidegree $(s,w)=(q+1,u_4+1)$ the complete initial May page consists of \begin{equation} acb B^{p-2},\qquad acybz B^{p-3}. \label{may:first-target-list} \end{equation} The first is not a splitting cycle and the second is a splitting cycle. In bidegree $(q+1,u_5+1)$ the complete list is given in Table~\ref{may:n5-table}. The four entries of exterior length three or five have no nonzero splitting cycle in their span. \end{lemma} \begin{proof} We prove exhaustion simultaneously for $w=u_n+1$, with $n=4,5$. Since $u_n+1p-2. \] Here $2p-19=1999$ and $p-2=1007$. This contradicts \eqref{may:L-range}. Thus the high positions form a single block $[l,r]$ with $1\leq l\leq r\leq n-2$. At the next position \eqref{may:carry-equations} reads $B_{r+1}=-H_{r+1}$, so \begin{equation} B_{r+1}=H_{r+1}=0. \label{may:gap-after-block} \end{equation} At the beginning of the block, $B_l=p+1-H_l\leq L$, giving $H_l\geq D+1$. If $l1$, at most one begins before $l$, since $H_{l-1}\leq1$, and at most one begins at $l$. Thus in either case $H_l\leq2$, contradicting $H_l\geq D+1\geq3$. We henceforth have $l1$, at most two exterior intervals cover both $l$ and $r$: they must end at $r+1$, and at most one starts before $l$, while at most one starts at $l$. At least one of the two intervals starting at zero misses both positions, since $H_{l-1}\leq1$. Consequently \[ H_l+H_r =m+\#\{\text{intervals meeting both}\} -\#\{\text{intervals meeting neither}\} \leq m+1. \] This contradicts $H_l+H_r\geq2D+1=m+2$. Therefore $l=1$. There are now at most two exterior intervals covering both endpoints, namely $[0,r+1)$ and $[1,r+1)$. Equality in the same counting identity is forced: both are present, every exterior interval meets position $1$ or position $r$, and \[ H_1=D+1,\qquad H_r=D. \] It follows that $B_1=B_r=L$. Every polynomial interval therefore covers both positions, and by \eqref{may:gap-after-block} every one is exactly $[1,r+1)$. No exterior or polynomial interval can occur after the block. The carry supplies the target digit at $r+1$; all further digits would be zero. Since the required remaining digits are all one, necessarily $r=n-2$. Thus \begin{equation} \text{polynomial factor }b_{n-2,0}^{p-D},\qquad H_0=2,\quad H_1=D+1,\quad H_2=\cdots=H_{n-2}=D, \quad H_{n-1}=0. \label{may:forced-profile} \end{equation} It remains to list a small, now forced set of exterior intervals. For $n=4$, the intervals $[0,3)$ and $[1,3)$ are mandatory. The other start-zero interval must be $[0,2)$; $[0,1)$ would meet neither endpoint. The only optional intervals are $[1,2)$ and $[2,3)$. The coverage requirements in \eqref{may:forced-profile} require both or neither. This is exactly \eqref{may:first-target-list}. For $n=5$, the mandatory intervals are $[0,4)$ and $[1,4)$. The remaining start-zero interval is $[0,t)$ with $t=2$ or $3$. The only optional intervals are $[1,2),[1,3),[2,4),[3,4)$; $[2,3)$ meets neither endpoint and is forbidden. Write their inclusion indicators as $e_{12},e_{13},e_{24},e_{34}\in\{0,1\}$, and let their sum be $o$. The three required coverage equations are \begin{equation} e_{12}+e_{13}=o/2,\qquad e_{24}+e_{34}=o/2,\qquad e_{13}+e_{24}+[t=3]=o/2, \label{may:four-indicators} \end{equation} where $o\in\{0,2,4\}$. For $t=2$ the solutions are no optional intervals, the pair $\{12,24\}$, the pair $\{13,34\}$, and all four. For $t=3$ the only solution is the pair $\{12,34\}$. This proves the exhaustive Table~\ref{may:n5-table}, including its length-seven entry. \begin{table}[htbp] \centering \begin{tabular}{c l l} \toprule $m$ & Exterior intervals (exclusive right endpoints) & Polynomial factor\\ \midrule $3$ & $02,04,14$ & $b_{3,0}^{p-2}$\\ $5$ & $03,04,12,14,34$ & $b_{3,0}^{p-3}$\\ $5$ & $02,04,13,14,34$ & $b_{3,0}^{p-3}$\\ $5$ & $02,04,12,14,24$ & $b_{3,0}^{p-3}$\\ $7$ & $02,04,12,13,14,24,34$ & $b_{3,0}^{p-4}$\\ \bottomrule \end{tabular} \caption{Every initial May monomial in bidegree $(q+1,u_5+1)$. The symbol $ab$ in this table denotes the interval $[a,b)$, not a product of the letters in \eqref{may:low-alphabet}.} \label{may:n5-table} \end{table} For completeness, the splitting calculation is also explicit. The length-three entry has, for example, a nonzero component on $01,04,12,14$. For the three length-five entries in their displayed order, take the following four rows, ordering exterior intervals lexicographically: \[ \begin{array}{c|rrr} 01,04,12,13,14,34&1&-1&0\\ 02,03,12,14,24,34&1&0&1\\ 02,04,12,13,24,34&0&1&1\\ 02,04,12,14,23,34&-1&-1&1 \end{array} \] The common polynomial factor is $b_{3,0}^{p-3}$. The minor using the first, second, and fourth rows is $3$, which is nonzero in $\mathbb F_{1009}$. Thus these three images are linearly independent. Finally, \eqref{may:small-splitting} shows that $acb$ has a nonzero splitting differential and that $acybz$ is closed: every term in the latter differential repeats an exterior factor. This finishes the lemma. \end{proof} \begin{proposition}[An injective multiplication] \label{may:injectivity} Multiplication by the primitive class $h_{1,4}$ is injective on actual cobar cohomology in bidegree $(q+1,u_4+1)$: \[ H^{q+1,u_4+1}(P)\xrightarrow{\ h_{1,4}\ } H^{q+2,u_5+1}(P). \] \end{proposition} \begin{proof} Use the filtration in Lemma~\ref{may:page} with the explicit choice $M=3p$. If the source cohomology is nonzero, any nonzero class has leading detector a nonzero scalar multiple of $acybz B^{p-3}$, by Lemma~\ref{may:first-list}. Its filtration is \[ F_0=3p^2-9p+13+M(2+p(p-3)). \] The cocycle $[t_1^{p^4}]$ represents $h_{1,4}$, with degree, weight, and filtration shifts $(1,p^4,1)$. The proposed leading product therefore has filtration \begin{equation} F_*=3p^2-9p+14+M(2+p(p-3)). \label{may:product-filtration} \end{equation} Every possible incoming May differential to this product has source cohomological degree $q+1$ and weight $u_5+1$. Table~\ref{may:n5-table} is its complete initial-page source list. The length-three and length-five source filtrations are respectively \[ 5p^2-10p+15+M(2+p(p-2)),\qquad 5p^2-15p+19+M(2+p(p-3)); \] all three length-five sources have the latter filtration. The length-three and length-five entries disappear under $d_1$, since their displayed splitting maps are injective. None can be a splitting boundary for the product itself, because its polynomial factor is a power of $B=b_{2,0}$, whereas their polynomial factor is a power of $b_{3,0}$. The length-seven source has filtration \[ F_7=5p^2-20p+23+M(2+p(p-4)). \] For $M=3p$ the difference is \[ F_7-F_*=2p^2-11p+9-pM=-p^2-11p+9<0. \] As all May differentials lower filtration, this last source cannot hit the proposed product on any later page. There are no other incoming sources. This argument does not require $B$ to represent a permanent class by itself. Choose an actual cocycle for the given nonzero source cohomology class and multiply that entire cocycle by $[t_1^{p^4}]$. The result is an actual cocycle. Its leading associated-graded cup product is the nonzero monomial $acybz B^{p-3}h_{1,4}$, since the new exterior letter is distinct from all the old ones. An actual cocycle has no outgoing May differential; its leading detector could disappear only by becoming a boundary, which the preceding incoming-source calculation excludes at every page. Its product cohomology class is therefore nonzero. \end{proof} \begin{proposition}[The first Adams--Novikov differential] \label{may:first-differential} In the Adams--Novikov spectral sequence for $Y$, one has $d_{q+1}(v_4)=0$. \end{proposition} \begin{proof} Equation~\eqref{may:specialized-obstruction} and Proposition~\ref{may:injectivity} give $\sigma(d_{q+1}(v_4))=0$. Lemma~\ref{may:target-bound} identifies this specialization with the original target group. Hence $d_{q+1}(v_4)=0$. \end{proof} \subsection{The remaining targets} \begin{proposition}[All later obstruction candidates] \label{may:remaining-list} For $k\geq2$ the complete initial May page in \eqref{may:obstruction-degrees} is Table~\ref{may:remaining-table}; all other values of $k$ give an empty page. Put $N_0=(p-1)^2$ and $N=p(p-1)-2$. \begin{table}[htbp] \centering \begin{tabular}{c l} \toprule $k$ & Initial May monomials\\ \midrule $2$ & $xac B A^{p-2}C^{p-2}$, $\quad xacyz A^{p-2}C^{p-2}$\\[2pt] $p-1$ & $y B^3 A^{N_0-3}$, $\quad b B^2 A^{N_0-2}$, $\quad ybz B A^{N_0-2}$\\[2pt] $p$ & $cyb A^{N+1}$, $\quad ayb B A^N$\\ \bottomrule \end{tabular} \caption{Every initial May monomial for the later $v_4$ obstruction bidegrees.} \label{may:remaining-table} \end{table} \end{proposition} \begin{proof} Lemma~\ref{may:target-bound} allows only positions $0,1,2$ and $2\leq k\leq p+2$. Thus \eqref{may:three-position-weight} applies. There are six exterior letters; $m$ is odd, so write \[ m=2f+1,\quad 0\leq f\leq2,\qquad L=(p-1)k-f. \] Reducing the weight modulo $p$ gives $H_0\equiv1+k\pmod p$, with $0\leq H_0\leq3$. The only possibilities are $k=2$ or $k=p+d$ with $-1\leq d\leq2$. When $k=2$, $H_0=3$ forces all of $x,a,c$. The other exterior letters are an even subset of $\{y,b,z\}$. Put $T=H_2+n_B+n_C$. Substitution in \eqref{may:three-position-weight} gives \begin{equation} pT-n_C+H_1=p^2-p+3+f. \label{may:k2-equation} \end{equation} For the four possible exterior subsets the value $e=H_1-3-f$ is $-2$ or $-1$. Therefore $n_C\equiv e\pmod p$. Since $0\leq n_C\leq L\leq2p-3$, the only possible lift is $n_C=p+e$: the lift $e$ is negative, and $2p+e\geq2p-2>L$. Equation~\eqref{may:k2-equation} now gives $T=p$. All the remaining counts follow as displayed: \[ \begin{array}{c|c|c|c|c|c|c} \text{extra exterior letters}&f&H_1&H_2&n_C&n_B&n_A\\ \hline \varnothing&1&2&1&p-2&1&p-2\\ y,b&2&4&2&p-1&-1&p-2\\ y,z&2&3&2&p-2&0&p-2\\ b,z&2&3&3&p-2&-1&p-1 \end{array} \] The negative counts exclude the second and fourth rows. The first and third are exactly the $k=2$ row of Table~\ref{may:remaining-table}. Now let $k=p+d$, $-1\leq d\leq2$. Then $H_0=1+d$, and the weight equation becomes \begin{equation} pT-n_C+H_1=(2-d)p+2+d+f. \label{may:pd-equation} \end{equation} Since $n_C\leq T$ and $H_1\geq0$, it follows that \[ (p-1)T\leq(2-d)(p-1)+(4+f). \] Here $4+f\leq63p+1$, it must have the form \[ f(D;g)=D^3f_3(g)+D^2f_2(g)+Df_1(g)+f_0(g). \] All these powers are in the finite module, since $N_0\geq3$. For a polynomial function $\varphi$ write $\delta\varphi(g,h)=\varphi(gh)-\varphi(g)-\varphi(h)$. Comparing the four coefficients in \eqref{frob:crossed-equation} gives \begin{align} \delta f_3&=0,\nonumber\\ \delta f_2&=3r^pf_3(h),\nonumber\\ \delta f_1&=2r^pf_2(h)+3r^{2p}f_3(h), \label{frob:crossed-coefficients}\\ \delta f_0&=r^pf_1(h)+r^{2p}f_2(h)+r^{3p}f_3(h). \nonumber \end{align} Homogeneity forces \begin{align*} f_3&=a_3\xi,\\ f_2&=b_2\eta+c_2\xi^{p+1},\\ f_1&=d_1\eta\xi^p+e_1\xi^{2p+1},\\ f_0&=r_0\xi^{3p+1}+s_0\eta\xi^{2p} +t_0\eta^2\xi^{p-1}. \end{align*} Here the coefficients $a_3,b_2,c_2,d_1,e_1,r_0,s_0,t_0$ lie in $\mathbb F_p$. The second equation of \eqref{frob:crossed-coefficients}, together with \[ \delta\eta=rs^p,\qquad \delta(\xi^{p+1})=r^ps+rs^p, \] forces $c_2=3a_3$ and $b_2=-3a_3$. The next equation has coefficient $2b_2$ on $r^pv$ and zero on $us^p$. On the other hand, \begin{equation}\label{frob:asymmetric-pair} \delta(\eta\xi^p) =us^p+vr^p+r^{p+1}s^p+rs^{2p}, \end{equation} and $\delta(\xi^{2p+1})$ contains neither $u$ nor $v$. Its left side must therefore have equal coefficients on $r^pv$ and $us^p$. Since $2b_2=-6a_3$ and $6\neq0$ in $\mathbb F_{1009}$, we obtain $a_3=0$, and then $b_2=c_2=0$. No term in $f_0$ can alter this coefficient-of-$D$ equation. Now $f_1$ is primitive. Equation~\eqref{frob:asymmetric-pair} first forces $d_1=0$, and the coefficient of $r^{2p}s$ in $\delta(\xi^{2p+1})$ then forces $e_1=0$. Finally $f_0$ is primitive. In its reduced coproduct the monomial $u^2s^{p-1}$ has coefficient $t_0$, so $t_0=0$. The monomial $us^{2p}$ then has coefficient $s_0$, so $s_0=0$. The coefficient of $r^{3p}s$ in $\delta(\xi^{3p+1})$ is one, giving $r_0=0$. Every normalized cocycle is thus zero, proving the required $H^1$ vanishing, including all lower-$D$ corrections. For $n=N_0-1$, the residual weight is $\mathcal R=3p+2$ and the desired quotient degree is three. The first coefficient differential comes from the term $2D\xi^p$ in \[ (D+\xi^p)^2-D^2=2D\xi^p+\xi^{2p}. \] Using the actual class $\upsilon$ of Lemma~\ref{frob:quotient-groups}, it is \begin{equation}\label{frob:h3-differential} D^2\upsilon\longmapsto\pm2D\upsilon z. \end{equation} The coefficient differential first prepends $z$. Its product $z\upsilon$ agrees with $\upsilon z$: both are actual cocycles with the same leading May monomial $ybz$, with coefficient $+1$ after two exterior interchanges, and the list in Equation~\eqref{frob:small-monomials} leaves no other initial May monomial or lower-filtration cohomology in this bidegree. Thus the product order in Equation~\eqref{frob:h3-differential} is justified directly. Both powers of $D$ lie in the finite module, since $n\geq3$. Equation~\eqref{frob:D-weight} gives quotient degree and weight $(2,p+2)$ for the source and $(3,2p+2)$ for the target. Both therefore have the required residual weight $3p+2$. All earlier coefficient differentials vanish, so source and target are still their actual quotient-cohomology classes on this first possible page. They are both nonzero by Lemma~\ref{frob:quotient-groups}. A same-page incoming differential to the source would require $D$-degree three and quotient $H^1$ of weight two, which is zero. The only same-page incoming source for the target is $D^2H^2(Q_2)_{p+2}$, the one-dimensional span displayed in \eqref{frob:h3-differential}. A same-page outgoing differential from the target is proportional to $\upsilon z^2=0$. Indeed $[\xi^p\mid\xi^p]=\tfrac12d[\xi^{2p}]$, so this last vanishing also follows directly in the cobar complex. The term $\xi^{2p}$ in the expansion of $D^2$ lowers two $D$-levels and cannot cancel its first differential. Since $2\neq0$ in $\mathbb F_p$, \eqref{frob:h3-differential} removes the target. There is no other degree-three coefficient-leading term. In fact, the possibilities for $D$-degrees $0,1,2,3$ require quotient degree three at respective weights $3p+2,2p+2,p+2,2$; by Lemma~\ref{frob:quotient-groups}, only the second has a candidate, namely $ybz$. Larger $D$-degrees exceed the residual weight. Thus there is no replacement in a lower coefficient filtration. The initial candidate table excludes every other kernel degree or exterior summand at this total bidegree. Both possible terms of \eqref{frob:lhss} consequently vanish, and the proposition follows. \end{proof} \begin{corollary}\label{frob:vanishing} The obstruction groups in bidegrees \[ (s,w)=(2q+1,u_4+2),\qquad (s,w)=(q(p-1)+1,u_4+p-1) \] vanish in the additive cobar calculation of Table~\ref{may:remaining-table}. Among its rows with $k\geq2$, only the row $k=p$ remains to be addressed. \end{corollary} \begin{proof} Apply Propositions~\ref{frob:k2} and~\ref{frob:kpminus1}, and the exhaustive list in Table~\ref{may:remaining-table}. \end{proof} ==== END A-finite-Smith-Toda-complex-V4-at-the-prime-1009//04-frobenius-elimination.tex ==== ==== BEGIN A-finite-Smith-Toda-complex-V4-at-the-prime-1009//05-final-obstruction.tex ==== \section{The final obstruction and the finite cofiber} \label{top:section} We retain the prime $p=1009$ and put \begin{equation} \label{top:constants} N=p(p-1)-2,\qquad K=p(p-2)+1=(p-1)^2, \qquad W=2p^2+4p+1. \end{equation} The spectrum $Y$ of Proposition~\ref{low:V3} has $BP_*Y=R/I_4$ and a binary two-sided unital multiplication $\mu:Y\wedge Y\longrightarrow Y$ with unit $\eta:S^0_{(p)}\to Y$. All spheres and smash products in this section are $p$-local. We shall use the product and convergence assertions of Proposition~\ref{fw:anss-pairing} and Proposition~\ref{fw:anss}. In this section the second superscript of $E_r^{s,w}$ is weight: it denotes the group of internal degree $qw$ in the earlier notation. The May comparison and the remaining-target calculation, Lemma~\ref{may:comparison} and Table~\ref{may:remaining-table}, give a surjection \begin{equation} \label{top:An-surjection} A^N\colon E_2^{5,W}(Y)\longrightarrow E_2^{qp+1,u_4+p}(Y). \end{equation} Indeed, every initial May monomial in the target is divisible by $A^N$, and the two degree identities are \[ 5+2N=qp+1,\qquad W+pN=u_4+p. \] For the source, $WW+1-6. \] Consequently $j\leq1012$ and \begin{equation} \label{top:small-stem-cutoff} w\leq2{,}041{,}212 W+1$, this improves the bound to $j\leq2$. For completeness, we enumerate this remaining finite calculation. Let $H_i$ count the exterior letters covering position $i$, and let $n_A,n_B,n_C$ be the polynomial exponents. The weight equation is \begin{equation} \label{top:weight-equation} w=H_0+p(H_1+L-n_C)+p^2(H_2+n_B+n_C). \end{equation} In \eqref{top:small-stem-degrees}, $m$ is even and reduction modulo $p$ gives $H_0=1+\delta+j$, whenever a solution exists. Put $\kappa_2=H_2+n_B+n_C$. All position contributions are nonnegative; \eqref{top:small-stem-cutoff} also gives $w<3p^2$, so $\kappa_2\leq2$ and $n_C\leq2$. Substituting $L=3+(p-1)j-m/2$ into \eqref{top:weight-equation} yields \[ p(\kappa_2+j-2)=1+j+m/2+n_C-H_1. \] For $0\leq j\leq2$, its right side lies between $-3$ and $8$. Its only multiple of $p=1009$ is zero. Hence \begin{equation} \label{top:small-stem-coverage} H_0=1+\delta+j,\qquad \kappa_2=2-j,\qquad H_1=1+j+m/2+n_C. \end{equation} For $j=2$, either $H_0=4$, which is impossible, or $H_0=3$ forces $c$ to occur whereas $\kappa_2=0$. For $j=1,\delta=1$, all of $x,a,c$ occur and $\kappa_2=1$ excludes $b,z$. Thus $H_1\leq3$, but the last equation of \eqref{top:small-stem-coverage}, with even $m\geq4$, requires $H_1\geq4$. For $j=1,\delta=0$, one has $H_0=2$. If $m=2$, the required inequality $H_1\geq3$ is impossible. If $m=4$, one must have $H_1=4$ and $n_C=0$; the four letters covering position one are exactly $a,c,y,b$, which have $H_2=2$, contrary to $\kappa_2=1$. Finally $m=6$ would give $H_0=3$. All positive-$j$ groups in \eqref{top:small-stem-degrees} therefore have empty initial May pages. It remains to examine $j=0,\delta=1$, where $H_0=2$ and $m$ is $2$ or $4$. For $m=2$ both letters start at zero, and \eqref{top:small-stem-coverage} permits only $ac$, followed by $n_B=1,n_C=0,n_A=1$. For $m=4$ choose two of $x,a,c$ and two of $y,b,z$. The equations become \[ n_C=H_1-3,\qquad n_B=2-H_2-n_C,\qquad n_A=1-n_B-n_C. \] The following table lists all nine choices; a dash means that at least one exponent is negative. \begin{center} \begin{tabular}{c c c c} \toprule Start-zero pair & $yb$ & $yz$ & $bz$\\ \midrule $xa$ & $xaybB$ & --- & ---\\ $xc$ & $xcybA$ & --- & ---\\ $ac$ & --- & $acyzA$ & ---\\ \bottomrule \end{tabular} \end{center} Thus the complete initial-page list at $(6,W+1)$ is \begin{equation} \label{top:four-terms} acAB,\qquad xaybB,\qquad xcybA,\qquad acyzA. \end{equation} With exterior order $x,a,c,y,b,z$ and $da=xy$, $db=yz$, $dc=xb+az$, their splitting differentials are \[ \begin{aligned} d(acAB)&=(xab-xcy)AB, &d(xaybB)&=0,\\ d(xcybA)&=-xaybzA, &d(acyzA)&=-xaybzA. \end{aligned} \] The first image has polynomial factor $AB$ and cannot cancel either of the other images. The two remaining cycle directions are boundaries, since \[ d(acyB)=-xaybB,\qquad d(acbA)=-xcybA+acyzA. \] The splitting cohomology of \eqref{top:four-terms} is zero. The May spectral sequence, finite in each of these weights, proves all the claimed $E_2$ vanishings. For orientation, the same equations also enumerate the source $(5,W)$. Here $H_0=1$, $m$ is odd, and $L=(5-m)/2$. They force $H_2+n_B+n_C=2$ and $H_1+L-n_C=4$. If $m=1$, then $H_1+L\leq1+2<4$; if $m=5$, then at least two of $x,a,c$ occur, contrary to $H_0=1$. Thus $m=3$, which forces $n_C=0$ and $H_1=3$. The only monomials are $aybB$ and $cybA$. We need no assertion that either one survives the May spectral sequence. No topological Adams--Novikov differential can hit filtration five: its length would have to be at least $q+1>5$. The outgoing target groups have just been proved zero. Thus every actual class of $E_2^{5,W}(Y)$ survives and is represented by a class $\gamma\in\pi_{qW-5}Y$. The vanishings for $\delta=1$ exhaust all filtrations in the stem $q(W+1)-6$; strong convergence proves \eqref{top:alpha-gamma-vanishing}. \end{proof} \subsection{Two sphere classes and an extended-power relation} \begin{lemma}[The sphere classes] \label{top:sphere-classes} There are sphere classes \[ \alpha\in\pi_{q-1}S^0_{(p)},\qquad \beta\in\pi_{pq-2}S^0_{(p)} \] detected respectively by $[t_1]$ and the divided coproduct $b_{1,0}$. The former generates a cyclic group of order $p$, and \[ \pi_iS^0_{(p)}=0\qquad(0