An _∞-structure on the cohomology ring of the symmetric group  with coefficients in
Stephan Schmid

Let $p$ be a prime. Let $\fp\!\Sp$ be the group algebra of the symmetric group over the finite field $\fp$ with $|\fp|=p$. Let $\fp$ be the trivial $\fp\!\Sp$-module. We present a projective resolution $\pres \fp$ of the module $\fp$ and equip the Yoneda algebra $\Ext^*_{\fp\!\Sp}(\fp,\fp)$ with an $\A_∞$-structure such that $\Ext^*_{\fp\!\Sp}(\fp,\fp)$ becomes a minimal model of the dg-algebra $\Hom^*_{\fp\!\Sp}(\pres \fp, \pres \fp)$.

[0]MSC 2010: 18G15.

§.§ Introduction

Let $R$ be a commutative ring.  Let $A$ be a $\Z$-graded $R$-module. Let $m_1:A→A$ be a graded map of degree $1$ with $m_1^2=0$, i.e. a differential on $A$. Let $m_2:A\otimes A→A$ be a graded map of degree $0$ satisfying the Leibniz rule, i.e.

\begin{align*}
m_1\circ m_2 = m_2\circ(m_1\otimes 1 + 1\otimes m_1).
\end{align*}

The map $m_2$ is in general not required to be associative. Instead, we require that for a morphism $m_3:A^{\otimes 3}→A$, the following identity holds.

\[m_2\circ (m_2\otimes 1 - 1\otimes m_2) = m_1\circ m_3 + m_3\circ(m_1\otimes 1^{\otimes 2} + 1\otimes m_1 \otimes 1 + 1^{\otimes 2} \otimes  m_1)\]

Following Stasheff, cf. [23], this can be continued in a certain way with higher multiplication maps to obtain a tuple of graded maps $(m_n:A^{\otimes n}→ A)_{n\geq 1}$ of certain degrees satisfying the Stasheff identities, cf. e.g. (<ref>). The tuple $(A, (m_n)_{n\geq 1})$ is then called an $\A_∞$-algebra.

A morphism of $\A_∞$-algebras from $(A',(m'_n)_{n\geq 1})$ to $(A,(m_n)_{n\geq 1})$ is a tuple of graded maps   of certain degrees satisfying the identities (<ref>). The first two of these are

\begin{align*}
\eqref{finfrel}[1]: && f_1 \circ m'_1 \eqs & m_1 \circ f_1\\
\eqref{finfrel}[2]: && f_1\circ m'_2 - f_2\circ(m'_1\otimes 1 + 1\otimes m'_1) \eqs& m_1\circ f_2 + m_2\circ (f_1\otimes f_1).
\end{align*}

So a morphism $f=(f_n)_{n\geq 1}$ of $\A_∞$-algebras from $(A',(m'_n)_{n \geq 1})$ to $(A,(m_n)_{n \geq 1})$ contains a morphism of complexes $f_1:(A',m_1')→(A,m_1)$. We say that $f$ is a quasi-isomorphism of $\A_∞$-algebras if $f_1$ is a quasi-isomorphism. Furthermore, there is a concept of homotopy for  $\A_∞$-morphisms, cf. e.g. <cit.> and <cit.>.

The history of $\A_∞$-algebras is outlined in [12] and [13].

As already mentioned, Stasheff introduced $\A_∞$-algebras in 1963.

If $R$ is a field, $\mathbb{F}:=R$, we have the following basic results on $\A_∞$-algebras, which are known since the early 1980s.

* Each quasi-isomorphism of $\A_∞$-algebras is a homotopy equivalence, cf. [21], [10], …

* The minimality theorem: Each $\A_∞$-algebra $(A,(m_n)_{n\geq 1})$ is quasi-isomorphic to an $\A_∞$-algebra  $(A',\{m'_n\}_{n \geq 1})$ with $m'_1=0$, cf. [9], [8], [21], [5], [7], [18], … . The $\A_∞$-algebra  $A'$  is then called a minimal model of $A$.

Suppose given an $\mathbb{F}$-algebra $B$ and suppose given an $B$-module $M$ together with a projective resolution $\pres M$ of $M$.
The homology of the dg-algebra $\Hom^*_B(\pres M,\pres M)$ is the Yoneda algebra $\Ext^*_B(M,M)$. By the minimality theorem, it is possible to construct an $\A_∞$-structure on $\Ext^*_B(M,M)$ such that $\Ext^*_B(M,M)$ becomes a minimal model of the dg-algebra $\Hom^*_B(\pres M,\pres M)$.

For the purpose of this introduction, we will call such an $\A_∞$-structure on $\Ext^*_B(M,M)$ the canonical $\A_∞$-structure on $\Ext^*_B(M,M)$, which is unique up to isomorphisms of $\A_∞$-algebras, cf. <cit.>.

This structure has been calculated or partially calculated in several cases.

Let $p$ be a prime.

For an arbitrary field $\mathbb{F}$, Madsen computed the canonical $\A_∞$-structure on  $\Ext^*_{\mathbb{F}[α]/(α^n)}(\mathbb{F},\mathbb{F})$, where $\mathbb{F}$ is the trivial $\mathbb{F}[α]/(α^n)$-module, cf. <cit.>. This can be used to compute the canonical $\A_∞$-structure on the group cohomology $\Ext^*_{\fp {\rm C}_{m}}(\fp,\fp)$, where $m∈\Z_{\geq 1}$ and ${\rm C}_m$ is the cyclic group of order $m$, cf. <cit.>.

Vejdemo-Johansson developed algorithms for the computation of minimal models [24]. He applied these algorithms to compute large enough parts of the canonical $\A_∞$-structures of the group cohomologies $\Ext^*_{\fs {\rm D}_8}(\fs,\fs)$ and $\Ext^*_{\fs {\rm D}_{16}}(\fs,\fs)$ to distinguish them, where ${\rm D}_8$ and ${\rm D}_{16}$ denote dihedral groups. He stated a conjecture on the complete $\A_∞$-structure on $\Ext^*_{\fs {\rm D}_8}(\fs,\fs)$. Furthermore, he computed parts of the canonical $\A_∞$-structure on $\Ext^*_{\fs {\rm Q}_8}(\fs, \fs)$ for the quaternion group ${\rm Q}_8$. He conjecturally stated the minimal complexity of such a structure. Based on this work, there are now built-in algorithms for the Magma computer algebra system. These are capable of computing partial $\A_∞$-structures on the group cohomology of $p$-groups.

In [25], Vejdemo-Johansson examined  the canonical $\A_∞$-structure $(m_n)_{n\geq 1}$  on the group cohomology $\Ext^*_{\fp({\rm C}_k\times {\rm C}_l)}(\fp, \fp)$ of the abelian group ${\rm C}_k\times {\rm C}_l$ for $k,l\geq 4$ such that $k,l$ are multiples of $p$. He showed that for infinitely many $n∈\Z_{\geq 1}$, the operation $m_n$ is non-zero.

In [14], Klamt investigated canonical $\A_∞$-structures in the context of the representation theory of Lie-algebras. In particular, given certain direct sums $M$ of parabolic Verma modules, she examined the canonical $\A_∞$-structure $(m'_k)_{k\geq 1}$ on $\Ext^*_{\mathcal{O}^{\mathfrak{p}}}(M, M)$. She proved upper bounds for the maximal $k∈\Z_{\geq 1}$ such that $m'_k$ is non-vanishing and computed the complete $\A_∞$-structure in certain cases.

The result

For $n∈\Z_{\geq 1}$, we denote by $\Sy_n$ the symmetric group on $n$ elements.

The group cohomology $\Ext^*_{\fp\!\Sp}(\fp,\fp)$ is well-known. For example, in <cit.>, it is calculated using group cohomological methods.

Here, we will construct the canonical $\A_∞$-structure on $\Ext^*_{\fp\!\Sp}(\fp, \fp)$.

We obtain homogeneous elements $ι,χ∈\Hom_{\fp\!\Sp}^*(\pres \fp, \pres \fp)=:A$ of degree $|ι|=2(p-1)=: l$ and $|χ|=l-1$ such that $ι^j,χ\circ ι^j=:χι^j$ are cycles for all $j∈\Z_{\geq 0}$ and such that their set of homology classes $\{\overline{ι^j} \mid j∈\Z_{\geq 0}\}\sqcup \{\overline{χι^j} \mid j∈\Z_{\geq 0}\}$ is an $\fp$-basis of $\Ext^*_{\fp\!\Sp}(\fp, \fp)=\Hm^*A$, cf. <ref>.

For all primes $p$, the canonical $\A_∞$-structure $(m'_n:(\Hm^*A)^{\otimes n}→\Hm^*A)_{n\geq 1}$ on $\Hm^*A$ is given as follows.

On the elements $\overline{χ^{a_1}ι^{j_1}}\otimes \cdots \otimes \overline{χ^{a_n}ι^{j_n}}$, $n∈\Z_{\geq 1}$, $a_i∈\{0,1\}$ and $j_i∈\Z_{\geq 0}$ for $i∈\{1,…,n\}$, the maps $m'_n$ are given as follows, cf. <ref>.

If there is an $i∈\{1,…,n\}$ such that $a_i=0$, then

\begin{align*}
m'_n(\overline{χ^{a_1}ι^{j_1}}\otimes \cdots \otimes \overline{χ^{a_n}ι^{j_n}}) \eqs& 0 \hphantom{\overline{χ^{a_1 + a_1}ι^{j_1 + j_2}}} \text{for $n\neq 2$ and}\\
m'_2(\overline{χ^{a_1}ι^{j_1}}\otimes  \overline{χ^{a_2}ι^{j_2}}) \eqs & \overline{χ^{a_1 + a_1}ι^{j_1 + j_2}}.% & \text{ and}\\
\end{align*}

If all $a_i$ equal $1$, then

\begin{align*}
m'_n(\overline{χι^{j_1}}\otimes \cdots \otimes \overline{χι^{j_n}}) \eqs&0  \hphantom{(-1)^p\overline{ι^{p-1+j_1+… + j_p}}}\text{for $n\neq p$ and }\\
m'_p(\overline{χι^{j_1}}\otimes \cdots \otimes \overline{χι^{j_p}}) \eqs  &%\begin{cases}
(-1)^p\overline{ι^{p-1+j_1+… + j_p}}. %&\text{ and}\\ %&&\text{ if $n=p$} \\
\end{align*}

In particular, we have $m'_n=0$ for all $n∈ \Z_{\geq 1}\setminus\{2,p\}$.

§.§ Outline

Section 1

The goal of <ref> is to obtain a projective resolution of
the trivial $\fp\!\Sp$-Specht module $\fp$. A well-known method for that is "Walking around the Brauer tree", cf. [4].

Instead, we use locally integral methods to obtain a projective resolution in an explicit and straightforward manner.

Over $ℚ$, the Specht modules are absolutely simple. Therefore we have a morphism of $\zp$-algebras

$r:\zp\!\Sp→\prod_{λ\dashv p} \End_{\zp} S^λ_{\zp}=:Γ$ induced by the operation of the elements of $\zp\!\Sp$ on the Specht modules $S^λ$ for  partitions $λ$ of $p$, which becomes an Wedderburn isomorphism when tensoring with $ℚ$. So $Γ$ is a product of matrix rings over $\zp$.

There is a well-known description of  $\im r=:Λ$,
which we use for $p\geq 3$ to obtain projective $Λ$-modules $\tilde P_k\subseteq Λ$, $k ∈[1,p-1]$, and to construct the indecomposible projective resolution $\pres\zp$ of the trivial $\zp\!\Sp$-Specht module $\zp$. The non-zero parts of $\pres\zp$ are periodic with period length $l=2(p-1)$. In <ref>, we reduce $\pres\zp$ modulo $p$ to obtain a projective resolution $\pres\fp$ of the trivial $\fp\!\Sp$-Specht module $\fp$.

Section 2

The goal of <ref> is to compute a minimal model of the dg-algebra $\Hom_{\fp\!\Sp}^*(\pres\fp, \pres\fp)=: A$ by equipping its homology $\Ext^*_{\fp\!\Sp}(\fp,\fp) = \Hm^*A$ with a suitable $\A_∞$-structure and  finding a quasi-isomorphism of $\A_∞$-algebras from $\Hm^* A$ to $A$.

Towards that end, we recall the basic definitions concerning $\A_∞$-algebras and some general results in <ref>.

While there does not seem to be a substantial difference between the cases $p=2$ and $p\geq 3$, we separate them to  simplify notation and argumentation.
Consider the case $p\geq 3$. In <ref>, we obtain a set of cycles $\{ι^j \mid j∈\Z_{\geq 0}\}\cup\{χι^j \mid j∈\Z_{\geq 0}\}$ in $A$ such that their homology classes are a graded basis of $\Hm^*A$.

In <ref>, we obtain a suitable $\A_∞$-structure on $\Hm^*A$ and a quasi-isomorphism of $\A_∞$-algebras from $\Hm^*A$ to $A$. For the prime $2$, both steps are combined in the short <ref>.

§.§ Notations and conventions

* For the remainder of this document, $p$ will be a prime with $p\geq 3$.

* Write $l:=2(p-1)$. This will give the period length of the constructed projective resolution of $\fp$ over $\fp\!\Sp$, cf. e.g. (<ref>), <ref>.

* Concerning "$∞$", we assume the set $\Z\cup\{∞\}$ to be ordered in such a way that $∞$ is greater than any integer, i.e. $∞> z$ for all $z∈\Z$, and that the integers are ordered as usual.

* For $a∈\Z$, $b∈\Z\cup \{∞\}$, we denote by $[a,b] := \{z∈\Z \mid a\leq z\leq b\} \subseteq \Z$ the integral interval. In particular, we have $[a,∞] = \{z∈\Z \mid  z\geq a\}\subseteq \Z$ for $a∈\Z$.

* For $n∈\Z_{\geq 0}$, $k∈\Z$, let the binomial coefficient $\binom{n}{k}$ be defined by the number of subsets of the set $\{1,…,n\}$ that have cardinality $k$. In particular, if $k<0$ or $k>n$, we have $\binom{n}{k}= 0$. Then the formula $\binom{n}{k-1} + \binom{n}{k} = \binom{n+1}{k}$ holds for all $k∈\Z$.

* For a commutative ring $R$, an $R$-module $M$ and $a,b∈M$, $c∈R$, we write

\begin{align*} b&\equiv_c a &:\Longleftrightarrow & & a-b ∈ cM.\end{align*}

Often we have $M=R$ as module over itself.

* Modules are right-modules unless otherwise specified.

* For sets, we denote by $\sqcup$  the disjoint union of sets.

* $|\cdot|$: For a homogeneous element $x$ of a graded module or a graded map $g$ between graded modules, we denote by $|x|$ resp. $|g|$ their degrees (This is not unique for $x=0$ resp. $g=0$). For $y$ a real number, $|y|$ denotes its absolute value.

Symmetric Groups
Let $n∈\Z_{\geq 1}$. We denote the symmetric group von $n$ elements by $\Sy_n$.

For a partition $λ\dashv n$, we denote the corresponding Specht module by $S^λ$.

Complexes Let $R$ be a commutative ring and $B$ an $R$-algebra.

* For a complex of $B$-modules

\[\cdots \rightarrow C_{k+1}\xrightarrow{d_{k+1}}C_k\xrightarrow{d_k} C_{k-1}\rightarrow\cdots\hphantom{,},\]

its $k$-th boundaries, cycles and homology groups are defined by $\text{B}^k:=\im d_{k+1}$, $\text{Z}^k:=\ker d_{k}$ and $\Hm^k:= \text{Z}^k/\text{B}^k$.

For a cycle $x∈\text{Z}^k$, we denote by $\overline{x}:=x+\text{B}^k∈\Hm^k$ its equivalence class in homology.

* Let
\begin{align*}
C\eqs&(\cdots \rightarrow C_{k+1}\xrightarrow{d_{k+1}}C_k\xrightarrow{d_k} C_{k-1}\rightarrow\cdots)\\
C'\eqs&(\cdots \rightarrow C'_{k+1}\xrightarrow{d'_{k+1}}C'_k\xrightarrow{d'_k} C'_{k-1}\rightarrow\cdots)
\end{align*}

be two complexes of $B$-modules.

Given $z∈\Z$, let

\begin{align*}
\Hom_B^z(C,C') := \prod_{i∈\Z} \Hom_B(C_{i+z},C'_i).
\end{align*}

For an additional complex $C''=(\cdots \rightarrow C''_{k+1}\xrightarrow{d''_{k+1}}C''_k\xrightarrow{d''_k} C''_{k-1}\rightarrow\cdots)$ and maps $h=(h_i)_{i∈\Z}∈\Hom_B^m(C,C')$, $h'=(h'_i)_{i∈\Z}∈\Hom_B^n(C',C'')$, $m,n∈\Z$, we define the composition by component-wise composition as

\begin{align*}
h'\circ h := (h'_{i}\circ h_{i+n})_{i∈\Z} ∈ \Hom_B^{m+n}(C,C'').
\end{align*}

We will assemble elements of $\Hom_B^z(C,C')$ as sums of their non-zero components, which motivates the following notations regarding "extensions by zero" and sums.

For a map $g:C_x→C'_y$ , we define $\ls g\rs_x^y∈\Hom_B^{x-y}(C,C')$  by

\begin{align*}
(\ls g\rs_x^y)_{i} := \begin{cases} g & \text{ for }i=y\\ 0 & \text{ for } i∈\Z\setminus \{y\} \end{cases}\, .
\end{align*}

Let $k∈\Z$. Let $I$ be a (possibly infinite) set. Let $g_i = (g_{i,j})_j∈\Hom_B^k(C,C')$ for $i∈I$  such that $\{i∈I \mid g_{i,j}\neq 0\}$ is finite for all $j∈\Z$.

We define the sum $\sum_{i∈I} g_i∈\Hom_B^k(C,C')$ by

\begin{align*}
\left(\dum_{i∈I} g_i\right)_j := \sum_{i∈I,g_{i,j}\neq 0} g_{i,j}\,.
\end{align*}

The graded $R$-module $\Hom_B^*(C,C') := \bigoplus_{k∈\Z} \Hom_B^k(C,C')$ becomes a complex via the differential $d_{\Hom_B^*(C,C')}$, which is defined on elements $g∈\Hom_B^k(C,C')$, $k∈\Z$ by

\begin{align*}
d_{\Hom_B^*(C,C')}(g) := d' \circ g -(-1)^k g\circ d ∈\Hom_B^{k+1}(C,C'),
\end{align*}

where $d := (d_{i+1})_{i∈\Z}= \sum_{i∈\Z}\ls d_{i+1}\rs^{i}_{i+1}∈\Hom_B^1(C,C)$ and analogously $d' := (d'_{i+1})_{i∈\Z}= \sum_{i∈\Z}\ls d'_{i+1}\rs^{i}_{i+1}∈\Hom_B^1(C',C')$.

An element $h∈\Hom_B^0(C,C')$ is called a complex morphism if it satisfies $d_{\Hom_B^*(C,C')}(h) = 0$, i.e. $d'\circ g = g\circ d$.

§ THE PROJECTIVE RESOLUTION OF FP OVER FPSP

§.§ A description of ZpSp

In this paragraph, we review results found e.g. in <cit.>. We use the notation of [6].

Let $n∈\Z_{\geq 1}$.

A partition of the form $λ^k := (n-k+1,1^{k-1})$, $k∈[1,n]$ is called a hook partition of $n$.

Suppose $λ\dashv n$, i.e. $λ$ is a partition of $n$.

Let $S^λ$ be the corresponding integral Specht module, which is a right $\Z\!\Sy_n$-module, cf. <cit.>.
Then $S^{λ}$ is finitely generated free over $\Z$, cf. <cit.>, having a standard $\Z$-basis consisting of the standard $λ$-polytabloids. We write $n_λ$ for the rank of $S^λ$.

For a tuple $b=(b_2,b_3,…,b_k)$, $k∈[1,n]$,  of pairwise distinct elements of $[1,n]$, let $\lpt b\rpt$ be the  $λ^k$-polytabloid generated by the $λ^k$-tabloid

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*       ⋯       *

[decorate,thick] (m-1-1.north west) – (m-1-3.north east);
[decorate,thick] (m-1-1.south west) – (m-1-3.south east);

where $*\cdots *$ are the elements of $[1,n]\setminus b$.
Any polytabloid of $S^{λ^k}$ can be expressed this way.

For such a tuple $b$ and distinct elements $y_1,…,y_s∈[1,n]\setminus b$, we denote by $(b,y_1,…,y_s)$ the tuple $(b_2,b_3,…,b_k,y_1,…,y_s)$.
Recall the notations for manipulation of tuples from <ref>.

The $λ^k$-polytabloid $\lpt b\rpt$ is standard iff $2\leq b_2 < b_3 < \cdots < b_k \leq n$, cf. <cit.>. This entails the following lemma.

For $k∈[1,n]$, the rank of $S^{λ^k}$ is given by $n_{λ^k}=\binom{n-1}{k-1}$.

Let $k∈[1,n-1]$. We have the $\Z$-linear box shift morphisms for hooks

\[\begin{array}{rcl}
S^{λ^k} & \overset{f_k}{\longrightarrow} & S^{λ^{k+1}}\\
\lpt b \rpt & \longmapsto & \sum_{s∈[2,n]\setminus b} \lpt (b,s)\rpt.
\end{array}\]

For $x∈S^{λ^k}$ and $ρ∈\Sy_n$, we have

\begin{align}
\label{modn}
f_k(x\cdot ρ) \equiv_n f_k(x)\cdot ρ.
\end{align}

I.e. the composite $(S^{λ^k} \xrightarrow{f_k}S^{λ^{k+1}} \xrightarrow{π} S^{λ^{k+1}}/nS^{λ^{k+1}})$, where $π$ is residue class map, is $\Z\!\Sy_n$-linear.

The following sequence of $\Z$-linear maps is exact.

\begin{align*}
0 \rightarrow S^{λ^1} \xrightarrow{f_1} S^{λ^2} \xrightarrow{f_2} \cdots \xrightarrow{f_{n-1}} S^{λ^n} \rightarrow 0
\end{align*}

We show that $\im f_k\subseteq \ker f_{k+1}$ for $k∈[1,n-2]$, i.e. that $f_{k+1}\circ f_k = 0$. Let $\lpt b\rpt ∈ S^{λ^k}$ be a polytabloid. We obtain

\begin{align*}
f_{k+1}f_k(\lpt b\rpt) \eqs& f_{k+1}\left( \sum_{s∈[2,n]\setminus b} \lpt (b,s)\rpt\right)
%= \sum_{s∈[2,n]\setminus b}\sum_{t∈[2,n]\setminus (b\cup \{s\})} \lpt ((b,s),t)\rpt \\
= \sum_{\substack{s,t∈ [2,n]\setminus b,\\s\neq t}} \lpt (b,s,t)\rpt\\
\eqs& \sum_{\substack{s,t ∈[2,n]\setminus b,\\s<t}} \big(\lpt (b,s,t)\rpt + \lpt(b,t,s)\rpt\big) \overset{\text{cf. \cite[4.3]{Ja78}}}{=} 0.
\end{align*}

Now we show the exactness of the sequence. For convenience, we set  and $f_n\colon S^{λ^n}→0$. We define $T^k$ for $k∈[1,n]$ to be the tuple of all tuples $b=(b_2,…,b_k)$ such that $2\leq b_2<b_3 < … < b_k\leq n-1$, where $T^k$  is ordered, say, lexicographically.

Then we set $B^k_{\rm b} := ( \lpt b \rpt \colon b∈T^k)$, which consists of standard $λ^k$-polytabloids.
We set $B^1_{\rm c} := ()$, which is the empty tuple, and for $k∈[2,n]$,

\begin{align*}
B^k_{\rm c} :\eqs& (f_{k-1}(x) \colon x∈B^{k-1}_{\rm b})\\
\eqs& \left(\sum_{s∈[2,n]\setminus b} \lpt (b,s)\rpt \colon b∈T^{k-1}\right)
= \left(\lpt (b,n)\rpt + \sum_{\mathclap{s∈[2,n-1]\setminus b}} \lpt (b,s)\rpt \colon b∈T^{k-1}\right).
\end{align*}

So $B^k_{\rm c} \subseteq \im f_{k-1}$ and thus $f_k(B^k_{\rm c})\subseteq \{0\}$ for $k∈[1,n]$.

By comparing $B^k_{\rm c} \sqcup B^k_{\rm b}$ with the standard basis, we observe that $B^k_{\rm c}\sqcup B^k_{\rm b}$ is a $\Z$-basis of $S^{λ^k}$ for $k∈[1,n]$.

For $k∈[1,n]$, we have

\begin{align*}
n_{\rm b}^k :\eqs& |B^k_{\rm b}| = \binom{n-2}{k-1}\\
n_{\rm c}^k :\eqs& |B^k_{\rm c}| =\left\{\begin{array}{ll} |B^{k-1}_{\rm b}| = \binom{n-2}{k-2} & \text{for }k∈[2,n] \\ 0 = \binom{n-2}{1-2} & \text{for }k=1 \end{array}\right\}
= \binom{n-2}{k-2}.
\end{align*}

For $k∈[1,n-1]$, the morphism $f_k$ maps $\langle B^k_{\rm b}\rangle_{\Z}$ bijectively to $\langle B^{k+1}_{\rm c}\rangle_{\Z}$ and $\langle B^{k}_{\rm c}\rangle_{\Z}$ to zero. So $\ker f_k = \langle B^{k}_{\rm c}\rangle_{\Z}$ and $\im f_k = \langle B^{k+1}_{\rm c}\rangle_{\Z}$. As $B^1_{\rm c} = () = B^n_{\rm b}$, we have also $\im f_0 = \langle B^1_{\rm c}\rangle_{\Z}$ and $\ker f_n = \langle B^n_{\rm c}\rangle_{\Z}$. So the sequence in question is exact.

We equip the Specht modules $S^{λ^k}$ of hook type with the ordered $\Z$-basis $B^k_{\rm c}\sqcup B^k_{\rm b}$. We equip all other Specht modules with the standard $\Z$-basis with an arbitrarily chosen total order. From now on each of these bases will be referred to as the basis of the respective Specht module.  We define the $\Z$-algebra

\begin{align*}
Γ^{\Z} := \prod_{λ\dashv n} \Z^{n_λ\times n_λ}.
\end{align*}

Let $λ\dashv n$ and let $B = (b_1,…,b_{n_λ})$ be the basis of $S^{λ}$.
For the multiplication with matrices, we identify $S^λ$ with $\Z^{1\times n_λ}$ via $B$.

Then $S^λ$ becomes a right $Γ^{\Z}$-module via $x\cdot ρ := x\cdot ρ^λ$ for $x∈S^λ$ and $ρ∈Γ^{\Z}$, where $ρ^λ$ is the $λ$-th component of $ρ$.

I.e. $ρ∈Γ^{\Z}$ operates by multiplication with the matrix $ρ^λ$ on the right  with respect to the basis $B$.

Similarly, $\bigoplus_{λ\dashv n} S^λ$ becomes a right $Γ^{\Z}$-module. Each $\Z$-endomorphism of  $\bigoplus_{λ\dashv n} S^λ$ is represented by the operation of a unique element of $Γ^{\Z}$. As the operation of $\Z\!\Sy_n$ defines such endomorphisms (cf. <cit.>), we obtain a $\Z$-algebra morphism $r^{\Z}:\Z\!\Sy_n → Γ^{\Z}$
such that $y\cdot r^{\Z}(x) = y\cdot x$ for all $λ\dashv n$, $y∈S^λ$, $x∈\Z\Sy_n$.

As the Specht modules give all irreducible ordinary  representations of $\Sy_n$, the map $r^{\Z}$ is injective. Because of (<ref>), the image of $r^{\Z}$ is contained in

\begin{align*}
Λ^{\Z} := \{ρ∈Γ^{\Z} \mid  f_k(xρ) \equiv_n f_k(x)ρ\, ∀_{k∈[1,n-1]}\, ∀_{x∈S^{λ^k}}\}  \subseteq Γ^{\Z}.
\end{align*}

As the basis $B_{\rm c}^k\sqcup B_{\rm b}^k$ of $S^{λ^k}$, $k∈[1,n]$, consists of two parts, we may split each $ρ^{λ^k}$ for a $ρ∈Γ^{\Z}$ into four blocks corresponding to the parts $B^k_{\rm c}$ and $B^k_{\rm b}$:

\begin{equation}
\label{eq:blockmotivation}
ρ^{λ^k} =
\xyprot{
\begin{tikzpicture}[decoration=brace,baseline]
\matrix (m) [matrix of math nodes,left delimiter=(,right delimiter=),
nodes={%rectangle,minimum width=3em,
outer sep=0pt},
row 1/.style={bottomrule},column 1/.style=rightrule] {
ρ^{λ^k}_{\rm cc}\vphantom{ρ^{λ^k}_{\rm bc}} & ρ^{λ^k}_{\rm bc} \\
ρ^{λ^k}_{\rm cb} & ρ^{λ^k}_{\rm bb} \\
\draw[decorate,transform canvas={xshift=1.5em},thick] (m-1-2.north east) -- node[right=2pt] {$n^k_{\rm c}$} (m-1-2.south east);
\draw[decorate,transform canvas={xshift=1.5em},thick] (m-2-2.north east) -- node[right=2pt] {$n^k_{\rm b}$} (m-2-2.south east);
\draw[decorate,transform canvas={yshift=0.5em},thick] (m-1-1.north west) -- node[above=2pt] {$n^k_{\rm c}$} (m-1-1.north east);
\draw[decorate,transform canvas={yshift=0.5em},thick] (m-1-2.north west) -- node[above=2pt] {$n^k_{\rm b}$} (m-1-2.north east);
\end{tikzpicture}
\end{equation}

Suppose given $k∈[1,n-1]$. We represent $f_k$ by a matrix $M_{f_k}$ with respect to the bases of $S^{λ^k}$ and $S^{λ^{k+1}}$, i.e. $f_k(x) = x\cdot M_{f_k}$ for $x∈S^{λ^k}$. As $f_k(B_{\rm b}^k) = B_{\rm c}^{k+1}$ and $f_k(B_{\rm c}^k)\subseteq \{0\}$, the matrix $M_{f_k}$  has the following block form:

\begin{equation*}
M_{f_k} =
\xyprot{
\begin{tikzpicture}[decoration=brace,baseline]
\matrix (m) [matrix of math nodes,left delimiter=(,right delimiter=),
nodes={%minimum width=\widthof{\tikz \node{ABCD};},%rectangle,minimum width=3em,
outer sep=0pt},
row 1/.style={bottomrule},column 1/.style=rightrule] {
{\hphantom{E_{n^k_{\rm b}}}\llap{0}} & 0 \\
E_{n^k_{\rm b}} & 0\vphantom{E_{n^k_{\rm b}}} \\
\draw[decorate,transform canvas={xshift=1.5em},thick] (m-1-2.north east) -- node[right=2pt] {$n^k_{\rm c}$} (m-1-2.south east);
\draw[decorate,transform canvas={xshift=1.5em},thick] (m-2-2.north east) -- node[right=2pt] {$n^k_{\rm b}$} (m-2-2.south east);
\draw[decorate,transform canvas={yshift=0.5em},thick] (m-1-1.north west) -- node[above=2pt] {$n^{k+1}_{\rm c}$} (m-1-1.north east);
\draw[decorate,transform canvas={yshift=0.5em},thick] (m-1-2.north west) -- node[above=2pt] {\rlap{$n^{k+1}_{\rm b}$}\hphantom{$n$}} (m-1-2.north east);

\end{tikzpicture}
\end{equation*}

Here $E_i$ is the $i\times i$-identity matrix for $i∈\Z_{\geq 1}$.

So for $x∈S^{λ^k}$, $ρ∈Γ^{\Z}$ we have

\begin{align*}
f_k(x)\cdot ρ \eqs & x\cdot M_{f_k} \cdot ρ^{λ^{k+1}}
x\cdot \left(\begin{array}{c|c} 0 & 0 \\ \hline E_{n^k_{\rm b}}& 0\end{array}\right)
\cdot \left(\begin{array}{c|c} ρ^{λ^{k+1}}_{\rm cc} & ρ^{λ^{k+1}}_{\rm bc} \\ \hline ρ^{λ^{k+1}}_{\rm cb} \ru{5} & ρ^{λ^{k+1}}_{\rm bb} \end{array}\right)
x\cdot  \left(\begin{array}{c|c} 0&0  \\ \hline \ru{5} ρ^{λ^{k+1}}_{\rm cc} & ρ^{λ^{k+1}}_{\rm bc} \end{array}\right)\\
f_k(x\cdot ρ) \eqs & x  \cdot ρ^{λ^{k}} \cdot M_{f_k} =
x\cdot \left(\begin{array}{c|c} ρ^{λ^k}_{\rm cc} & ρ^{λ^k}_{\rm bc} \\ \hline ρ^{λ^k}_{\rm cb} \ru{5} & ρ^{λ^k}_{\rm bb} \end{array}\right)
\cdot \left(\begin{array}{c|c} 0 & 0 \\ \hline  E_{n^k_{\rm b}}& 0\end{array}\right)
= x\cdot \left(\begin{array}{c|c} ρ^{λ^k}_{\rm bc} & 0 \\ \hline \ru{5} ρ^{λ^k}_{\rm bb} & 0 \end{array}\right).
\end{align*}

This way we have $f_k(x\cdot ρ)\equiv_n f_k(x)\cdot ρ$ for all $x∈S^{λ^k}$ if and only if , $ρ_{\rm bc}^{λ^k} \equiv_n 0$ and $ρ_{\rm bc}^{λ^{k+1}} \equiv_n 0$. So

\begin{align}
\label{lambdazex}
Λ^{\Z} = \{ρ∈Γ^{\Z} \mid (ρ^{λ^k}_{\rm bb} \equiv_n  ρ^{λ^{k+1}}_{\rm cc}\text{ for } k∈[1,n-1])\text{ and } (ρ^{λ^k}_{\rm bc} \equiv_n 0\text{ for } k∈[1,n])\}.
\end{align}

We have (cf. e.g. <cit.>)

\[|Γ^{\Z}/Λ^{\Z}|=n^{\frac{1}{2}\sum_{k∈[1,n]} \binom{n-1}{k-1}^2},\]

which is proven by counting the congruences in  (<ref>):

\begin{align*}
|Γ^{\Z}/Λ^{\Z}| \eqs& n^{\sum_{k=1}^{n-1} (n_{\rm b}^k)^2 + \sum_{k=1}^{n} n_{\rm b}^k\cdot n_{\rm c}^k}\\
\ovs{n_{\rm b}^n=0}& n^{\sum_{k∈[1,n]} ((n_{\rm b}^k)^2 + n_{\rm b}^k\cdot n_{\rm c}^k)} = n^{\sum_{k∈[1,n]} n_{\rm b}^k(n_{\rm b}^k + n_{\rm c}^k)} \\
\sum_{k∈[1,n]} n_{\rm b}^k(n_{\rm c}^k + n_{\rm b}^k)\eqs& \sum_{k∈[1,n]} \tbinom{n-2}{k-1}\left(\tbinom{n-2}{k-2}+\tbinom{n-2}{k-1}\right)\\
\eqs& \frac{1}{2}\sum_{k∈[1,n]} \left(\tbinom{n-2}{k-1}\tbinom{n-2}{k-2}+\tbinom{n-2}{k-1}^2\right)
+ \frac{1}{2}\sum_{k∈[1,n]} \left(\tbinom{n-2}{k-1}\tbinom{n-2}{k-2}+\tbinom{n-2}{k-2}^2\right)\\
\eqs& \frac{1}{2}\sum_{k∈[1,n]} \tbinom{n-2}{k-1}\left(\tbinom{n-2}{k-2}+\tbinom{n-2}{k-1}\right)
\eqs&\frac{1}{2}\sum_{k∈[1,n]} \tbinom{n-2}{k-1}\tbinom{n-1}{k-1} +\tbinom{n-2}{k-2}\tbinom{n-1}{k-1} =\frac{1}{2}\sum_{k∈[1,n]} \tbinom{n-1}{k-1}^2\\
\end{align*}

Recall that $p\geq 3$ is a prime. Let $n=p$.
We have the commutative diagram of $\Z$-modules

\begin{align}
\begin{aligned}
\label{diagrz}
\xyprot{
\xymatrix@!C{
\Z \Sp\ar@{^{(}->}[d]^{r^{\Z}}\ar@{^{(}->}[r]^{ι^{\Z}\circ r^{\Z}} & Γ^{\Z}\ar@{->>}[r] \ar[d]^{\rotatebox{90}{=}} & Γ^{\Z}/(ι^{\Z}\circ r^{\Z}(\Z \Sp)) \ar@{->>}[d]^{s^{\Z}} \\
Λ^{\Z} \ar@{^{(}->}[r]^{ι^{\Z}} & Γ^{\Z}  \ar@{->>}[r] &  Γ^{\Z}/Λ^{\Z}
\end{aligned}
\end{align}

The map $ι^{\Z}$ is the inclusion of $Λ^{\Z}$ in $Γ^{\Z}$.
The maps from $Γ^{\Z}$ to $Γ^{\Z}/(ι^{\Z}\circ r^{\Z}(\Z \Sp))$ and to $Γ^{\Z}/Λ^{\Z}$ are the residue class maps. As $r^{\Z}(\Z \Sp)\subseteq Λ^{\Z}$, we have an unique surjective map $s^{\Z}:Γ^{\Z}/(ι^{\Z}\circ r^{\Z}(\Z \Sp))→ Γ^{\Z}/Λ^{\Z}$ such that the right rectangle is commutative. By construction, the rows of the diagram are short exact sequences.
Note that the morphisms of the left rectangle are in fact $\Z$-algebra morphisms.

We will need the following result on the localization of rings.

Let $A$ be a commutative ring. Let $P\subseteq R$ a prime ideal of $A$. Let $A_P$ be the localization of $A$ at $P$. Then
$A_P$ is a flat $A$-module, that is, the functor $\funcblc\underset{A}{\otimes} \leftidx{_A}{(A_P)}{_{A_P}}$ from the category of $A$-modules to the category of $A_P$-modules is exact.

We denote by $\zp$ the localization of $\Z$ at the prime ideal $(p):=p\Z$. We apply the functor $\funcblc\underset{\Z}{\otimes } \zp$ to obtain a commutative diagram (<ref>) of the following form:

\begin{align}
\label{diagr}
\begin{aligned}
\xyprot{
\xymatrix@!C{
\zp\!\Sp\ar@{^{(}->}[d]^{r}\ar@{^{(}->}[r]^{ι \circ r} & Γ\ar@{->>}[r] \ar[d]^{\rotatebox{90}{=}} & Γ/(ι\circ r(\zp\!\Sp)) \ar@{->>}[d]^{s} \\
Λ \ar@{^{(}->}[r]^{ι} & Γ  \ar@{->>}[r] &  Γ/Λ
\end{aligned}
\end{align}

By <ref>, the functor $\funcblc\underset{\Z}{\otimes } \zp$ is exact, so the short exact sequences are mapped to short exact sequences, monomorphisms to monomorphisms and epimorphisms to epimorphisms. So the rows of diagram (<ref>) are exact and we have mono-/epimorphism as indicated by the arrows. We identify $\Z \Sp \underset{\Z}{\otimes } \zp$ with $\zp\!\Sp$. We identify $Γ^{\Z}\underset{\Z}{\otimes } \zp$ with

\begin{align*}
Γ := \prod_{λ\dashv n} \zp^{n_λ\times n_λ}.
\end{align*}

The map $ι$ realizes $Λ:=Λ^{\Z}\underset{\Z}{\otimes } \zp$ as the following subset of $Γ$, for which we will use notation analogous to (<ref>):

\begin{align*}
Λ = \{ρ∈Γ \mid (ρ^{λ^k}_{\rm bb} \equiv_p  ρ^{λ^{k+1}}_{\rm cc}\text{ for } k∈[1,p-1])\text{ and } (ρ^{λ^k}_{\rm bc} \equiv_p 0 \text{ for } k∈[1,p])\}
\end{align*}

As the rows are exact, we identify $(Γ^{\Z}/(ι^{\Z}\circ r^{\Z}(\Z \Sp)) )\underset{\Z}{\otimes } \zp$ with $Γ/(ι\circ r(\zp\!\Sp)$ and $(Γ^{\Z}/Λ^{\Z})\underset{\Z}{\otimes } \zp$ with $Γ/Λ$.

By the classification of finitely generated $\Z$-modules, each finite $\Z$-module $M$ is isomorphic to a finite direct sum of modules of the form $\Z/q^a \Z$, where $q$ is a prime and $a\in \Z_{\geq 0}$. If $q\neq p$ then $(\Z/q^a \Z) \underset{\Z}{\otimes } \zp \cong (0)$. Otherwise $(\Z/p^a \Z) \underset{\Z}{\otimes } \zp \cong \zp/p^a\zp$ and $|(\Z/p^a \Z) \underset{\Z}{\otimes } \zp| = p^a = |\Z/p^a\Z|$.
For $x=p^{a_p} \cdot \prod_{\natop{q \text{ prime}}{q\neq p}} q^{a_q} ∈\Z_{\geq 1}$, we set

\[(x)_p := p^{a_p}.\]

So for finite $M$, we have $|M\underset{\Z}{\otimes } \zp| = (|M|)_p$.

By the total index formula (cf. e.g. <cit.>), we have

\begin{align*}
|Γ^{\Z}/(ι^{\Z}\circ r^{\Z}(\Z \Sp))| = \sqrt{\frac{p!^{p!}}{\prod_{λ \dashv p} n_λ^{n_λ^2}}}\,.
\end{align*}

By the hook formula (cf. <cit.>, <cit.>), we have for $λ\dashv p$

\begin{align*}
(n_λ)_p = \begin{cases} 1 & \text{if $λ$ is a hook-partition} \\ p & \text{ otherwise} \end{cases}\,.
\end{align*}

\begin{align*}
|Γ/(i\circ r(\zp\!\Sp))| \eqs& \left(\sqrt{\frac{p!^{p!}}{\prod_{λ \dashv p} n_λ^{n_λ^2}}}\right)_{\mathllap{p}}
= \sqrt{\frac{p^{p!}}{\prod_{\natop{λ \dashv p}{λ\text{ not a hook}}} (n_λ)_p^{n_λ^2}}}\\
\eqs& \sqrt{\frac{\prod_{λ\dashv p} p^{n_λ^2}}{\prod_{\natop{λ \dashv p}{λ\text{ not a hook}}} p^{n_λ^2}}}
= \sqrt{\prod_{k∈[1,n]} p^{n_{λ^k}^2}} = p^{\frac{1}{2}\sum_{k∈[1,n]}\binom{p-1}{k-1}^2}\\
\eqs& |Γ^{\Z}/Λ^{\Z}| = (|Γ^{\Z}/Λ^{\Z}|)_p=  |Γ/Λ|.
\end{align*}

By the pigeon-hole-principle, $s$ is an isomorphism as it is surjective. As (<ref>) has exact rows, $r$ needs to be an isomorphism as well. Note that the functor $\funcblc\underset{\Z}{\otimes } \zp$ transforms morphisms of $\Z$-algebras into morphisms of $\zp$-algebras. In particular, the left rectangle in (<ref>) consists of morphisms of $\zp$-algebras and $r:\zp\!\Sp→Λ$ is an isomorphism of $\zp$-algebras. We have proven the

Recall that $p\geq 3$ is a prime.

For $R$ a ring and $λ\dashv p$ a partition of $p$, the $R\!\Sp$-Specht module $S^λ$ is finitely generatey free over $R$ with dimension independent of $R$, cf. <cit.>. We denote this dimension by $n_λ$.

A partition of the form $λ^k := (p-k+1,1^{k-1})$, $k∈[1,p]$ is called a hook partition of $p$.

Over the valuation ring $\zp$, there is a well-known description of the group algebra $\zp\!\Sp$,  cf. e.g. <cit.> (using [19]), cf. also <cit.>:

$n^{k}_{\rm b} = \binom{p-2}{k-1}$, $n^{k}_{\rm c} = \binom{p-2}{k-2}$. Then $n^{k}_{\rm b} + n^{k}_{\rm c} = \binom{p-1}{k-1} = n_{λ^k}$. Set

$ Γ := \prod_{λ \dashv p}  \zp^{n_λ\times n_λ} $.

For $ρ∈Γ$, and $λ\dashv p$, we denote by $ρ^λ$ the $λ$-th component of $ρ$. For $λ=λ^k$, $k∈[1,p]$, a hook partition, we name certain subblocks of $ρ^{λ^k}$ as follows.

\begin{equation*}
ρ^{λ^k} =
\xyprot{
\begin{tikzpicture}[decoration=brace,baseline]
\matrix (m) [matrix of math nodes,left delimiter=(,right delimiter=),
nodes={%rectangle,minimum width=3em,
outer sep=0pt},
row 1/.style={bottomrule},column 1/.style=rightrule] {
ρ^{λ^k}_{\rm cc} & ρ^{λ^k}_{\rm bc} \\
ρ^{λ^k}_{\rm cb} & ρ^{λ^k}_{\rm bb} \\
\draw[decorate,transform canvas={xshift=1.5em},thick] (m-1-2.north east) -- node[right=2pt] {$n^k_{\rm c}$ rows} (m-1-2.south east);
\draw[decorate,transform canvas={xshift=1.5em},thick] (m-2-2.north east) -- node[right=2pt] (ru) {$n^k_{\rm b}$ rows} (m-2-2.south east);
\draw[decorate,transform canvas={yshift=0.5em},thick] (m-1-1.north west) -- node[above=2pt] {$n^k_{\rm c}$,} (m-1-1.north east);
\draw[decorate,transform canvas={yshift=0.5em},thick] (m-1-2.north west) -- node[above=2pt] {$n^k_{\rm b}$ \rlap{columns}} (m-1-2.north east);

\node[transform canvas={xshift=1.5em}] at (ru.south east) {.};
\end{tikzpicture}
\end{equation*}

We have the following $\zp$-subalgebra $Λ$ of $Γ$.

\begin{align*}
% \end{align*}
% given by
% \begin{align*}
Λ &:= \{ρ∈Γ \mid  ρ^{λ^k}_{\rm bb} \equiv_p ρ^{λ^{k+1}}_{\rm cc}\text{ for }k∈[1,p-1]\text{ and } ρ^{λ^k}_{\rm bc} \equiv_p 0\text{ for }k∈[1,p]\}
\end{align*}

Now there is an isomorphism of $\zp$-algebras

\[\begin{array}{rccc}
r:&\zp\!\Sp & \xrightarrow{\phantom{X}\sim\phantom{X}}& Λ.
\end{array}
\]

such that $ρ∈Λ$ acts on the trivial $\zp\!\Sp$-module $\zp$ by multiplication with its ($1\times 1$ / scalar-)component $ρ^{λ^1}$, i.e. for $x∈\zp\!\Sp$ and for $y∈\zp$, we have $yx = y\cdot r(x)^{λ^1}$.

FIXME: Dieses Bsp. rausmachen.
For $p=3$, the ring $\Z_{(3)}\Sy_3$ is isomorphic to the subring $Λ$ of $Γ=\Z_{(3)}^{1\times 1} \times \Z_{(3)}^{2\times 2} \times \Z_{(3)}^{1\times 1}$
described as

[inner xsep=0.1mm, inner ysep=0.0mm,
space/.style=minimum height=1.0em,matrix of nodes,row sep=0.0mm,column sep=1mm,

(m1) [space, below delimiter=}0$\begin{matrix*}[l]\scriptstyle{\times}\\ \scriptstyle\times \\  \scriptstyle{\times}\end{matrix*}$]

(m2) [right=of m1, space, below delimiter=}0$\begin{matrix*}[l]\scriptstyle\times \\  \scriptstyle{\times}{\times}\end{matrix*}$]

_(3)     3_(3)

_(3)     _(3)

(m3) [right=of m2,space, below delimiter=}0$\begin{matrix*}[l] \scriptstyle{\times}{\times}{\times}\end{matrix*}$]

[draw=black, fit=(m1-1-1)](B12) ;
[draw=black, fit=(m2-1-1)](B21) ;
[draw=black, fit=(m2-2-2)](B22) ;
[draw=black, fit=(m3-1-1) ](B31) ;
[black] (B12.east) –  node[sloped,above]0.5mm$\scriptscriptstyle 3$ (B21.west);
[black] (B22.east) –  node[sloped,above]0.5mm$\scriptscriptstyle 3$ (B31.west);

An entry in this tuple of matrices indicates that an element of $Λ$ must have its corresponding entry in the indicated set. A relation "  (n1) ; (n2) [right=of n1] ;
[black] (n1.east) –  node[sloped,above]$\scriptscriptstyle p$ (n2.west);
" between (equal sized) subblocks indicates that these subblocks  are equivalent modulo $p$, i.e. the difference of corresponding entries is an element of $p\Z_{(p)}$. The blocks are labeled with the diagrams of the corresponding partitions.
Alternatively, $Λ$ is the $\Z_{(3)}$-span of

\begin{align*}
\left( 3, \begin{pmatrix} 0 & 0 \\ 0& 0\end{pmatrix}, 0 \right)=:u, &&
\left( 1, \begin{pmatrix} 1 & 0 \\ 0& 0\end{pmatrix}, 0 \right)=:\tilde e_1, &&
\left( 0, \begin{pmatrix} 0 & 3 \\ 0& 0\end{pmatrix}, 0 \right)=: v_1,\\
\left( 0, \begin{pmatrix} 0 & 0 \\ 1& 0\end{pmatrix}, 0 \right)=: v_2,&&
\left( 0, \begin{pmatrix} 0 & 0 \\ 0& 1\end{pmatrix}, 1 \right)=:\tilde e_2, &&
\left( 0, \begin{pmatrix} 0 & 0 \\ 0& 3\end{pmatrix}, 0\right)=: w.
\end{align*}

We have an orthogonal decomposition $1=\tilde e_1 + \tilde e_2$ into primitive idempotents. Thus we have a decomposition $Λ=\tilde P_1 \oplus \tilde P_2$
into indecomposable projective right modules, where

\begin{align*}
\tilde P_1 :\eqs& \tilde e_1 Λ = \langle u, \tilde e_1, v_1\rangle_{\Z_{(3)}}, &
\tilde P_2 :\eqs& \tilde e_2 Λ = \langle v_2, \tilde e_2, w \rangle_{\Z_{(3)}}.
\end{align*}

A more pictoral description of $\tilde P_i$, $i∈[1,2]$ is that the elements of $\tilde P_i⊂Λ$ are exactly the elements of $Λ$ whose non-zero entries are all in the box in FIXME.

In this case all partitions of $3$ are of hook-type. Thus there appear no full matrix algebras as direct factors of $Λ$.

For $p=5$, the $\zp$-algebra $\Z_{(5)}\Sy_5$ is isomorphic to the subalgebra $Λ$ of $Γ=\Z_{(5)}^{1\times 1}\times \Z_{(5)}^{4\times 4}\times \Z_{(5)}^{6\times 6}\times \Z_{(5)}^{4\times 4}\times \Z_{(5)}^{1\times 1}\times  \Z_{(5)}^{5\times 5}\times \Z_{(5)}^{5\times 5}$ described as

[inner xsep=0.1mm, inner ysep=0.0mm,node distance=4mm,

space/.style=minimum height=1.0em,matrix of nodes,row sep=1.0mm,column sep=1mm,

(m1) [space, below delimiter=}0$\begin{matrix*}[l]\scriptstyle{\times}\\ \scriptstyle\times \\ \scriptstyle\times  \\ \scriptstyle\times \\ \scriptstyle{\times}\end{matrix*}$]

(m2) [right=of m1, space, below delimiter=}0$\begin{matrix*}[l]\scriptstyle\times \\ \scriptstyle\times  \\ \scriptstyle\times \\ \scriptstyle{\times}{\times}\end{matrix*}$]

_(5)     5_(5)     5_(5)     5_(5)

_(5)     _(5)     _(5)      _(5)

_(5)     _(5)     _(5)     _(5)

_(5)     _(5)     _(5)     _(5)

(m3) [right=of m2, space, below delimiter=}0$\begin{matrix*}[l]\scriptstyle\times  \\ \scriptstyle\times \\ \scriptstyle{\times}{\times}{\times}\end{matrix*}$]

_(5)     _(5)     _(5)     5_(5)     5_(5)     5_(5)

_(5)     _(5)     _(5)     5_(5)     5_(5)     5_(5)

_(5)     _(5)     _(5)     5_(5)     5_(5)     5_(5)

_(5)     _(5)     _(5)     _(5)     _(5)     _(5)

_(5)     _(5)     _(5)     _(5)     _(5)     _(5)

_(5)     _(5)     _(5)     _(5)     _(5)     _(5)

(m4) [right=of m3, space, below delimiter=}0$\begin{matrix*}[l]\scriptstyle\times  \\  \scriptstyle{\times}{\times}{\times}{\times}\end{matrix*}$]

_(5)     _(5)     _(5)     5_(5)

_(5)     _(5)     _(5)     5_(5)

_(5)     _(5)     _(5)     5_(5)

_(5)     _(5)     _(5)     _(5)

(m5) [right=of m4,space, below delimiter=}0$\begin{matrix*}[l] \scriptstyle{\times}{\times}{\times}{\times}{\times}\end{matrix*}$]

[draw=black, fit=(m1-1-1)](B12) ;
[draw=black, fit=(m2-1-1)](B21) ;
[draw=black, fit=(m2-2-2.north west) (m2-4-4.south east)](B22) ;
[draw=black, fit=(m3-1-1.north west) (m3-3-3.south east)](B31) ;
[draw=black, fit=(m3-4-4.north west) (m3-6-6.south east)](B32) ;
[draw=black, fit=(m4-1-1.north west) (m4-3-3.south east)](B41) ;
[draw=black, fit=(m4-4-4)] (B42) ;
[draw=black, fit=(m5-1-1)] (B51) ;
[black] (B12.east) –  node[sloped,above,pos=0.6]0.5mm$\scriptscriptstyle 5$ (B21.west);
[black] (B22.east) –  node[sloped,above,pos=0.6]0.5mm$\scriptscriptstyle 5$ (B31.west);
[black] (B32.east) –  node[sloped,above,pos=0.6]0.5mm$\scriptscriptstyle 5$ (B41.west);
[black] (B42.east) –  node[sloped,above,pos=0.6]0.5mm$\scriptscriptstyle 5$ (B51.west);

[left=] (C011) at (m1-1-1.south west) ;
[below=] (C11) at (C011) ;
[below=] (C12) at (m1-1-1.south east) ;
[below=] (C13) at (m2-1-1.south west) ;
[below=] (C14) at (m2-1-4.south east) ;
[above=] (C15) at (m2-1-4.north east) ;
[above=] (C16) at (m2-1-1.north west) ;
[above=] (C17) at (m1-1-1.north east) ;
[left=] (C018) at (m1-1-1.north west) ;
[above=] (C18) at (C018) ;
[red,rounded corners=2pt] (C11.center) – (C12.center) – (C13.center) – (C14.center) – (C15.center) –
node[above]1.5mm$\tilde P_1$ (C16.center)  – (C17.center) – (C18.center) – cycle;

[below=] (C21) at (m2-2-1.south west) ;
[below=] (C22) at (m2-2-4.south east) ;
[below=] (C23) at (m3-1-1.south west) ;
[below=] (C24) at (m3-1-6.south east) ;
[above=] (C25) at (m3-1-6.north east) ;
[above=] (C26) at (m3-1-1.north west) ;
[above=] (C27) at (m2-2-4.north east) ;
[above=] (C28) at (m2-2-1.north west) ;
[red,rounded corners=2pt] (C21.center) – (C22.center) – (C23.center) – (C24.center) – (C25.center) – node[above]1.5mm$\tilde P_2$ (C26.center)  – (C27.center) – (C28.center) – cycle;

[below=] (C31) at (m3-4-1.south west) ;
[below=] (C32) at (m3-4-6.south east) ;
[below=] (C33) at (m4-1-1.south west) ;
[below=] (C34) at (m4-1-4.south east) ;
[above=] (C35) at (m4-1-4.north east) ;
[above=] (C36) at (m4-1-1.north west) ;
[above=] (C37) at (m3-4-6.north east) ;
[above=] (C38) at (m3-4-1.north west) ;
[red,rounded corners=2pt] (C31.center) – (C32.center) – (C33.center) – (C34.center) – (C35.center) – node[above]1.5mm$\tilde P_3$ (C36.center)  – (C37.center) – (C38.center) – cycle;

[below=] (C41) at (m4-4-1.south west) ;
[below=] (C42) at (m4-4-4.south east) ;
[below=] (C43) at (m5-1-1.south west) ;
[right=](C044) at (m5-1-1.south east) ;
[below=] (C44) at (C044) ;
[right=](C045) at (m5-1-1.north east) ;
[above=] (C45) at (C045) ;
[above=] (C46) at (m5-1-1.north west) ;
[above=] (C47) at (m4-4-4.north east) ;
[above=] (C48) at (m4-4-1.north west) ;
[red,rounded corners=2pt] (C41.center) – (C42.center) – (C43.center) – (C44.center) – (C45.center) –
node[above]1.5mm$\tilde P_4$ (C46.center)  – (C47.center) – (C48.center) – cycle;

[inner xsep=0.1mm, inner ysep=0.0mm,node distance=4mm,
space/.style=minimum height=1.0em,matrix of nodes,row sep=0.0mm,column sep=1mm,

(mx) [space, below delimiter=}0$\begin{matrix*}[l] \scriptstyle{\times}{\times} \\ \scriptstyle{\times}{\times}{\times}  \end{matrix*}$]

_(5)     _(5)     _(5)     _(5)     _(5)

_(5)     _(5)     _(5)     _(5)     _(5)

_(5)     _(5)     _(5)     _(5)     _(5)

_(5)     _(5)     _(5)     _(5)     _(5)

_(5)     _(5)     _(5)     _(5)     _(5)

(my) [space,right=of mx, below delimiter=}0$\begin{matrix*}[l]\scriptstyle{\times}\\ \scriptstyle{\times}{\times} \\ \scriptstyle{\times}{\times}  \end{matrix*}$]

_(5)     _(5)     _(5)     _(5)     _(5)

_(5)     _(5)     _(5)     _(5)     _(5)

_(5)     _(5)     _(5)     _(5)     _(5)

_(5)     _(5)     _(5)     _(5)     _(5)

_(5)     _(5)     _(5)     _(5)     _(5)

An entry in this tuple of matrices indicates that an element of $Λ$ must have its corresponding entry in the indicated set. A relation "  (n1) ; (n2) [right=of n1] ;
[black] (n1.east) –  node[sloped,above]$\scriptscriptstyle 5$ (n2.west);
" between (equal sized) subblocks indicates that these subblocks  are equivalent modulo $5$, i.e. the difference of corresponding entries is an element of $5\Z_{(5)}$. The blocks are labeled with the diagrams of the corresponding partitions.

The right ideals $\tilde P_i = \tilde e_iΛ$, $i∈[1,4]=[1,p-1]$ (cf. the definitions below) are framed with red lines.

§.§ A projective resolution of Zp over ZpSp

Recall that $p\geq 3$ is a prime.

Recall from <ref> that $Λ$ is a subring of $Γ=\prod_{λ\dashv p} \zp^{n_λ\times n_λ}$.

For $λ\dashv p$ and $i,j∈[1,n_λ]$, we set $η_{λ,i,j}$ to be the element of $Γ$ such that $(η_{λ,i,j})^{\tilde{λ}} = 0$ for $\tilde{λ}\neq λ$ and $(η_{λ,i,j})^{λ}∈\Z^{n_λ\times n_λ}$ has entry $1$ at position $(i,j)$ and zeros elsewhere.

Let $k∈[1,p-1]$. We obtain the idempotent

\begin{align*}
\tilde e_k :=  η_{λ^k,n_{\rm c}^{k}+1,n_{\rm c}^{k}+1} + η_{λ^{k+1},1,1} ∈ Λ.
\end{align*}

We define corresponding projective right $Λ$-modules

\begin{align*}
\tilde P_k :\eqs& \tilde e_k Λ \quad \text{ for $k∈[1,p-1]$.}
\end{align*}

Let $A$ be an $R$-algebra and let $e,e'∈A$ be two idempotents. For the right modules $eA$, $e'A$, we have the isomorphism of $R$-Modules

\[ \begin{array}{rcl}
\Hom_A(eA, e'A) &\underset{\sim}{\overset{T_{e',e}}{\longrightarrow}}& e'Ae\\
f &\longmapsto& T_{e',e}(f) := f(e)\\
T_{e',e}^{-1}(e'be):=(ea \mapsto e'bea) &\reflectbox{$\longmapsto$}&
e'be \phantom{:= f(e)X}.
\end{array}\]

Thus given $m∈e'Ae$, the morphism $T_{e',e}^{-1}(m)$ acts on elements $x∈eA$ by the multiplication of $m$  on the left: $(T_{e',e}^{-1}(m))(x) = m\cdot x$.

Given idempotents $e,e',e''∈A$, and elements $f∈\Hom_A(eA,e'A)$, $g∈A(e'A,e''A)$, we have $T_{e'',e}(g\circ f) = g(f(e)) = g(e'f(e)) = g(e')\cdot f(e)= T_{e'',e'}(g)\cdot T_{e',e}(f)$.

We define via <ref>

\begin{equation*}
\begin{array}{lclll}
\hat e_{k}   &:\eqs& T_{\tilde e_k, \tilde e_k}^{-1}(\tilde e_k)				&∈\Hom_Λ(\tilde P_k,\tilde P_k) & \text{for }k∈[1,p-1]\\
\hat e_{1,1}    &:\eqs& T_{\tilde e_1,\tilde e_1}^{-1}(pη_{λ^1,1,1})			&∈\Hom_Λ(\tilde P_1,\tilde P_1) \\
\hat e_{p-1,p-1}&:\eqs& T_{\tilde e_{p-1},\tilde e_{p-1}}^{-1}(pη_{λ^p,1,1})			&∈\Hom_Λ(\tilde P_{p-1},\tilde P_{p-1})\\
\hat e_{k+1,k}  &:\eqs& T_{\tilde e_{k+1},\tilde e_k}^{-1}(η_{λ^{k+1},n_{\rm c}^{k+1}+1,1})	&∈\Hom_Λ(\tilde P_k,\tilde P_{k+1}) & \text{for }k∈[1,p-2]\\
\hat e_{k,k+1}  &:\eqs& T_{\tilde e_k,\tilde e_{k+1}}^{-1}(pη_{λ^{k+1},1,n_{\rm c}^{k+1}+1})	&∈\Hom_Λ(\tilde P_{k+1},\tilde P_k) & \text{for }k∈[1,p-2].
\end{array}
\end{equation*}

Note that $\hat e_k$ is the identity map on $\tilde P_k$ for $k∈[1,p-1]$.

Moreover, we define the $\zp\!\Sp$-linear map $\hat{ε}:\tilde P_1 → \zp$, $\hat{ε}(ρ) := ρ^{λ^1}$.

It is straightforward to show the following lemma.

We have

\begin{equation*}
\begin{array}{lclr}
\hat e_{1,1} + \hat e_{1,2}\circ \hat e_{2,1} &=& p\hat e_1\\
\hat e_{k,k-1}\circ\hat e_{k-1,k} + \hat e_{k,k+1}\circ\hat e_{k+1,k} &=& p\hat e_k &\text{ for }k∈[2,p-2]\\
\hat e_{p-1,p-2}\circ\hat e_{p-2,p-1} + \hat e_{p-1,p-1} &=& p\hat e_{p-1}\\
\hat{ε}\circ \hat e_{1,1} &=& p\hat{ε}.
\end{array}
\end{equation*}

Furthermore, it is straightforward to check that we obtain a projective resolution of $\zp$ as follows.
We set

\begin{align*}
\tilde{\pr}_i :\eqs& \begin{cases} \tilde P_{ω(i)} & i\geq 0 \\ 0 & i<0 \end{cases},
\end{align*}

where the integer $ω(i)$ is given by the following construction: Recall the stipulation $l:=2(p-1)$. We have $i = jl+r$ for some $j∈\Z$ and $0\leq r\leq l-1$. Then

\begin{align}
\label{omega}
ω(i) := \begin{cases} r+1 & \text{ for }0\leq r \leq p-2\\ l-r = 2(p-1) -r & \text{ for } p-1 \leq r \leq 2(p-1)-1 = l-1 \end{cases}.
\end{align}

So $ω(i)$ increases by steps of one from $1$ to $p-1$ as $i$ runs from $jl$ to $jl+(p-2)$ and $ω(i)$ decreases from $p-1$ to $1$ as $i$ runs from $jl+(p-1)$ to $jl + (l-1)$. Finally we set

\begin{align*}
\hat d_i := \begin{cases} \hat e_{ω(i-1),ω(i)}: \tilde P_{ω(i)} → \tilde P_{ω(i-1)} & i\geq 1\\ 0 & i\leq 0 \end{cases}.
\end{align*}

Now we have the projective resolution of $\zp$

\begin{align}
\label{presform1}
\cdots\xrightarrow{\hat d_3} \tilde{\pr}_2\xrightarrow{\hat d_2}\tilde{\pr}_1\xrightarrow{\hat d_1}\tilde{\pr}_0\xrightarrow{0=\hat d_0} 0\rightarrow\cdots\,,
\end{align}

written more explicitly as

\begin{align*}
\cdots\rightarrow &\tilde P_2\xrightarrow{\hat e_{1,2}} \tilde P_1 \xrightarrow{\hat e_{1,1}} \tilde P_1 \xrightarrow{\hat e_{2,1}} \tilde P_2 \rightarrow …
\rightarrow \tilde P_{p-2} \xrightarrow{\hat e_{p-1,p-2}}\tilde P_{p-1} \nonumber \\
&\xrightarrow{\hat e_{p-1,p-1}} \tilde P_{p-1} \xrightarrow{\hat e_{p-2,p-1}}\tilde P_{p-2}\rightarrow … \rightarrow \tilde P_2\xrightarrow{\hat e_{1,2}} \tilde P_1  \rightarrow 0\rightarrow \cdots\,,
\end{align*}

with augmentation $\hat{ε}:\tilde P_1 → \zp$.

Then let

* $\mathscr{B}^{\Leftrightarrow}:=\{β_{k,x,y}^{\Leftrightarrow} \mid k∈[1,p-1], x,y∈[1,n_{\rm b}^{k}]\}$, where  $β_{k,x,y}^{\Leftrightarrow} := η_{λ^k,n_{\rm c}^{k}+x,n_{\rm c}^{k}+y} + η_{λ^{k+1},x,y}$.

* $\mathscr{B}^{\Leftarrow}:=\{β_{k,x,y}^{\Leftarrow} \mid k∈[1,p-1], x,y∈[1,n_{\rm b}^{k}]\}$, where .

* $\mathscr{B}^{\Rightarrow}:=\{β_{k,x,y}^{\Rightarrow} \mid k∈[1,p-1], x,y∈[1,n_{\rm b}^{k}]\}$, where .

* $\mathscr{B}^{\leftarrow} := \{β_{k,x,y}^{\leftarrow} \mid k∈[1,p], x∈[1,n_{\rm b}^{k}],y∈[1,n_{\rm c}^{k}]\}$,

* $\mathscr{B}^{\rightarrow} := \{β_{k,x,y}^{\rightarrow} \mid k∈[1,p], x∈[1,n_{\rm c}^{k}],y∈[1,n_{\rm b}^{k}]\}$,
where  .

* $\mathscr{B}^*:=\{η_{λ,x,y} \mid λ\dashv p\text{ not a hook partition}, x,y∈[1,n_λ]\}$.

We have two $\zp$-bases $\mathscr{B}^{\Leftrightarrow} \sqcup \mathscr{B}^{\Leftarrow} \sqcup \mathscr{B}^{\leftarrow} \sqcup \mathscr{B}^{\rightarrow} \sqcup \mathscr{B}^*$ and   of $Λ$.

[$p=3$, continuation of <ref>]
The only of the $β_{a,b,c}^d$ that are defined above and that are not shown in <ref> are the following elements.

\begin{align*}
\left( 0, \begin{pmatrix} 3 & 0 \\ 0& 0\end{pmatrix}, 0 \right)= β_{1,1,1}^⇒, &&
\left( 0, \begin{pmatrix} 0 & 0 \\ 0& 0\end{pmatrix}, 3 \right)= β_{2,1,1}^⇒
\end{align*}

$\mathscr{B}^*$ is empty since all partitions are hook partitions.

Once more, see <ref> for an illustration of the case $p=3$.

* $\mathscr{B}_k^{\Leftrightarrow} := (β_{k,1,y}^{\Leftrightarrow}\colon y∈(1,…,n_{\rm b}^{k}))
= (η_{λ^k,n_{\rm c}^{k}+1,n_{\rm c}^{k}+y} + η_{λ^{k+1},1,y}\colon y∈(1,…,n_{\rm b}^{k}))$

* $\mathscr{B}_k^{\Leftarrow} := (β_{k,1,y}^{\Leftarrow}\colon y∈(1,…,n_{\rm b}^{k}))
= (pη_{λ^k,n_{\rm c}^{k}+1,n_{\rm c}^{k}+y}\colon y∈(1,…,n_{\rm b}^{k}))$

* $\mathscr{B}_k^{\Rightarrow} := (β_{k,1,y}^{\Rightarrow}\colon y∈(1,…,n_{\rm b}^{k}))
= (pη_{λ^{k+1},1,y} \colon y∈(1,…,n_{\rm b}^{k}))$

* $\mathscr{B}_k^{\leftarrow} := (β_{k,1,y}^{\leftarrow}\colon y∈(1,…,n_{\rm c}^{{k}}))
=(η_{λ^{k},n_{\rm c}^{k}+1,y} \colon y∈(1,…,n_{\rm c}^k))$

* $\mathscr{B}_k^{\rightarrow} := (β_{k+1,1,y}^{\rightarrow}\colon y∈(1,…,n_{\rm b}^{k+1}))
=(pη_{λ^{k+1},1,n_{\rm c}^{k+1}+y}\colon y∈(1,…, n_{\rm b}^{k+1}))$

Similarly to the bases of $Λ$, the tuples $\mathscr{B}_k^{\Leftrightarrow}\sqcup \mathscr{B}_k^{\Leftarrow} \sqcup \mathscr{B}_k^{\leftarrow} \sqcup \mathscr{B}_k^{\rightarrow}$ and  are  $\zp$-bases of  $\tilde P_k$.

Let $k∈[1,p-1]$. The idempotent $\tilde e_k$ is actually a primitive idempotent and thus $\tilde P_k$ is an indecomposable projective $Λ$-right module: Assume $\tilde e_k=c+c'$ for some idempotents $0\neq c,c'∈Λ$ that are orthogonal, that is $c\cdot c' = c'\cdot c = 0$. Then $\tilde e_k \cdot c = (c+c')c = c^2 = c = c(c+c') = c\cdot \tilde e_k$. Similarly, we have $\tilde e_k \cdot c' = c' = c'\cdot \tilde e_k$.  Thus $c,c' ∈\tilde e_kΛ\tilde e_k$. The $\zp$-algebra

\[\tilde e_kΛ\tilde e_k = \langle e_k, β_{k,1,1}^⇐\rangle_{\zp} = \langle e_k, β_{k,1,1}^⇒\rangle_{\zp}\]

is isomorphic to the $\zp$-algebra

\begin{equation*}
\begin{tikzpicture}
\node (n1) {$\zp$};
\node (n2) [right=of n1] {$\zp$};
\draw[black] (n1.east) --  node[above]{$\scriptstyle p$} (n2.west);
\node%[transform canvas={xshift=1.5em}]
at (n1.west) {$\mathllap{J:=}$};
\end{tikzpicture}
\end{equation*}

consisting of elements $\{(a,b)∈\zp\times \zp \mid a\equiv_p b\}$. The only idempotents in $\zp\times \zp$ are $(0,0)∈J$, $(1,1)∈J$, $(1,0)\notin J$ and $(0,1)\notin J$. Thus the identity element $(1,1)$ of $J$ cannot be decomposed into non-trivial idempotents and the same holds for $\tilde e_k$.

We have

\begin{align*}
&{\rm(a)}&&\ker \hat e_{k+1,k} = \langle \mathscr{B}_k^{\leftarrow} \sqcup \mathscr{B}_k^{⇐}\rangle_{\zp},&&  \im \hat e_{k+1,k} = \langle  \mathscr{B}_{k+1}^{\leftarrow} \sqcup \mathscr{B}_{k+1}^{⇐}\rangle_{\zp} \,\, \text{ for $k∈[1,p-2]$},\\
&{\rm(b)}&&\ker \hat e_{k,k+1} = \langle \mathscr{B}_{k+1}^→\sqcup \mathscr{B}_{k+1}^⇒\rangle_{\zp}, &&\im \hat e_{k,k+1} = \langle \mathscr{B}_k^{→}\sqcup \mathscr{B}_k^⇒ \rangle_{\zp}\,\,\text{ for $k∈[1,p-2]$},\\
&{\rm(c)}&&\ker \hat e_{p-1,p-1} {=} \langle  \mathscr{B}_{p-1}^{\leftarrow} \sqcup \mathscr{B}_{p-1}^⇐ \rangle_{\zp}, &&\im \hat e_{p-1,p-1} {=} \langle \mathscr{B}_{p-1}^⇒ \sqcup \mathscr{B}_{p-1}^→\rangle_{\zp},\\
&{\rm(d)}&&\ker \hat e_{1,1} = \langle \mathscr{B}_1^⇒ \sqcup \mathscr{B}_1^→ \rangle_{\zp},&& \im \hat e_{1,1} = \langle \mathscr{B}_1^⇐ \sqcup \mathscr{B}_1^{\leftarrow} \rangle_{\zp}.
\end{align*}

\begin{array}[t]{rcl}
\hat e_{k+1,k}(\mathscr{B}_k^{⇔}) &\svs{\text{R.}\ref{rem:moralg}}& (η_{λ^{k+1},n_{\rm c}^{k+1}+1,1} η_{λ^{k+1},1,y} \colon y∈(1,…,n_{\rm b}^{k}))\\
&\sqs & (η_{λ^{k+1},n_{\rm c}^{{k+1}}+1,y} \colon y∈(1,…,n_{\rm c}^{{k+1}})) = \mathscr{B}_{k+1}^{\leftarrow}\\
\hat e_{k+1,k}(\mathscr{B}_k^{→}) &\svs{\text{R.}\ref{rem:moralg}}& (η_{λ^{k+1},n_{\rm c}^{k+1}+1,1} pη_{λ^{k+1},1,n_{\rm c}^{{k+1}}+y} \colon y∈(1,…,n_{\rm b}^{k+1}))\\
&\sqs& (pη_{λ^{k+1},n_{\rm c}^{k+1}+1,n_{\rm c}^{k+1}+y} \colon y∈(1,…,n_{\rm b}^{k+1})) = \mathscr{B}_{k+1}^{⇐}\\
\hat e_{k+1,k}(\mathscr{B}_k^{\leftarrow}) &\svs[\subseteq]{\text{R.}\ref{rem:moralg}} & \{0\}\\
\hat e_{k+1,k}(\mathscr{B}_k^{⇐}) &\svs[\subseteq]{\text{R.}\ref{rem:moralg}} & \{0\}
\end{array}%\\
% \begin{array}{lrclrcl}
% ⇒& \ker \hat e_{k+1,k} &=& \langle \mathscr{B}_k^{\leftarrow} \sqcup \mathscr{B}_k^{⇐}\rangle,&  \im \hat e_{k+1,k} &=& \langle  \mathscr{B}_{k+1}^{\leftarrow} \sqcup \mathscr{B}_{k+1}^{⇐}\rangle
% \end{array}

Thus by <ref>, assertion (a) holds.

\begin{array}[t]{rcl}
\hat e_{k,k+1}(\mathscr{B}_{k+1}^{⇔}) &\svs{\text{R.}\ref{rem:moralg}}& (pη_{λ^{k+1},1,n_{\rm c}^{k+1}+1} η_{λ^{k+1},n_{\rm c}^{{k+1}}+1,n_{\rm c}^{{k+1}}+y} \colon y∈(1,…,n_{\rm b}^{{k+1}})) \\
&\sqs& (pη_{λ^{k+1},1,n_{\rm c}^{{k+1}}+y} \colon y∈(1,…,n_{\rm b}^{{k+1}})) = \mathscr{B}_k^{→}\\
\hat e_{k,k+1}(\mathscr{B}_{k+1}^{\leftarrow}) &\svs{\text{R.}\ref{rem:moralg}}& (pη_{λ^{k+1},1,n_{\rm c}^{{k+1}}+1} η_{λ^{k+1},n_{\rm c}^{{k+1}}+1,y} \colon y∈(1,…,n_{\rm c}^{{k+1}})) \\
&\sqs& (pη_{λ^{k+1},1,y} \colon y∈(1,…,n_{\rm b}^{k})) = \mathscr{B}_k^⇒\\
\hat e_{k,k+1}(\mathscr{B}_{k+1}^→) &\svs[\subseteq]{\text{R.}\ref{rem:moralg}} & \{0\}\\
\hat e_{k,k+1}(\mathscr{B}_{k+1}^⇒) &\svs[\subseteq]{\text{R.}\ref{rem:moralg}} & \{0\}
\end{array}

Thus by <ref>, assertion (b) holds.

\begin{array}[t]{rcl}
\hat e_{p-1,p-1}(\mathscr{B}_{p-1}^⇔) &\svs{\text{R.}\ref{rem:moralg}}& (pη_{λ^p,1,1}η_{λ^{(p-1)+1},1,y} \colon y∈(1,…,n_{\rm b}^{{p-1}})) \\
&\sqs& (pη_{λ^{(p-1)+1},1,y} \colon y∈(1,…,n_{\rm b}^{{p-1}})) = \mathscr{B}_{p-1}^⇒\\
\mathscr{B}_{p-1}^→ &\sqs& () \text{ as $n_{\rm b}^{p} = 0$}\\
\hat e_{p-1,p-1}(\mathscr{B}_{p-1}^{\leftarrow}) &\svs[\subseteq]{\text{R.}\ref{rem:moralg}} & \{0\}\\
\hat e_{p-1,p-1}(\mathscr{B}_{p-1}^⇐) &\svs[\subseteq]{\text{R.}\ref{rem:moralg}} & \{0\}
\end{array}

Thus by <ref>, assertion (c) holds.

\begin{array}[t]{rcl}
\hat e_{1,1}(\mathscr{B}_1^⇔) &\svs{\text{R.}\ref{rem:moralg}}& (pη_{λ^1,1,1}η_{λ^1,n_{\rm c}^{1}+1,n_{\rm c}^{1}+y} \colon y∈(1,…,n_{\rm b}^{1}))\\
&\svs{n_{\rm c}^{1}=0}& (pη_{λ^1,n_{\rm c}^{1}+1,n_{\rm c}^{1}+y} \colon y∈(1,…,n_{\rm b}^{1})) = \mathscr{B}_1^⇐\\
\mathscr{B}_1^{\leftarrow} &\sqs& ()\text{ as $n_{\rm c}^{1}=0$}\\
\hat e_{1,1}(\mathscr{B}_1^⇒) &\svs[\subseteq]{\text{R.}\ref{rem:moralg}} & \{0\}\\
\hat e_{1,1}(\mathscr{B}_1^→) &\svs[\subseteq]{\text{R.}\ref{rem:moralg}} & \{0\}
\end{array}

Thus by <ref>, assertion (d) holds.

The trivial $\zp\!\Sp$-module $\zp$ becomes a $Λ$-module via the isomorphism of $\zp$-algebras $r:\zp\!\Sp→Λ$ described in <ref>. We want to construct a projective resolution of $\zp$ over $Λ$.

$Γ$ is a right $Λ$-module as $Λ$ is a subalgebra of $Γ$. The set $Γ^{λ^1}:=\{ρ∈Γ \mid ρ^λ = 0 \text{ for }λ\neq λ^1\}$ is a right $Λ$-submodule of $Γ$. As $n_{\rm c}^{k}= 0$ and $n_{\rm b}^k=1$, $Γ^{λ^1}$ is free over $\zp$ with basis $\{η_{λ^1,1,1}\}$.

Given a partition $λ\dashv p$, the operation of an element $x∈\zp\!\Sp$ on the Specht module corresponding to $λ$ is multiplication  with the matrix $r(x)^λ$ with respect to a certain basis of  that Specht module, cf. the definition of $r^{\Z}$ in the proof of <ref>.

As $\zp$ is the Specht module corresponding to the trivial partition $λ^1$ of $p$, and as $\zp$ is one-dimensional, the operation of $x∈\zp\!\Sp$ on $\zp$ is multiplication with the scalar $r(x)^{λ^1}$. Thus an element $ρ∈Λ$ operates on $\zp$ via multiplication with the scalar $ρ^{λ^1}$ and we haven an isomorphism of right $Λ$-modules by

\begin{align*}
\begin{array}{rlcl}
\hat{ε}^1 : &Γ^{λ^1} & \longrightarrow & \zp\\
& η_{λ^1,1,1} & \longmapsto & 1.
\end{array}
\end{align*}

We have the morphism of right $Λ$-modules

\begin{align*}
\begin{array}{rlclr}
\hat{ε}^0: &\tilde P_1 &\longrightarrow & Γ^{λ^1}\\
&\tilde e_1x &\longmapsto & η_{λ^1,1,1} \tilde e_1x= η_{λ^1,1,1}x & \text{ for }x∈Λ
\end{array}
\end{align*}

We have $\hat{ε}^0(\tilde e_1) = \hat{ε}^0(η_{λ^1,1,1} + η_{λ^2,1,1}) = η_{λ^1,1,1}$, thus
$\hat{ε}^0$ is surjective as
$\{η_{λ^1,1,1}\}$ is a $\zp$-basis of $Γ^{λ^1}$.

Given $x∈\tilde P_1$, we have
$\hat e_{1,1}(x) = p \hat{ε}^0(x)$ as elements of $Γ$. Thus the maps $\hat e_{1,1}$ and $\hat{ε}^0$ have the same kernel. Concatenation with the isomorphism $\hat{ε}^1$ yields the surjective morphism of right $Λ$-modules

\[ \hat{ε} := \hat{ε}^1 \circ \hat{ε}^0 : \tilde P_1 \longrightarrow \zp,\]

for which we have $\ker \hat{ε} = \ker \hat e_{1,1}$.

With these properties of $\hat{ε}$ and <ref>,
we are able to directly formulate a projective resolution of $\zp$:

We set

\begin{align*}
\tilde{\pr}_i :\eqs& \begin{cases} \tilde P_{ω(i)} & i\geq 0 \\ 0 & i<0 \end{cases},
\end{align*}

where the integer $ω(i)$ is given by the following construction: Recall the stipulation $l:=2(p-1)$. We have $i = jl+r$ for some $j∈\Z$ and $0\leq r\leq l-1$. Then

\begin{align}
\label{omega}
ω(i) := \begin{cases} r+1 & \text{ for }0\leq r \leq p-2\\ l-r = 2(p-1) -r & \text{ for } p-1 \leq r \leq 2(p-1)-1 = l-1 \end{cases}.
\end{align}

So $ω(i)$ increases by steps of one from $1$ to $p-1$ as $i$ runs from $jl$ to $jl+(p-2)$ and $ω(i)$ decreases from $p-1$ to $1$ as $i$ runs from $jl+(p-1)$ to $jl + (l-1)$. Finally we set

\begin{align*}
\hat d_i := \begin{cases} \hat e_{ω(i-1),ω(i)}: \tilde P_{ω(i)} → \tilde P_{ω(i-1)} & i\geq 1\\ 0 & i\leq 0 \end{cases}.
\end{align*}

Now <ref> gives the projective resolution of $\zp$

\begin{align}
\label{presform1}
\cdots\xrightarrow{\hat d_3} \tilde{\pr}_2\xrightarrow{\hat d_2}\tilde{\pr}_1\xrightarrow{\hat d_1}\tilde{\pr}_0\xrightarrow{0=\hat d_0} 0\rightarrow\cdots\,.
\end{align}

More explicitly, we have

\begin{align*}
\cdots\rightarrow &\tilde P_2\xrightarrow{\hat e_{1,2}} \tilde P_1 \xrightarrow{\hat e_{1,1}} \tilde P_1 \xrightarrow{\hat e_{2,1}} \tilde P_2 \rightarrow …
\rightarrow \tilde P_{p-2} \xrightarrow{\hat e_{p-1,p-2}}\tilde P_{p-1} \nonumber \\
&\xrightarrow{\hat e_{p-1,p-1}} \tilde P_{p-1} \xrightarrow{\hat e_{p-2,p-1}}\tilde P_{p-2}\rightarrow … \rightarrow \tilde P_2\xrightarrow{\hat e_{1,2}} \tilde P_1  \rightarrow 0\rightarrow \cdots\,.
\end{align*}

The corresponding extended projective resolution is

\begin{align*}
\cdots \rightarrow &\tilde P_2\xrightarrow{\hat e_{1,2}} \tilde P_1 \xrightarrow{\hat e_{1,1}} \tilde P_1 \xrightarrow{\hat e_{2,1}} \tilde P_2 \rightarrow …
\rightarrow \tilde P_{p-2} \xrightarrow{\hat e_{p-1,p-2}}\tilde P_{p-1} \nonumber \\
&\xrightarrow{\hat e_{p-1,p-1}} \tilde P_{p-1} \xrightarrow{\hat e_{p-2,p-1}}\tilde P_{p-2}\rightarrow … \rightarrow \tilde P_2\xrightarrow{\hat e_{1,2}} \tilde P_1 \xrightarrow{\hat{ε}} \zp \rightarrow 0\rightarrow\cdots\,,
\end{align*}

which is an exact sequence.

We have proven the

tm]tm:przp Recall that $p\geq 3$ is a prime.

The sequence (<ref>) is a projective resolution of $\zp$, with augmentation $\tilde{\pr}_0=\tilde P_1 \xrightarrow{\hat{ε}}\zp$.

Recall that $p\geq 3$ is a prime. We have

\begin{equation*}
\begin{array}{lclr}
\hat e_{1,1} + \hat e_{1,2}\circ \hat e_{2,1} &=& p\hat e_1\\
\hat e_{k,k-1}\circ\hat e_{k-1,k} + \hat e_{k,k+1}\circ\hat e_{k+1,k} &=& p\hat e_k &\text{ for }k∈[2,p-2]\\
\hat e_{p-1,p-2}\circ\hat e_{p-2,p-1} + \hat e_{p-1,p-1} &=& p\hat e_{p-1}\\
\hat{ε}\circ \hat e_{1,1} &=& p\hat{ε}.
\end{array}
\end{equation*}

We have by <ref>

\begin{align*}
T_{\tilde e_1,\tilde e_1}(&\hat e_{1,1} + \hat e_{1,2}\circ\hat e_{2,1}) = T_{\tilde e_1,\tilde e_1}(\hat e_{1,1}) + T_{\tilde e_1,\tilde e_2}(\hat e_{1,2})T_{\tilde e_2,\tilde e_1}(\hat e_{2,1})\\
&= pη_{λ^1,1,1} + pη_{λ^2,1,n_{\rm c}^2+1}η_{λ^2,n_{\rm c}^2+1,1} = p(η_{λ^1,1,1}+η_{λ^2,1,1}) = T_{\tilde e_1,\tilde e_1}(p\hat e_{1})\\
T_{\tilde e_k,\tilde e_k}(&\hat e_{k,k-1}\circ\hat e_{k-1,k} + \hat e_{k,k+1}\circ\hat e_{k+1,k})\\
&= T_{\tilde e_k,\tilde e_{k-1}}(\hat e_{k,k-1})T_{\tilde e_{k-1},\tilde e_k}(\hat e_{k-1,k}) + T_{\tilde e_k,\tilde e_{k+1}}(\hat e_{k,k+1})T_{\tilde e_{k+1},\tilde e_k}(\hat e_{k+1,k})\\
&= η_{λ^k,n_{\rm c}^k+1,1}pη_{λ^k,1,n_{\rm c}^k+1} + pη_{λ^{k+1},1,n_{\rm c}^{k+1}+1}η_{λ^{k+1},n_{\rm c}^{k+1}+1,1}\\
&= p(η_{λ^k,n_{\rm c}^k+1,n_{\rm c}^k+1} + η_{λ^{k+1},1,1}) = T_{\tilde e_k,\tilde e_k}(p\hat e_{k})\\
T_{\tilde e_{p-1},\tilde e_{p-1}}(&\hat e_{p-1,p-2}\circ\hat e_{p-2,p-1} + \hat e_{p-1,p-1})\\
&= T_{\tilde e_{p-1},\tilde e_{p-2}}(\hat e_{p-1,p-2})T_{\tilde e_{p-2},\tilde e_{p-1}}(\hat e_{p-2,p-1}) + T_{\tilde e_{p-1},\tilde e_{p-1}}(\hat e_{p-1,p-1})\\
&= η_{λ^{p-1},n_{\rm c}^{p-1}+1,1}pη_{λ^{p-1},1,n_{\rm c}^{p-1}+1} + pη_{λ^p,1,1}\\
&= p(η_{λ^{p-1},n_{\rm c}^{p-1}+1,n_{\rm c}^{p-1}+1} + η_{λ^p,1,1}) = T_{\tilde e_{p-1},\tilde e_{p-1}}(p\hat e_{p-1}).
\end{align*}

Finally for $x∈\tilde P_1$, we have

\begin{align*}
(\hat{ε}^0\circ \hat e_{1,1})(x) \eqs& η_{λ^1,1,1}\cdot  pη_{λ^1,1,1}\cdot x = pη_{λ^1,1,1}\cdot x
= p \hat{ε}^0(x),
\end{align*}

thus $\hat{ε}\circ \hat e_{1,1} = \hat{ε}^1\circ \hat{ε}^0\circ \hat e_{1,1} = p\hat{ε}^1\circ \hat{ε}^0 = p\hat{ε}$.

§.§ A projective resolution of Fp over FpSp

We have $\zp\!\Sp/(p\zp\!\Sp) \simeq \fp\!\Sp$. Since $\zp\!\Sp \simeq Λ$, we identify $\fp\!\Sp = Λ/(pΛ)$.
We obtain the desired projective resolution by reducing the projective resolution of $\zp$ "modulo $p$":

§.§.§ Reduction modulo $I$

Let $R$ be a principal ideal domain.
Let $(A,ρ)$ be an $R$-algebra. Let $I$ be an ideal of $R$. We set $\bar{R}:=R/\!I$.

As $R$ is a principal ideal domain, $ρ(I)A$ is an additive subset of $A$. As $ρ(I)$ is a subset of the center of $A$, $ρ(I)A$ is an ideal of $A$ and $A/(ρ(I)A)=:\bar{A}$ is an $\bar{R}$-algebra.

We regard a right $A$-module $M_A$ as a right $R$-module $M_R$ via $m\cdot r := m\cdot ρ(r)$ for $m∈M$, $r∈R$.

The functors $-\underset{A}{\otimes} \bar{A}$ and $-\underset{R}{\otimes} \bar{R}$ from $\Mod$-$A$ to $\Mod$-$R$ are naturally isomorphic. The natural isomorphism $-\underset{A}{\otimes} \bar{A}→-\underset{R}{\otimes} \bar{R}$ is given at the module $M_A$ by

\begin{equation*}
\begin{array}{rcl}
M_A \underset{A}{\otimes} \leftidx{_A}{\bar{A}}{} &\overset{\sim}{\longrightarrow}& M_R\underset{R}{\otimes} \leftidx{_R}{\bar{R}}{}\\
m\otimes (a+ ρ(I)A) &\longmapsto & ma \otimes (1+I)\\
m\otimes (r+ρ(I)A)&\longmapsfrom & m\otimes (r+I).
\end{array}
\end{equation*}

By the universal property of the tensor product, the two maps given above are well-defined and $R$-linear. Straightforward calculation gives that they invert each other and that we have a natural transformation.

The functor $-\underset{A}{\otimes}\leftidx{_A}{\bar{A}}{_{\bar{A}}}$ from $\Mod$-$A$ to $\Mod$-$\bar A$ maps exact sequences of right $A$-modules that are free and of finite rank as $R$-modules to exact sequences of right $\bar{A}$-modules.

Because $-\underset{A}{\otimes}\leftidx{_A}{\bar{A}}{_{\bar{A}}}$  is an additive functor, it maps complexes to complexes. For considerations of exactness, we may compose our functor with the forgetful functor from $\Mod$-$\bar A$ to $\Mod$-$ R$. This composite is $-\underset{A}{\otimes}\leftidx{_A}{\bar{A}}{}$.

By the natural isomorphism given in <ref>, it suffices to show that $-\underset{R}{\otimes}\leftidx{_R}{\bar{R}}{}$ transforms exact sequences of right $A$-modules that are free and of finite rank as $R$-modules into exact sequences.

Let $\cdots\xrightarrow{d_{i+1}}M_i\xrightarrow{d_i}M_{i-1}\xrightarrow{d_{i-1}}\cdots$ be an exact sequence of right $A$-modules that are free and of finite rank as $R$-modules. Then $\im d_i$ is a submodule of the free $R$-module $M_{i-1}$. As $R$ is a principal ideal domain, $\im d_i$ is free. Hence the short exact sequence $\im d_{i+1} \rightarrow M_i \rightarrow \im d_i$ splits. Now the additive functor $-\underset{R}{\otimes}\leftidx{_R}{\bar{R}}{}$ maps split short exact sequences to (split) short exact sequences and the proof is complete.

§.§.§ Reduction modulo $p$

The isomorphism $r:\zp\!\Sp \rightarrow Λ$ from <ref> induces an isomorphism of $\fp$-algebras $\fp\!\Sp = \zp\!\Sp/(p\zp\!\Sp) \xrightarrow{\bar{r}} Λ/(pΛ)=:\bar{Λ}$.

For the sake of simplicity in the next step, we identify $\bar{Λ}$ and $\fp\!\Sp$ along $\bar{r}$.

Recall that $p\geq 3$ is a prime. Applying the functor $-\underset{Λ}{\otimes} \leftidx{_Λ}{\bar{Λ}}{_{\bar{Λ}}}$ , we obtain

* the projective modules $P_k:= \tilde P_k\underset{Λ}{\otimes} \leftidx{_Λ}{\bar{Λ}}{_{\bar{Λ}}}$
for $k∈[1,p-1]$,

* $\fp:=\zp\underset{Λ}{\otimes} \leftidx{_Λ}{\bar{Λ}}{_{\bar{Λ}}}$ (the trivial $\fp\!\Sp$-module),

\begin{array}[t]{lclll}
e_k&:=&\hat e_k\underset{Λ}{\otimes} \leftidx{_Λ}{\bar{Λ}}{_{\bar{Λ}}}&∈\Hom_{\fp\Sp}(P_k,P_k)&\text{for $k∈[1,p-1]$,} \\
e_{1,1}&:=& \hat e_{1,1}\underset{Λ}{\otimes} \leftidx{_Λ}{\bar{Λ}}{_{\bar{Λ}}}&∈\Hom_{\fp\Sp}(P_1,P_1),\\
e_{p-1,p-1} &:=& \hat e_{p-1,p-1}\underset{Λ}{\otimes} \leftidx{_Λ}{\bar{Λ}}{_{\bar{Λ}}}&∈\Hom_{\fp\Sp}(P_{p-1},P_{p-1}), \\
e_{k+1,k}&:=& \hat e_{k+1,k}\underset{Λ}{\otimes} \leftidx{_Λ}{\bar{Λ}}{_{\bar{Λ}}}&∈\Hom_{\fp\Sp}(P_k,P_{k+1})&\text{for $k∈[1,p-2]$,} \\
e_{k,k+1} &:=& \hat e_{k,k+1}\underset{Λ}{\otimes} \leftidx{_Λ}{\bar{Λ}}{_{\bar{Λ}}}&∈\Hom_{\fp\Sp}(P_{k+1},P_k) &\text{for $k∈[1,p-2]$,} \\
ε &:=& \hat{ε}\underset{Λ}{\otimes} \leftidx{_Λ}{\bar{Λ}}{_{\bar{Λ}}}&∈\Hom_{\fp\Sp}(P_1,\fp) .
\end{array}

The complex

\begin{align}
\label{prfp}
\pres \fp := (\pres \zp)\underset{Λ}{\otimes} \leftidx{_Λ}{\bar{Λ}}{_{\bar{Λ}}}= (\cdots\xrightarrow{d_3} {\pr}_2\xrightarrow{ d_2}{\pr}_1\xrightarrow{ d_1}{\pr}_0\xrightarrow{0=d_0} 0\rightarrow\cdots),
\end{align}

\begin{align*}
{\pr}_i :\eqs& \begin{cases} P_{ω(i)} & i\geq 0 \\ 0 & i<0 \end{cases} &
d_i :\eqs& \begin{cases} e_{ω(i-1),ω(i)}: P_{ω(i)} →P_{ω(i-1)} & i\geq 1\\ 0 & i\leq 0, \end{cases}
\end{align*}

is a projective resolution of $\fp$ with augmentation $ε:P_1→\fp$.
More explicitly, $\pres \fp$ is

\begin{align*}
…\rightarrow & \underbrace{P_2}_{\mathclap{l+1}}\xrightarrow{ e_{1,2}}  \underbrace{P_1}_{\mathclap{l=2(p-1)}} \xrightarrow{ e_{1,1}}  \underbrace{P_1}_{\mathclap{(p-2)+p-1}} \xrightarrow{ e_{2,1}}  \underbrace{P_2}_{\mathclap{(p-2)+p-2}} \rightarrow …
\rightarrow  \underbrace{P_{p-2}}_{\mathclap{p=(p-2)+2}} \xrightarrow{ e_{p-1,p-2}} \underbrace{P_{p-1}}_{\mathclap{(p-2)+1}} \nonumber \\
&\xrightarrow{ e_{p-1,p-1}}  \underbrace{P_{p-1}}_{\mathclap{p-2}} \xrightarrow{ e_{p-2,p-1}} \underbrace{P_{p-2}}_{p-3}\rightarrow … \rightarrow  \underbrace{P_2}_{1} \xrightarrow{ e_{1,2}}  \underbrace{P_1}_0  \rightarrow 0.
\end{align*}

Recall that $p\geq 3$ is a prime.

* We have the relations

\begin{equation*}
\begin{array}{lclr}
e_{1,1} +  e_{1,2} \circ e_{2,1} &=& 0\\
e_{k,k-1}\circ e_{k-1,k} +  e_{k,k+1}\circ e_{k+1,k} &=& 0 &\text{ for } k∈[2,p-2]\\
e_{p-1,p-2} \circ e_{p-2,p-1} +  e_{p-1,p-1} &=& 0\\
ε \circ e_{1,1} &=& 0
\end{array}
\end{equation*}

and $e_k$ is the identity on $P_k$ for $k∈[1,p-1]$.

* Given $k∈[2,p-1]$, we have $\Hom_{\fp\!\Sp}(P_k,\fp) = \{0\}$.

* Given $k,k'∈[1,p-1]$ such that $|k-k'|>1$, we have  $\Hom_{\fp\!\Sp}(P_k,P_{k'}) = \{0\}$.

* The set $\{ε\}$ is an $\fp$-basis of $\Hom_{\fp\!\Sp}(P_1,\fp)$.

Assertion (a) results from <ref>.

Assertions (b), (c) and (d) are derived from corresponding assertions over $\zp\!\Sp$ using $\Hom_{\fp\!\Sp}(P/pP, M/pM) \simeq \Hom_{\zp\!\Sp}(P,M)/p\Hom_{\zp\!\Sp}(P,M)$ for $\zp\!\Sp$-modules $P$ and $M$, where $P$ is projective.

§ A(OO)-ALGEBRAS

§.§ Definitions, General theory

In this subsection, we review results presented in [12] and we fix notation.

Let $R$ be a commutative  ring.
We understand linear maps between $R$-modules to be $R$-linear. Tensor products are tensor products over $R$. By graded $R$-modules we understand $\Z$-graded $R$-modules.

A graded $R$-module $V$ is a $R$-module of the form $V = \oplus_{q∈\Z} V^q$.
An element $v_q∈V^q$, $q∈\Z$ is said  to be of degree $q$. An element $v∈V$ is called homogeneous if there is an integer $q∈\Z$ such that $v∈V^q$. For homogeneous elements $v$ resp. graded maps $g$ (see below), we denote their degrees by $|v|$ resp. $|g|$.

Let $A = \oplus_{q∈\Z}A^q$, $B = \oplus_{q∈\Z}B^q$ be two graded  $R$-modules. A graded map of degree $z∈\Z$ is a linear map $g:A→B$ such that  for $q∈\Z$.

Let $A=\oplus_{q∈\Z}A^q$, $B=\oplus_{q∈\Z}B^q$ be two graded $R$-modules. We have

\begin{align*}
A\otimes B \eqs& \bigoplus_{z_1,z_2∈\Z} A^{z_1}\otimes B^{z_2}
= \bigoplus_{q∈\Z}\left(\bigoplus_{z_1+z_2=q} A^{z_1}\otimes B^{z_2}\right).
\end{align*}

As we understand the direct sums to be internal direct sums in $A\otimes B$ and understand $A^{z_1}\otimes B^{z_2}$ to be the linear span of the set $\{a\otimes b∈A\otimes B \mid a ∈ A^{z_1},b∈A^{z_2}\}$, we have equations in the above, not just isomorphisms.

We then set $A\otimes B$ to be graded by $A\otimes B = \bigoplus_{q∈\Z}(A\otimes B)^q$, where $(A\otimes B)^q:= \bigoplus_{z_1+z_2=q} A^{z_1}\otimes B^{z_2}$.

Moreover, we grade the direct sum

\begin{align*}
A\oplus B \eqs& \bigoplus_{q∈\Z} (A^q\oplus B^q)
%\bigoplus_{z_1,z_2∈\Z} A^{z_1}\oplus B^{z_2}
%= \bigoplus_{q∈\Z}\left(\bigoplus_{z_1+z_2=q} A^{z_1}\oplus B^{z_2}\right)
\end{align*}

by $(A\oplus B)^q := A^q\oplus B^q$.

In the definition of the tensor product of graded maps, we implement the Koszul sign rule: Let $A_1,A_2,B_1,B_2$ be graded $R$-modules and $g:A_1→B_1$, $h:A_2→B_2$ graded maps. Then we set for homogeneous elements $x∈A_1,y∈A_2$

\begin{align}
\label{koszulraw}
(g\otimes h)(x\otimes y) := (-1)^{|h|\cdot|x|} g(x)\otimes h(y).
\end{align}

It is known that for graded $R$-modules $A,B,C$, the map

\begin{align}
\label{theta}
\begin{array}{rccc}
Θ:&(A\otimes B) \otimes C& \longrightarrow & A\otimes (B\otimes C)\\
&(a\otimes b) \otimes c &\longmapsto & a\otimes (b\otimes c)
\end{array}
\end{align}

is an isomorphism of $R$-modules. Because of the following, $Θ$ is homogeneous of degree $0$.

\begin{align*}
((A\otimes B)\otimes C)^q \eqs& \bigoplus_{\mathclap{y+z_3=q}} (A\otimes B)^y \otimes C^{z_3} = \bigoplus_{y+z_3=q}\bigoplus_{z_1+z_2= y} (A^{z_1}\otimes B^{z_2})\otimes C^{z_3}\\
\eqs& \bigoplus_{\mathclap{z_1+z_2+z_3=q}} (A^{z_1}\otimes B^{z_2})\otimes C^{z_3}\\
(A\otimes (B\otimes C))^q \eqs& \bigoplus_{\mathclap{z_1+y=q}} A^{z_1}\otimes (B\otimes C)^y = \bigoplus_{z_1+y=q}\bigoplus_{z_2+z_3=y} A^{z_1}\otimes (B^{z_2}\otimes C^{z_3})\\
\eqs& \bigoplus_{\mathclap{z_1+z_2+z_3=q}} A^{z_1}\otimes (B^{z_2}\otimes C^{z_3})
\end{align*}

Let $A_1,A_2,B_1,B_2,C_1,C_2$ be graded  $R$-modules, $f:A_1→A_2$, $g:B_1→B_2$, $h:C_1→C_2$ graded maps. For homogeneous elements $x∈A_1
$, $y∈B_1$, $z∈C_1$, we have

\begin{align*}
((f\otimes g)\otimes h)((x\otimes y)\otimes z) \eqs& (-1)^{|x\otimes y|\cdot|h|}((f\otimes g)(x\otimes y))\otimes h(z)\\
\eqs& (-1)^{(|x|+|y|)|h|+ |x|\cdot |g|} (f(x)\otimes g(y))\otimes h(z)\\
(f\otimes(g\otimes h))(x\otimes (y\otimes z)) \eqs&  (-1)^{|x|\cdot |g\otimes h|} f(x)\otimes((g\otimes h)(y\otimes z))\\
\eqs& (-1)^{|x|(|g|+|h|) + |y|\cdot|h|} f(x)\otimes(g(y)\otimes h(z)) \\
\eqs& (-1)^{(|x|+|y|)|h|+ |x|\cdot |g|} f(x)\otimes(g(y)\otimes h(z)).
\end{align*}

Thus we have the following commutative diagram ($Θ_1$ and $Θ_2$ are derived from (<ref>))

\begin{align*}
\xymatrix@C=2pc{
(A_1\otimes B_1)\otimes C_1\ar[r]^{Θ_{1}}\ar[d]^{(f\otimes g)\otimes h} & A_1\otimes (B_1\otimes C_1)\ar[d]^{f\otimes (g\otimes h)}\\
(A_2\otimes B_2)\otimes C_2\ar[r]^{Θ_{2}} & A_2\otimes (B_2\otimes C_2)
\end{align*}

It is therefore valid to use $Θ$ as an identification and to omit the brackets for the tensorization of graded $R$-modules and the tensorization of graded maps.

Concerning the signs in the definition of $\A_∞$-algebras and $\A_∞$-morphisms, we follow the variant given e.g. in [16] and [8].

Let $n∈\Z_{\geq 0}\cup \{∞\}$.

* Let $A$ be a graded $R$-module. A pre-$\A_n$-structure on $A$ is a family of graded maps $(m_k:A^{\otimes k}→A)_{k∈[1,n]}$ with $|m_k|=2-k$ for $k∈[1,n]$. The tuple $(A,(m_k)_{k∈[1,n]})$ is called a pre-$\A_n$-algebra.

* Let $A$, $A'$ be graded $R$-modules. A pre-$\A_n$-morphism from $A'$ to $A$ is a family of graded maps $(f_k:A'^{\otimes k}→A)_{k∈[1,n]}$ with $|f_k|=1-k$ for $k∈[1,n]$.

Let $n∈\Z_{\geq 0}\cup\{∞\}$.

An $\A_n$-algebra is a pre-$\A_n$-algebra $(A,(m_k)_{k∈[1,n]})$ such that for

\begin{align}
\sum_{\natop{k=r+s+t,}{r,t\geq 0, s\geq 1}} (-1)^{rs+t}m_{r+1+t}\circ (1^{\otimes r}\otimes m_s\otimes 1^{\otimes t})=0 .
\tag*{(\theequation)[$k$]}
\end{align}

In abuse of notation, we sometimes abbreviate $A = (A,(m_k)_{k\geq 1})$ for $\A_∞$-algebras.

Let $(A',(m'_k)_{k∈[1,n]})$ and $(A,(m_k)_{k∈[1,n]})$ be $\A_n$-algebras. An $\A_n$-morphism or morphism of $\A_n$-algebras from $(A',(m'_k)_{k∈[1,n]})$ to $(A,(m_k)_{k∈[1,n]})$ is a pre-$\A_n$-morphism $(f_k)_{k∈[1,n]}$ such that for $k∈[1,n]$, we have

\begin{align}
%\sum_{\mathclap{n=r+s+t}} (-1)^{r+st} f_u(1^{\otimes r}\otimes m_s\otimes 1^{\otimes t}) = \sum_{\mathclap{\natop{1\leq r\leq n}{i_1+…+i_r=n}}} (-1)^s m_r(f_{i_1}\otimes f_{i_2}\otimes … \otimes f_{i_r})
\sum_{\mathclap{\substack{k=r+s+t\\r,t\geq 0,s\geq 1}}} (-1)^{rs+t} f_{r+1+t}\circ (1^{\otimes r}\otimes m'_s\otimes 1^{\otimes t}) = \sum_{\mathclap{\substack{1\leq r\leq k\\ i_1+…+i_r=k\\ i_s\geq 1}}} (-1)^v m_r\circ (f_{i_1}\otimes f_{i_2}\otimes … \otimes f_{i_r}),
\tag*{(\theequation)[$k$]}
\end{align}

$v :=  %\sum_{2\leq s\leq r}\left((1-i_s)\sum_{1\leq t\leq s} i_t\right) \equiv_2
\sum_{1\leq t < s\leq r}(1-i_s)i_t$.

bsp]ex:dg Let $(A,(m_k)_{k\geq 1})$ be an $\A_\infty$-algebra.
If $m_n=0$ for $n\geq 3$ then $A$ is called  a differential graded algebra or dg-algebra. In this case the equations (<ref>)[$n$] for $n\geq 4$ become trivial: We have $(r+1+t)+s = n+1$ ⇒ $(r+1+t)+s\geq 5$ ⇒ $m_{r+1+t}=0$ or $m_s=0$. So all summands in (<ref>)[$n$] are zero for $n\geq 4$. Here are the equations for $n∈\{1,2,3\}$:

\begin{align*}
\eqref{ainfrel}[1]: && 0 \eqs& m_1\circ m_1\\
\eqref{ainfrel}[2]: && 0 \eqs& m_1\circ m_2 - m_2\circ (m_1\otimes 1+1\otimes m_1)\\
\eqref{ainfrel}[3]: && 0 \eqs& m_1\circ m_3 + m_2\circ (1\otimes m_2-m_2\otimes1)\\
&&& + m_3\circ (m_1\otimes 1^{\otimes 2}+1\otimes m_2\otimes 1+1^{\otimes 2}\otimes m_1)
\\&&
\ovs{m_3=0}& m_2\circ (1\otimes m_2-m_2\otimes1)
\end{align*}

So (<ref>)[$1$] ensures that $m_1$ is a differential. Moreover, (<ref>)[$3$] states that $m_2$ is an associative binary operation, since for homogeneous $x,y,z∈A$ we have  $0=m_2\circ \mbox{$(1\otimes m_2-m_2\otimes1)(x\otimes y\otimes z)$} = m_2(x\otimes m_2(y\otimes z) - m_2(x\otimes y)\otimes z)$, where because of $|m_2|=0$ there are no additional signs caused by the Koszul sign rule. Equation (<ref>)[$2$] is the Leibniz rule.

[$\A_n$-morphisms induce complex morphisms]
bsp]ex:cc $ $

Let $n∈\Z_{\geq 1}\cup \{∞\}$. Let $(A', (m'_k)_{k∈[1,n]})$ and $(A, (m_k)_{k∈[1,n]})$ be two $\A_n$-algebras and let $(f_k)_{k∈[1,n]}: (A', (m'_k)_{k∈[1,n]}) → (A, (m_k)_{k∈[1,n]})$ be an $\A_n$-morphism.

By (<ref>)[$1$], $(A',m'_1)$ and $(A,m_1)$ are complexes. Equation (<ref>)[$1$] is

\begin{align*}
%\eqref{finfrel}[1]: &&
f_1\circ m'_1 \eqs& m_1\circ f_1.
\end{align*}

Thus $f_1:(A',m'_1)→(A,m_1)$ is a complex morphism.

For $n\geq 2$, we have also (<ref>)[$2$]:

\begin{align}
\label{finfrel2}
%\eqref{finfrel}[2]: &&
f_1\circ m'_2 - f_2\circ(m'_1\otimes 1 + 1\otimes m'_1) \eqs& m_1\circ f_2 + m_2\circ (f_1\otimes f_1)
\end{align}

Recall the conventions concerning $\Hom_B^k(C,C')$.

Let  $B$ be an (ordinary)  $R$-algebra and $M=((M_i)_{i∈\Z}, (d_i)_{i∈\Z})$ a complex of $B$-modules, that is a sequence $(M_i)_{i∈\Z}$ of $B$-modules and $B$-linear maps $d_i:M_i→M_{i-1}$ such that $d_{i-1}\circ d_i = 0$ for all $i∈\Z$.

\begin{align*}
\Hom^i_B(M,M):\eqs& \prod_{z∈\Z}\Hom_B(M_{z+i},M_z)\\
\eqs& \{g=(g_z)_{z∈\Z} \mid  g_z∈\Hom_B(M_{z+i},M_z)\text{ for }z∈\Z\}.
\end{align*}

\begin{align*}
A=\Hom^*_B(M,M) := \bigoplus_{i∈\Z} \Hom^i_B(M,M)
\end{align*}

is a graded  $R$-module.
We have $d := (d_{z+1})_{z∈\Z}=\sum_{i∈\Z} \ls d_{i+1}\rs_{i+1}^{i} ∈\Hom^1_B(M,M)$. We define $m_1:=d_{\Hom^*(M,M)}:A→A$, that is  for homogeneous $g∈A$ we have

\begin{align*}
m_1(g) = d \circ g -(-1)^{|g|}  g\circ d.
\end{align*}

We define $m_2:A^{\otimes 2}→A$ for homogeneous $g,h∈A$ to be composition, i.e.

\begin{align*}
m_2(g\otimes h) := g \circ h.
\end{align*}

For $n\geq 3$ we set $m_n:A^{\otimes n}→A$, $m_n=0$. Then $(m_n)_{n\geq 1}$ is an $\A_\infty$-algebra structure on $A=\Hom^*_B(M^*,M^*)$. More precisely, $(A, (m_n)_{n\geq 1})$ is a dg-algebra.

Because of $|d|=1$ we have $|m_1| = 1 = 2-1$. The graded map $m_2$ has  degree $0 = 2-2$. The other maps $m_n$ are zero and have therefore automatically correct degree. As discussed in <ref> we only need to check (<ref>)[$n$] for $n=1,2,3$.
Equation (<ref>)[$1$] holds  because for homogeneous $g∈A$ we have

\begin{align*}
m_1(m_1(g)) \eqs& m_1[d\circ g - (-1)^{|g|}g\circ d]\\
\eqs& d\circ[d\circ g - (-1)^{|g|}g\circ d] - (-1)^{|g|+1}[d\circ g - ({-}1)^{|g|}g\circ d]\circ d\\
\ovs{d^2=0}& -(-1)^{|g|}d\circ g\circ d - (-1)^{|g|+1}d\circ g\circ d = 0.
\end{align*}

Concerning (<ref>)[$2$], we have for homogeneous $g,h∈A$

\begin{align*}
(m_2\circ (m_1\otimes 1+&1\otimes m_1))(g\otimes h)= m_2(m_1(g)\otimes h + (-1)^{|g|}g\otimes m_1(h))\\
\eqs& (d\circ g - (-1)^{|g|}g\circ d)\circ h + (-1)^{|g|}g\circ(d\circ h - (-1)^{|h|}h\circ d)\\
\eqs& d\circ g \circ h - (-1)^{|g|+|h|}g\circ h\circ d\\
\eqs& (m_1\circ m_2 )(g\otimes h).
\end{align*}

The map $m_2$ is induced by the composition of morphisms which is associative. As discussed in <ref>, equation (<ref>)[$3$] holds.

In $\Hom^*(\pres\fp,\pres\fp)$ we have (cf. (<ref>))

\begin{align*}
d \eqs& \sum_{i\geq 0} \ls e_{ω(i),ω(i+1)}\rs_{i+1}^i\,.
\end{align*}

[Homology of $\A_∞$-algebras, quasi-isomorphisms, minimality, minimal models]
As $m_1^2=0$ (cf. (<ref>)[$1$]) and

$|m_1|=1$, we have the complex

\[\cdots → A^{i-1}\xrightarrow{m_1|_{A^{i-1}}}A^{i}\xrightarrow{m_1|_{A^i}} A^{i+1}→\cdots\,.\]

We define $\Hm^k A:= \ker(m_1|_{A^k})/\im(m_1|_{A^{k-1}})$ and $\Hm^* A := \bigoplus_{k∈\Z} \Hm^k A$, which gives the homology of $A$ the structure of a graded $R$-module.

A morphism of $\A_∞$-algebras $(f_k)_{k\geq 1}: (A', (m'_k)_{k\geq 1}) → (A, (m_k)_{k\geq 1})$ is called a quasi-isomorphism if the morphism of complexes $f_1:(A', m'_1)→(A,m_1)$ (cf. <ref>) is a quasi-isomorphism.

An $\A_∞$-algebra is called minimal, if $m_1=0$. If $A$ is an $\A_∞$-algebra and $A'$ is a minimal $\A_∞$-algebra quasi-isomorphic to $A$, then $A'$ is called a minimal model of $A$.

The existence of minimal models is assured by the following theorem.
(minimality theorem, cf. [13] (history), [9], [8], [21], [5], [7], [18], … )
Let $(A,(m_k)_{k\geq 1})$ be an $\A_\infty$-algebra such that the homology $\Hm^* A$ is a projective $R$-module. Then there exists an $\A_\infty$-algebra structure $(m'_k)_{k\geq 1}$ on $\Hm^*A$ and a quasi-isomorphism of $\A_∞$-algebras $(f_k)_{k\geq 1}:(\Hm^* A,(m'_k)_{k\geq 1})→(A,(m_k)_{k\geq 1})$, such that

* $m'_1 = 0$ and

* the complex morphism $f_1:(\Hm^*A,m'_1)→(A, m_1)$ induces the identity in homology. I.e. each element $x∈\Hm^* A$, which is a homology class of $(A,m_1)$, is mapped by $f_1$ to a representing cycle.

For constructing $\A_∞$-structures induced by another $\A_∞$-algebra, we have the following

Let $n∈\Z_{\geq 1}\cup \{∞\}$. Let $(A',(m'_k)_{k∈[1,n]})$ be a pre-$\A_n$-algebra. Let $(A,(m_k)_{k∈[1,n]})$ be an $\A_n$-algebra. Let $(f_k)_{k∈[1,n]}$ be a pre-$\A_n$-morphism from $A'$ to $A$ such that (<ref>)$[k]$ holds for $k∈[1,n]$. Suppose $f_1$ to be injective.

Then $(A',(m'_k)_{k∈[1,n]})$ is  an $\A_n$-algebra and $(f_k)_{k∈[1,n]}$ is a morphism of $\A_n$-algebras from $(A',(m'_k)_{k∈[1,n]})$ to $(A,(m_k)_{k∈[1,n]})$.

This results from the bar construction and a straightforward induction on $n$.

We have the corresponding pre-$\A_n$-triple $((m'_k)_{k∈[1,n]}, (b'_k)_{k∈[1,n]}, b')$, the corresponding pre-$\A_n$-triple  $((m_k)_{k∈[1,n]}, (b_k)_{k∈[1,n]}, b)$ and the corresponding pre-$\A_n$-morphism triple  $((f_k)_{k∈[1,n]}, (F_k)_{k∈[1,n]}, F)$.
It suffices to prove by induction  on $k∈[0,n]$ that $(b')^2\big|_{TSA'_{\leq k}} = 0$, cf. <ref>.

For $k=0$, there is nothing to prove.

For the induction step, suppose that $b'\,^2\big|_{TSA'_{\leq k}} = 0$. Then by <ref>(i), we have $\im(b'\,^2 \circ ι'_{k+1})\subseteq SA$. Thus $0 = b^2\circ F \circ ι'_{k+1} \overset{\text{L.}\ref{lem:stasheff2}}{=} F\circ b'\,^2 \circ ι'_{k+1} = F_1\circ b'\,^2 \circ ι'_{k+1}$. As the injectivity of $f_1$ implies the injectivity of  $F_1$, we have $b'\,^2\circ ι'_{k+1} =0$ and thus $b'\,^2\big|_{TSA'_{\leq k+1}} = 0$.

Let $R$ be a commutative ring and $(A,(m_n)_{n\geq 1})$ be a dg-algebra (over $R$). Suppose given a graded $R$-module $B$ and graded maps $f_n:B^{\otimes n}→A$, $m'_n:B^{\otimes n}→B$ for $n\geq 1$. Suppose given $k\geq 1$ such that
we have $f_i=0$ for $i\geq k$, we have $m'_i=0$ for $i\geq k+1$,

and  (<ref>)$[n]$ is satisfied for $n∈[1,2k-2]$. Then (<ref>)$[n]$ is satisfied for all $n\geq 1$.

§.§ The homology of Hom*(FpSp)(PResFp,PResFp)

We need a well-known result of homological algebra in a particular formulation:

Let $F$ be a field. Let  $B$ be an $F$-algebra. Let $M$ be a $B$-module. Let $Q = (\cdots \rightarrow Q_2 \xrightarrow{d_2} Q_1 \xrightarrow{d_1} Q_0 → 0→
\cdots)$ be a projective resolution of $M$ with augmentation $ε:Q_0→M$. Then we have maps for $k∈\Z$

\begin{align*}
Ψ_k:\Hom^k_B(Q,Q) &\xrightarrow{}\Hom^k_B(Q, M) := \Hom_B(Q_k,M)\\
(g_i:Q_{i+k}→Q_i)_{i∈\Z} &\mapsto ε\circ g_0
\end{align*}

The right side is equipped with the differentials (dualization of $d_k$)

\begin{align*}
(d_k)^*:\Hom_B(Q_k,M) &\rightarrow \Hom_B(Q_{k+1},M)\\
g & \mapsto (-1)^k g\circ d_k
\end{align*}

and the left side is equipped with the differential $m_1$ of its dg-algebra structure, cf. <ref>.

Then $(Ψ_k)_{k∈\Z}$ becomes a complex morphism from the complex  $\Hom^*_B(Q,Q)$ to the complex $\Hom^*_B(Q, M)$ that induces isomorphisms $\bar{Ψ}_k$ of $F$-vector spaces on the homology

\begin{align*}
\bar{Ψ}_k:\Hm^k\Hom^*_B(Q,Q) & \xrightarrow{\simeq}\Hm^k \Hom^*_B(Q, M)\\
\overline{(g_i:Q_{i+k}→Q_i)_{i∈\Z}}&\mapsto \overline{ε\circ g_0}
\end{align*}

<ref> is  <cit.> applied to the quasi-isomorphism induced by the augmentation, cf. <cit.>.

Recall the notation $\ls x\rs_y^z$ for the description of elements of $\Hom_B^k(C,C')$.

Recall that $p\geq 3$ is a prime and $l=2(p-1)$.
Write   .

\begin{align*}
ι:\eqs&\dum_{i\geq 0} \ls e_{ω(i)} \rs_{i+l}^i =\dum_{i\geq 0}\dum_{k=0}^{l-1}\ls e_{ω(k)} \rs_{(i+1)l+k}^{il+k} ∈ A^l
\\
% \end{align*}
% \begin{align*}
χ :\eqs& \dum_{i\geq 0}\left(\ls e_1 \rs_{il+l-1}^{il} + \left\lgroup\dum_{k=1}^{p-2}\ls e_{k+1,k}\rs_{il+l-1+k}^{il + k}\right\rgroup\right.\\
&\left. +
\ls e_{p-1}\rs_{il+l-1 +(p-1)}^{il+(p-1)}
+ \left\lgroup\dum_{k=1}^{p-2}\ls e_{p -k-1,p-k}\rs_{il+l-1+(p-1)+k}^{il +(p-1)+ k}\right\rgroup\right)∈A^{l-1}.
\end{align*}

* For $j\geq 0$, we have
ι^j = \dum_{i\geq 0} \ls e_{ω(i)} \rs_{i+jl}^i= \dum_{i\geq 0}\dum_{k=0}^{l-1}\ls e_{ω(k)} \rs_{(i+j)l+k}^{il+k}\,.

* Suppose given $y\geq 0$. Let $h∈A^y$ be $l$-periodic, that is
$h=\dum_{i\geq 0} \dum_{k=0}^{l-1} \ls h_k\rs_{il+k+y}^{il + k}$.
Then for $j\geq 0$, we have

\begin{align*}
h\circ ι^j = ι^j\circ h = \dum_{i\geq 0} \dum_{k=0}^{l-1} \ls h_k\rs_{(i+j)l + k +y}^{il+k}∈A^{y+jl}.
\end{align*}

* Suppose given $y∈\Z$. For $h∈A^y$ and $j\geq 0$, we have $m_1(h\circ ι^j) = m_1(h)\circ ι^j$.

* For $j\geq 0$, we have $m_1(ι^j)=0$. Thus $ι^j$ is a cycle.

* For $j\geq 0$, we have

\begin{align*}
χι^j :\eqs& χ\circ ι^j = ι^j\circ χ\\
\eqs&  \dum_{i\geq 0}\left(\ls e_1 \rs_{(i+j+1)l-1}^{il} + \left\lgroup\dum_{k=1}^{p-2}\ls e_{k+1,k}\rs_{(i+j+1)l-1+k}^{il + k}\right\rgroup\right.\\
&\left. + \ls e_{p-1}\rs_{(i+j+1)l-1 +(p-1)}^{il+(p-1)}
+ \left\lgroup\dum_{k=1}^{p-2}\ls e_{p -k-1,p-k}\rs_{(i+j+1)	l-1+(p-1)+k}^{il +(p-1)+ k}\right\rgroup\right)∈A^{jl+l-1}.
\end{align*}

For convenience, we also define $χ^0ι^j:= ι^j$ and $χ^1ι^j:=χι^j=χ\circ ι^j$ for $j\geq 0$.

* For $j\geq 0$, we have $m_1(χι^j) = 0$. Thus $χι^j$ is a cycle.

Suppose given $k∈\Z$. A $\fp$-basis of $\Hm^kA$ is given by

\begin{align*}
\{\overline{ι^j}\} & \text{ if $k=jl$ for some $j\geq 0$}\\
\{\overline{χι^j}\} & \text{ if $k=jl+l-1$ for some $j\geq 0$} \\
∅ &\text{ else}.
\end{align*}

Thus the set $\mathfrak{B}:=\{\overline{ι^j}\mid j\geq 0\}\sqcup \{\overline{χι^j}\mid j\geq 0\}$ is an $\fp$-basis of $\Hm^*A = \bigoplus_{z∈\Z} \Hm^zA$.

Before we proceed we display $ι$ and $χ$ for the case $p = 5$ as an example:

The period is of length $l = 2p - 2 = 2\cdot 5 - 2 = 8$. The terms inside circles denote the degrees.
For $χ$, the differentials on the right side are negated to obtain a commutative diagram (We have $0=m_1(χ) = d\circ χ + χ\circ d$ because $|χ|$ is odd)

\[
\xymatrix@C=0.5pc@!R=0.43pc{
& \pres(\ff)[8]\ar[r]^-\iota \ar@{}[d]^*--{\rotatebox{90}{\hspace{3mm}$=$\ru{2}}}  & \pres(\ff)  \ar@{}[d]^*--{\rotatebox{90}{\hspace{3mm}$=$\ru{2}}}\\
& \rotatebox{90}{$\cdots)$}\ar[d]              & \rotatebox{90}{$\cdots)$}\ar[d]         \\
*+[F-:<5pt>]{ 8} & P_1\ar[r]^{e_1}    \ar[d]_{e_{1,1}} & P_1\ar[d]^{e_{1,1}} \\
*+[F-:<5pt>]{ 7} & P_1\ar[r]^{e_1}    \ar[d]_{e_{2,1}} & P_1\ar[d]^{e_{2,1}} \\
*+[F-:<5pt>]{ 6} & P_2\ar[r]^{e_2}    \ar[d]_{e_{3,2}} & P_2\ar[d]^{e_{3,2}} \\
*+[F-:<5pt>]{ 5} & P_3\ar[r]^{e_3}    \ar[d]_{e_{4,3}} & P_3\ar[d]^{e_{4,3}} \\
*+[F-:<5pt>]{ 4} & P_4\ar[r]^{e_4}    \ar[d]_{e_{4,4}} & P_4\ar[d]^{e_{4,4}} \\
*+[F-:<5pt>]{ 3} & P_4\ar[r]^{e_4}    \ar[d]_{e_{3,4}} & P_4\ar[d]^{e_{3,4}} \\
*+[F-:<5pt>]{ 2} & P_3\ar[r]^{e_3}    \ar[d]_{e_{2,3}} & P_3\ar[d]^{e_{2,3}} \\
*+[F-:<5pt>]{ 1} & P_2\ar[r]^{e_2}    \ar[d]_{e_{1,2}} & P_2\ar[d]^{e_{1,2}} \\
*+[F-:<5pt>]{ 0} & P_1\ar[r]^{e_1}    \ar[d]_{e_{1,1}} & P_1\ar[d]            \\
*+[F-:<5pt>]{-1} & P_1\ar[r]          \ar[d]_{e_{2,1}} & 0  \ar[d]            \\
*+[F-:<5pt>]{-2} & P_2\ar[r]          \ar[d]_{e_{3,2}} & 0  \ar[d]            \\
*+[F-:<5pt>]{-3} & P_3\ar[r]          \ar[d]_{e_{4,3}} & 0  \ar[d]            \\
*+[F-:<5pt>]{-4} & P_4\ar[r]          \ar[d]_{e_{4,4}} & 0  \ar[d]            \\
*+[F-:<5pt>]{-5} & P_4\ar[r]          \ar[d]_{e_{3,4}} & 0  \ar[d]            \\
*+[F-:<5pt>]{-6} & P_3\ar[r]          \ar[d]_{e_{2,3}} & 0  \ar[d]            \\
*+[F-:<5pt>]{-7} & P_2\ar[r]          \ar[d]_{e_{1,2}} & 0  \ar[d]            \\
*+[F-:<5pt>]{-8} & P_1\ar[r]          \ar[d]           & 0  \ar[d]            \\
*+[F-:<5pt>]{-9} & 0  \ar[r]\ar[d]                     & 0  \ar[d]            \\
& \rotatebox{-90}{$\cdots)$}         & \rotatebox{-90}{$\cdots)$}         \\
\hspace{2cm}
\xymatrix@C=0.5pc@!R=0.43pc{
& \pres(\ff)[8-1]\ar[r]^-\chi \ar@{}[d]^*--{\rotatebox{90}{\hspace{3mm}$=$\ru{2}}}  & \pres(\ff)  \ar@{}[d]^*--{\rotatebox{90}{\hspace{3mm}$=$\ru{2}}}\\
& \rotatebox{90}{$\cdots)$}\ar[d]              & \rotatebox{90}{$\cdots)$}\ar[d]         \\
*+[F-:<5pt>]{ 9} & P_1\ar[r]^{e_{2,1}}\ar[d]_{e_{1,1}} & P_2\ar[d]^{-e_{1,2}} \\
*+[F-:<5pt>]{ 8} & P_1\ar[r]^{e_1}    \ar[d]_{e_{2,1}} & P_1\ar[d]^{-e_{1,1}} \\
*+[F-:<5pt>]{ 7} & P_2\ar[r]^{e_{1,2}}\ar[d]_{e_{3,2}} & P_1\ar[d]^{-e_{2,1}} \\
*+[F-:<5pt>]{ 6} & P_3\ar[r]^{e_{2,3}}\ar[d]_{e_{4,3}} & P_2\ar[d]^{-e_{3,2}} \\
*+[F-:<5pt>]{ 5} & P_4\ar[r]^{e_{3,4}}\ar[d]_{e_{4,4}} & P_3\ar[d]^{-e_{4,3}} \\
*+[F-:<5pt>]{ 4} & P_4\ar[r]^{e_4}    \ar[d]_{e_{3,4}} & P_4\ar[d]^{-e_{4,4}} \\
*+[F-:<5pt>]{ 3} & P_3\ar[r]^{e_{4,3}}\ar[d]_{e_{2,3}} & P_4\ar[d]^{-e_{3,4}} \\
*+[F-:<5pt>]{ 2}  & P_2\ar[r]^{e_{3,2}}\ar[d]_{e_{1,2}} & P_3\ar[d]^{-e_{2,3}} \\
*+[F-:<5pt>]{ 1}  & P_1\ar[r]^{e_{2,1}}\ar[d]_{e_{1,1}} & P_2\ar[d]^{-e_{1,2}} \\
*+[F-:<5pt>]{ 0}  & P_1\ar[r]^{e_1}    \ar[d]_{e_{2,1}} & P_1\ar[d]            \\
*+[F-:<5pt>]{-1}  & P_2\ar[r]          \ar[d]_{e_{3,2}} & 0  \ar[d]            \\
*+[F-:<5pt>]{-2}  & P_3\ar[r]          \ar[d]_{e_{4,3}} & 0  \ar[d]            \\
*+[F-:<5pt>]{-3}  & P_4\ar[r]          \ar[d]_{e_{4,4}} & 0  \ar[d]            \\
*+[F-:<5pt>]{-4}  & P_4\ar[r]          \ar[d]_{e_{3,4}} & 0  \ar[d]            \\
*+[F-:<5pt>]{-5}  & P_3\ar[r]          \ar[d]_{e_{2,3}} & 0  \ar[d]            \\
*+[F-:<5pt>]{-6}  & P_2\ar[r]          \ar[d]_{e_{1,2}} & 0  \ar[d]            \\
*+[F-:<5pt>]{-7}  & P_1\ar[r]          \ar[d]           & 0  \ar[d]            \\
*+[F-:<5pt>]{-8} & 0  \ar[r]\ar[d]                     & 0  \ar[d]            \\
& \rotatebox{-90}{$\cdots)$}         & \rotatebox{-90}{$\cdots)$}         \\
\]

The element $ι$ is well-defined since $ω(y) = ω(l+y)$ for $y\geq 0$.

In the definition of $χ$ we need to check that the "$\ls * \rs_*^*$" are well defined. This is easily proven by calculating the $ω(y)$ where $y$ is the lower respective upper index of  "$\ls * \rs_*^*$".

(a): As $\pr_i = \{0\}$ for $i<0$, the identity element of $A$  is given by
$ι^0 = \dum_{i\geq 0}\ls e_{ω(i)}\rs_{i}^i$,
which agrees with the assertion in case $j=0$. So we have proven the induction basis for induction on $j$. So now assume that for some $j\geq 0$ the assertion holds. Then

\begin{align*}
ι^{j+1} \eqs& ι\circ ι^j = \left(\dum_{i\geq 0} \ls e_{ω(i)} \rs_{i+l}^i\right)\circ\left(\dum_{i'\geq 0} \ls e_{ω(i')} \rs_{i'+jl}^{i'}\right) \\
\eqs& \dum_{i\geq 0} \ls e_{ω(i)}\circ e_{ω(i+l)} \rs_{i+l+jl}^i = \dum_{i\geq 0} \ls e_{ω(i)}\rs_{i+(j+1)l}^i\,.
\end{align*}

Thus the proof by induction is complete.

We have

\begin{align*}
ι^j \circ h \eqs& \left(\sum_{i\geq 0} \sum_{k=0}^{l-1}\ls e_{ω(il+k)} \rs_{(i+j)l+k}^{il+k}\right) \circ \left(\sum_{i'\geq 0}\sum_{k'=0}^{l-1}\ls h_{k'}\rs_{i'l+k'+ y}^{i'l+k'}\right)
\overset{\natop{i'\leadsto i+j}{k'\leadsto k}}{=} \sum_{i\geq 0} \sum_{k=0}^{l-1} \ls h_k\rs_{(i+j)l+k+y}^{il+k}\\
h\circ ι^j \eqs& \left(\sum_{i\geq 0}\sum_{k=0}^{l-1}\ls h_k\rs_{il+k+y}^{il+k}\right) \circ
\left(\sum_{i'\geq 0} \ls e_{ω(i')} \rs_{i'+jl}^{i'}\right)
= \sum_{i\geq 0}\sum_{k=0}^{l-1} \ls h_k\rs_{(i+j)l + k+y}^{il+k}\, .
\end{align*}

So we have proven (b).

(c): The differential $d$ of $\pres\fp$ is $l$-periodic (cf. <ref>) and thus

\begin{align*}
m_1(h)\circ ι^j \eqs& (d\circ h - (-1)^y h\circ d)\circ ι^j\\
\ovs{(b), |ι^j|\equiv_2 0}& d\circ h\circ ι^j - (-1)^{y+|ι^j|} h\circ ι^j\circ d = m_1(h\circ ι^j).
\end{align*}

(d): We have
m_1(ι^j) \overset{(c)}{=} m_1(ι^0)\circ ι^j = (d\circ ι^0 - (-1)^0 ι^0 d)\circ ι^j = (d - d)\circ ι^j = 0.

(e) is implied by (b) using the fact that $χ$ is $l$-periodic.

(f): Because of (c) we have $m_1(χι^j) = m_1(χ)\circ ι^j$. Because $|χ|=l-1$ is odd we have

\begin{align*}
m_1&(χ) = d\circ χ - (-1) χ\circ d = χ\circ d + d\circ χ\\
\ovs{\text{R.}\ref{bem:di}}& \Big(\dum_{i\geq 0}\left(\ls e_1 \rs_{il+l-1}^{il} + \left\lgroup\dum_{k=1}^{p-2}\ls e_{k+1,k}\rs_{il+l-1+k}^{il + k}\right\rgroup+ \ls e_{p-1}\rs_{il+l-1 +(p-1)}^{il+(p-1)}\right.\\
+ \left\lgroup\dum_{k=1}^{p-2}\ls e_{p -k-1,p-k}\rs_{il+l-1+(p-1)+k}^{il +(p-1)+ k}\right\rgroup\right)\Big)
\circ \left(\dum_{y\geq 0} \ls e_{ω(y),ω(y+1)}\rs_{y+1}^y\right)\\
&+\left(\dum_{y\geq 0} \ls e_{ω(y),ω(y+1)}\rs_{y+1}^y\right) \circ \Big(
\dum_{i\geq 0}\left(\ls e_1 \rs_{il+l-1}^{il} + \left\lgroup\dum_{k=1}^{p-2}\ls e_{k+1,k}\rs_{il+l-1+k}^{il + k}\right\rgroup\right.\\
&\left. + \ls e_{p-1}\rs_{il+l-1 +(p-1)}^{il+(p-1)}
+ \left\lgroup\dum_{k=1}^{p-2}\ls e_{p -k-1,p-k}\rs_{il+l-1+(p-1)+k}^{il +(p-1)+ k}\right\rgroup\right)\Big)\\
\eqs&
\dum_{i\geq 0}\Big(
\ls e_1\circ e_{1,1}\rs_{il+l}^{il} + \left\lgroup\dum_{k=1}^{p-2}\ls e_{k+1,k}\circ e_{k,k+1}\rs_{il+l+k}^{il+k}\right\rgroup\\
&+\ls e_{p-1}\circ e_{p-1,p-1}\rs_{il+l+(p-1)}^{il+(p-1)}
+ \left\lgroup\dum_{k=1}^{p-2}\ls e_{p-k-1,p-k}\circ e_{p-k,p-k-1}\rs_{il+l+(p-1)+k}^{il+(p-1)+k}\right\rgroup\Big)\\
&+\dum_{i\geq 1} \ls e_{1,1}\circ e_1\rs_{il+l-1}^{il-1}
+ \dum_{i\geq 0}\Big( \left\lgroup\dum_{k=1}^{p-2}\ls e_{k,k+1}\circ e_{k+1,k}\rs_{il+l+k-1}^{il+k-1}\right\rgroup\\
&+ \ls e_{p-1,p-1}\circ e_{p-1}\rs_{il+l-1+(p-1)}^{il-1+(p-1)}
+ \left\lgroup\dum_{k=1}^{p-2}\ls e_{p-k,p-k-1}\circ e_{p-k-1,p-k}\rs_{il+l-1+(p-1)+k}^{il-1+(p-1) + k}\right\rgroup\Big)\\
\overset{*}{\eqs}& \dum_{i\geq 0}\Big(
\ls e_{1,1} + e_{1,2}\circ e_{2,1}\rs_{il+l}^{il} + \left\lgroup\dum_{k=1}^{p-3}\ls e_{k+1,k}\circ e_{k,k+1}+ e_{k+1,k+2}\circ e_{k+2,k+1}\rs_{il+l+k}^{il+k}\right\rgroup\\
&+ \ls e_{p-1,p-2}\circ e_{p-2,p-1} + e_{p-1,p-1}\rs_{il+l+p-2}^{il+p-2}
+ \ls e_{p-1,p-1}+e_{p-1,p-2}\circ e_{p-2,p-1}\rs_{il+l+p-1}^{il+p-1}\\
&+ \left\lgroup\dum_{k=1}^{p-3}\ls e_{p-k-1,p-k}\circ e_{p-k,p-k-1} + e_{p-k-1,p-k-2}\circ e_{p-k-2,p-k-1}\rs_{il+l+p-1+k}^{il+p-1+k}\right\rgroup\\
&+ \ls e_{1,2}\circ e_{2,1} + e_{1,1}\rs_{(i+1)l+l-1}^{(i+1)l-1}\Big)
\overset{\text{L.}\ref{lem:relfp}(a)}{=} 0
\end{align*}

In the step marked by "$*$" we sort the summands by their targets. Note that when splitting sums of the form $\dum_{k=1}^{p-2}(…)_k$ into $(…)_1+\dum_{k=2}^{p-2}(…)_k$ or into $(…)_{p-2}+\dum_{k=1}^{p-3}(…)_k$, the existence of the summand that is split off is ensured by $p\geq 3$.

(g): We first show that the differentials of the complex $\Hom^*(\pres \fp, \fp)$ (cf. <ref>) are all zero:
By <ref>, $\{ε\}$ is an $\fp$-basis of $\Hom_{\fp\!\Sp}(P_1,\fp)$, and for $k∈[2,p-1]$ we have $\Hom_{\fp\!\Sp}(P_k,\fp) = 0$. So the only non-trivial $(d_k)^*$ are those where $\pr_k = \pr_{k+1} = P_1$. This is the case only when $k=lj + l-1$ for some $j\geq 0$. Then $d_k= e_{1,1}$. For $ε∈\Hom(P_1,\fp)$, we have $(d_k)^* (ε) = (-1)^kε\circ e_{1,1} \overset{\text{L.}\ref{lem:relfp}(a)}{=} 0$. As $ \Hom(P_1,\fp)= \langle ε\rangle_\fp$, we have $(d_k)^* = 0$.

So $\Hm^k \Hom^*(\pres \fp, \fp) = \Hom^k(\pres \fp, \fp)$. We use <ref>.

For $k=jl$, $j\geq 0$, we have $\bar{Ψ}^k(\overline{ι^j}) \overset{(a)}{=} ε$, and $\{ε\}$ is a basis of $\Hm^k \Hom^*(\pres \fp, \fp)$.

For $k=jl+l-1$, $j\geq 0$, we have $\bar{Ψ}^k(\overline{χι^j})\overset{(e)}{=}ε$, and $\{ε\}$ is a basis of $\Hm^k \Hom^*(\pres \fp, \fp)$.

Finally, for $k=jl+r$ for some $j\geq 0$ and some $r∈[1,l-2]$ and for $k<0$, we have $\Hm^k \Hom^*(\pres \fp, \fp)=\{0\}$.

§.§ An A(oo)-structure on Ext*(FpSp)(Fp,Fp) as a minimal model of Hom*(FpSp)(PResFp,PresFp)

Recall that $p\geq 3$ is a prime. Write $A := \Hom^*_{\fp\!\Sp}(\pres\fp,\pres\fp)$, which becomes an $\A_∞$-algebra $(A, (m_n)_{n\geq 1})$ over $R=\fp$ via <ref>. We implement $\Ext^*_{\fp\!\Sp}(\fp,\fp)$ as $\Ext^*_{\fp\!\Sp}(\fp,\fp):=\Hm^*A$.

Our goal in this section is to construct an $\A_\infty$-structure $(m'_n)_{n\geq 1}$ on $\Hm^*A$ and a morphism of $\A_\infty$-algebras $f=(f_n)_{n\geq 1}:(\Hm^*A,(m'_n)_{n\geq 1})→(A, (m_n)_{n\geq 1})$ which satisfy the statements of <ref>. I.e. we will construct a minimal model of $A$.

In preparation of the definitions of the $f_n$ and $m'_n$, we name and examine certain elements of $A$:

Suppose given $k∈[2,p-1]$. We set

\begin{align*}
γ_k:= \dum_{i\geq 0}\left( \ls e_k \rs_{k(l-1)+li}^{k-1+li}
+\ls e_{p-k} \rs_{k(l-1) + (p-1) + li}^{k-1+(p-1)+li}\right)∈A^{k(l-2)+1}.
\end{align*}

For $j\geq 0$, we have

\begin{align*}
γ_kι^j := γ_k\circ ι^j = ι^j\circ γ_k = \dum_{i\geq 0}\left( \ls e_k \rs_{k(l-1)+l(i+j)}^{k-1+li}
+ \ls e_{p-k} \rs_{k(l-1) + (p-1) + l(i+j)}^{k-1+(p-1)+li}\right) ∈A^{k(l-2)+1+jl}
\end{align*}

\begin{align*}
m_1(γ_kι^j)\eqs&\dum_{i\geq 0}\left( \ls e_{k-1,k} \rs_{k(l-1)+l(i+j)}^{k-2+li}
+ %\dum_{i\geq 0}
\ls e_{p-k+1,p-k}\rs_{k(l-1) + (p-1) + l(i+j)}^{k-2+(p-1)+li}\right.\\
&\left.+%\dum_{i\geq 0}
\ls e_{k,k-1}\rs_{k(l-1)+1+l(i+j)}^{k-1+li}
+ %\dum_{i\geq 0}
\ls e_{p-k,p-(k-1)}\rs_{k(l-1) + p + l(i+j)}^{k-1+(p-1)+li}\right).
\end{align*}

We need to prove that $γ_k$ is well-defined. Let $i\geq 0$.

We consider the first term. The complex $\pres \fp$ (cf. (<ref>), (<ref>)) has entry $P_k$ at position $k(l-1)+li$ and at position $k-1+li$:
We have $k(l-1)+li = (k-1+i)l + l - k$. So $ω(k(l-1)+li) = l - (l - k) = k$ since  $p-1\leq l-k \leq l-1$.
We have $ω(k-1+li) = (k-1)+1 = k$ since $0\leq k-1 \leq p-2$. As $k(l-1)+li, k-1+li\geq 0$, we have $\pr_{k(l-1)+li}=P_{ω(k(l-1)+li)}=P_k$ and $\pr_{k-1+li} = P_{ω(k-1+li)} = P_k$. So the first term is well-defined.

Now consider the second term. The complex $\pres \fp$ has entry $P_{p-k}$ at position $k(l-1) + (p-1) + li$ and at position $k-1+(p-1)+li$:
We have $k(l-1) + (p-1) + li = (i+k)l + (p-1)-k$, so $ω(k(l-1) + (p-1)+li) = (p-1)-k + 1 = p-k$ since  $0\leq (p-1)-k \leq p-2$. We have $ω(k-1+(p-1)+li) = 2(p-1) - (k-1) -(p-1) = p-k$ since $p-1\leq k-1 + (p-1) \leq 2(p-1)-1$. As $k(l-1) + (p-1) + li, k-1+(p-1)+li\geq 0$, we have  $\pr_{k(l-1)+(p-1)+li} = P_{ω(k(l-1)+(p-1)+li)} = P_{p-k}$ and $\pr_{k-1+(p-1)+li} = P_{ω(k-1+(p-1)+li)} = P_{p-k}$. So the second term is well-defined.

The degree of the tuple of  maps is computed to be $(k(l-1) + li) - (k-1+li) = k(l-2) + 1 = (k(l-1)+(p-1)+li) - (k-1+(p-1)+li)$.

The explicit formula for $γ_kι^j$ is an application of <ref>(b).

The degree $|γ_kι^j| = k(l-2) + 1$ is odd, so

\begin{align*}
m_1(γ_kι^j)\,\, \ovs{\text{L.}\ref{lem:cai}}& d \circ γ_kι^j + γ_kι^j \circ d\\
\ovs{\text{R.}\ref{bem:di}}& \dum_{i\geq 0} \ls e_{ω(k-2),ω(k-1)}\rs_{k-1+li}^{k-2+li} \circ \dum_{i\geq 0} \ls e_k \rs_{k(l-1)+l(i+j)}^{k-1+li} \\
&+ \dum_{i\geq 0} \ls e_{ω(p-1+k-2),ω(p-1+k-1)}\rs_{k-1+(p-1)+li}^{k-2+(p-1)+li}\circ \dum_{i\geq 0} \ls e_{p-k} \rs_{k(l-1) + (p-1) + l(i+j)}^{k-1+(p-1)+li} \\
&+\dum_{i\geq 0} \ls e_k \rs_{k(l-1)+l(i+j)}^{k-1+li} \circ \dum_{i\geq 0} \ls e_{ω(l-k),ω(l-k+1)}\rs_{k(l-1)+1+l(i+j)}^{k(l-1)+l(i+j)}\\
&+ \dum_{i\geq 0} \ls e_{p-k} \rs_{k(l-1) + (p-1) + l(i+j)}^{k-1+(p-1)+li}\circ
\dum_{i\geq 0} \ls e_{ω(p-1-k),ω(p-k)}\rs_{k(l-1) + p + l(i+j)}^{k(l-1) + (p-1) + l(i+j)}\\
\eqs&\dum_{i\geq 0} \ls e_{k-1,k} \rs_{k(l-1)+l(i+j)}^{k-2+li}
+ \dum_{i\geq 0} \ls e_{p-k+1,p-k}\rs_{k(l-1) + (p-1) + l(i+j)}^{k-2+(p-1)+li}\\
&+\dum_{i\geq 0} \ls e_{k,k-1}\rs_{k(l-1)+1+l(i+j)}^{k-1+li}
+ \dum_{i\geq 0} \ls e_{p-k,p-(k-1)}\rs_{k(l-1) + p + l(i+j)}^{k-1+(p-1)+li}\\
\end{align*}

Note that in the second line  $k-2 +li\geq 0$ as $i\geq 0$ and $k\geq 2$.

For $j,j'\geq 0$, we have
χι^j \circ χι^{j'} = m_1(γ_2ι^{j+j'}).

It suffices to prove that $χ\circ χ = m_1(γ_2)$ since then $χι^j \circ χι^{j'} \overset{\text{P.}\ref{pp:iota}(e)}{=} χ\circ χ\circ ι^{j+j'} = m_1(γ_2)\circ ι^{j+j'} \overset{\text{P.}\ref{pp:iota}(c)}{=} m_1(γ_2ι^{j+j'})$.

To determine when a composite is zero, we will need the following.  For , we examine the condition

\begin{align}
\label{mulccg}
il +l - 1 + k = i'l + k'.
\end{align}

If $k=0$ then (<ref>) holds iff $i=i'$ and $k' = l-1$.

If $k\geq 1$ then (<ref>) holds iff $i+1=i'$ and $k'=k-1$.

\begin{align*}
χ\circ χ \ovs{p\geq 3\vphantom{I_{I_y}}}&%\overset{p\geq 3\vphantom{I_{I_y}}}{=}
\left(\dum_{i\geq 0}\left(\ls e_1 \rs_{il+l-1}^{il} + \ls e_{2,1}\rs_{il+l}^{il+1}+\left\lgroup\dum_{k=2}^{p-2}\ls e_{k+1,k}\rs_{il+l-1+k}^{il + k}\right\rgroup\right.\right.\\
& + \ls e_{p-1}\rs_{il+l-1 +(p-1)}^{il+(p-1)}
+ \ls e_{p-2,p-1}\rs_{il+l+p-1}^{il+p}
+\left.\left.\left\lgroup\dum_{k=2}^{p-2}\ls e_{p -k-1,p-k}\rs_{il+l-1+(p-1)+k}^{il +(p-1)+ k}\right\rgroup\right)\right)\\
&\circ\left(\dum_{i'\geq 0}\left(\ls e_1 \rs_{i'l+l-1}^{i'l} + \left\lgroup\dum_{k'=1}^{p-3}\ls e_{k'+1,k'}\rs_{i'l+l-1+k'}^{i'l + k'}\right\rgroup\right.\right.
+ \ls e_{p-1,p-2}\rs_{i'l+l+p-3}^{i'l+p-2}\\
& + \ls e_{p-1}\rs_{i'l+l-1 +(p-1)}^{i'l+(p-1)}
\left.\left.+ \left\lgroup\dum_{k'=1}^{p-3}\ls e_{p -k'-1,p-k'}\rs_{i'l+l-1+(p-1)+k'}^{i'l +(p-1)+ k'}\right\rgroup+ \ls e_{1,2}\rs_{i'l+l+2(p-2)}^{i'l+l-1}\right)\right)\\
\eqs&
\dum_{i\geq 0}\Big(\ls e_1\circ e_{1,2}  \rs_{il+l+2(p-2)}^{il} + \ls e_{2,1}\circ e_1\rs_{il+2l-1}^{il+1}\\
& +\left\lgroup\dum_{k=2}^{p-2}\right.\!\underbrace{\ls e_{k{+}1,k}\circ e_{k,k{-}1}\rs_{il+2l-1+k-1}^{il + k}}_{=0\text{ by L.\ref{lem:relfp}(c)}}\left.\vphantom{\dum_k^p}\!\!\right\rgroup + \ls e_{p-1}\circ e_{p-1,p-2}\rs_{il+2l+p-3}^{il+(p-1)}\\
&+ \ls e_{p-2,p-1}\circ e_{p-1}\rs_{il+2l+p-2}^{il+p}
+\left\lgroup\dum_{k=2}^{p-2}\right.\underbrace{\ls e_{p -k-1,p-k}\circ e_{p-k,p-k+1}\rs_{il+2l-1+p-1+k-1}^{il +(p-1)+ k}}_{=0\text{ by L.\ref{lem:relfp}(c)}}\left.\vphantom{\dum_k^p}\!\right\rgroup\Big)\\
\eqs& \dum_{i\geq 0}\Big(\ls e_{1,2}  \rs_{(i+2)l-2}^{il} + \ls e_{2,1}\rs_{(i+2)l-1}^{il+1}
+\ls e_{p-1,p-2}\rs_{(i+2)l+p-3}^{il+p-1} + \ls e_{p-2,p-1}\rs_{(i+2)l+p-2}^{il+p}
\Big)\\
\ovs{\text{L.}\ref{lem:gamma}}& m_1(γ_2)
\end{align*}

Below are the definitions which will give a minimal $\A_∞$-algebra structure on $\Hm^*A$ and a quasi-isomorphism of $\A_∞$-algebras $\Hm^*A → A$.

Recall from <ref> that $\mathfrak{B}= \mbox{$\{\overline{ι^j}\mid j\geq 0\}\sqcup \{\overline{χι^j}\mid j\geq 0\}$} = \{\overline{χ^aι^j} \mid j\geq 0, a∈\{0,1\}\}$ is a basis of $\Hm^*A$. For $n∈\Z_{\geq 1}$, we set
\begin{align*}
\mathfrak{B}^{\otimes n} :\eqs& \{\overline{χ^{a_1}ι^{j_1}}\otimes … \otimes \overline{χ^{a_n}ι^{j_n}}  ∈ (\Hm^*A)^{\otimes n} \mid  a_i∈\{0,1\} \text{ and } j_i∈\Z_{\geq 0} \text{ for all }i∈[1,n] \},%\\
% \eqs& \{b_1\otimes … \otimes b_n \mid b_1,…,b_n∈\mathfrak{B}\},
\end{align*}

which is a basis of $(\Hm^* A)^{\otimes n}$ consisting of homogeneous elements.

For $n\geq 1$, we define the $\fp$-linear map $f_n:(\Hm^*A)^{\otimes n}→ A$ as follows:

Case $n=1$: $f_1$ is given on $\mathfrak{B}$ by $f_1(\overline{ι^j}):=ι^j$ and $f_1(\overline{χι^j}):= χι^j$.

Case $n∈[2,p-1]$: $f_n$ is given on elements of $\mathfrak{B}^{\otimes n}$

\begin{align*}
f_n(\overline{χ^{a_1}ι^{j_1}}\otimes … \otimes \overline{χ^{a_n}ι^{j_n}}) := \begin{cases}
0 & \text{if }∃i∈[1,n]: a_i = 0 \\
(-1)^{n-1}γ_nι^{j_1+…+j_n} & \text{if } 1=a_1 = a_2 = … = a_n
\end{cases}
\end{align*}

Case $n\geq p$: We set $f_n:= 0$.

For $n\geq 1$, we define the $\fp$-linear map $m'_n:(\Hm^*A)^{\otimes n}→ \Hm^*A$ by defining it on elements  $\overline{χ^{a_1}ι^{j_1}}\otimes … \otimes \overline{χ^{a_n}ι^{j_n}}∈\mathfrak{B}^{\otimes n}$:

Case $∃i∈[1,n]: a_i = 0$: $ $

$m'_n(\overline{χ^{a_1}ι^{j_1}}\otimes … \otimes \overline{χ^{a_n}ι^{j_n}}) := 0$ for $n\neq 2$ and

m'_2(\overline{χ^{a_1}ι^{j_1}} \otimes \overline{χ^{a_2}ι^{j_2}}) := \overline{χ^{a_1+a_2}ι^{j_1+j_2}}% \text{ note $a_1+a_2∈[0,1]$}
$ (Note that $a_1+a_2∈\{0,1\}$).

Case $a_1 = a_2 = … = a_n=1$: $ $

$m'_n(\overline{χι^{j_1}}\otimes … \otimes \overline{χι^{j_n}}):= 0$ for $n\neq p$ and

$m'_p(\overline{χι^{j_1}}\otimes … \otimes \overline{χι^{j_p}}) := (-1)^p\overline{ι^{p-1+j_1+…+j_p}}=-\overline{ι^{p-1+j_1+…+j_p}}$.

Note that since $p\geq 3$, we have $m'_2(\overline{χι^{j_1}} \otimes \overline{χι^{j_2}}) = 0$ for $j_1,j_2\geq 0$.

The pair $(\Hm^*A, (m'_n)_{n\geq 1})$ is a minimal $\A_∞$-algebra.

The tuple  $(f_n)_{n\geq 1}$ is an quasi-isomorphism of $\A_∞$-algebras from $(\Hm^*A, (m'_n)_{n\geq 1})$ to $(A, (m_n)_{n\geq 1})$. More precisely,  induces the identity in homology.

The proof of <ref> will take the remainder of <ref>. We will use <ref>.

The maps $f_n$ and $m'_n$ have degree $|f_n| = 1-n$ and $|m'_n| = 2-n$.
I.e. $(f_n)_{n\geq 1}$ is a pre-$A_∞$-morphism from $\Hm^* A$ to $A$, and $(\Hm^*A, (m'_n)_{n\geq 1})$ is a pre-$\A_∞$-algebra.

We have $|f_1|=0$ as $|\overline{ι^j}| = |ι^j|$ and $|\overline{χι^j}|=|χι^j|$. For $n\geq p$ the map $f_n$ is  of degree $1-n$ as $f_n=0$.  For $n∈[2,p-1]$  the statement $|f_n|=1-n$ is proven by checking the degrees for the elements of the basis $\mathfrak{B}^{\otimes n}$ whose image under $f_n$ is non-zero:

\begin{align*}
|f_n(\overline{χι^{j_1}}\otimes … \otimes \overline{χι^{j_n}})| \eqs&
\overset{\text{L.}\ref{lem:gamma}}{=} (j_1+…+j_n)l + n(l-1) + 1-n\\
\eqs& 1-n+\dum_{x=1}^n |\overline{χι^{j_x}}| = 1-n + |\overline{χι^{j_1}}\otimes … \otimes \overline{χι^{j_n}}|
\end{align*}

Thus $|f_n|=1-n$ for all $n$ and we have proven the first statement.

Now we show $|m'_n|=2-n$. As before, we only need check the degrees for basis elements whose image is non-zero: For $\overline{χ^{a_1}ι^{j_1}}\otimes\overline{χ^{a_2}ι^{j_2}}$, $j_1,j_2\geq 0$, $a_1,a_2∈\{0,1\}$, $0∈\{a_1,a_2\}$, we have

\begin{align*}
|m'_2(\overline{χ^{a_1}ι^{j_1}}\otimes\overline{χ^{a_2}ι^{j_2}})| \eqs& |\overline{χ^{a_1+a_2}ι^{j_1+j_2}}| = (a_1+a_2)(l-1)+l(j_1+j_2)\\
\eqs& a_1(l-1) + j_1l + a_2(l-1) + j_2l = |\overline{χ^{a_1}ι^{j_1}}\otimes\overline{χ^{a_2}ι^{j_2}}| + (2-2).
\end{align*}

For $\overline{χι^{j_1}}\otimes \cdots \otimes \overline{χι^{j_p}}$, $j_x\geq 0$ for $x∈[1,p]$, we have

\begin{align*}
|m'_p(\overline{χι^{j_1}}\otimes \cdots \otimes \overline{χι^{j_p}})| \eqs& |\overline{ι^{p-1+j_1+…+j_p}}| = l(p-1+j_1+…+j_p)\\
\eqs& lp - l + l(j_1+…+j_p) = lp - 2p+2 + l(j_1+…+j_p)\\
\eqs& p(l-1) + l(j_1+…+j_p) + 2-p
= |\overline{χι^{j_1}}\otimes \cdots \otimes \overline{χι^{j_p}}| + 2-p
\end{align*}

We have $m'_1=0$. The equation (<ref>)$[1]$ holds. The complex morphism  $f_1:(A',m'_1)→(A,m_1)$ is a quasi-isomorphism inducing the identity in homology.

The equality $m'_1=0$ follows immediately from the definition. Thus $m_1\circ f_1 = 0 = f_1\circ m'_1$. Moreover $f_1$ is a quasi-isomorphism inducing the identity in homology by construction, cf. <ref>(g).

The map $f_1$ is injective.

The set $X:=\{ χ^aι^j \mid a∈\{0,1\}, j∈\Z_{\geq 1}\}\subseteq A$ is linearly independent, since it consists of nonzero elements of different summands of the direct sum $A=\bigoplus_{k∈\Z}\Hom^k(\pres\fp,\pres\fp)$.

The set $\mathfrak{B}$, which is a basis of $\Hm^* A$, is mapped bijectively to $X$ by $f_1$, so $f_1$ is injective.

The equation (<ref>)$[2]$ holds.

As $m'_1=0$, equation (<ref>)[$2$] is equivalent to (cf. (<ref>))

\begin{align*}
%&&f_1\circ m'_2 -f_2\circ (1\otimes m'_1) -f_2\circ (m'_1\otimes 1) \eqs& m_1\circ f_2 + %m_2\circ (f_1\otimes f_1)\\
f_1\circ m'_2 \eqs& m_1\circ f_2 + m_2\circ (f_1\otimes f_1).
\end{align*}

We check this equation on $\mathfrak{B}^{\otimes 2}$: Recall <ref>.

\begin{align*}
f_1m'_2(\overline{ι^j}\otimes \overline{ι^{j'}}) \eqs& ι^{j+j'}
= m_2(f_1\otimes f_1)(\overline{ι^j}\otimes \overline{ι^{j'}})
= (m_1\circ f_2 + m_2\circ (f_1\otimes f_1))(\overline{ι^j}\otimes \overline{ι^{j'}})\\
f_1m'_2(\overline{ι^j}\otimes \overline{χι^{j'}})\eqs& χι^{j+j'}
= m_2(f_1\otimes f_1)(\overline{ι^j}\otimes \overline{χι^{j'}})\\
\eqs& (m_1\circ f_2 + m_2\circ (f_1\otimes f_1)) (\overline{ι^j}\otimes \overline{χι^{j'}})\\
f_1m'_2(\overline{χι^j} \otimes \overline{ι^{j'}}) \eqs& χι^{j+j'}
= m_2(f_1\otimes f_1)(\overline{χι^j} \otimes \overline{ι^{j'}})\\
\eqs& (m_1\circ f_2 + m_2\circ (f_1\otimes f_1))(\overline{χι^j} \otimes \overline{ι^{j'}})\\
f_1m'_2(\overline{χι^j}\otimes \overline{χι^{j'}}) \eqs &0 \overset{\text{L.}\ref{lem:chichi}}{=}
-m_1(γ_2ι^{j+j'}) + m_2(f_1\otimes f_1)(\overline{χι^j}\otimes \overline{χι^{j'}})\\
\eqs& (m_1\circ f_2 + m_2\circ (f_1\otimes f_1))(\overline{χι^j}\otimes \overline{χι^{j'}})
\end{align*}

Note that there are no additional signs due to the Koszul sign rule since $|f_1|=0$.

The following results directly from <ref>.

For $n\geq 2$ and  $a_1,…,a_n∈\{0,1\}$, $j_1,…,j_n\geq 0$, we have

\begin{align*}
f_n(\overline{χ^{a_1}ι^{j_1}}\otimes … \otimes \overline{χ^{a_n}ι^{j_n}})= f_n(\overline{χ^{a_1}}\otimes … \otimes \overline{χ^{a_n}}) \circ ι^{j_1+…+j_n}.
\end{align*}

If there is additionally an $x∈[1,n]$ with $a_x=0$ then

\[f_n(\overline{χ^{a_1}ι^{j_1}}\otimes … \otimes \overline{χ^{a_n}ι^{j_n}})=0.\]

Equation (<ref>)[$n$] can be reformulated as

\begin{align*}
&f_1\circ m'_n+\underbrace{\sum_{\substack{n=r+s+t \\ r,t\geq 0,s\geq 1\\ s\leq n-1}} (-1)^{rs+t} f_{r+1+t}\circ (1^{\otimes r}\otimes m'_s\otimes 1^{\otimes t})}_{=:Φ_n}\\
&= m_1\circ f_n +\underbrace{\sum_{\substack{2\leq r\leq n \\ i_1+…+i_r=n\\ i_s\geq 1}} (-1)^v m_r\circ (f_{i_1}\otimes f_{i_2}\otimes … \otimes f_{i_r})}_{=:Ξ_n}\,,
\end{align*}

where $v= \sum_{1\leq t < s\leq r}(1-i_s)i_t$.

A term of the form $f_{r+1+t}\circ (1^{\otimes r}\otimes m'_s\otimes 1^{\otimes t})$, $s\geq 3$, $r+t\geq 1$, is zero because of <ref> and the definition of $m'_p$. Also recall $m'_1=0$. Thus

\begin{align}
\label{eqphi}
Φ_n %\eqs& \sum_{\mathclap{n=r+2+t}} (-1)^{r+2t}f_{n-1}(1^{\otimes r}\otimes m'_2\otimes 1^{\otimes t}) \nonumber \\
% \eqs& \sum_{r=0}^{n-2} (-1)^r f_{n-1}(1^{\otimes r}\otimes m'_2\otimes 1^{\otimes n-r-2})
\eqs&\sum_{\mathclap{\substack{n=r+2+t\\ r,t\geq 0}}} (-1)^{2r+t}f_{n-1}\circ (1^{\otimes r}\otimes m'_2\otimes 1^{\otimes t}) = \sum_{r=0}^{n-2} (-1)^{n-r} f_{n-1}\circ (1^{\otimes r}\otimes m'_2\otimes 1^{\otimes n-r-2}).
\end{align}

Because of $m_k=0$ for $k\geq 3$, we have

\begin{align}
\label{eqxi}
Ξ_n %\eqs& \sum_{i_1+i_2 = n}(-1)^s m_2(f_{i_1}\otimes f_{i_2}) = \sum_{i=1}^{n-1}(-1)^{i-1}m_2(f_i\otimes f_{n-i})
\eqs& \sum_{\substack{i_1+i_2 = n \\ i_1,i_2\geq 1}}(-1)^{(1-i_2)i_1} m_2\circ (f_{i_1}\otimes f_{i_2}) = \sum_{i=1}^{n-1}(-1)^{ni}m_2\circ (f_i\otimes f_{n-i}).
\end{align}

We have proven:

For $n\geq 1$, condition (<ref>)$[n]$ is equivalent to $f_1\circ m'_n + Φ_n = m_1\circ f_n + Ξ_n$ where $Φ_n$ and $Ξ_n$ are as in (<ref>) and (<ref>).

Condition (<ref>)$[n]$ holds for $n\geq 3$ and arguments $\overline{χ^{a_1}ι^{j_1}}\otimes…\otimes\overline{χ^{a_n}ι^{j_n}}∈\mathfrak{B}^{\otimes n}=\{\overline{χ^{a_1}ι^{j_1}}\otimes … \otimes \overline{χ^{a_n}ι^{j_n}}  ∈ (\Hm^*A)^{\otimes n} \mid  a_i∈\{0,1\} \text{ and } j_i∈\Z_{\geq 0} \text{ for all }i∈[1,n] \}$ where $0∈\{a_1,…,a_n\}$.

Because of <ref> it is sufficient to show

\[Φ_n(\overline{χ^{a_1}ι^{j_1}}\otimes…\otimes\overline{χ^{a_n}ι^{j_n}}) = Ξ_n(\overline{χ^{a_1}ι^{j_1}}\otimes…\otimes\overline{χ^{a_n}ι^{j_n}})\]

if at least one $a_x$ equals $0$.

Case 1 At least two $a_x$ equal $0$:

To show $Φ_n(\overline{χ^{a_1}ι^{j_1}}\otimes…\otimes\overline{χ^{a_n}ι^{j_n}})=0$, we show

$f_{n-1}(1^{\otimes r}\otimes m'_2\otimes 1^{\otimes n-r-2})(\overline{χ^{a_1}ι^{j_1}}\otimes…\otimes\overline{χ^{a_n}ι^{j_n}})=0$ for $r∈[0,n-2]$: In case both components of the argument of $m'_2$ are of the form $\overline{χ^0ι^j}$, the result of $m'_2$ is of the form $\overline{ι^{j'}}$ (see <ref>). Since $2\leq n-1$, <ref> implies the result of $f_{n-1}$ is zero. Otherwise at least one of the components of the argument of $f_{n-1}$ must be of the form $\overline{ι^j}$ and the result of $f_{n-1}$ is zero as well. So $Φ_n(\overline{χ^{a_1}ι^{j_1}}\otimes…\otimes\overline{χ^{a_n}ι^{j_n}})=0$.

To show $Ξ_n(\overline{χ^{a_1}ι^{j_1}}\otimes…\otimes\overline{χ^{a_n}ι^{j_n}})=0$, we show $m_2(f_i\otimes f_{n-i})(\overline{χ^{a_1}ι^{j_1}}\otimes…\otimes\overline{χ^{a_n}ι^{j_n}})=0$ for $i∈[1,n-1]$:

* Suppose given $i∈[2,n-2]$: The statements $a_1=…=a_i=1$ and $a_{i+1}=…=a_n=1$ cannot be true at the same time, so $f_i(…)=0$ or $f_{n-i}(…)=0$ and we have $m_2(f_i\otimes f_{n-i})(\overline{χ^{a_1}ι^{j_1}}\otimes…\otimes\overline{χ^{a_n}ι^{j_n}})=0$.

* Suppose that $i=1$. Because at least two $a_x$ equal $0$ the statement $a_2=…=a_n=1$ cannot be true. Since $n-1\geq 2$, we have $f_{n-1}(…)=0$ and $m_2(f_1\otimes f_{n-1})(\overline{χ^{a_1}ι^{j_1}}\otimes…\otimes\overline{χ^{a_n}ι^{j_n}})=0$.

* The case $i=n-1$ is analogous to the case $i=1$.

So we have $Φ_n(\overline{χ^{a_1}ι^{j_1}}\otimes…\otimes\overline{χ^{a_n}ι^{j_n}}) = 0=Ξ_n(\overline{χ^{a_1}ι^{j_1}}\otimes…\otimes\overline{χ^{a_n}ι^{j_n}})$.

Case 2a Exactly one $a_x$ equals $0$, where $x∈[2,n-1]$.

We have $Φ_n(\overline{χ^{a_1}ι^{j_1}}\otimes…\otimes\overline{χ^{a_n}ι^{j_n}})=0$: In case $n\geq p+1$, it follows from $f_{n-1}=0$. Let us check the case $n∈[3,p]$: Because of <ref>, we have  unless $r∈\{x-2,x-1\}$. So

\begin{align*}
\eqs& (-1)^{n-x+2}f_{n-1}(1^{\otimes x-2}\otimes m'_2\otimes 1^{\otimes n-x} - 1^{\otimes x-1}\otimes m'_2\otimes 1^{n-x-1})\\
\eqs& (-1)^{n-x}f_{n-1}(\overline{χι^{j_1}}\otimes …\otimes\overline{χι^{j_{x-2}}}\otimes m'_2(\overline{χι^{j_{x-1}}}\otimes \overline{ι^{j_x}})\otimes \overline{χι^{j_{x+1}}}\otimes…\otimes \overline{χι^{j_n}}\\
&- \overline{χι^{j_1}}\otimes …\otimes\overline{χι^{j_{x-1}}}\otimes m'_2(\overline{ι^{j_{x}}}\otimes \overline{χι^{j_{x+1}}})\otimes \overline{χι^{j_{x+2}}}\otimes…\otimes \overline{χι^{j_n}})\\
\eqs& (-1)^{n-x}f_{n-1}(\overline{χι^{j_1}}\otimes …\otimes\overline{χι^{j_{x-2}}}\otimes \overline{χι^{j_{x-1}+j_x}}\otimes \overline{χι^{j_{x+1}}}\otimes…\otimes \overline{χι^{j_n}}\\
&- \overline{χι^{j_1}}\otimes …\otimes\overline{χι^{j_{x-1}}}\otimes \overline{χι^{j_{x}+j_{x+1}}}\otimes \overline{χι^{j_{x+2}}}\otimes…\otimes \overline{χι^{j_n}})\\
\eqs& (-1)^{n-x}((-1)^{n-2}γ_{n-1}ι^{j_1+…+j_n} - (-1)^{n-2}γ_{n-1}ι^{j_1+…+j_n}) = 0
\end{align*}

To show $Ξ_n(\overline{χ^{a_1}ι^{j_1}}\otimes…\otimes\overline{χ^{a_n}ι^{j_n}})=0$, we show $m_2(f_i\otimes f_{n-i})(\overline{χ^{a_1}ι^{j_1}}\otimes…\otimes\overline{χ^{a_n}ι^{j_n}})=0$ for $i∈[1,n-1]$: The element $χ^{a_x}ι^{j_x}$ is a tensor factor of the argument of $f_i$ or of $f_{n-i}$. We write $y=i$ or $y=n-i$ accordingly. Then $y\geq 2$ since $x\notin\{1,n\}$, so $f_y(…) = 0$ and thus $m_2(f_i\otimes f_{n-i})(\overline{χ^{a_1}ι^{j_1}}\otimes…\otimes\overline{χ^{a_n}ι^{j_n}})=0$.

So $Φ_n(\overline{χ^{a_1}ι^{j_1}}\otimes…\otimes\overline{χ^{a_n}ι^{j_n}}) = 0=Ξ_n(\overline{χ^{a_1}ι^{j_1}}\otimes…\otimes\overline{χ^{a_n}ι^{j_n}})$.

Case 2b Only $a_1=0$, all other $a_x$ equal $1$.

We have $f_{n-1}(1^{\otimes r}\otimes m'_2\otimes 1^{\otimes n-r-2})(\overline{χ^{a_1}ι^{j_1}}\otimes…\otimes\overline{χ^{a_n}ι^{j_n}})=0$ unless $r=0$. So

\begin{align*}
\eqs&(-1)^nf_{n-1}(1^{\otimes 0}\otimes m'_2\otimes 1^{\otimes n-2})(\overline{χ^{a_1}ι^{j_1}}\otimes…\otimes\overline{χ^{a_n}ι^{j_n}})\\
\eqs&(-1)^n f_{n-1}(m'_2(\overline{ι^{j_1}}\otimes \overline{χι^{j_2}})\otimes \overline{χι^{j_3}}\otimes …\otimes \overline{χι^{j_n}})\\
\eqs&(-1)^n f_{n-1}(\overline{χι^{j_1+j_2}}\otimes \overline{χι^{j_3}}\otimes …\otimes \overline{χι^{j_n}})\\
%\eqs& s_{n-1}γ_{n-1}ι^{j_1+…+j_n}
\eqs& \begin{cases} γ_{n-1}ι^{j_1+…+j_n} & 3\leq n\leq p\\
0 & n\geq p+1 \end{cases}
\end{align*}

We have $(f_i\otimes f_{n-1})(\overline{χ^{a_1}ι^{j_1}}\otimes…\otimes\overline{χ^{a_n}ι^{j_n}})=0$ if $i\geq 2$. So

\begin{align*}
\eqs& (-1)^{1\cdot n}m_2(f_1\otimes f_{n-1})(\overline{χ^{a_1}ι^{j_1}}\otimes…\otimes\overline{χ^{a_n}ι^{j_n}})\\
\ovs{\eqref{koszulraw}}& (-1)^nm_2\left( (-1)^{n\cdot |ι^{j_1}|}f_1(\overline{ι^{j_1}})\otimes f_{n-1}(\overline{χι^{j_2}}\otimes…\otimes\overline{χι^{j_n}})\right)\\
\eqs& (-1)^n m_2\left(ι^{j_1}\otimes f_{n-1}(\overline{χι^{j_2}}\otimes…\otimes\overline{χι^{j_n}})\right)\\% =  s_{n-1}γ_{n-1}ι^{j_1+…+j_n}
\eqs& \begin{cases} (-1)^nm_2(ι^{j_1}\otimes (-1)^{n-2}γ_{n-1}ι^{j_2+…+j_n}) & 3\leq n\leq p\\
0 & n\geq p+1 \end{cases}\\
\eqs& \begin{cases} γ_{n-1}ι^{j_1+…+j_n} & 3\leq n\leq p\\
0 & n\geq p+1 \end{cases}
% = ι^{j_1}\circ f_{n-1}(χι^{j_2}\otimes…\otimesχ^{j_n})
\end{align*}

So $Φ_n(\overline{χ^{a_1}ι^{j_1}}\otimes…\otimes\overline{χ^{a_n}ι^{j_n}}) =Ξ_n(\overline{χ^{a_1}ι^{j_1}}\otimes…\otimes\overline{χ^{a_n}ι^{j_n}})$.

Case 2c Only $a_n=0$, all other $a_x$ equal $1$.

Argumentation analogous to case 2b gives

\begin{align*}
\eqs&(-1)^{2}f_{n-1}(1^{\otimes n-2}\otimes m'_2\otimes 1^{\otimes 0})(\overline{χ^{a_1}ι^{j_1}}\otimes…\otimes\overline{χ^{a_n}ι^{j_n}})\\
\ovs{|m'_2|=0}&f_{n-1}(\overline{χι^{j_1}}\otimes …\otimes \overline{χι^{j_{n-2}}}\otimes m'_2(\overline{χι^{j_{n-1}}}\otimes \overline{ι^{j_n}}))\\
\eqs& \begin{cases} (-1)^{n-2}γ_{n-1}ι^{j_1+…+j_n} & 3\leq n\leq p\\
0 & n\geq p+1 \end{cases}
\end{align*}

\begin{align*}
\eqs& (-1)^{n(n-1)}m_2(f_{n-1}\otimes f_{1})(\overline{χ^{a_1}ι^{j_1}}\otimes…\otimes\overline{χ^{a_n}ι^{j_n}})\\
\ovs{|f_1|=0}& m_2\left(f_{n-1}(\overline{χι^{j_1}}\otimes…\otimes\overline{χι^{j_{n-1}}})\otimes f_{1}(\overline{ι^{j_n}})\right)\\
%\eqs& m_2( s_{n-1}γ_{n-1}ι^{j_1+…+j_{n-1}} \otimes ι^{j_n})\\
\eqs& \begin{cases} (-1)^{n-2}γ_{n-1}ι^{j_1+…+j_n} & 3\leq n\leq p \\
0& n\geq p+1 \end{cases}
\end{align*}

So $Φ_n(\overline{χ^{a_1}ι^{j_1}}\otimes…\otimes\overline{χ^{a_n}ι^{j_n}}) =Ξ_n(\overline{χ^{a_1}ι^{j_1}}\otimes…\otimes\overline{χ^{a_n}ι^{j_n}})$.

Now we examine the cases where $a_1=…=a_n=1$:

For $n\geq 3$, we have $Φ_n(\overline{χι^{j_1}}\otimes…\otimes\overline{χι^{j_n}})=0$ for $\overline{χι^{j_1}}\otimes…\otimes\overline{χι^{j_n}}∈\mathfrak{B}^{\otimes n}=\{\overline{χ^{a_1}ι^{j_1}}\otimes … \otimes \overline{χ^{a_n}ι^{j_n}}  ∈ (\Hm^*A)^{\otimes n} \mid  a_i∈\{0,1\} \text{ and } j_i∈\Z_{\geq 0} \text{ for all }i∈[1,n] \}$.

We have $Φ_n(\overline{χι^{j_1}}\otimes…\otimes\overline{χι^{j_n}})=0$ since $Φ_n=\sum_{r=0}^{n-2} (-1)^{n-r} f_{n-1}(1^{\otimes r}\otimes m'_2\otimes 1^{\otimes n-r-2})$ and the argument of $m'_2$ is always of the form $\overline{χι^x}\otimes \overline{χι^y}$, whence its result is zero.

Condition (<ref>)$[n]$ holds for $n∈[3,p-1]$ and arguments $\overline{χι^{j_1}}\otimes…\otimes\overline{χι^{j_n}}∈\mathfrak{B}^{\otimes n}=\{\overline{χ^{a_1}ι^{j_1}}\otimes … \otimes \overline{χ^{a_n}ι^{j_n}}  ∈ (\Hm^*A)^{\otimes n} \mid  a_i∈\{0,1\} \text{ and } j_i∈\Z_{\geq 0} \text{ for all }i∈[1,n] \}$.

For computing $Ξ_n$, we first show that $m_2(f_k\otimes f_{n-k})(\overline{χι^{j_1}}\otimes…\otimes\overline{χι^{j_n}})=0$ for $k∈[2,n-2]$. We will need the following congruence.

\begin{align}
\label{gradesmismatch}
\underbrace{(k(l-1)+l(i+x))}_{\equiv_{p-1} k(l-1)+(p-1)+l(i+x)} - \underbrace{(n-k-1+li')}_{\equiv_{p-1} n-k-1+(p-1)+li'} &\equiv_{p-1} -k+k-n+1 = -(n-1) \nonumber \\&
\not\equiv_{p-1} 0
\end{align}

The last statement results from $2\leq n\leq p-1$.
We set "$\pm$" as a symbol for the (a posteriori irrelevant) signs in the following calculation. For $k∈[2,n-2]$, we have

\begin{align*}
\hphantom{XXX}m_2&(f_k\otimes f_{n-k})(\overline{χι^{j_1}}\otimes…\otimes\overline{χι^{j_n}})\\
\eqs& \pm m_2((-1)^{k-1} γ_kι^{j_1+…+j_k}\otimes (-1)^{n-k-1}γ_{n-k}ι^{j_{k+1}+…+j_n})\\
\ovs{\substack {j_1+…+j_k=:x,\\j_{k+1}+…+j_n=:y}}&
\pm γ_kι^x\circ γ_{n-k}ι^{y}\\
\eqs& \pm \left(\dum_{i\geq 0} \ls e_k \rs_{k(l-1)+l(i+x)}^{k-1+li}
+ \dum_{i\geq 0} \ls e_{p-k} \rs_{k(l-1) + (p-1) + l(i+x)}^{k-1+(p-1)+li}\right)\\
&\circ \left(\dum_{i'\geq 0} \ls e_{n-k} \rs_{(n-k)(l-1)+l(i'+y)}^{n-k-1+li'}
+ \dum_{i'\geq 0} \ls e_{p-n+k} \rs_{(n-k)(l-1) + (p-1) + l(i'+y)}^{n-k-1+(p-1)+li'}\right)\\
\ovs{\eqref{gradesmismatch}} 0.
\end{align*}

\begin{align*}
\eqs& m_2((-1)^nf_1\otimes f_{n-1}+(-1)^{n(n-1)}f_{n-1}\otimes f_1)(\overline{χι^{j_1}}\otimes…\otimes\overline{χι^{j_n}})\\
\eqs& m_2((-1)^{n+n|\overline{χι^{j_1}}|}f_1(\overline{χι^{j_{1}}})\otimes f_{n-1}(\overline{χι^{j_2}}\otimes…\otimes\overline{χι^{j_n}})\\
&+ f_{n-1}(\overline{χι^{j_1}}\otimes…\otimes\overline{χι^{j_{n-1}}})\otimes f_1(\overline{χι^{j_n}})) \\
\eqs&  m_2(χι^{j_1}\otimes (-1)^{n-2}γ_{n-1}ι^{j_2+…+j_n} + (-1)^{n-2}γ_{n-1}ι^{j_1+…+j_{n-1}}\otimes χι^{j_n})\\
\eqs& (-1)^n (χι^{j_1}\circ γ_{n-1}ι^{j_2+…+j_n} + γ_{n-1}ι^{j_1+…+j_{n-1}}\circ χι^{j_n})\\
\ovs{\text{P.}\ref{pp:iota}(e),\text{L.}\ref{lem:gamma}}& (-1)^n (χ\circ γ_{n-1} + γ_{n-1}\circ χ)\circ ι^{j_1+…+j_n}
\end{align*}

\begin{align*}
χ\circ γ_{n-1}
\eqs& \Big(\dum_{i\geq 0}\left(\ls e_1 \rs_{(i+1)l-1}^{il} + \left\lgroup\dum_{k=1}^{p-2}\ls e_{k+1,k}\rs_{(i+1)l-1+k}^{il + k}\right\rgroup\right.\\
& \left.  + \ls e_{p-1}\rs_{(i+1)l-1 +(p-1)}^{il+(p-1)}
+ \left\lgroup\dum_{k=1}^{p-2}\ls e_{p -k-1,p-k}\rs_{(i+1)l-1+(p-1)+k}^{il +(p-1)+ k}\right\rgroup\right)\Big) \\
&\circ\Big(\dum_{i'\geq 0} \ls e_{n-1} \rs_{(n-1)(l-1)+li'}^{n-2+li'}
+ \dum_{i'\geq 0} \ls e_{p-n+1} \rs_{(n-1)(l-1) + (p-1) + li'}^{n-2+(p-1)+li'}\Big)\\
\overset{\hphantom{\displaystyle{=}}\mathllap{3\leq n\leq p-1}}{\underset{\hphantom{\displaystyle{=}}\mathllap{\natop{k\leadsto n-1}{i'\leadsto i+1}}}{=}}\eqsp&
\dum_{i\geq 0} \ls e_{n,n-1}\circ e_{n-1} \rs_{(n-1)(l-1)+l(i+1)}^{il+n-1}\\
&+ \dum_{i\geq 0} \ls e_{p-n,p-n+1}\circ e_{p-n+1} \rs_{(n-1)(l-1) + (p-1) + l(i+1)}^{il+p-1+n-1}
\\
\eqs&\dum_{i\geq 0}\left(\ls e_{n,n-1} \rs_{n(l-1) + 1 +li}^{il+n-1}+ \ls e_{p-n,p-n+1}\rs_{n(l-1)+p+li}^{il+p-1+n-1}\right)\\
γ_{n-1}\circ χ \eqs&\Big(\dum_{i'\geq 0} \ls e_{n-1} \rs_{(n-1+i'-1)l+2(p-1)-(n-1)}^{n-2+li'}
+ \dum_{i'\geq 0} \ls e_{p-n+1} \rs_{(n-1+i')l-n + p}^{n-2+(p-1)+li'}\Big) \\
&\circ\Big(\dum_{i\geq 0}\left(\ls e_1 \rs_{(i+1)l-1}^{il} + \left\lgroup\dum_{k=1}^{p-2}\ls e_{k+1,k}\rs_{(i+1)l-1+k}^{il + k}\right\rgroup\right.\\
& \left.  + \ls e_{p-1}\rs_{(i+1)l-1 +(p-1)}^{il+(p-1)}
+ \left\lgroup\dum_{k=1}^{p-2}\ls e_{p -k-1,p-k}\rs_{(i+1)l-1+(p-1)+k}^{il +(p-1)+ k}\right\rgroup\right)\Big)\\
\ovs{\substack{k\leadsto p-n}}& \dum_{i'\geq 0} \ls e_{n-1} \circ e_{n-1,n} \rs_{(n-1+i')l-1+(p-1)+(p-n)}^{n-2+li'}  \\
&+ \dum_{i'\geq 0} \ls e_{p-n+1}\circ e_{p-n+1,p-n} \rs_{(n+i')l-1+p-n}^{n-2+(p-1)+li'}\\
\eqs&  \dum_{i'\geq 0} \ls e_{n-1,n} \rs_{n(l-1)+ i'l}^{n-2+li'} + \dum_{i'\geq 0} \ls e_{p-n+1,p-n} \rs_{n(l-1) +(p-1)+i'l }^{n-2+(p-1)+li'}
\end{align*}

So $χ\circ γ_{n-1} + γ_{n-1}\circ χ = m_1(γ_n)$ by <ref>. Therefore

\begin{align*}
\eqs& (-1)^n  m_1(γ_n)\circ ι^{j_1+…+j_n}
\overset{\text{P.}\ref{pp:iota}(c)}{=} (-1)^n m_1(γ_n ι^{j_1+…+j_n})\\
\eqs& - m_1((-1)^{n-1}γ_n ι^{j_1+…+j_n})\\
\eqs& - m_1\circ f_n(\overline{χι^{j_1}}\otimes…\otimes\overline{χι^{j_n}}).
\end{align*}

We conclude using <ref> by

\begin{align*}
(f_1\circ m'_n + Φ_n)(\overline{χι^{j_1}}\otimes…\otimes\overline{χι^{j_n}}) \overset{\text{L.}\ref{lem:phichi},\text{D.}\ref{defall}}{=} 0 = (m_1\circ f_n + Ξ_n)(\overline{χι^{j_1}}\otimes…\otimes\overline{χι^{j_n}}).
\end{align*}

Condition (<ref>)$[p]$ holds for arguments $\overline{χι^{j_1}}\otimes…\otimes\overline{χι^{j_p}}∈\mathfrak{B}^{\otimes p}=\{\mbox{$\overline{χ^{a_1}ι^{j_1}}\otimes … \otimes \overline{χ^{a_p}ι^{j_p}}$}  ∈ (\Hm^*A)^{\otimes p} \mid  a_i∈\{0,1\} \text{ and } j_i∈\Z_{\geq 0} \text{ for all }i∈[1,p] \}$.

Recall that $|ι|=l=2(p-1)$ is even, $|χ|=l-1$ is odd and $|f_i| = 1-i$ by <ref>. We have

\begin{align*}
Ξ_p(\overline{χι^{j_1}}\otimes…&\otimes\overline{χι^{j_p}}) = \dum_{i=1}^{p-1}(-1)^{pi}m_2(f_i\otimes f_{p-i})
\eqs& \dum_{i=1}^{p-1}(-1)^{pi+i(1-(p-i))}m_2(f_i(\overline{χι^{j_1}}\otimes…\otimes\overline{χι^{j_i}})\otimes f_{p-i}(\overline{χι^{j_{i+1}}}\otimes…\otimes\overline{χι^{j_p}}))\\
\eqs& \dum_{i=1}^{p-1}f_i(\overline{χι^{j_1}}\otimes…\otimes\overline{χι^{j_i}})\circ f_{p-i}(\overline{χι^{j_{i+1}}}\otimes…\otimes\overline{χι^{j_p}})\\
% The last equality involves $(-1)^{(i-1)+i(1-(p-i))} = (-1)^{-1-i(p-i)}$; as $p$ is odd, $i(p-i)$ is even so there is left only $(-1)^{-1} = -1$.\\
% We will need the fact that
% \begin{align}
% \label{spme}
% s_is_{p-i} \eqs& s_{p-1}  & 1\leq i\leq p-1
% \end{align}
% The induction base $i=1$ is trivial.\\
% Now let $s_ks_{p-k}=s_{p-1}$ for some $1\leq k<p-1$. As $s_n = (-1)^{n-1}s_{n-1}$, $n\geq 2$ or equivalently $s_{n+1} = (-1)^n s_n$, $n\geq 1$
% \begin{align*}
% s_{p-1}\eqs& s_ks_{p-k} = (-1)^ks_{k+1}(-1)^{p-k-1}s_{p-k-1}\\
% \eqs& (-1)^{p-1} s_{k+1} s_{p-(k+1)} \overset{p\text{ odd}}{=} s_{k+1}s_{p-(k+1)}
% \end{align*}
% and we have proven the above statement.
% Now
% \begin{align*}
% Ξ_p(\overline{χι^{j_1}}\otimes…&\otimes\overline{χι^{j_p}})= - \dum_{i=1}^{p-1}f_i(\overline{χι^{j_1}}\otimes…\otimes\overline{χι^{j_i}})\circ f_{p-i}(\overline{χι^{j_{i+1}}}\otimes…\otimes\overline{χι^{j_p}})\\
\ovs{p\geq 3\vphantom{I_{I_I}}}& χι^{j_1}\circ (-1)^{p-2}γ_{p-1}ι^{j_2+…+j_p} + (-1)^{p-2}γ_{p-1}ι^{j_1+…+j_{p-1}}\circ χι^{j_p}\\
& + \dum_{i=2}^{p-2} (-1)^{i-1}γ_iι^{j_1+…+j_i}\circ (-1)^{p-i-1}γ_{p-i}ι^{j_{i+1}+…+j_p}\\
\ovs{\text{P.}\ref{pp:iota}(b)}&
(-1)^p\left(χ\circ γ_{p-1} + γ_{p-1}\circ χ + \dum_{k=2}^{p-2} γ_k\circ γ_{p-k}\right)\circ ι^{j_1+…+j_p}
\end{align*}

\begin{align*}
χ\circγ_{p-1}\eqs& \Big(\dum_{i\geq 0}\left(\ls e_1 \rs_{(i+1)l-1}^{il} + \left\lgroup\dum_{k=1}^{p-2}\ls e_{k+1,k}\rs_{(i+1)l-1+k}^{il + k}\right\rgroup\right.\\
&\left. + \ls e_{p-1}\rs_{(i+1)l-1 +(p-1)}^{il+(p-1)}
+ \left\lgroup\dum_{k=1}^{p-2}\ls e_{p -k-1,p-k}\rs_{(i+1)l-1+(p-1)+k}^{il +(p-1)+ k}\right\rgroup\right)\Big)\\
&\circ\left(\dum_{i'\geq 0} \ls e_{p-1} \rs_{(p-1)(l-1)+li'}^{(p-1)-1+li'}+ \dum_{i'\geq 0} \ls e_{1} \rs_{(p-1)(l-1) + (p-1) + li'}^{-1+2(p-1)+li'}\right)\\
\eqs& \dum_{i\geq 0} \ls e_{p-1} \rs_{(p-1)(l-1)+l(i+1)}^{il+(p-1)}+ \dum_{i\geq 0} \ls e_{1} \rs_{(p-1)(l-1) + (p-1) + li}^{il}\\
\eqs& \dum_{i\geq 0} \ls e_{p-1} \rs_{(p+i-1)l+(p-1)}^{il+(p-1)}+ \dum_{i\geq 0} \ls e_{1} \rs_{(p+i-1)l}^{il}
\end{align*}

\begin{align*}
γ_{p-1}\circ χ\eqs& \left(\dum_{i'\geq 0} \ls e_{p-1} \rs_{(p+i'-2)l+(p-1)}^{(p-1)-1+li'}+ \dum_{i'\geq 0} \ls e_{1} \rs_{(p+i'-1)l}^{-1+2(p-1)+li'}\right) \\
&\circ\Big(\dum_{i\geq 0}\left(\ls e_1 \rs_{(i+1)l-1}^{il} + \left\lgroup\dum_{k=1}^{p-2}\ls e_{k+1,k}\rs_{(i+1)l-1+k}^{il + k}\right\rgroup\right.\\
&\left. + \ls e_{p-1}\rs_{(i+1)l-1 +(p-1)}^{il+(p-1)}
+ \left\lgroup\dum_{k=1}^{p-2}\ls e_{p -k-1,p-k}\rs_{(i+1)l-1+(p-1)+k}^{il +(p-1)+ k}\right\rgroup\right)\Big)\\
\eqs& \dum_{i'\geq 0} \ls e_{p-1} \rs_{(p+i'-1)l-1+(p-1)}^{(p-1)-1+li'}+ \dum_{i'\geq 0} \ls e_{1} \rs_{(p+i')l-1}^{-1+2(p-1)+li'}\\
\eqs& \dum_{i'\geq 0} \ls e_{p-1} \rs_{(p+i'-1)l+p-2}^{p-2+i'l}+ \dum_{i'\geq 0} \ls e_{1} \rs_{(p+i'-1)l+l-1}^{i'l+l-1}
\end{align*}

\begin{align*}
γ_k\circ γ_{p-k}\eqs& \left(\dum_{i\geq 0} \ls e_k \rs_{(i+k-1)l+l-k}^{k-1+li}
+ \dum_{i\geq 0} \ls e_{p-k} \rs_{(i+k)l+ (p-1)-k}^{k-1+(p-1)+li}\right)\\
&\circ\left(\dum_{i'\geq 0} \ls e_{p-k} \rs_{(p-k)(l-1)+li'}^{p-k-1+li'}
+ \dum_{i'\geq 0} \ls e_{k} \rs_{(p-k)(l-1) + (p-1) + li'}^{-k+2(p-1)+li'}\right)\\
\eqs&\dum_{i\geq 0} \ls e_k \rs_{(p-k)(l-1)+(p-1)+l(i+k-1)}^{k-1+li}
+ \dum_{i\geq 0} \ls e_{p-k} \rs_{(p-k)(l-1)+l(i+k)}^{k-1+(p-1)+li}\\
\eqs&\dum_{i\geq 0} \ls e_k \rs_{(p-k+i+k-1)l -(p-k)+(p-1)}^{k-1+li}
+ \dum_{i\geq 0} \ls e_{p-k} \rs_{(p-k+i+k)l-(p-k)}^{k-1+(p-1)+li}\\
\eqs&\dum_{i\geq 0} \ls e_k \rs_{(p+i-1)l + k-1}^{k-1+li}
+ \dum_{i\geq 0} \ls e_{p-k} \rs_{(p+i-1)l+ k-1 + (p-1)}^{k-1+(p-1)+li}\,.
\end{align*}

\begin{align*}
χ\circ γ_{p-1} +& γ_{p-1}\circ χ + \dum_{k=2}^{p-2} γ_k\circ γ_{p-k}\\
\eqs&\dum_{i\geq 0} \dum_{k=0}^{p-2}\left(\ls e_{k+1} \rs_{(p+i-1)l + k}^{k+li}
+  \ls e_{p-k-1} \rs_{(p+i-1)l+ k + (p-1)}^{k+(p-1)+li}\right)\\
\eqs& \dum_{i\geq 0} \dum_{k'=0}^{l-1} \ls e_{ω(k')}\rs_{(p-1+i)l+k'}^{k'+li}
\overset{\text{P.}\ref{pp:iota}(a)}{=} ι^{p-1}
\end{align*}

\begin{align*}
Ξ_p(\overline{χι^{j_1}}\otimes…\otimes\overline{χι^{j_p}})=(-1)^pι^{p-1+j_1+…+j_p}\,.% = -ι^{p-1+j_1+…+j_n}.
\end{align*}

So we conclude using <ref> by

\begin{align*}
\begin{array}{rcl}
(f_1\circ m'_p + Φ_p)(\overline{χι^{j_1}}\otimes…\otimes\overline{χι^{j_p}}) &\overset{\text{L.}\ref{lem:phichi},\text{D.}\ref{defall}}{=} &(-1)^p ι^{p-1+j_1+…+j_p} \\
&\overset{\text{D.}\ref{defall}}{=}& (m_1\circ f_p + Ξ_p)(\overline{χι^{j_1}}\otimes…\otimes\overline{χι^{j_p}}).
\end{array}
\end{align*}

Condition (<ref>)$[n]$ holds for $n∈[p+1,2(p-1)]$ and arguments

$\mbox{$\overline{χι^{j_1}}\otimes…\otimes\overline{χι^{j_n}}$} ∈\mathfrak{B}^{\otimes n}=\{\overline{χ^{a_1}ι^{j_1}}\otimes … \otimes \overline{χ^{a_n}ι^{j_n}}  ∈ (\Hm^*A)^{\otimes n} \mid  a_i∈\{0,1\} \text{ and } j_i∈\Z_{\geq 0} \text{ for all }i∈[1,n] \}$.

As $f_k=0$ for $k\geq p$, we have

\begin{align*}
Ξ_n&(\overline{χι^{j_1}}\otimes…\otimes\overline{χι^{j_n}}) = \dum_{k=n-p+1}^{p-1} (-1)^{nk}m_2(f_k\otimes f_{n-k})
\end{align*}

The right side is a linear combination of terms of the form $γ_k\circ γ_{n-k}$ for $k∈[n-p-1,p-1]$. We have

\begin{align*}
γ_k\circ γ_{n-k} \eqs& \left(\dum_{i\geq 0} \ls e_k \rs_{k(l-1)+li}^{k-1+li}
+ \dum_{i\geq 0} \ls e_{p-k} \rs_{k(l-1) + (p-1) + li}^{k-1+(p-1)+li}\right)\\
&\circ\left(\dum_{i'\geq 0} \ls e_{n-k} \rs_{(n-k)(l-1)+li'}^{n-k-1+li'}
+ \dum_{i'\geq 0} \ls e_{p-n+k} \rs_{(n-k)(l-1) + (p-1) + li'}^{n-k-1+(p-1)+li'}\right)
\end{align*}

A necessary condition for that term  to be non-zero is  $k(l-1)\equiv_{p-1}n-k-1$ as $l=2(p-1)$. We have

\begin{align*}
k(l-1)-(n-k-1) \equiv_{p-1}\eqsp& -k-n+k+1 = 1-n
\not\equiv_{p-1} 0,
\end{align*}

since $p\leq n-1\leq 2(p-1)-1$. So $γ_k\circ γ_{n-k}= 0$ and $Ξ_n(\overline{χι^{j_1}}\otimes…\otimes\overline{χι^{j_n}})=0$.
We conclude using <ref> by

\begin{align*}
(f_1\circ m'_n + Φ_n)(\overline{χι^{j_1}}\otimes…\otimes\overline{χι^{j_n}}) \overset{\text{L.}\ref{lem:phichi}, \text{D.}\ref{defall}}{=} 0 \overset{\text{D.}\ref{defall}}{=} (m_1\circ f_n + Ξ_n)(\overline{χι^{j_1}}\otimes…\otimes\overline{χι^{j_n}}).
\end{align*}

One could formulate a lemma similar to <ref> for the case $n> 2(p-1)$ as then the sum $ \sum_{k=n-p+1}^{p-1} (-1)^{nk} m_2(f_k\otimes f_{n-k})
(\overline{χι^{j_1}}\otimes…\otimes\overline{χι^{j_n}})$ is in fact empty.
Instead we use  <ref> to prove (<ref>)[$n$] for $n>2p-2$:

<ref> ensure that (<ref>)[$n$] holds for . Then <ref> with $k=p$ proves that (<ref>)[$n$] holds for all $n∈[1,∞]$, cf. also <ref>. By <ref>, $f_1$ is injective. By <ref>, the degrees are as required in <ref>. <ref> proves that $(\Hm^* A, (m'_n)_{n\geq 1})$ is an $\A_∞$-algebra and $(f_n)_{n\geq 1}$ is an $\A_∞$-morphism from  $(\Hm^* A, (m'_n)_{n\geq 1})$ to $(A, (m_n)_{n\geq 1})$.
By <ref>, we have $m'_1=0$. Thus $(\Hm^* A, (m'_n)_{n\geq 1})$ is a minimal $\A_∞$-algebra. By <ref>, the complex morphism $f_1:(\Hm^* A,m'_1)→(A,m_1)$ is a quasi-isomorphism which induces the identity in homology. So the $\A_∞$-morphism $(f_n)_{n\geq 1}: (\Hm^* A, (m'_n)_{n\geq 1}) → (A, (m_n)_{n\geq 1})$ is a quasi-isomorphism and the proof of <ref> is complete.

§.§ At the prime 2

We examine the case at the prime $2$.

We use a direct approach.
Note that $\Sy_2$ is a cyclic group so the theory of cyclic groups applies as well.