2. The group algebra and its cohomology

We assume from now on, without loss of generality, that $G$ has a normal cyclic Sylow $p$-subgroup $P=C_G(P)$, with inertial index $q=|G:P|$. We shall assume that $q>1$, which then forces $p$ to be odd. For notation, let

\[ G=\langle g,s\mid g^{p^{n}}=1, s^{q}=1, sgs^{{-}1}=g^{\gamma}\rangle\cong \mathbb{Z}/p^{n}\rtimes\mathbb{Z}/q, \]

where $\gamma$ is a primitive $q$th root of unity modulo $p^{n}$. Let $P=\langle g\rangle$ and $H=\langle s\rangle$ as subgroups of $G$.

In this section, we introduce a grading on $kG$. This comes from the fact that the radical filtration of $kG$ is isomorphic to its associated graded in this case. This is a somewhat rare phenomenon, but when it happens, it induces an extra grading on mod $p$ cohomology of $BG$ and homology of $\Omega BG{{}^{{}^{\wedge }}_p}$, that we can exploit to good effect.

Let $k$ be a field of characteristic $p$. The action of $H$ on $kP$ by conjugation preserves the radical series, and since $|H|$ is not divisible by $p$, there are invariant complements. Thus, we may choose an element $U\in J(kP)$ such that $U$ spans an $H$-invariant complement of $J^{2}(kP)$ in $J(kP)$. It can be checked that

\[ U = \sum_{\substack{1\leqslant j \leqslant p^{n}-1, \\j^{p}\equiv j\pmod{p}}} g^{j}/j \]

is such an element, and that $sUs^{-1}=\gamma U$. This gives us the following presentation for $kG$:

\[ kG = k\langle s,U \mid U^{p^{n}}=0, s^{q}=1, sU = \gamma Us \rangle. \]

We shall regard $kG$ as a $\mathbb {Z}[{1}/{q}]$-graded algebra with $|s|=0$ and $|U|=1/q$. Then the bar resolution is doubly graded, and taking homomorphisms into $k$, the cochains $C^{*}(BG;k)$ inherit a double grading. The differential decreases the homological grading and preserves the internal grading. Thus, the cohomology $H^{*}(G,\,k)=H^{*}(BG;k)$ is doubly graded:

\[ H^{*}(BG;k) = k[x] \otimes \Lambda(t) \]

where $|x|=(-2q,\,-p^{n})$, $|t|=(-2q+1,\,-h)$, and $h=p^{n}-(p^{n}-1)/q$. Here, the first degree is homological, the second internal. The Massey product $\langle t,\,t,\,\ldots ,\,t\rangle$ ($p^{n}$ repetitions) is equal to $-x^{h}$. This may easily be determined by restriction to $P$, where it is well known that the $p^{n}$-fold Massey product of the degree one exterior generator is a non-zero degree two element. The usual convention is to make the constant $-1$, because this Massey product is minus the $n$th Bockstein of $t$ [Reference Kraines10, Theorem 19].

3. $A_\infty$-algebras

An $A_{\infty }$-algebra over a field is a $\mathbb {Z}$-graded vector space $A$ with graded maps $m_n: A^{\otimes n}\rightarrow A$ of degree $n-2$ for $n\geq 1$ satisfying

\[ \sum_{r+s+t=n}({-}1)^{r+st}m_{r+1+t}(id^{{\otimes} r}\otimes m_s \otimes id^{{\otimes} t})=0 \]

for $n\geq 1$. The map $m_1$ is therefore a differential, and the map $m_2$ induces a product on $H_*(A)$.

A theorem of Kadeishvili [Reference Kadeishvili7] (see also Keller [Reference Keller8, Reference Keller9] or Merkulov [Reference Merkulov13]) may be stated as follows. Suppose that we are given a differential graded algebra $A$, over a field $k$. Let $Z^{*}(A)$ be the cocycles, $B^{*}(A)$ be the coboundaries, and $H^{*}(A)=Z^{*}(A)/B^{*}(A)$. Choose a vector space splitting $f_1\colon H^{*}(A) \to Z^{*}(A) \subseteq A$ of the quotient. Then this gives by an inductive procedure an $A_\infty$ structure on $H^{*}(A)$ so that the map $f_1$ is the degree one part of a quasi-isomorphism of $A_\infty$-algebras.

If $A$ happens to carry auxiliary gradings respected by the product structure and preserved by the differential, then it is easy to check from the inductive procedure that the maps in the construction may be chosen so that they also respect these gradings. It then follows that the structure maps $m_i$ of the $A_\infty$ structure on $H^{*}(A)$ also respect these gradings.

Let us apply this to $H^{*}(BG;k)$. We examine the elements $m_i(t,\,\ldots ,\,t)$. By definition, we have $m_1(t)=0$ and $m_2(t,\,t)=0$. The degree of $m_i(t,\,\ldots ,\,t)$ is $i$ times the degree of $t$, increased in the homological direction by $i-2$. This gives

\[ |m_i(t,\ldots,t)| = i({-}2q+1,h)+(i-2,0) =({-}2iq+2i-2,ih). \]

The homological degree is even, so if $m_i(t,\, \ldots ,\, t)$ is non-zero then it is a multiple of a power of $x$. Comparing degrees, if $m_i(t,\,\ldots ,\,t)$ is a non-zero multiple of $x^{\alpha }$ then we have

\[ 2iq-2i+2=2\alpha q,\quad ih = \alpha p^{n}. \]

Eliminating $\alpha$, we obtain $(iq-i+1)p^{n}=ihq$. Substituting $h=p^{n}-(p^{n}-1)/q$, this gives $i=p^{n}$. Finally, since the Massey product of $p^{n}$ copies of $t$ is equal to $-x^{h}$, it follows that $m_{p^{n}}(t,\,\ldots ,\,t)=\epsilon (p^{n}) x^{h}$, where the sign is as defined in Remark 1.2 ([Reference Lu, Palmieri, Wu and Zhang11, Theorem 3.1], corrected in [Reference Buijs, Moreno-Fernández and Murillo4, Theorem 3.2]). Thus, we have

\[ m_i(t,\ldots,t) = \begin{cases} \epsilon (p^{n}) x^{h} & i=p^{n} \\ 0 & \text{otherwise.} \end{cases} \]

We shall elaborate on this argument in a more general context in the next section, where we shall see that the rest of the $A_\infty$ structure is also determined in a similar way.

4. $A_\infty$ structures on a polynomial tensor exterior algebra

In this section, we shall examine the following general situation. Our goal is to establish that there are only two possible $A_\infty$ structures satisfying Hypothesis 4.1, and that the Koszul dual also satisfies the same hypothesis with the roles of $a$ and $b$, and of $h$ and $\ell$ reversed.

Proof. The argument is the same as in the last section. The degree of $m_i(t,\,\ldots ,\,t)$ is $i|t| +(i- 2,\,0) = (-2ib - 2,\, -ih)$. Since the homological degree is even, if $m_i(t,\,\ldots ,\,t)$ is non-zero then it is a multiple of some power of $x$, say $x^{\alpha }$. Then we have

\[ 2ib+2 = 2\alpha a,\qquad ih = \alpha\ell. \]

Eliminating $\alpha$ gives $(ib+1)\ell =iha$, and so using $ha-\ell b =1$ we have $i=\ell$. Substituting back gives $\alpha = h$.

Elaborating on this argument gives the entire $A_\infty$ structure. If $m_\ell (t,\,\ldots ,\,t)$ is non-zero, then by rescaling the variables $t$ and $x$ if necessary we can assume that $m_\ell (t,\,\ldots ,\,t)=\epsilon (\ell ) x^{h}$ (note that we can even do this without extending the field, since $\ell$ and $h$ are coprime).

Proof. All monomials live in different degrees, so we do not need to consider linear combinations of monomials. Suppose that $m_i(x^{j_1}t^{\varepsilon _1},\,\ldots ,\,x^{j_i}t^{\varepsilon _i})$ is some constant multiple of $x^{j}t^{\varepsilon }$, where each of $\varepsilon _1,\,\ldots ,\,\varepsilon _i,\,\varepsilon$ is either zero or one. Then comparing degrees, we have

\[ (j_1+\cdots+j_i)|x| + (\varepsilon_1+\cdots+\varepsilon_i)|t| + (i-2,0) = j|x| + \varepsilon|t|. \]

Setting

\[ \alpha=j_1+\cdots+j_i-j,\quad \beta = \varepsilon_1+\cdots+\varepsilon_i-\varepsilon \]

we have $\beta \leqslant i$, and

\[ \alpha({-}2a,-\ell) + \beta({-}2b-1,-h) + (i-2,0) = 0. \]

Thus

\[ 2\alpha a + 2\beta b + \beta + 2 - i = 0, \quad \alpha \ell +\beta h = 0. \]

Eliminating $\alpha$, we obtain

\[{-}2\beta ha + 2\beta \ell b + \beta \ell + 2 \ell - i \ell = 0. \]

Since $ha-\ell b=1$, this gives $\beta = \ell (i-2) /(\ell -2)$. Combining this with $\beta \leqslant i$ gives $i\leqslant \ell$. If $i<\ell$ then $\beta$ is not divisible by $\ell$, and so $\alpha \ell + \beta h = 0$ cannot hold. So we have $\beta =i=\ell$, $\varepsilon _1=\dots =\varepsilon _\ell =1$, $\varepsilon =0$, $\alpha =-h$, and $j=h+j_1+\cdots +j_\ell$. Finally, the identities satisfied by the $m_i$ for an $A_\infty$ structure show that all the constant multiples have to be the same, hence all equal to zero or after rescaling, all equal to $\epsilon (\ell )$.

Theorem 4.5 Under Hypothesis 4.1, if $\ell > 2$ then there are two possible $A_\infty$ structures on $A$. There is the formal one, where $m_i$ is equal to zero for $i>2$, and the non-formal one, where after replacing $x$ and $t$ by suitable multiples, the only non-zero $m_i$ with $i>2$ is $m_\ell$, and the only non-zero values on monomials are given by

\[ m_\ell(x^{j_1}t,\ldots,x^{j_\ell}t)=\epsilon(\ell)x^{h+j_1+\cdots+j_\ell}. \]

5. Loops on $BG{{}^{{}^{\wedge }}_p}$

In general, for a finite group $G$, the classifying space $BG$ is $p$-good, see Proposition VII.5.1 of Bousfield and Kan [Reference Bousfield and Kan3]. So its $p$-completion $BG{{}^{{}^{\wedge }}_p}$ is $p$-complete. This space and its loop space have been the subject of considerable study, beginning with the work of Cohen and Levi [Reference Cohen and Levi5]. We have $H^{*}(BG{{}^{{}^{\wedge }}_p};k)\cong H^{*}(BG;k)=H^{*}(G,\,k)$ and $\pi _1(BG{{}^{{}^{\wedge }}_p})=G/O^{p}(G)$, the largest $p$-quotient of $G$. In our case, $G=\mathbb {Z}/p^{n}\rtimes \mathbb {Z}/q$ with $q>1$, we have $G=O^{p}(G)$ and so $BG{{}^{{}^{\wedge }}_p}$ is simply connected. So the Eilenberg–Moore spectral sequence converges to the homology of its loop space:

\[ \mathsf{Ext}^{*,*}_{H^{*}(BG;k)}(k,k) \cong \mathsf{Cotor}^{*,*}_{H_*(BG;k)}(k,k) \Rightarrow H_*(\Omega BG{{}^{{}^{{\wedge}}}_p};k) \]

(Eilenberg–Moore [Reference Eilenberg and Moore6], Smith [Reference Smith14]). The internal grading on $C^{*}(BG;k)$ gives this spectral sequence a third grading that is preserved by the differentials, and $H_*(\Omega BG{{}^{{}^{\wedge }}_p};k)$ is again doubly graded. Since $H^{*}(G,\,k)=k[x] \otimes \Lambda (t)$ with $|x|=(-2q,\,-p^{n})$ and $|t|=(-2q+1,\,-h)$, it follows that the $E^{2}$ page of this spectral sequence is $k[\tau ] \otimes \Lambda (\xi )$ where $|\xi |=(-1,\,2q,\,p^{n})$ and $|\tau |=(-1,\,2q-1,\,h)$ (recall $h=p^{n}-(p^{n}-1)/q$). Provided that we are not in the case $h=2$, which only happens if $p^{n}=3$, ungrading $E^{\infty }$ gives

\[ H_*(\Omega BG{{}^{{}^{{\wedge}}}_p};k) = k[\tau] \otimes \Lambda(\xi) \]

with $|\tau |=(2q-2,\,h)$ and $|\xi |=(2q-1,\,p^{n})$.

In the exceptional case $h=2$, $p^{n}=3$, we have $q=2$, and the group $G$ is the symmetric group $\Sigma _3$ of degree three. An explicit computation (for example by squeezed resolutions [Reference Benson2]) gives

\[ H_*(\Omega (B\Sigma_3){}^{{}^{{\wedge}}}_3;k) = k[\tau,\xi]/(\xi^{2}+\tau^{3}) \]

with $|\tau |=(2,\,2)$ and $|\xi |=(3,\,3)$, and the two gradings collapse to a single grading.

Applying Theorem 4.5, and using the fact that either formal case is Koszul dual to the other, we have the following.

Theorem 5.1 Suppose that $p^{n}\ne 3$. Then the $A_\infty$ structure on $H_*(\Omega BG{{}^{{}^{\wedge }}_p};k)=k[\tau ]\otimes \Lambda (\xi )$ is given by

\[ m_h(\tau^{j_1}\xi,\ldots,\tau^{j_h}\xi)=\epsilon (h) \tau^{p^{n}+j_1+\cdots+j_h}, \]

and for $i>2$, all $m_i$ on all other $i$-tuples of monomials give zero.

Again using [Reference Lu, Palmieri, Wu and Zhang11] corrected in [Reference Buijs, Moreno-Fernández and Murillo4], we may reinterpret this in terms of Massey products.

Note that the exceptional case $p^{n}=3$ also fits the corollary, if we interpret a $2$-fold Massey product as an ordinary product.