A minimal A∞-model whose ternary operation realizes a triple Massey product

The product is associative on triples involving the unit by the unit axiom. On nonunit basis elements, every nonzero pair product is one of $a,b,c$, each of which annihilates every nonunit. The only triple for which both adjacent pair-products are nonzero is $(x,y,z)$; both $(xy)z=az=0$ and $x(yz)=xb=0$. If just one adjacent pair-product is nonzero, the iterated product is zero whichever pair is multiplied first. Hence the bilinear product is associative.

The differential has degree $+1$: it sends $u,v$ of degree $-1$ to $a,b$ of degree zero and is zero on the other homogeneous basis elements. Since $d(a)=d(b)=0$, $d^2(u)=d^2(v)=0$, and $d^2$ is zero on every other basis element as well.

Define $f_1:H^{\ast }(A)\to A$ by the listed cocycle representatives, define $f_2([x],[y])=-u$ and $f_2([y],[z])=-v$, and set $f_2=0$ on all other basis pairs, in particular on pairs containing $[1_A]$. Put $f_n=0$ for $n\ge 3$. The degrees are $\mathrm{\mid }{f}_1\mathrm{\mid }=0$ and $\mathrm{\mid }{f}_2\mathrm{\mid }=-1$, as required by [F5].

For the unit pairs, the signed Leibniz rule follows from $d(1_A)=0$. For the three nonzero nonunit products, both sides agree: $d(xy)=0=d(x)y+xd(y)$, $d(yz)=0=d(y)z+yd(z)$, and $d(xv)=0=d(x)v+xd(v)=xb$. If a nonunit product is zero and one factor is $u$ or $v$, its differential is $a$ or $b$, which annihilates every nonunit; the remaining term is zero because the other differential is zero. If neither factor is $u$ nor $v$, both differentials vanish. Thus the signed Leibniz rule holds on basis pairs and hence, by bilinearity, on all homogeneous elements.

The map $d:A^{-1}\to A^0$ sends $u\mapsto a$, $v\mapsto b$, and $c\mapsto 0$. Its kernel is $kc$, its image is $k\{a,b\}$, and there is no degree $-2$ term. Also $d$ vanishes on $A^0$. Therefore $H^{-1}(A)=k[c]$, $H^0(A)=k[1_A]\oplus k[x]\oplus k[y]\oplus k[z]$, and all other cohomology groups are zero.

By steps 1.1, 1.2 and 2.1 the multiplication and differential satisfy every DGA axiom in [F1], so $A$ is a cohomological DGA. Its associated A∞-structure is the one of [F3].

The DGA product descends to these cohomology classes: replacing a degree-zero cycle by a boundary changes it by a multiple of $a$ or $b$, which multiplies every nonunit cycle to zero and multiplies the unit to a boundary; products involving $[c]$ and a nonunit cycle are zero. The unit acts as $[1_A]$. The only nonzero products among the chosen nonunit degree-zero representatives are $xy=a=du$ and $yz=b=dv$, so their classes are zero. Thus every product in the reduced cohomology span $k\{[x],[y],[z],[c]\}$ is zero.

The triple $(x,y,z;u,v)$ is valid Massey data because $xy=du=a$ and $yz=dv=b$. The fixed expression of [F7] is $-xv+uz=-c+0=-c$. Every representative of $[x]$, $[y]$, or $[z]$ differs from that basis vector by an element of $k\{a,b\}$. Their adjacent products remain $a$ and $b$, since $a,b$ annihilate nonunit basis elements. Every null-homotopy of $a$ is $u+\lambda c$ and every null-homotopy of $b$ is $v+\mu c$ for scalars $\lambda ,\mu \in k$. All correction products involving $a,b,c$ in the expression vanish, so every datum representing these three classes gives $-c$. The triple Massey product is exactly $\{-[c]\}$.

On $H^{\ast }(A)$, take $m_1=0$ and let $m_2$ be the induced product. Define $m_3([x],[y],[z])=-[c]$, let $m_3$ be zero on every other ordered triple of the displayed basis, and set $m_n=0$ for $n\ge 4$. The only nonzero value of $m_3$ has degree $0+0+0-1=-1$, as required by [F4].

The arity-one morphism identity says $d{f}_1=f_1{m}_1$, and both sides are zero because the chosen representatives are cycles. In arity two, the identity is $f_1{m}_2^H=d{f}_2+\mu (f_1\otimes f_1)$. On the pairs $([x],[y])$ and $([y],[z])$, its right side is respectively $-a+a=0$ and $-b+b=0$; on all other pairs of nonunit basis classes both terms are zero. If an input is $[1_A]$, $f_2$ is zero and $f_1$ preserves the unit, so the equality holds there too.

For arities $n=1,2$, the Stasheff sums vanish because $m_1=0$. For $n=3$, the identity is associativity of $m_2$: it holds on the unit by the unit law, and on three nonunit basis elements because their pairwise products are zero. Thus the identities hold through arity three.

In arity three, on $([x],[y],[z])$ the left side is $f_1{m}_3=-c$, and the right side is $\mu (f_1\otimes f_2)-\mu (f_2\otimes f_1)=x(-v)-(-u)z=-c$. On every other triple of nonunit basis classes both sides are zero: $m_3$ is zero there, and the only nonzero $f_2$ values have products $uz=0$ or $xv=c$ in precisely the triple already checked. If one input is the unit, the normalized condition $f_2(1_A,-)=f_2(-,1_A)=0$ makes the identity reduce to the unit law. For $(1_A,a,b)$ both sides equal $f_2(a,b)$; for $(a,1_A,b)$ both sides are zero; and for $(a,b,1_A)$ both sides equal $-f_2(a,b)$. With at least two units every $f_2$ term is zero.

In arity four, the only terms are insertions of $m_2$ into $m_3$ and of $m_3$ into $m_2$. On four nonunit basis inputs, the first terms vanish because $m_2$ is zero on nonunit pairs; the second vanish because $m_3$ outputs a multiple of $[c]$ and $m_2([c],h)=m_2(h,[c])=0$ for nonunit $h$. If a unit occurs but the other three inputs are not $[x],[y],[z]$ in that order, every $m_3$ term is zero. For the four insertions of a unit into $([x],[y],[z])$, the cancellations in the arity-four sum are respectively $(-[c])+[c]$, $(-[c])+[c]$, $[c]-[c]$, and $(-[c])+[c]$: they come from $m_3(m_2(1_A,x),y,z)$ paired with $-m_2(1_A,m_3(x,y,z))$, $m_3(m_2(x,1_A),y,z)$ paired with $-m_3(x,m_2(1_A,y),z)$, $-m_3(x,m_2(y,1_A),z)$ paired with $m_3(x,y,m_2(1_A,z))$, and $m_3(x,y,m_2(z,1_A))$ paired with $-m_2(m_3(x,y,z),1_A)$. With two or more units every term vanishes by the prescribed zero values of $m_3$. Hence the arity-four sum is zero on every basis quadruple and therefore on all inputs.

In arity four, the only possible left terms apply $f_2$ to an $m_3$ output, which is a multiple of $[c]$ and is sent to zero by $f_2$. The only possible right term with nonzero Taylor maps is a product of two $f_2$ outputs; all products among $u,v$ are zero. For $n\ge 5$, the left side requires an input operation of arity at most three and an outer $f$ of arity at most two, so $n\le 4$; the right side requires a target operation of arity at most two and each $f_i$ of arity at most two, again so $n\le 4$. All morphism identities in arities $n\ge 4$ therefore hold.

The map $f_1$ sends the basis $[1_A],[x],[y],[z],[c]$ of $H^{\ast }(A)$ to representatives of the same cohomology basis computed in step 2.2. It therefore induces an isomorphism on cohomology, so $f$ is an A∞ quasi-isomorphism.

In arity five, the only possible nonzero composition is $m_3$ inserted into $m_3$; its inner output is a multiple of $[c]$, on which the outer $m_3$ is zero. For $n\ge 6$, a nonzero summand would require both inner and outer arities at most three, forcing $n$ to be at most five. Thus every higher Stasheff sum vanishes. Together with the arity-four verification, these establish the stated minimal A∞-structure on $H^{\ast }(A)$.

By [F6], the A∞-algebra $H^{\ast }(A)$ with $m_1=0$ and the morphism $f$ is a minimal A∞-model for $A$. Its value $m_3([x],[y],[z])=-[c]$ is the unique triple Massey product computed in step 3.3.