Using: <https://arxiv.org/pdf/math/9910179.pdf> as a reference...


My question involves spelling out explicitly the comment in 4.2 -


"Equivalently, the datum of an $A\_\infty$-structure on a graded space $M$ is the datum of a differential $m\_1$ on $M$ and of a morphism of $A\_\infty$-algebras from $A$ to the opposite of the dg-algebra $Hom\_k(M,M)$ (cf. 3.3)."


$Q1$: What is the $A\_\infty$ structure on $Hom\_k(M,M)^\text{op}$? (Is it just $m\_1$ is the usual differential and $m\_2$ is composition in opposite order with all higher $m\_i$ vanishing for $i>2$..?) Could they possibly mean any other $A\_\infty$-structure?


$Q2$: If my answer for Q1 is correct how does this imply that an $A\_\infty$-module structure on $M$ is the same as an $A\_\infty$ map $\alpha:A \to Hom\_k(M,M)^\text{op}$ given that:


a map of $A\_\infty$-algebras is the same as having 


$$
 \alpha\_m:A^{\otimes m}\to Hom\_k(M,M)^\text{op}
$$


for all $m>0$ s.t.


$$
\sum(-1)^{r+st}\alpha\_u(1^{\otimes r}\otimes m^A\_s \otimes 1^{\otimes t})=\sum(-1)^{s}m^M\_r(\alpha\_{i\_1} \otimes \cdots \otimes \alpha\_{i\_r} )
$$


where the left hand side is a sum over all decompositions $m=r+s+t$ where $u=r+1+t$, $s\geq1$, and $r,t\geq 0$. The right hand side is a sum over all $1\leq r \leq m$ and over all $i\_1+\cdots +i\_r=m$ and the sign on the right is given in the paper in 3.4.


$m^M\_r$ are the higher multiplications on $Hom\_k(M,M)$ that are referenced in Q1.


Ps. Obviously, there will have to be some use of tensor-hom adjunction, I just can't quite spell it out.



A-infinity modules
Let $(A,(m\_i)\_i)$ be an $A\_\infty$-algebra and $(M,d\_M)$, a (co)chain complex.


$Q1$: We endow $L:=Hom\_k^\*(M,M)$ with an $A\_\infty$-structure where


$$
m\_1^L(f):=d\_M\circ f - (-1)^{\text{deg} f} f\circ d\_M
$$


$$
m\_2^L(f\_1\otimes f\_2):=f\_2\circ f\_1
$$


$$
m\_i^L=0 \text{ for all }i\geq3
$$


Now we show that Q2 puts an $A$-$A\_\infty$-module structure on $M$.


$Q2$: Assume we have an $A\_\infty$ map $\alpha:A\to L$.


This is equivalent to having maps for all $n\geq 1$:


$$
\alpha\_n:A^{\otimes n}\to L
$$
of (cohomological) degree $1-n$ such that (\*):


$$
\sum(-1)^{r+st}\alpha\_u(1^{\otimes r} \otimes m\_s \otimes 1^{\otimes t})=
m\_1^L(\alpha\_n) + \sum\_{i+j=n}(-1)^{i-1} m\_2^L(\alpha\_i \otimes \alpha \_j)$$
$$
=d\_M\circ \alpha\_n - (-1)^{1-n} \alpha\_n \circ d\_M +\sum\_{i+j=n}(-1)^{i-1}\alpha\_j \circ \alpha\_i
$$ as maps from $A^{\otimes n}$ to $L$.


Now if we redefine our maps so that $\beta\_1=d\_M$ and for all $n\geq 2$
 $$\beta\_n:M\otimes A^{\otimes n-1}\to M$$


$$
m\otimes a\_1\otimes \cdots \otimes a\_{n-1} \mapsto [\alpha\_{n-1}(a\_1\otimes \cdots \otimes a\_{n-1})](m)
$$


then from the equation (\*) above: 


$$
\sum(-1)^{r+st}\beta\_{u+1}(m\otimes a\_1 \otimes \cdots \otimes a\_r \otimes m\_s(a\_{r+1}\otimes \cdots \otimes a\_{r+s})\otimes a\_{r+s+1} \otimes \cdots \otimes a\_{n-1})
$$


$$
=\beta\_1(\beta\_n(m\otimes a\_1 \otimes \cdots \otimes a\_{n-1})) - (-1)^{1-n} \beta\_n(\beta\_1(m)\otimes a\_1 \otimes \cdots \otimes a\_{n-1}) + 
\sum\_{i+j=n-1} (-1)^{i-1} \beta\_{j+1} ( \beta\_{i+1}(m\otimes a\_1 \otimes \cdots \otimes a\_{i})\otimes a\_{i+1} \otimes \cdots \otimes a\_{n-1})
$$
for all $m\in M$ and all $a\_1,\ldots,a\_{n-1}\in A$.


This shows that $M$ has the structure of a right $A$-$A\_\infty$-module.


