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.


1 Answer 1


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.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .