I'm reading Keller's Introduction to A-infinity algebras and modules to learn about $A_\infty$-algebras. For reference, an $A_\infty$-algebra $A$ is a graded $k$-vector space $A = \bigoplus_{i\in\mathbb{Z}}A^i$ equipped with linear maps $m_n:A^{\otimes n}\to A$ each of degree $2-n$ satisfying the Stasheff identities

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

The map $m_1$ is a differential on $A$ and $m_2$ is a product on $A$ satisfying the graded Leibniz identity, making it a (not necessarily associative) differential graded algebra.

When describing the Stasheff identities for low values of $n$, it's written in the document I linked above (and other places I've read defining $A_\infty$-algebras) that for $n=3$ the identity

$$m_2(\boldsymbol{1}\otimes m_2 - m_2\otimes \boldsymbol{1}) = m_1m_3 + m_3(m_1 \otimes\boldsymbol{1}\otimes\boldsymbol{1}+\boldsymbol{1}\otimes m_1\otimes \boldsymbol{1} + \boldsymbol{1}\otimes \boldsymbol{1}\otimes m_1)$$

and the fact that

$$ m_1m_3 + m_3(m_1 \otimes\boldsymbol{1}\otimes\boldsymbol{1}+\boldsymbol{1}\otimes m_1\otimes \boldsymbol{1} + \boldsymbol{1}\otimes \boldsymbol{1}\otimes m_1) = \delta(m_3)$$

where $\delta$ is the differential of $\mbox{Hom}(A^{\otimes 3}, A)$, implies that $m_2$ is "associative up to homotopy". I've been trying to find a precise description of what this actually means but I've not been able to find anything. My assumption is that it means $m_2$ is homotopic to an associative product, but that somehow doesn't feel completely correct.

I realise that $m_2(\boldsymbol{1}\otimes m_2 - m_2\otimes \boldsymbol{1})$ is the associator of $m_2$, so that when it's equal to $0$ then this means that $m_2$ is an associative product, but I'm not sure how this coupled with the fact that $m_1m_3 + m_3(m_1 \otimes\boldsymbol{1}\otimes\boldsymbol{1}+\boldsymbol{1}\otimes m_1\otimes \boldsymbol{1} + \boldsymbol{1}\otimes \boldsymbol{1}\otimes m_1)$ is a coboundary has anything to do with homotopies.

If someone could help me understand I would really appreciate it!

  • $\begingroup$ The third Stasheff identity says that the associator of $m$ is a boundary for $\delta=m_1$: it is zero up to a boundary. Maybe you will be happier reading about the topological side of the story, i.e. $A_\infty$-spaces. Then the statement is that $m(m,1)$ and $m(1,m)$ are (literally) homotopic as continuous maps. The name originates there! It's explained in 2.2. $\endgroup$
    – Pedro
    Commented Jun 9, 2021 at 20:28
  • $\begingroup$ @PedroTamaroff Thanks for responding! But I don't understand how it being zero up to a boundary implies that it's associative up to homotopy, because I don't really know what associative up to homotopy means. The topological viewpoint is fine, that makes sense, but what I really want to understand is the algebraic viewpoint. $\endgroup$
    – SeraPhim
    Commented Jun 9, 2021 at 21:08
  • 2
    $\begingroup$ If $f,g:X\to Y$ are topological maps that are homotopic, then the maps they induce $C_*(X)\to C_*(Y)$ are chain homotopic, meaning that $f_* - g_* = \delta(h)$ for some element $h$. Thus the algebraic equation you're looking at is the direct translation of what it means for $m(1,m) \simeq m(m,1)$ as topological maps. $\endgroup$
    – Pedro
    Commented Jun 9, 2021 at 21:27

1 Answer 1


You asked about the algebraic point of view. If $(C, d_C)$ and $(D, d_D)$ are chain complexes with differentials of degree 1 and $f, g: C \to D$ are chain maps, then they are chain homotopic, or just homotopic, if there is a sequence of maps $H: C_n \to D_{n-1}$ such that $$ f-g = d_D H + H d_C. $$ This is an equivalence relation, and one consequence is that $f$ and $g$ induce the same map on homology.

That's the situation here: one chain complex is $(A, m_1)$ and the other is $(A \otimes A \otimes A, m_1 \otimes 1 \otimes 1 + 1 \otimes m_1 \otimes 1 + 1 \otimes 1 \otimes m_1)$, and the two chain maps $m_2(m_2 \otimes 1)$ and $m_2(1 \otimes m_2)$. The Stasheff identity for $n=2$ says that $m_2$ induces a multiplication on the homology of $A$ with respect to $m_1$, and the $n=3$ identity implies that at the level of homology, the multiplication is associative. (Being an $A_\infty$-algebra is much stronger than just being associative at the level of homology, though.)

  • $\begingroup$ Oh of course, I can't believe I didn't see that. I actually know all the points you've made here but somehow didn't put two and two together that $m_3$ was a chain homotopy :') Thanks for clarifying! $\endgroup$
    – SeraPhim
    Commented Jun 10, 2021 at 13:04

You must log in to answer this question.

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