# What does "associative up to homotopy" mean for $A_\infty$-algebras?

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!

• 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.
– Pedro
Commented Jun 9, 2021 at 20:28
• @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. Commented Jun 9, 2021 at 21:08
• 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.
– Pedro
Commented Jun 9, 2021 at 21:27

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.)
• 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! Commented Jun 10, 2021 at 13:04