# Heuristic behind $A_{\infty}$ - algebras

How to think about the $A_{\infty}$-algebras ?

I am looking at the Bernhard Keller's introduction, he says a few words about the topological origin (not in details) and motivates by stating two problems in homological algebra but what I am looking for is the intuitive idea and some prototypical examples to keep in mind and justify the axioms of $A_{\infty}$- algebra and their morphisms.

Thanks

-

I'd like to correct one mistake in the answers to this question. $A_{\infty}$ spaces and algebras are ones that are associative up to all higher homotopies; $E_{\infty}$ spaces and algebras are ones that are associative and commutative up to all higher homotopies. Neither notion has anything to do with groups, a priori. When connected, an $A_{\infty}$ space is equivalent to a loop space (Stasheff) and an $E_{\infty}$ space is equivalent to an infinite loop space (Boardman and Vogt, May). In the non-connected case, both assertions remain true provided that the monoid of components is a group. The $A$ can be thought of as standing for "Associative'', the $\infty$ standing for "up to the infinitude of relevant higher homotopies''. The $E$, due to Boardman, stands for homotopy Everything'', the $\infty$ again standing for "up to the infinititude of relevant higher homotopies''. In both cases there are notions of $A_n$ spaces and $E_n$ spaces and many examples of $A_n$-spaces that are not $A_{n+1}$ spaces. This is more striking for $E_n$-spaces, since those are equivalent to n-fold loop spaces (with the same proviso on components in the non-connected case). It is correct that $A_{\infty}$ spaces preceded operads. A key motivation for operads is to encode the infinitude of homotopies in the $E_{\infty}$ case since, unlike the $A_{\infty}$ case, it seems impossible to specify all of the relevant homotopies explicitly. The DG world works in exactly the same way, and passage to chains transports the topological examples to algebraic ones.

-
I did not know the $E$ stood for Everything. Thanks! – Mariano Suárez-Alvarez Mar 7 '11 at 2:40

If you have a differential algebra $A$, a complex $B$ and a isomorphism of complexes $f:A\to B$, then you can transport the structure (i.e., the multiplication) on $A$ to turn $B$ into a differential algebra. If you only have a homotopy equivalence $f:A\to B$, then that does not work: but $f$ is enough to construct a $A_\infty$-algebra structure on $B$.

«$A_\infty$-algebra structures are what you get from transporting algebra structures along homotopy equivalences» is not a bad slogan.

(Notice that this really gives you a slogan for $X_\infty$ for most $X$s... We can talk about $\mathrm{Lie\\ triple\\ systems}\_\infty$ things, for example)

-

The standard example is the based loop space $\Omega M$ thought of as the space of continuous maps from $[0,1]$ to $M$ where 0 and 1 are mapped to the base point. This has a product on it given by taking the loop you get from going around one loop and then the other. In other words, if you have loops $f$ and $g$, if $t\in [0,1/2]$, $(fg)(t) = g(2t)$, and if $t\in [1/2,1]$, $(fg)(t) = f(2t-1)$.

It's easy to see that this product is not associative, but it is associative up to a reparametrization of the circle. Thus, there's a homotopy between $f(gh)$ and $(fg)h$ which is a map

$$[0,1] \times \Omega M^{{}\times 3} \to \Omega M$$

For four loops, you can draw a pentagon of homotopies: $$f(g(hi)) \sim f((gh)i) \sim (f(gh))i \sim ((fg)h)i \sim (fg)(hi) \sim f(g(hi))$$

Let $K_4$ be the pentagon. This is a map:

$$\partial K_4 \times \Omega M^{{}\times 4} \to \Omega M$$

These homotopies are coherent which means that this extends to a map

$$K_4 \times \Omega M^{{}\times 4} \to \Omega M$$

This pattern continues and gives Stasheff's Associahedra. A space, $H$, possessing a set of maps (and my memory's not so great here, so I'll assume $i>1$ which might not be correct)

$$K_i \times H^{{}\times i} \to H$$

where $K_2 = pt$, $K_3 = [0,1]$, $K_4$ is as above, etc. is called an $A_\infty$-space. It's a theorem of Stasheff that any connected $A_\infty$-space is homotopic to a based loop space.

Now, pass to chains on the space and you get an $A_\infty$ algebra. The theorem of Stasheff is then the statement that the bar construction on an $A_\infty$ algebra is a dg-coalgebra.

This can all be thought of in terms of operads, of course, but I think this all came first. I don't know which introduction of Keller's you're using (he has several), but I believe this is all in this one.

-

There is one aspect of $A_\infty$-algebras I haven't seen mentioned yet: Massey products and related structures.

Suppose you have a differential graded algebra $(A,d)$: a chain complex with a product such that the differential satisfies the Leibnitz rule: $d(xy) = d(x)y + (-1)^{|x|} x d(y)$, where $x$ is in degree $|x|$ in the chain complex. Then its cohomology $H^*(A,d)$ has the structure of an algebra. Suppose further that the product in the chain complex is associative (this is not strictly necessary, but it makes the rest of this easier to write down). Then $H^*(A,d)$ has the structure of an $A_\infty$-algebra. There are two important features:

• There is an $A_\infty$-algebra isomorphism between $A$ and $H^*(A,d)$. This is despite the fact that there doesn't have to be any nontrivial honest map of associative algebras between these two.

• $H^*(A,d)$ is associative on the nose, not just up to homotopy, so the $A_\infty$-structure conveys other information: it has to do with Massey products. Massey products are complicated, but they are an important piece of the structure in the cohomology of any differential graded algebra. To define them: suppose that $a_1$, $a_2$, and $a_3$ are cocycles, representing cohomology classes $\alpha_1$, $\alpha_2$, and $\alpha_3$. Suppose also that $a_1 a_2$ is a boundary: $d(b) = a_1 a_2$ for some $b$, and similarly $d(c) = a_2 a_3$. This means that in cohomology, the products $\alpha_1 \alpha_2$ and $\alpha_2 \alpha_3$ are zero. Then with an appropriate sign, $a_1 c \pm b a_3$ is a cocycle, because $d(a_1 c) = \pm a_1 a_2 a_3$ and the same for $d(b a_3)$. The element $a_1 c \pm ba_3$ represents a cohomology class, an element of the Massey product $\langle \alpha_1, \alpha_2, \alpha_3 \rangle$. (You can modify $b$ by adding any cocycle to it, and similarly for $c$, and the Massey product is the resulting set of cohomology classes.) You can also define Massey products of length 4, 5, etc., as long as suitable shorter products vanish. It turns out that the $A_\infty$ structure on $H^*(A,d)$ captures at least some of this information: the higher product $m_3(\alpha_1 \otimes \alpha_2 \otimes \alpha_3)$ is an element of $\langle \alpha_1, \alpha_2, \alpha_3 \rangle$, and similarly for higher products. The axioms for an $A_\infty$-algebra says that the $A_\infty$-structure on $H^*(A,d)$ is a way of choosing elements from all of the different Massey products according to some compatibility requirement.

-

I like to think of definition of $A_\infty$ algebras as what you get when you take a space with an action of Stasheff's $A_\infty$ operad of associahedra and you take cellular chains. The algebraic definition follows directly from the cellular structure of the associahedra - they are convex polytopes with poset of faces isomorphic to the opposite of the poset of planar rooted trees and edge contractions.

-
Similar to this, the view of $A_\infty$ that I enjoy the most (as opposed to the one that I find most productive as a mental model) is that it is what you get if you want to do homological algebra on algebraic theories: The $A_\infty$ operad is the free resolution of the associative operad. – Mikael Vejdemo-Johansson Mar 6 '11 at 22:48

Abstractly, you can think of an $A_\infty$-space (an $A_\infty$-algebra in the category of topological spaces) as being a homotopy theorist's substitute for a space with a product which is strictly associative (a monoid). An $E_\infty$-space would then be a homotopy theorist's (edit: abelian) group (this analogy is quoted from Classifying spaces made easy by John Baez). and an $L_\infty$-space $L_\infty$-algebra would be a homotopy theorist's Lie algebra.
More concretely, the archetype for an $A_\infty$-space is a loop space, I think. Indeed, a space $X$ is a loop-space if and only if it is an $A_\infty$-space whose $\pi_0$ is a group. For more on this point of view, I recommend Infinite Loop Spaces by J.F. Adams.

-
Isn't $L_\infty$ homotopy theorists Lie algebras, rather (whatever that may mean...) – Mariano Suárez-Alvarez Mar 7 '11 at 2:34
@Mariano: corrected – Daniel Moskovich Mar 7 '11 at 2:39
I'm confused about a couple of things here. Aren't $E_\infty$ spaces a homotopy theorists (very) commutative group? the closest to a "group" is a grouplike $A_\infty$ space, i.e. as you point out, a loop space. Also is there such a thing as an $L_\infty$-space? I was under the impression that $L_\infty$ is a stable notion (operad), and doesn't make sense in unstable settings like spaces. (In the stable setting it's the Koszul dual to $E_\infty$, but I don't think that notion makes sense unstably - then again would be very happy to learn otherwise.) – David Ben-Zvi Mar 7 '11 at 3:12
@David You're right- I wrote rubbish. Hopefully now it's correct. – Daniel Moskovich Mar 7 '11 at 3:49

A source of $A_\infty$-algebras structures is the Hochschild cochain complex $C(A,A)$ of a ($\mathbb Z$)-graded vector space $A$ over a field $\mathbb k$ of characteristic zero (for simplicity); funnily enough the so produced $A_\infty$-structures are in general curved.

Skipping details about Koszul signs, one introduces

$$C^{n,m}(A,A):=\operatorname{Hom}^m_{\mathbb k}(A^{\otimes n}, A):= \prod_i \operatorname{Hom}_{\mathbb k}((A^{\otimes n})^i, A^{i+m}),$$ for all $m\in\mathbb Z$, $n\geq 0$, with $A^{\otimes 0}:=\mathbb k$ and shows that $C(A,A)[1]$ is a Lie algebra with Lie bracket $[f,g]:=f\circ g\pm g\circ f$; $\circ$ denotes the composition of cochains.

An $A_\infty$-algebra structure on $A$ is the datum of a cochain $m\in C(A,A)[1]$ of total degree $1$ s.t. $[m,m]=0$. The curvature $m_0$ is the element $m_0\in C^{0,2}(A,A)=\operatorname{Hom}_{\mathbb k}(\mathbb k, A^{2})$. The ground field $\mathbb k$ is supposed to be concentrated in degree $0$.

-