I found it useful to thoroughly go through and understand why the differential in Legendrian Contact Homology squares to 0 and so did some others, so I’m going to continue and discuss where higher A-infinity relations come from. A-infinity things seem intimidating because of all the little details necessary to define them. I hope it helps to have a geometric interpretation. It also helps to open your mind to universal algebra and operads, but I won’t go into that here.
Recall the definition of an A-infinity algebra. Let be a graded -vector space. Then there exists an infinite family of maps
that satisfy the A-infinity relations, which are a nightmare to state. I’ll describe them pictorially as follows. Each can be represented by a box with k strands entering on the top and 1 strand leaving on the bottom, i.e. there are k inputs and 1 output. (Some people use trees to visualize this as well).
Then, we some over all possible ways to combine two of these maps into a map that has inputs and 1 output:
and require that this sum is 0, (if for simplicity we ignore signs and assume ). So, we sum over all such that and we can move the map left and right so that it takes as inputs any adjacent -tuple.
The simplest condition is that , since there is only one way to combine two maps in a manner that takes 1 input and yields 1 output.
As a consequence, this means that the map is a differential.
The new two conditions are .
Using Morse theory, Fukaya proved that the cohomology ring of a real analytic manifold is actually an A-infinity algebra. In the case of proving , we used the fact that each term in corresponds to a union of two flowlines, called a broken flow.
Fukaya studied gradient flow trees. Let T be the tree in figure 1, with 4 vertices and 3 edges in a Y pattern. Now, pick 3 Morse functions such that their difference functions are generic. This means that the (un)stable manifolds of all difference functions intersect transversely. To define the map, we are going to look at the moduli space of flow trees corresponding to Y. That is, each edge will parametrize a flow line of some . Let denote the moduli space of gradient flow trees from to for a critical point of , a critical point of and a critical point of ,:
One way to get our hands on this space of trees is to take . For each point in this space, there is a unique triple of oriented flow lines connecting it to and . Together, these form a tree of the form we are looking for. The dimension of the moduli space is , which can easily be checked because this is assumed to be a transverse intersection.
Then the map can be defined as
We need to show that this satisfies the A-$\infty$ relation
In Morse theory, all moduli spaces, of flow lines and trees and of all dimensions, can be compactified. We just need to know how trees/flows degenerate as they head off to the open end. But as always, the principle here is that it can only degenerate into a union of trees you already know about.
For instance, take the Y. Since everything is finite dimensional, any open end of the moduli space must come from an open end of the moduli of the individual flow lines. So degeneration for trees looks exactly like degeneration for flow lines. Any of the three edges could break into pieces.
So, now we have a union of two trees, a segment and another Y. But the segment is just a flow line from to , and so algebraically shows up in the differential. And the Y is another tree corresponding to an map.
Let’s work the other way. The each of term of corresponds to a pair of a rigid flow line from to some $w$ and a rigid tree connecting to . Similarly, each term of corresponds to a pair of a rigid flow line from to some and a rigid tree connecting to . Finally, each term of corresponds to a pair of a rigid tree connecting to some and a rigid flow line from to .
As with the differential, these pairs can be glued together into a 1-dimensional tree and each pair corresponds to the endpoint of some 1-dimensional component of the moduli space of trees from to . Since this 1-dimensional space can be compactified in such a way that if we look at the boundary of all 1-dimensional moduli of trees, we get pairs as in the figure above.
Thus, and we know that our chain complex is at least an -algebra.
Now, we have been using 3 different Morse functions and then three other difference functions. These critical points live in different chain complexes. This is ok. We already know that the chain homotopy type of the Morse complex is independent of the Morse function. So the algebraic structure of the map is the same in all three chain complexes. So it’s OK to think of this relation as living on a single chain complex
Fukaya also shows how this map descends to the familiar cup product on cohomology. First, recall the chain homotopy equivalence between Morse homology and singular homology. In one direction, the descending manifold of an index critical point, which is topologically a disk, is a singular -chain, a continuous map of an -dimensional simplex into the manifold . In the other direction, given a singular -chain, its image will (generically) intersect the stable manifolds of an index critical points in a finite number of points. Summing over all such intersections gives the algebraic image of the chain in the Morse complex. In more suggestive terms, one can think of flowing the image of the singular chain down by the descending gradient flow. The singular chain will “hang” on some of the index critical points and summing over these points gives the corresponding algebraic chain in the Morse complex.
In Morse cohomology (N.B the differential increases the grading, so we look at rigid, ascending flow lines), the cohomology complex is still generated by the critical points and their Poincare duals can be represented by the singular chains given by their ascending manifolds. Thus, take two basis elements , . These have Poincare duals which are disks of dimension . The Poincare dual of is , which has dimension . To determine which element this is in Morse cohomology, we flow this intersection upward by the gradient vector field and see which index critical points it gets caught by. In terms of gradient trajectories, we look for all the gradient flow lines from to some critical point of index . There are a finite number of such flow lines, each of which corresponds to a unique tree connecting to . In the chain complex, this is exactly the map and so passing to cohomology we recover the cup product from .