Morse Homotopy and A-infinity; Part 1

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 A be a graded k-vector space.  Then there exists an infinite family of maps \{m_k \}

m_k: A^{\otimes k} \rightarrow A

that satisfy the A-infinity relations, which are a nightmare to state.  I’ll describe them pictorially as follows.  Each m_k 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).

m_k map

Then, we some over all possible ways to combine two of these maps into a map that has l inputs and 1 output:

a_infinity relation

and require that this sum is 0, (if for simplicity we ignore signs and assume k = \mathbb{F}_2).  So, we sum over all i,j such that i + j -1 = l and we can move the m_i map left and right so that it takes as inputs any adjacent i-tuple.

The simplest condition is that m_1 \circ m_1 = 0, since there is only one way to combine two maps in a manner that takes 1 input and yields 1 output.

m_1 relation

As a consequence, this means that the m_1 map is a differential.

The new two conditions are m_2 ( d(x), y) + m_2 ( x,d(y)) + d( m_2 (x,y)) = 0.

m_2 relation

and m_2(m_2(x,y),z) + m_2(x,m_2(y,z)) + d(m_3(x,y,x)) + m_3(d(x),y,z) + m_3(x,d(y),z) + m_3(x,y,d(z)) = 0

m_3 relation

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 d^2=0, we used the fact that each term in d^2 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 f_1,f_2,f_3 such that their difference functions f_i - f_j are generic.  This means that the (un)stable manifolds of all difference functions intersect transversely.  To define the m_2 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 f_i-f_j.  Let \widetilde{\mathcal{M}}(x,y;z) denote the moduli space of gradient flow trees from x,y to z for x a critical point of f_1 - f_2, y a critical point of f_2 - f_3 and z a critical point of f_1 - f_3,:

m_2 source tree

One way to get our hands on this space of trees is to take W_u(x) \cap W_u(y) \cap W_s(z).  For each point in this space, there is a unique triple of oriented flow lines connecting it to x,y and z.  Together, these form a tree of the form we are looking for.  The dimension of the moduli space is I(x) + I(y) - I(z) - n, which can easily be checked because this is assumed to be a transverse intersection.

Then the m_2 map can be defined as

m_2:C(f_1 - f_2) \otimes C(f_2 - f_3) \rightarrow C(f_1 - f_3)
m_2(x,y) = \sum_{z: I(z) = I(x) + I(y) - n} |\widetilde{\mathcal{M}}(x,y;z)| z

We need to show that this satisfies the A-$\infty$ relation
d(m_2(x,y)) + m_2(d(x),y) + m_2(x,d(y)) = 0

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.

m_2 broken trees

So, now we have a union of two trees, a segment and another Y.  But the segment is just a flow line from x to w, and so algebraically shows up in the differential.  And the Y is another tree corresponding to an m_2 map.

Let’s work the other way.  The each of term of m_2(d(x),y) corresponds to a pair of a rigid flow line from x to some $w$ and a rigid tree connecting w,y to z.  Similarly, each term of m_2(y,d(y)) corresponds to a pair of a rigid flow line from y to some w and a rigid tree connecting x,w to z.  Finally, each term of d(m_2(x,y)) corresponds to a pair of a rigid tree connecting x,y to some w and a rigid flow line from w to z.

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 x,y to z.  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, d(m_2(x,y)) + m_2(d(x),y) + m_2(x,d(y)) = 0 and we know that our chain complex is at least an A_2-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 d 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 m_2 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 i critical point, which is topologically a disk, is a singular i-chain, a continuous map of an i-dimensional simplex into the manifold M.  In the other direction, given a singular i-chain, its image will (generically) intersect the stable manifolds of an index i 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 i critical points and summing over these points gives the corresponding algebraic chain in the Morse complex.

Morse-Singular equivalence

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 x \in H^i(M,\partial_{\text{Morse}}), y \in H^j(M,\partial_{\text{Morse}}).  These have Poincare duals X,Y which are disks of dimension n-i,n-j.  The Poincare dual of x \wedge y is X \cap Y, which has dimension n-i-j.  To determine which element this is in Morse cohomology, we flow this intersection upward by the gradient vector field and see which index i+j critical points it gets caught by.  In terms of gradient trajectories, we look for all the gradient flow lines from X \cap Y to some critical point z of index i + j.  There are a finite number of such flow lines, each of which corresponds to a unique tree connecting x,y to z.  In the chain complex, this is exactly the m_2 map and so passing to cohomology we recover the cup product from m_2.


Leave a comment

Filed under Uncategorized

Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out /  Change )

Google+ photo

You are commenting using your Google+ account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )


Connecting to %s