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The integral Morse differential squares to zero

Statement

Assume the Axiom of Choice. Let $(f, X)$ be Morse--Smale on a closed manifold and fix an orientation of every unstable manifold. Then the signed Morse differential of The signed Morse differential over the integers satisfies $\partial_{k - 1} \circ \partial_{k} = 0$ for every $k$ . Equivalently, the integral Morse complex is a chain complex over $Z$ , whose homology is the integral Morse homology of $(f, X)$ .

Facts & Assumptions

Given: The Axiom of Choice, a Morse--Smale pair $(f, X)$ on a closed manifold, orientations of all unstable manifolds, and an integer $k$ .

[A1]

The Axiom of Choice; the finiteness of the index-one moduli spaces and the oriented boundary-count lemma are supplied through it, the latter with ${A C}_{ω}$ via the bridge (The Axiom of Choice, AC implies DC implies countable choice, The Axiom of Countable Choice ( ${A C}_{ω}$ )).

[F1]

On a basis element the signed differential is $\partial_{k} p = \sum_{q \in {Crit}_{k - 1} (f)} (\sum_{γ \in M (p, q)} ϵ (γ)) q$ , with finite inner and outer sums and signs supplied by the unstable orientations (The signed Morse differential over the integers, The integers as equivalence classes of pairs of naturals).

[F2]

For $λ (p) - λ (q) = 2$ the compactification $\overline{M} (p, q)$ is a compact oriented one-manifold with boundary whenever the unstable orientations are fixed, and its boundary is the disjoint union of the products $M (p, r) \times M (r, q)$ over $r$ of index $λ (p) - 1$ ; every such boundary point is once-broken (The index-two compactification is a compact one-manifold with boundary, Breaking length is bounded by the index drop, Broken Morse trajectories).

[F3]

With the orientation of the compactification restricting to the flow-first orientation of the interior and the outward-normal-first orientation on the boundary, the boundary sign of a once-broken point is $- ϵ (γ_{1}) ϵ (γ_{2})$ (Boundary orientation of the compactified one-dimensional Morse moduli space).

[F4]

Under ${A C}_{ω}$ , the sum of the outward-normal-first boundary signs of a compact oriented smooth one-manifold vanishes; on each interval component the two endpoints carry opposite signs and circle components contribute nothing (Oriented boundary counts of a compact oriented 1-manifold cancel).

[F5]

A chain complex over $Z$ is a family of $Z$ -modules with degree $- 1$ endomorphisms squaring to zero (Chain complex in an abelian category).

Proof

technique · direct, by evaluating on basis elements

1.1F1algebra

Let $p \in {Crit}_{k} (f)$ and $q \in {Crit}_{k - 2} (f)$ . Expanding [F1], the coefficient of $q$ in $\partial_{k - 1} (\partial_{k} p)$ is the finite sum $\sum_{r \in {Crit}_{k - 1} (f)} \sum_{γ_{1} \in M (p, r), γ_{2} \in M (r, q)} ϵ (γ_{1}) ϵ (γ_{2})$ : only intermediate points of index $k - 1$ can contribute, both index drops are equal to one, and the two inner sums are finite by [F1].

2.1F2F3step 1.1

The terms of that sum are indexed by the once-broken trajectories $(γ_{1}, γ_{2})$ with $λ (p) - λ (r) = λ (r) - λ (q) = 1$ , which by [F2] are exactly the boundary points of the compact oriented one-manifold $\overline{M} (p, q)$ ; and by [F3] the summand attached to $(γ_{1}, γ_{2})$ is the negative of its outward-normal-first boundary sign. Hence the coefficient of $q$ in $\partial^{2} p$ is the negative total signed boundary count of $\overline{M} (p, q)$ .

3.1A1F4step 2.1

By [F4] the total signed boundary count of a compact oriented one-manifold vanishes; hence the coefficient of $q$ in $\partial^{2} p$ is zero. For a critical point not of index $k - 2$ the coefficient is zero by the definition of the chain groups, so $\partial_{k - 1} \circ \partial_{k} = 0$ on basis elements and hence on all of $C M_{k} (f, X; Z)$ by linearity.

4.1F5step 3.1∎

Since this holds for every $k$ , the integral Morse complex satisfies the defining condition of a chain complex over $Z$ by [F5]; its homology is the integral Morse homology of $(f, X)$ by definition.

Depends on

Used by

A naive signed count without the quotient orientation can fail to square to zero Counterexample
Morse homology of a Morse--Smale pair Definition
Changing an unstable orientation changes two sets of basis signs Example
The Morse complex of the circle Example
Integral Morse homology does not require orientability of the manifold Remark

Dependency tree · two levels

56 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Liviu I. Nicolaescu, An Invitation to Morse Theory, 2nd ed., complete PDF (standard reference, not scraped)
Alberto Abbondandolo and Pietro Majer, Lectures on the Morse Complex for Infinite-Dimensional Manifolds, complete PDF (standard reference, not scraped)
Udhav Fowdar, A Functional Analytic Approach to Morse Homology (UCL 4th-year project, 2016, supervised by C. Wendl), complete PDF (standard reference, not scraped)