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A naive signed count without the quotient orientation can fail to square to zero

Statement refuted

Assume AC. An arbitrary assignment of signs to the index-one trajectories of a Morse--Smale pair yields a differential squaring to zero. Unstable orientations give a gluing-compatible convention that does square to zero. Other assignments can accidentally square to zero; this counterexample refutes the universal assertion, rather than characterizing all assignments that work.

Facts & Assumptions

Given: AC and the explicit normalized torus model of Broken trajectories in an index-two torus moduli space: critical points $a$ of index $2$ , $b, c$ of index $1$ and $d$ of index $0$ , with each of $M (a, b)$ , $M (a, c)$ , $M (b, d)$ , $M (c, d)$ of cardinality two, and the naive signs $\tilde{ϵ} (γ) : = + 1$ for every index-one trajectory $γ$ .

[F1]

The eight once-broken trajectories from $a$ to $d$ are the boundary points of the compact one-manifold $\overline{M} (a, d)$ , which is the disjoint union of four closed intervals; each interval has two boundary points, and each is a product of one trajectory of index drop one from $a$ to a saddle with one from that saddle to $d$ (Broken trajectories in an index-two torus moduli space, Unparametrized Morse trajectory moduli space).

[F2]

The coherent signed differential sums the comparison signs $ϵ (γ)$ determined by the unstable orientations: $\partial a = \sum_{γ} ϵ (γ) q_{γ}$ over the four trajectories out of $a$ , and $\partial b = \partial c = 0$ , since the two halves of each oriented unstable interval have opposite flow directions and the same minimum co-orientation, hence opposite coherent signs, over the integers (The signed Morse differential over the integers, The integers as equivalence classes of pairs of naturals).

[F3]

At each boundary point of the compactified one-manifold the outward-normal-first boundary sign is the negative product $- ϵ (γ_{1}) ϵ (γ_{2})$ of the two comparison signs, so the two ends of each of the four intervals carry opposite products and the total signed boundary vanishes (Boundary orientation of the compactified one-dimensional Morse moduli space, The integral Morse differential squares to zero).

[F4]

The mod-two differential counts the same finite sets without signs, so it is unaffected by any sign assignment (The mod-two Morse differential).

Counterexample

technique · direct, by an explicit computation with the naive signs

1.1F1given

The explicit product Morse function and normalized field of [F1] have the four critical points and four adjacent-index moduli spaces stated in the Given data; each such moduli space has two points, and the index-two compactification has four intervals and eight endpoints. Thus this is realized Morse--Smale data, rather than an assumed counting diagram.

1.2F1given

With $\tilde{ϵ} \equiv + 1$ the naive signed count of the four index-one moduli spaces gives $\tilde{\partial} a = 2 b + 2 c$ , $\tilde{\partial} b = 2 d$ , $\tilde{\partial} c = 2 d$ and $\tilde{\partial} d = 0$ , because each of the four moduli spaces has exactly two elements.

2.1F1F3step 1.1

By [F1] the four relative-delay intervals have the eight once-broken endpoints. The coherent sign identity in [F3] makes the products at their two ends opposite.

2.2step 1.2algebra

Hence ${\tilde{\partial}}^{2} a = 2 \tilde{\partial} b + 2 \tilde{\partial} c = 4 d + 4 d = 8 d \neq 0$ in $Z$ . The naive assignment of signs therefore fails to square to zero: it is not a differential.

3.1F3step 2.2

The coherent convention behaves differently. By [F3] the two ends of each of the four compactified intervals carry opposite products $ϵ (γ_{1}) ϵ (γ_{2})$ ; summing over the four intervals, the coefficient of $d$ in $\partial^{2} a$ is the negative total signed boundary count of $\overline{M} (a, d)$ , which is zero. Thus $\partial^{2} = 0$ with the orientation-induced signs, and the failure of step 2.2 is a failure of the sign assignment, not of the Morse complex.

4.1F2F3F4step 2.2step 3.1∎

The comparison signs of [F2] are determined by the chosen orientations of the unstable manifolds through the boundary-orientation identity, not chosen per trajectory; the all-plus assignment $\tilde{ϵ}$ is not of that form, since it makes both ends of an interval contribute with the same product. The mod-two differential avoids the issue because signs are invisible in $Z / 2$ by [F4]. Hence a sign convention compatible with the compactified moduli spaces — not an arbitrary assignment of signs to trajectories — is what makes the integral Morse complex a complex.

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Sources

Udhav Fowdar, A Functional Analytic Approach to Morse Homology (UCL 4th-year project, 2016, supervised by C. Wendl), 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)
Alexander F. Ritter, Part III Morse Homology (Cambridge lecture notes), Lectures 17-19, complete combined PDF (standard reference, not scraped)