Morse Trajectory Moduli Spaces and the Morse Differential — Examples

Choose a smooth positive periodic function $\mu (\theta )$ equal to $4\mathrm{/}(1+\cos \theta )$ near $0$ and $4\mathrm{/}(1-\cos \theta )$ near $\pi$. It exists by [F1]: use disjoint cutoff neighbourhoods of the poles and take the convex combination of these positive local functions with the constant $2$ outside them. Put $Y(\theta )=\mu (\theta )\sin \theta$ and $X=(Y(u),Y(v))$. Then $df(X)=-\mu (u){\sin }^{2}u-2\mu (v){\sin }^{2}v<0$ off the four critical points. Near $0$, the signed coordinate $w=\mathrm{sgn}(\theta )\sqrt{1-\cos \theta }$ is smooth and satisfies $\dot{w}=2w$; near $\pi$, $w=\mathrm{sgn}(\theta -\pi )\sqrt{1+\cos \theta }$ satisfies $\dot{w}=-2w$. Multiplying the second coordinate by $\sqrt{2}$ gives exactly the Morse coordinates for the weighted second cosine. Thus $X$ satisfies [F2]. The Hessian $\mathrm{diag}(-\cos u,-2\cos v)$ has indices $2,1,1,0$ at $a,b,c,d$.

In each coordinate, stable and unstable sets are a pole or the circle with the other pole removed. Their products in $M$ are transverse: every nonempty coincidence has, in each coordinate, at least one full tangent direction; the only potential point/point intersection for distinct equilibria is empty. The smooth field is complete on compact $M$ by [F2], so the pair is Morse--Smale. For each adjacent-index pair, one coordinate is constant and the other follows either of the two complementary arcs, giving exactly two orbit classes. These are all adjacent-index pairs.

On each of the four open rectangles between the coordinate separatrices, both coordinates run from $0$ to $\pi$. On either chosen arc the time coordinate $\tau (\theta )={\int }_{{\theta }_{\ast }}^{\theta }d\eta \mathrm{/}Y(\eta )$ is a diffeomorphism onto $\mathbb{R}$: its derivative has the sign of $Y$, and the simple zeros of $Y$ make the endpoint times infinite. Every trajectory is therefore $u(t)={\tau }_u^{-1}(t-A)$, $v(t)={\tau }_v^{-1}(t-B)$. Common time translation changes $A$ and $B$ equally, leaving the relative delay $A-B\in \mathbb{R}$ as the unique parameter. Its two infinite ends break through $b$ and $c$, respectively, with the arc choices fixed. Hence there are four open intervals, each with its two distinct broken endpoints.

By [F3] these are exactly the compactifying endpoints, with one collar branch each; the fixed arc choices identify each broken pair with the appropriate rectangle, so no endpoints are identified across intervals. There are $2\cdot 2+2\cdot 2=8$ pairs in $(\mathcal{M}(a,b)\times \mathcal{M}(b,d))\bigsqcup (\mathcal{M}(a,c)\times \mathcal{M}(c,d))$, two per closed interval. By [F4] the differential is $\mathrm{\partial }a=2b+2c=0$, $\mathrm{\partial }b=\mathrm{\partial }c=2d=0$ and $\mathrm{\partial }d=0$.

Choose smooth positive periodic $\mu$ equal to $4\mathrm{/}(1+\cos \theta )$ near $p=0$ and $4\mathrm{/}(1-\cos \theta )$ near $q=\pi$, using disjoint cutoff neighbourhoods and the positive constant $2$ elsewhere. In the signed Morse coordinates $\sqrt{1-\cos \theta }$ near $p$ and $\sqrt{1+\cos \theta }$ near $q$, with signs chosen across each pole, $X=\mu \sin \theta {\mathrm{\partial }}_{\theta }$ is respectively $2w{\mathrm{\partial }}_w$ and $-2w{\mathrm{\partial }}_w$. Also $df(X)=-\mu {\sin }^2\theta <0$ off the poles, and the Hessians are $-1$ and $+1$. The smooth field is complete by [F2]. Stable and unstable sets are the poles and their complementary open intervals, whose nonempty intersections are transverse. On each of the two arcs $X$ never vanishes, so $\int d\theta \mathrm{/}(\mu \sin \theta )$ gives a time coordinate onto $\mathbb{R}$, with endpoints $p$ backward and $q$ forward. Thus each arc is exactly one orbit class, and $\mathcal{M}(p,q)=\{{\gamma }_1,{\gamma }_2\}$.

Modulo two, [F4] gives $\mathrm{\partial }p=n_2(p,q)q$ with $n_2(p,q)=\mathrm{\#}\mathcal{M}(p,q)<mrow><mi\; mathvariant="normal">m</mi><mi\; mathvariant="normal">o</mi><mi\; mathvariant="normal">d</mi></mrow>2=2<mrow><mi\; mathvariant="normal">m</mi><mi\; mathvariant="normal">o</mi><mi\; mathvariant="normal">d</mi></mrow>2=0$, so $\mathrm{\partial }p=0$; and $\mathrm{\partial }q=0$ because there is no critical point of index $-1$.

For the signed differential, [F5] says that the single orientation of $W^u(p)\cong \mathbb{R}$ orients both arcs, and the flow direction along the two arcs is opposite with respect to it: traversing $S^1$ from $p$ to $q$ along one arc and back along the other reverses the direction. Hence $ϵ({\gamma }_1)=-ϵ({\gamma }_2)$, and [F4] gives $\mathrm{\partial }p=(ϵ({\gamma }_1)+ϵ({\gamma }_2))q=0$, while $\mathrm{\partial }q=0$.

Both complexes therefore have zero differentials with one generator in degree $1$ and one in degree $0$; their homology is $\mathbb{Z}\mathrm{/}2$ in degrees $0$ and $1$ for the mod-two complex and $\mathbb{Z}$ in degrees $0$ and $1$ for the integral complex, with all other graded pieces zero. The integral differential is a chain complex differential by The integral Morse differential squares to zero, consistent with the computation $\mathrm{\partial }p=0$.

In polar coordinates $f=\cos \theta$ and the round downward gradient is $\sin \theta {\mathrm{\partial }}_{\theta }$. Multiply it by a smooth positive function of $f$ equal to $4\mathrm{/}(1+f)$ near $N$ and $4\mathrm{/}(1-f)$ near $S$, using cutoffs from [F3] and the positive constant $2$ elsewhere. In the Cartesian radial Morse coordinates of radius $\sqrt{1-f}$ at $N$ and $\sqrt{1+f}$ at $S$, the resulting field is $2w$ and $-2w$, respectively; explicitly they are $w=(x,y)\mathrm{/}\sqrt{1+z}$ near $N$ and $w=(x,y)\mathrm{/}\sqrt{1-z}$ near $S$, hence smooth local Cartesian coordinates with nonsingular derivative at the pole. Thus $X$ satisfies the normalized local model and strictly decreases $f$ elsewhere. The only critical points are $N,S$, with Hessians negative and positive definite and hence indices $2,0$. The unstable set of $N$ and stable set of $S$ are the complementary-pole open disks; the other two sets are single points, so all nonempty stable--unstable intersections are transverse. The smooth field is complete by [F3], so the pair is Morse--Smale. The chain groups in degree one are free on the empty set and are zero.

The differentials out of and into degree one vanish identically: ${\mathrm{\partial }}_2:C{M}_2\to C{M}_1$ has zero target and ${\mathrm{\partial }}_1:C{M}_1\to C{M}_0$ has zero source. The remaining differentials ${\mathrm{\partial }}_0$ and ${\mathrm{\partial }}_3$ have zero target and zero source respectively. So all differentials are zero.

Although every meridian from $N$ to $S$ is a connecting trajectory, its index drop is two. The differential definition in [F2] counts index drop one only, so these trajectories supply no coefficient.

With zero differentials and one generator in degree $2$ and one in degree $0$, the mod-two complex has homology $\mathbb{Z}\mathrm{/}2$ in degrees $0$ and $2$ and zero elsewhere, and the integral complex has homology $\mathbb{Z}$ in degrees $0$ and $2$ and zero elsewhere.

The normalized positive multiple of the round circle gradient in [F5] preserves the two arcs, and its flow directions are opposite relative to one orientation of the unstable interval. Thus their comparison signs are opposite and their signed sum is zero.

By [F2], reversing $o{r}_r$ reverses the comparison sign of every trajectory in the two families $\mathcal{M}(x,r)$ with $\lambda (x)=\lambda (r)+1$ and $\mathcal{M}(r,y)$ with $\lambda (y)=\lambda (r)-1$, and of no other trajectory: every other moduli space is built from orientations of unstable manifolds different from $r$. Consequently the coefficient $n(x,r)$ of $r$ in $\mathrm{\partial }x$ and the coefficient $n(r,y)$ of $y$ in $\mathrm{\partial }r$ change sign, by [F3], while all other coefficients are unchanged. This proves claims (1) and (2).

Let $T$ be the linear automorphism of the integral chain groups sending the basis element $r$ to $-r$ and every other basis element to itself; it is invertible with $T^{-1}=T$. Compare $T\mathrm{\partial }{T}^{-1}$ with ${\mathrm{\partial }}^{\mathrm{\prime }}$ on basis elements. On $r$: $T\mathrm{\partial }{T}^{-1}(r)=-T(\mathrm{\partial }r)=-\mathrm{\partial }r$, because $\mathrm{\partial }r$ has no $r$-component (its terms are critical points of index one less than $\lambda (r)$) and $T$ fixes every other basis element, while $\mathrm{\partial }r=T(\mathrm{\partial }r)$ holds as $T$ also fixes $r$'s own absent component; by step 1.2, ${\mathrm{\partial }}^{\mathrm{\prime }}r=-\mathrm{\partial }r$. On a basis element $x$ with $\lambda (x)=\lambda (r)+1$: $T\mathrm{\partial }(x)=T({\sum }_{y}n(x,y)y)={\sum }_{y\mathrm{\ne }r}n(x,y)y-n(x,r)r$, which is ${\mathrm{\partial }}^{\mathrm{\prime }}x$ by step 1.2. On every other basis element $x$: $\mathrm{\partial }x$ has no $r$-component, so $T\mathrm{\partial }(x)=\mathrm{\partial }x={\mathrm{\partial }}^{\mathrm{\prime }}x$ by step 1.2 and claim (2). Hence the two linear maps agree on a basis.

Since $T$ is an invertible linear map, ${\mathrm{\partial }}^{\mathrm{\prime }2}=T\mathrm{\partial }{T}^{-1}T\mathrm{\partial }{T}^{-1}=T{\mathrm{\partial }}^2{T}^{-1}$, so ${\mathrm{\partial }}^{\mathrm{\prime }2}=0$ by [F4] and $T$ restricts to an isomorphism of the homologies of $\mathrm{\partial }$ and ${\mathrm{\partial }}^{\mathrm{\prime }}$. This proves claim (3).

For the circle instance, take $r=p$, the maximum. The two arcs ${\gamma }_1,{\gamma }_2$ of $\mathcal{M}(p,q)$ both use $o{r}_p$, so by step 1.2 both comparison signs flip and their sum remains zero by [F5]; hence the integral differential still vanishes and the homology is unchanged, in agreement with the general statement.

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.

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

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.

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

The coherent convention behaves differently. By [F3] the two ends of each of the four compactified intervals carry opposite products $ϵ({\gamma }_1)ϵ({\gamma }_2)$; summing over the four intervals, the coefficient of $d$ in ${\mathrm{\partial }}^{2}a$ is the negative total signed boundary count of $\bar{\mathcal{M}}(a,d)$, which is zero. Thus ${\mathrm{\partial }}^2=0$ with the orientation-induced signs, and the failure of step 2.2 is a failure of the sign assignment, not of the Morse complex.

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 $\widetilde{ϵ}$ 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 $\mathbb{Z}\mathrm{/}2$ 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.