Morse Homology Continuation and Comparison — Examples

By [F1] the only critical point of the adapted function is the interior point of index $k$; hence the relative Morse chain groups of [F2] are $\mathrm{\Lambda }$ in degree $k$ and zero in every other degree, since they are free on the interior critical points.

The only differential that could be nonzero is ${\mathrm{\partial }}_k:C{M}_k\to C{M}_{k-1}$, whose target is the group of the critical points of index $k-1$; there are none, so ${\mathrm{\partial }}_k=0$, and all other differentials have zero source or target. Therefore the relative Morse complex is $\mathrm{\Lambda }$ concentrated in degree $k$.

The homology of that complex is $\mathrm{\Lambda }$ in degree $k$ and zero otherwise, which is the displayed formula for $H{M}_i(W,M_0;\mathrm{\Lambda })$.

By [F3] the handle pair has the relative homology of $(D^k,S^{k-1})$, again $\mathrm{\Lambda }$ in degree $k$ and zero otherwise; the two computations agree, and the single relative generator is the one recorded by the handle attachment.

Apply the CW supplier of [F2] to $(-f,-X)$, which is Morse--Smale because its stable and unstable manifolds are those of $X$ interchanged. Its dual CW structure has two zero-cells, one one-cell and one two-cell. The one-skeleton is connected: attaching a two-disk along its connected boundary circle cannot join distinct components, whereas the final sphere is connected. Hence the unique edge joins the two distinct vertices. Its two half-orbits, reversed back to $X$, are exactly the two stable saddle branches, each tending to a different original maximum. Thus each maximum supplies exactly one orbit to the saddle, proving the geometric assertion of [F1]. By [F1] the only index-drop-one trajectory moduli with source a maximum are $\mathcal{M}(p_1,a)$ and $\mathcal{M}(p_2,a)$, each a single point; thus $\mathrm{\partial }{p}_i=\pm a$ in the integral case and $\mathrm{\partial }{p}_i=a$ over $\mathbb{Z}\mathrm{/}2$.

By [F2] the compactified unstable manifold of the saddle is an interval whose boundary consists of the two trajectories to the minimum, and the boundary orientation makes their signs cancel; hence $\mathrm{\partial }a=0$ over $\mathbb{Z}$, and over $\mathbb{Z}\mathrm{/}2$ the count is $2\equiv 0$.

There are no other critical points, so the Morse complex is $\mathrm{\Lambda }\{p_1,p_2\}\stackrel{\mathrm{\partial }}{\to }\mathrm{\Lambda }\{a\}\stackrel{\mathrm{\partial }=0}{\to }\mathrm{\Lambda }\{r\}$ with $\mathrm{\partial }{p}_i=\pm a$; by [F3] the cellular boundary matrix coincides with this one up to the index normalization, so $[e_{p_i}:e_a]=\pm 1$ and $[e_a:e_r]=0$, realizing the coefficient comparison in a case with a nonzero coefficient.

The homology of this complex is $\mathrm{\Lambda }$ in degree $0$ (generated by $r$), $\mathrm{\Lambda }$ in degree $2$ (generated by the class $p_1-p_2$ or $p_1+p_2$ according to the signs), and zero in degree $1$, since the image of $\mathrm{\partial }$ on $\mathrm{\Lambda }\{p_1,p_2\}$ is the whole of $\mathrm{\Lambda }\{a\}$; by [F4] this agrees with $H_{\ast }(S^2;\mathrm{\Lambda })$ and with the cellular complex of the decomposition.

The two ends are valid: by [F1] and [F2] both pairs are complete Morse--Smale pairs with exactly one nondegenerate critical point of index $1$, so both Morse chain complexes are $\mathrm{\Lambda }$ concentrated in degree $1$ with zero differential.

Rigidity of the upper end. Let $x$ solve the continuation equation with ${\lim }_{s\to +\mathrm{\infty }}x(s)=x_0$. On $s\ge 1+\delta$ the equation is $\dot{x}=F(x)$, and every solution with $x(1+\delta )>x_0$ increases and converges to $+\mathrm{\infty }$, while every solution with $x(1+\delta )<x_0$ decreases, crosses $0$ (where $F(0)=-1$) and converges to $-\mathrm{\infty }$; hence necessarily $x(1+\delta )=x_0$, and then $x(s)=x_0$ for all $s\ge 1+\delta$.

Backward blow-up. Extend the solution of step 2.1 backward from $x(1+\delta )=x_0$. On the interpolation region the field is $G_s(x)=-\frac{d}{dx}((1-\theta (s))\frac{1}{3}{x}^3+\theta (s)f^{+}(x))=(1-\theta (s))(-x^2)+\theta (s)F(x)$ with $0\le \theta \le 1$, and at $x=x_0$ this equals $-(1-\theta (s))x_0^2<0$ whenever $\theta (s)<1$, In backward time the vector field at $x_0$ is nonnegative, so uniqueness (or the scalar differential inequality for the negative part of $x-x_0$) makes $x_0$ a lower barrier. On $[x_0,2x_0]$ all fields $G_s$ are bounded in absolute value by a common $C>0$. Choosing $\delta <x_0\mathrm{/}C$, the integral bound $\mathrm{\mid }x(s)-x_0\mathrm{\mid }\le C\delta$ precludes a first exit through $2x_0$; the lower barrier precludes a first exit below $x_0$. Thus the backward trajectory exists throughout this short interpolation and stays in $[x_0,2x_0]$. Entering the plateau at $s=1$ with $x(1)\in [x_0,2x_0]$, its backward evolution there obeys $\frac{dx}{d\tau }=x^2$ in the backward time $\tau =1-s$, with solution $x(\tau )=x(1)\mathrm{/}(1-x(1)\tau )$, which blows up at $\tau =1\mathrm{/}x(1)$; since $x(1)\ge x_0>\frac{1}{2}$ we have $1\mathrm{/}x(1)<2$, so the blow-up occurs strictly inside the plateau $[-1,1]$ and the backward trajectory is not defined beyond it.

Consequently no solution of the continuation equation has both the lower limit $p=0$ and the upper limit $q=x_0$: a solution with upper limit $x_0$ must satisfy $x(1+\delta )=x_0$ by step 2.1, and its backward continuation blows up in finite time by step 3.1, so it has no lower limit at all. Hence $\mathcal{C}(p,q)=\mathrm{\varnothing }$ although $\mathrm{ind}(p)=\mathrm{ind}(q)=1$, and the continuation count is the zero map $\mathrm{\Lambda }\to \mathrm{\Lambda }$, which is not an isomorphism.

The claim is therefore false, and the failure is exactly the mechanism recorded in Flow and compactness hypotheses for noncompact Morse homology: the end pairs have a single critical point each, but the interpolating field $\dot{x}=-x^2$ on the plateau is not complete, so the connecting trajectory escapes to spatial infinity in finite time, the energy identity has no finite-energy solution to apply to, and no compactness-up-to-breaking statement is available.

By [F2] the differential of $f^{+}$ is ${\mathrm{\partial }}^{+}c=\pm b$ and ${\mathrm{\partial }}^{+}b=0$ on the new generators, with all other coefficients as in the differential of $f^{-}$; hence the Morse complex of $f^{+}$ is the direct sum of the Morse complex of $f^{-}$ and the two-term complex $\mathrm{\Lambda }c\stackrel{\pm 1}{\to }\mathrm{\Lambda }b$.

By [F3] the new summand is acyclic in both coefficient cases: $H_{k+1}=\ker (\pm 1)=0$ and $H_k=\mathrm{\Lambda }b\mathrm{/}\mathrm{im}(\pm 1)=0$; therefore the homology of the Morse complex of $f^{+}$ equals the homology of the complex of $f^{-}$, so the birth--death pair does not change the Morse homology.

By [F4] the continuation map is a chain map and an isomorphism on homology; this is consistent with step 2.1 and shows that the continuation across the birth--death pair adds the two new generators without changing the homology.

For the polynomial bookkeeping: by [F5] the number of generators in degrees $k$ and $k+1$ each increases by one, while the only differential of changed rank is ${\mathrm{\partial }}_{k+1}^{+}$, whose rank increases by one; hence the Morse polynomial satisfies $c^{+}(t)=c^{-}(t)+t^k+t^{k+1}$ and the correction polynomial satisfies $Q^{+}(t)=Q^{-}(t)+t^k$, which is exactly the addition of $t^k(1+t)=(1+t)t^k$ to the decomposition $c(t)=b(t)+(1+t)Q(t)$.

For $f_1$, enumerate the critical points in circular order as $p,q_1,c,q_2$, with $p,c$ maxima and $q_1,q_2$ minima. Orient both unstable intervals in the direction of increasing angle. The two outgoing arcs at each maximum then give $\mathrm{\partial }p=q_1-q_2$ and $\mathrm{\partial }c=q_2-q_1$ (over $\mathbb{Z}\mathrm{/}2$ replace minus by plus). In the integral bases $P=p+c$, $C=c$, $Q=q_1$, $B=q_2-q_1$, one has $\mathrm{\partial }P=0$, $\mathrm{\partial }C=B$. These are invertible basis changes over $\mathbb{Z}$ and over the stated coefficient rings. The pair $C\mapsto B$ has contracting homotopy $B\mapsto C$, so the complex is the direct sum of the minimal zero-differential complex on $P,Q$ and this acyclic pair.

By [F1] both pairs are Morse--Smale, so the Morse complexes are defined. By [F3] the Morse complex of $f_0$ is $\mathrm{\Lambda }p\to \mathrm{\Lambda }q$ with $\mathrm{\partial }p=0$, because the only differential is the signed count of the two arcs from $p$ to $q$ and the two signs cancel.

Hence $H_1(f_0)=\ker {\mathrm{\partial }}_1=\mathrm{\Lambda }p$ and $H_0(f_0)=\mathrm{\Lambda }q\mathrm{/}\mathrm{im}{\mathrm{\partial }}_1=\mathrm{\Lambda }q$, so $H{M}_0(f_0)=\mathrm{\Lambda }=H{M}_1(f_0)$ and $H{M}_k(f_0)=0$ for $k\mathrm{\ne }0,1$.

By [F4] the birth--death pair adds an acyclic summand, so the homology of the complex of $f_1$ is the same: $H{M}_{\ast }(f_1)\cong H{M}_{\ast }(f_0)$, and the continuation isomorphism realizes the identification.

By [F5] both computations agree with the singular homology of the circle, $H_0(S^1;\mathrm{\Lambda })=H_1(S^1;\mathrm{\Lambda })=\mathrm{\Lambda }$ and $H_k=0$ otherwise, so the canonical Morse homology of $S^1$ is $\mathrm{\Lambda }$ in degrees $0$ and $1$; this is the claimed comparison of the minimal and the stabilised function.