Handle Decompositions Duality and Rearrangement — Examples

The projection $\pi :W\to [0,1]$, $\pi (x,t)=t$, is smooth, and its differential is $dt$, which is nowhere zero; hence $\pi$ has no critical point and is a Morse function with empty critical set, and the condition of excellence is vacuous. It satisfies ${\pi }^{-1}(0)=M\times \{0\}=M_0$ and ${\pi }^{-1}(1)=M_1$, and it is constant on each face, so it is adapted (with the boundary collar containing no critical point since there are none).

For every $\epsilon >0$, the map $W\to M_0\times [0,\epsilon ]$, $(x,t)\mapsto ((x,0),\epsilon t)$, is a diffeomorphism of manifolds with boundary, with inverse $((x,0),s)\mapsto (x,s\mathrm{/}\epsilon )$. It fixes $M_0$ pointwise and carries the projection to the rescaled collar coordinate. Thus the initial collar stage already presents the whole cylinder up to the required relative diffeomorphism.

By [F2] the empty ordered list is an allowed handle decomposition: it presents the collar $M_0\times [0,\epsilon ]$ itself. By step 2.1 the cylinder is that collar, so the empty list is a handle decomposition of $W$ relative to $M_0$ in which no handle is attached; and by [F3] this is exactly the critical-point-free presentation whose Morse function is the projection.

Start with a disk. In each of $g$ repetitions, attach an orientable band along two arcs of the current single boundary circle so that the boundary splits into two circles; attach a second orientable band between these two circles. The boundary is again one circle and the surface has acquired one punctured-torus piece. This is the usual genus-$g$ orientable surface with one disk removed. Capping its final circle gives ${\mathrm{\Sigma }}_g$, using exactly one disk, $2g$ bands and one cap.

Read the disk, the bands and the cap as handles of indices $0,1,2$. By [F1] the resulting finite handle presentation is realized by an adapted excellent $h:{\mathrm{\Sigma }}_g\to [0,1]$ with one minimum, $2g$ saddles of distinct values and one maximum. No arbitrary embedded height function is being assumed excellent or already in the adapted range. For $g=0$ the construction is two disks glued along their circle.

Patch a background metric from [F8] to Euclidean metrics in smaller disjoint Morse charts of the realizing function of step 2.1, using the finite chart cutoffs. Its negative gradient is strictly descending off the critical points and equals $(2u,-2v)$ in these charts. It is complete by [F8] because the closed surface is compact. Thus it is an adapted field, and [F4] applies to this pair. The negated function $1-h$ is adapted excellent with the same $2g+2$ critical points, and the indices are exchanged by $k\mapsto 2-k$: the maximum of $h$ has index $0$ for $1-h$, the $2g$ saddles keep index $1$, and the minimum of $h$ has index $2$ for $1-h$ (this is the general fact that negating a function changes the index of a nondegenerate critical point from $k$ to $n-k$, here $n=2$). By [F1] applied to $1-h$, the dual presentation has one $0$-handle, $2g$ $1$-handles and one $2$-handle.

Identify the handles of the two presentations through [F3]: the dual $0$-handle is the original $2$-handle, the $2g$ one-handles are self-dual since $2-1=1$, and the dual $2$-handle is the original $0$-handle; the order of attachment is reversed and attaching and belt spheres are interchanged. For $g=0$ this says that the dual presentation of the sphere's two-handle presentation is the same pair of handles read in reverse order, which agrees with the explicit picture of two disks glued along their boundary circle.

The index exchange is verified in every surface degree: $0\leftrightarrow 2$ and $1\leftrightarrow 1$, so no handle of the dual presentation has an index outside $\{0,1,2\}$ and the numbers of handles of each index are $1,2g,1$ in both presentations. This is exactly the surface case of the duality theorem.

The disk $D^2$ is the $0$-handle of [F3], with boundary the circle $\mathrm{\partial }{D}^2$. The two $1$-handles are attached along the pairs of disks $D_1$ and $D_2$, which are disjoint, so the attaching regions of the two handles are disjoint subsets of the level $\mathrm{\partial }{D}^2$; the index of both is $1$.

In either order the same two attachments are performed along the same disjoint attaching regions, and each attachment adds a handle body homeomorphic to $D^1\times D^1$; the surface produced is the disk with two bands attached, a compact surface with two bands, in both cases.

By [F1] the two equal-index handles may be attached simultaneously or in either order with the same result up to diffeomorphism relative to the lower stage $D^2$; hence $M_{12}$ and $M_{21}$ are diffeomorphic by a diffeomorphism fixing the disk and identifying each labelled handle with the same labelled handle. The dimension count of [A1] records the reason: the two attaching spheres of the $1$-handles are $0$-dimensional in a $1$-dimensional level and can be made disjoint, so no trajectory or intersection obstruction to the reordering exists.

The comparison also holds at the level of homotopy types: by [F4] each of the two attachments is, up to homotopy of pairs, the attachment of a $1$-cell along a pair of points, so both orders produce the homotopy type of a wedge of two circles, in accordance with the disk with two bands. Corner rounding does not affect the conclusion, by [F5].

Its only critical points are $p,q$, with values $3\mathrm{/}4,1\mathrm{/}4$ and indices $1,0$. Near $p$ take the Morse coordinate $u=\sin (\theta \mathrm{/}2)\mathrm{/}\sqrt{2}$, so $f=3\mathrm{/}4-u^2$ and $Xu=2u$; near $q$ take $v=\sin ((\theta -\pi )\mathrm{/}2)\mathrm{/}\sqrt{2}$, so $f=1\mathrm{/}4+v^2$ and $Xv=-2v$. Elsewhere $df(X)=-a{\sin }^2\theta \mathrm{/}4<0$. Thus this is an exact downward gradient-like field, rather than merely a descending round-metric gradient.

On each of the two open arcs the field is nonzero and points from $p$ to $q$. Its solutions are full trajectories: near either endpoint the smooth field has a simple linear zero with slope $\pm 2$, so reaching it requires infinite time (equivalently the separated time integral has logarithmic divergence). Consequently each arc has past limit $p$ and future limit $q$.

If $X$ is downward gradient-like for a new function $g$ with these same critical points, then $g$ is strictly decreasing on either arc trajectory. For finite $t_1<t_2$, continuity at the endpoints gives $g(p)\ge g(\gamma (t_1))>g(\gamma (t_2))\ge g(q)$; hence $g(p)>g(q)$. Reversing their values while retaining $X$ is impossible. This refutes the stated universal interchange without the no-connection hypothesis.

The lower point has index zero and the upper point index one, so the separation hypothesis requiring lower index at least upper index is absent here. Perturbation cannot be promised for every connecting pair; the index hypothesis is exactly what licenses it in the rearrangement argument. The counterexample establishes the fixed-field obstruction independently of such a perturbation.

The initial stage of any presentation relative to $M_0=\mathrm{\varnothing }$ is empty, so its boundary is empty as well, and the first attaching embedding $h_1$ must map into it; hence the first handle must have empty attaching region. By [A1] this happens exactly for $k=0$: the attaching region of a $0$-handle is $S^{-1}\times D^n=\mathrm{\varnothing }$, while a $k$-handle with $k\ge 1$ has nonempty attaching region and cannot be attached to the empty initial boundary. Therefore every presentation begins with at least one $0$-handle.

A connected manifold with empty incoming boundary needs at least one $0$-handle, since the first stage is empty and only a $0$-handle creates a component by [F2]; and by [F4] exactly the $0$-handles needed to create the components of $W$ remain, which for connected $W$ is one $0$-handle. Hence the elimination of $0$-handles is impossible when $M_0=\mathrm{\varnothing }$, and the hypothesis $M_0\mathrm{\ne }\mathrm{\varnothing }$ in [F4] is necessary.

The sphere example: the closed $n$-sphere is the union of two closed disks glued along their common boundary sphere, $S^n=D^n{\cup }_{S^{n-1}}{D}^n$. Read the first disk as a $0$-handle and the second as an $n$-handle attached along its whole boundary $S^{n-1}$, which by [F3] fills the whole boundary sphere and produces $S^n$. This presentation has exactly one $0$-handle and one $n$-handle, and no other handles, in agreement with the fact that a connected manifold with $M_0=\mathrm{\varnothing }$ keeps exactly one $0$-handle by step 2.1 and that the $n$-handle closes the remaining boundary sphere.

By [F5] the presentation of step 3.1 is realized by a Morse function on the triadic description of $S^n$ with two critical points, of indices $0$ and $n$; this is the standard round-sphere height function with a minimum and a maximum. In particular the sphere carries a presentation with exactly one $0$-handle as claimed, and the presentation of any connected triad with empty incoming boundary must begin with a $0$-handle, so the elimination proposition cannot be applied without change in that case.