Sublevel Deformation and the Handle Attachment Theorem — Examples

A critical point has the vertical vector normal to the sphere, so the only critical points are the two poles. In horizontal coordinates $z$ at those poles, height is respectively $-\sqrt{1-\mathrm{\mid }z{\mathrm{\mid }}^2}$ and $\sqrt{1-\mathrm{\mid }z{\mathrm{\mid }}^2}$; their Hessians at zero are $+I$ and $-I$. The indices are $0$ and $n$, and their values are $-1$ and $1$.

Stereographic coordinates from the north pole identify $S^n\setminus \{\text{north}\}$ with ${\mathbb{R}}^n$ and give height $(\mathrm{\mid }w{\mathrm{\mid }}^2-1)\mathrm{/}(\mathrm{\mid }w{\mathrm{\mid }}^2+1)$. For $-1<a<1$ the sublevel is therefore $\mathrm{\mid }w{\mathrm{\mid }}^2\le (1+a)\mathrm{/}(1-a)$, a closed disk. The values at and beyond the poles give the point, empty set and whole sphere stated above.

The sphere is compact, and each band crossing only one pole satisfies the handle theorem. The lower change adds a disjoint disk; the upper change caps its boundary by the whole-boundary attachment. At $n=1$ the cap attaches along two endpoints, as required by the endpoint qualification.

The critical equations are $\sin \theta =0$ and $2\sin \varphi =0$. Modulo $2\pi$ there are four solutions: $(0,0),(\pi ,0),(0,\pi ),(\pi ,\pi )$. The Hessian is $\mathrm{diag}(\cos \theta ,2\cos \varphi )$, giving respectively indices $0,1,1,2$ and values $-3,-1,1,3$.

The torus is compact. Choose successive regular levels, for example $-4,-2,0,2,4$. Each intervening closed band contains exactly one of these nondegenerate points. Applying the handle theorem to each band gives the stated counts from the empty sublevel to the full torus.

A surface handle is a rectangle, attached by its two opposite end edges. Boundary interval parametrizations can be straightened in disk collars: extend their increasing one-dimensional coordinate changes across an annular collar by interpolating a lifted circle coordinate, whose derivative stays positive. If necessary reflect an entire disk or the rectangle to normalize an end orientation. Thus the two-disk attachment is represented by two rectangular disks joined end to end by a rectangular strip. Their union is a longer rectangle before compatible corner rounding, hence a disk afterward.

For the one-disk orientable attachment, represent the disk as the rectangle $[0,1]\times [0,1]$ cut from an annulus along one radial interval. Its two radial edges are the prescribed attaching intervals after the same boundary straightening. Glue in a second rectangle bridging these edges with the orientation-compatible identifications; in coordinates the result is $([0,2]\mathrm{/}(0\sim 2))\times [0,1]$, an annulus. Reversing just one end identification instead reverses the transverse interval after one circuit, so this coordinate description no longer gives the orientable annulus. The claim explicitly excludes that twist.

The critical equations are $\sin \theta =\sin \varphi =0$. The Hessian is $\mathrm{diag}(\cos \theta ,\cos \varphi )$. At $(0,0)$ it is positive definite with value $-2$; at $(\pi ,\pi )$ it is negative definite with value $2$. At $(\pi ,0)$ and $(0,\pi )$ it has one negative entry and value zero.

For $0<\epsilon <2$, both endpoints are regular and the compact band contains precisely the two saddle points. The simultaneous-attachment proposition gives two disjoint index-one handles, with no need to perturb their equal values. The restrictions on epsilon exclude both the collapsed band and endpoints through the extrema.

Using the closed-band notation, the differential is $df=dx$, nonzero at every point of $M$. The level at $-1$ is a copy of $\mathbb{R}$, while the level at zero is $\{0\}\times (\mathbb{R}\setminus \{0\})$, with two connected components. A level-preserving product would restrict to homeomorphisms from one fixed fiber onto both, which is impossible.

The band is noncompact, for it contains the unbounded sequence $(0,j)$ for positive integers $j$. The normalized gradient is ${\mathrm{\partial }}_x$, whose trajectory from $(-1,0)$ is $(-1+t,0)$ for $t<1$. At $t=1$ its only possible limit in ${\mathbb{R}}^2$ is the removed point, so it cannot extend as a trajectory in $M$. This is exactly the missing compact-band hypothesis in the regular-interval theorem, not a counterexample to that theorem.

The collar-excision argument for a single critical point replaces the sublevel pair in relative homology by the standard handle pair with $k=2,n=4$. It uses an open collar thickening of the lower sublevel before excision, so the closure-in-interior requirement is met.

The explicit pair homotopy $(u,v)\mapsto (u,(1-t)v)$ contracts the second disk factor. The standard-pair calculation then identifies $H_2(D^2,S^1;G)$ with $H_1(S^1;G)=G$. In degree zero the map from the circle to the disk is the identity on $G$, so the relative degree-zero group and degree-one group vanish; all higher groups except degree two vanish as well. This includes $G=0$.