The Smooth H Cobordism Theorem — Examples

The projections $r_i:W\to M_0\times \{i\}$, $r_i(x,t)=(x,i)$, are continuous and satisfy $r_i({\iota }_i(x))=r_i(x,i)=(x,i)={\iota }_i(x)$ for every $x\in M_0$, so by [F1] each $r_i$ is a retraction onto the corresponding face.

Read the product presentation through [F3]: $W$ has the empty handle list relative to $M_0\times \{0\}$, and no handle is attached.

The linear homotopy $H_s(x,t)=(x,(1-s)t+si)$ is continuous with $H_0={\mathrm{id}}_W$ and $H_1={\iota }_i\circ r_i$, and it satisfies $H_s(x,i)=(x,i)$ for all $s$; hence by [F1] each face is a deformation retract of $W$, in particular each inclusion is a homotopy equivalence with homotopy inverse $r_i$. The product is a compact smooth triad with its product collars and dimension $n+1\ge 2$, so by [F2] the triad $(W;M_0\times \{0\},M_0\times \{1\})$ is an h-cobordism.

The relative homology now vanishes by Relative homology of an h-cobordism vanishes at both ends, applied to step 2.1. Therefore the product cobordism has an empty presentation and vanishing relative homology; when $M_0$ is simply connected of dimension $n\ge 5$, the later h-cobordism theorem applied to this h-cobordism returns precisely the product it started from, so the product is the trivial model that the theorem's conclusion describes.

The given configuration is exactly the hypothesis of [F1]: the handles $h^k$, $h^{k+1}$ are consecutive, the attaching sphere of $h^{k+1}$ meets the belt sphere of $h^k$ transversely in exactly one point, so $h^k,h^{k+1}$ form a geometrically cancelling pair.

Suppose $M_0$ is oriented and $1\le k\le n-2$, and use the induced orientations on the handles and middle level. Restrict to the connected collar component containing the prescribed boundary disc: its outgoing boundary before the pair is connected, as required by the matrix definition, and every other component has an empty handle list. This component’s two-handle presentation has a single $k$-handle and a single $(k+1)$-handle, so by the matrix definition the intersection matrix is the $1\times 1$ matrix with the single entry given by the oriented intersection number of the two spheres; by [F5] that entry is $\pm 1$ equal to the local intersection sign of the unique transverse point, and the finiteness underlying the count is [F6]. Without orientations [F5] supplies instead the mod-two entry $1$. Reorienting the core of $h^k$ or the cocore of $h^{k+1}$ reverses the induced orientation of the corresponding sphere and hence the sign of the entry alone, so the matrix is normalisable to $(1)$.

Conversely, by [F3] a geometrically cancelling consecutive pair with the standard complementary configuration may be inserted at any prescribed disc of the outgoing boundary of any presentation without changing the presented manifold relative to the incoming boundary; hence the displayed configuration is exactly the inverse of a trivial local modification of the collar.

Apply the cancellation theorem to the connected collar component containing the given outgoing disk, and leave all other collar components fixed. For the geometrically cancelling pair of step 1.1, the manifold $W=M_0\times [0,1]+h^k+h^{k+1}$ is diffeomorphic to the collar $M_0\times [0,1]$ relative to $M_0\times \{0\}$, and the diffeomorphism may be taken supported in a collar of the affected disc and the two handles; the cancelled manifold is presented relative to $M_0\times \{0\}$ by the empty handle list, which by [F4] is diffeomorphic to $M_0\times [0,\epsilon ]$, and the collar reparametrization gives $M_0\times [0,1]$.

The diffeomorphism $F:W\to M_0\times [0,1]$ of step 2.1 is the identity on $M_0\times \{0\}=M_0$ and carries the second face of $W$ onto $M_0\times \{1\}$, so it conjugates the inclusion $M_0\hookrightarrow W$ to the standard inclusion of the face $M_0\times \{0\}$ and the other face inclusion to the standard inclusion of $M_0\times \{1\}$ composed with a diffeomorphism of that face; both standard inclusions are homotopy equivalences (via the projections $(x,t)\mapsto (x,i)$ and the linear homotopies). Hence both face inclusions of $W$ are homotopy equivalences and, by [F7], $W$ is an h-cobordism.

Therefore an elementary cancelling pair attached to a collar builds a genuine product: $W\cong M_0\times [0,1]$ relative to $M_0\times \{0\}$, so $W$ is an h-cobordism, the connected component containing the pair has two-index intersection matrix $(\epsilon )$ with $\epsilon =\pm 1$ normalisable to $(1)$ when $M_0$ is oriented and $1\le k\le n-2$ (without orientations, in the same middle range, use the mod-two entry), and by step 1.3 the configuration is precisely the inverse of the trivial local modification that inserts a cancelling pair. This is the local model of the cancellation step of the h-cobordism theorem.

(The $1$-handlebody.) Let $Y$ be the $0$-handle with the two $1$-handles attached orientation-compatibly. It is compact, connected and oriented, and has the homotopy type of the wedge of two circles by [F1], so ${\pi }_1(Y)$ is freely generated by $a,b$. Its boundary is connected: each $1$-handle deletes two $5$-disks from the connected boundary and joins their boundary spheres by the connected cylinder $[-1,1]\times S^4$. The dual decomposition of $(Y,\mathrm{\partial }Y)$ has only indices $5,6$ by [F4]. A relative CW pair with cells only in dimensions at least $3$ induces a fundamental-group isomorphism: cellular approximation moves loops into the boundary and homotopies of loops into the boundary because their domains have dimensions $1,2$. Hence ${\pi }_1(\mathrm{\partial }Y)\to {\pi }_1(Y)$ is an isomorphism and the words $w_1,w_2$ can be represented by loops in $\mathrm{\partial }Y$.

(Embedded framed attaching loops.) Apply [F2] first to smooth representatives and then to their disjoint union $S^1\bigsqcup S^1\to \mathrm{\partial }Y$. The target has dimension $5\ge 2\cdot 1+1$, so a homotopic embedding gives two disjoint embedded circles representing the words (up to basepoint conjugacy, which leaves their normal closure unchanged). Each oriented circle in the oriented $5$-manifold has an oriented rank-four normal bundle. Pull that bundle back to $[0,1]$ by cutting the circle at one point: finite successive local trivializations give a smooth oriented frame over the interval, arranged near the endpoints to be the pullbacks of one fixed seam trivialization up to constant matrices. The endpoint gluing is then a constant $A\in {\mathrm{G}\mathrm{L}}^{+}(4,\mathbb{R})$. By [F3] join the identity to $A^{-1}$ by a smooth path, made constant near its endpoints, and multiply the interval frame by that path. The adjusted frames agree under endpoint gluing and are smooth across the seam, giving a global framing. Tubular neighbourhoods supply disjoint attaching regions $S^1\times D^4$, so attach two orientation-compatible $2$-handles to obtain a compact oriented smooth $6$-manifold $X$.

(Boundary connectedness and group.) Removing the two circle cores from $\mathrm{\partial }Y$ leaves it path connected: a path between two points can be perturbed relative to endpoints to be transverse to those circles, and $1+1<5$ forces the perturbed path to miss them. Radially pushing the punctured normal $4$-disks outward shows that deleting the interiors of small tubular neighbourhoods also leaves a connected complement. Each $2$-handle replaces $S^1\times D^4$ by the connected $D^2\times S^3$, glued along the connected $S^1\times S^3$; hence $N=\mathrm{\partial }X$ is connected. It is a closed smooth $5$-manifold. The dual decomposition of $(X,N)$ has indices $4,5,6$, so the same relative cellular-approximation argument as in step 1.1 gives ${\pi }_1(N)\cong {\pi }_1(X)$.

(Acyclicity.) By [F1], $X$ has one $0$-cell, two $1$-cells and two $2$-cells with attaching words $w_1,w_2$. Traversing a letter contributes its signed exponent to the incidence degree on the corresponding $1$-cell; the exponent vectors are $(2,-3)$ for $w_1$ and $(-3,5)$ for $w_2$, since $2-5=-3$ for the exponent of $a$ in $w_2$. Thus the cellular chain complex is $0\to {\mathbb{Z}}^2\stackrel{d_2}{\to }{\mathbb{Z}}^2\stackrel{0}{\to }\mathbb{Z}\to 0$, with $d_2=\left(\begin{array}{cc}2& -3\\ -3& 5\end{array}\right)$. Its determinant is $10-9=1$ and its integral inverse is $\left(\begin{array}{cc}5& 3\\ 3& 2\end{array}\right)$. Consequently $H_0(X;\mathbb{Z})=\mathbb{Z}$ and ${\tilde{H}}_j(X;\mathbb{Z})=0$ for every $j$.

(A nontrivial fundamental group.) Van Kampen gives ${\pi }_1(X)=⟨a,b\mid a^2{b}^{-3}=1,a^2(b^{-1}a{)}^{-5}=1⟩$. Map $a$ to $(12)(34)$ and $b$ to $(135)$ in the permutation group on five letters, composing right to left. Then $a^2=b^3=1$, and direct composition gives $b^{-1}a=(12534)$, of order $5$, so both relators map to the identity. This defines a homomorphism from ${\pi }_1(X)$ whose value on $a$ is nonidentity, proving ${\pi }_1(X)\mathrm{\ne }1$; step 3.1 also gives ${\pi }_1(N)\mathrm{\ne }1$.

(Puncturing and the open cover.) Choose one interior coordinate disk $j:B^6\hookrightarrow \mathrm{int}X$, let $D=j(B^6)$ and $S=\mathrm{\partial }D\cong S^5$, and put $W=X\setminus \mathrm{int}D$. It is a compact oriented smooth manifold with boundary $N\bigsqcup S$. Set $D_0=j(\{\mathrm{\mid }x\mathrm{\mid }\le 1\mathrm{/}2\})$, $U=X\setminus D_0$, $V=\mathrm{int}D$. These are open in $X$, cover it, and overlap in $j(\{1\mathrm{/}2<\mathrm{\mid }x\mathrm{\mid }<1\})\cong S^5\times (1\mathrm{/}2,1)$. The overlap retracts onto its radius-$3\mathrm{/}4$ sphere, and $U$ strongly deformation retracts onto $W$ by pushing radius $r$ to $(1-t)r+t$ on $1\mathrm{/}2<r\le 1$ and fixing $W$. The radius-$3\mathrm{/}4$ sphere inclusion and $S$ inclusion in $U$ are homotopic by interpolating radii. The open ball $V$ contracts radially to its centre. The reduced degree-zero Mayer--Vietoris sequence, with connected overlap, connected $V$ and connected $X$, gives ${\tilde{H}}_0(U)=0$; since $U$ is a manifold it is locally path connected and therefore path connected, as is $W$.

(The sphere end is a homology equivalence.) For $j\ge 1$, the Mayer--Vietoris map $H_j(U\cap V)\to H_j(U)\oplus H_j(V)$ is an isomorphism because both adjacent positive-degree groups of $X$ vanish by step 3.2 and $H_j(V)=0$. Via the radial homotopies in step 4.2 this is exactly the inclusion-induced map $H_j(S;\mathbb{Z})\to H_j(W;\mathbb{Z})$. In degree zero it is an isomorphism since $S,W$ are nonempty and connected. The long exact sequence of $(W,S)$ therefore gives $H_j(W,S;\mathbb{Z})=0$ for all $j\ge 0$.

(The other end is a homology equivalence.) The relative singular chain complex is free over $\mathbb{Z}$ (its basis is the singular simplices not lying wholly in $S$). Its homology vanishes by step 5.1, so the universal coefficient sequence in [F7] has zero Hom and Ext terms and gives $H^p(W,S;\mathbb{Z})=0$ for every $p$. Duality with $\mathrm{\partial }W=N\bigsqcup S$ then gives $H_j(W,N;\mathbb{Z})=H^{6-j}(W,S;\mathbb{Z})=0$ for $0\le j\le 6$; for $j>6$, apply the Morse/handle correspondence of [F4] to the compact triad $(W;N,S)$ and the CW comparison of [F1]: its relative cells have indices at most $6$, so those higher relative homology groups also vanish. Thus the long exact sequence of $(W,N)$ makes its inclusion an integral homology isomorphism in every degree.

(Failure of the h-cobordism condition.) In the cover of step 4.2, $V$ is contractible and $U\cap V\simeq S^5$ is simply connected; van Kampen gives ${\pi }_1(U)\cong {\pi }_1(X)$. The radial deformation retraction identifies ${\pi }_1(W)\cong {\pi }_1(U)\mathrm{\ne }1$ by step 4.1. But ${\pi }_1(S)={\pi }_1(S^5)=1$ by [F5], so $S\hookrightarrow W$ cannot be a homotopy equivalence by [F8]. Both end inclusions are integral homology isomorphisms by steps 5.1 and 6.1, yet one fails the defining homotopy-equivalence condition. Hence $W$ is a homology cobordism which is not an h-cobordism, refuting the claim.

The proof route fails in boundary dimension four: by [F1], a clean embedded Whitney disk in a four-dimensional middle level cannot generally be produced, since the general-position dimension count for complementary sheets of dimensions $2$ and $2$ gives no room for an embedded disk; the Whitney-trick step of the theorem is therefore unavailable exactly when $n=4$.

The conclusion also fails, so the failure is not merely a gap in the proof: by [F2] there are smooth orientable simply connected 4-manifolds that are smoothly s-cobordant and homeomorphic but not diffeomorphic, and an s-cobordism is in particular an h-cobordism between them; were such an h-cobordism diffeomorphic to the product relative to its incoming boundary, [F3] would make its two faces diffeomorphic, contradicting the choice of the pair.

Therefore the hypothesis $n\ge 5$ of the smooth h-cobordism theorem cannot be weakened to $n=4$: the dimension count at the Whitney step stops working and the conclusion itself is false in general for simply connected closed 4-manifolds, while the parallel topological statement for good fundamental groups is a genuinely different theorem. No smooth four-dimensional Poincaré or disk conclusion follows from the theorem of this page, whose statement is restricted to boundary dimension at least five by [F4].

The determinant of $A$ is $2\cdot 1-1\cdot 1=1$, so by [F1] the matrix $A$ is invertible over $\mathbb{Z}$ with $A\in {\mathrm{GL}}_2(\mathbb{Z})$ and inverse $\left(\begin{array}{cc}1& -1\\ -1& 2\end{array}\right)$, whose product with $A$ is the identity.

The displayed operations transform $A$ into $I_2$: interchanging the rows gives $\left(\begin{array}{cc}1& 1\\ 2& 1\end{array}\right)$; subtracting twice the first row from the second gives $\left(\begin{array}{cc}1& 1\\ 0& -1\end{array}\right)$; subtracting the first column from the second gives $\left(\begin{array}{cc}1& 0\\ 0& -1\end{array}\right)$; multiplying the second row by $-1$ gives $\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right)=I_2$. Each step is a row or column addition, an interchange or a sign change, so the sequence is exactly the reduction of [F2].

If a two-index h-cobordism presentation with matrix $A$ satisfies the stated dimension, simple-connectivity and countable-choice hypotheses, [F3] realizes exactly these four operations by slides, relabeling and reorientation, preserving the manifold relative to the incoming face. Its matrix becomes $I_2$. Then The Whitney trick realizes algebraic middle-handle cancellation geometrically and Middle-handle pairs with one geometric intersection cancel give geometric cancellation and the product presentation. This is conditional on such a handle presentation; invertibility of an arbitrary integer matrix alone is not an existence construction of a cobordism.