Concordance of decompositions given by defining sequences

If $X$ has no boundary, then the statement follows from Theorem 2 and Proposition 3 in [Da86, Chapter 34]. If $X$ has non-empty boundary, then the argument is also similar. ∎

We have to show that the quotient space

is an $n$-dimensional homology manifold with boundary the homology manifold $\pi(\partial X)$. Take the closed manifold

where $\varphi\colon\thinspace\partial X\to\partial X$ is the identity map.

The decomposition space $X^{\prime}/\mathcal{D}^{\prime}$ (that is the part of the decomposition space $X/\mathcal{D}$ which is obtained from $X^{\prime}$) is finite dimensional by Lemma 2.3. The doubling of the decomposition $\mathcal{D}$ on $X\cup_{\varphi}X$ yields a finite dimensional quotient space, we get this by using estimations for the covering dimension, see [HW41] and [Da86, Corollary 2.4A]. So the decomposition space $P$ obtained by factorizing $X\cup_{\varphi}X$ by the double of $\mathcal{D}$ is a closed finite dimensional homology manifold by [DV09, Proposition 8.5.1]. Since a small neighbourhood of a singleton results an open set in $P$ homeomorphic to $\mathbb{R}^{n}$, it is $n$-dimensional. We obtain the space $X/\mathcal{D}$ by cutting $P$ into two pieces along $\pi(\partial X)$. Because of a similar argument the space $\pi(\partial X)$ is a closed $(n-1)$-dimensional homology manifold. The set $\pi(\partial X)$ is closed in the decomposition space $X/\mathcal{D}$ since $\mathcal{D}$ is upper semi-continuous and $\partial X$ is closed. Also, the homology group $H_{n}(X/\mathcal{D};X/\mathcal{D}-\{p\})$ is equal to $0$ for every $p\in\pi(\partial X)$. So $\pi(\partial X)$ is the boundary of $X/\mathcal{D}$.

Moreover the space $X/\mathcal{D}$ is a locally compact separable metric space because $X$ is so. By [DV09, Corollary 7.4.8] the space $X/\mathcal{D}$ is an ANR so it follows that it is an ENR. The same holds for $\pi(\partial X)$. ∎

If there are two resolutions $f_{1}\colon\thinspace M_{1}\to N$ and $f_{2}\colon\thinspace M_{2}\to N$ of a closed $n$-dimensional homology manifold $N$, then as in the proof of [Qu82, Theorem 2.6.1] take a resolution

of the double mapping cylinder $X_{f_{1}}\cup X_{f_{2}}$ of the maps $f_{1}$ and $f_{2}$ by applying [Qu83, Theorem 1.1] and [Qu87]. This resolution exists because $X_{f_{1}}\cup X_{f_{2}}$ is an $(n+1)$-dimensional ENR homology manifold and $i(X_{f_{1}}\cup X_{f_{2}})=1$. Let

be the natural map of the double mapping cylinder onto $N\times[-1,1]$, where the target $N$ of the two mapping cylinders is mapped onto $N\times\{0\}$.

It follows that the composition

is a resolution, moreover by [Qu83, Theorem 1.1] the cell-like map $Y\to X_{f_{1}}\cup X_{f_{2}}$ can be chosen so that it is a homeomorphism over the boundary hence $Y$ is a cobordism between $M_{1}$ and $M_{2}$. ∎

Checking the conditions (1)-(4) in Definition 2.13 is obvious, details are left to the reader. ∎

Suppose that $C_{0},C_{1},C_{2},\ldots$ and $D_{0},D_{1},D_{2},\ldots$ are admissible defining sequences for a decomposition $(S^{3},\mathcal{D},A,C_{0})$ such that $C_{0}=D_{0}$. Of course

We use an algorithm applied in [Sh68, Proof of Theorem 2]. We restrict ourselves to one component of $C_{0}$ and to the components of the defining sequences in it, the following argument works the same way for the other components. We can suppose that $\partial C_{1}\cap\partial D_{1}$ is a closed $1$-dimensional submanifold of $S^{3}$. Suppose some component $P$ of $\partial C_{1}\cap\partial D_{1}$ bounds a $2$-dimensional disk $Q\subset\partial D_{1}$. Also suppose that $P$ is an innermost component of $\partial C_{1}\cap\partial D_{1}$ in $\partial D_{1}$ so $\mathrm{int}Q\cap\partial C_{1}=\emptyset$. By (4) in Definition 2.13 if $P$ does not bound a disk $Q^{\prime}$ in $\partial C_{1}$, then $P$ is not homotopic to constant in the complement of $A$ but then $P$ cannot bound the disk $Q\subset\partial D_{1}$. Hence $P$ bounds a disk $Q^{\prime}\subset\partial C_{1}$ as well. Then the interior of the sphere $Q\cup Q^{\prime}$ does not intersect $A$ because of (3) in Definition 2.13. So we can modify $C_{1}$ by pushing $Q^{\prime}$ through the sphere $Q\cup Q^{\prime}$ by a self-homeomorphism of the complement of $A$ and hence we obtain fewer circles in the new $\partial C_{1}\cap\partial D_{1}$. After repeating these steps finitely many times we obtain a new $C_{1}$ such that $\partial C_{1}\cap\partial D_{1}$ contains no circles which bound disks on $\partial C_{1}\cup\partial D_{1}$. Similarly, by further adjusting $C_{1}$ in the complement of $A$ as written on [Sh68, page 1198] in order to eliminate the circles in $\partial C_{1}\cap\partial D_{1}$ which bound annuli we finally obtain a $C_{1}$ such that

• •

  the intersection $\partial C_{1}\cap\partial D_{1}$ is empty,

• •

  no component of $C_{1}$ is disjoint from all the components of $D_{1}$ and vice versa,

• •

  each component of $C_{1}$ is inside a component of $D_{1}$ or it contains some components of $D_{1}$.

Then we can see that there is a bijection between the number of components of $C_{1}$ and $D_{1}$ because of the following.

If a component of $C_{1}$ is in $\mathrm{int}D_{1}$ and it is homotopic to constant in $\mathrm{int}D_{1}$, then all the other components of $C_{1}$ are in the same component of $\mathrm{int}D_{1}$ by (2) in Definition 2.13. This would result that no part of $A$ is in other components of $D_{1}$, which would contradict to (1) in Definition 2.13 so no component of $C_{1}$ in $\mathrm{int}D_{1}$ is homotopic to constant in $\mathrm{int}D_{1}$. The same holds if we switch the roles of $C_{1}$ and $D_{1}$. This means that

• •

  the winding number of a component $T$ of $C_{1}$ in the component of $D_{1}$ which contains $T$ is not equal to $0$ and the same holds for $D_{1}$ and $C_{1}$ with opposite roles.

Furthermore suppose that $T$ is some component of $D_{1}$ and $T$ contains at least two components $T_{1}$ and $T_{2}$ of $C_{1}$. Then $T$ is linking with other component $T^{\prime}$ of $D_{1}$ by (1) in Definition 2.13 with algebraic linking number non-zero. Let $T_{3}$ be a component of $C_{1}$ such that $T_{3}\subset T^{\prime}$ or $T^{\prime}\subset T_{3}$. If $T_{3}\subset T^{\prime}$, then $T_{1}$ and $T_{2}$ are linking with $T_{3}$ with algebraic linking number non-zero. If $T^{\prime}\subset T_{3}$, then $T$ is not in $T_{3}$ because for example $T_{1}$ cannot be in $T_{3}$. But then $T$ is linking with $T_{3}$ with algebraic linking number non-zero since the same holds for $T$ and $T^{\prime}$. So again we obtained that $T_{1}$ and $T_{2}$ are linking with $T_{3}$ with linking number non-zero. Now, there is a $T^{\prime\prime}$ component of $D_{1}$ which is linking with $T^{\prime}$ with linking number non-zero and which is disjoint from all the previously mentioned tori ($T^{\prime},T^{\prime\prime}\subset T_{3}$ is impossible because then both of $T^{\prime},T^{\prime\prime}$ are linking with $T$ and also with each other and this contradicts to (1) in Definition 2.13). Let $T_{4}$ be a component of $C_{1}$ such that $T_{4}\subset T^{\prime\prime}$ or $T^{\prime\prime}\subset T_{4}$. There are a number of cases to check. If $T_{3}\subset T^{\prime}$ and $T_{4}\subset T^{\prime\prime}$, then $T_{3}$ is linking with $T_{4}$. If $T_{3}\subset T^{\prime}$ but $T^{\prime\prime}\subset T_{4}$, then since $T_{4}$ cannot contain $T$ or $T^{\prime}$, we have again that $T_{3}$ is linking with $T_{4}$. Finally, if $T^{\prime}\subset T_{3}$, then since $T^{\prime\prime}$ cannot be in $T_{3}$, we have that $T_{4}\subset T^{\prime\prime}$ implies that $T_{3}$ and $T_{4}$ are linking and $T^{\prime\prime}\subset T_{4}$ implies that since $T_{4}$ is disjoint from all the other tori, again $T_{4}$ is linking with $T_{3}$. So we obtain that $T_{1}$ and $T_{2}$ are linking with $T_{3}$ and $T_{3}$ is linking with $T_{4}$ resulting that $T_{3}$ is linking with three other components of $C_{1}$ which contradicts to (1) in Definition 2.13. Summarizing, we obtained the following.

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  The intersection $\partial C_{1}\cap\partial D_{1}$ is empty,

• •

  no component of $C_{1}$ is disjoint from all the components of $D_{1}$ and vice versa,

• •

  every component of $C_{1}$ contains one component of $D_{1}$ or is contained in one component of $D_{1}$,

• •

  no component of $C_{1}$ contains more than one component of $D_{1}$ and vice versa.

All of these imply that the number of components of $C_{1}$ is equal to the number of components of $D_{1}$. We repeat the same line of arguments for the components of $C_{2}$ and $D_{2}$ lying in each component of $C_{1}$ or $D_{1}$ separately, where we perform the previous algorithm in the larger component which contains the smaller one, and so on, in this way we get the result. ∎

If decompositions with defining sequences $C_{0},C_{1},\ldots$ and $D_{0},D_{1},\ldots$ are cylindrically related, then for every $m\geq 0$ the pairs of components of $C_{m}$ and the pairs of corresponding components of $D_{m}$ have the same algebraic linking numbers. Suppose for a decomposition there are two admissible defining sequences $C_{0},C_{1},\ldots$ and $D_{0},D_{1},\ldots$ such that $C_{0}=D_{0}$, we have to show that the linking numbers are equal to $0$ simultanously for both of them (for the components of $C_{0}$ and $D_{0}$ this is obviously true). Of course we know that the components are in bijection with each other by the proof of Proposition 3.2 and in every component of $C_{0}$ after some deformation we have that

• •

  the intersection $\partial C_{1}\cap\partial D_{1}$ is empty,

• •

  no component of $C_{1}$ is disjoint from all the components of $D_{1}$ and vice versa,

• •

  every component of $C_{1}$ contains one component of $D_{1}$ or is contained in one component of $D_{1}$,

• •

  no component of $C_{1}$ contains more than one component of $D_{1}$ and vice versa.

If a component $T$ of $C_{1}$ is linked with a component $T^{\prime}$ of $C_{1}$ with linking number $0$, then any knot in $T^{\prime}$ is linked with $T$ with linking number $0$. Also, if a knot in $T^{\prime}$ is linked with $T$ with linking number $0$, then $T$ and $T^{\prime}$ are linked with linking number $0$. For every $m\geq 1$ after a finite number of iterations of the algorithm in the proof of Proposition 3.2 we get the result. ∎

For every element $(l_{0},l_{1},\ldots)\in\mathbb{Z}_{2}^{\mathbb{N}}$, where $l_{0}=0$, we have an Antoine decomposition representing a class $x$ such that $\nu([x])=(l_{0},l_{1},\ldots)$. Hence we get uncountably many different classes in the concordance group. ∎

Take $M\times[0,1]$ and consider the cell-like decomposition $\mathcal{D}$ of $M$ which results the homology manifold $N$. If $\mathcal{S}(X)$ denotes the collection of singletons in a space $X$, then $\mathcal{D}\times\mathcal{S}([0,1/2])$ union $\mathcal{S}(M\times(1/2,1])$ is a cell-like decomposition of $M\times[0,1]$, denote it by $\mathcal{E}$. We have to show that the quotient space

is an $(n+1)$-dimensional homology manifold with boundary homology manifolds $N$ and $M$. Take the closed manifold

where $\varphi\colon\thinspace\partial(M\times[0,1])\to\partial(M\times[0,1])$ is the identity map. Since $M/\mathcal{D}$ is $n$-dimensional, the doubling of the decomposition $\mathcal{E}$ on $M\times[0,1]\cup_{\varphi}M\times[0,1]$ yields a finite dimensional quotient space, we get this by using estimations for the covering dimension, see [HW41] and [Da86, Corollary 2.4A]. So the decomposition space $P$ obtained by factorizing $M\times[0,1]\cup_{\varphi}M\times[0,1]$ by the double of $\mathcal{E}$ is a closed finite dimensional homology manifold by [DV09, Proposition 8.5.1]. Since this space has an open set homeomorphic to $\mathbb{R}^{n+1}$, it is $(n+1)$-dimensional. We obtain the space $M\times[0,1]/\mathcal{E}$ by cutting $P$ into two pieces along two subsets homeomorphic to $M$ and $N$. This means that $M$ and $N$ are cobordant in $\mathfrak{N}_{n}^{E}$. ∎

Let $M_{1}$ and $M_{2}$ be closed manifolds. If the two cobordism classes $[M_{1}]$ and $[M_{2}]$ in $\mathfrak{N}_{n}^{E}$ coincide, then since $M_{i}$ are manifolds, we have $i(M_{i})=1$ hence a cobordism in $\mathfrak{N}_{n}^{E}$ between $M_{1}$ and $M_{2}$ also has index $1$ so this cobordism is resolvable. By [Qu83, Theorem 1.1] and Lemma 2.8 there is a manifold cobordism between $M_{1}$ and $M_{2}$. ∎