Concordance implies isotopy for smooth structures on 3-manifolds?

Revision as of 07:11, 8 January 2019

1 Problem

A $== Problem == <wikitex>; A $-manifold has a smooth structure which is essentially unique. One question is what does ``essentially unique" mean. In high dimensions you can classify smooth structures either up to concordance or up to isotopy. In high dimensions concordance implies isotopy. If you have a topological manifold $M^n$ and a smooth manifold $W^{n+1}$ homeomorphic to $M \times I$, you can ask if that smooth manifold $W$ is diffeomorphic to a product, $$W \xrightarrow{diffeo} \partial_+ W \times I$$ This ``concordance implies isotopy" theorem of Kirby-Siebenmann and Hirsch is: \begin{itemize} \item true in high dimensions, $n \geq 5$. If you have a topological manifold $M^n$, you can classify smooth structures as homotopy classes of maps $$\mathcal{S}^{TOP/O}(M^n) = [M^n, TOP/O]$$ The structure set $\mathcal{S}^{TOP/O}(M^n)$ is the set of classes of smoothings on $M^n$. But it is important to decide what is meant by the ``structure" and there are two possible equivalence relations: concordance or isotopy. A smoothing is a homeomorphism from a smooth manifold $\Sigma$ to $M.$ $$\Sigma \xrightarrow{homeo} M.$$ Two smoothings $\Sigma_0 \xrightarrow{homeo} M^n$ and $\Sigma_1 \xrightarrow{homeo} M$ are \textit{concordant} if there is a smooth manifold $W^{n+1}$ and a homeomorphism to $M \times I$ that restricts to these two smoothings. \medskip Two smoothings $\Sigma_0 \xrightarrow{homeo} M$ and $\Sigma_1 \xrightarrow{homeo} M$ are isotopic if there is a smooth manifold $\Sigma$ and a level-preserving homeomorphism $H: \Sigma \times I \to M \times I$, $$(H(s, t) = (F(s, t), t))$$ inducing $\Sigma_i \xrightarrow{homeo} M \times \{ i \}$ for $i = 0, 1.$ \begin{theorem} If dim $M \geq 5$, then concordant structures are isotopic (and hence diffeomorphic). \end{theorem} \item false for simply-connected $-manifolds by the failure of the $h$-cobordism theorem. Cappell and Shanneson proved in the non-simply connected case that there are counterexamples to the $h$-cobordism theorem. \item we don't know the answer for $n=3$. \end{itemize} aa </wikitex> == References == {{#RefList:}} [[Category:Questions]] [[Category:Research questions]]3$-manifold has a smooth structure which is essentially unique. One question is what does "essentially unique" mean.

In high dimensions, smooth structures can be classified either up to concordance or up to isotopy. In high dimensions concordance implies isotopy.

Let $M^n$ be a topological manifold and let $W^{n+1}$ be a smooth manifold homeomorphic to $M \times I$, one can ask if that smooth manifold $W$ is diffeomorphic to a product,

$$\displaystyle W \xrightarrow{diffeo} \partial_+ W \times I$$

This "concordance implies isotopy" theorem of Kirby-Siebenmann and Hirsch is: \begin{itemize} \item true in high dimensions, $n \geq 5$.

If you have a topological manifold $M^n$, you can classify smooth structures as homotopy classes of maps

$$\displaystyle \mathcal{S}^{TOP/O}(M^n) = [M^n, TOP/O]$$

The structure set $\mathcal{S}^{TOP/O}(M^n)$ is the set of classes of smoothings on $M^n$. But it is important to decide what is meant by the ``structure" and there are two possible equivalence relations: concordance or isotopy.

A smoothing is a homeomorphism from a smooth manifold $\Sigma$ to $M.$

$$\displaystyle \Sigma \xrightarrow{homeo} M.$$

Two smoothings $\Sigma_0 \xrightarrow{homeo} M^n$ and $\Sigma_1 \xrightarrow{homeo} M$ are \textit{concordant} if there is a smooth manifold $W^{n+1}$ and a homeomorphism to $M \times I$ that restricts to these two smoothings.

\medskip

Two smoothings $\Sigma_0 \xrightarrow{homeo} M$ and $\Sigma_1 \xrightarrow{homeo} M$ are isotopic if there is a smooth manifold $\Sigma$ and a level-preserving homeomorphism $H: \Sigma \times I \to M \times I$,

$$\displaystyle (H(s, t) = (F(s, t), t))$$

inducing $\Sigma_i \xrightarrow{homeo} M \times \{ i \}$ for $i = 0, 1.$

\item false for simply-connected $4$-manifolds by the failure of the $h$-cobordism theorem. Cappell and Shanneson proved in the non-simply connected case that there are counterexamples to the $h$-cobordism theorem. \item we don't know the answer for $n=3$. \end{itemize}

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2 References