isotopy inverse embeddings vs. diffeomorphisms - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T03:08:08Z http://mathoverflow.net/feeds/question/100498 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/100498/isotopy-inverse-embeddings-vs-diffeomorphisms isotopy inverse embeddings vs. diffeomorphisms Ricardo Andrade 2012-06-24T02:13:31Z 2012-06-27T02:33:35Z <p>I would like to find an example, if one exists, of manifolds $M$ and $N$ with embeddings $f:M\to N$ and $g:N\to M$ such that $f\circ g$ and $g\circ f$ are both isotopic (i.e. homotopic through embeddings) to the respective identities, yet the interiors of $M$ and $N$ are not diffeomorphic. Obviously, $M$ and $N$ cannot be closed. You may assume that the manifolds have no boundary, but I would also be interested in compact examples.</p> <p>By the way: in the other direction, are there conditions under which $M$ and $N$ are necessarily diffeomorphic if $f$ and $g$ as above exist?</p> http://mathoverflow.net/questions/100498/isotopy-inverse-embeddings-vs-diffeomorphisms/100660#100660 Answer by Agol for isotopy inverse embeddings vs. diffeomorphisms Agol 2012-06-26T06:13:43Z 2012-06-27T02:33:35Z <p>Here I'll prove that $M$ and $N$ must be diffeomorphic under your hypotheses (I'll assume that $M$ and $N$ are open, i.e. no boundary). </p> <p>Consider the direct limit (see below) $X= M \overset{f}{\to} N \overset{g}{\to} M \overset{f}{\to} N \overset{g}{\to}\cdots$. By this, I mean the union $M \times \mathbb{N} \sqcup N\times \mathbb{N}$ modulo the identifications <code>$(m,k)\sim (f(m),k), (n,k)\sim (g(n),k+1)$</code>, for <code>$m\in M, n\in N, k\in \mathbb{N}$</code>. Then $X$ is a smooth manifold, since it is a nested union of smooth manifolds. Similarly, we have the direct limit $Y=N \overset{g}{\to} M \overset{f}{\to} N \overset{g}{\to}\cdots$. Clearly, $X \cong Y$, since the direct limit depends only on the tail of the sequence. Also, $X$ (and therefore $Y$) is diffeomorphic to the direct limits $ M \overset{g\circ f}{\to} M \overset{g\circ f}{\to} M \overset{g\circ f}{\to}\cdots$ and $N\overset{f\circ g}{\to} N\overset{f\circ g}{\to} N\overset{f\circ g}{\to}\cdots$. </p> <p><strong>Claim:</strong> If $F:M\to M$ is isotopic to the identity, then the direct limit $M\overset{F}{\to} M\overset{F}{\to} M\overset{F}{\to} \cdots \cong M$. </p> <p>By the above discussion, this implies that $M\cong N$. </p> <p>To prove the claim, let $F_t:M\to M, t\in [0,1]$ be an isotopy of $F$ to the identity, so $F_1=F$, and $F_0=Id$. Take an exhaustion of $M$ by smooth compact submanifolds with boundary $K_1\subset K_2\subset K_3 \subset \cdots$, so that $F([0,1]\times K_i)\subset int(K_{i+1})$, <code>$K_i\subset int(K_{i+1})$</code>, and $\cup K_i=M$. The we see that $X_t = M\overset{F_t}{\to} M\overset{F_t}{\to}M\overset{F_t}{\to}\cdots$ is diffeomorphic to $Y_t= K_2 \overset{F_t}{\to} K_4 \overset{F_t}{\to} K_6 \overset{F_t}{\to} \cdots$ for all $t$ (where we really meant $F_{t| K_{2i}}$ in the maps of this direct limit and topologized by the inclusion of interiors). To see this, note that $X_t$ is a quotient of $M\times \mathbb{N}$, which is exhausted by <code>$\{(K_k,i), k,i\in\mathbb{N}\}$</code>. Since $Y_t\subset X_t$, we need only show that each point of $X_t$ is in $Y_t$. Suppose $x\in X_t$, then $x$ is the image of some $x'\in (K_k,i)$ for some $k,i$. If $k\leq 2i$, then $x'\in (K_{2i},i)$, so we see that $x\in Y_t$. Otherwise, $k>2i$ and we see that <code>$(K_k,i) \overset{F_t}{\subset}(K_{k+1},i+1) \overset{F_t}{\subset} \cdots \overset{F_t}{\subset} (K_{2k-2i},k-i)$</code> after taking the quotient, so $x\in Y_t$. </p> <p>By the <a href="http://books.google.com/books?id=iSvnvOodWl8C&amp;lpg=PP1&amp;dq=hirsch%2520differential%2520topology&amp;pg=PA180#v=onepage&amp;q&amp;f=false" rel="nofollow">Isotopy Extension Theorem</a>, one may see that $Y_0 \cong Y_1$. This gives the desired diffeomorphism $M\cong M\overset{F}{\to} M\overset{F}{\to} M\overset{F}{\to} \cdots $. </p> <p>We actually use the isotopy extension theorem by induction. We want to find a sequence of diffeomorphisms $G^i: K_{2i}\to K_{2i}$ so that <code>$G^{i+1}_{|K_{2i}} = F\circ G^i$</code>. This immediately gives the diffeomorphism </p> <p><code>$$\begin{matrix} Y_1= K_2 &amp; \overset{F}{\to} &amp; K_4 &amp; \overset{F}{\to} &amp; K_6 &amp; \overset{F}{\to} &amp; \cdots\\ G^1 \uparrow &amp; &amp;G^2\uparrow &amp; &amp;G^3\uparrow &amp; &amp; \\ Y_0= K_2 &amp; \hookrightarrow &amp; K_4 &amp; \hookrightarrow &amp; K_6 &amp; \hookrightarrow &amp; \cdots \end{matrix}$$</code></p> <p>We will actually construct a diffeotopy $G^i_t:K_{2i}\to K_{2i}, t\in [0,1]$ such that <code>$G^{i+1}_{t|K_{2i}}=F_t\circ G^i_t$</code>, where $G^i_0=Id$, and then set $G^i=G^i_1$ for the desired diffeomorphism. </p> <p>Let $G^1_t=Id$. Suppose we have constructed $G^i_t$ by induction, so that $G^i_0=Id$, <code>$G^i_{t|K_{2i-2}}=F_t \circ G^{i-1}_t$</code>, $t\in [0,1]$. Then $F_t\circ G^i_t: K_{2i}\to int(K_{2i+2})$ is an isotopy from $F_0\circ G^i_0=Id$ to $F_1\circ G^i_1=F\circ G^i_1$. By the isotopy extension theorem, there exists a diffeotopy <code>$G^{i+1}_t: K_{2i+2}\to K_{2i+2}$</code> such that <code>$G^{i+1}_0=Id$</code>, and <code>$G^{i+1}_{t|K_{2i}}=F_t\circ G^i_t$</code>, with compact support in <code>$int(K_{2i+2})$</code>. This completes the proof. </p> <p><strong>Addendum:</strong> I'll add some explanation of direct limits. Consider a sequence of smooth embeddings of open manifolds $X_i \overset{f_i}{\to} X_{i+1}$. By the direct limit $X$ of <code>$X_1\overset{f_1}{\to} X_2 \overset{f_2}{\to} X_3 \cdots$</code>, I mean the quotient of $X_1\sqcup X_2 \sqcup X_3 \sqcup \cdots$ with respect to the equivalence relation generated by $x_i \sim f_i(x_i)$ for all $x_i \in X_i, i\in \mathbb{N}$. Then there is a natural embedding of $X_i \hookrightarrow X$ for all $i$, in such a way that $X= \cup_i X_i$, and one gives $X$ the structure of a smooth manifold by taking the atlas of charts generated by the charts of each $X_i$. </p> <p>The space $X$ is determined by the direct limit of any subsequence <code>$\{ X_{i_j}\}$</code>, with maps $F_j: X_{i_j}\to X_{i_{j+1}}$ defined by $F_j=f_{i_{j+1}-1}\circ \cdots \circ f_{i_j}$. One may therefore verify that two direct limits are diffeomorphic if there are compatible diffeos. between subsequences, which justifies some of the arguments above. </p>