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We can embed $S^2\times I$ into $\mathbb{R}^3$ by taking a compact 3-ball and removing an open 3-ball from its interior. Taking the boundary gives an embedding $i: S^2\sqcup S^2\hookrightarrow\mathbb{R}^3$, as "a sphere contained inside another sphere". Now it's intuitively clear that this embedding is not ambient-isotopic to the embedding $j$ given by putting these two spheres "side-by-side" in $\mathbb{R}^3$. That is, there is no isotopy $F_t:\mathbb{R}^3\to \mathbb{R}^3$ with $F_0=1_{\mathbb{R}^3}$ and $F_1\circ i=j$. At least, this looks visually obvious. Assuming my intuition isn't betraying me and this isn't false, what's an elegant way to prove this?

Also, (why) does there (not) exist an embedding $S^2\times I\hookrightarrow\mathbb{R}^3$ whose boundary is ambient-isotopic to the "side-by-side" embedding? How about for $S^{n-1}\times I\hookrightarrow\mathbb{R}^n$?

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(Edited math markup to stay in the margins. This also draws more attention to your later questions, which seem trickier ;) –  Andrew Critch Dec 10 '09 at 8:01

2 Answers 2

up vote 7 down vote accepted

For your first question, look at complements (a "fundamental" technique in analyzing ambient isotopies/ambient homeomorphisms; e.g. the "knot group"):

Let the union of the initial two spheres be $S$, and the union of the final two spheres be $T$. An isotopy on $\mathbb{R}^3$ taking $S$ to $T$ in particular ends with a homeomorphism from $\mathbb{R}^3$ to itself taking $S$ to $T$, hence producing a homeomorphism from $U=\mathbb{R}^3\setminus S$ to $V=\mathbb{R}^3\setminus T$. But $U$ has only one of its three connected components contractible (the inside of the inner sphere), whereas $V$ has two of its three components contractible (the insides of the two spheres), a contradicting that they be homeomorphic.

$\Big($One way to prove the other components are not contractible is that they have non-trivial second homotopy and homology groups, as exhibited by the elements represented by the spheres themselves.$\Big)$

For your third (and second) question, in generality, here's an argument that there is no embedding from $M=S^{n-1}\times I$ to $\mathbb{R}^n$ with boundary ambient isotopic to the "side-by-side" embedding of
$S_0\sqcup S_1$. By any letter $S$ I denote a copy of $S^{n-1}$. For comfort of the imagination, think $n=2$:

(1) Up to ambient isotopy, there are only three embeddings of $S_0\sqcup S_1$ in $\mathbb{R}^n$ : the "avocado$^+$ type" embeddings $A^+$ with $S_0$ inside $S_1$, the "avocado$^-$ type" embeddings $A^-$ with $S_1$ inside $S_0$, and the "side-by-side type" embeddings $B$ with $S_0$ and $S_1$ "next to" each other.

$\Big($That they fall into these three types is just a case analysis; that any two embeddings in the same "type" are ambient isotopic is also easy: (a) first translate and dilate until the $S_1$ embeddings match up, and then being in the same case means the two $S_2's$ are embedding in the same component of the compliment of $S_1$, so (b) you can move them around in that component until they match up, too.$\Big)$

(2) $A^\pm$ are not ambient isotopic to $B$, by counting contractible components of their complements, as in the answer to your first question: $A^\pm$ has one, $B$ has two.

(3) An embedding from $M=S\times I$ to $R^n$ cannot have $\partial M = B$ up to ambient isotopy.

$\Big($ Why? Here's one proof, which I found kind of fun: Suppose $\partial M = B$, i.e. is in the "side-by-side" arrangement. Since $M$ is connected, $int(M)=M\setminus \partial M$ must lie outside both spheres of $\partial M$, since if it lies inside one, it can't touch the other. Let $x$ be the center of of the $S_1$ component of $\partial M$. Now $M$ is compact, so it has a point $p$ which is of maximal distance from $x$. Such a $p$ must be a boundary point of
$M$, since at an interior point, we could move slightly in a direction away from $x$ and stay inside $M$. But $p$ can't be on $S_1$ or $S_2$ either, since their outer sides are "padded" by the open set $int(M)$... this a Euclidean geometry argument which I'll stop rigorizing here. So $p$ is nowhere, a contradiction.$\Big)$

(4) By process of elimination, $\partial M$ must be ambient isotopic to $A^\pm$, which by (2) is not ambient isotopic to the "side-by-side" embedding $B$, which is what you wanted.

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slaps forehead Thanks, can't believe I didn't notice that. I'm still curious about the (non-)existence of an embedding with that described property. –  Brad Hannigan-Daley Dec 10 '09 at 7:38
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I've heard that a proof is a convincing argument, so compliments are obviously good technique. –  Sam Nead Dec 10 '09 at 8:44

Edit: This is meant to answer the question of why we can't have an embedding $\mathbb S^{n-1}\times I\hookrightarrow\mathbb R^n$ such that the boundary is two side-by-side spheres rather than two nested spheres. I said "the second question" but it changed.

It seems to me that if you have an embedding of $S^{n-1}\times I$ into $\mathbb R^n$, this is the same as an ambient isotopy of one copy of $S^{n-1}$ to another. (I found this easier to visualize when thinking of the fact that you can't embed a cylinder in $\mathbb R^2$ except as an annulus of concentric circles.)

In particular, you're trying to get one sphere to bound the same disk as the other sphere does, by a homotopy whose image at any time never intersects the image at another time. You've already noted that two concentric spheres bound the same disk: so you're trying to get the first sphere $A$ to surround the other sphere $B$. Since $B$ cuts off a component of $\mathbb R^n$, we can remove that component and know the isotopy will not pass through it without first passing through $B$, which is not allowed. Therefore remove a ball $C$ or a point from the interior of $B$. $A$ is contractible in $\mathbb R^n-C$, and $B$ is not: $B$ generates the $n-1$st homology group of the resulting manifold. This also is a nice way to see why it will work in $\mathbb R^{n+1}$, though of course in that case you really do have a "cylinder" (take $I$ to run in the $n+1$st dimension with each $S^{n-1}$ in an $n$-hyperplane).

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