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I'm reading Hatcher's proof of Alexander's theorem in his 3 manifolds notes. The statement is the following:

Let $\Sigma \subset \mathbb{R}^3$ be an embedded $2$-sphere; then $\Sigma$ bounds a $3$-ball.

To me, this means the following: given a smooth embedding $\phi:S^2 \to \mathbb{R}^3$ with $\Sigma = \phi(S^2)$, we have that $\mathbb{R}^3 \setminus \Sigma$ has two connected components; one is unbounded, and the other is a bounded region $B$ such that $\partial B = \Sigma$ and there's a diffeomorphism $\psi: B^3 \to B$, where $B^3$ is the standard open unit ball in $\mathbb{R}^3$.

The proof proceeds as follows. One first considers the height function $h:\mathbb{R}^3 \to \mathbb{R}$ given by projection onto the third coordinate. This is a smooth function. Given $\epsilon > 0$, I believe we should be able to find a smooth function $\tilde{h}:\mathbb{R}^3 \to \mathbb{R}$ so that $\tilde{h}|_{\Sigma}$ is a Morse function all of whose critical points assume distinct critical values and a smooth homotopy $H:\mathbb{R}^3 \times I \to \mathbb{R} $ from $h$ to $\tilde{h}$ such that $||H(*, t) - H(*, s)||_{C^2} < \epsilon$ for all $t, s \in I$.

Question 1: I believe Hatcher then deduces that if $\epsilon$ is chosen small enough, the map $F:\mathbb{R}^3 \times I \to \mathbb{R}^3$ given by $F(x,y,z, t) = (x,y, H(x,y,z,t))$ restricts on $\mathbb{S}^2 \times I$ to an isotopy from $\Sigma$ to an embedded $2$-sphere $\tilde{\Sigma}$ such that the height function on $\tilde{\Sigma}$ is $\tilde{h}$ is a Morse function with all of the above properties. I cannot see why this is true, nor am I actually sure this is what is going on.

And so it's enough to show that the theorem holds for $\tilde{\Sigma}$ since, if we can embed a $3$-ball into this, we can just reverse it under the isotopy to get it into $\Sigma$.

So from now on we assume that the height function $h:\Sigma \to \mathbb{R}$ is a Morse function so that all critical points assume distinct values.

Question 2: Consider the following VERY simple example: suppose that $h$ has exactly 2 critical points, say $p_0$ and $p_1$, which are necessarily a global minimum and global maximum respectively. Then, I should be able to write $\Sigma$ as basically a bottom cap which locally admits coordinates $(u_1, u_2)$ satisfying $h(u_1, u_2) = h(p_0) + u_1^2 + u_2^2$, a top cap which locally admits coordinates $(v_1, v_2)$ satisfying $h(v_1, v_2) = h(p_1) - v_1^2 - v_2^2$, and then a thin strip in the middle where the two caps meet which can be made arbitrarily thin.

I feel like it should be obvious in this case to explicitly write down a map $\psi:B^3 \to \mathbb{R}^3$ as above using these coordinate parametrizations, but I honestly cannot come up with anything. I can certainly draw a picture of this and it looks like the surface clearly bounds a $3$-ball. I don't have any doubt about it. But, push comes to shove, I cannot leverage these coordinate representations to actually write the map down.

Question 3: At this point, I feel like what I should do is induct on the number of saddle points $h$ has. In the presence of a single saddle point, I feel like I should again pick local representations around each of the 4 critical points (either 2 mins, a max, and a saddle or 2 maxs, a min, and a saddle) and fill up each area locally with part of a ball; the regions where these pieces will connect can be taken to be disks (by the lower dimensional version of Alexander's theorem) and so piecing together the balls along disks is easy.

However I am unsure of how to write down explicitly the required diffeomorphism in terms of the local coordinates here.

In general, given $n$ saddle points, I feel like I'd just want to cut along a disk of the surface to separate it into two pieces, each having fewer than $n$ saddles and cap it off with a disk (and possibly smooth things out if necessary, which will not introduce an extra saddle); appealing to the inductive hypothesis, I'd fill in each piece with a ball; since the regions can be put back together along a bounding disk, I fill the whole thing in with a ball.

I apologize for the length of this post, but if someone could tell me if I've got a good idea of what's going on and help me clear up some of these issues, I'd appreciate it a lot.

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First, you're on the right track.

For question 1, without the notes handy, I can't tell you anything, but what you wrote sure seems plausible.

Q 2: I'm assuming that you can show that near the min, you have something that's topologically a half-disk: send $(x, y, z)$ in the antarctic region of $S^2$ to the point $(x', y', z')$ of $3-space$, where $z' = h(p_0) + z$, and $(x', y', z')$ is a point with $uv$-coordinates $(x, y)$.

The same idea works for the arctic. And in between, the slices of your surface are all circles, which each bound a disk in the slice-plane, so you have a nice map from $D^2 \times I$ to this space. Now there's also a nice map from $D^2 \times I$ to the temperate zone of $S^2$, and you're done.

Q3: This does sound like what you should do, but Alexander's argument is quite a lot more subtle. It's worth reading the original article in the Proc. of the Nat'l Acad of Sciences (I think!). He argues that in any non-critical slice, you have a collection of circles, and that there's an "innermost circle" which must bound a disk in the slice plane that meets no other slice curve. You cut the surface along the innermost circle and glue in two copies of the disk -- one slightly above and one slightly below the plane. Now you have TWO surfaces, each with simpler slices than before. They must (by induction) both bound balls, and their union (along the two disks) therefore bounds a ball.

I know this is kind of informal, but Alexander's argument is even more so.

What's neat is that Alexander also showed that an embedded torus bounds a $D^2 \times S^1$ on at least one side. But of course a 2-holed torus need not do so, and it's a good exercise to see where his argument breaks down in that case.

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  • $\begingroup$ Forgive me for being dense, but I really don't understand what you've written for question 2. I need to produce a map from the 3 ball to the region bounded by my topological disk. It looks like you're taking a point $(x,y,z)$ in the lower part of the 3 ball and mapping it onto a point $(x',y',z')$ in the region bounded by the topological disk so that $(x',y'z')$ has $(u,v)$ coordinates $(x,y)$; if this is true, then $(x',y',z')$ lies on the surface of the disk, not inside of it. $\endgroup$
    – anonymous
    Mar 3, 2015 at 15:07
  • $\begingroup$ math.cornell.edu/~hatcher/3M/3M.pdf specifically the proof begins on page 1 of this document $\endgroup$
    – anonymous
    Mar 3, 2015 at 15:09
  • $\begingroup$ I apologize. I was unclear and incoherent and just plain wrong. The idea is to map the antarctic region of a 3-ball (NOT a 2-sphere) to a region near the critical point. At height $k$ above the critical point, your surface slice consists of points with $uv$-coords satisfying $k = u^2 + v^2$; this slice bounds a region that's a topological disk in the slice plane. For each point $P = (x, y, h(p_0)+k)$ in that disk, you trace a ray downwards until you hit the surface, at a point $(x, y, z)$ for some $z$. That point has $uv$ coords, say, $u_0, v_0$. You map $P$ to $(u_0, v_0, z-h(p_0)) \in B^3$. $\endgroup$ Mar 3, 2015 at 23:11
  • $\begingroup$ ... For points on the boundary of the disk, your third coord will be exactly $k$; for points inside the disk, your third coord will be between $0$ and $k$. In short, the image will be an antarctic solid (i.e., all points of the solid ball with $-1 \le z \le k$. $\endgroup$ Mar 3, 2015 at 23:13

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