# Warm up to the Whitehead contractable 3-manifolds…

Eventually I want to talk about proper knot theory: this is the theory of proper embeddings of the real line into open three manifolds.
Note: a proper embedding is one in which the infinities of the real line go to ends of the manifold; the inverse image of the embedded real line intersect a compact set in the manifold has to have a compact pre image. Example of a non-proper embedding of the real line:

Let $f(x) = (arctan(x), 0, 0) \in R^3$; if $M$ is any compact set containing $(\frac{pi}{2}, 0, 0)$ in its interior then $f^{-1}(M)$ is unbounded and therefore not compact.

So it would be helpful to have some nice open manifolds to work with and one of the nicest examples of such gadgets are the Whitehead Manifolds.

This diagram represents three stages of the construction.

Here is the first stage: (imagine thickening the knot inside the torus)

If one keeps iterating these embeddings to get nested solid tori $W_1, W_2, W_3....$ where $W_i$ is situated in $W_{i-1}$ as shown, and then one gets $X = \cap^{\infty}_{i=1}W_i$ which is called the Whitehead continuum. Then $S^3 - X$ is the required manifold.

One can also view this as an increasing union as well: in the “simple” diagram think of the inside torus as stage 1 and the outer torus as stage 2; that is, the inner torus is $W_1$ and the outer is $W_2$. Now there is a homeomorophism $f:S^3 \rightarrow S^3$ where $f(W_1) = W_2$. Then define $W_k = f^k(W_1) = f \circ f \circ.......f \circ f (W_1)$ (composed $k$ times). Then the Whitehead manifold is $W = \cup^{\infty}_{i=1}W_i$.

It is well known that $W$ is open, contractable, not homeomorphic to $R^3$ and that $R^3 \times R$ is homeomorphic to $R^4$; there is also a new result by Gabai that says that $W$ is the union of two copies of $R^3$. The eventual goal of this note is to show that $R^3 \times R$ is homeomorphic to $R^4$. To do that, I’ll do a bit of warm up and then work through this paper by Glimm, which can be freely accessed here.

Warm up: why does a smooth knot unknot in 4-space?
When we work through the proof that $W \times R$ is homeomorphic to $R^4$ we will be using the “4’th dimension” to lift the inner torus so as to unwrap it and then put it back in a trivial fashion (the inner torus would lie in a ball of the outer torus) and a product of these things is easy to visualize. So we need some 4-dimensional intuition to get the pictures clear in our mind.

(to you experts: yes, I know that $X$ can be thought of a cellular in $S^4$ and therefore $S^4 - X$ is homeomorphic to $R^4$, but I am trying to get an intuitive visualization here)

So, let’s see how a knot unknots in $R^4$. Sure, there is an easy way: let $K \subset R^3 \subset R^4$ where $R^3$ is identified with the $(x, y, z, 0)$ hyperplane in $R^4$. Now pick a point $z = (0, 0, 0, 1)$ and let $D$ be the cone of $K$ from $z$. $D$ is a piecewise linear disk but is NOT locally flat (and therefore isn’t a famous “slice disk”) because the link of the cone point is a copy of the knot $K$, but this is enough to show that $K$ unknots. But that is not very satisfying; let’s construct a nice locally flat disk instead.

Start with a knot diagram; the knot $K$ consists of a finite number of arcs in the $(x, y, 0, 0)$ plane with a finite number of “overpasses” which leaves the $(x, y, 0, 0)$ plane but stays in the $(x, y, z, 0)$ hyperplane.

Here is a view from the top with the z-axis coming at you and the w-axis just somewhere else. 🙂

Now any p. l. (or smooth) knot can be changed into the unknot by changing a finite number of crossings (the above example: we can get the unknot by changing any of the crossings). The minimum number that have to be changed is called the “unknotting number”; in general this is hard to calculate. But that doesn’t matter for this exercise. That that means for us is that there is a homotopy $F: K \times [0,1] \rightarrow R^3$ for which $F(K, 0) = K$,

$F(K, 1)$ is the unknot and if $A$ is the collection of arcs of $K$ on the plane, $F((x,y,z), t) = (x,y,z)$ for all $(x, y, z) \in A$. That is, this homotopy keeps the plane arcs fixed and moves the necessary overpasses into underpasses so as to effect the necessary crossing changes. Note: of course, the knot has to pass through itself in $R^3$; we do get double points at a certain stage. In fact, we might say this homotopy affects the $z$ coordinate; think of a homotopy that drives the “unknotting overpasses” straight down and fixes the rest of the knot. So let’s describe the homotopy (in $R^3$ ) in two parts: $F_1((x, y, z(t)),t), t \in [0,1]$ and then $F_2 ((x, y, z(t)), t)=F_1((x, y, -z(1-t)),1-t)$. Now the ambient isotopy in $R^4$ can be described by $G_1((x(t), y(t), z(t), w(t), t) = (x, y, 0, 0) = G_2(x(t), y(t), z(t), w(t), t)$ for points on $A$ (the set of arcs in the plane) and
$G_1((x, y, z, w, t) = (x, y, z(t), t, t)$ for $t \in [0,1]$ and $G_2(x, y, z, w, t) = (x, y, -z(1-t), 1-t, t), t \in [0,1]$.

So the upshot: this isotopy swings the offending overpasses into the next dimension and then returns them underneath the plane; in the final $R^3$ slice, the knot has been taken to the unknot.

So, what does this have to do with the Whitehead manifold? Look at the simple picture of the first stage: one does exactly the same isotopy described in $W_2 \times [-2, 2]$ to change the single crossing into an under crossing; this puts $W_1$ in $W_2$ in a trivial manner; hence we have a homeomorphism from $W_2 \times [-2, 2]$ to $R_2 \times [-2, 2]$ where the latter is a trivial torus pair (one torus embedded in the larger torus in a geometrically trivial way (the inner torus lies in a ball contained in the larger torus).

So we obtain a homeomorphism between the pair $(W_2, W_1) \times [-2,2] \rightarrow (T_2, T_1) \times [-2,2]$ which is the identity on the boundary $W_1 \times [-2,2]$ and the latter is just a pair of solid tori where $T_2$ lies in a ball in $T_1$. The reference linked to above (Glimm) shows that such homeomorphisms can be pieced together to yield a homeomorphism $W \times R \rightarrow \cup^{\infty}_{i=1}T_i \times R$ and the latter is known to be homeomorphic to $R^4$.

Basically, the $R$ factor gives one enough “room” to unlink the Whitehead type link in the torus.

This best is simply connected; in fact it is contractable. But it has a very weird property: if one, say, constructs a smooth simple closed curve by, say, running along the center line of one of the defining tori, the complement of this curve (the knot group, if you well) is NOT finitely generated! So this beast is NOT $R^3$.

# Wild Arcs with simply connected complements: two examples.

As we saw in this previous post: this arc contains just one wild point; hence its complement is simply connected. But we can say more than that: this arc $K$ is “cellular”: that means that there is a collection of smooth 3-balls $B_1 \supset B_2 \supset B_3....$ where $\cap^{\infty}_{i=1}B_i = K$. That isn’t too hard so see: start by taking a small round 3-ball that encloses the wild point and taking a union of that ball with a tubular neighborhood of the rest of the arc outside of the ball. That first gadget is “almost” a ball; it might have a few handles which can be filled in. Then take a smaller ball around the wild point and repeat the process with a smaller tubular neighborhood…and repeat.

In fact, we can do this with ANY arc that has just one wild point.

Therefore $S^3 - K = S^3 - \cap^{\infty}_{i=1}B_i = \cup^{\infty}_{i=1}(S^3 - B_i)$ which is homeomorphic to $R^3$; hence this wild arc complement is homeomorphic to the complement of a tame arc.
Aside: this shows that arcs are NOT determined by their complements whereas smooth (or p. l. ) knots are.

But this arc is still wild; it has penetration index 3 near the wild point.

Now consider this example

This is like the wild Fox-Artin arc talked about in the previous post in this series, except that “one stitch has been missed. It’s complement is simply connected; the proof of this fact is suggested by this diagram:

The small loop represents a generator of the fundamental group; note how it bounds a disk in the complement of the arc.

However the complement is NOT $R^3$! There are different ways to see that; here is a fun one:

This shows $K$ with a smooth simple closed curve $J$ in its complement. Note: $J$ is compact (it is a smooth knot). Therefore there are two different ways to use $J$ to show that $S^3 - K$ is NOT homeomorphic to $R^3$:

1. In $R^3$, every compact set lies inside of some ball $B$. But one can show that there is no ball $B$ that contains $J$ and misses $K$
2. One can use algebraic topology: the fundamental group of $(S^3 -K)-J$ is NOT finitely generated; in $R^3$, the complement of any smooth knot is finitely generated.

This gives an example of an open, simply connected manifold that is not homemorphic to $R^3$.

The interested reader can consult the following references for more detail:

R. Daverman and G. Venema: Embeddings in Manifolds, American Mathematics Society Graduate Studies in Mathemtics, Vol. 106, 2009: Section 2.8
T. B. Rushing, Topological Embeddings, Academic Press, (Pure and Applied Mathematics, Volume 52) 1973, Section 2.4.