# 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. ðŸ™‚

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.

# Introduction to Wild Arcs: the interesting stuff that can happen

By “arc” I mean a topological embedding of the unit arc $[0,1]$ into $R^3$ or $S^3$.
If the embedding is chosen to be smooth or piecewise linear, there isn’t much to do since all such embeddings are ambient isotopic to a standard, boring linear line segment.

But if we even allow for as much as a single wild point, then weird things can happen.

It is probably best to procede with some examples.

1. Not all “wild looking” arcs are wild.

The arc in the upper left hand corner IS wild and the arc in the lower right hand corner (which is supposed to represent an infinite series of trefoil knots, each tied smaller and smaller and converging to a single point) is NOT wild.

I’ll sketch out a proof as to why it is tame: imagine a series of round 3-balls surrounding the “potentially bad” endpoint: arrange these balls so that they are concentric about the endpoint and there is one trefoil knot between the boundaries of each of these balls (as shown)

Denote these balls by $B_1, B_2, B_3.....$

Now for each trefoil between each $B_k, B_{k+1}$ there is an ambient isotopy which fixes ALL points outside of $B_k$ and inside $B_{K+1}$ and removes the trefoil knot; this is called the “lamp chord trick”:

The ONLY points that are moved are those between the boundaries of $B_k$ and $B_{k+1}$.

Denote this orientation preserving homeomorphism by $h_k$.

We can now define a homemorphism of 3-space by $h(x) = h_k(x)$ if $x \in B_k - int(B_{k+1})$ and $h(x) = x$ otherwise.
This takes this arc to a straight arc.

Note: if we took this “almost wild” arc and added even a slight arc “past” the endpoint, then the above procedure would fail. A wild arc that is the union of two tame arc is called mildly wild.

2. An arc that has only one wild point has a simply connected complement, even though the arc is wild.
An example of this is the arc in the upper left and picture. It turns out that if one encloses the “potentially bad” endpoint of the arc by a smooth 2-sphere, that sphere will intersect that arc in at least 3 points. That seems “obvious” but requires proof. And the proof, while not requiring sophisticated techniques (e. g. esoteric algebraic or differential topology), isn’t easy. I know from experience. ðŸ™‚

The idea of a sphere surrounding a wild point having to hit the arc in a set number of points is called the “penetration index” of the arc. More on that for a subsequent post; a good reference is: 24), : W. R. Alford and B. J. Ball, Some almost polyhedral wild arcs, Duke Math. J. 30 (1963), 33-38. MR 26 # 1858.

Note: some wild arcs have a wild point with an infinite penetration index.

On the other hand, any closed loop in the complement of the arc bounds a disk that misses the arc. Here is a sketch as to why:

Let $\alpha$ be a closed loop (say, a smooth or piecewise linear one) that misses the arc. Two compact sets in a metric space that miss must remain at least some set distance $\epsilon$ from each other. So enclose the endpoint of the arc by a ball of radius $\frac{\epsilon}{2}$ and it is easy to see that $\alpha$ bounds a disk that misses that ball. But that disk might hit the arc in a finite number of points. But that is no problem because those points are tame points of the arc; they are outside of the ball that contains the only wild point.

That part of the arc can be enclosed by a tube (called a tubular neighborhood of the arc) that runs to the non-wild endpoint of the arc. That neighborhood can be used to construct “caps” that run around the tame endpoint of the arc; in that way the original disk that $\alpha$ bounded can be replaced by a new disk which has “feelers” which push past the tame endpoint of the arc.

Now this proof works for an arc whose wild point is at an endpoint; the same idea works if the solitary wild point is not an end point.

3. There are arcs whose complement is NOT simply connected. Here is an example of one:

Note: the fundamental group of the arc complement maps non-trivially onto a subgroup of the alternating group $A_5$ (the same one you studied in your abstract algebra class).

A good reference for this:

Fox, Ralph H.; Artin, Emil (1948), “Some wild cells and spheres in three-dimensional space”, Annals of Mathematics. Second Series 49: 979â€“990, ISSN 0003-486X, JSTOR 1969408, MR 0027512

Note: this arc has two wild points.

4. It is possible for two disjoint wild arcs to “link”: that is, be situated in a way so that the arcs are disjoint but also so that there is no sphere which encloses one arc but not the other.

Example: image two copies of the Fox-Artin arc shown above, but in the middle “stitch”, link the two middle strands.

Now if there were a 2-sphere $S$ separating the two arcs, it would have to miss each arc (and therefore the 4 wild points) by some distance $2\epsilon$. So surround each of the 4 wild points with a smooth sphere of radius, say, $\epsilon$. Then one can obtain two disjoint graphs; each of these graphs is equal to the respective wild arc outside of the ball bounded by the sphere, and inside the balls we can just connect the non-wild parts of the arcs (the parts outside of the balls) by smooth arcs inside of the balls; hence in a sense, we have approximated the two disjoint wild arcs by two disjoint smooth graphs which are equal to the wild arcs outside of a small neighborhood of the wild points.

It is easy to see (via standard algebraic topology or by classical knot theory techniques) that these two smooth graphs are NOT splittable. Hence the two wild arcs weren’t splittable either.

Note: we have just skimmed the surface of wild arc theory; there is a lot out there.

A good place to start would be the Fox-Artin paper mentioned above along with the two books:

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. The edition I have is out of print; I linked to the newer one.

Note: arcs can be very pathologically embedded in higher dimensions. That doesn’t make sense to me, but check out this paper by D. Wright: (link is to an open access PDF copy)

Abstract. Let A and B be arcs in E^3, Euclidean 3-space. Then A can be
“slipped” off B; i.e., there exists a homeomorphism of ^E3 onto itself,
arbitrarily close to the identity, such that h(A) n B = 0. The purpose of
this note is to show that arcs in E^n (n > 4) do not always enjoy this
property. The examples depend heavily on a recent result of McMillan.

1. Introduction. If X and Y are subsets of E^n, Euclidean Â«-space, we say
that X can be slipped off Y in E ” if for each e > 0 there is an (epsilon)-homeomorphism h: E” -Â» E” such that h{X) n Y = 0; otherwise, we say X
cannot be slipped off Y. Results of Armentrout [1] and McMillan [5] show that
if A and B are arcs in E^3, then A can be slipped off B. We show that this is
false in higher dimensions by proving the following
Theorem. There exist cellular arcs A and B in E^n (n > 4) such that A
cannot be slipped off B.

This strikes me as very non-intuitive; you’d think that higher dimensional space would provide “more room” for maneuver but it also provides more room to introduce pathologies as well.

To see the beast of an arc that Wright is talking about, here is D. McMillan’s article (open access PDF) about an arc that is so pathological, it doesn’t have a neighborhood inside a manifold that embedds in any $S^n$. Of course, this neighborhood cannot be a manifold because of the Whitney embedding theorem.