An embedded circle separates the sphere into two connected components; this is the Jordan curve theorem. A strengthening of this fact, called the Jordan-Schoenflies theorem, says that the two components are disks; i.e. every embedded circle in the sphere bounds a disk on both sides.

One dimension higher, Alexander proved that every *smoothly embedded* 2-sphere in the 3-sphere bounds a ball on both sides. However the hypothesis of smoothness cannot be removed; in two three-page papers which appeared successively in the same volume of the Proceedings of the National Academy of Science, Alexander proved his theorem, and gave an example of a topological sphere that does not bound a ball on one side (a modified version bounds a ball on neither side). This counterexample is usually called the *Alexander Horned Sphere*; the `bad’ side is called a *crumpled cube*. For a picture of Alexander’s sphere, see this post (the `bad’ side is the outside in the figure). The horned sphere is wild; it has a Cantor set of bad points where the sphere does not have a collar; it can’t be smooth at these points.

Let’s denote the horned sphere by and the crumpled cube (i.e. the `bad’ complementary region) by . The interior of is a manifold with perfect infinitely generated fundamental group. itself is not a manifold, but it is simply connected; its `boundary’ is the topological 2-sphere . We can *double* to produce ; i.e. we glue two copies of together along their common boundary . It is by no means obvious how to analyze the topology of , but Bing famously proved that is . . . homeomorphic to the 3-sphere! I find this profoundly counterintuitive; on the face of it there seems to be no reason to expect is a manifold at all.

There is an obvious involution on which simply switches the two sides; it follows that there is a involution on the 3-sphere whose fixed point set is a wild 2-sphere. Bing’s proof appeared in the Annals of Mathematics; see here. This is an extremely important paper, historically speaking; it introduces for the first time Bing’s `shrinkability criterion’ for certain quotient maps to be approximable by homeomorphisms, and the ideas it introduces are a key part of the proof of the double suspension theorem and the 4-dimensional (topological) Poincare conjecture (more on this in a later post).

The paper is nine pages long, and the heart of the proof is only a couple of pages, and depends on an ingenious inductive construction. However, in Bing’s paper, this construction is indicated only by a series of four hand-drawn figures which in the first place do not obviously satisfy the property Bing claims for them, and in the second place do not obviously suggest how the sequence is to be continued. I spent several hours staring at Bing’s paper without growing any wiser, and decided it was easier to come up with my own construction than to try to puzzle out what Bing must have actually meant. So in the remainder of this blog post I will try to explain Bing’s idea, what his mysterious sequence of figures is supposed to accomplish, and say a few words about how to make this more precise and transparent.

**1. The crumpled cube**

First we give a precise description of the crumpled cube.

Start with the 3-ball . We will realize the crumpled cube as a subset of obtained by removing a subset defined by an infinite process.

Let denote an open solid cylinder, which we can think of technically as a 1-handle running between the centers of the disks at either end of .

We think of as a product . By the *middle third* of we mean the solid cylinder ; we denote this . Inside we insert two 1-handles . We attach along two disks contained in the bottom disk of , and we attach along two disks contained in the top disk of . These two 1-handles are `linked’ in as follows:

If we replace by then the union looks like this:

Denote the middle third of and by and , and replace each middle third by a pair of linked 1-handles and to obtain

And so on. Thus the crumpled cube is equal to where the index ranges over all finite strings in the alphabet . As the length of an index goes to infinity, the diameter of goes to zero, and these cylinders accumulate on a Cantor set indexed by the set of infinite binary strings. The boundary of is a 2-sphere; this is obtained from the 2-sphere by inductively cutting out disks and gluing back the side of a cylinder and a disk at the other end, together with the limiting Cantor set

**2. The crumpled cube as a quotient**

The next step is to give a description of as a quotient of . Formally this is quite easy. Instead of replacing the middle third of with the 1-handles and so on, simply replace the *entire* solid cylinder .

In other words, we let be a pair of 1-handles attached along the boundary disks of . Note that this conflicts with our notation from the previous section. Now define and let be the intersection of this infinite family of nested solid cylinders:

The limit is a Cantor set worth of tame arcs embedded in , each running from a point on the boundary of to the corresponding point of .

By abuse of notation we can think of as a union of arcs where is an infinite binary string, obtained as obtained over all finite binary strings which are a prefix of .

To go from to we simply shrink push the boundary of into itself along the arcs , so that every arc of is pushed down to its endpoint. We start by pushing in from either end a third of the way, then push in each of from either end a third of the way, and so on; the result is evidently and it exhibits where is the equivalence relation which crushes each arc of to a point.

**3. Double the picture**

Now let’s double this picture.

We replace with its double . We think of the 3-sphere as together with a point at infinity, and we think of the dividing 2-sphere as the plane together with infinity. The involution acts in coordinates by taking the coordinate to its negative. The solid cylinder is doubled to a solid torus with core an unknot which we imagine as a round circle in the plane.

The solid cylinders and double to solid tori and with cores and . These are an unlink on two components; together with the core of the complement they form the three components of the Borromean rings.

In general, given any knot there is an operation which thickens the knot to a solid torus, and inserts two new knots in this solid torus, clasped as are clasped in the solid torus ; this operation is known as *Bing doubling*. So we can say that are obtained by Bing doubling . Inductively, we obtain by thickening which are obtained by Bing doubling , and similarly for . Bing doubling in the obvious way produces a family of nested solid tori obtained by doubling the solid cylinder , which nest down to a Cantor set of tame arcs obtained by doubling . We obtain the double of the crumpled cube as a quotient where crushes each arc of to a point.

In order not to make the pictures too complicated, we draw the shadow of each solid torus in the plane (in a rather schematic fashion). The three figures below show, successively, the torus , then inside that the shadow of , then inside that, the shadow of .

If we proceed in this way, each core has length approximately equal to , and consists roughly of two `arcs’, each of which goes half way around the core of .

**4. The magic isotopy**

How do we show that is homeomorphic to the 3-sphere? Bing’s idea is the following one. The arrangement of the thickened links is such that the diameter of each component in the 3-sphere is pretty big, and we must perform a quotient in the limit (which collapses the components of to points) to get . Suppose we could find a sequence of isotopies of the 3-sphere and a sequence of numbers with the following properties:

- each is supported in
- if we define then each component of with has diameter

If we could find such , then the sequence of homeomorphisms would converge to a map taking to a Cantor set in such a way that is a homeomorphism. In particular, it would descend (after taking quotients) to a homeomorphism from to .

Each isotopy, roughly speaking, `slides’ the components of with around inside the with ; if this is done judiciously, the components can be individually moved so that their diameters are smaller than in the original configuration, and in the limit, the diameters go to zero.

As an example, we indicate how to slide inside so it only goes `a quarter’ of the way around the core of :

**5. Some notation**

Let’s restrict the rules of the game. We use the notation to denote the union of all with ; i.e. the union of solid tori at `depth’ . We idealize each component of as a slightly thickened circle; by abuse of notation we use the same notation to refer to the component and its core circle, assuming it is clear from context which is meant at any given time. Each of the two components of inside is idealized as a circle that starts at some point of , goes exactly half way around it, then turns around, and retraces its path to the start where it closes up. The other component starts at the same point of , but heads out in the opposite direction. Because itself is zigzagging back and forth inside its own thickened tubes, the actual image of each circle of for large jitters like crazy, and though all curves have the same length, it is conceivable that their diameters can eventually get small.

We need a bit of notation to get started. can be thought of as a single solid unknot in in which all the successive are nested. Let’s agree that we only really need to give the angular coordinate of the core of each component of projected onto the core of (i.e. we only really care how much it `winds around the original circle’). As measured in terms of this angular coordinate, each component of each has the same length, which we normalize to . I will describe each projection by a cyclic word in the alphabet , as follows: if has length then each letter describes a segment of length which winds positively or negatively around the core of according to whether the letter is or . Thus (trivially) is given by the string , since it just winds positively once around itself.

This notation is ambiguous; it defines the image under radial projection relatively but not absolutely; it is well-defined up to the choice of a starting point. But this notation does let us compute the total angular length of the projection to the core of , which will be a good proxy for diameter. So, for example, a component associated to the word has projection angular length .

Now, suppose we have a component of some , encoded by a cyclic word , and suppose are the components of inside . We think of the letters of as segments of the loop . To build the cores of the we break into two segments each of half the length; write . Then has the same projection as and similarly for where the asterisk means the same segment with opposite orientation. We restrict ourselves to two possibilities:

- the endpoint of is at the endpoint of some segment corresponding to a letter of ; or
- the endpoint of is in the middle of some segment corresponding to a letter of .

In the first case we have a decomposition of parallel to as where are words of length . If is a word in the alphabet let be the word obtained by interchanging with *and* reversing the order of letters. Thus for example . With this notation, the cyclic words associated to the are and . We call the operation of replacing with the pair a *split*.

In the second case we must first *subdivide*; this means replacing by a new string by the substitution ; i.e. each letter is doubled successively. Note that by our convention and define the same radial projection (up to translation). Then as above we decompose and form and .

**6. An inductive lemma**

OK, we are nearly done. The initial torus corresponds to the string consisting of a single letter . We subdivide to form and then decompose to form each with angular projection of length . We subdivide again to form and decompose to form each with angular projection of length . So far so good. But now after subdivision we have cyclic conjugates of and no matter how we split this into we will get words with some or string.

The `best’ strings are those of the form with total angular projection . Say a string is cubeless if it has no or . If is cubeless, so are the strings obtained by any split.

In a cubeless string, the only `bad’ subwords are (disjoint) substrings of the form or ; we call these *runs*. Our goal is to produce strings with as few runs as follows. The only strings with no runs at all are ; we call these *tight*.

We imagine a binary rooted tree of cyclic strings, whose node is , and such that the two children of each are obtained either from a split of or a split of . We will never double a string before splitting unless it is tight; every other string will be successively split (without doubling) until all its descendants are tight.

It is clear that Bing’s claim is proved if we can show that there is an infinite tree of this form which is a union of finite trees so that every leaf of is the cyclic string .

To prove the existence of such a tree inductively, we start at a vertex with the label and generate the part of that lies below it. That this can be done follows immediately from a lemma:

**Lemma:** Let be a cubeless string of even length. Then either for some , or there is a split so that each of the terms in the split have fewer runs than .

*Proof:*Just choose any subdivision into strings of half the length so that each of the has fewer than half of the runs of (i.e. at least one run of must be split in half by the subdivision) That this can be done follows e.g. from the intermediate value theorem. QED

You should post more.

here must most as say prof dr mircea orasanu