## Bing’s wild involution

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

There is an obvious involution on $DM$ 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 $B$. We will realize the crumpled cube $M$ as a subset of $B$ obtained by removing a subset defined by an infinite process.

Let $C$ 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 $C$.

We think of $C$ as a product $[0,1] \times D$. By the middle third of $C$ we mean the solid cylinder $[1/3,2/3] \times D$; we denote this $C'$. Inside $C'$ we insert two 1-handles $C_0,C_1$. We attach $C_0$ along two disks contained in the bottom disk $[1/3]\times D$ of $C'$, and we attach $C_1$ along two disks contained in the top disk $[2/3]\times D$ of $C'$. These two 1-handles are linked’ in $C'$ as follows:

If we replace $C'$ by $C_0 \cup C_1$ then the union $(C-C') \cup C_0 \cup C_1$ looks like this:

Denote the middle third of $C_0$ and $C_1$ by $C_0'$ and $C_1'$, and replace each middle third by a pair of linked 1-handles $C_{00} \cup C_{01}$ and $C_{10} \cup C_{11}$ to obtain

And so on. Thus the crumpled cube $M$ is equal to $B - \cup_I (C_I - C_I')$ where the index $I$ ranges over all finite strings in the alphabet $\lbrace 0,1\rbrace$. As the length of an index $I$ goes to infinity, the diameter of $C_I-C_I'$ goes to zero, and these cylinders accumulate on a Cantor set $\mathcal{C}$ indexed by the set of infinite binary strings. The boundary of $M$ is a 2-sphere; this is obtained from the 2-sphere $\partial B$ 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 $\mathcal{C}$

2. The crumpled cube as a quotient

The next step is to give a description of $M$ as a quotient of $B$. Formally this is quite easy. Instead of replacing the middle third $C'$ of $C$ with the 1-handles $C_0\cup C_1$ and so on, simply replace the entire solid cylinder $C$.

In other words, we let $C_0\cup C_1$ be a pair of 1-handles attached along the boundary disks of $C$. Note that this conflicts with our notation from the previous section. Now define $K_n = \cap_{j=0}^n \cup_{|I|=j} C_I$ and let $K:=K_{\infty}$ be the intersection of this infinite family of nested solid cylinders:

The limit $K$ is a Cantor set worth of tame arcs embedded in $B$, each running from a point on the boundary of $B$ to the corresponding point of $\mathcal{C}$.

By abuse of notation we can think of $K$ as a union of arcs $C_I$ where $I$ is an infinite binary string, obtained as $C_I = \cap_{J \subset I} C_J$ obtained over all finite binary strings $J$ which are a prefix of $I$.

To go from $B$ to $M$ we simply shrink push the boundary of $B$ into itself along the arcs $K$, so that every arc of $K$ is pushed down to its endpoint. We start by pushing in $C$ from either end a third of the way, then push in each of $C_0,C_1$ from either end a third of the way, and so on; the result is evidently $M$ and it exhibits $M = B/\sim$ where $\sim$ is the equivalence relation which crushes each arc of $K$ to a point.

3. Double the picture

Now let’s double this picture.

We replace $B$ with its double $S^3$. We think of the 3-sphere as $\mathbb{R}^3$ together with a point at infinity, and we think of the dividing 2-sphere as the $y,z$ plane together with infinity. The involution acts in coordinates by taking the $x$ coordinate to its negative. The solid cylinder $C$ is doubled to a solid torus $T$ with core an unknot $L$ which we imagine as a round circle in the $x,y$ plane.

The solid cylinders $C_0$ and $C_1$ double to solid tori $T_0$ and $T_1$ with cores $L_0$ and $L_1$. These are an unlink on two components; together with the core of the complement $S^3 - T$ 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 $L_0 \cup L_1$ are clasped in the solid torus $T$; this operation is known as Bing doubling. So we can say that $L_0\cup L_1$ are obtained by Bing doubling $L$. Inductively, we obtain $T_{00}\cup T_{01}$ by thickening $L_{00}\cup L_{01}$ which are obtained by Bing doubling $L_0$, and similarly for $L_1$. Bing doubling in the obvious way produces a family of nested solid tori $T_I$ obtained by doubling the solid cylinder $C_I$, which nest down to a Cantor set of tame arcs $DK$ obtained by doubling $K$. We obtain the double of the crumpled cube $DM$ as a quotient $DM = S^3/\sim$ where $\sim$ crushes each arc of $DK$ to a point.

In order not to make the pictures too complicated, we draw the shadow of each solid torus in the $x,y$ plane (in a rather schematic fashion). The three figures below show, successively, the torus $T$, then inside that the shadow of $T_0$, then inside that, the shadow of $T_{01}$.

If we proceed in this way, each core $L_I$ has length approximately equal to $2\pi$, and consists roughly of two arcs’, each of which goes half way around the core of $T$.

4. The magic isotopy

How do we show that $DM$ is homeomorphic to the 3-sphere? Bing’s idea is the following one. The arrangement of the thickened links $T_I$ 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 $DK$ to points) to get $DM$. Suppose we could find a sequence of isotopies $i_n$ of the 3-sphere and a sequence of numbers $\epsilon_n \to 0$ with the following properties:

• each $i_n$ is supported in $\cup_{|I|=n} T_I$
• if we define $j_n:=i_n \circ i_{n-1} \circ \cdots \circ i_1$ then each component of $j_n(T_I)$ with $|I|=n$ has diameter $\le \epsilon_n$

If we could find such $i_n$, then the sequence of homeomorphisms $j_n$ would converge to a map $j:S^3 \to S^3$ taking $DK$ to a Cantor set $X$ in such a way that $j:S^3-L \to S^3 - X$ is a homeomorphism. In particular, it would descend (after taking quotients) to a homeomorphism from $DM$ to $S^3$.

Each isotopy, roughly speaking, slides’ the components of $T_I$ with $|I|=n$ around inside the $T_J$ with $|J|=n-1$;  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 $T_{01}$ inside $T_0$ so it only goes a quarter’ of the way around the core of $T$:

5. Some notation

Let’s restrict the rules of the game. We use the notation $T_n$ to denote the union of all $T_I$ with $|I|=n$; i.e. the union of $2^n$ solid tori at depth’ $n$. We idealize each component $\gamma$ of $T_i$ 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 $\alpha_1,\alpha_2$ of $T_{i+1}$ inside $\gamma$ is idealized as a circle that starts at some point of $\gamma$, 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 $\gamma$, but heads out in the opposite direction. Because $\gamma$ itself is zigzagging back and forth inside its own thickened tubes, the actual image of each circle of $T_n$ for large $n$ 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.  $T_0$ can be thought of as a single solid unknot in $S^3$ in which all the successive $T_n$ are nested. Let’s agree that we only really need to give the angular coordinate of the core of each component of $T_n$ projected onto the core of $T_0$ (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 $T_n$ has the same length, which we normalize to $1$. I will describe each projection by a cyclic word $S$ in the alphabet $\lbrace L,R\rbrace$, as follows: if $S$ has length $m$ then each letter describes a segment of length $1/m$ which winds positively or negatively around the core of $T_0$ according to whether the letter is $L$ or $R$. Thus (trivially) $T_0$ is given by the string $L$, 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 $T_0$, which will be a good proxy for diameter. So, for example, a component associated to the word $LRLLRLRR$ has projection angular length $2/8 = 1/4$.

Now, suppose we have a component $\gamma$ of some $T_n$, encoded by a cyclic word $S$, and suppose $\alpha_1,\alpha_2$ are the components of $T_{n+1}$ inside $\gamma$. We think of the letters of $S$ as segments of the loop $\gamma$. To build the cores of the $\alpha_j$ we break $\gamma$ into two segments each of half the length; write $\gamma=\gamma_1\gamma_2$. Then $\alpha_1$ has the same projection as $\gamma_1 \gamma_1^*$ and similarly for $\alpha_2$ where the asterisk means the same segment with opposite orientation. We restrict ourselves to two possibilities:

• the endpoint of $\gamma_1$ is at the endpoint of some segment corresponding to a letter of $S$; or
• the endpoint of $\gamma_1$ is in the middle of some segment corresponding to a letter of $S$.

In the first case we have a decomposition of $S$ parallel to $\gamma$ as $S = U_1U_2$ where $U_1,U_2$ are words of length $|S|/2$. If $U$ is a word in the alphabet $\lbrace L,R\rbrace$ let $U^*$ be the word obtained by interchanging $L$ with $R$ and reversing the order of letters. Thus for example $(LRLLR)^*=LRRLR$. With this notation, the cyclic words associated to the $\alpha_i$ are $U_1U_1^*$ and $U_2U_2^*$. We call the operation of replacing $S$ with the pair $U_1U_1^*, U_2U_2^*$ a split.

In the second case we must first subdivide; this means replacing $S$ by a new string $DS$ by the substitution $L \to LL, R \to RR$; i.e. each letter is doubled successively. Note that by our convention $S$ and $DS$ define the same radial projection (up to translation). Then as above we decompose $DS = U_1U_2$ and form $U_1U_1^*$ and $U_2U_2^*$.

6. An inductive lemma

OK, we are nearly done. The initial torus $T_0$ corresponds to the string consisting of a single letter $L$. We subdivide to form $LL$ and then decompose to form $T_1 = \lbrace LR,RL\rbrace$ each with angular projection of length $1/2$. We subdivide again to form $LLRR,RRLL$ and decompose to form $T_2 = LRLR,RLRL,RLRL,LRLR$ each with angular projection of length $1/4$. So far so good. But now after subdivision we have cyclic conjugates of $LLRRLLRR$ and no matter how we split this into $U_1U_2$ we will get words with some $LL$ or $RR$ string.

The best’ strings are those of the form $(LR)^{2^k}$ with total angular projection $2^{-k}$. Say a string is cubeless if it has no $LLL$ or $RRR$. If $S$ is cubeless, so are the strings obtained by any split.
In a cubeless string, the only `bad’ subwords are (disjoint) substrings of the form $LL$ or $RR$; 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 $(LR)^n$; we call these tight.

We imagine a binary rooted tree of cyclic strings, whose node is $L$, and such that the two children of each $S$ are obtained either from a split of $S$ or a split of $DS$. We will never double a string $S$ 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 $t_k$ so that every leaf of $t_k$ is the cyclic string $(LR)^{2^k}$.

To prove the existence of such a tree inductively, we start at a vertex with the label $(LR)^{2^k}$ and generate the part of $t_{k+1}$ that lies below it. That this can be done follows immediately from a lemma:

Lemma: Let $S$ be a cubeless string of even length. Then either $S = (LR)^m$ for some $m$, or there is a split so that each of the terms in the split have fewer runs than $S$.

Proof:Just choose any subdivision $S =U_1U_2$ into strings of half the length so that each of the $U_i$ has fewer than half of the runs of $S$ (i.e. at least one run of $S$ must be split in half by the subdivision) That this can be done follows e.g. from the intermediate value theorem. QED

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### 4 Responses to Bing’s wild involution

1. You should post more.

2. dedusuiu says:

here must most as say prof dr mircea orasanu

Geometry and the imagination: SRT spacetime (4D)
Minkowski Light cone and an antique sand watch ( hourglass )
—–
Minkowski explained the spacetime by using the ”Light cone” scheme.
Minkowski light cone
”Light cone in 2D space plus a time dimension.. . , .
A light cone is the path that a flash of light, . . . through spacetime”
(light travel from an enormous  past light cone through a place
of the very tiny present to an enormous future light cone)
/ look the scheme /
https://en.wikipedia.org/wiki/Light_cone
#
Antique  sand watch ( hourglass )
Sand in hourglass flows from the upper vessel (place of a past)
through very tiny hole (place of the short present life) to the lower vessel
( place of the future )    / look the picture /
#
We can turn over the hourglass and the time will flow vice versa.
Similar: . . .  the light in an absolute Minkowski  spacetime can travel
backward in time, according to ”The law of conservation and
transformation of energy-mass”  and  the entropy principle.
#
The Minkowski scheme of Light cone has three systems
of coordinate: past, present, future . . . for light traveling
with constant speed the time is ”frozen” . . . the present
state is the border between past and future . . . light takes
an important place in the present system . . . .
(from photosynthesis . . . to atoms, cells, living creatures  . . .)
To go from past to future Light must change its parameters
in the present system according to ”The law of conservation
and transformation of energy-mass”. The concrete changes
of quantum of light in the present time were described
by the ”Lorentz laws of transformation”
————-
Practically Minkowski ”cone” is a flat, homogeneous, isotropic.
Mathematically Minkowski ”cone” is an abstract construction.
Practically, according to the WMAP (2013 measurement) the
Cosmic Space is ”pretty flat” to within 0,4% – 0,5%
#
Minkowski’s  kamuflage.
The ”time” in Einstein’s SRT was negative.
Minkoski saw that mathematically it is ”ugly” and he
changed negative time into positive time by the beautiful
mathematical construction ”an absolute spacetime-4D.
Minkowski did not create a new theory, he only masked the negative
time problem, he only masked the reference frame for ”spacetime”.
Where can we see the negative time and spacetime in nature?
The unity of space and time we can see in the cold cosmic vacuum.
The structure of the cold cosmic vacuum doesn’t have ”time”
My conclusion:
Einstein’s SRT (1905) has only one absolute reference frame.
This absolute reference frame. is a cold , flat, homogeneous,
isotropic cosmic vacuum.
All other reference frames are relative systems.
——–
Best wishes