## Bing’s wild involution

Everyone knows that Alexander proved that the 3-sphere is irreducible; i.e. that every smoothly embedded 2-sphere bounds a ball on both sides. Everyone also knows that he showed that one cannot remove the hypothesis smooth’ altogether, since at about the same time he discovering his famous “horned” sphere, which does not bound a ball on one side (a modified version bounds a ball on neither side). 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 bad’ complementary region by $M$. Not everyone knows that Bing famously proved that the double $DM$ of $M$ along $S$ is . . . homeomorphic to the 3-sphere! That’s right: there is an involution on the 3-sphere (namely the involution which switches the two sides of the double) 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.

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. $M$ as a quotient

We must start by following Bing.

Start with the 3-ball $B$. We will realize the open “bad” manifold $M$ as the complement of a nested sequence of thickened tangles. Let $A_1$ be the union of two 1-handles which clasp each other in $B$. Topologically, the complement of $A_1$ is a genus 2 solid handlebody. Each 1-handle of $A_1$ is attached to the boundary of $B$ along two disks.

Now, we obtain $A_2$ from $A_1$ as follows: in each 1-handle of $A_1$ insert the union of two 1-handles which clasp each other, so that each 1-handle of $A_2$ is attached along two disks which are contained in the same disk of a single 1-handle of $A_1$; see the figure. Continue indefinitely, so that $A_n$ is obtained by putting two clasped 1-handles in each 1-handle of $A_{n-1}$, and let $A$ be the intersection of all the $A_n$. Then $M = B - A$.

Where is the wild sphere? It is obtained by collapsing each component of $A$ to a point. To see this, suppose we push’ the attaching disks of $A_1$ into $B$ by sliding them into the interior of the 1-handles, before attaching the handles of $A_2$. After attaching these handles, push’ the attaching disks along the handles of $A_2$ into the interior, and continue in this way, so that at each stage the boundary of $B$ is growing horns, and then horns on horns, and then horns on horns on horns . . . along the cores of the $A_n$, growing itself into the standard picture of the horned sphere at the infinite stage.

2. Double the picture

Now let’s double this picture. We replace $A_1$ with a thickened 2-component unlink $L_1$ clasped like two components of the Borromean rings. Then $L_2$ is obtained from $L_1$ by inserting two thickened knots in each component of $L_1$ so that the two components together with the core of the complement are a Borromean rings. Define $L_n$ inductively by the same construction, and let $L$ be the intersection of all the $L_n$. By the discussion above, the space obtained by collapsing each component of $L$ to a point is the double $DM$.

3. 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 $L_n$ 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 $L$ 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 $L_n$
• if we define $J_n:=I_n \circ I_{n-1} \circ \cdots \circ I_1$ then each component of $J_n(L_n)$ has diameter $\epsilon_n$

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 $L$ to a cantor set $C$ in such a way that $J:S^3 - L \to S^3 - C$ 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 $L_n$ around inside $L_{n-1}$; if this is done judiciously, the components can be individually moved so that their diameters are smaller than in the original configuration.

So this is the whole game! Can we successively slide the components of $L_n$ around in each $L_{n-1}$ in such a way that the diameters of the components go to zero as $n \to \infty$? Bing says yes, and gives’ an inductive procedure to do so.

4. Some notation

Let’s restrict the rules of the game. We idealize each component $\gamma$ of $L_i$ 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 $\alpha_1,\alpha_2$ 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 $L_n$ 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. The inductively defined sequence $L_n$ can actually be backed up: $L_0$ can be thought of as a single solid unknot in $S^3$ in which all the successive $L_n$ are nested. Let’s agree that we only really need to give the angular coordinate of the core of each component of $L_n$ projected onto the core of $L_0$ (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 $L_n$ 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 $L_0$ according to whether the letter is $L$ or $R$. Thus (trivially) $L_0$ 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 $L_0$, 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 $L_n$, encoded by a cyclic word $S$, and suppose $\alpha_1,\alpha_2$ are the components of $L_{n+1}$ 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 $S = T_1T_2$ where $T_1,T_2$ are words of length $|S|/2$. If $T$ is a word in the alphabet $\lbrace L,R\rbrace$ let $T^*$ 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 $T_1T_1^*$ and $T_2T_2^*$. We call the operation of replacing $S$ with the pair $T_1T_1^*, T_2T_2^*$ 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 $DS = T_1T_2$ and form $T_1T_1^*$ and $T_2T_2^*$.

5. An inductive lemma

OK, we are nearly done. We have $L_0=\lbrace L \rbrace$, a single component whose angular projection has length 1. We subdivide to form $LL$ and then decompose to form $L_1 = \lbrace LR,RL\rbrace$ each with angular projection of length $1/2$. We subdivide again to form $LLRR,RRLL$ and decompose to form $L_2 = LRLR,RLRL,RLRL,LRLR$ 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 $T_1T_2$ 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 $S =T_1T_2$ into strings of half the length so that each of the $T_i$ 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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