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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