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.cylinder_0.jpg

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:

K_1            K_2           K_3

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

This entry was posted in 3-manifolds, Uncategorized and tagged , , . Bookmark the permalink.

4 Responses to Bing’s wild involution

  1. dedusuiu says:

    here must most as say prof dr mircea orasanu

  2. Israel Socratus Sadovnik says:

    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 /
    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
    Israel Sadovnik Socratus
    ”You cannot be a physicist, if you don’t understand
    the beauty of the Minkowski mathematical construction.”
        / a professor to the students /

  3. great blog
    really informative and educative

Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )

Connecting to %s