WildWildWhitehead - American Mathematical Society · 2019. 3. 7. · SHORTSTORIES WildWildWhitehead...
3
SHORT STORIES Wild Wild Whitehead Danny Calegari Take a knot. Take a tube around the knot. Put a new knot in the tube, twisted around and clasping itself as in Fig- ure 1. The new knot goes once around the tube, and then “doubles back” and clasps itself. The new knot is the White- head double of the old. Figure 1. The green knot is the Whitehead double of the black knot. Let’s call the first knot 0 and its Whitehead double 1 , and let’s call the tube around 0 (actually a solid torus) 0 . It makes sense to take the Whitehead double of any knot, but in Figure 1, 0 is a trivial knot; i.e., it bounds an embedded disk in 3 . In this case, 1 is trivial too: it bounds an embedded disk in 3 . But 1 is knotted in 0 . Any embedded disk that 1 bounds must go outside 0 . On the other hand, 1 is homotopically trivial in 0 ; i.e., it bounds an immersed disk, one that crosses itself, but does not cross 0 . To see this, just push one of the clasps of 1 through the other one; this undoes the knotting, and the Danny Calegari is a professor of mathematics at the University of Chicago. His email address is [email protected]. For permission to reprint this article, please contact: [email protected]. DOI: https://doi.org/10.1090/noti1837 result can be shrunk down to a point. The track of the knot 1 during this process sweeps out an immersed disk. In terms of the fundamental group, a knot in a space determines a conjugacy class [] in the fundamental group 1 (). Now, 1 ( 0 )=ℤ, and since 1 bounds an immersed disk, [ 1 ] is trivial in 1 ( 0 ). We can keep going. Let 1 be a tube around 1 , thin enough to fit in 0 , and let 2 ⊂ 1 be the Whitehead double of 1 . And so on. Each bounds an embedded disk in 3 , but each of these disks must go (many times!) all the way outside 0 . The tubes get thinner and thinner as we go, and longer and longer. Consequently, the knots must get longer and longer too: Each must wind back and forth at least 2 times around 0 , clasping itself in a complicated way at the end. The infinite intersection ⋂ is called the Whitehead continuum, which we write Wh, see Figure 2. The Whitehead continuum is connected but not path connected. It has an entangled dyadic Cantor set of “strands” that wind around 0 . The complement 3 − Wh is an open 3-manifold called the Whitehead manifold. It turns out that 3 − Wh is con- tractible but not homeomorphic to a 3-ball. Let’s see why. The outside of 0 in 3 is another solid torus ′ 0 , whose core is a knot ′ 0 linking 0 in a Hopf link. [ ′ 0 ] is the generator of 1 ( 3 − 0 )= 1 ( ′ 0 )=ℤ. The knots ′ 0 and 1 together form a 2-component link called the Whitehead link. This link is symmetric: we can isotope it around and interchange ′ 0 and 1 ; see Figure 3. Since [ 1 ] is trivial in 1 ( 0 )= 1 ( 3 − ′ 0 ), it follows by symmetry that [ ′ 0 ] is trivial in 1 ( 3 − 1 ). Consequently the inclusion map 3 − 0 → 3 − 1 APRIL 2019 NOTICES OF THE AMERICAN MATHEMATICAL SOCIETY 581
Transcript of WildWildWhitehead - American Mathematical Society · 2019. 3. 7. · SHORTSTORIES WildWildWhitehead...
SHORT STORIES
Wild Wild WhiteheadDanny Calegari
Take a knot. Take a tube around the knot. Put a new knotin the tube, twisted around and clasping itself as in Fig-ure 1. The new knot goes once around the tube, and then“doubles back” and clasps itself. The new knot is theWhite-head double of the old.
Figure 1. The green knot is the Whitehead double of the blackknot.
Let’s call the first knot 𝐾0 and its Whitehead double𝐾1,and let’s call the tube around 𝐾0 (actually a solid torus)𝑁0. It makes sense to take the Whitehead double of anyknot, but in Figure 1, 𝐾0 is a trivial knot; i.e., it boundsan embedded disk in 𝑆3. In this case, 𝐾1 is trivial too: itbounds an embedded disk in 𝑆3. But 𝐾1 is knotted in 𝑁0.Any embedded disk that 𝐾1 bounds must go outside 𝑁0.
On the other hand, 𝐾1 is homotopically trivial in 𝑁0; i.e.,it bounds an immersed disk, one that crosses itself, but doesnot cross 𝑁0. To see this, just push one of the clasps of 𝐾1through the other one; this undoes the knotting, and the
Danny Calegari is a professor of mathematics at the University of Chicago. Hisemail address is [email protected].
For permission to reprint this article, please contact:[email protected].
DOI: https://doi.org/10.1090/noti1837
result can be shrunk down to a point. The track of the knot𝐾1 during this process sweeps out an immersed disk.
In terms of the fundamental group, a knot 𝐾 in a space𝑋 determines a conjugacy class [𝐾] in the fundamentalgroup 𝜋1(𝑋). Now, 𝜋1(𝑁0) = ℤ, and since 𝐾1 boundsan immersed disk, [𝐾1] is trivial in 𝜋1(𝑁0).
We can keep going. Let 𝑁1 be a tube around 𝐾1, thinenough to fit in 𝑁0, and let 𝐾2 ⊂ 𝑁1 be the Whiteheaddouble of 𝐾1. And so on. Each 𝐾𝑖 bounds an embeddeddisk in 𝑆3, but each of these disks must go (many times!)all the way outside 𝑁0.
The tubes get thinner and thinner as we go, and longerand longer. Consequently, the knots must get longer andlonger too: Each 𝐾𝑛 must wind back and forth at least2𝑛 times around 𝑁0, clasping itself in a complicated wayat the end. The infinite intersection ⋂𝑖 𝑁𝑖 is called theWhitehead continuum, whichwewriteWh, see Figure 2. TheWhitehead continuum is connected but not path connected.It has an entangled dyadic Cantor set of “strands” that windaround 𝑁0.
The complement 𝑆3−Wh is an open 3-manifold calledthe Whitehead manifold. It turns out that 𝑆3 −Wh is con-tractible but not homeomorphic to a 3-ball. Let’s see why.
The outside of𝑁0 in𝑆3 is another solid torus𝑁′0, whose
core is a knot 𝐾′0 linking 𝐾0 in a Hopf link. [𝐾′
0] is thegenerator of 𝜋1(𝑆3 − 𝑁0) = 𝜋1(𝑁′
0) = ℤ. The knots𝐾′
0 and 𝐾1 together form a 2-component link called theWhitehead link. This link is symmetric: we can isotope itaround and interchange 𝐾′
0 and 𝐾1; see Figure 3.Since [𝐾1] is trivial in 𝜋1(𝑁0) = 𝜋1(𝑆3 − 𝑁′
0), itfollows by symmetry that [𝐾′
0] is trivial in 𝜋1(𝑆3 − 𝑁1).Consequently the inclusion map 𝑆3 − 𝑁0 → 𝑆3 − 𝑁1
APRIL 2019 NOTICES OF THE AMERICAN MATHEMATICAL SOCIETY 581
Short Stories
Figure 2. Successive approximations converge to Wh.
Figure 3. The Whitehead link is symmetric.
induces the zero map on 𝜋1. Each 𝑁𝑖 is unknotted in𝑆3, and each 𝑁𝑖+1 sits in 𝑁𝑖 the same way that 𝑁1 sitsin 𝑁0. So each 𝜋1(𝑆3 − 𝑁𝑖) = ℤ, and each inclusion𝑆3 −𝑁𝑛 → 𝑆3 −𝑁𝑛+1 induces the zero map on 𝜋1. Tak-ing a direct limit, 𝜋1(𝑆3 −Wh) itself is trivial. A similarargument shows that all the homotopy groups of 𝑆3−Whvanish, and it is contractible.
On the other hand, the complement 𝑆3 −Wh−𝑁′0 =
𝑁0 − Wh has an infinitely generated fundamental group;each𝜋1(𝑁𝑖−𝑁𝑖+1) is complicated (it contains free groupsof every rank), and all of them include as subgroups of𝜋1(𝑁0 −Wh). This shows that 𝑆3 −Wh is not a ball.
Now let’s compactify 𝑆3 −Wh by adding a single pointat infinity. This compact space can also be thought of asthe quotient space 𝑆3/Wh that we get by crushingWh to asingle point. Because 𝑆3 −Wh is not a ball, 𝑆3/Wh is nota manifold. However—remarkably—it is a manifold factor:the product (𝑆3/Wh) × ℝ is homeomorphic to 𝑆3 ×ℝ!
How can this possibly be???First, each 𝑁1 slice can be unknotted by a tiny perturba-
tion in 𝑁0 × ℝ. To distinguish the ℝ factor and, for thesake of brevity, we refer to it as the “time” coordinate (thisis purely a notational convenience). In this language, weunclasp 𝑁1 from itself by nudging one clasp very slightlyforward into the future, and the other very slightly backinto the past. After the nudge, 𝑁1 will not clasp itself, butit will clasp a “future” 𝑁1 on one side, and a “past” 𝑁1on the other. Instead of 𝑁1 clasping itself in a circle, weget a chain of successive 𝑁1’s, each clasping the next, in aslowly ascending spiral. Let’s let 𝜖/4 be the size of the per-turbations of each clasp in the time direction, so that theprojection of each𝑁1 to the time coordinate after it’s beennudged has total length 𝜖/2.
Nudging adjusts points in𝑁1×ℝ by sliding each point×ℝ slightly backward or forward in time. Nudging ex-tends to a self-homeomorphism 𝜈 of 𝑁0 × ℝ, fixed onthe boundary.
Figure 4. Folding the clasps of each 𝑁1 back and forth in timenudges the union of all 𝑁1’s into a collection of spirals.
By the way, there’s not just one spiral—there’s a circle’sworth of them, filling the whole of 𝑁1 × ℝ. Two slices𝜈(𝑁1 ×𝑡), 𝜈(𝑁1 ×𝑠) are in the same spiral if and only if𝑡 − 𝑠 is an integer multiple of 𝜖/2.
After nudging, the next move will straighten out thisand every other spiral so that its projection to the 𝑆3 factoris small (let’s say for concreteness it has diameter < 𝜖/2)without affecting the projection to the ℝ factor.
The cylinder 𝐾0 × ℝ ⊂ 𝑆3 × ℝ has polar coordinates(𝜃, 𝑡)where𝜃 ∈ ℝ/ℤ. Extend these polar coordinates to asmall tubular neighborhood of𝐾0×ℝ containing𝑁1×ℝ,with closure contained in the interior of 𝑁0 ×ℝ.
We can “untwist” every spiral simultaneously by themap
(𝜃, 𝑡) → (𝜃− 2𝑡/𝜖, 𝑡)
on our small tubular neighborhood. Twisting extends toa self-homeomorphism 𝜏 of 𝑁0 ×ℝ, once again fixed onthe boundary.
In summary, first we nudge, then we twist. After doingthis, every 𝜏𝜈(𝑁1) slice projects to subsets of diameter atmost 𝜖/2 in both the ℝ and the 𝑆3 directions. So 𝜏𝜈(𝑁1)has diameter at most 𝜖.
In other words, ℎ1 ∶= 𝜏𝜈 simultaneously shrinks allthe 𝑁1 slices in 𝑆3 ×ℝ as small as we like, while keeping(𝑆3 −𝑁0) × ℝ fixed pointwise.
582 NOTICES OF THE AMERICAN MATHEMATICAL SOCIETY VOLUME 66, NUMBER 4
Figure 5. Screw the top and bottom in opposite directions likeyou’re taking the lid off a pickle jar.
Take a sequence 𝜖𝑖 → 0, and repeat this operation foreach 𝑖 > 1 in place of 1 with 𝜖𝑖 in place of 𝜖. We get a se-quence of self-homeomorphisms ℎ𝑖 ∶ 𝑆3 × ℝ → 𝑆3 ×ℝ, each supported in 𝑁𝑖−1 × ℝ, as a composition of anudge-and-twist ℎ𝑖 ∶= 𝜏𝑖𝜈𝑖. Each 𝑁𝑖 slice gets smallerand smaller in diameter as we apply consecutive ℎ𝑖’s. Un-der application of successive ℎ𝑖’s, the orbit of every pointis a Cauchy sequence, and the infinite composition
ℎ ∶= lim𝑖→∞
ℎ𝑖 ℎ𝑖−1 ⋯ℎ1 ∶ 𝑆3 ×ℝ → 𝑆3 ×ℝ
is well-defined and continuous.For any compact subset 𝑋 of 𝑆3 − Wh the restriction
of ℎ to 𝑋 is the composition of finitely many homeomor-phisms, so the restriction of ℎ to (𝑆3−Wh)×ℝ is a home-omorphism. On the other hand, each Wh slice is succes-sively shrunk smaller and smaller by successive ℎ𝑖, so inthe end ℎ crushes each Wh slice to a point, and ℎ factorsas ℎ = 𝑔𝜋,
𝑆3 ×ℝ 𝜋−→ (𝑆3/Wh) × ℝ 𝑔−→ 𝑆3 ×ℝ,where 𝜋 ∶ 𝑆3 ×ℝ → (𝑆3/Wh) × ℝ is the quotient map,and 𝑔 is the homeomorphism we’ve been looking for.
AUTHOR’S NOTE. The main theorem in this articleand its proof are both well-known, and not due to me!They are due to J. Andrews and L. Rubin, Bull. Amer.Math. Soc. 71(1965), 675-677.
Credits
All images are courtesy of Danny Calegari.
GET MORE FROM YOUR JOURNALS
Learn about AMS MathViewer:
An interactive, dual-panel reading experience for journals published by the American Mathematical Society.
AMS MathViewer’s responsive HTML format perfectly renders mathematical content on desktop, tablet, or phone. Click on embedded links to formulas, theorems, � gures, and references to view them in the second panel, in tandem with the original text.
Keep an eye out for additional journals that will become available throughout the year.
www.ams.org/AMSMathViewer
CURRENTLY AVAILABLE FOR:
Transactions of the AMS—Series BProceedings of the AMS—Series BJournal of the American Mathematical Society
Mathematics of Computation(Starting with 2019 volumes andselect articles from past issues.)
APRIL 2019 NOTICES OF THE AMERICAN MATHEMATICAL SOCIETY 583
cav
Rectangle
