Folklore, the Borromean rings, the icosahedron, and three dimensions
FFolklore, the Borromean rings, the icosahedron,and three dimensions
Dave Auckly
Mathematics Department, Kansas State University, Manhattan, KS 66506 [email protected]
Introduction
There are many bits of mathematical folklore – results that have been known for along time – that are not covered in a typical undergraduate curriculum. We will givea proof that the Borromean rings are linked, but our proof will not be the shortestpossible. Indeed, we give a number of unnecessary, but loosely related, stories toexpose more folklore. In particular we will see that the Borromean rings are relatedto the icosahedron and something called the Poincar´e homology sphere. We beginwith a cautionary tale.Around 1900 Poincar´e made many important contributions to topology. He wasconsidering spaces that looked a bit like 3-space, but were not necessarily just R .The same phenomena may be seen with shapes that look like 2-space. For instance asmall neighborhood on the surface of the earth looks like R even though the surfaceof the Earth is not just a flat plane. The surface of the Earth is better modeled bythe 2-sphere S := { v ∈ R | | v | = 1 } . The same could be said about the surface ofa donut, neighborhoods of it look like R but it isn’t. Shapes that look like R arecalled 2-manifolds and shapes that look like R are called 3-manifolds.Poincar´e noticed that a compact, roughly speaking “does not extend to infinity,”2-manifold that has no holes in it must be the two-sphere, S . He conjectured thatthe same thing was true for 3-manifolds where the measure of holes in a manifold M is given by something that we now call the first homology, H ( M ). In 1904 hefound a counter-example to this conjecture. To see this he introduced the notionof the fundamental group of a space, π ( M ), and constructed a counter-examplecalled the Poincar´e homology sphere that may be described using an icosahedron.He then revised his conjecture to state that the only compact 3-manifold with trivialfundamental group is S [3]. More than one hundred years later Perelman gave thefirst correct proof of this conjecture [2].In fact, before the correct proof a number of incorrect proofs were given of thePoincar´e conjecture. In many incorrect proofs the only hypothesis that is used is thatthe homology is trivial. Of course any such proof is doomed to fail as the Poincar´ehomology sphere shows. John Stallings wrote a nice paper about this called, HowNot to Prove the Poincar´e Conjecture [5]. Had these authors payed a bit moreattention to the folklore about the difference between homology and fundamental1 a r X i v : . [ m a t h . HO ] S e p roup, there may have been fewer false proofs. The first homology H ( M ) is theabelianization of the fundamental group, π ( M ). The difference between trivialhomology and trivial fundamental group, while understood, can be subtle. We willexplain the fundamental group in greater detail in the first section below.We note that the electronic version of this paper has color graphics that may beeasier to follow. The fundamental group
Imagine you are walking a slightly skittish dog on a leash in a forest. Most of thetime the dog will stay by your side. Now the dog thinks it detects a squirrel andsniffs its way around a tree twice. You are stuck until you can convince the dog tounwind from the tree. There is a group here. The dog could sniff 2 times aroundthe tree, or 13 times, or could go the other way to wrap − X be a topological space, i.e., a space for which the notion of continuous is defined, andlet x be a point in X . Elements of the fundamental group are equivalence classesof continuous paths, γ : [0 , → X with γ (0) = γ (1) = x . Two paths, γ and γ are equivalent if there is a deformation ( homotopy ) H : [0 , × [0 , → X so that H (0 , s ) = γ ( s ), H (1 , s ) = γ ( s ), H ( t,
0) = x , and H ( t,
1) = x . In the example ofthe dog walk, the space X is all of the ground that is not covered by the tree. Thebase point x is the point that you are standing on. The leash may be viewed as acontinuous path γ . In the equivalence relation, one can view the parameter t in thefunction H as time. Thus H (0 , s ) = γ ( s ) requires that the leash is along the path γ at time zero. Fixing t , the function H ( t, · ) represents the position of the leashat time t . The condition H ( t,
0) = x corresponds to the handle of the leash beingwith you at all time, and the condition H ( t,
1) = x corresponds to your skittishdog sitting on your foot for all time. This is captured in the definition below. π ( X, x ) := { γ : [0 , → X | γ (0) = γ (1) = x } / homotopy . The group operation is called concatenation . Let γ represent the position of theleash after one short exploration of the dog, and let δ represent the position of theleash after another short exploration. The product, γ ∗ δ is what would happen ifthe dog first did γ , and then did δ . The formula is:( γ ∗ δ )( t ) = (cid:40) γ (2 t ) if t ∈ [0 , / ,δ (2 t −
1) if t ∈ [1 / , . Going further on this walk you now approach a pair of close trees, and there is asquirrel in them. In complete excitement your dog races counterclockwise around2he left tree L back to you, then clockwise around the tree on the right R − andback to you, clockwise around the tree on the left L − , and counterclockwise aroundthe tree on the right R . A figure showing the leash and how it may be pulled abit tighter to an equivalent (homotopic) path is shown below. This path will bedenoted by LR − L − R = L ∗ R − ∗ L − ∗ R .Figure 1: Tangled leashThere is something interesting about this particular example. In total the dog haswrapped around the left tree zero times, so the leash would pull free if the righttree was not there. In addition the dog walked around the right tree a total of zerotimes. Yet as long as the squirrel continues to scare your dog and your dog does notleave your foot, the leash will be linked with the trees. This is exactly the differencebetween the first homology and the fundamental group. The first homology is theabelianization of the fundamental group (in other words XY = Y X in homology).In homology LR − L − R = LL − R − R = 1. We typically write the operation in thefundamental group multiplicatively ( XY ) and operation in the homology additively( X + Y ) to remind us that the homology is abelian.The fundamental group of the complement of two disjoint disks in the plane (theforest floor away from the two trees) is what is known as a free group on twogenerators. The elements of this group are a trivial element 1 and all finite “words”that may be bade from the letters L , R , L − , R − with all cancelations of X nextto X − . Multiplication is just the process used to create a compound word. Thusthe product of LRLR − with RRRLL is LRLRRLL .The false proofs of the Poincar´e conjecture asserted that when every loop in acompact 3-manifold wrapped zero times around any hole an the abelian sense asmeasured in homology, the manifold had to be trivial. This turned out to be false.Now that we have a new trick for our dog, let’s find a better place to tangle ourleash.
The Borromean rings
Start by making a roughly 34 inch by 55 inch rectangle. Really a 34 by 34 · (1+ √ / √ / golden ratio . It is the length of the diagonalof a sidelength 1 pentagon. These numbers are close to adjacent numbers in theFibonacci sequence (34 and 55) and this is a nice size for public display. We could3repare and make two rectangles as in Figure 2a and later move them into positionas in Figure 2b.A more precise description of the right hand side starts with three of these of theserectangles. We put one with center at the origin and line of symmetry parallel tothe long side (call this the rectangle axis) matching the x -axis of the x − y plane.Similarly, we could put one with the axis matching the y -axis of the y − z planeand one with the axis matching the z -axis of the z − x plane. The result is the veryspecific model of the Borromean rings in Figure 3b. The Borromeo family used aversion of this link on its coat of arms. The version from the coat of arms is thetraditional image of the Borromean rings, and it is displayed in Figure 5 later. (a) Loose (b) In position Figure 2: Two unlinked rectanglesThis brings us to a question:
Question:
If we set our three rectangles out on the floor as in Figure 3a,would we be able to move them into the desired configuration in Figure3b without taking one apart?The answer is that it is not possible to assemble three rectangles into the Borromeanconfiguration without breaking one of the rectangles. We will ultimately give a proofof this fact using the fundamental group of the complement of the rectangles.A link is a disjoint collection of topological circles in 3-space. Given a picture of alink, one may label some loops by putting arrows crossing under various strands ofthe link as in the Figure 4a. We just discovered that your dog is a super dog. Yourdog now flies up to a point above a crossing. Each arrow in Figure 4 representsthe loop of leash generated when your dog starts in the air (at the base point) fliesto the tail of the arrow, follows the arrow and returns to the floating base point.The arrows labeled L and R are below the horizontal strand but above the heightof the vertical strand in Figure 4a.Thus the loops represented by arrows labeled L and R are equivalent because one may be deformed into the other. However, theloops labeled F and B are not equivalent, because the horizontal component of the4 a) Unlinked (b) Linked? Figure 3: Assemble three rectangleslink gets in the way of the deformation. See Figure 4b. The inverse of a loop, is thesame loop, with reversed orientation. (a) Top view (b) Super leash (c) Deformed
Figure 4: Loops Near a CrossingSuperdog will now show us a trick while trying to tie its leash around the strands.Superdog flies1. down then along the arrow on the right,2. back up, back down,3. follows the arrow in the back in the opposite direction,4. back up, then down,5. follows the arrow on the left in the opposite direction,6. back up, then down,7. follows the front arrow in the indicated direction, and8. finally fies back up to the base point.The result is the product RB − L − F displayed in Figure 4b. However, this is thetrivial element of the fundamental group. To see this, note that the portions of the5eash generated by the ”back up, then down” portions 2, 4, and 6 of the flight maybe deformed down to give the representative in Figure 4c. From here the squareportion of the leash loop on the bottom can shrink and the entire loop can be pulledback up to the base point. It is a fact that the fundamental group of the complementof any link may be described as the group with one generator for each strand andone relation similar to the one we just described for each crossing. This is called the Wirtinger presentation . See the book by Dale Rolfsen for a proof of this, [4].How could this possibly help us? We are trying to show that we can not move theunlinked rectangles from Figure 3a into the configuration in Figure 3b. Imaginethat we could, and fill up the space around the rectangles with honey. When onepushes the set of rectangles from Figure 3a over to the position in Figure 3b, allof the honey will move as well. The correspondence between the starting locationof a molecule of honey and the ending location of that molecule of honey gives ahomeomorphism between the complement of the configuration in Figure 3a and thecomplement of the configuration in Figure 3b. Thus the fundamental groups of thecomplements of these two configurations would be the same. The fancy name forthis idea is the isotopy extension theorem.The fundamental group of the complement of the trivial 3-component link in Fig-ure 3a is the group with three generators and no relations (one generator for eachcomponent, and no relations because there are no crossings): π (Complement of unlink) = (cid:104) P, Q, R (cid:105) . The three rectangles in the configuration in Figure 3b may be continuously deformed,without crossing, to the configuration in Figure 5a. Thus the fundamental groupsof the complements of these configurations are the same.To compute the fundamental group of the complement of the Borromean rings welabel arrows under the strands in Figure 5a with
T, U, V, X, Y, Z . The strand on theupper right of this figure has an X arrow running under it. The loop represented bythis arrow starts at a base point above the figure (say your nose), travels down tothe tail of the arrow, follows the arrow and returns to the basepoint. This loop maybe slid (homotoped) to other positions along this strand as long as one does nottry to slide past an undercrossing. We label two loops equivalent to X near the topcrossing to make it easier to see the relation arising from this crossing. Following theprocedure in Figure 4 starting at the head of the T and prodeeding clockwise we readthe relation T − XY X − = 1. Thus T = XY X − . Similarly, the outside crossing onthe left shows that U = Y ZY − and the outside crossing on the right shows that V = ZXZ − . Using the three inner crossings leads to the presentation: π (Complement of the Borromean rings) = (cid:104) X, Y, Z | Y ZY − XY Z − Y − ZX − Z − = 1 ,XY X − ZXY − X − Y Z − Y − = 1 ,ZXZ − Y ZX − Z − XY − X − = 1 (cid:105) . a) Generators (b) Glue curve Figure 5: Borromean Rings.Indeed, the relation coming from the inner crossing closest to arrow Y going counter-clockwise starting with the U on the top reads:1 = U XU − V − = ( Y ZY − ) X ( Y Z − Y − )( ZX − Z − ) . To prove that the two links are different, we could just show that these two groupsare different. Instead we will use a slightly weird argument. We choose a weirdargument because it will allow us to discuss the relationship between the Borromeanrings, the Poincar´e homology sphere and the icosahedron. Sometimes the long wayaround is more scenic.
From the Borromean Rings to the Poincar´e Sphere
This is the most technical section of this paper. A quick summary is that if weadd the relation P = 1 and the corresponding relations for the other components( Q = R = 1) to the fundamental group of the complement of the trivial 3-componentlink, we would get a trivial group. If the trivial 3-component link was equivalent(isotopic) to the Borromean rings the relation corresponding to P = 1 would be X − U − Z = 1. Adding this to the fundamental group of the complement of theBorromean rings together with the analogous relations for the other two componentsresults in a non-trivial group so the two links are not equivalent. Right now youshould probably have no idea where the relation X − U − Z = 1 comes from. We willturn this around and start with the relation X − U − Z = 1 and see that it leads to P = 1. We will also construct a space with the corresponding fundamental group –the Poincar´e homology sphere. 7e can perform something called surgery on the trivial 3-component link and on theBorromean rings. If these two links were the same, the result of the correspondingsurgeries would be the same. We begin with a warm-up. Consider the two “links” inFigure 6 below. Each is two points put into a space consisting of two 2-spheres.Figure 6: The pointsIt there was a deformation taking the trivial 2-point link on the left to the 2-pointlink on the right, there would be a homeomorphism taking the complement of thetwo points on the left to the complement of the two points on the right. (This is theisotopy extension theorem again. Fill the complement with honey and see wherethe honey molecules would go after pushing the one link to the other.)We will now glue the same thing to each complement, namely a cylinder. Notice that S × (0 ,
1) is homeomorphic to the open unit ball with the origin deleted. Indeed, justconsider the coordinate in S as an angle and the coordinate in (0 ,
1) as a radius and S × (0 ,
1) will be the punctured open ball expressed in polar coordinates. Similarly S × ( − ,
0) is also homeomorphic to a deleted open ball. Thus we can glue theopen cylinder, S × ( − ,
1) to the complement of either link. The homeomorphismthat we get by assuming that the links are equivalent extends and would imply thatthe result of gluing in the open cylinder to the left side would be homeomorphicth the result of gluing the open cylinder to the right side. However, we displaythe result of this gluing in Figure 7 below, and can tell that the two sides are nothomeomorphic because one is disconnected, and the other is connected. This processis called surgery because we are cutting something out and sewing something elseback in, a bit like an organ transplant.Figure 7: Surgery on the pointsWe are now going to do the analogous thing to the Borromean rings.We cut out therings and glue in solid tori (donuts). This is displayed in Figure 8 below.To specify how an open torus is glued to the complement of the Borromean rings, wewill keep track of one curve. The thin black curve on the left side of the boundaryof the solid torus on the left of Figure 8 bounds a disk. However it is not part ofthe open solid torus. To be clear the open solid torus is given by { ( v, w ) ∈ R × R | | V | < , | w | = 1 } . a) Donut (b) Rectangular glue curve Figure 8: The Poincar´e homology sphere.The red curve is parallel to the thin black curve. When we glue the open solid torusto the complement of the Borromean rings we will make sure that the thin (red inelectronic) curves in Figure 8 match. The thin red curve in the complement of theBorromean rings is called a surgery curve. The reason we keep track of this curve isbecause the corresponding loop will be trivial in the new manifold. We do the samegluing with each of the other components.The result of attaching three open solid tori to the complement of theBorromean rings is equivalent to removing one point from the Poincar´ehomology sphere. We can take this as the definition of the Poincar´ehomology sphere. Figure 9: Two ballsWe could fill in the missing point by gluing in a hemisphere from the 3-dimensionalsphere, i.e. the set of points one unit from the origin in R . A schematic of this isdisplayed in Figure 9. Here the left hemisphere with a small copy of the Borromeanrings represents the result of removing the Borromean rings from a solid ball andgluing in donuts. The right hemisphere represents a second solid ball. Gluing thesetogether yields the Poincar´e homology sphere. We denote it by Σ. To compute9he fundamental group of Poincar´e homology sphere we just need to add relationsindicating that the loop corresponding to the thin (red) circle and the analogousloops are trivial. The thin (red) loop in Figure 8 corresponds to the loop in theFigure 5b. Reading the relation we get X − U − Z = X − Y Z − Y − Z = 1. Addingthe analogous relations for the other two components gives: π (Σ) = (cid:104) X, Y, Z | Y ZY − XY Z − Y − ZX − Z − = 1 ,XY X − ZXY − X − Y Z − Y − = 1 ,ZXZ − Y ZX − Z − XY − X − = 1 ,X − Y Z − Y − Z = 1 ,Y − ZX − Z − X = 1 ,Z − XY − X − Y = 1 (cid:105) . This is the fundamental group of the Poincar´e homology sphere. We can simplifythe expression for this group. The last relation in π (Σ) implies Z = XY − X − Y .Substituting this into the other relations allows us to write the group without usingthe generator Z . Using this, the relation Y − ZX − Z − X = 1 is seen to be equivalentto XY − X − Y X − Y − X = 1, and the relation X − Y Z − Y − Z = 1 is seen to beequivalent to Y X − Y − XY − X − Y = 1. The other relations all follow from thesetwo and the expression for Z . Thus, π (Σ) = (cid:104) X, Y | XY − X − Y X − Y − X = 1 , Y X − Y − XY − X − Y = 1 (cid:105) . To write this in an interesting way set X = A − BA − B and Y = A − B . Notice thatthis implies that A = Y X − Y and B = Y X − Y so we can either use generators X and Y to describe the group or generators A and B . Using this substitution therelation Y X − Y − XY − X − Y = 1 becomes AB − A = 1 or just B = A . Now XY − X − Y X − Y − X = 1 becomes A − BA − B − A B − A B − A − B = 1. Using A = B A − this becomes ( A − B ) B − ( A − B ) = 1 or ( A − B ) = B . Thus wehave π (Σ) = (cid:104) A, B | ( A − B ) = A = B (cid:105) . Now imagine that the Borromean rings could be deformed (isotoped is the technicalword) to the trivial link. Where would the surgery curves go? Notice that thesurgery curve is on the boundary of a solid torus centered on one of the rings. Itwould still be on the boundary of such a solid torus after the deformation. Thesurgery curve is parallel to the link component in this solid torus. It would still beparallel after deformation. It also links the component once. After the deformation,it would still have to link once.In fact one way to define how many times a loop links a simple ring is to look atthe class of the loop in the fundamental group of the complement. The fundamentalgroup of the complement of a single ring with no crossings has just one generatorand no relations. Call the generator F . The only words that may be made in this10roup are F n so it is isomorphic to the integers, and we say a loop homotopic to F n links the ring n times.The tail of the loop represented by the surgery curve in Figure 5 might get deformedinto something CRAZY , so the resulting loop that would have to be killed in thefundamental group of the complement of the trivial 3-component link would havethe form (
CRAZY ) P ± ( CRAZY ) − and this will be trivial exactly when P ± = 1,i.e., P = 1.Thus if the Borromean rings could be deformed into the trivial 3-component link,the fundamental group of the Poincar´e homology sphere would have to be (cid:104) P, Q, R | P = Q = R = 1 (cid:105) . In other words, it would be the trivial group.Let’s see, we can abelianize the fundamental group of the Poincar´e homology sphere.This means we are assuming that XY = Y X , etc.. Using this in the last threerelations in π (Σ) as expressed via X , Y , and Z generators, gives X − = 1, Y − = 1and Z − = 1 and this implies that the abelianization of the fundamental group ofthe Poincar´e homology sphere is trivial. Maybe we should assume that this meansthat this space is just a 3-sphere??? We wouldn’t be the first to make this guess.Perhaps it is possible to unlink the Borromean rings after all. Hmm, our yappyfriend is reminding us of a leash that represented a trivial loop in an abelianization,but was not trivial in the fundamental group. We need to think.Well, the fundamental group of the Poincar´e sphere does not look trivial. Is this goodenough? Consider a different group presentation. The following group is trivial, butit takes work to prove it. (Try.) (cid:104) C, D | C − D C = D , D − C D = C (cid:105) . To see that the fundamental group of the deleted Poincar´e homology sphere is nottrivial, it would suffice to construct a surjective homomorphism from it to some othernon-trivial group. We could do so now and end the paper. However, sometimes itis entertaining when the old man on the porch starts telling stories. This remindsus of something related to an icosahedron, so we’ll tell a few more stories.
The icosahedron
Now return to the three linked rectangles form Figure 3 and add 34-inch strutsconnecting the corner of each rectangle to the four corners of the other rectanglesthat are closest to it. The result is the structure displayed in Figure 10a. If oneremoves the long edges of each original rectangle, the remaining figure is called anicosahedron. It is the structure represented by the MAA logo, and it is displayed inFigure 10b. 11 a) Rectangles in an Icosahedron (b) Icosahedron
Figure 10: Icosahedron around three rectanglesThe icosahedron is one of the five Platonic solids. These solids have captured theimaginations of people for many years. One early model of the orbits of the planetswas based on placing one Platonic solid inside of the next. It started by placingthe octahedron in the icosahedron. Why don’t we try to do the same thing in oneway?The convex hull of three congruent, mutually perpendicular line segments meetingat their centers is a regular octahedron. The axes of the three original rectanglesexactly meet this condition, so we can add the edges of a regular regular octahedronto our figure by connecting the mid points of the short edges of the original threerectangles. This is displayed in Figure 11a with just the octahedron in Figure11b.This is fun, why stop. We started with three rectangles, and thus had six shortedges. We put the six vertices of a red octahedron at the midpoints of these first sixedges. We then added more short edges until each rectangle corner met a total offive edge ends. This gave a total of 4 corners per rectangle times 3 original rectanglestimes 5 (original short edge plus four new short edges) short edge ends. Thus thereare 60 short edge ends and 30 short edges. It looks like we can fit 5 = 30 / a) Both (b) Octahedron Figure 11: Octahedron in an Icosahedron5. a blue octahedron.It works. The result is an icosahedral compound of octahedra as displayed in Fig-ure 13. Figure 12: Compound of Octahedra in and out of an IcosahedronConsider the orientation-preserving symmetries of the icosahedron. These are allrotations, since they must fix the center of the icosahedron. There is a 1 / / / /
5, 2 /
5, 3 /
5, and 4 / × ×
12 triangle corners and 60 / / / / − = (135) and (153) = (1) . Similarly,(135) ◦ (12)(34) = (12345) ((12)(34)) = (1) and (12345) = (1) . Thus, (cid:0) (153) − (12)(34) (cid:1) = (153) = ((12)(34)) . The set of all functions permuting the numbers 1 , ..., G and has 5! = 120 elements. Thelabeling scheme demonstrates that the orientation-preserving symmetry group of theicosahedron is isomorphic to the index-2 subgroup of G known as the alternatinggroup of 5 symbols, A . 14h, this reminds us where we were going. We were going to explain why the fun-damental group of the Poincar´e homology sphere is non-trivial. We just saw that a = (153) and b = (12)(34) satisfy the same relations as A and B in the simplifiedpresentation of the fundamental group of the Poincar´e homology sphere. Recallthese were ( A − B ) = A = B . Yet the alternating group on 5-symbols is nottrivial. In fact this shows that the fundamental group of the Poincar´e homologysphere surjects onto the symmetry group of the icosahedron.The Poincar´e homology sphere is a remarkably beautiful space that appears inmany different contexts. Read about several descriptions of it in Eight Faces ofthe Poincar´e Homology Sphere , [1]. To learn more about computing fundamentalgroups of knot and link complements, cutting out tubes and gluing back in donuts,and many other foundations of low-dimensional topology, Rolfsen’s book is a goodplace to start, [4].The Author would like to thank the referees and Bob Burckel for very helpful com-ments on an earlier draft.
References [1] R. C. Kirby and M. G. Scharlemann. Eight faces of the Poincar´e homology3-sphere. In
Geometric topology (Proc. Georgia Topology Conf., Athens, Ga.,1977) , pages 113–146. Academic Press, New York-London, 1979.[2] G. Perelman. The entropy formula for the ricci flow and its geometric applica-tions. Preprint, 2002.[3] H. Poincar´e. Second Complement a l’Analysis Situs.
Proc. London Math. Soc. ,S1-32(1):277, 1900.[4] Dale Rolfsen.
Knots and links , volume 7 of
Mathematics Lecture Series . Publishor Perish, Inc., Houston, TX, 1990. Corrected reprint of the 1976 original.[5] John Stallings. How not to prove the Poincar´e conjecture. In
Topology Seminar,Wisconsin, 1965 , volume 60 of
Ann. of Math. Stud. , pages 83–88. PrincetonUniv. Press, Princeton, NJ, 1966.