Flat-Knotted Ribbons

Transcription

1 Flat-Knotted Ribbons Shivani ryal and Shorena Kalandarishvili (Dated: September 14, 009) This paper makes conjectures about the least ribbonlength required to tie the 4-stick unknot and the (, p)-torus knot. We find that the ribbonlength of the 4-stick unknot is less than or equal to 4 and the ribbonlength of the (, p)-torus knot is less than or equal to ( + p + ). 1. INTRODUTION We learned basic concepts of knot theory and applied them to develop an understanding of various problems involving knots formed of folded-flat ribbons. flat-knotted ribbon is a knot or a link constructed from a rectangular ribbon hich is folded flat in the plane. flat ribbon of idth is a rectangle of fixed idth framing the knot diagram and satisfying the folloing conditions: The over-under information of the knot diagram is retained. The folds behave as a mirror hich reflects the core curve. There are no extra tists. We mostly orked ith closed knots, rather than the truncated versions. Intuitively, in a truncated knotted ribbon, one edge is cut and the ribbon is shortened appropriately. Our objective as to calculate minimizing lengths for different classes of knots. To do so, e defined ribbonlength R(K) of a flat-knotted ribbon knot K as the length to idth ratio. Therefore, R(K) = len(k). Notice that e can minimize the ribbonlength of a flat-ribbon knot in to ays. First, by fixing the length and miximizing the idth; second, by fixing the idth and minimizing the length. lthough these are functionally equivalent, e chose the latter. In his paper [4], Kauffman introduces these ideas and conjectures that The shape of the minimum-length of a truncated flat-ribbon trefoil is a pentagon and its ribbonlength is The shape of the minimum-length truncated flat-ribbon figure-8 is a hexagon and its ribbonlength is Brennan, Mattman, Raya and Tating [] use Kauffman s ideas to find the folded ribbonlength of torus knots. They make a conjectures that Smith ollege

2 The family T q+1,q has ribbonlength (q + 1) cot( π q+1 ). The family T p,, here p is odd has ribbonlength p cot( π p ). The family T q+1,q, here q 1 has ribbonlength (q + 1) cot( π (q+1) ). The family T q+,q and T q+4,q here q 1 and is odd, has ribbonlength n cot( π n ) here n = q + or q + 4 respectively.. DEFINITIONS knot may be described as a closed curve in space that never intersects itself. The folloing defintions are standard and may be found in many introductory knot theory texts, for example: [1], [3], [5]. Definition 1. knot is a simple closed curve in R 3. polygonal knot consists of a finite number of edges joined at their vertices. FIG. 1: Figure-8 knot. Figure 1 illustrates the figure-8 knot. This is among the simplest non-trivial knots. Figure shos the Hopf link, hich is one of the simplest forms of a link. Definition. link L is a finite disjoint union of knots here each knot in the union is called the component of the link. Ho to tell to knots apart? is the big question in knot theory. To knots may look completely different at first but, hen played around ith, they might turn out to be the same knot. For example: many embeddings of the circle that look nothing like a round circle are actually the unknot. The geometry of a folded ribbon is best modeled by polygonal knots. From no on, e restrict out attention to polygonal knots and links. Figure 3 illustrates a polygonal trefoil knot. Let L be a polygonal link embedded inr 3 and let be a triangle such that:

3 3 FIG. : Hopf link. FIG. 3: polygonal trefoil knot. L does not meet the interior of L meets one or to sides of The vertices of L in L are also vertices of Definition 3. -move on a link L is defined as follos: replace L ith (L (L )) ( L). Figure 4 shos that a -move changes the path of a polygon. Definition 4. To polygonal links L 1 and L are equivalent if there is a finite sequence of -moves that transforms L 1 into L. Definition 5. The unknot, or trivial knot, is equivalent to a standard circle. Definition 6. knot invariant, I(K), is a quantity assigned to a knot that is the same for equivalent knots. It means, if K 1 = K then I(K 1 ) = I(K ).

4 4 FIG. 4: Effect of a -move. FIG. 5: To different trivial knots. Remark 1. lso, if I(K 1 ) I(K ) then K 1 is not necessarily equivalent to K. Knot invariants and the methods to calculate them are examples of global problems in knot theory. Knot invariants range from easy to define and difficult to calculate to difficult to define and easy, but time-consuming, to calculate. One of the simplest knot invariants is crossing number. The image of a knot K in three space to the plane is called the projection of the knot K. knot diagram is a modified form of the projection of a knot K, here gaps are left in the draings of the projections to indicate hich parts of the knot pass under other parts. Definition 7. knot projection is called a regular projection if no three points on the knot project to the same point, and no vertex projects to the same point as any other point on the knot. regular diagram is stable so that even if the direction of projection is changed by a small amount, the projected image is essentially unchanged. For example, hen e do not have to edges projecting into the same line segment, the projection as clear as possible. Definition 8. The crossing number of a knot K, r(k), is the minimal number of crossings in any diagram of the knot. Example. For an unknot r(o 1 ) = 0, for a trefoil r(3 1 ) = 3, and for a figure eight r(4 1 ) = 4. The 5 crossing knots have r(5 1 ) = r(5 ) = 5, but these knots are not equivalent. These knots sho that equal invariants does not necessarily imply equivalence of knots.

5 5 FIG. 6: regular projection of the trefoil knot. Definition 9. Given a polygonal knot K the stick number, denoted by s(k), is the smallest number of straight edges needed to form K. For example, the stick number of the unknot is 3, the stick number of the trefoil is 6. Figure?? shos the 3-stick unknot and the 4-stick unknot. In this project, e mostly orked the unknot, the trefoil and (, p) torus knots. Because of its irregular diagram, e did not pay much attention to the unknot made from sticks. The three- and four-stick unknots had more applicability to our purposes. Figure 7 illustrates the three- and the four-stick unknots. FIG. 7: 3-stick unknot (left) and 4-stick unknot (right). Definition 10. torus knot is a special kind of knot hich lies on the surface of an unknotted torus in R 3. (p, q) torus knot inds p times around a circle inside the torus (hich goes all the ay around the torus), and q times around a line through the hole in the torus (hich passes once through the hole). (, p) torus knot is the one that inds tice around a circle inside the torus and p times around a line through the hole in the torus. (4, 3) torus knot is shon in Figure 8.

6 6 FIG. 8: (4,3) Torus knot.1. Flat-Knotted Ribbons Intuitively, if you take a piece of paper, tie a knot and squash it into the plane, you ill get a folded flat ribbon knot. This is illustrated in Figure.1. folded ribbon follos a path of connected straight line segments, just like a polygonal knot. We use polygonal knot diagrams around hich e construct flat ribbon knots. more formal definition follos. FIG. 9: Flat-ribbon knot. Definition 11. flat-ribbon knot is a rectangle of fixed idth framing a knot diagram of K and satisfying the folloing conditions: The ribbon retains the over-under information of the knot diagram, Each fold behaves like a mirror: the knot and its ribbon are reflected in the fold (see Lemma 1).

7 7 FIG. 10: We disallo a construction ith to identical folds placed adjacent to each other, as this distorts the perception of constant idth. a 1 a D α α F a 3 O E B b 1 b b 3 FIG. 11: The knot behaves like a ray of light reflected in a plane mirror. Extra tists are disalloed (see Figure 10). Figure 11 shos the ay the knot acts like a light ray reflected on the surface of a plane mirror. It obeys one of the major properties of reflection: the angle of incidence equals the angle of reflection. Remark. For a ribbon of idth and fold angle α (0 < α < π/) as shon in Figure 11, lines a 1, a and a 3 are pairise parallel and lines b 1, b and b 3 are pairise parallel. The fold line B is a transversal for the parallel lines a 1,a. Lines a and b represent the core curve and is their intersection point in the fold line B. Note that by construction of the fold, F B = BO. Lemma 1. Given a ribbon of idth and a fold of angle α such that 0 < α < π/ (as shon in Figure 11), OB = OB = D = BE. Proof. s lines a 1, a and a 3 are parallel, F B = D = BO (alternate interior angles). Lines b and b 3 are parallel, therefore, OB = BE (corresponding angles). By Remark, F B = BO, hence the result. We no give a number of useful corollaries to Lemma 1. orollary. Given a ribbon of idth and a fold angle α such that 0 < α < π/ (as shon in Figure.1), the core curve, fold and the transversals form to equal isosceles triangles D and EB as ell as the rhombus DEO.

8 8 a 1 a D α a 3 O E B b 1 b b 3 FIG. 1: ribbon ith a fold of angle α forms isosceles triangles and a rhombus. a 1 a α D a 3 O E B b 1 b b 3 FIG. 13: ribbon ith a fold angle of α creates many line segments ith equal length and many congruent triangles. Proof. Lemma 1 shos that OB = OB = D = BE. The definition of a ribbon tells us that = B. This shos that both D and EB are isosceles. So, D = D and E = EB. No, D is similar to OB and =1/ B. Hence, D = DO and OE = EB. Hence D = DO = OE = EB. Therefore, ODE is a rhombus. orollary 3. Given a ribbon of idth and a fold angle α such that 0 < α < π/ (as shon in Figure 13), the isosceles triangle BO, bounded by the fold and ribbon edges, can be divided into four equal smaller isosceles triangles; each has area 1 4 of the area of the BO. Proof. We have the same information provided as in the previous examples. dditionally, e dra diagonal DE inside the rhombus ODE to form 4 triangles inside the isosceles OB. We have proved already that D = EB. s lines b and b 3 are parallel, D = DE (alternate interior angles). We also kno D = D = E. Thus, D = ED = EB (by S..S). Diagonal DE divides rhombus ODE into to equal triangles. Thus, D = ED = EB = DOE. Finally, the area of each small isosceles triangle is 1 4 rea( OB). We no calculate a number of lengths hich ill be useful in the later sections of this paper.

9 9 a 1 a D α F (90 α) α a 3 O E H 1 H B b 1 b b 3 FIG. 14: Details of a ribbon fold angle α used for the calculation of lengths in Lemma 4. Lemma 4. Given a ribbon of idth folded ith angle α, such that 0 < α < π/ (as shon in Figure.1). The length of the fold B is The idth created by the fold is B = O = sin α. (1) cos α. () The transversals of a ribbon have the same length as the core curve inside the fold are O = OB = D + E = Proof. (1) onsider right triangle H 1 B in Figure.1. B = sin α. sin α. (3) Observe that H 1 =, hence () In the isosceles triangle OB, O bisects O, and is a median and height. Hence, = B = B O =. In right-triangle OB, tan α = sin α B. Hence, O = tan α B = tan α sin α = cos α. (3) In right-triangle OB, cos α = B OB. Hence, OB = O = B cos α = sin α cos α.

10 10 So far in the paper e have just assumed that the ribbon folds ith angle α such that 0 < α < π/. It is interesting to see ho the ribbon behaves hen α is 0 or π/ and ho this compares to Lemma 4. Remark 3. Intuition tells us that hen α = 0 a flat ribbon of constant idth has no folds, so it doesn t make sense to talk about the idth and length of the fold. ccording to Lemma 4, hen α = 0, the idth of the fold O = /. Figure 14 shos that OH = α. s α 0, segment O moves closer and closer to the segment H. Thus, O H = /. Mathematically, lim α 0 sin α =. Note that as α 0, the point B moves to the right along line a 3 and so B (hich is the length of the folded edge) goes to. Remark 4. When α = π/, the ribbon is folded over itself. This means the length of the fold is but it doesn t make sense to talk about the idth of the fold. ccording to Lemma 4, B =, hence hen hen α = π/, B =. s α π/, point O moves to the left sin α along the line a 3 and so O ( the idth of the fold ) goes to. Lemma 5. To consecutive folds sharing a common ribbon edge have their folding angles sum up to π/. α α B β β E O D FIG. 15: To consecutive folds ith α+β = π/. Proof. Given a ribbon of idth as shon in Figure 15, folds of angle α and β are made along EB and BD respectively. ccording to the construction, BE = EBO = α and BD = DBO = β. The angle should add up to π: α + β = π (as these consecutive angles form straight line B). Therefore, α + β = π/.

11 STIK UNKNOTS In this section e consider unknots made from 4 sticks. The diagram of a stick unknot is a bi-gon and it has ribbonlength equal to tice the length of one stick. Thus its ribbonlength can be made arbitrarily small. The ribbon of a 3 stick unknot is topologically a Möbius band. regular diagram of a convex 4 stick unknot is topologically an annulus. Initially, e decided to sho that a rectangular 4 stick unknot framed ith a ribbon hich has a shape of an octagon is minimized into a square 4 stick unknot framed ith a ribbon hich has a shape of a square. This is illustrated in Figure 16. FIG. 16: Flat-ribbons for the 3-stick unknot (left) and 4-stick unknot (right). H G x F P S B Q R E y z D H G x P S F y z D H x P F y z FIG. 17: Minimization of the 4-stick unknot from the shape of the octagon to the shape of a square. Look at the left-most picture in Figure 17. Because the 4-stick unknot is a rectangle (red curve), the folding angle α = π/4: π π/ = π 4. The folds of the ribbon are H, B, DE and GF. There are to congruent isosceles trapezoids, HB = GF ED. Let, P and BQ, ER and F S be the heights of the trapezoids HB and GF ED respectively. By the definition of a flat-ribbon knot, B = 0.

12 1 We can cut out the segments of the ribbon: BQP and GDF E. Points and B ill merge into, P and Q ill merge into P, S and R ill merge into S, and F and E ill merge into F to produce the center picture in Figure 17. The right-most picture in Figure 17 shos that e can reduce the ribbonlength further. In the center figure of Figure 17, segments DSP and HP SG can be cut out, leaving us ith the center picture of Figure 17, here H and G merge into H, and and D merge into. Notice that all the segments of the unknot under the folds are equal. (This is equation (3) in Lemma 4). Hence xy = yz = z = x. Equation 1 of Lemma 4 shos that the length of the folded edges are also equal. By assumption, the fold angles are all π/4 (for example HF = F = π/4). Hence, F H is also a square. Let us no calculate the ribbonlength using Lemma 4 ith folding angle α = π/4. The length of a knot under the fold is / sin α = / sin(π/) =. Since e have four identical folds, the total length of the knot is 4. Therefore, the ribbonlength equals 4. Let us take random shape of a 4-stick unknot and sho, in a similar ay, the that minimum ribbonlength is achieved hen the 4-stick unknot has the shape of a square. J a b E α B π H 1 H π F β d e K D FIG. 18: Random shape of a 4-stick unknot.

13 13 onsidering the top segment of the ribbon ith folded edges B and D. Fold B has folding angle 0 < α < π/4, and fold D has folding angle π/ > β > π/4. Points E and F are the points of intersection inside the ribbon; and BH 1 and F H are heights respectively. Small letters denote lengths of the knot segments. The length of the to knot segments under fold B is a + b, under fold D is d + e. Let c be the length of the segment in beteen b and d, so xy = b + c + d. Let x be the length of the knot segment in beteen BH 1 and F H. Segment x is the part of segment c that may be removed hen minimizing ribbonlength. B = sin α ; E = EB = a + b = sin α. D = The triangle BH 1 E is right. sin β ; F = F D = d + e = sin β. BEH 1 = α EBH 1 = π/ α, sin( EBH 1 ) = EH 1 EB EH 1 = EB sin( EBH 1 ), EH 1 = sin(π/ α) sin α cos α = cos α sin α = cot α In a similar ay, H = ( cot(β)). c = EH 1 + x + H = cot(α) + x + cot(β), and a = b = d = e = sin(α), and sin(β). We ant to minimize a, c and d, for a constant idth, x 0, and $0 α, β < π/.

14 14 Let a = b = g(, α) = sin(α). Since the idth of the ribbon is arbitrary, e need to maximize the denominator sin(α) to minimize the length of a and b. The maximum value for sin(α) = 1, here α = π/4. The same arguement holds for β, hence β = π/4. Let c = f(, α, β, x) = (cot(α) cot(β)) + x. ssume x = 0, e ant c = 0. Therefore, cot(α) = 0, and cot(β) = 0 α = π/4, and β = π/4 α = π/4 and β = π/4 If e repeat the same procedure on the bottom segment of the ribbon shon in Figure 18, e ill get the same result that the folding angles should be π/4. When the folding angles are π/4 the knot and the ribbon around are both squares and the ribbonlength equals 4. FIG. 19: Expanded ribbon of a 4 stick unknot 4. (, P) TORUS KNOTS With the hope of making conjectures about a variety of classes of knots, e decided to explore about the -bridge knots. Formal definitions related to this class of knots follos [1]. Definition 1. Given a projection of a knot to a plane, an overpass is a subarc of the knot that goes over at least one crossing but never goes under a crossing.

15 15 Remark 5. maximal overpass is an overpass that cannot be made any longer. It forms a bridge over the rest of the knot. Definition 13. The bridge number of the projection is the number of maximal overpasses in the projection. The bridge number of a knot K, denoted by b(k), is the least bridge number of all of the projections of the knot K. Definition 14. Knots that have bridge number are a special class of knots, knon as -bridge knots. If e cut a -bridge knot along the projection plane, e ould be left ith to unknotted untangled arcs from the knot above the plane, corresponding to the to maximal overpasses, and to unknotted untangled arcs from the knot belo the plane. We looked at a special case of to-bridge knot hich e initially, for our purposes, named as n tist knots. The primary knot that e dealt ith is shon in Figure 0. FIG. 0: The 5-tist knot or (, 5) torus knot. This is a 5-tist knot as it has a total of 5 half-tists. To find the minimum ribbonlength of this knot, e first constructed it ith to strands of paper having constant idth and then joined them together at the end. We did this by eaving or tining one strand of paper ith the other and joining the ends together. We had to be careful ith the overlapping and underlapping of the strands of paper as e joined the ends together in the end. more defined description of the construction follos. s shon in the Figure 1, e placed the strand of paper on our left hand on top of the strand of paper on our right hand. This no creates a cross ith an intersection point of the core curves of the to strands. Let us label the ends of the paper strands after e have put them in the cross position as north-east, north-est, south-east and south-est ends. To construct the knot, first, the south-east end of the cross as brought over the intersection of the cross to make a fold. Next, the south-est end of the paper as brought over at the intersection point to make another fold. Note that by no e already have 3 tines of the strands of papers. To add to more tines, the south-east end of the paper (hich is no at the north-est end) is brought back to the south-east end. This step has created the fourth tine. To make the fifth tine, the south-est end of the paper (hich is no at the north-east end) is brought back to the south-est end. fter this, the ends are

16 16 FIG. 1: onstruction of the 5-tist knot. joined by folding them backards. The south-east end is joined ith the south-est end and the north-east end is joined ith the north-est end. During our research, e realized that these tist-knots are actually the (, p) torus knots. More precisely, our 5-tist knot as a (, 5) torus knot. Note that all (, p) torus knots can be constructed by the method described above. Given a ribbon of length constructed as shon in Figure 1, e kno that in the section at hich the strands join together, the core curves cross each other at an angle of π/ ( NOM). detailed knot diagram of this setting is found in Figure. Lemma 6 shos that the shortest length of the hypotenuse NM is and OM.

17 17 I O J V π X R B N G S P Q T H M E D W Y FIG. : The loer half of the flat-ribbon constructed as shon in Figure 1. Remark 6. Figure 3 illustrates the result e get by tightening the flat-knotted ribbon in Figure. s e make our knot tighter, e cannot move B more forard than hat is shon in Figure 3 In other ords, B (the top edge of the ribbon for NM) does not go beyond O (the point of intersection of the knots). If it did, the ribbons ould fold over themselves at points I and J, contradicting the definition of a fold. (In the tight configuration, point merges ith point J and B ith I.) Since = R = RB = B = and triangle B is a right-angled, B =. Lemma 6. Let a ribbon of idth satisfy the conditions described above, in Remark 6, and as illustrated Figure. Then OMN is a right isosceles triangle and NM = and OM = ON =. Proof. The Figure shos the bottom half of the knot that e constructed by the description mentioned above. In OMN, B = OM + GB + H. From Remark 6, B =. Since GB = H, = OM + GB. Hence; OM = GB. If e look at BSN, SBN = 3π 8, tan 3π 8 = GN GB. Hence,

18 18 I J B O π/ N M E R D FIG. 3: Tighter version of Figure. GB = cot 3π 8. It follos that OM = (1 1 cot 3π 8 ) =. Remark 7. Note that OM in Figure is decreasing as e are making the knot tighter. But as this happens the tightest form of the knot e can get to is Figure 3. Fold BE cannot move further toards fold BR, else e ould fold over a fold e already have, contradicting the definition of flat-knotted ribbons. This means in particular, that ED cannot go to zero. Mathematically, ED = OM cot 3π 8. Replacing OM =, e find ED = ( ) > 0. Proposition 7. The ribbonlength for the given (,5) torus knot ( 5 tist-knot ) having idth is ( + 6). Proof. To calculate the total ribbon length, notice that the total length of the core curve at the bottom half of the knot as illustrated in Figure and 3 is: ON + NM + MO + OI = Similarly, the total length of the core curve of the top half of the knot ill be the same: The remaining folds in the middle have the length +. Therefore, the total ribbonlength is the total of the top, bottom and the middle sections of the knot is ( + 6).

19 19 Proposition 8. The ribbonlength for a given (,3) torus knot or the 3 tist-knot having idth is ( + 4). Proof. The proof is the similar to the proof for Proposition 7; except the trefoil does not have the middle section. Therefore, the ribbon length ( + 4). Theorem 9. The ribbonlength for a given (, p + 1) torus knot or the ith idth is ( + p + ). Proof. We use the method of induction to prove the above theorem. Base case (hen p = 1) is the (, 3) torus knot. In Proposition 7 e found that the ribbonlength for the (, 3) torus knot is ( ) = + 4. When p =, e have the (, 5) torus knot. Its ribbonlength is ( + + ) = ( + 6) (true from Proposition 8). Let us assume that our given statement is true for p = k. This means the ribbon length for (, k + 1) torus knot is ( + k + ). We no need to sho that for p = k + 1, i.e., for (, k + 3) torus knot, the ribbon length ( + (k + 1)+) = ( + k + 4). For the induction step p = k + 1, e construct a (k + 3) torus knot by starting ith a (k + 1) torus knot constructed as above. Then, e fold to more times in the middle section. Thus, the ribbonlength of this torus knot diagram is more units than the (, k + 1) torus knot. In other ords, the ribbonlength of the (, k + 3)-torus knot, is ( + k + ) +. But, ( + k + ) + = ( + k + 4), hence the result. Note that according to [] the ribbonlengths for (, 3) torus knot and (, 5) torus knot ccording to Theorem 9 the ribbonlengths for (, 3) torus knot and (, 5) torus knot In this ay, e can keep increasing p to calculate and compare the ribbonlengths of (, p + 1)- torus knots given by [] and our paper. s seen above for (, 3)- and (, 5)-torus knots, e ill find that the ribbonlengths given by the formula in our paper is smaller than the ribbonlength given by []. 5. FUTURE WORK We ould like to extend our ork for the 4-stick unknots. We ould like to prove that the repetition of the procedure on the 4-stick unknot as shon in Section 3 indeed leads to the same result that the folding angles should be π 4. Further, e ould like to extend our research to a ider variety of flat-knotted ribbons. 6. KNOWLEDGMENTS We ould like to thank the Smith SURF program hich funded our research in Summer 009 and the Schultz Foundation for funding the travel expenses. We give special thanks to Sarah Meyer

20 0 ho as also a SURF student on this project and ho took part in the research discussions described in this report, particularly those on the 3 stick unknot. Some of the figures came from [6], [7]. We ould also like to thank our advisor Prof. Elizabeth Denne for her immense support and guidance throughout. [1]. dams. The Knot Book. W.H Freeman and ompany (1994) [] B. Brennan; T.W. Mattman; R. Raya; D. Tating Ribbonlength of torus knots. J. Knot Theory Ramifications 17 (008), no. 1, [3] P. romell. Knots and Links ambridge University Press (004). [4] L. Kauffman. Minimal flat knotted ribbons. Physical and numerical models in knot theory , Ser. Knots Everything, 36, World Sci. Publ., Singapore (005). [5]. Livingston. Knot Theory. Mathematical ssociation of merica (1993). [6] Department of Mathematics, ornell University. [7] Wikipedia ommons.