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Contemporary Mathematics

Characterizing Polygons in R3 Jorge Alberto Calvo Abstract. For a given positive integer n ≥ 3, the collection of n-sided polygons embedded in three-space, together with their deformations, gives a “geometric” theory of knots. In fact, considering (i) deformations which change the length of the edges of the polygons, versus (ii) deformations which preserve these lengths in equilateral polygons, yield two distinct theories. Here we present a series of open questions pertaining to these geometric knot theories. Some are broad in scope and lie at the core of the theory. Others are more computational, and present the next steps in a program towards understanding the geometric nature of these theories.

Suppose that we take a collection of n line segments and glue them end to end, forming an embedded piecewise linear loop in R3 . The simple closed curve formed by the union of the line segments will be a polygonal realization of some (possibly trivial) knot K. By taking a large enough number of segments, we can construct polygonal representatives of any knot type. However, if we look at polygons with only a small number of edges, then we will restrict the types of knots we can construct, in one sense obtaining only the simplest of knots. By insisting that all deformations of these knotted polygons preserve both the piecewise linearity and the number of edges of a knot, we obtain a stricter notion of knot equivalence than traditional knot theory allows. We will call two n-sided polygons “geometrically equivalent” if one can be transformed into the other by a deformation of this type. The additional requirement that the lengths of the edges of a polygon remain constant throughout a deformation produces yet another “knot theory.” In the specific case of equilateral polygons, we say two knotted polygons are “equilaterally equivalent” if there is a deformation transforming one knot into the other which only passes through equilateral polygons. The two fundamental questions that we ask when faced with these “geometric knot theories” are: Question 1. What is the smallest number of edges required to build a polygon of a particular knot type K? Question 2. How many “different” polygons with a given number n of edges represent K? 1991 Mathematics Subject Classification. 57M25. Key words and phrases. Polygonal knots, space polygons, knot spaces. c

2002 Jorge Alberto Calvo

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In the following pages we present several approaches for answering these questions. Most of the results at hand are preliminary in nature and prompt more questions than they answer. One of the most intriguing approaches to finding a (partial) answer to Question 1 involves finding relationships between the number of edges required to build polygonal representatives of a particular knot type and other classical knot invariants such as the bridge number or the superbridge number. We begin by taking a closer look at the association between a knot’s crossing number and its “polygonal index.” Let K be a knot. Recall that its minimal crossing number c(K) is defined as the least number of crossings in any generic projection of K to a plane or sphere. In a similar way, we define the minimal polygonal index p(K) of a knot K as the smallest number of segments in any polygonal realization of K. If a minimal polygon representing K is projected on a plane, then each edge can cross every other edge except its two neighbors. Thus a quick computation reveals that p(K)(p(K) − 3) . 2 By choosing a plane perpendicular to an edge, we can restrict our attention to projections with no more than p − 1 edges. Thus (1)

c(K) ≤

(p(K) − 1)(p(K) − 4) . 2 By solving (2) for p and utilizing results from graph theory, Negami [6] obtains the following well-known inequalities: p 5 + 8 c(K) + 9 (3) ≤ p(K) ≤ 2 c(K). 2 The upper bound in (2) is hardly ever minimal. In fact, if p is odd, a (p − 1)sided knot projection will actually have at most 12 (p − 2)(p − 5) + 2 crossings. The situation can be slightly improved (at least for knots with small polygonal indices) by the following theorem, which we prove by considering generic projections to spheres.

(2)

c(K) ≤

Theorem 1. [4] For any knot K, (4)

c(K) ≤

(p(K) − 3)(p(K) − 4) . 2

Consequently, (5)

7+

p

8 c(K) + 1 ≤ p(K) ≤ 2 c(K). 2

Proof. Let P = hv1 , v2 , v3 , . . . , vn i be a polygonal realization of the knot K. By relabeling the vertices of P in sequence, if necessary, we can assume that v 1 is a point on the boundary of the convex hull spanned by the vertices of P. In this case, we can find a plane P1 which intersects P only at the vertex v1 , with P lying entirely on one side of P1 . Let S be a large sphere centered at v1 and enclosing all of P, and consider the image of the radial projection π : P − {v1 } → S. By our choice of v1 , this image lies entirely in a hemisphere of S cut by the equator S ∩ P1 . Furthermore, note that the interiors of edges v1 v2 and v1 vn are respectively mapped to the single

CHARACTERIZING POLYGONS IN R3

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points π(v2 ) and π(vn ). Thus, by picking a generic polygon P , we can assume that Γ = π(P − {v1 }) is a graph consisting of a chain of n − 2 great circular arcs on S intersecting in four-valent crossings. Suppose that Γ has c crossings. Since Γ is contained in a single hemisphere of S, a pair of its arcs will intersect at most once. Furthermore adjacent arcs cannot intersect, so each one of the n − 4 interior arcs π(v3 v4 ), . . . , π(vn−2 vn−1 ) can intersect at most n − 5 other arcs, while each of the extreme arcs π(v2 v3 ) and π(vn−1 vn ) can intersect at most n − 4 other arcs. Hence 1 (n − 3)(n − 4) . (6) c≤ (n − 4)(n − 5) + 2(n − 4) = 2 2 Unfortunately, Γ is not a knot projection for K. However, we can deform K in a small neighborhood of v1 so that we do obtain a (spherical) knot projection for K with no more than 21 (n − 3)(n − 4) crossings. Let > 0 be small enough that the closed -ball B centered at v1 intersects the polygon P in exactly two small segments of the edges v1 v2 and vn v1 , as shown in Figure 1(a). Suppose that the edge v1 v2 intersects the sphere ∂B at the point

(a)

(c)

(b)

(d)

Figure 1. Deformation of P inside a small -ball about v1 .

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JORGE ALBERTO CALVO

q1 . Furthermore, let q2 be the point where the equator ∂B ∩ P1 intersects the half-plane containing v2 and bounded by the line determined by v1 and v3 . Then we can deform the segment q1 v1 so that it curves along a great circle path α1 from q1 to q2 and then proceeds in a straight line path to v1 . See Figure 1(b). Note that since the arc α1 lies on the same plane through v1 as v2 v3 , then π(α1 ) ∪ π(v2 v3 ) forms a single great circle trajectory on S from π(v3 ) to π(q2 ). Thus, after this deformation, the upper bound on the total number of crossings given in (6) still holds. Similarly, let q3 be the point of intersection between the equator ∂B ∩ P1 and the half-plane which contains vn and is bounded by the line determined by v1 and vn−1 , and let q4 be the point at which the edge vn v1 intersects ∂B . Then the segment v1 q4 can be deformed so that it travels in a straight line path from v1 to q3 and then curves along a great circle path α3 from q3 to q4 . See Figure 1(c). As before, the arc α3 lies on the same plane through v1 as vn−1 vn , so π(vn−1 vn )∪π(α3 ) forms a single great circle trajectory on S from π(vn−1 ) to π(q3 ). Therefore the upper bound in (6) still holds after this deformation. Finally, isotope P by moving v1 into the interior of the triangle 4q2 v1 q3 while curving the segments q2 v1 and v1 q3 until they coincide with a single arc α2 along the equator ∂B ∩ P1 , as in Figure 1(d). This final transformation turns P into a non-polygonal embedding of the same knot type; the new embedding agrees with P outside of the ball B but completely avoids its interior. In the meanwhile, the image under π of this embedding is a spherical knot projection Γ0 consisting of the n − 2 arcs of Γ (with its ends extended by π(α1 ) and π(α3 )), together with an (n − 1)th arc π(α2 ) running along the equator S ∩ P1 and joining the endpoints π(q2 ) and π(q3 ). Since Γ is contained entirely on one side of the equator, π(α2 ) does not cross any other arcs. Hence the new projection has no more crossings than it did before the last deformation. Supposing that P is a minimal polygon representing K, this proves (4). Completing the square and solving for p(K) then gives the lower bound in (5). The lower bound in (5) does not provide much of an improvement over (3); the difference is always less than one. On the other hand, (4) does provide some useful information, at least for small polygonal indices. For instance, if p(K) ≤ 5 then c(K) ≤ 1. Since the minimal crossing number of any non-trivial knot is at least three, this shows that any quadrilateral or pentagonal configuration must be unknotted. Furthermore, since the hexagon with coordinates h(1, 0, 0), (4, 0, 0), (1, 6, 2), (0, 2, −5), (5, 2, 5), (4, 6, −2)i is a realization of a right-handed trefoil, then we have p(31 ) = 6. On the other hand, if p(K) = 6 then c(K) = 3. Thus, every hexagonal knot is either a trefoil or an unknot. Since the coordinates h(1, 0, 0), (7, 0, 0), (3, 4, 1), (7, 4, −2), (4, 1, 2), (0, 5, −7), (6, 5, 0)i give a heptagonal realization of a figure-eight knot, then we also have p(4 1 ) = 7. For polygonal indices higher than 6, the upper bound in (4) becomes less and less effective, as the rigidity of the straight edges prevents the construction of knots with high crossing number. For example, even though the upper bound does not exclude the possibility for heptagonal knots with crossing number 5 or 6, these are in fact impossible to construct. Let c(n) be the largest minimal crossing number


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Table 1. A comparison between the predicted and observed crossing numbers among knots with polygonal index p or less. p 3 4 5 6 7 8 9

(p−3)(p−4) 2

0 0 1 3 6 10 15

c(p) 0 0 0 3 4 8 ???

observed among all of the knots with polygonal index p or less. Table 1 gives the values predicted by the upper bound above together with the actual values of c(p) for each polygonal index p. Question 3. Is there a better upper bound for c(p)? Notice that all of the upper bounds presented are Θ(p2 ). 1 Theorem 8.2 in [1] asserts that the polygonal index for the (n, n − 1)-torus knot is p = 2n; since this knot has crossing number c = n(n − 2), then c(p) must grow like a quadratic at least when p is even. Question 4. Is c(p) = Θ(p2 )? To show that the figure-eight knot is the only knot with polygonal index seven, we consider the spherical projections Γ0 obtained in the proof of Theorem 1. Suppose that K is a heptagonal knot with minimal crossing number 5 or 6. Using the same notation as in the proof of Theorem 1, we can deform K in a neighborhood of v1 so that its projection Γ0 on S consists of six great circular arcs, one of which forms part of the equator S ∩ P1 . By bumping π(α2 ) slightly up from the equator and projecting from v1 to a plane parallel to P1 , we can view Γ0 as a planar diagram consisting of five straight segments and one large circular arc. Figure 2 shows one such configuration. Up to mirror images, relabeling of vertices, and choice in over-crossings, there are only four such diagrams obtained in this way having five or six crossings. Careful scrutiny over the possible configurations shows that only unknots, trefoils, and figure eight knots are possible. In particular, the configuration in Figure 2 is impossible since its edges weave over and under each other more than piecewise linearity allows. To see this, let P345 be the plane defined by v3 , v4 , and v5 . Since edge v5 v6 crosses over v3 v4 , vertex v6 must live above P345 . Similarly, v2 must lie above P345 since v2 v3 crosses over v4 v5 . This means that edge v6 v7 starts above the plane P345 at v6 , intersects the plane once before crossing underneath v3 v4 , and then intersects the plane a second time before crossing over v2 v3 . It is fairly simple to build every five or six crossing knot (including the square and granny knots), as well as the knots 819 and 820 , out of eight segments. We will show that these are the only possible octagonal knots. In particular, the knot 8 18 pictured in Figure 3 is impossible to construct. We begin with two lemmas. 1Θ(f (n)) refers to the class of functions that is bounded both above and below by a constant multiple of f (n). Thus, g(n) = Θ(n2 ) if there are constants b1 , b2 for which b1 n2 ≤ g(n) ≤ b2 n2 for all n.

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Figure 2. Planar projection of the graph Γ0 corresponding to a heptagonal knot 62 .

Figure 3. This octagonal embedding of the knot 818 cannot be constructed with straight edges. Lemma 2. [2] (i) If K is a non-trivial octagonal knot, then it is one of the following knots: 31 , 41 , 51 , 52 , 61 , 62 , 63 , 31 + 31 , 31 − 31 , 818 , 819 , or 820 . (ii) If K is an octagonal 818 knot, then its associated graph Γ0 must look (up to a mirror image reflection or a reversal in the order of the vertex labels) like the one pictured in Figure 4. Proof. The details, which appear in [2], consist of an enumeration of all possible configurations of graphs Γ0 leading to knots with crossing number greater than 7, and the elimination of all those which are clearly impossible. Of those that remain, only the configuration in Figure 4 leads to a knot which has not been previously observed as an octagon.


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Figure 4. Planar projection of the graph Γ0 corresponding to an octagonal knot 818 . Lemma 3. If the polygonal index of the knot 818 is eight then we can construct an immersed octagon G0 = hw1 , w2 , w3 , . . . , w8 i with exactly seven transverse selfintersections, as indicated below:

(7)

w2 w3 w7 w8 w4 w5 w1 w2 w6 w7 w3 w4 w8 w1

intersects

w5 w6 w2 w3 w7 w8 w4 w5 . w1 w2 w6 w7 w3 w4

Additionally, we can arrange it so the two quadruples of points {w1 , w5 , w6 , w8 } and {w1 , w2 , w4 , w8 } are in general position. Proof. Consider an octagonal realization G = hv1 , v2 , v3 , . . . , v8 i of the knot 818 . By Lemma 2, we can assume that the vertices of G are labelled as in the projection in Figure 4. Notice that both v4 v5 and v5 v6 cross under the edge v2 v3 . Thus both of these edges intersect the interior of the triangular disc 4123 . Similarly, both v2 v3 and v3 v4 cross under the edge v7 v8 , so they both pass through the interior of 4781 . Replace v8 v1 v2 by a single edge v8 v2 . Then v4 v5 must cross under v8 v2 ; otherwise let P be the plane containing v7 , v8 , and v2 . Then v2 v3 will have to bend as it leaves the plane P at v2 , travels above this plane when it crosses over v4 v5 , and then ducks below the plane as it crosses under v7 v8 . See Figure 5(a). Notice that this means that v3 v4 also crosses under v8 v2 ; otherwise v4 will lie above the plane P and v4 v5 will have to bend as it leaves v4 , travels below P as it passes under v8 v2 and v2 v3 , and then jumps above P as it crosses over v7 v8 . See Figure 5(b). Therefore both v3 v4 and v4 v5 cross through the interior of the triangular disc 4812 . By shifting the vertex labels and projecting from v1 once again, we obtain the same picture with the same labels. In particular, shifting by s2 : hv1 , v2 , v3 , . . . , v8 i 7→ hv3 , . . . , v8 , v1 , v2 i

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will replace v1 by v3 , which clearly lies on the boundary of the convex hull of the vertices. See Figure 4. The label shift replaces the disc 4781 with 4123 and the two-edge linkage v2 v3 v4 with v4 v5 v6 , so that we see the same two edges intersect the same triangular disc. Therefore if we project from v3 , we will obtain exactly the same graph Γ0 with exactly the same set of intersections. This means that v8 v1 and v1 v2 cross through the interior of 4567 while v1 v2 and v2 v3 cross through the interior of 4678 . Continuing in this fashion, we see that in G:

(8)

v4 v5 , v5 v6 , v6 v7 , v7 v8 , the edges v8 v1 , v1 v2 , v2 v3 , v3 v4 ,

v5 v6 v6 v7 v7 v8 v8 v1 v1 v2 v2 v3 v3 v4 v4 v5

4123 4234 4345 4456 intersect the open disc 4567 4678 4781 4812

transversely.

Consider the deformation Gt = hv1 , (1 − t)v2 + tv1 , v3 , . . . , v8 i which starts at G0 = G and pushes v2 along a straight-line path towards v1 . This homotopy will give embedded octagons of the same knot type as G until it introduces a selfintersection, say at time t = a. Let w2 be the second vertex in Ga . Note that Gt will be a realization of the knot 818 for every t ∈ [0, a), and therefore will have the same intersections as indicated in (8). Furthermore, since Ga is arbitrarily close to an embedded 818 , it, too, will have the same intersections as in (8), except of course for w2 v3 which will intersect v4 v5 , v5 v6 , or both. See Figure 6. Suppose that w2 v3 intersects v4 v5 , so that v5 lies on the same plane as w2 , v3 , and v4 , on the exterior of the triangle 4234 . Recall that for t ∈ [0, a), the edge v5 v6 must pierce through the interior of 4234 . Thus in Ga , this edge must intersect the closure of this disc. This implies that w2 , v3 , v4 , v5 , and v6 are all coplanar. See Figure 6(b). However, this implies that in G, edge v5 v6 does not pierce the interior of 4234 , giving us a contradiction. Therefore Ga has a single self-intersection between edges v2 v3 and v5 v6 .

(a)

(b)

Figure 5. Two impossible configurations.


(a)

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(b)

Figure 6. The edge w2 v3 will intersects either v5 v6 or v4 v5 . We can continue performing local deformations. Arguing as above, we can show that a homotopy which pushes v2 v7 v4 v1 v6 v3 v8

v1 v6 v3 toward v8 v5 v2 v7

will create a single intersection between the edges

v2 v3 , v7 v8 , v4 v5 , v1 v2 , v6 v7 , v3 v4 , v8 v1 ,

v5 v6 v2 v3 v7 v8 v4 v5 . v1 v2 v6 v7 v3 v4

Furthermore, performing these homotopies in sequence will not remove any of the pre-existing self-intersections, and each of these homotopies will preserve the intersections in (8), with the caveat that (with the exception of v8 v1 ) edges on the left-hand column actually intersect the boundary of the corresponding disc. Let wi denote the final position of the ith vertex after we move it, i.e. the position at which it first creates a self-intersection; also, let w1 = v1 . Then, once we have performed all seven deformations, we will have introduced the seven desired self-intersections. Finally, note that since the deformations preserved the transversality in (8), w8 w1 and w3 w4 still have transverse intersections with 4456 and 4812 , respectively. The first of these intersections occurs in the interior of the triangle, so {w1 , w5 , w6 , w8 } is in general position. The second of these intersections occurs along the edge w8 w1 , so {w1 , w2 , w4 , w8 } is also in general position. We now appeal to the contrapositive of this lemma, showing that such a singular octagon – and hence the embedded octagonal 818 shown in Figure 3 – is impossible to construct. Theorem 4. p(818 ) = 9. Proof. First of all, we note that it is a trivial exercise to construct a ninesided knot 818 from Figure 3 by adding a vertical edge at one of the outer “points” of the star. Thus p(818 ) ≤ 9. Suppose, then, that p(818 ) = 8 and let G0 be the immersed octagon from Lemma 3. Since the quadruple of vertices {w1 , w2 , w4 , w8 } is in general position, w1 −w4 , w2 −w4 , and w8 −w4 are linearly independent vectors. Thus, by translating

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in R3 and applying an appropriate linear transformation, we can assume that w4 = (0, 0, 0), w1 = (1, 0, 0), w2 = (0, 1, 0), and w8 = (0, 0, 1). Recall that w3 w4 intersects w8 w1 . Thus w3 must lie in the plane containing w4 , w1 , and w8 : y = 0. Therefore w3 = (a, 0, b) for an appropriate choice of parameters a and b. Similarly, w4 w5 intersects w1 w2 , so w5 must lie in the same plane as w4 , w1 , and w2 : z = 0. This means w5 = (c, d, 0) for an appropriate choice of parameters c and d. Now w7 w8 intersects w4 w5 so w7 must lie in the plane containing w4 , w5 , and w8 : d x = c y. Simultaneously, w7 w8 intersects w2 w3 so w7 must lie in the plane containing w2 , w3 , and w8 : (1 − b) x + a y + a z = a. Thus for some parameter e, we have w7 = (ace, ade, 1 − ce + bce − ade). Finally, consider the coordinates for the vertex w6 . Since w5 w6 intersects w2 w3 , w6 must lie in the plane containing w2 , w3 , and w5 : (bd − b) x − bc y + (a − ad − c) z = −bc. Additionally, w6 w7 intersects w1 w2 , so w6 must also lie in the plane containing w1 , w2 , and w7 : (1 − ce + bce − ade) x + (1 − ce + bce − ade) y + (1 − ace − ade) z

= 1 − ce + bce − ade.

Finally, w6 w7 also intersects w3 w4 , so w6 must lie in the plane containing w3 , w4 , and w7 : bde x + (1 − ce − ade) y − ade z = 0.

It is then easy to check that these three equations are linearly independent and hence determine the coordinates for w6 . Now consider the plane containing w1 , w5 , and w8 : d x + (1 − c) y + d z = d. The matrix   bd − b −bc a − ad − c −bc 1 − ce + bce − ade 1 − ce + bce − ade 1 − ace − ade 1 − ce + bce − ade      bde 1 − ce − ade −ade 0 d 1−c d d

has determinant zero and rank 3, so the fourth row is a linear combination of the previous three. However this means that w6 lies in the same plane as w1 , w5 , and w8 . But this contradicts the condition from Lemma 3 that {w1 , w5 , w6 , w8 } is in general position. Therefore neither G0 nor any octagonal 818 can be constructed. As we mentioned earlier, we are sometimes interested in restricting our attention to equilateral polygons. This leads to a new notion of polygonal index, which we might call the equilateral polygonal index of a knot K, pEqu (K), equal to the


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least number of edges required for an equilateral representation of K. Immediately we obtain (9)

p(K) ≤ pEqu (K).

No examples are known for which equality does not hold. However, we do have (approximate) coordinates for equilateral hexagonal trefoils and equilateral heptagonal figure eight knots. Since the level of inaccuracy in these approximations is many times smaller than what would be required to change knot type, we know that our examples are very near a real equilateral knots. See, for example Proposition 3 and Corollary 2 in [5]. Thus, the best hope of finding a simple counterexample to the equality in (9) falls in the world of octagons. Here we do have equilateral examples for nearly all octagonal knot types except 819 . Question 5. What is pEqu (819 )? Question 6. Are there other simple examples of knots whose equilateral polygonal index is strictly greater than their polygonal index? Perhaps one can obtain a thorough enough analysis of octagonal 819 ’s to fashion an argument similar to the one used for 818 above. However, this argument must be significantly more complex, since edge length will have to play a significant role. In order to answer the second of our fundamental questions, we turn to the notion of “moduli spaces.” (See, for example, [7] in this volume). The main idea here is to identify each n-sided polygon with the 3n-tuple of vertex coordinates which define it. This gives a bijective correspondence between points in R3n and (potentially immersed) polygons in R3 . The collection of polygons which are not embedded corresponds to a closed semi-algebraic set Σ(n) . More specifically, Σ(n) is the closure of the union of a finite number of pieces, each described as the locus of a system of equations and strict inequalities corresponding to a single intersection between two non-adjacent edges; for example, those polygons hv1 , v2 , . . . , vn−1 , vn i in which v1 v2 intersects v3 v4 can be described by the system (v2 − v1 ) × (v3 − v1 ) · (v4 − v1 ) = 0

(v2 − v1 ) × (v3 − v1 ) · (v2 − v1 ) × (v4 − v1 ) < 0

(v4 − v3 ) × (v1 − v3 ) · (v4 − v3 ) × (v2 − v3 ) < 0. By excluding these singular points, we are left with an open set Geo(n) = R3n −Σ(n) , the points of which correspond to embedded polygons in Rn . Paths in Geo(n) determine isotopies of polygons, so path-components are in bijective correspondence with the geometric knot types available with n edges. We call Geo(n) the space of geometric knots. Let Equ(n) be the subspace of Geo(n) corresponding to embedded polygons having all unit-length edges. This space is the preimage of the smooth map 2 2 2 2 (10) hv1 , v2 , v3 , . . . , vn i 7→ |v1 − v2 | , |v2 − v3 | , . . . , |vn−1 − vn | , |vn − v1 | at the regular value (1, 1, . . . 1), and hence forms a smooth 2n-dimensional manifold whose path-components correspond to the equilateral knot types present with n edges. We call this the space of equilateral knots.

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In this framework, then, Question 2 can be answered by determining the number of homotopy classes of maps from S 0 to one of these spaces mapping 1 to a given point (say, the standard equilateral planar n-gon). Question 7. Compute π0 (Geo(n) ) and π0 (Equ(n) ) for all n. Since every triangle, quadrilateral, and pentagon can be deformed into a standard unknot with the appropriate number of edges, Geo(n) is connected for n ≤ 5. These deformations can be completed without ever changing the length of any of the edges, so Equ(n) is also connected for n ≤ 5. In [3], we show that both Geo(6) and Equ(6) have a total of five components, corresponding to one geometric type of unknot, two geometric types of right-handed trefoils, and two geometric types of left-handed trefoils. In fact, we find that two hexagons are equilaterally equivalent exactly when they are geometrically equivalent. Thus Equ(6) intersects each component of Geo(6) exactly once. However, the following theorem implies that this correspondence is not a homotopy equivalence, since the inclusion i : Equ(6) ,→ Geo(6) has a nontrivial kernel in fundamental group. Theorem 5. Let T be a component of Geo(6) consisting of trefoils. Then π1 (T ) = Z2 while π1 (T ∩ Equ(6) ) is infinite. Proof. (i) Note that mirror reflection and sequential relabeling of vertices give isometries of both Geo(n) and Equ(n) , so we can freely choose which component T we wish to investigate. Suppose, then, that H = hv1 , v2 , v3 , v4 , v5 , v6 i is a hexagonal right-handed trefoil in T . We can translate H through R3 so that v1 lies at the origin. A solid rotation of R3 then can place v3 on the positive x-axis and v5 on the upper-half xy-plane. By relabeling vertices and repositioning H if necessary, we can assume that v2 lies above the xy-plane, so that its z-coordinate is positive. In this case, it is an easy exercise to show that v4 and v6 also have will have positive z-coordinates; otherwise, we can find an open triangular disc 4(i−1)i(i+1) which does not intersect H and can thus provide an isotopy deforming this trefoil into an unknotted pentagon. Through combinatorial argument (see Corollary 10 in [3]) we can also show that, since H is right-handed, (11)

v4 v5 the edge v6 v1 v2 v3

intersects the open discs

4612 , 4123 4234 , 4345 4456 , 4561

transversely;

furthermore, these are the only intersections between edges and triangles in H. Any geometric deformation of H must keep v2 , v4 , and v6 above the xy-plane; thus any trefoil in the same component of Geo(6) as H can also be placed, through solid motions of R3 , into this standard position, with v1 on the origin, v3 on the positive x-axis, v5 on the upper-half xy-plane, and all even-index vertices above this plane. Furthermore, the intersections described by (11) will hold for all trefoils in T . (ii) The group action by solid motions in R3 gives T the structure of a principal bundle with fibre R3 o SO(3). Placing hexagons in the standard position described in (i) above gives a section σ : T /R3 o SO(3) → T into the bundle, so that π1 (T ) ∼ = π1 (T /R3 o SO(3)) ⊕ π1 (R3 o SO(3)) ∼ = π1 (T /R3 o SO(3)) ⊕ Z2 .

Suppose that Ht : [0, 1] → T is a loop contained in the image of the section σ. Let At be the linear transformation taking v3 (t) to (1, 0, 0), v5 (t) to (0, 1, 0), and


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Figure 7. Ht = h(0, 0, 0), (0, 0, 1), (1, 0, 0), v4(t), (0, 1, 0), v6 (t)i. v2 (t) to (0, 0, 1). Left multiplication by sAt + (1 − s)I describes a deformation of R3 taking Ht into a loop of hexagons with v1 = (0, 0, 0), v2 = (0, 0, 1), v3 = (1, 0, 0), and v5 = (0, 1, 0), as in Figure 7. Since this transformation is non-degenerate, we can assume that Ht = h(0, 0, 0), (0, 0, 1), (1, 0, 0), v4(t), (0, 1, 0), v6 (t)i. Let φ6 be the smallest angle which the xz-plane makes with the plane through the x-axis determined by v6 (t), and set P6 to be the plane at this angle. Then the edge v5 v6 intersects this plane for all t. Since v2 v3 is the only edge which pierces through the interior of the triangular disc 4561 , we can deform Ht by pushing v6 → along the ray − v− 6 v5 until it lies on P6 . Thus we can assume that Ht has v6 (t) on the plane P6 for all t. Let λ6 be the largest value of kv6 (t)k in the loop. For each t, we → can push v6 along the ray − v− 1 v6 until kv6 (t)k = λ6 , so that we can assume that v6 (t) only moves along part of some circular arc on P6 . Finally, let ω6 be the smallest measure of the angle ∠v3 v1 v6 . We can move v6 (t) inside the plane P6 along the circle of radius λ6 about the origin until ]v3 v1 v6 = ω6 . In particular, there is no danger that v1 v6 intersect v3 v4 since the former is contained in P6 and the latter crosses P6 only at v3 . Therefore, we can assume that v6 (t) remains constant for all t, so that the loop Ht consists of trefoils with five fixed vertices. Now, let φ4 be the largest angle which the xz-plane makes with the plane containing v4 (t) and the x-axis, and let P4 be the plane realizing this angle. For → each t, v4 (t) can be pushed along the ray − v− 5 v4 until v4 (t) lies on P4 . Let λ4 be the smallest value that the function kv4 (t)k takes on. Then v4 (t) can be pushed in a straight-line path towards the origin until kv4 (t)k = λ4 for all t. This deformation will create no self-intersections since the edges v3 v4 and v4 v5 will move through the interior of the triangular discs 4134 and 4145 , respectively, but Ht never pierces either of these discs. Thus, we can assume that v4 (t) moves along a piece of some circular arc on P4 . A contraction of this arc to a point will fix v4 (t) for all values

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of t, so that Ht is a null-homotopic loop in T . Therefore π1 (T /R3 o SO(3)) ∼ =1 and π1 (T ) ∼ = Z2 , as claimed. (iii) Suppose that H = hv1 , v2 , v3 , v4 , v5 , v6 i is an equilateral hexagonal righthanded trefoil in T which has been placed in standard position, as in (i). Let a, b, and c be the distances between v1 and v3 , v3 and v5 , and v5 and v1 , respectively. Furthermore, set p d(a, b, c) = 2a2 b2 + 2a2 c2 + 2b2 c2 − a4 − b4 − c4 . Let α, β, and γ be the dihedral angles between the xy-plane and the triangles 4123 , 4345 , and 4561 , respectively. Note that since the even vertices of H live above the xy-plane, these angles take values in the interval (0, π). Then H can be parameterized by the formulae (12) v1 = (0, 0, 0) a 1p 1p v2 = ( , 4 − a2 cos α, 4 − a2 sin α) 2 2 2 v3 = (a, 0, 0) d p 3a2 − b2 + c2 − v4 = ( 4 − b2 cos β, 4a 4ab 1p a2 + b 2 − c 2 p d 4 − b2 cos β, 4 − b2 sin β) − 4a 4ab 2 a2 − b 2 + c 2 d v5 = ( , , 0) 2a 2a 2 2 2 d p a −b +c + 4 − c2 cos γ, v6 = ( 4a 4ac a2 − b 2 + c 2 p 1p d − 4 − c2 cos γ, 4 − c2 sin γ). 4a 4ac 2 Recall that by (11) the triangular discs 4123 , 4345 , and 4561 are pierced by the edges v4 v5 , v1 v6 , and v2 v3 , respectively. First consider the disc 4345 and the edge v1 v6 . Since the even vertices of H live above the xy-plane, the line through v6 and v1 intersects 4345 in a positive direction. Therefore, (v6 − v1 ) × (v4 − v1 ) · (v3 − v1 ) = v6 × v4 · v3 > 0. Using (12), this inequality reduces to p a2 − b 2 + c 2 p 4 − c2 cos γ) 4 − b2 sin β c p a2 + b 2 − c 2 p − (d − 4 − b2 cos β) 4 − c2 sin γ > 0. b Now consider the other two triangles and the other two edges. This amounts to cyclically permuting the parameters a, b, and c, and the angles α, β, and γ in the formula for f1 . Thus, f1 = (d −

f2 = (d −

p a2 + b 2 − c 2 p 4 − a2 cos α) 4 − c2 sin γ a p −a2 + b2 + c2 p − (d − 4 − c2 cos γ) 4 − a2 sin α > 0 c


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and f3 = (d −

p −a2 + b2 + c2 p 4 − b2 cos β) 4 − a2 sin α b p a2 − b 2 + c 2 p 4 − a2 cos α) 4 − b2 sin β > 0. − (d − a

Now let τ : Equ(6) → (0, 2) × (0, 2) × (0, 2) be the map taking the equilateral hexagon H to the triple (kv3 − v1 k, kv5 − v3 k, kv1 − v5 k). In addition, let D = {(x, x, x) : x ∈ (0, 2)} be the diagonal across the open cube (0, 2)×(0, 2)×(0, 2). Then τ −1 D consists of those equilateral hexagons whose odd vertices form an equilateral triangle. Suppose that a right-handed trefoil H in T is parameterized as above by the sextuple (a, b, c, α, β, γ) and has τ H ∈ D. Consider the function √ √ √ 4 − a2 f1 sin α + 4 − b2 f2 sin β + 4 − c2 f3 sin γ √ √ √ . µ= 2 4 − a2 4 − b2 4 − c2 sin α sin β sin γ

Since {α, β, γ} ∈ (0, π) and f1 , f2 , and f3 are all positive, µ > 0. However, a simple algebraic manipulation shows that

c2 − b 2 a2 − c 2 b2 − a 2 cot α + cot β + cot γ. a b c Since τ H ∈ D, we have a = b = c and therefore µ = 0, giving a contradiction. In particular, we find that the triangle formed by the odd vertices of any hexagonal trefoil cannot be equilateral. Thus, the image of the subset of equilateral trefoil knots under τ is contained in the open solid torus (0, 2) × (0, 2) × (0, 2) − D. (iv) Let H1 ∈ T ∩ Equ(6) have a < b < c. For example, suppose that µ=

H1 = h(.886375, .276357, .371441), (.125043, −.363873, .473812), (.549367, .461959, .845227), (.818041, 0, 0),

(.4090205, −.343939, .845227), (0, 0, 0), i so that τ H1 = (0.610324, 0.818027, 0.914936). By cyclically shifting vertex labels twice, we obtain a hexagon H2 with b < c < a; a second shift yields a hexagon H3 with c < a < b. Note that all three sets of vertices have the same underlying hexagon. Furthermore, in all three cases the even vertices lie above the plane determined by the odd vertices. By Theorem 13 in [3], every equilateral right-handed hexagonal trefoil whose even vertices lie on the positive side of the plane determined by its odd vertices lives in the same component of Equ(6) . This means that there exists a path h1 in Equ(6) from H1 to H2 . By cyclically shifting vertex labels in h1 (·) twice and four times, we obtain two paths h2 and h3 from H2 to H3 and from H3 to H1 , respectively. Let τ be defined as in (iii). Then τ (h1 ·h2 ·h3 ) is a generator for the fundamental group of the solid torus (0, 2)×(0, 2)×(0, 2)−D. Therefore h1 ·h2 ·h3 is an essential loop of infinite order in T ∩ Equ(6) . In particular, π1 (T ∩ Equ(6) ) is an infinite group. Question 8. We know that π1 (T ∩ Equ(6) ) has elements of infinite order and of order 2. Compute π1 (T ∩ Equ(6) ). Question 9. More generally, what is the fundamental group of any component of Geo(n) or Equ(n) ?

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Figure 8. An essential loop in Geo(7) .


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The last question is relevant because understanding the fundamental group of the components of Geo(n) or Equ(n) may yield critical information about the number of components in each space. As a case in point, consider Geo(7) , which consists of five components: one of unknots, one of each type of trefoil (left-handed and right-handed), and two of figure-eight knots [2, 4]. In each component of figureeight knots, there is an essential loop taking a heptagonal figure-eight knot to it’s mirror image and back. See Figure 8. Heptagons in adjacent frames of this “movie” differ by the motion of one vertex passing through the interior of a triangular disc. Empirical evidence seems to indicate that this is an impossible motion in Equ(7) . Question 10. Are there examples of figure-eight knots in Equ(7) with three consecutive edges which are coplanar? If the answer to this last question is “no,” then Equ(7) must have at least 31 components, leaving us with the first example of a knot with more equilateral representatives then geometric ones. This brings us to our last question. Question 11. Are there examples of knots for which the number of geometric representatives is not the same as the number of equilateral representatives? References 1. C. C. Adams, B. M. Brennan, D. L. Greilsheimer, and A. K. Woo, Stick numbers and composition of knots and links, Journal of Knot Theory and its Ramifications 6 (1997), no. 2, 149–161. 2. J. A. Calvo, Geometric knot theory: the classification of spatial polygons with a small number of edges, Ph.D. thesis, University of California, Santa Barbara, 1998. , The embedding space of hexagonal knots, Topology and its Applications 112 (2001), 3. no. 2, 137–174. 4. , Geometric knot spaces and polygonal isotopy, Journal of Knot Theory and its Ramifications 10 (2001), no. 2, 245–267. 5. K. C. Millett and E. Rawdon, Energy, ropelength, and other physical aspects of equilateral knots, submitted for publication, 2001. 6. S. Negami, Ramsey theorems for knots, links, and spatial graphs, Transactions of the American Mathematical Society 324 (1991), no. 2, 527–541. 7. R. Randell, The space of piecewise-linear knots, Physical Knots: a study of knotting, linking, and folding of geometric objects in 3-dimensional space, 2002. Department of Mathematics, North Dakota State University, Fargo, ND 58105 E-mail address: [email protected]