Inscribed Trefoil Knots
Abstract
Let be a knot type for which the quadratic term of the Conway polynomial is nontrivial, and let be an analytic periodic function with nonvanishing derivative which parameterizes a knot of type in space. We prove that there exists a sequence of numbers so that the polygonal path obtained by cyclically connecting the points by line segments is a trefoil knot.
1 Introduction
We investigate the existence of piecewise linear trefoils inscribed in a knot. In particular, we consider the problem of finding six points on a given knot which form a Trefoil knot when cyclically connected by straight line segments in the same order in which they appear. Our main result is as follows.
Theorem 1.
Let be a knot type for which the quadratic term of the Conway polynomial is nontrivial, and let be an analytic periodic function with nonvanishing derivative which parameterizes a knot of type in space. Then there exists a sequence of numbers so that the polygonal path obtained by cyclically connecting the points by line segments is a trefoil knot.
For the proof, we will begin by defining a submanifold of configuration space which lies inside the closure of the set of 6tuples which form trefoils. Then, we will use some intersection theory to prove that if the quadratic term of the Conway polynomial of our knot is nontrivial then a 1parameter family of 6tuples of points on our knot lies in this submanifold. Finally, we will use some geometric arguments relying on the analyticity of the parameterization to show that we can perturb one of those 6tuples to make a trefoil.
It appears quite challenging to remove the requirement that the curve be analytic. The same sort of difficulties that arise in removing regularity requirements in the Toeplitz inscribed square problem also arise in the problem of inscribed trefoils. A simple limiting argument fails to work because the trefoils cannot be guaranteed not to degenerate to planar configurations, much like how when one approximates a Jordan curve by smoothings, the inscribed squares cannot be guaranteed not to shrink to zero. In order to relax the regularity requirements in these kind of problems, we likely need new geometric insights. [8]
2 A Submanifold of Configuration Space
If is a manifold, then will denote the space of all ntuples of distinct points of . This is called the nth (labeled) configuration space of . In this section we will construct a submanifold of with some interesting properties.
Consider a small geometrically spherical 4ball in real projective 4space, . Given a point not inside of , we can consider the lines in that pass through as well as some part of . These lines either intersect at two distinct points, or they lie tangent to , intersecting it at exactly one point. For fixed , let be the set of points in at which some line passing through and tangent to intersects . The set is a 2sphere that divides into two regions. Furthermore, for any line that passes through and intersects at two distinct points, the two intersection points will lie on different sides of . This means that if are three distinct lines that each pass through and each intersect at two points, then there is a welldefined partition of the six intersection points into two groups of three which is determined by grouping together intersection points that lie on the same side of . Let us define a graph with the six intersection points as its vertices and and edge between two vertices if they are on the same side of or they lie on the same line from the set . We will call this graph . Regardless of which three lines we pick, this graph will be isomorphic to the edge graph of a triangular prism. One interesting property of this graph is that its compliment graph is cyclic. Therefore, if we define to be the graph with vertices and edges between any two numbers that differ by 1 mod 6, then we may select a graph isomorphism from the compliment graph to the fixed cyclic graph . We now define to be the moduli space for the following data:

A point .

An unordered triple of distinct lines that each pass through and each intersect the 3sphere at exactly two points.

A specified graph isomorphism .
Alternatively, we could have simply defined as the subset of consisting of tuples so that (when we identify with ) the lines going through the pairs of points , , and all meet at a single point in . This makes it clear that admits an embedding into by taking a point in to the 6tuple of intersection points ordered via the labeling induced by the map to . It should be noted that when stereographically projected to , an element of generically consists of the set of vertices for a triangular prism inscribed in a sphere.
Proposition 1.
The manifold is diffeomorphic to .
Proof.
The tangent bundle of is trivial, so there is a choice of diffeomorphism, continuously depending on , from to the space of lines that pass through and intersect at exactly two points. Furthermore, the data of a specified graph isomorphism is equivalent to a specified ordering of the three lines along with a specified side of the partition determined by . To see why this equivalence holds, we see that the ordering of allows us to determine an orientation for the graph , and the specified side of the partition determined by allows us to select a single base point out of the six intersection points by choosing the intersection point of the first line which lies in the specified side of the partition. The base point along with the orientation defines an isomorphism. Now, we have enough information to say that is diffeomorphic to a double cover for . Therefore, to complete the proof we simply need to check that acts nontrivially on the orientation of determined by the specified isomorphism to . This is straightforward to check, as this action corresponds to a degree rotation of our configuration of points in , and this reverses the orientation of the cyclic order. ∎
Corollary 1.
The manifold is dimensional and orientable.
Proof.
is dimensional and orientable. ∎
Proposition 2.
The manifold is properly embedded in .
Proof.
We can define three functions via the diffeomorphism to that was constructed in the proof of Proposition 1. The first function, which we will call , will take a point to the absolute value of the component of that point. The second function will take a point in to the reciprocal of the minimum distance between two of the three points in from the component. The third function will take a point in to the maximum distance from zero of the three points in from the component. These functions have the property that any subset of on which all of these functions are bounded has compact closure. Therefore, we see that if is a sequence of points in that eventually leaves every compact set, then this sequence has a subsequence such that for some , the sequence goes to . Therefore, to show that is properly embedded in , it suffices to prove that if and is a sequence of points in with going to , then does not have a limit in . For , such a sequence will have approaching which will cause three of the intersection points to become arbitrarily close to each other. For , the minimum distance between a pair of intersection points will go to zero because two lines will get arbitrarily close. For , the distance between the two intersection points from some line will go to zero because some line will get arbitrarily close to being tangent to . Therefore, we see that must be properly embedded. ∎
Perhaps the most interesting aspect of is how it interacts with knots. Let be a periodic parameterization of a knot. We define to be the set of all 6tuples of the form such that . We will see that the oriented intersection class of with is welldefined in despite the fact that these two manifolds are noncompact.
Proposition 3.
is diffeomorphic to .
Proof.
We can select a point of by first selecting , and then selecting positive real numbers such that and then setting . The only redundancy is the choice of which only matters mod . This demonstrates that a point of is determined uniquely by a point in and a point in the interior of the 5simplex, meaning that is diffeomorphic to . ∎
Corollary 2.
The manifold is dimensional and orientable
Proof.
is dimensional and orientable. ∎
Proposition 4.
The manifold is properly embedded in .
Proof.
If a sequence of points in eventually leaves every compact set, then using the notation from Proposition 3, the minimum value of must go to zero. This means that the minimum distance between two points must go to zero in . ∎
Let be a periodic smooth parameterization of a knot with nonvanishing derivative. If we think of as the unit sphere in , we can define the thickness to be the infimum radius of all of the circles in that lie in and pass through at least three points on the knot. For a smooth parameterization with nonvanishing derivative, the thickness is always a positive real number, and it varies continuously with respect to the smooth topology.
We will now prove a couple of of facts about thickness.
Proposition 5.
If and are two points on a knot with a distance strictly less than , then, letting denote the 2sphere in in which and are antipodal points, the knot intersects transversely, and intersects only at the points and . Thus, of the two pieces of the knot on either side of the points and , one piece lies entirely on one side of and the other piece lies entirely on the other side of .
Proof.
To demonstrate that and are the only intersection points with , we simply note that if a third point lied on then the circle passing through would have radius less than which contradicts our assumptions. Now, to see that the knot must intersect transversely, observe that if the tangent line of the knot at (for or ) were to be tangent to , then taking to be on the knot and very close to , the circle through these points will have a radius less than which contradicts our assumptions.
∎
Proposition 6.
Given a finite set of points on a knot , if some pair of the points are within distance of each other, then the pair with minimal distance are adjacent with respect to the cyclic order of the points on the knot (in the sense that there is a path on the knot between them that does not contain any other point from the set).
Proof.
From the previous proposition, we see that if have minimal distance and that distance is less than , then the sphere with and as its antipodes has a path in passing through the inside of this sphere and going from to . If any point from lied on this path, then that point would be closer to than and this would contradict minimality. ∎
Proposition 7.
Let be a smooth parameterization of a knot with nonvanishing derivative. Then, considering as the unit sphere in , for any 6tuple of points and any from to , we have .
Proof.
We will derive a contradiction from assuming that there exists a point such that the pair of indices with minimal have . By Proposition 6 we have that and are adjacent with respect to the cyclic ordering of the knot. Recalling the construction of , we have that and are from different lines, and they lie on different sides of the partition . This means that if is the other point on the same line as and is the other point on the same line as , then the four points lie on a circle in such a way that and are nonadjacent. However, the closest pair of points on a cyclic polygon must be adjacent. This means that the pair of points of minimal distance from of the set are not and , which contradicts our assumption of minimality.∎
Proposition 8.
Despite and being noncompact, the oriented intersection homology class is welldefined and invariant with respect to isotopies of .
Proof.
Let denote the compact subset of consisting of tuples for which each pair of points are no less than a distance of apart. From Proposition 7, we see that lies inside , and therefore lies strictly within the interior of . This means that we can make transverse to by applying a small isotopy that only modifies the inside of . Let be the perturbation of we get after applying such an isotopy. Then is an oriented 1dimensional manifold that lies inside of , and it therefore defines a homology class which we denote . First, we must check that this homology class does not depend on . Let and be two choices for . Then we let be an isotopy so that and , and agrees with outside of . Then for can be thought of as a submanifold of , and we can perturbe this manifold slightly to make a submanifold of that intersects transversely. The intersection then yields a cobordism from to over . This demonstrates that the homology class does not depend on the choice of . Now, we must show that the homology class we have defined is an isotopy invariant of . Take an isotopy . As the manifolds vary with we get a submanifold of . Now, let , which exists and is positive because is continuously dependent on . The intersection is inside the interior of the compact set so we can let be a perturbation of which is unperturbed outside of and is transverse to . Then, is a cobordism in from a manifold in representing to a manifold in representing . Furthermore, is homotopy equivalent via the inclusions to both and to . This demonstrates that the homology class in is isotopy invariant with respect to , and is therefore a knot invariant. ∎
We have a canonical, continuously dependent diffeomorphism from the proof of Proposition 3. Therefore, the knot invariant can essentially be thought of as an integer, but with sign dependent on our choice of orientations for , , and .
3 Identifying the Intersection Class
The invariant constructed in Proposition 8 will be denoted by . In this section, we will identify as the quadratic term of the Conway polynomial multiplied by some fixed element of .
Proposition 9.
is a rank 2 Vassiliev invariant.
Proof.
Fix a stereographic inclusion .
Let be a smooth periodic parameterization with nonvanishing derivative for an immersed circle in with exactly three transverse selfintersection points . Now, let be nonzero vectors such that both tangent lines to at are orthogonal to for each . Take small disjoint closed spherical neighborhoods centered around the points respectively. We can assume that these neighborhoods are small enough that for each ball , the curve intersects the boundary transversely at exactly four points, and the two intervals in between these points map to intersecting strands of the curve , and at no point in is the tangent line of parallel to . Lastly, at each point , we make an arbitrary distinction between the two intersecting strands in , calling them strand 1 and strand 2. We choose six smooth periodic functions, labeled where ranges from 1 to 3, and corresponds to the indices of the selfintersection points, and ranges from 1 to 2, denoting the strand. We require that these functions have support in the preimage of their corresponding strand, and are nonzero at the points that map to . Now, given an , we define . We will think of as a function . We will fix to be small enough that the following properties hold.

If then .

For each , if and are both in , and and we have , then and .
We will call the resolution of the self intersecting curve . To prove that is a rank 2 invariant, we must prove that the following formula holds.
(1) 
Select a point on away from the neighborhoods . If denotes the submanifold of consisting of tuples where the first term is , then as a consequence of Proposition 8, we see that the oriented intersection number of and , which we will denote by , equals . To prove our proposition, we will describe a way to perturb the manifolds so that they overlap in a nice way that makes Equation 1 obvious.
Let be a real number with . Now, if we fix a 5tuple of real numbers, we define to be “doubled” if the two intersecting segments of passing through each contain at least one point from . Now, let be smooth functions which map tuples of real numbers into the closed interval , such that if is not doubled with respect to , then , and if both points in that map to are represented in , then .
We are now ready to define, for any triple , a perturbation of . We will call our perturbation and we produce it by moving each tuple of the form to the 6tuple
(2) 
via the isotopy , where
For convenience, we will write as shorthand to denote the 6tuple in (2).
One way to describe what is geometrically happening here is to think of this isotopy as bringing the submanifold as close to the subset of 6tuples that lie on as we possibly can. Indeed, if there are no doubled neighborhoods, then our perturbed 6tuple consists only of points that lie in . In order to avoid points coinciding with each other, we have introduced the notion of doubled neighborhoods, and selected our perturbation so that when a doubled neighborhood occurs, the points move back away from to the resolved arc. We will now describe how this perturbation causes our submanifolds to overlap nicely.
By the reverse pigeonhole principle, every 5tuple has at least one nondoubled neighborhood . Suppose are such that we have and for . Then since is not doubled, we have that . This demonstrates that our perturbed manifolds overlap. If intersects transversely at the point then it also intersects transversely at the point , and because the parameterization by determines the orientation, the sign of these intersections will be the same. Furthermore, so the signs for the sum in Equation 1 cancel. Therefore, all we need to do to prove Equation 1 is justify the claim that the isotopy described above preserves the oriented intersection number of each perturbed manifold with .
For convenience, we will write
To prove that the oriented intersection number is preserved, we can use the same argument as in Propositions 7 and 8. All we need for this argument to work is to show that if two terms of a 6tuple of the form are sufficiently close together, then the two terms of minimal distance are adjacent with respect to the cyclic ordering. To prove this, observe that the derivative of is always nonzero, and when . Now suppose for the sake of contradiction we have a sequence of pairs each of which are the pair of minimal distance in some 6tuple , and each of which are cyclically non adjacent in that 6tuple, and . Let and be such that
and
Now, if the distance between and goes to zero as , then the limit points of will be points where the derivative of is zero for some which yields a contradiction. Otherwise, there will be a subsequence with the distance between and bounded below, and by compactness, we get with which yields a contradiction. ∎
Proposition 10.
If is an unknot, .
Proof.
By Proposition 8, we only need to prove that for being a great circle. For this choice of , the manifold does not intersect at all because if three lines intersecting outside of in each pass through two points on a great circle in , at least one of those lines will have its intersection points adjacent along the great circle, which means the order of the points will be wrong and the 6tuple will not be in . ∎
Proposition 11.
If is a trefoil knot, is a nontrivial element of .
Proof.
By Proposition 8, we only need to prove that is a nontrivial element of when is the embedding of the trefoil in given by
Let . Then we want to compute the oriented intersection number of with . First, notice that contains the 6tuple , and the intersection is transverse at this point. We claim that the remaining points, if they exist, must have an even contribution to . To prove this, we first define two actions of on . Our actions will be denoted by and , where negates the second and fourth coordinates in , and negates the third and fourth coordinates in . Now, we extend these two actions to . The action is first applied termwise to each term in the 6tuple, and then we apply the permutation . The action is first applied termwise, and then we apply the permutation . Combining and , we get a action on both and . Inside , the intersection is mapped to itself under . Let denote the subset of which consists of fixed points under . We see that any such fixed point has as the 4th term in the 6tuple. This lets us conclude that is mapped to itself under . Now, observe that if a point is in and is fixed by both and , then it is the 6tuple . If we take a small perturbation of to which only modifies at points where the action of is free and makes the intersection with transverse at these points, then the intersection is transverse with finitely many points. These intersection points are acted on by the involution . The fixed points of are acted on by , which has only one fixed point. Therefore, . This proves that . ∎
It is very likely, and intuitive from a geometric perspective, that is the only element of . Although counting the points of is a computationally finite exercise which simply involves solving some trigonometric equations, the equations in question are highly complicated. Without some clever algebraic or geometric trick which simplifies the computation, it is likely that any proof that will be computerassisted. I suspect that such a trick exists, but it has eluded me.
Proposition 12.
is equal to the quadratic term of the Conway polynomial multiplied by some fixed nontrivial element of .
Proof.
Nontrivial multiples of the quadratic term of the Conway polynomial are characterized by being rank 2 invariants which are trivial on the unknot and nontrivial on the trefoil. [3] ∎
4 Constructing the Inscribed Trefoil
In this section, we restrict our attention to a fixed analytic parameterization of a knot which we can think of as a map by composing with the stereographic projection. Furthermore, we will assume the knot parameterized by has a nontrivial quadratic term for its Conway polynomial, and thus satisfies the criteria for Theorem 1.
We define to be the group of Möbius transformations of . This is the group of transformations (which are defined but we think of as transformations on defined at all but at most one point) generated by translation, scaling by nonzero real constants, and the operation which consists of mapping to under the stereographic projection, applying an isometric rotation, and then projecting back to .
We define a spherical trefoil knot to be any tuple of points in that all lie on some sphere, and when the points are cyclically connected by line segments, produce a trefoil knot.
Proposition 13.
Let be a spherical trefoil knot lying on a sphere . Let be a Möbius transformation that does not take to a flat plane. Then is also a spherical trefoil knot. (But the handedness might be different from .)
Proof.
First we prove the result for Möbius transformations that do not turn the sphere inside out. The set of such Möbius transformations is connected, so we just need to show that as a Möbius transformation varies and the point of inversion does not pass through , the knot type of the polygonal path generated by a sequence of points on does not change. If the knot type were to change, there would need to be a time at which two line segments in the polygonal path cross. For this to happen, a 4tuple of points would need to be coplanar when they were not previously. However, the point of inversion is not on which implies that our coplanar 4tuple lies on a circle. However, Möbius transformations preserve circles so we arrive at a contradiction because the 4tuple would also have to lie on a circle in and be coplanar originally. The same argument demonstrates that we now only need to show that there is some Möbius transformation that turns inside out and produces a spherical trefoil. The mirror image transformation is such an example. ∎
We define a closed subset to be the subset of consisting of tuples for which all six points lie on some circle in . We also define a “stereographic trefoil” to be a tuple of points in which lie on some 2sphere and which form a spherical trefoil knot under some choice of stereographic projection to . Due to Proposition 13, any stereographic trefoil yields either a spherical trefoil knot or a coplanar set of points when stereograpically projected to under an arbitrary choice of stereographic projection.
Proposition 14.
For any neighborhood of the manifold , there is a small isotopic perturbation inside such that and coincide exactly on and every point in is a stereographic trefoil.
Proof.
A point in is determined by a point in and an ordered triple of affinely independent lines passing through at points , as well as a specified side of the partition determined by (see the proof of Proposition 1). Then the affine 3space generated by the three lines passes through to form a 2sphere containing all points of the 6tuple. The 2spheres and intersect at a circle in , and we are given a specified side of induced by the specified side of , as well as a specified orientation of . This orientation of is induced by the intersection of the orientation of which comes from the specified side, and the orientation of the span of the lines induced by their order. Now, for any , we can define by isometrically rotating (by angle ) the specified side of in the direction of the orientation of . For sufficiently small , this will yield a stereographic trefoil. See Figure 2. We select a small smoothly dependent on a point in and equal to zero exactly on . Now, applying the transformation at all points in yields the desired perturbation . ∎
Proposition 15.
Suppose the analytic curve with nonvanishing derivative is tangent to a plane at the point . Let denote the distance between and . Let denote the angle between the plane and the tangent line of at . Then there exist positive constants so that if we have then we have .
Proof.
Let be the minimal natural number so that the th degree Taylor approximation of at does not lie in . Then, there exists a constant so that for sufficiently close to , we have . From the derivative, we see that there exists a constant so that for sufficiently close to , we have . We then have the inequality for sufficiently close to , so we see that for some constant . Letting , and letting be the required closeness of and , we have the desired inequality. ∎
Proposition 16.
Let be a plane in , and let be a triple of affinely independent points in . Let be the angle between and the plane spanned by . For , let be the distance between and . Let be the radius of the incircle for the triangle with vertices . Then we have
Proof.
First, let be the plane spanned by and . If and are parallel, then the proposition is trivial, so assume and intersect at some line . Then, for let be the distance in between and . We have that . Furthermore, so . Therefore, it suffices to prove that . This inequality becomes obvious when one considers the orthogonal projection of onto a line orthogonal to . In such a projection, the triangle with vertices maps to an interval of length less than or equal to , and this interval contains the projection of the incircle of the triangle which is of length , therefore, we have the inequality . Multiplying both sides by , and using , we get the desired inequality. ∎
We define to be the set of all triples of affinely independent points on such that, at each of the points in the triple, is tangent to the plane passing through the triple.
Proposition 17.
Every point of is isolated.
Proof.
As a corollary of this proposition, is finite because it can be expressed as the set of zeroes of a multivariate analytic function, and such a set is locally connected [4] [1]. However, in the hopes that minimizing the use of analyticity will make it easier to generalize our arguments to larger classes of curves, we will avoid using the finiteness of . The only fact we will use about about analytically parameterized curves is that they intersect every plane at only finitely many points, and at those points, Proposition 15 is satisfied.
Proposition 18.
Let be the subset of