Cubulated moves and discrete knots^{†}^{†}thanks: 2010 Mathematics Subject Classification. Primary: 57M25. Secondary: 57M27, 57Q45. Key Words. Cubic knots, discrete knots.
Abstract
In this paper, we prove that given two cubic knots , in , they are isotopic if and only if one can pass from one to the other by a finite sequence of cubulated moves. These moves are analogous to the Reidemeister moves for classical tame knots. We use this fact to describe a cubic knot in a discrete way, as a cyclic permutation of contiguous vertices of the lattice (with some restrictions); moreover, we describe a regular diagram of a cubic knot in terms of such cyclic permutations.
Dedicated to Francisco González Acuña,
Fico, on the occasion of his birthday.
1 Introduction
In [5] it was shown that any smooth knot can be deformed isotopically into the skeleton
of the canonical cubulation of and this isotopic copy is called cubic knot. In particular, every classical smooth knot
is isotopic to a cubic knot.
There are two elementary “cubulated moves”. The first one (M1) is obtained by dividing each cube of the original cubulation of into cubes, which means that each
edge of the knot is subdivided into equal segments. The second one (M2) consists in exchanging a connected set of edges in a face of the cubulation (or a subdivision
of the cubulation) with the complementary edges in that face.
If two cubic knots and are such that we can convert into using a finite sequence of cubulated moves then we
say that they are equivalent via cubulated moves, notated .
This allows us to prove the following:
Theorem 1.
Given two cubic knots and in , they are isotopic if and only if is equivalent to by a finite sequence of cubulated moves; i.e., .
Theorem 1 is analogous to the Reidemeister moves of classical tame knots for cubic knots.
Since a cubic knot is given by a sequence of edges whose boundaries are in the canonical lattice of points with integer coefficients in , i.e., the abelian group , each knot is determined by a cyclic permutation (with some restrictions), . In section we describe a regular diagram of a cubic knot in terms of such cyclic permutations by projecting onto a plane , such that it is injective when restricted to the lattice and the image of the lattice, , is dense. More precisely, the projection of each knot is determined by a cyclic permutation (with some restrictions), . This fact allows us to develop algorithms to compute some invariants of cubic knots.
Remark 1.1.
Professor Scott Baldridge has brought to our attention the fact that our main theorem follows also from
Theorem 1.1 of his paper with Adam Lowrance [3]. The methods we use are different from theirs
and we remark that our moves have a natural higher dimensional counterpart. If is a cubic knot contained in the skeleton of the canonical cubulation
of ,
then given a cube , for a sufficiently fine subdivision of , the union of all dimensional faces of contained in ,
if nonempty, is homeomorphic to a cubulated disk, hence the closure of , is also a cubulated disk. The movement (M2) consists in replacing
by . We conjecture that our theorem is also valid in this higher dimensional situation and the proof should be similar to ours.
Another motivation for our theorem is to use it to construct dynamically defined wild knots as in [6].
Remark 1.2.
There are several mathematicians working with cubic knots and cube diagrams and their higherdimensional analogous. For example, Ben McCarty was the first to use cube knots to give the easiest known proof that the lefthand and righthand trefoils are not isotopic. Also there are several papers which consider lattices and knots. See [16], [17],[4],[12],[13].
2 Cubulations for
A cubulation of is a decomposition of into a collection of dimensional
hypercubes such that any two of its hypercubes are
either disjoint or meet in one common face of dimension . This provides with the structure of a cubic
complex.
In general, the category of cubic complexes and cubic maps is similar to the simplicial category. The main
difference consists in using cubes instead of simplexes. In this context,
a cubulation of a manifold is specified by a cubical complex PLhomeomorphic to the
manifold (see [8], [11], [15]).
The canonical cubulation of is the decomposition of into hypercubes which are the images of the unit cube
by translations by vectors with integer coefficients.
Definition 2.1.
The skeleton of , denoted by , consists of the union of the skeletons of the cubes in , i.e., the union of all cubes of dimension contained in the faces of the cubes in . We will call the canonical scaffolding of .
Any cubulation of is obtained from the canonical cubulation by applying to a conformal transformation
where is a real number, and .
Consider the homothetic transformation given by
,
where is an integer. The set is called a subcubulation or cubical subdivision of .
In [5], we proved the following theorem which is central to this paper.
Theorem 2.2.
Let be the canonical cubulation of . Let be a smooth knot of dimension . There exists a knot continuously isotopic to , which is contained in the scaffolding of of . The cubulation of the knot admits a subdivision by simplexes and with this structure the knot is PLequivalent to the sphere with its canonical PLstructure.
3 Cubulated isotopy
Recall that a smooth parametrized knot is a smooth embedding . As is usual,
we will at times identify the embedding with its image, which we call a geometrical knot.
Given two smooth parametrized dimensional knots , we say they are smoothly isotopic if there exists a smooth isotopy such that
and is an embedding of for .
Definition 3.1.
We will say is the isotopy cylinder of and . Note that is a smooth submanifold of codimension two in .
Let be the canonical cubulation of . Let be the projection onto the last coordinate.
Definition 3.2.
If is a connected subset of such that or is connected for all , we say that is sliced by connected level sets of .
Observe that there is no restriction on the dimension of .
Note that is sliced by connected sets. The goal of this section is to prove that there exists an isotopic copy of contained
in the skeleton of the canonical cubulation of which is sliced by connected level sets of .
We need some preliminary results:
Proposition 3.3.
Let be a closed subset of , such that is a proper function and is nonempty for all . Assume is sliced by connected level sets of . Let be the union of all cubes of the canonical cubulation intersecting . Then is also sliced by connected level sets of .
Proof. Let be any number in . Then belongs to the closed interval for some number .
Let be the intersection , so is compact in and is sliced by connected level sets. Assume is not connected, so
where , are nonempty, disjoint compact subsets and both and are also sliced by connected level sets of .
Then , and and are compact sets, hence is not connected, which is a contradiction. Therefore is connected.
Claim 1: is connected.
Suppose
that , where , are nonempty disjoint closed sets. As is connected, we have that either
or . We assume that . But intersects each of the hypercubes belonging to , since each
hypercube is connected, it follows that , which is a contradiction. Therefore is connected.
Claim 2: is sliced by connected level sets of .
This is a consequence of the above claim and the fact that the cubes intersecting and are the same cubes, since is the canonical cubulation.
Recall that , are two smoothly isotopic knots in with the smooth isotopy between them, which gives us the isotopy cylinder .
Lemma 3.4.
The isotopy cylinder is sliced by connected level sets of .
Proof. Since is the image of an embedding of , it is connected. Hence the result follows.
Lemma 3.5.
Let be the union of all cubes which intersect . Then is sliced by connected level sets of .
Proof. Note that is clearly proper, therefore this is a consequence of Proposition 3.3.
Theorem 3.6.
The isotopy cylinder can be cubulated. In other words, there exists an isotopic copy of contained in the skeleton of , hence is contained in the skeleton of the canonical cubulation of . Moreover can be chosen to be sliced by connected level sets of .
Proof. Theorem 2.9 in [5] implies that can be cubulated. More precisely in [5] it is shown the following:
let be a closed tubular neighborhood of , then there
exists a sufficiently small subcubulation
of , such that the union of all cubes which intersect , is a bicollar neighborhood of .
We can deform to any of its boundary components via an adapted flow, which is nonsingular in and is transverse to .
Let be this isotopic copy. Observe that contains an
isotopic copy of , say . It was also shown in [5] that can be deformed into the skeleton of . This cubic manifold is contained in the subcubulation , hence
to obtain we rescale back the subcubulation to the original one .
Finally, notice that is contained in the skeleton of , hence the only obstruction for to be sliced by connected level sets arises if is not sliced by connected level sets, but this contradicts Lemma 3.5.
4 Cubulated moves
We say that a knot is a cubic knot, if is contained in the scaffolding of the
canonical cubulation of .
Definition 4.1.
The following are the allowed cubulated moves:
 M1

Subdivision: Given an integer , consider the subcubulation of . Since , then is contained in the scaffolding (the 1skeleton) of and as a cubic complex, each edge of is subdivided into equal edges.
 M2

Face Boundary Move: Suppose that is contained in some subcubulation of the canonical cubulation of . Let be a cube such that contains an edge. We can assume, up to applying the elementary move , that consists of either one or two edges that are connected and are part of the boundary of a face . Thus is an arc contained in the boundary of and is divided by into two cubulated arcs. One of them is and we denote the other by . Observe that both arcs share a common boundary. The move consists in replacing by .
Remark 4.2.
Notice that the face boundary move can be extended to an ambient isotopy of .
Definition 4.3.
Given two cubic knots and in . We say that is equivalent to by cubulated moves, denoted by , if we can transform to by a finite number of cubulated moves.
4.1 Main theorem
We are now ready to prove Theorem 1.
Theorem 1.
Given two cubic knots and in , then they are isotopic if and only if is equivalent to by cubulated moves; i.e., .
Proof.
First, note that if and are equivalent by cubulated moves, then these knots must clearly be isotopic. Hence, what
remains to be proved is that two cubic knots that are isotopic must also be equivalent by cubulated moves.
Our strategy is as follows. First, for , we will smooth each to obtain , and then
cubulate these two knots to obtain in such a way that a) and b) we can show
are equivalent by cubulated moves.
Given a cubic knot , there exists a smooth knot isotopic to such that is
arbitrarily close to . This is because
we can round the corners at the vertices of in an arbitrarily small neighborhoods of them (see [9]).
Let be the isotopy cylinder (defined above) of and .
Then is a smooth submanifold of codimension two in .
By Theorem 3.6, there exists an isotopic copy of , say , contained in the skeleton of the canonical
cubulation of . Recall that is sliced by connected level sets of .
Furthermore, note that there exist integer numbers
and such that for all and
for all , where and are cubic knots which are isotopic to and , respectively.
Now, we will use the following results which will be proved in Sections 4.1.1 and 4.1.2, respectively.
Lemma 4.4.
Given a cubic knot we can choose a small cubulation fine enough that is a closed tubular neighborhood of and is equal to either a vertex, one edge, or two edges sharing a vertex (neighboring edges). We can also choose isotopic to such that is arbitrarily close to and . Let be an isotopic copy of contained in , then ; i.e., we can go from to by a finite sequence of cubulated moves.
Remark 4.5.
Theorem 4.6.
Given two cubic knots and , we obtain and as in Lemma 4.4. Then there exists a finite sequence of cubulated moves that carries into . In other words, is equivalent to by cubulated moves.
Thus by Lemma 4.4, there exists a finite sequence of cubulated moves that carries into and also a finite sequence of cubulated moves that carries into . By Theorem 4.6, there exists a finite sequence of cubulated moves that carries into . As a consequence there exists a finite sequence of cubulated moves that converts into .
4.1.1 Cubic case
Let be a cubic knot. We can choose a cubulation fine enough such that the subcollection
satisfies that
is a closed tubular neighborhood of and
is equal to either a vertex, one edge, or two neighboring edges (see algo Remark 4.5). We can also choose isotopic to
such that is
arbitrarily close to such that for any cube the intersection is nonempty,
and .
Observe that is also a closed tubular neighborhood of . By Theorem 3.1 in [5],
there exists an isotopic copy of contained in the
1skeleton of .
In this section, we will prove that there exists a finite sequence of cubulated moves that carries into .
Lemma 4.4.
There exists a finite sequence of cubulated moves that carries into .
Proof.
Observe the following. The knots and are isotopic, both are contained in the 1skeleton of , but and
. Our goal is to construct a surface such that it will be contained in the 2skeleton of and its boundary
will consist of two connected components, namely and .
Let .
Since all cubes in intersect , consists
of a finite number of cubes, say , such that all of them intersect . By orienting we can enumerate the cubes in in such a way that
consecutive numbers belong to neighboring cubes (cubes
sharing a common face) and the cube is next to the first one.
To construct the surface , we will look at all possible cases of
and find pieces , which will be consist of either one face or two neighboring faces of and whose union will be .
The boundary of these ’s will
intersect both and , hence faces corresponding to neighboring cubes will share a common edge.
Claim 1: Let . If consists of a vertex and consists of either a vertex or an edge, then
. In other words, is superfluos.
Proof of Claim 1. We will look at all cases.

Each intersection ( and ) consists of a vertex. This implies that each knot passes through two edges sharing the corresponding vertex and these edges are contained in cubes belonging to .

consists of a vertex and consists of an edge or vice versa (see Figure 1). Suppose that both intersections are contained in the same face (see Figure 1(a)). Notice that passes through two edges sharing the vertex . These two edges belong to neighbor cubes of . By construction of both and , we can assume that these cubes belong to . This implies that one of these two neighbor cubes contains both and .
Now, suppose that the vertex is opposed to , i.e., there is no edge of joining to any of the endpoints of (see Figure 1(b)). As in the previous case, passes through two edges sharing the vertex and these two edges belong to neighbor cubes of such that intersects them. We have that one of these cubes must contain both and one of the endpoints of . However this configuration implies that does not intersect any of the remaining neighbor cubes of such that . This is a contradiction, hence this configuration is not possible. Therefore, the claim 1 follows.
For simplicity, we will assume that does not contained superfluous cubes.
Claim 2: Let . Then, using face boundary moves if necessary, we can assume that
is a path consisting of the union of at most three edges.
Proof of Claim 2. We will prove that can be reduced using face boundary moves if necessary, to a path consisting
of the union of at most three edges.
We will consider all possible cases such that consists of the union of at least four edges. Remember that
is equal to either a vertex, one edge or two neighboring edges.
 Case 1.

Suppose that consists of a vertex . Since , it follows that can be contained in either one, two or three faces of and these faces do not contained . By a combinatorial analysis, we have that is a path consisting of at most 6 edges.

consists of a six edges path. In this case, three edges are contained in a face , two disjoint edges lie on a neighbor face of and one edge lies on a neighbor face of (see Figure 2). Now, we apply the face boundary move (M2) on to obtain a four edges path, where three of them are contained in . Next, we apply again an (M2)move on to obtain that can be reduced to a two edges path.

consists of a five edges path. In this case, three edges are contained in a face and the remaining two edges lie on a neighbor face . Then, applying an (M2)move on we obtain that can be reduced to a three edges path (see Figure 3).

consists of a four edges path. If three edges are contained in a face and one edge lies on a neighbor face , then applying an (M2)move on , we obtain that can be reduce to a two edges path (see Figure 4).
Now, if two edges are contained in a face and two edges lie on a neighbor face , then we apply an (M2)move on to obtain a new path consisting of four edges. Next we apply again an (M2)move on to obtain that can be reduced to a two edges path (see Figure 5).
 Case 2.

Suppose that consists of an edge. Then may be contained in either 1 or 2 faces.

Three edges of are contained in a face and two edges lie on the neighbor face , then we apply an (M2)move on to obtain that can be reduced to a one edge path (see Figure 6).

Three edges of are contained in a face and one edge lies on the neighbor face . Then we apply an (M2)move on to obtain that can be reduced to a two edges path (see Figure 7).
 Case 3.

Suppose that consists of two neighboring edges. In this case, is contained in one face, hence is equal to a path consisting of at most three edges. Then, applying an (M2)move if necessary, we can assume that consists of either one or two edges.
This proves claim 2.
By claim 2, we will assume that is equal to a path consisting of at most three edges. Remember that by hypothesis, consists of a path of at most two neighboring edges. Next, we will construct considering all possible cases of both and .
 Case 1.

Suppose that consists of a vertex . By claims 1 and 2, we have that consists of a path of either two or three edges.

is a three edges path which can not be reduced by (M2)move (see Figure 8). In this case, up to a face boundary move, we may assume that and two neighboring edges of lie on the same face . Then . Notice that the remaining edge of is contained in a neighbor cube of which must belong to , hence this edge will be lie on the corresponding .

is a two edges path. We have the following possible cases.

Both and lie on the same face . Then . See Figure 9.

and lie on opposite faces. We have two possibilities which are equivalent via an (M2)move (see Figure 10).
Now, we have that there exist neighbor cubes of such that they intersect both and . Notice that this can only happens in the configuration described on Figure 11. Then applying (M2)moves on these neighbor cubes, we have that is superfluous, hence .

The last case appears in Figure 12. By the same argument of claim 1 (b), this configuration is not possible.

 Case 2.

Suppose that consists of one edge. By claim 2, we can assume that consists of a path of either one, two or three edges.

is a three edges path which can not be reduced by an (M2)move to a path consisting of either one or two edges. Since , then must be contained in two neighboring faces of which intersect (see Figure 13). So is the union of these two neighboring faces.

is a two edges path.

Two edges of lie on a face . Then (see Figure 14). Notice that the remaining edge is contained in a neighbor cube of which must belong to , hence this edge will be lie on the corresponding .


is an one edge path.

Both and lie on the same face . Then (see Figure 16).

and are two opposite edges of , i.e., consider the vertices , and the vertices , , then , lie on the face and , lie on the opposite face ; such that , are opposite vertices on , and , are opposite vertices on . See Figure 17. In other words, the edge is opposite to the edge .
Observe that there exists a finite sequence of neighbor cubes, , keeping this configuration. However, at the cubes and this configuration changes. This implies that each intersection , consists of two neighboring edges. We can assume, applying induction if necessary, that this happens in the neighbor cubes of ; i.e. and . Let and .
Consider the cube . Then we have two possible configurations.

Let , , , , be vertices satisfying the following. , , , , , the edge is opposite to the edge and the vertices , , lie on the face opposite to (see Figure 18).
Notice that we have the same configuration described on case . Applying an (M2)move on , we get that the edges are replaced by the edges , hence is exactly the face of containing . Now, consider the cube , so it has the same configuration analyzed on . Then, we apply an (M2)move to obtain .

Suppose that . Then we apply an (M2)move on (see case), so the edges are replaced by the edges . Now, there exists another cube such that , , ; hence is superfluous (see Claim 1). Observe that has the same configuration described on , then applying an (M2)move, we obtain . Notice that .


The last possible configuration appears in Figure 19. As we can see, it is very similar to the case 1. Then, by the same argument .

 Case 3.

Suppose that consists of two neighboring edges. Then lies on a face .

is a three edges path. Since , then lies on five possible edges, four of them belong to the opposite face of , and the remaining edge . Hence consists of and two edges (see Figure 20). Applying an (M2)move if necessary, we can assume that and belong to a face and and an edge of are contained in a neighbor face . Notice that the remaining edge of is contained in a neighbor cube of which belongs to , where it is considered. Then .

is a two edges path.

lies on the opposite face of , in such a way that there is only one face such that both and consists on an edge, respectively. Notice that the remaining edges of and belong to neighboring cubes of , where they are considered. Then . See Figure 23
Let . By the construction, consists of either one face or two neighboring faces which are contained in the 2skeleton of ,
hence is contained in the 2skeleton of .
As a consequence of the previous analysis, we know that if we apply an (M2)move on a cube , then this movement does not alter the configuration on its neighbor cubes; in other words,
the choice of depends only of the configuration given on . Now the boundary of each is composed by edges belonging to and and
two disjoint edges and that do not belong to
neither nor . Thus, if we take the cubes , and we construct and , then
the intersection is one of these two edges , ; moreover, since and are closed, connected 1manifolds, then
each of these two edges belong exactly to two cubes.
This implies that is a surface whose boundary
consists of two connected components, namely and .
Now, we will carry the knot onto the knot via a finite number of cubulated moves. Notice that is the union
of squares which are enumerated according to the numbering provided by the set . In particular, the square is a neighbor square of for all ,
i.e., consists of an edge .
We will use induction on . Consider . We apply an (M2)move on in such a way that the edges belonging to replace the others (by construction and ). Next, we consider . By the previous step, and share an edge belonging to . Then we apply again an (M2)move, so the edges belonging to replace the others. We continue this finite process in thi way. Notice that if then , thus is not a common edge of two squares and belonging to , hence if is replaced by an edge belonging to , then this replacement will be kept in the following steps. Therefore, the result follows.
4.1.2 Smooth case
Suppose that and are isotopic smooth 1knots in .
Let be the isotopy cylinder of and .
Then is a smooth submanifold of codimension two in .
By Theorem 3.6, there exists an isotopic copy of , say , contained in the skeleton of the canonical
cubulation of , such that is sliced by connected level sets of . Moreover there exist integer numbers
and and cubic knots and isotopic to and , respectively;
such that for all and
for all .
Our goal it to prove that is equivalent to by cubulated moves,
Let be the projection onto the last coordinate. Thus is an affine hyperplane parallel to the space .
Lemma 4.7.
Each hyperplane has a canonical cubulation.
Proof. Since is an affine hyperplane parallel to the hyperplane and has a canonical cubulation given by the restriction of the cubulation of to it; i.e., is the decomposition into cubes which are the images of the unit cube by translations by vectors with integer coefficients whose last coordinate is zero. Then has a canonical cubulation coming from the translation of the cubulation by the vector .
Definition 4.8.
A cell (face) of the canonical cubulation of is called horizontal if is a fixed integer number. Otherwise is called vertical.
Definition 4.9.
Let be a cubulated surface (a surface contained in the skeleton of ) and be a hyperplane in . We say that and intersect transversally, denoted by , if is a polygonal curve.
Lemma 4.10.
Let , be an affine hyperplane. Then intersects transversally.
Proof. Let , . By Theorem 3.6, is connected. Let . Then , where is a 2face of the cubulation . Notice that is a vertical face, since . So, we have two possibilities: either or belongs to an edge of .

If then is a linear segment parallel to an edge contained in .

If belongs to an edge of , then there exists another vertical face such that . Thus is a linear segment parallel to an edge, and by the same argument is also a linear segment , and .
Therefore, the result follows.
Corollary 4.11.
Let be a real number such that . Then the set is a knot.
Proof. By the above, is a polygonal connected curve.
Now, for each we define
and
Observe that and are cubic knots.
Let be the set of squares (cells) belonging to . Consider the spaces
and . By construction
and .
Let . Notice that , where Cl denotes closure.
Lemma 4.12.
The space is homeomorphic to .
Proof. By construction, is a compact connected submanifold of . Since is homeomorphic to , and has two connected components, then the result follows.
Lemma 4.13.
has the homotopy type of .
Proof. Consider the set . By Lemma 4.12, we know that , where
and .
Hence is homeomorphic to . By Alexander duality, using reduced homology and cohomology groups,
we have that . One has that has two connected components, so
. Since , we have that either
or , but , hence .
Therefore, has the homotopy type of .
Next, we are going to describe the subset . Notice that the edges of are of four types, which we will denote by , , and .

An edge belongs to if but .

An edge belongs to if but .

An edge belongs to if