Cubulated moves for 2-knots
Juan Pablo Díaz Gabriela Hinojosa Alberto Verjovsky





Cubulated moves for 2-knots.2010 Mathematics Subject Classification. 
Primary: 57M25. Secondary: 57M27, 57Q45.
Key Words: Cubulated moves, Discrete knots, 2-knots.
     Juan Pablo DíazThis work was partially supported by CONACyT (México), FORDECYT 265667,
 Gabriela HinojosaThis work was partially supported by CONACyT (México), CB-2009-129939., Alberto Verjosvky, This work was partially supported by
CONACyT (México), CB-2009-129280 and  PAPIIT (Universidad
Nacional Autónoma de México) #IN 106817.
    August 21, 2017
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 In this paper, we prove that given two cubical links of dimension two in ℝ^4,
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 and Roseman moves for classical tame knots of dimension one and two, respectively. 








§ INTRODUCTION


In <cit.> it was shown that any smooth knot K^n:𝕊^n↪ℝ^n+2 can be deformed isotopically into the n-skeleton  
of the canonical cubulation of  ℝ^n+2 and this isotopic copy is called cubical n-knot. In particular, every smooth 1-knot 
𝕊^1⊂ℝ^3 is isotopic to a cubical knot.   

There are two types of elementary “cubulated moves”. The first one (M1) is obtained by dividing each cube of the original cubulation of ℝ^3 into m^3 cubes, 
which means that each edge of the knot is subdivided into m equal segments. The second one (M2) consists in exchanging a connected set of edges in a face of the 
cubulation 

with the complementary edges in that face. If two cubical knots K_1 and K_2 are such that we can convert K_1 into K_2 using a finite sequence of cubulated moves then we say that they are equivalent via cubulated moves and is denoted by  K_1c∼ K_2.   

In <cit.> it was  proved the following:


Given two cubical knots K_1 and K_2  in ℝ^3, K_1≅ K_2 ≅𝕊^1, 
they are isotopic if and only if K_1 is equivalent to K_2 by a finite sequence of cubulated moves; i.e., K_1∼  K_2  K_1c∼ K_2.


Theorem 0 is analogous to the Reidemeister moves of classical tame knots for cubical knots.

 

Notice that cubulated moves can be extended to cubical 2-knots  in a natural way: 
The first one (M1) is obtained by dividing each hypercube of the original cubulation of 
ℝ^4 into m^4 hypercubes, and the second one (M2) consists in exchanging a connected set of squared faces homeomorphic to a 
disk 𝔻^2 in a cube of the cubulation (or a subdivision of the cubulation) with the complementary faces in that cube.  

The study of 2-knots in ^4 has been considered by various authors, for instance in 
<cit.>, <cit.>, <cit.> and <cit.>.  


Our goal is to extend the Theorem 0 for cubical knots of dimension two:


Given two cubical 2-knots K^2_1 and K^2_2  in ℝ^4,
 then they are isotopic if and only if K^2_1 is equivalent to K^2_2 by cubulated moves; i.e., 
    K^2_1∼  K^2_2  K^2_1c∼  K^2_2.








§ PRELIMINARIES




 §.§ Cubulations of ℝ^4

The regular hypercubic honeycomb whose Schläfli symbol is {4,3,3,4}, is called a cubulation of ℝ^4. In other words, a cubulation of  ℝ^4 is a decomposition into a collection of right-angled 4-dimensional
hypercubes {4,3,3} called  cells such that any two are
either disjoint or meet in one common k-face of some dimension k. This provides  ℝ^4 with the structure of a cubical
complex whose category is similar to the simplicial category PL.   

The combinatorial structure of the regular Euclidean honeycomb {4,3,3,4} is the following: around each vertex  there are 8 edges, 24 squares, 32 cubes and 16 hypercubes. Around each edge there are 6 squares, 12 cubes and 8 hypercubes. Around each square there are 4 cubes and 4 hypercubes. Finally around each cube there are  2 hypercubes.    





The canonical hypercubic honeycomb C of ℝ^4 is its decomposition into hypercubes which are the images of the unit hypercube:

    {4,3,3}=I^4=[0,1]^4={(x_1,x_2,x_3,x_4)∈ℝ^4  | 0≤ x_i≤ 1}
 by translations by vectors with integer coefficients. Then all vertices of C have integer coordinates.   

Any regular hypercubic honeycomb {4,3,3,4} or cubulation of ℝ^4 is obtained from the canonical cubulation by applying a conformal transformation  to the canonical cubulation. Remember that a conformal transformation is of the form  
x↦λA(x)+a, where λ≠0,  a∈ℝ^4,   A∈SO(4). 

 The k-skeleton of C, denoted by 𝒮^k, consists of the union of the k-skeletons of the hypercubes in  C,
i.e., the union of all cubes of dimension k contained in the faces of the 4-cubes in  C.
We will call the 2-skeleton 𝒮^2 of C the canonical scaffolding of ℝ^4.


Notice that all the previous definitions can be extended  in a natural way to ℝ^n+2.



 §.§ Cubical and smooth 2-knots


In classical knot theory, a subset K of a space X is a knot if K is homeomorphic to a  p-dimensional sphere 
𝕊^p embedded in either the Euclidean n-space ℝ^n or the n-sphere 𝕊^n=ℝ^n∪{∞}, where p<n. Two knots K_1, K_2 are equivalent or isotopic if there is a homeomorphism h:X↪ X such that h(K_1)=K_2;
in other words (X,K_1)≅ (X,K_2). However, a knot K is sometimes defined to be an embedding
K:𝕊^p↪𝕊^n≅ℝ^n∪{∞} (see <cit.>, <cit.>).
We shall also find this convenient at times and will use the same symbol to
denote either the map K or its image K(𝕊^p) in 𝕊^n.


Let K^2 be a 2-dimensional knot in ℝ^4. If K^2 is contained in  𝒮^2, we say that K^2 is a cubical knot. 



 If K^2 is a smooth knot and K^2 denotes a cubical knot which is isotopic to K^2, i.e., K^2  ∼K^2, we say that K^2 is a cubical diagram of the knot K^2.



Given two parametrized 2-dimensional smooth knots K^2_1, K^2_2:𝕊^2↪ℝ^4 we say they are smoothly isotopic if there exists a 
smooth isotopy H:𝕊^2×ℝ→ℝ^4 such that


    H(x,t)= {[ K^2_1(x)        ,; K^2_2(x)        , ].


and H( · ,t) is an embedding of 𝕊^2 for all t∈ℝ.  


We will say 
J^3={(H(x,t),t)∈ℝ^5 | x∈𝕊^2,  t∈ℝ} is the isotopic cylinder of K^2_1 and K^2_2. 


Note that J^3 is a smooth noncompact submanifold of codimension two in ℝ^5.  


Let p:ℝ^5↪ℝ be the projection onto the last coordinate. Let M be a connected subset of ℝ^5 such that p^-1(c)∩ M (or p|_M^-1(c)) is connected for all c∈ℝ,
we say that M  is sliced by connected level sets of p. 


Observe that there is no restriction on the dimension of M.  


In  <cit.> it was proved the following result.


The isotopic cylinder J^3 can be cubulated. In other words, there exists an isotopic copy J^3 of J^3 contained
in the 3-skeleton of the canonical cubulation of ℝ^5. 
Moreover J^3 can be chosen to be sliced by connected level sets of p.

 
Consider J^3 as above; so J^3 is a cubical 3-manifold and  there exist integer numbers 
m_1 and m_2 and cubical knots K^2_1 and K^2_2 isotopic to K^2_1 and K^2_2, respectively;
such that p|_J^3^-1(t)= K^2_1 for all t≤ m_1 and
p|_J^3^-1(t)= K^2_2 for all t≥ m_2.  



 §.§ Cubulated moves


 The following are the allowed cubulated moves:


  M1 Subdivision: Given an integer m>1, consider the subcubulation  C_m of C by subdividing each k-dimensional cell of C in 
m^k congruent k-cells in C_m, in particular each hypercube in C is subdivided in m^4 congruent hypercubes in C_m. Moreover as a cubical complex, each k-dimensional face of the 2-knot K^2 is subdivided into m^k congruent 
k-faces. Since C⊂C_m, then
K^2 is contained in the scaffolding S^2_m (the 2-skeleton) of C_m. 
 

  M2 Face Boundary Moves: Suppose that K^2 is contained in some subcubulation C_m of  
the canonical cubulation C of ℝ^4. 
Let Q^4∈C_m be a 4-cube such that
A^2=K^2∩ Q^4 contains a 2-face. We can assume, up to applying the elementary (M1)-move if necessary, that A^2 consists of either one, 
two, or three
squares such that it is a connected surface and it is contained in the boundary of a 3-cube
F^3⊂ Q^4; in other words A^2 is a cubical disk contained in the boundary of
F^3. The boundary ∂ F^3 is divided by ∂ A^2 into two cubulated surfaces, one of them is A^2 and we denote the other by B^2. Observe
that both  cubulated surfaces share a common circle boundary. The face boundary move  consists in replacing A^2 by B^2 (see Figure <ref>). There are three types of face boundary moves depending of the number of 2-faces in each A^2 and B^2. If A^2 has p squares then B^2 has 6-p squares. 

 




It can be easily shown that the (M2)-move can be extended to an ambient isotopy of ℝ^4 .



Given two cubical 2-knots K^2_1 and K^2_2 in ℝ^4. We say that
K^2_1 is equivalent to K^2_2 by cubulated moves, denoted by 
K^2_1c∼ K^2_2, if we can transform K^2_1 into K^2_2 by a finite number
of cubulated moves. 





§ MAIN THEOREM


We are now ready to prove our main theorem. Notice that this proof is a generalization of the one given in <cit.> for the one dimensional case.


Given two cubical 2-knots K^2_1 and K^2_2  in ℝ^4,
 then they are isotopic if and only if K^2_1 is equivalent to K^2_2 by cubulated moves; i.e., 
    K^2_1∼  K^2_2  K^2_1c∼  K^2_2.




 First, note that if K^2_1 and K^2_2 are equivalent by cubulated moves, then
these 2-knots are isotopic by Remark <ref>.  Hence, what remains to be proved is
that two cubical 2-knots that are isotopic must also be equivalent by cubulated
moves.  

Our strategy is like the one used for cubic knots of dimension one (see <cit.>).  First, for i ∈{1,2}, we will smooth each K^2_i to obtain K^2_i, and then cubulate these two 2-knots to obtain K^2_i in such a way that 

A) K^2_i c∼K^2_i and 

B) K^2_1c∼K^2_2.   

In <cit.> was proved the following.    

Theorem. Any compact, closed, oriented, cubical surface M^2 in ℝ^4 is smoothable. More precisely, M^2
admits a global continuous transverse field of 2-planes and therefore by a theorem of J. H. C. Whitehead there is an arbitrarily small topological 
 isotopy that moves M^2 onto a smooth surface M^2 in ^4 (see  <cit.>, <cit.>).  


As a consequence, we have that  given a cubical knot K^2, there exists a smooth knot K^2 isotopic to
K^2 such that K^2 is C^0-arbitrarily close to K^2.  

Let J^3 be the isotopic cylinder (see Definition <ref>) of K^2_1 and K^2_2.
Then J^3 is a smooth submanifold of codimension two in ℝ^5. 
By Theorem <ref>, there exists an isotopic copy of J^3, say J^3, contained in the 3-skeleton of the canonical 
cubulation C of ℝ^5. Recall that J^3 is sliced by connected level sets of p and  there exist integer numbers 
m_1 and m_2 such that p^-1(t)∩J^3 =K^2_1 for all t≤ m_1 and
p^-1(t)∩J^3 = K^2_2 for all t≥ m_2, where K^2_1 and K^2_2 are cubical knots which are isotopic to K^2_1 and K^2_2, respectively.  

Now, we will use the following two results which will be proved in Sections <ref> and <ref>, respectively.  

 
Given a cubical 2-knot K^2, we can choose a small cubulation C_m  fine enough so that 
N^4_K={Q^4∈C_m | Q^4∩ K^2≠∅} is a closed tubular neighborhood of K^2 and
Q^4∩ K^2 is equal to either a vertex, one square, two squares sharing an edge (neighboring 2-faces) or three neighboring squares (two by two neighboring 2-faces). 
We can also choose K^2 isotopic to K^2 such that K^2 is 
C^0-arbitrarily close to K^2 and K^2⊂(N^4_K). Let 
K^2 be an isotopic copy of K^2 contained in ∂ N^4_K, then K^2c∼K^2; i.e., we can
go from K^2 to K^2 by a finite sequence of cubulated moves.





  * The existence of K^2 is proved on Theorem 3.1 in <cit.> (see also the proof of Theorem <ref>).

  * We may assume, using a subdivision move if necessary, that the intersection Q^4∩ K^2 is equal to either a vertex, one square, two squares sharing an edge (neighboring 2-faces) or three neighboring squares (two by two neighboring 2-faces). 



 
Given two cubical knots K^2_1 and K^2_2, we obtain K^2_1 and K^2_2 as in Lemma A. Then
there exists a finite sequence
of cubulated moves that carries K^2_1 into K^2_2. In other words, K^2_1 is equivalent to K^2_2 by cubulated moves: K^2_1c∼K^2_2


If we go back to the proof of the Theorem 1, we have by Lemma A that there exist a finite sequence of cubulated moves that carries K^2_1 into K^2_1 and also a finite
sequence of cubulated moves that transforms K^2_2 into K^2_2. By Lemma B,
 there exists a finite sequence of cubulated moves that converts K^2_1 into K^2_2. As a consequence there exists a finite sequence
of cubulated moves that carries K^2_1 onto K^2_2. 




 §.§ K^2_i is equivalent to K^2_i by cubulated moves


Our goal is to prove Lemma A.

 
Given a cubical 2-knot K^2, we can choose a small cubulation C_m  fine enough so that 
N^4_K={Q^4∈C_m | Q^4∩ K^2≠∅} is a closed tubular neighborhood of K^2 and
Q^4∩ K^2 is equal to either a vertex, one square, two squares sharing an edge (neighboring 2-faces) or three neighboring squares (two by two neighboring 2-faces). 
We can also choose K^2 isotopic to K^2 such that K^2 is 
C^0-arbitrarily close to K^2 and K^2⊂(N^4_K). Let 
K^2 be an isotopic copy of K^2 contained in ∂ N^4_K, then K^2c∼K^2; i.e., we can
go from K^2 to K^2 by a finite sequence of cubulated moves.



Proof. 
The knots K^2 and K^2 are isotopic and both are contained in the 2-skeleton of N^4_K; however K^2⊂(N^4_K)  and K^2⊂∂ N^4_K. Our goal is to construct a connected 3-manifold M^3  such that it will be contained in the 3-skeleton of N^4_K and its boundary 
will consist of two connected components, namely K^2 and K^2.  

Let B^4={Q^4⊂ N^4_K |   Q^4∩K^2≠∅}.   

Since all hypercubes in N^4_K intersect K^2, B^4 consists
of a finite number of hypercubes, say m, such that all of them also intersect K^2. 
By orienting K^2 we can enumerate the cubes in B^4 in such a way that
consecutive numbers belong to neighboring hypercubes (hypercubes sharing a common 3-face). 

To construct the 3-manifold M^3, we will look at all possible cases of Q^4_j∈ B^4 
and find pieces M_j^3 which will be consist of the union of some  3-faces  of Q^4_j such that they are two by two neighboring faces. The union of all M_j^3 will be M^3. 
Notice that the boundary of these M_j^3's will 
intersect both K^2 and K^2, hence 3-faces corresponding to neighboring 4-cubes will share a common face.   

In order to describe the cubes M_j^3's in this proof, we will use the following notation for the 3-faces of  a given hypercube Q^4_j.



Next, we will construct M^3 considering
 all possible cases of both K^2∩ Q^4_j and K^2∩ Q^4_j.  

Claim 1: Let Q^4_j∈ B^4 and suppose that K^2∩ Q^4_j consists of  three squares (two by two neighboring 2-faces). 
Then, using face boundary moves if necessary, we can assume that 
K^2∩ Q^4_j is a connected 2-disk consisting of the union of at most three squares.  

Proof of Claim 1. Suppose that K^2∩ Q^4_j consists of three neighboring 2-faces F_1^2, F_2^2 and F_3^2. Observe that 
F_1^2∪ F_2^2∪ F_3^2 
intersects fifteen distinct 2-faces of Q^4_j, hence K^2∩ Q^4_j must be contained in the remaining six 2-faces. Notice that 
these six 2-faces are contained in a 3-face C^3 of Q^4_j and applying  (M2)-moves if necessary, we can assume that at most three 2-faces of 
K^2∩ Q^4_j lie on C^3. 
Therefore K^2∩ Q^4_j consists of at most three 2-faces. See Figure <ref>.1.


This proves claim 1. □  


Case 1. Suppose that K^2∩ Q^4_j consists of three neighboring 2-faces F_1^2, F_2^2 and F_3^2. By the above claim, we have that
K^2∩ Q^4_j is a connected 2-disk consisting of the union of at most three 2-faces; therefore K^2∩ Q^4_j 
consists of at most three neighboring faces E_1^2, E_2^2 and E_3^2 such that E_i^2 is parallel
to F_i^2 (i=1,2,3).  Then the only possibility up to a face boundary moves (M2) is shown in Figure <ref>.2. Each K^2∩ Q^4_j  and  K^2∩ Q^4_j consist of three neighboring 2-faces.
Thus  M_j^3=M_3^3∪ M_-1^3∪ M_-4^3.  

Claim 2: Let Q^4_j∈ B^4 and suppose that K^2∩ Q^4_j consists of  two squares sharing an edge (neighboring 2-faces). Then, 
using face boundary moves if necessary, we can assume that K^2∩ Q^4_j is a connected 2-disk consisting of the union of at most four faces.  

Proof of Claim 2. Suppose that K^2∩ Q^4_j consists of two neighboring 2-faces F_1 and F_2. Observe that the union 
F_1∪ F_2 intersects
 fifteen distinct 2-faces of Q^4_j, then K^2∩ Q^4_j must be contained in the remaining seven 2-faces. 
 Notice that six of these seven 2-faces are contained in a 3-face C^3 of Q^4_j and
applying  (M2)-moves if necessary, we can assume that three 2-faces of K^2∩ Q^4_j lie on C^3. See Figure <ref>.1.
Therefore K^2∩ Q^4_j consists of at most four 2-faces. 

 
This proves claim 2. □  
 
Case 2. Suppose that K^2∩ Q^4_j=F_1^2∪ F_2^2 where F_1^2 and F_2^2 are two neighboring 2-faces. Since K^2∩ Q^4_j is a connected 2-disk consisting of the union of at most four 2-faces, and K^2∩K^2=∅; we have that K^2∩ Q^4_j is contained in the union of seven 2-faces
of Q^4_j, such that F_1^2, F_2^2, …, F_6^2 belong to a 3-face C^3 and F_7^2  belongs to the opposite 3-face C̅^̅3̅. See Figure <ref>.1. Since K^2 intersects some  neighbor cubes of Q^4_j, we have three possibilities:


  (a) Suppose that K^2∩ Q^4_j consists of the union of four squares.  Hence F_7^2 belongs to K^2∩ Q^4_j. Thus, there exist three 3-faces M_1^3, M_-1^3 and M_-4^3 that
contained two 2-faces of (K ^2∪K^2)∩ Q^4_j (see Figure <ref>.2). 
 Then M_j^3=M_1^3∪ M_-1^3∪ M_-4^3.



  (b) Suppose that K^2∩ Q^4_j consists of the union of three neighboring faces. Then considering all the possibilities 
satisfying that K^2∩ Q^4_j can be extended, we have that  K^2∩ Q^4_j consists of F_1^2, F_2^2 and F_3^2 such that 
F_1^2 and F_3^2 are disjoint,  F_1^2 and F_2^2 share an edge and so does
 F_2^2 and F_3^2 (see Figure <ref>.3). So M_j^3=M_1^3∪ M_-1^3∪ M_-4^3.


  (c) Suppose that K^2∩ Q^4_j consists of the union of two neighboring faces. So considering all the options which K^2∩ Q^4_j can be extended, we have two possibilities for K^2∩ Q^4_j: 


  (i) K^2∩ Q^4_j is the union of the two 2-faces F_7^2 and F_4^2 (see Figure <ref>.4). Then there exist two 3-faces M_4^3 and M_-2^3 containing two 2-faces of
(K ^2∪K^2)∩ Q^4_j. So M_j^3=M_4^3∪ M_-2^3.


  (ii) K^2∩ Q^4_j is the union of the two 2-faces  F_2^2 and F_3^2 (see Figure <ref>.5). Then 
there exist two 3-faces M_-1^3 and M_-4^3 containing two 2-faces: one from K ^2∩ Q^4_j and the other from  K^2∩ Q^4_j. 
So M_j^3=M_-1^3∪ M_-4^3.

 
 
 
Claim 3: Let Q^4_j∈ B^4 and suppose that K^2∩ Q^4_j consists of one 2-face F^2. Then, using face boundary moves if necessary, we can assume that 
K^2∩ Q^4_j is a connected 2-disk consisting of the union of at most five 2-faces.  

Proof of Claim 3. Observe that F^2 intersects twelve distinct 2-faces of Q^4_j, then K^2∩ Q^4_j must be contained
in the remaining eleven 2-faces. These eleven 2-faces are distributed in the following way: six of them are contained in a 3-face C^3 and the
remaining faces lie on a neighboring 3-face D^3 such that C^3∩ D^3 consists of one 2-face (see Figure <ref>.1). Since K^2∩ Q^4_j is a connected disk, 
then using (M2)-moves if necessary, we can assume that three 2-faces of K^2∩ Q^4_j lie on C^3 and two 2-faces lie on D^3.
 
This proves claim 3. □  

 
 Case 3. Suppose that K^2∩ Q^4_j consists of one 2-face F^2. By claim 3,   we can assume that at most three 2-faces  
 F_1^2, F_2^2, F_3^2 of K^2∩ Q^4_j lie on C^3 and two 2-faces F_4^2, F_5^2 lie on D^3. We have five possibilities:


  (a) Suppose that K^2∩ Q^4_j consists of the union of five 2-faces. Then we have only one possibility for K^2∩ Q^4_j 
(see Figure <ref>.2).
There exist three 3-faces M_1^3, M_-2^3 and M_-1^3 which
contain two 2-faces of (K ^2∪K^2)∩ Q^4_j. Take M_j^3=M_1^3∪ M_-2^3∪ M_-1^3.


  (b) Suppose that K^2∩ Q^4_j consists of the union of four 2-faces. In this case, we have three possibilities:


  (i) K^2∩ Q^4_j is the union of the four 2-faces F_1^2, F_2^2, F_3^2 and F_4^2. Then we have only one possibility (see Figure <ref>.3). 
Let M_-2^3 be the 3-face of Q^4_j such that (K^2∩ Q^4_j)∪ F_2^2⊂ M_-2^3 and  let M_1^3 and M_-1^3 be the 3-faces of Q^4_j such that each of them is a 
neighboring 3-face of M_-2^3 and  F_1^2⊂ M_1^3 and F_4^2⊂ M_-1^3. Then M_j^3=M_1^3∪ M_-1^3∪ M_-2^3.


  (ii) K^2∩ Q^4_j is the union of the 2-faces F_1^2, F_2^2 and F_4^2, F_5^2. Thus the only possible configuration is shown in 
Figure <ref>.4. Let M_1^3, M_-2^3 and M_-1^3 be as above and let M_-4^3 be the neighboring 3-face such that (K^2∩ Q^4_j)∪ F_5^2⊂ M_-4^3 
so M_j^3=M_1^3∪ M_-2^3∪ M_-1^3∪ M_-4^3.


  (iii) This configuration is similar to the one described in case 3b (i), but we exchange the cubes C^3 and D^3. Then, we get only one possibility 
(see Figure <ref>.5 ). Let M_j^3=M_1^3∪ M_-1^3∪ M_-4^3.



  (c) Suppose that K^2∩ Q^4_j consists of the union of three 2-faces. Let F_1^2, F_2^2, F_3^2 be 2-faces contained in a
cube C^3 and let F_4^2, F_5^2, F_6^2 be 2-faces contained in the 3-cube D^3. Then we have three possibilities:


  (i) K^2∩ Q^4_j is the union of the 2-faces F_1^2, F_2^2 and F_3^2.  
Let M_1^3, M_-1^3 and M_-2^3 be as above, so M_j^3=M_1^3∪ M_-1^3∪ M_-2^3 (see Figure <ref>.6).


  (ii) K^2∩ Q^4_j is the union of the 2-faces F_2^2, F_3^2 and F_6^2. Thus M_j^3=M_1^3∪ M_-2^3
(see Figure <ref>.7).


  (iii) K^2∩ Q^4_j is the union of the 2-faces F_4^2, F_5^2 and F_6^2. Let M_2^3 be the 3-face which contains the three squares K^2∩ Q^4_j. Then M_j^3= M_2^3∪ M_-4^3 
(see Figure <ref>.8).



  (d) Suppose that K^2∩ Q^4_j consists of the union of two neighboring faces. Let F_1^2, F_2^2, F_3^2 be 2-faces contained in the
cube C^3 and let F_4^2, F_5^2, F_6^2 be 2-faces contained in the cube D^3. We have two possibilities:


  (i) K^2∩ Q^4_j is the union of the 2-faces F_1^2 and F_2^2. Thus 
M_j^3=M_1^3∪ M_-2^3 
(see Figure <ref>.9).


  (ii) K^2∩ Q^4_j is the union of the 2-faces F_3^2 and F_4^2.  Thus 
M_j^3=M_2^3∪ M_-4^3 (see Figure <ref>.10).



  (e) Suppose that K^2∩ Q^4_j consists of  one square. Then we have two possibilities:


  (i) K^2∩ Q^4_j is the square F_4^2. Thus
M_j^3=M_-2^3∪ M_4^3
(see Figure <ref>.11).


  (ii) K^2∩ Q^4_j  is the 2-face F_2^2. Thus
M_j^3=M_-2^3
(see Figure <ref>.12).





Claim 4: Let Q^4_j∈ B^4 and suppose that K^2∩ Q^4_j consists of   an edge e. Then, using face boundary moves if necessary, we can assume that 
K^2∩ Q^4_j is a connected 2-disk consisting of the union of at most six faces.  

Proof of Claim 4. Suppose that K^2∩ Q^4_j consists of an edge e. See Figure <ref>. Then K^2 must turn on Q^4_j; in other words,  
K^2∩ Q^4_j must contain two
faces F_1^2 and F_2^2 such that F_1^2=v+{ae_i_1+be_j : 0≤ a, b≤ 1} and F_2^2=v+{ce_i_2+de_j : 0≤ c, d≤ 1}, where e_i_1, e_i_2 and
e_j are distinct canonical vectors and v is a vector with integer coordinates. Notice that e_j is parallel to the edge e and only 
nine 2-faces of Q^4_j satisfy both they do not intersect K^2 and they have an edge parallel to e_j; 
hence K^2∩ Q^4_j must be contained in the union of these nine 2-faces. Since
K^2∩ Q^4_j is a connected disk it follows,  up to applying face boundary moves (M2) if necessary, that it consists of the union of at most six faces. 
This proves claim 4. □  



 Case 4. Suppose that K^2∩ Q^4_j consists of an edge e. By claim 4, 
K^2∩ Q^4_j is contained in three 3-faces M_3^3, M_-2^3 and M_-4^3 such that any two of them share a  2-face.


  (a) K^2∩ Q^4_j consists of the union of six 2-faces. Then, applying  (M2)-moves if necessary, we have that K^2∩ Q^4_j consists of the 
union of two 2-faces contained in M_3^3, three 2-faces  in M_-3^3 and one 2-face in M_-4^3 (see Figure <ref>.2). Thus M_j^3=M_3^3∪ M_-3^3∪ M_-4^3.


  (b) K^2∩ Q^4_j consists of the union of five 2-faces. Then, applying  (M2)-moves if necessary, as in the previous case
we have that M_j^3=M_3^3∪ M_-3^3∪ M_-4^3. (see Figure <ref>.3). 


  (c) K^2∩ Q^4_j consists of the union of four 2-faces. We have five possibilities:


  (i) K^2∩ Q^4_j consists of two 2-faces contained in M_4^3 and three 2-faces  in M_-3^3. 
Thus M_j^3=M_-3^3∪ M_2^3∪ M_-4^3 (see Figure <ref>.4).


  (ii) K^2∩ Q^4_j consists of one 2-face contained in M_-4^3 and three 2-faces  in M_-3^3. 
 Then M_j^3=M_-3^3∪ M_-4^3 (see Figure <ref>.5).


  (iii) K^2∩ Q^4_j consists of one square contained in M_-4^3 and three 2-faces  in M_-3^3 (see Figure <ref>.6). Then M_j^3=M_-3^3∪ M_-4^3.


  (iv) K^2∩ Q^4_j consists of two 2-faces contained in M_-3^3 and two 2-faces  in M_2^3 (see Figure <ref>.7). 
Then M_j^3=M_-3^3∪ M_2^3∪ M_-4^3.


  (v) K^2∩ Q^4_j consists of three 2-faces contained in M_2^3 and one square in M_-4^3 (see Figure <ref>.8). 
So M_j^3= M_2^3∪ M_-4^3.




  (d) K^2∩ Q^4_j consists of the union of three 2-faces. We have two possibilities:


  (i) K^2∩ Q^4_j consists of one square contained in M_2^3 and two 2-faces  in M_-3^3 (see Figure <ref>.9). 
 Then M_j^3=M_2^3∪ M_-3^3∪ M_-4^3.

  (ii) K^2∩ Q^4_j consists of two 2-faces contained in M_-3^3 and one 2-face  in M_-4^3 (see Figure <ref>.10). 
 Then M_j^3=M_-3^3∪ M_-4^3.




  (e) K^2∩ Q^4_j consists of the union of two neighboring 2-faces. We have two possibilities:


  (i) K^2∩ Q^4_j consists of two 2-faces contained in M_4^3  (see Figure <ref>.11). 
 Then M_j^3=M_4^3∪ M_3^3.


  (ii) K^2∩ Q^4_j consists of two 2-faces contained in M_-4^3  (see Figure <ref>.12). 
Thus M_j^3=M_-4^3.




 
Claim 5: Let Q^4_j∈ B^4 and suppose that K^2∩ Q^4_j consists of  a vertex v. Then, using face boundary moves if 
necessary, we can assume that 
K^2∩ Q^4_j is a connected 2-disk consisting of the union of at most six 2-faces.  

Proof of Claim 5. 
 Since K^2∩ Q^4_j consists of a vertex v, then v must be a corner of K^2 i.e. v is a common vertex of three 
 neighboring faces of 
K^2.  Hence K^2 must have a corner at some vertex of Q^4_j. Since K^2∩K^2 =∅,
it follows that K^2∩ Q^4_j can be contained in either one, two or three 3-faces of Q^4_j such that
these 3-faces do not contained v. Suppose that K^2∩ Q^4_j is a connected disk consisting of at least six 2-faces, 
then by a combinatorial analysis considering all the possible
descriptions of  K^2, we should have that
four of these 2-faces are contained in some 3-face of Q^4_j; hence applying a face boundary move (M2), we get that 
K^2∩ Q^4_j can be reduced to at most six 2-faces. (see Figure <ref>.1.)
  This proves claim 5. □  

Case 5. Suppose that K^2∩ Q^4_j consists of a vertex v. By the above claim,
 K^2∩ Q^4_j is the union of at least two 2-faces of Q^4_j and its boundary must be
contained in the union of the three neighboring 3-faces containing v.


  (a) K^2∩ Q^4_j is the union of six faces. Then K^2∩ Q^4_j 
is contained in three 3-faces M_-2^3, M_3^3 and M_4^3 of Q^4_j (see Figure <ref>.2). 
Since  K^2∩ Q^4_j is homeomorphic to a disk, we have that 
each 3-face  M_i^3 must contain two 2-faces of it and two of these three 3-faces contain v. 
Then M_j^3=M_-2^3 ∪ M_3^3∪ M_4^3.


  (b) K^2∩ Q^4_j is the union of five 2-faces. Then K^2∩ Q^4_j 
is contained in the union of two 3-faces M_-4^3 and  M_3^3 of Q^4_j (see Figure <ref>.3). So M_j^3= M_3^3∪ M_-4^3.



  (c) K^2∩ Q^4_j is the union of four 2-faces. Again, K^2∩ Q^4_j 
is contained in a 3-face M_4^3 of Q^4_j and its boundary must be
contained in the union of the three neighboring 3-faces containing v. Hence, one 3-face M_3^3, M_-1^3, M_-2 must contain one 2-face.  The only possibility is that four 2-faces lie in some 3-face M_4^3 of  Q^4_j (see Figure <ref>.4). 

 In this case M_j^3=M_4^3.


  (d) K^2∩ Q^4_j is the union of three neighboring 2-faces. These three 2-faces must be contained in some 3-face M_-4^3. 
Then M_j^3=M_-4^3 
(see Figure <ref>.5).


  (e) K^2∩ Q^4_j is the union of two neighboring 2-faces. These two 2-faces must be contained in some 3-face M_-4^3. 
Then M_j^3=M_-4^3.



Next, we will construct our 3-manifold M^3.  

Let  M^3=∪M_j^3. By construction, each M_j^3 consists of the union of 3-faces contained in the  3-skeleton of N_K^4,
hence M^3 is contained in the  3-skeleton of N_K^4.  


As we mention before,  if we apply an (M2)-move on a hypercube Q^4_j, then this movement does not affect the corresponding choice of M_l^3 on its neighbor hypercube Q^4_l; in other words,
the choice of M_j^3 depends only of the configuration of K^2∩ Q^4_j and K^2∩ Q^4_j in Q^4_j. Observe that the boundary of each M_j^3 is composed by  2-faces belonging to  K^2 and K^2  and 
some disjoint faces F^j_i (i=1,2,… ,s_j) that do not belong to
neither K^2 nor K^2. Thus, if we take a neighbor hypercube of Q^4_j, say Q^4_j_l∈ B^4, and we construct M_j^3 and M_j_l^3, then 
the intersection M_j^3∩ M_j_l^3 consists of the union of some F^j_r_i; notice that each of these 2-faces F^j_i belongs to a neighbor hypercube of Q^4_j, 
Q^4_j_l_s∈ B. Hence M^3 is a 3-manifold whose boundary
consists of two connected components, namely K^2 and K^2.  


Now, we will carry the knot K^2 onto the knot K^2 via a finite number of cubulated moves. Notice that M^3 is 
the union 
of m components M_1^3, … M_m^3 which are enumerated in such a way that   M_n+1^3 is a 
neighbor of M_n^3 for all n, i.e., M_n+1^3∩ M_n^3 consists of a 2-face F_n^2.  

We will use induction on m. Consider M_1^3. We apply  (M2)-moves on M_1^3 in such a way that the faces (of any dimension)
belonging to K^2 are replaced by 
those belonging to K^2 (by construction M_1^3∩ K^2≠∅ and M_1^3∩K^2≠∅).
Next, we consider M_2^3. By the previous step, M_1^3 and M_2^3 share 2-faces belonging to K^2. Then we apply again (M2)-moves, 
the faces (of any dimension) belonging to K^2 are replaced by those belonging to K^2. We continue this finite process in this way. 
Notice that if l⊂  K^2  
then l⊂∂M^3, thus l is not a common 2-face of some pair M_i^3, M_j^3 belonging to M^3, hence if 
l is replaced by a 2-face belonging to K^2, then this replacement will be kept in the following steps. Therefore, the result follows. □




 §.§ K^2_1 is equivalent to K^2_2 via cubulated moves


In this section, we shall prove the Lemma B which says that  K^2_1 is equivalent to K^2_2 by cubulated moves.  

Consider again the projection p:ℝ^5↪ℝ on the last coordinate. We call horizontal hyperplane to an affine hyperplane
parallel to the space ℝ^4×{0}. 
Thus p^-1(t)=ℝ^4_t is a horizontal hyperplane. Observe that each hyperplane 
ℝ^4_t has a canonical cubulation
given by the restriction of the canonical cubulation of ℝ^5 to it.  


 A 2-cell (2-face) F^2 of the canonical cubulation C of ℝ^5  is called horizontal if p(F^2) is 
 a constant number in ℕ.
A 2-cell F^2 is vertical if p(F^2)=[m,m+1] for some m∈ℕ.



 Let Σ^3 be a cubulated 3-manifold and P^4 be a horizontal hyperplane in ℝ^5.
We say that P^4  intersects transversally to Σ^3, denoted by  P^4⋔Σ^3, if P^4 intersects transversally each k-cube of Σ^3, k≥ 1.




 Let ℝ^4_t ⊂ℝ^5, t∉ℤ be an affine hyperplane. Then  ℝ^4_t intersects J^3 transversally.


Proof. By Theorem <ref>,  
ℝ^4_t∩J^3 is connected. Let x∈R^4_t∩J^3. Then
x∈ Q^3_i, where Q^3_i is a 3-face of the cubulation of ℝ^5. Notice that Q^3_i is a vertical 3-face, since t∉ℤ. 
So, we have two possibilities:
either x∈(Q^3_i) or  x∈ F^2∖∂ F^2 where F^2 is a 2-face of Q^3_i, or x belongs to an edge.


  * If x∈(Q^3_i) then ℝ^4_t∩ Q^3_i=E^2, where E^2 is a square parallel to a 2-face and x∈ E^2.

  * If x belongs to F^2∖∂ F^2, then there exists another vertical 3-face Q^3_j such that x∈ Q^3_i∩ Q^3_j. Thus
ℝ^4_t∩ Q^3_i must be a square E_i^2 such that it is parallel to a 2-face,  and analogously ℝ^4_t∩ Q^3_j is also a square E_j^2,
thus x∈ E_i^2∩ E_j^2. 

  * If x belongs to an edge l, then there exist  vertical 3-faces Q^3_i_r r=1,2,3 such that x∈ Q^3_j∩_r=1^3 Q^3_i_r. As in the previous case,
ℝ^4_t∩ Q^3_i_r must be a square E_i_r^2 parallel to a 2-face,  and analogously ℝ^4_t∩ Q^3_j is also a square E_j^2, hence
 x∈ E_j^2∩_r=1^4 E_i_r^2. 


Therefore, the result follows. □  


For x∉ℤ, the set p^-1(x)∩J^3 is a 2-knot.


Proof. By the above, p^-1(x)∩J^3 is a cubulated compact connected surface. Since
J^3 is homeomorphic to 𝕊^2×ℝ, it follows that p^-1(x)∩J^3 is homeomorphic to 𝕊^2. □  

Now, for each n∈ℕ we define 

    K^2_n-:= p^-1(n-1/2)∩J^3
 
and 

    K^2_n+:= p^-1(n+1/2)∩J^3.

Observe that K^2_n- and K^2_n+ are cubical 2-knots.  


Let C^3 be the set of 3-cubes (3-cells) belonging to the canonical cubulation of ℝ^5. Consider the three spaces:

    M^3_n-:={Q^3∈𝒞 ^3  | Q^3 ∩ K^2_n-≠∅},



    M^3_n+:={Q^3∈𝒞 ^3  | Q^3 ∩ K^2_n+≠∅}

and  M^3_n:= p^-1(n)∩J^3. 

By construction M^3_n-= K^2_n-× [0,1]
and M^3_n+= K^2_n+× [0,1].  

Let M^3:=M^3_n-∪ M^3_n∪ M^3_n+. Hence
M^3=(p^-1(n-1,n+1)∩J^3), where Cl denotes closure.


 The space M^3 is homeomorphic to 𝕊^2× [0,1].


Proof. Since M^3 is a compact submanifold of J^3, then by Lemma <ref>, it is also connected. Now J^3 is
homeomorphic to 𝕊^2×ℝ, and J^3-M^3 has two connected components, hence the result follows. □ 


 M^3_n has the homotopy type of 𝕊^2.


Proof. Consider the set M^3. Then by the previous Lemma, we have that  M^3≅𝕊^2× [0,1]. Notice that
K^2_n-×{0}≅𝕊^2×{0} and K^2_n+×{0}≅𝕊^2×{1}. 
Hence M^3=M^3/(K^2_n-×{0})∪ (K^2_n+×{1}) is homeomorphic to 𝕊^3. 
By Alexander duality,
we have that H_0(M^3-M^3_n,ℤ)=H^2 (M_n^3,ℤ), but M^3-M_n^3 has two simply connected components, so
H_0 (M^3,ℤ)≅ℤ. Since M_n^3⊂ M^3≅𝕊^2× [0,1], we have that either
π_2 (M_n^3)≅{0} or  π_2 (M_n^3)≅ℤ, but H̃^2(M_n^3,ℤ)≅ℤ, hence π_2 (M_n^3)≅ℤ. 

Again by Alexander duality, we have that H_1(M^3-M_n^3,ℤ)=H^1 (M_n^3,ℤ), but
H_1(M^3-M_n^3,ℤ)≅{0} hence H^1 (M_n^3,ℤ)≅{0}. This implies that 
π_1 (M_n^3)≅{0}.
Therefore, M_n^3 has the homotopy type of 𝕊^2. □  


The space M^3 retracts strongly to M_n^3.


Proof. Since M^3=M^3_n-∪ M^3_n∪ M^3_n+ is homeomorphic to 𝕊^2× [0,1], 
and M^3_n-= K^2_n-× [0,1] and M^3_n+= K^2_n+× [0,1], 
we have that 
M^3_n-= K^2_n-× [0,1] retracts strongly to K^2_n-×{1} and
M^3_n+= K^2_n+× [0,1] retracts strongly to K ^2_n+×{0}. Now
∂ M_n^3=(K^2_n-×{1})∪ (K^2_n+×{0}). Therefore, the result follows. □  


Next, we are going to describe the subset M_n^3. Notice that the squares of M_n^3 are of four types, which we will denote 
by T_-, T_+, T_± and T.



  * A  square F^2⊂ M_n^3 belongs to T_- if  F^2⊂ M^3_n- but F^2⊄M^3_n+.


  * A square F^2⊂ M_n^3 belongs to T_+ if F^2⊂  M^3_n+ but F^2⊄M^3_n-.


  * A square F^2⊂ M_n^3 belongs to T_± if F^2⊂  M^3_n-∩ M^3_n+.


  * A square F^2⊂ M_n^3 belongs to T if F^2⊄M^3_n+∪ M^3_n-.




By the above Lemma, there are copies of K^2_n- and K^2_n+ contained in ∂ M_n^3. By abuse of notation we will
denote them in the same way. Notice that K^2 _n- is the union of squares of types T_- and T_±, and
K^2_n+ is the union of squares of types T_+ and T_±.


K^2_n-c∼  K^2_n+: There exists a finite sequence of cubulated moves that carries the  2-knot K^2_n- into the 2-knot K^2_n+.


Proof. We will show it by cases.  
Case 1. Suppose that K^2_n-=K^2_n+.  Clearly, the result is true.  

Case 2. Suppose that K^2_n-∩ K^2_n+=∅. In other words, K^2_n- and 
K^2_n+
 do not have squares of type T_±.   
Remember that M_n^3 is a cubical compact 3-manifold whose fundamental group is isomorphic to ℤ. Thus  ∂ M_n^3
has two connected components; namely K^2_n- and K^2_n+, such that their intersection is empty.
Hence M_n^3 is the union of a finite number of cubes (3-faces) belonging to the 3-skeleton of the cubulation C, whose 2-faces are of any of the types T_-, T_+ and T.  

Next, we will carry the 2-knot K^2_n- onto the 2-knot K^2_n+ via a finite number of cubulated moves; i.e. we will
carry the squares of type T_- onto the squares of type T_+. Let Q^3 be a cube contained in M_n^3. We can assume, up to (M1)-move, that if a square
F^2⊂ Q^3 belongs to T_-, then Q^3∩ K^2_n+=∅  and
Q^3∩ K^2_n-  consists of either a square, two neighboring 2-faces or three neighboring 2-faces. Analogously, if
 F^2⊂ Q^3 belongs to T_+, then Q^3∩ K^2_n-=∅ and
Q^3∩ K^2_n+ consists of a square, two neighboring 2-faces or three neighboring 2-faces.  

The compact space M_n^3 is the union of a finite number of cubes, say m. We will enumerate them by levels in the following way. Remember that M_n^3 is a cubical compact 3-manifold, so we may assume that M_n^3⊂ℝ_+^3. 
Let q:ℝ^3↪ℝ be the projection
on the last coordinate. We define  the level k, M_k^3:=M_n^3∩ q^-1([k,k+1]), for k∈ℕ. Notice that  there exists k_1 and k_2 positive integers such that M_k^3=∅ for k_1≤ k≤ k_2. Then we start enumerating the cubes Q^3∈ M^3 by levels. We start at the level k_1. The first cube
Q_1^3 contains a 2-face of type T_-, and given the cube Q_n^3, the cube  Q_n+1^3 shares a 2-face F_n^2 with  Q_n^3 and whenever 
it is possible, we choose Q_n+1^3 in such a way that F_n^2 is parallel to F_n-1^2; otherwise we choose  Q_n+2^3 such that F^2_n+1 is parallel
to F^2_n-1, at the end we continue on the level k_1+1 and so on.  

We will use induction on m. Consider the cube Q_1^3. We apply the (M2)-move to Q_1^3 replacing the 2-faces of type T by 2-faces of type  T_-.
We consider Q_2^3. Observe that Q_1^3 and Q_2^3 share a 2-face of type T_-. Then we apply again the (M2)-move replacing the 2-faces of type 
T by 2-faces of type  T_-. We continue inductively. 

Notice that if F^2⊂  M_n^3 is a 2-face of type T_+ then F^2⊂∂ M_n^3;
so F^2 is not a common 2-face of two cubes Q_i^3 and Q^3_j in M_n^3; hence if  F^2 is replaced by a 2-face of type T_- then this replacement is not modified in 
any other next step. Therefore, the result follows.  

Case 3. Suppose that the intersection K^2_n-∩ K^2_n+ contains a finite number of 2-faces.  
The 3-manifold M_n^3 consists of connected components C_i^3, i=1,…,r such that each C_i^3 is the union of cubes
Q_i_1^3, …, Q_i_m_i^3∈C and the intersection C_i^3∩ C_j^3 is either empty or a square belonging to 
K^2_n-∩ K^2_n+. Therefore, we apply the previous argument to each C_i^3.  


Case 4. Suppose that the intersection K^2_n-∩ K^2_n+ contains a square of type T_±. The 3-manifold M_n^3 consists of 3-dimensional connected components C_i^3, i=1,…,r
 and cubical 2-disks γ_ij. As before, each C_i^3 is a union of cubes Q_i_1^3, …, Q_i_m_i^3∈C, and γ_ij is a 
 cubical disk (or edge) joining the component C_i^3 with the component C_j^3.
Observe that if γ_ij is the union of 2-faces of type T_±, hence γ_ij⊂ K^2_n-∩ K^2_n+.
Moreover K^2_n-∩ K^2_n+= γ_ij and  ∂ M_n^3=K^2_n-∪ K^2_n+.  

Since π_2 (M_n^3)≅ℤ, then  C_i^3  the homotopy type either the 2-sphere or the 3-ball. 
Suppose that C_i^3 has the homotopy type of the 2-sphere,
then by hypothesis ∂ C_i^3=∂ M_n^3=K^2_n-∪ K^2_n+ contains a face F^2 of type T_±, but F^2 does not belong to
any cube Q^3 of M_n^3; so F^2 does not belong to C_i^3. This is a contradiction, hence  C_i^3 has the homotopy type of the 3-ball.  

By the above, ∂ C_i^3 is homeomorphic to 𝕊^2 and consists of 2-faces of type T_- and T_+. Moreover, ∂ C_i^3 consists of two disks
D_-^2 and D_+^2, such that D_-^2 is the union of faces of type T_- and D_+^2 is the union of faces of type T_+. Now we apply the
argument of the case 2, so D_+^2 is replaced by D_-^2. Since we have a finite number of components C_i^3,
the result follows. □

 
K^2_1c∼K^2_2: There exists a finite sequence
of cubulated moves that carries K^2_1 into  K^2_2. In other words, K^2_1 is equivalent to  K^2_2 by cubulated moves. 

Proof of Lemma B.  Recall that there exist integer numbers 
m_1 and m_2 such that p^-1(t)∩J^3 = K^2_1 for all t≤ m_1 and
p^-1(t)∩J^3 = K^2_2 for all t≥ m_2. Consider the integer m_1+1. By 
Lemma <ref> there exists a finite number of cubulated moves that
carries the 2-knot K^2_1 into the 2-knot K^2_(m_1+1)-. We continue inductively, and again by 
Lemma <ref>
there exists a finite number of cubulated moves that
carries the knot K^2_(m_2-1)+ into the knot K^2_2. Since, we have a finite number of integers contained in the interval [m_1,m_2], then
there exists a finite sequence of cubulated moves that carries K^2_1 into  K^2_2. □





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J. P. Díaz. Instituto de Matemáticas, Unidad Cuernavaca. Universidad Nacional Au­tó­no­ma de México.
Av. Universidad s/n, Col. Lomas de Chamilpa. Cuernavaca, Morelos, México, 62209.

E-mail address: juanpablo@matcuer.unam.mx
.3cm
G. Hinojosa. Centro de Investigación en Ciencias. Instituto de Investigación en Ciencias Básicas y Aplicadas. Universidad Autónoma del Estado de Morelos. Av. Universidad 1001, Col. Chamilpa.
Cuernavaca, Morelos, México, 62209. 

E-mail address: gabriela@uaem.mx 

.3cm
A. Verjovsky. Instituto de Matemáticas, Unidad Cuernavaca. Universidad Nacional Au­tó­no­ma de México.
Av. Universidad s/n, Col. Lomas de Chamilpa. Cuernavaca, Morelos, México, 62209.

E-mail address: alberto@matcuer.unam.mx


