# Doubly Semiequivelar Maps on Torus and Klein Bottle

Authors: Anand K. Tiwari; Amit Tripathi; Yogendra Singh; Punam Gupta
Journal: Journal of Mathematics
 (2020)

---

## Abstract

A tiling of the Euclidean plane, by regular polygons, is called 2-uniform tiling if it has two orbits of vertices under the action of its symmetry group. There are 20 distinct 2-uniform tilings of the plane. Plane being the universal cover of torus and Klein bottle, it is natural to ask about the exploration of maps on these two surfaces corresponding to the 2-uniform tilings. We call such maps as doubly semiequivelar maps. In the present study, we compute and classify (up to isomorphism) doubly semiequivelar maps on torus and Klein bottle. This classification of semiequivelar maps is useful in classifying a category of symmetrical maps which have two orbits of vertices, named as 2-uniform maps.


---

## Body

## 1. Introduction

Equivelar and semiequivelar maps are generalizations of the maps on the surfaces of well-known Platonic solids and Archimedean solids to the closed surfaces other than the 2-sphere, respectively. A substantial literature is available for such maps (see [1–8]).Tilings of the plane are a great source of polyhedral maps on the surfaces of torus and Klein bottle, as the plane is the universal cover of these two surfaces. A tiling of the plane, by regular polygons, is called ak-uniform tiling if it has k orbits of vertices under its symmetry. The k-uniform tilings have been completely enumerated for k ≤ 6. There are 11 1-uniform, 20 2-uniform, 61 3-uniform, 151 4-uniform, 332 5-uniform, and 673 6-uniform tilings on the plane. For a detailed study on such tilings, readers are referred to see [9–11].The 11 1-uniform tilings of the plane are also called Archimedean tilings. Out of these, 3 are regular and 8 are semiregular tilings. The 3 regular tilings provide equivelar maps of types [36], [44], and [63] and 8 semiregular tilings provide semiequivelar maps of types [34, 6], [33, 42], [32, 4, 3, 4], [3, 4, 6, 4], [3, 6, 3, 6], [3, 122], [4, 6, 12], and [4, 82] on torus and Klein bottle. Altshuler [12] has given a construction for a map of the type [36] and [63] on the torus. Kurth [13] has enumerated maps of the types [36], [44], and [63] on the torus. In [2], Datta and Nilakantan classified map of type [36] and [44] on at most 11 vertices. In continuation of this, Datta and Upadhyay [14] classified these type of maps for n vertices with 12 ≤ n ≤ 15. In [15], Brehm and Kuhnel have classified these three types equivelar maps on the torus using a different approach. In [16], Tiwari and Upadhyay have classified the 8 types semiequivelar maps on at most 20 vertices. Recently, Maity and Upadhyay [17] have presented a way to classify the eight types of semiequivelar maps on the torus for arbitrary number of vertices.Analogues to the Archimedean tilings, here we initiate the theory of maps on torus and Klein bottle corresponding to the 2-uniform tilings. We call such maps as doubly semiequivelar map(s) or briefly DSEM(s). The present work provides a new class of polyhedra which have two classes of vertices in terms of the arrangement of polygons around the vertices. Polyhedra play an important role in human life. It has extensive application in ornament designing, architectural designing, cartography, computer graphics etc. (see [18–20]).This article is organized as follows: In Section2, we give basic definitions and notations used in the present work. In Section 3, we define doubly semiequivelar map (DSEM) and describe a methodology to enumerate a doubly semiequivelar map on torus and Klein bottle. In Section 4, we compute and classify DSEMs on torus and Klein bottle. In Section 5, we present the results obtained from the computation and classification. A tabular form of the results is shown in Table 1. In Section 6, we present discussion and future scope of the DSEMs followed by some concluding remarks.Table 1
DSEMs of face-size 4 on torus and Klein bottle on ≤15 vertices.
S. No.
Map type
|V|
No. of maps
On torus
On Klein bottle
(1)
9
4
T
1
3,6
3
6
:
3
3
,
4
2
1, T23,636:33,421
K
1
3,6
3
6
:
3
3
,
4
2
1, K23,636:33,421
3
6
:
3
3
,
4
2
1
12
6
T
3
4,8
3
6
:
3
3
,
4
2
1, T44,836:33,421, T54,836:33,421
K
3
4,8
3
6
:
3
3
,
4
2
1, K44,836:33,421, K54,836:33,421
15
5
T
6
5,10
3
6
:
3
3
,
4
2
1, T75,1036:33,421, T85,1036:33,421
K
6
5,10
3
6
:
3
3
,
4
2
1, K75,1036:33,421
(2)
3
6
:
3
3
,
4
2
2
12
3
T
1
6,6
3
6
:
3
3
,
4
2
2, T26,636:33,422
K
1
6,6
3
6
:
3
3
,
4
2
2
(3)
[36 : 32, 4, 3, 4]
14
1
—
K
1(2,12)[36 : 32, 4, 3, 4]
(4)
3
3
,
4
2
:
3
2
,
4,3,4
1
12
1
T
1
4,8
3
3
,
4
2
:
3
2
,
4,3,4
1
—
(5)
3
3
,
4
2
:
3
2
,
4,3,4
2
12
1
—
K
1
6,6
3
3
,
4
2
:
3
2
,
4,3,4
2
(6)
9
3
T
1
3,6
3
3
,
4
2
:
4
4
1, T23,633,42:441
K
1
3,6
3
3
,
4
2
:
4
4
1
3
3
,
4
2
:
4
4
1
12
3
T
3
4,8
3
3
,
4
2
:
4
4
1, T44,833,42:441
K
2
4,8
3
3
,
4
2
:
4
4
1
15
5
T
5
5,10
3
3
,
4
2
:
4
4
1, T65,1033,42:441, T75,1033,42:441
K
3
5,10
3
3
,
4
2
:
4
4
1, K45,1033,42:441
(7)
3
3
,
4
2
:
4
4
2
12
3
T
1
6,6
3
3
,
4
2
:
4
4
2, T26,633,42:442
K
1
6,6
3
3
,
4
2
:
4
4
2


## 2. Basic Definitions and Notations

For graph theory related terminologies, we refer [21]. A p-cycle, denoted as Cp, is a 2-regular graph with p vertices. We denote Cp explicitly as Cp(v1,…,vp), where the vertex set V(Cp) = {v1,…,vp} and edge set E(Cp) = {v1v2,…,vn−1vn,vnv1}.A surface (closed surface)F is a connected, compact 2-manifold without boundary. A surface F is either sphere, sphere with g handles (also called orientable surface of genus g, denoted as Sg), or sphere with g cross caps (also called nonorientable surface of genus g, denoted as Ng). To a surface, we associate a unique integer called its Euler characteristic χ and is defined as χ(Sg) = 2 − 2g and χ(Ng) = 2 − g. The surfaces S1 and N2 of Euler characteristic 0 are called torus and Klein bottle, respectively.An embedding of a connected, simple graphG into a surface F is called 2-cell embedding if the closure of each connected component of F\G is a 2-disk Dp. These components are called faces of the embedding. The vertices and edges of G are called the vertices and edges of the embedding. A map (polyhedral) M on a surface F is a 2-cell embedding such that the nonempty intersection of any two faces is either a vertex or an edge [22]. The face size of a map M is p, if p is the largest positive integer such that M has a face Dp.Two mapsM1 and M2, with vertex sets V(M1) and V(M2), respectively, are said to be isomorphic if there is a bijective map f: V(M1) ⟶ V(M2) which preserves the incidence of edges and incidence of faces. An isomorphism from a map M to itself is also called an automorphism. A collection Aut(M) of all the automorphisms of a map M forms a group under the composition of maps, called the automorphism group of M. A map M is called vertex-transitive if it has a unique orbit of vertices under the action of Aut(M).The face-sequence [7] of a vertex v, denoted as f−seqv, in a map M is a finite cyclic sequence p1n1,…,pknk, where p1, …, pk ≥ 3 and n1, …, nk ≥ 1, such that the face cycle at v is Dp1,…,n1times,…,Dpk,…,nktimes. A map is called semiequivelar of type p1n1,…,pknk if the face-sequence of each vertex is p1n1,…,pknk. A semiequivelar map of type [pn] is also called equivelar map.LetDp1,…,Dpk be the face cycle at a vertex v in a map M. Let Cpi denote the boundary cycles of these Dpi. Then, the link of v, denoted as lkv, is a cycle in M consisting of all the vertices of these Cpi’s except v and all the edges of these Cpi’s except which has one end vertex v. If v is a vertex with lkv=Ckv1,…,vk, the face-sequence of lkv is a cyclically ordered sequence (f − seq(v1), …, f − seq(vk)).Letv be a vertex with the face-sequence p1n1,…,pknk. The combinatorial curvature of v, denoted by ϕv, is defined as ϕv=1−∑i=1kni/2+∑i=1kni/pi.


## 3. Definition of the Problem and Description of Method

LetM be a map with two distinct face-sequences f1 and f2. We say that M is a doubly semiequivelar map, in short DSEM, if (i) the sign of ϕ(v) is same for all v ∈ M and (ii) vertices of same type face-sequence also have links of the same face-sequence up to a cyclic permutation. A doubly semiequivelar map M is called 2-uniform if it has 2 orbits of vertices under the action of its automorphism group. We denote the M of type f1f11,…,f1r1:f2f21,…,f2r2, where f1i or f2j is f1 or f2, for 1 ≤ i ≤ r1 and 1 ≤ j ≤ r2, if vertices of the face-sequence f1 have links of face-sequence f11,…,f1r1 and vertices of the face-sequence f2 have links of face-sequence f21,…,f2r2, respectively.There are 20 types of 2-uniform tilings of the plane denoted as36:33,421, 36:33,422, [36: 32, 4, 3, 4], 33,42:32,4,3,41, 33,42:32,4,3,42, 33,42:441, 33,42:442, 36:34,61, 36:34,62, [36: 32, 4, 12], [36: 32, 62], [34, 6: 32, 62], 33,42:3,4,6,41, [32, 4, 3, 4, 42: 3, 4, 6, 4], [32, 62: 3, 6, 3, 6], [3, 4, 3, 12 : 3, 122], [3, 42, 6 : 3, 4, 6, 4], 3,42,6:3,6,3,61, 3,42,6:3,6,3,62, [3, 4, 6, 4 : 4, 6, 12] (see [11]). Out of these, the first seven types have p-gons, with p ≤ 4 (see Figure 1).Figure 1
2-uniform tilings of types:36:33,421, 36:33,422, 36:32,4,3,4, 33,42:441, 33,42:32,4,3,41, 33,42:32,4,3,42, 33,42:442.
(a)
(b)
(c)
(d)
(e)
(f)
(g)


### 3.1. 2-Uniform Tilings of the Plane

We classify the DSEMs on torus and Klein bottle corresponding to the above seven tilings. We abbreviate the types of DSEMs by the same notation as used for the respective tilings (see Table2).Table 2
Tabulated list of DSEMs of face-size 4.
S No.
Abbreviated form
DSEM type
(1)
3
6
:
3
3
,
4
2
1
3
6
3
6
,
3
3
,
4
2
,
3
3
,
4
2
,
3
6
,
3
3
,
4
2
,
3
3
,
4
2
:
3
3
,
4
2
3
6
,
3
6
,
3
3
,
4
2
,
3
3
,
4
2
,
3
3
,
4
2
,
3
3
,
4
2
,
3
3
,
4
2
(2)
3
6
:
3
3
,
4
2
2
3
6
3
6
,
3
6
,
3
6
,
3
6
,
3
3
,
4
2
,
3
3
,
4
2
:
3
3
,
4
2
3
6
,
3
6
,
3
3
,
4
2
,
3
3
,
4
2
,
3
3
,
4
2
,
3
3
,
4
2
,
3
3
,
4
2
(3)
[36 : 32, 4, 3, 4]
3
6
3
2
,
4,3,4
,
3
2
,
4,3,4
,
3
2
,
4,3,4
,
3
2
,
4,3,4
,
3
2
,
4,3,4
,
3
2
,
4,3,4
:
3
2
,
4,3,4
3
6
,
3
2
,
4,3,4
,
3
2
,
4,3,4
,
3
2
,
4,3,4
,
3
2
,
4,3,4
,
3
2
,
4,3,4
,
3
2
,
4,3,4
(4)
3
3
,
4
2
:
3
2
,
4,3,4
1
3
3
,
4
2
3
3
,
4
2
,
3
2
,
4,3,4
,
3
2
,
4,3,4
,
3
2
,
4,3,4
,
3
2
,
4,3,4
,
3
2
,
4,3,4
,
3
2
,
4,3,4
:
3
2
,
4,3,4
3
3
,
4
2
,
3
2
,
4,3,4
,
3
2
,
4,3,4
,
3
2
,
4,3,4
,
3
3
,
4
2
,
3
3
,
4
2
,
3
2
,
4,3,4
(5)
3
3
,
4
2
:
3
2
,
4,3,4
2
3
3
,
4
2
3
3
,
4
2
,
3
2
,
4,3,4
,
3
2
,
4,3,4
,
3
2
,
4,3,4
,
3
3
,
4
2
,
3
3
,
4,3,4
,
3
3
,
4,3,4
:
3
2
,
4,3,4
3
3
,
4
2
,
3
3
,
4
2
,
3
3
,
4
2
,
3
2
,
4,3,4
,
3
3
,
4
2
,
3
3
,
4
2
,
3
2
,
4,3,4
(6)
3
3
,
4
2
:
4
4
1
3
3
,
4
2
3
3
,
4
2
,
3
3
,
4
2
,
3
3
,
4
2
,
3
3
,
4
2
,
4
4
,
4
4
,
4
4
:
4
4
3
3
,
4
2
,
3
3
,
4
2
,
3
3
,
4
2
,
4
4
,
3
3
,
4
2
,
3
3
,
4
2
,
3
3
,
4
2
,
4
4
(7)
3
3
,
4
2
:
4
4
2
3
3
,
4
2
3
3
,
4
2
,
3
3
,
4
2
,
3
3
,
4
2
,
3
3
,
4
2
,
4
4
,
4
4
,
4
4
:
4
4
3
3
,
4
2
,
3
3
,
4
2
,
3
3
,
4
2
,
4
4
,
4
4
,
4
4
,
4
4
,
4
4


### 3.2. Methodology

Each doubly semiequivelar map, out of the seven types (listed in Table2), contains two types of face-sequences among the four types (36), (33, 42), (32, 4, 3, 4), and (44) around the vertices. We use the following notations frequently to denote a vertex with specific type face-sequence in the computation. Here lk(v) means link of vertex v.(i)
The notation lk(v) = C6(a, b, c, d, e, f) means the face-sequence of v is (36), i.e., the triangular faces [v, b, c], [v, c, d], [v, d, e], [v, e, f], [v, f, a], and [v, b, a] are incident at v.(ii)
lk(v) = C7(a, b, [c, d, e, f, g]) means the face-sequence of v is (33, 42), i.e., the triangular faces [v, a, g], [v, a, b], and [v, b, c] and quadrangular faces [v, c, d, e] and [v, e, f, g] are incident at v.(iii)
lk(v) = C7(a, [b, c, d], [e, f, g]) means the face-sequence of v is (32, 4, 3, 4), i.e., the triangular faces [v, a, b], [v, a, g], and [v, d, e] and quadrangular faces [v, b, c, d] and [v, e, f, g] are incident at v.(iv)
lk(v) = C8(a, b, c, d, e, f, g, h) means the face-sequence of v is (44), i.e., the quadrangular faces [v, a, b, c], [v, c, d, e], [v, e, f, g], and [v, g, h, a] are incident at v.Since a doubly semiequivelar map contains two types of vertices, in terms of face-sequences, to distinguish these vertices, we denote vertices of one type face-sequence byn and the other type by an, for some n∈ℕ. We describe a methodology to compute and classify the DSEMs listed in Table 2. Without loss of generality, we illustrate the methodology for type 36:33.421. The same procedure is used for the remaining six types.LetM be a DSEM of type 36:33.421 with vertex set V on a surface of Euler characteristic 0 (i.e., on torus or Klein bottle). Let V36 and V33,42 denote the set of vertices with face-sequence type (36) and (33, 42), respectively. Here V36 and V33,42 denote the cardinality of the sets V36 and V33,42, respectively. Then, it is easy to see that the number of triangular faces is 4V36 or 2V33,42. Thus, if the map exists, then 2V36=V33,42. Therefore, we have V=V36∪V33,42=a1,a2,…,aV36,1,2,…,V33,42 such that 2V36=V33,42. Now we use the following steps to enumerate DSEM M for this V=V36∪V33,42.Steps to enumerate DSEMs of type36:33.421:Step 1:(1)
Without loss of generality, let us start with a vertexv1 having face-sequence type (36). Let lk(a1) = C6(a2, 1, 2, a3, 3, 4).(2)
This implies lk(a2) = C6(1, a1, 4, n2, x1, n1) with several choices for the triplet (n1, x1, n2) in V33,42×V36×V33,42 or lk4=C7a1,a2,n1,n2,n3,n4,3 with several choices for n1,n2,n3,n4∈V33,42×V33,42×V33,42×V33,42 (see Figure 2).(3)
Again among lk(a2) and lk(4), without loss of generality, we proceed with lka2=C61,a1,4n2,x1,n1. For each choice of (n1, x1, n2) we have distinct possibility for lk(a2). Out of these possibilities of lk(a2), we qualify those ones which preserves the face-sequence types of vertices. The similar procedure may be adopted for lk(4) (if required).Step 2: we continuously repeat Step 1 until we do not get the links of remaining vertices fromV.Step 3: the computation involved in Step 1 and Step 2 is case by case and exhaustive covering all possible scenarios.Step 4: we explore isomorphism between the maps obtained in Step 1 and Step 2, which leads to the enumeration of DSEMs of type36:33.421.Figure 2
Illustration of the methodology. (a) DSEM type: [36: 33, 42]1. (b) lk(a1) = C6(a2, 1, 2, a3, 3, 4) ⇒ lk(a2) or lk(4).
(a)
(b)To show that two mapsM1 and M2 are non-isomorphic, we compute the characteristic polynomials p(EG(M1)) and p(EG(M2)) of adjacency matrices associated the edge graphs EG(M1) and EG(M2) of the maps M1 and M2, respectively. The edge graph of M is a graph EG(M) consisting of vertices and edges of the map. Clearly, if p(EG(M1)) ≠ p(EG(M2)), M1 ≇ M2. However, if p(EG(M1)) = p(EG(M2)), we cannot say anything.


## 3.1. 2-Uniform Tilings of the Plane

We classify the DSEMs on torus and Klein bottle corresponding to the above seven tilings. We abbreviate the types of DSEMs by the same notation as used for the respective tilings (see Table2).Table 2
Tabulated list of DSEMs of face-size 4.
S No.
Abbreviated form
DSEM type
(1)
3
6
:
3
3
,
4
2
1
3
6
3
6
,
3
3
,
4
2
,
3
3
,
4
2
,
3
6
,
3
3
,
4
2
,
3
3
,
4
2
:
3
3
,
4
2
3
6
,
3
6
,
3
3
,
4
2
,
3
3
,
4
2
,
3
3
,
4
2
,
3
3
,
4
2
,
3
3
,
4
2
(2)
3
6
:
3
3
,
4
2
2
3
6
3
6
,
3
6
,
3
6
,
3
6
,
3
3
,
4
2
,
3
3
,
4
2
:
3
3
,
4
2
3
6
,
3
6
,
3
3
,
4
2
,
3
3
,
4
2
,
3
3
,
4
2
,
3
3
,
4
2
,
3
3
,
4
2
(3)
[36 : 32, 4, 3, 4]
3
6
3
2
,
4,3,4
,
3
2
,
4,3,4
,
3
2
,
4,3,4
,
3
2
,
4,3,4
,
3
2
,
4,3,4
,
3
2
,
4,3,4
:
3
2
,
4,3,4
3
6
,
3
2
,
4,3,4
,
3
2
,
4,3,4
,
3
2
,
4,3,4
,
3
2
,
4,3,4
,
3
2
,
4,3,4
,
3
2
,
4,3,4
(4)
3
3
,
4
2
:
3
2
,
4,3,4
1
3
3
,
4
2
3
3
,
4
2
,
3
2
,
4,3,4
,
3
2
,
4,3,4
,
3
2
,
4,3,4
,
3
2
,
4,3,4
,
3
2
,
4,3,4
,
3
2
,
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## 3.2. Methodology

Each doubly semiequivelar map, out of the seven types (listed in Table2), contains two types of face-sequences among the four types (36), (33, 42), (32, 4, 3, 4), and (44) around the vertices. We use the following notations frequently to denote a vertex with specific type face-sequence in the computation. Here lk(v) means link of vertex v.(i)
The notation lk(v) = C6(a, b, c, d, e, f) means the face-sequence of v is (36), i.e., the triangular faces [v, b, c], [v, c, d], [v, d, e], [v, e, f], [v, f, a], and [v, b, a] are incident at v.(ii)
lk(v) = C7(a, b, [c, d, e, f, g]) means the face-sequence of v is (33, 42), i.e., the triangular faces [v, a, g], [v, a, b], and [v, b, c] and quadrangular faces [v, c, d, e] and [v, e, f, g] are incident at v.(iii)
lk(v) = C7(a, [b, c, d], [e, f, g]) means the face-sequence of v is (32, 4, 3, 4), i.e., the triangular faces [v, a, b], [v, a, g], and [v, d, e] and quadrangular faces [v, b, c, d] and [v, e, f, g] are incident at v.(iv)
lk(v) = C8(a, b, c, d, e, f, g, h) means the face-sequence of v is (44), i.e., the quadrangular faces [v, a, b, c], [v, c, d, e], [v, e, f, g], and [v, g, h, a] are incident at v.Since a doubly semiequivelar map contains two types of vertices, in terms of face-sequences, to distinguish these vertices, we denote vertices of one type face-sequence byn and the other type by an, for some n∈ℕ. We describe a methodology to compute and classify the DSEMs listed in Table 2. Without loss of generality, we illustrate the methodology for type 36:33.421. The same procedure is used for the remaining six types.LetM be a DSEM of type 36:33.421 with vertex set V on a surface of Euler characteristic 0 (i.e., on torus or Klein bottle). Let V36 and V33,42 denote the set of vertices with face-sequence type (36) and (33, 42), respectively. Here V36 and V33,42 denote the cardinality of the sets V36 and V33,42, respectively. Then, it is easy to see that the number of triangular faces is 4V36 or 2V33,42. Thus, if the map exists, then 2V36=V33,42. Therefore, we have V=V36∪V33,42=a1,a2,…,aV36,1,2,…,V33,42 such that 2V36=V33,42. Now we use the following steps to enumerate DSEM M for this V=V36∪V33,42.Steps to enumerate DSEMs of type36:33.421:Step 1:(1)
Without loss of generality, let us start with a vertexv1 having face-sequence type (36). Let lk(a1) = C6(a2, 1, 2, a3, 3, 4).(2)
This implies lk(a2) = C6(1, a1, 4, n2, x1, n1) with several choices for the triplet (n1, x1, n2) in V33,42×V36×V33,42 or lk4=C7a1,a2,n1,n2,n3,n4,3 with several choices for n1,n2,n3,n4∈V33,42×V33,42×V33,42×V33,42 (see Figure 2).(3)
Again among lk(a2) and lk(4), without loss of generality, we proceed with lka2=C61,a1,4n2,x1,n1. For each choice of (n1, x1, n2) we have distinct possibility for lk(a2). Out of these possibilities of lk(a2), we qualify those ones which preserves the face-sequence types of vertices. The similar procedure may be adopted for lk(4) (if required).Step 2: we continuously repeat Step 1 until we do not get the links of remaining vertices fromV.Step 3: the computation involved in Step 1 and Step 2 is case by case and exhaustive covering all possible scenarios.Step 4: we explore isomorphism between the maps obtained in Step 1 and Step 2, which leads to the enumeration of DSEMs of type36:33.421.Figure 2
Illustration of the methodology. (a) DSEM type: [36: 33, 42]1. (b) lk(a1) = C6(a2, 1, 2, a3, 3, 4) ⇒ lk(a2) or lk(4).
(a)
(b)To show that two mapsM1 and M2 are non-isomorphic, we compute the characteristic polynomials p(EG(M1)) and p(EG(M2)) of adjacency matrices associated the edge graphs EG(M1) and EG(M2) of the maps M1 and M2, respectively. The edge graph of M is a graph EG(M) consisting of vertices and edges of the map. Clearly, if p(EG(M1)) ≠ p(EG(M2)), M1 ≇ M2. However, if p(EG(M1)) = p(EG(M2)), we cannot say anything.


## 4. Computation and Classification of DSEMs

In this section, we compute and classify the seven types DSEMs (listed in Table2) using the methodology given in Section 3. For the sake of computation, we consider the number of vertices ≤15.


### 4.1. Computation and Classification for Type36:33,421

Consider the following DSEMs of type36:33,421, in Figure 3, on torus and Klein bottle denoted by Tin,2n36:33,421, for i ∈{1, …, 8}, and Kin,2n36:33,421, for i ∈{1, …, 6}, respectively.Figure 3
Doubly semiequivelar maps on torus and Klein bottle of type36:33,421. (a) T1(3, 6) [36: 33, 42]1. (b) T2(3, 6) [36: 33, 42]1. (c) K1(3, 6) [36: 33, 42]1. (d) K2(3,6) [36: 33, 42]1. (e) T3(4,8) [36: 33, 42]1. (f) T4(4,8) [36: 33, 42]1. (g) K3(4,8) [36: 33, 42]1. (h) K4(4,8) [36: 33, 42]1. (i) K5(4,8) [36: 33, 42]1. (j) T5(4,8) [36: 33, 42]1. (k) T6(5,10) [36: 33, 42]1. (l) T7(5,10) [36: 33, 42]1. (m) K6(5,10) [36: 33, 42]1. (n) K7(5,10) [36: 33, 42]1. (o) T8(5,10) [36: 33, 42]1.
(a)
(b)
(c)
(d)
(e)
(f)
(g)
(h)
(i)
(j)
(k)
(l)
(m)
(n)
(o)Claim 1.
For the maps above, we have the following:(a)
T13,636:33,421≇T23,636:33,421.(b)
T34,836:33,421≇T44,836:33,421≇T54,836:33,421.(c)
K34,836:33,421≇K44,836:33,421≇K54,836:33,421.(d)
T65,1036:33,421≇T75,1036:33,421≇T85,1036:33,421.(e)
K65,1036:33,421≇K75,1036:33,421.Proof.
Letp(EG(M)) denote the characteristic polynomial of adjacency matrix associated with the edge graph of M. Then, the proof follows from the following polynomials:pEGT13,636:33,421=x9−24x7−42x6+63x5+138x4−72x3−144x2+48x+32pEGT23,636:33,421=x9−24x7−36x6+45x5+48x4−21x3−18x2+3x+2pEGT34,836:33,421=x12−32x10−40x9+254x8+440x7−628x6−1400x5+105x4+1000x3+300x2pEGT44,836:33,421=x12−32x10−32x9+254x8+224x7−932x6−448x5+1673x4+96x3−1156x2+160x+192pEGT54,836:33,421=x12−32x10−48x9+254x8+656x7−292x6−2352x5−2167x4+624x3+2044x2+1120x+192pEGK34,836:33,421=x12−31x10−39x9+227x8+377x7−561x6−1129x5+416x4+1283x3+92x2−492x−144pEGK44,836:33,421=x12−32x10−40x9+254x8+440x7−644x6−1400x5+457x4+1640x3+156x2−640x−192pEGK54,836:33,421=x12−32x10−48x9+258x8+640x7−364x6−220x5−1635x4+496x3+684x2−32x−64pEGT65,1036:33,421=x15−40x13−40x12+515x11+754x10−282x9−4940x8+6790x7+13430x6−668x5−15340x4+975x3+5490x2+1755x+162pEGT75,1036:33,421=x15−38x13−51x12+462x11+1033x10−1049x9−5533x8−4681x7+2905x6+6351x5+2282x4−1046x3−680x2+32x+48pEGT85,1036:33,421=x15−40x13−60x12+485x11+1374x10−985x9−7910x8−9955x7−1010x6+7623x5+7030x4+2820x3+570x2+55x+2pEGK65,1036:33,421=x15−39x13−58x12+462x11+1309x10−916x9−7455x8−9096x7−203x6+6562x5+3147x4−909x3−761x2−97x−3pEGK75,1036:33,421=x15−40x13−48x12+497x11+1010x10−1973x9−6234x8−111x7+12010x6+9531x5−3294x4−7264x3−3410x2−637x−38Claim 2.
K
1
3,6
3
6
:
3
3
,
4
2
1
≇
K
2
3,6
3
6
:
3
3
,
4
2
1
.Proof.
Note thatpEGK13,636:33,421=pEGK23,636:33,421=x9−24x7−38x6+51x5+78x4−44x3−24x2, but the maps are non-isomorphic. To see this, we use geometric argument as follows: define a basis {a, b} at any vertex vi (for 1 ≤ i ≤ 6), where a and b are minimal nontrivial loops (i.e., nontrivial cycle with minimum number of vertices); now, if we consider K13,636:33,421, then at each vi, we get a and b with length 3 (for example, at v1, a = C3(v1,v6,u2) and b = C3(v1,v2,v5)) while in K23,636:33,421, at each vi, we get a of length 3 and b of length 4 (for example, at v1, we see that a = C3(v4,v1,u1) or a = C3(v4,v1,u2) and b = C4(v1,v2,v6,v4)). Hence, K13,636:33,421≇K23,636:33,421.


#### 4.1.1. Computation

LetM be a DSEM of type 36:33,421 with the vertex set V. Let V36 and V33,42 denote the sets of vertices with face-sequence types (36) and (33, 42), respectively. Then, we see that the number of triangular faces in M is 4V36 or 2V33,42. This implies 2V36=V33,42. Thus, for V=V33,42+V33,42≤15, we let V=a1,a2,…,aV36,1,2,…,2|V36|, where V36|≤5. Without loss of generality, we may assume lk(a1) = C6(a2, 1, 2, a3, 3, 4). This implies lk(a2) = C6(a1, 1, n1, x1, n2, 4), lk(1) = C7(a1, a2, [n1, n3, n4, n5, 2]), lk2=C7a1,a3, n6,n7,n5,n4,1, lka3=C6a1,3, n8,x2,n6,2, lk3=C7a1,a3,n8,n9,n10,n11,4, and lk4=C7a1,a2,n2,n12,n11,n10,3 for some x1,x2∈V36 and n1,n2,…,n12∈V33,42.Now considering lk(a2), we see that n1≠2 or 3 (for n1 = 2, the set {2, 1, a2} forms triangular face in lk(a1) but not in lk(a2); for n1 = 3, we get deg(3) > 5). Similarly, we see that n2∉{2, 3}. From these observations, we have (n1, x1, n2) ∈{(5, a3, 6), (5, a4, 6)}.Case 1. if (n1, x1, n2) = (5, a3, 6), then lk(a3) = C6(a1, 3, 6, a2, 5, 2) or lka3=C6a1,3,5,a2,6,2.Whenlka3=C6a1,3,5,a2,6,2, considering lk(2), we have n4,n5,n7∈4,3,5,5,3,4. If n4,n5,n7=5,3,4, then lk(1) is a cycle of length 5, a contradiction. On the other hand, if n4,n5,n7=4,3,5, then lk2=C7a1,a3,6,5,3,4,1,lk1=C7a1,a2,5,6,4,3,2; completing successively, we get lk4=C7a1,a2,6,5,1,2,3, lk3=C7a1,a3,5,6,2,1,4, lk5=C7a2,a3,3,2,6,4,1, and lk6=C7a2,a3,2,3,5,1,4. Then, we get M≅K23,636:33,421 by the map i↦vi, aj↦uj, 1 ≤ i ≤ 6, 1 ≤ j ≤ 3.When lk(a3) = C6(a1, 3, 6, a2, 5, 2), considering lk(2), we get (n4, n5, n7) ∈{(4, 3, 6), (6, 3, 4), (3, 4, 6), (6, 4, 3), (3, 6, 4), (4, 6, 3)}. Observe that (3, 4, 6) ≅ (6, 3, 4) by the map (1, 5, 2) (3, 4, 6) (a1, a2, a3), (3, 6, 4) ≅ (6, 4, 3) by the map (1, 5) (4, 6) (a1, a3), and (4, 6, 3) ≅ (6, 3, 4) by the map (1, 2, 5) (3, 6, 4) (a1, a3, a2). So, we need to search only for (n4, n5, n7) ∈ {(4, 3, 6), (6, 3, 4), (6, 4, 3)}.In case (n4, n5, n7) = (4, 3, 6), completing successively, we get lk2=C7a1,a3,5,6,3,4,1, lk3=C7a1,a3,6,5,2,1,4, lk4=C7a1,a2,6,5,1,2,3, lk1=C7a1,a2,5,6,4,3,2, lk5=C7a2,a3,2,3,6,4,1, lk6=C7a2,a3,3,2,5,1,4. This gives M≅T13,636:33,421 by the map i↦vi, aj↦uj, 1 ≤ i ≤ 6, 1 ≤ j ≤ 3.Proceeding similarly as above, for (n4, n5, n7) = (6, 3, 4), we get M≅K13,636:33,421 by the map i↦vi, aj↦uj, where 1 ≤ i ≤ 6, 1 ≤ j ≤ 3.For (n4, n5, n7) = (6, 4, 3), M≅T23,636:33,421, by the map i↦vi, aj↦uj, for 1 ≤ i ≤ 6, 1 ≤ j ≤ 3.Case 2. For (n1, x1, n2) = (5, a4, 6), considering lk(a3), we get x2 ∈{a4, a5}.Subcase 2.1. If x2 = a4, then lk(a3) = C6(a1, 2, 7, a4, 8, 3). This implies lk(a4) = C6(a2, 5, 7, a3, 8, 6) or lk(a4) = C6(a2, 5, 8, a3, 7, 6).When lk(a4) = C6(a2, 5, 7, a3, 8, 6), then, up to isomorphism, we see that (n13, n14, n15) ∈ {(2, 1, 5), (5, 1, 2), (1, 2, 7), (7, 2, 1), (7, 5, 1)}. Now doing computation for these cases, we see the following:If (n13, n14, n15) = (2, 1, 5), M≅K34,836:33,421 by i↦vi, aj↦uj, 1 ≤ i ≤ 8, 1 ≤ j ≤ 4.Ifn13,n14,n15=5,1,2,M≅T34,836:33,421 by i↦vi, aj↦uj, 1 ≤ i ≤ 8, 1 ≤ j ≤ 4.Ifn13,n14,n15=1,2,7,M≅T44,836:33,421 by i↦vi, aj↦uj, 1 ≤ i ≤ 8, 1 ≤ j ≤ 4.Ifn13,n14,n15=7,2,1,M≅K44,836:33,421 by i↦vi, aj↦uj, 1 ≤ i ≤ 8, 1 ≤ j ≤ 4.Ifn13,n14,n15=7,5,1,M≅T54,836:33,421 by i↦vi, aj↦uj, 1 ≤ i ≤ 8, 1 ≤ j ≤ 4.On the other hand when lk(a4) = C6(a2, 5, 8, a3, 7, 6), we get (n13, n14, n15) ∈{(2, 1, 5), (1, 5, 8), (3, 8, 5), (5, 1, 2), (5, 8, 3), (8, 5, 1)}. If (n13, n14, n15) = (2, 1, 5) and (1, 5, 8), then lk(7) is a cycle of length 5 and 6 respectively, which is not possible. If (n13, n14, n15) = (3, 8, 5) and (5, 1, 2), then we see easily that lk(7) and lk(8) cannot be completed, respectively. If (n13, n14, n15) = (5, 8, 3), then completing lk(7) we get lk(1) of length 5, again a contradiction. If n13,n14,n15=8,5,1,M≅K54,836:33,421 by i↦vi, aj↦uj, 1 ≤ i ≤ 8, 1 ≤ j ≤ 4.Subcase 2.2. When x2 = a5, successively, we get lk(a3) = C6(a1, 2, 7, a5, 8, 3) and lk(a4) = C6(a2, 6, 9, a5, 10, 5). This implies lk(a5) = C6(a4, 9, 7, a3, 8, 10) or lk(a5) = C6(a4, 9, 8, a3, 7, 10).In case lk(a5) = C6(a4, 9, 7, a3, 8, 10), considering lk(1), we get (n3, n4, n5) ∈ {(3, 4, 6), (3, 8, 10), (4, 3, 8), (4, 6, 9), (6, 4, 3), (6, 9, 7), (7, 9, 6), (8, 3, 4), (9, 6, 4), (10, 8, 3)}. But a small calculation shows that no map exists for these cases, except for (n3, n4, n5) = (6, 4, 3). For (n3, n4, n5) = (6, 4, 3), we get M≅K65,1036:33,421 by the map i↦vi, aj↦uj, 1 ≤ i ≤ 10, 1 ≤ j ≤ 5.On the other hand, when lk(a5) = C6(a4, 9, 8, a3, 7, 10), considering lk(1), up to isomorphism, we get (n3, n4, n5) ∈ {(3, 4, 6), (3, 8, 9), (4, 3, 8), (6, 4, 3)}. Now doing computation for these cases, we see the following:If (n3, n4, n5) = (3, 4, 6), M≅K75,1036:33,421 by i↦vi, aj↦uj, 1 ≤ i ≤ 10 and 1 ≤ j ≤ 5.If (n3, n4, n5) = (3, 8, 9), M≅T65,1036:33,421 by i↦vi, aj↦uj, 1 ≤ i ≤ 10 and 1 ≤ j ≤ 5.If (n3, n4, n5) = (4, 3, 8), M≅T75,1036:33,421 by i↦vi, aj↦uj, 1 ≤ i ≤ 10 and 1 ≤ j ≤ 5.If (n3, n4, n5) = (6, 4, 3), M≅T85,1036:33,421 by i↦vi, aj↦uj, 1 ≤ i ≤ 10 and 1 ≤ j ≤ 5. This completes computation for the number of vertices ≤15 and we obtain the following results.


#### 4.1.2. Results

Lemma 1.
LetM be a DSEM of type 36:33,421 with number of vertices ≤15. Then, M is isomorphic to one of the following: T13,636:33,421, T23,636:33,421, K13,636:33,421, K23,636:33,421, T34,836:33,421, T44,836:33,421, T54,836:33,421, K34,836:33,421, K44,836:33,421, K54,836:33,421, T65,1036:33,421, T75,1036:33,421, T85,1036:33,421, K65,1036:33,421, and K75,1036:33,421, as shown in Figure 3.Combining Lemma1 together with Claims 1 and 2, we get the following theorem.Theorem 1.
There are exactly 15 DSEMs of type36:33,421 on the surfaces of Euler characteristic 0 with number of vertices ≤15. Out of these, 8 are on the torus and remaining 7 are on the Klein bottle.


### 4.2. Computation and Classification for Type36:33,422

Consider the following DSEMs of type36:33,422, shown in Figure 4, on torus and Klein bottle denoted by Ti6,636:33,422, for i = 1, 2, and K16,636:33,422, respectively.Figure 4
Doubly semiequivelar maps on torus and Klein bottle of type36:33,422. (a) K1(6,6) [36: 33, 42]2. (b) T1(6,6) [36: 33, 42]2. (c) T2(6,6) [36: 33, 42]2.
(a)
(b)
(c)Claim 3.
T
1
6,6
3
6
:
3
3
,
4
2
2
≇
T
2
6,6
3
6
:
3
3
,
4
2
2
.Proof.
The proof follows from the following polynomials:pEGT16,636:33,422=x12−33x10−44x9+258x8+432x7−682x6−1032x5+957x4+560x3−789x2+276x−32pEGT26,636:33,422=x12−33x10−44x9+252x8+456x7−568x6−1296x5+348x4+1328x3+108x2−432x−128


#### 4.2.1. Computation

LetM be a map of the type 36:33,422 with the vertex set V. Let V36 and V33,42 denote the sets of vertices with face-sequence types (36) and (33, 42), respectively. Observe that M has the number of edges = 5V36+V33,42/2, number of triangular faces = 3V36, and number of quadrangular faces = V33,42/2. Now by the Euler characteristic equation, we get V36+V33,42−5V36+V33,42/2+3V36+V33,42/2=0. This implies V36=V33,42. Also, considering the number of quadrangular faces, it is evident that the cardinality of both the sets should be positive even integer. Thus, for |V| ≤ 15, we let V=a1,a2,…,aV36,1,2,…,V36, where V36=2k for k ≤ 3.Without loss of generality, assume lk(a1) = C6(a2, a3, a4, a5, 1, 2). This implies lk(1) = C7(a1, a5, [n1, n2, n3, n4, 2]), lk(2) = C7(a2, a1, [1, n3, n4, n6, n5]) and lk(a2) = C6(a1, a3, x1, x2, 3, 2) for some n1, …, n6 ∈ Vm and x1, x2 ∈ Vl. It is easy to see that (x1, x2) ∈{(a5, a4), (a6, a5)}.Case 1. When (x1, x2) = (a5, a4), then successively we get lk(a2) = C6(a1, a3, a5, a4, 3, 2), lk(a5) = C6(a1, a4, a2, a3, 4, 1), lk(a3) = C6(a4, a1, a2, a5, 4, 5) and lk(a4) = C6(a3, a1, a5, a2, 3, 5). Now considering lk(1), we see that (n2, n3, n4) has no value for the V so that the links of remaining vertices can be completed. So, (x1, x2) ≠ (a5, a4).Case 2. When (x1, x2) = (a6, a5), then successively we get lk(a2) = C6(a1, a3, a6, a5, 3, 2), lk(a5) = C6(a1, a4, a6, a2, 3, 1), lk(a4) = C6(a3, a1, a5, a6, 5, 4), lk(a3) = C6(a4, a1, a2, a6, 6, 4), lk(a6) = C6(a3, a2, a5, a4, 5, 6). Considering lk(1), it is easy to see that (n2, n3, n4) ∈{(4, 5, 6), (4, 6, 5), (5, 4, 6), (5, 6, 4), (6, 4, 5), (6, 5, 4)}. Observe that (4, 5, 6) ≅ (6, 4, 5) by the map (1, 3, 2) (5, 6, 4) (a1, a5, a2) (a3, a4, a6), (4, 5, 6) ≅ (5, 6, 4) by the map (1, 2, 3) (4, 6, 5) (a1, a2, a5) (a3, a6, a4), and (5, 4, 6) ≅ (6, 5, 4) by the map (1, 3) (4, 6) (a1, a2) (a4, a6). So, we search for (n2, n3, n4) ∈ {(4, 5, 6), (4, 6, 5), (5, 4, 6)}. Now doing computation for these cases, we see the following:Ifn2,n3,n4=4,5,6,M≅K16,636:33,422 by the map i↦vi, ai↦ui, 1 ≤ i ≤ 6.Ifn2,n3,n4=4,6,5,M≅T16,636:33,422 by the map i↦vi, ai↦ui, 1 ≤ i ≤ 6.Ifn2,n3,n4=5,4,6,M≅T26,636:33,422 by the map i↦vi, ai↦ui, 1 ≤ i ≤ 6.This completes computation for ≤15 vertices. From this we get following results.


#### 4.2.2. Results

Lemma 2.
LetM be a DSEM of type 36:33,422 on the surfaces of Euler characteristic 0 with ≤15 vertices. Then, M is isomorphic to one of K16,636:33,422, T16,636:33,422, and T16,636:33,422, given in Figure 4.Combining Lemma2 with Claim 3, we get the following theorem.Theorem 2.
There are exactly 3 non-isomorphic DSEMs of type36:33,422 with number of vertices ≤15. Out of these, 2 are on torus and the remaining one is on Klein bottle.


### 4.3. Computation and Classification for Type [36: 32, 4, 3, 4]

Consider the following DSEM of type [36: 32, 4, 3, 4], given in Figure 5 on Klein bottle, denoted by K1(2,12)[36: 32, 4, 3, 4].Figure 5
Doubly semiequivelar maps on Klein bottle of typeK1(2,12) [36: 32, 4, 3, 4].


#### 4.3.1. Computation

LetM be a map of the type [36: 32, 4, 3, 4] with the vertex set V. Let V36 and V32,4,3,4 denote the sets of vertices with face-sequence types (36) and (32, 4, 3, 4), respectively. It is easy to see that 6V36=V32,4,3,4. Thus, for |V| ≤ 15, we let V=a1,a2,…,aV36,1,2,…,6V36, where V36≤2. Without loss of generality, we assume lk(a1) = C6(1, 2, 3, 4, 5, 6). Then, successively, we have lk(1) = C7(a1, [2, n1, n2], [n3, n4, 6]), lk(2) = C7(a1, [1, n2, n1], [n5, n6, 3]), lk(3) = C7(a1, [2, n5, n6], [n7, n8, 4]), lk(4) = C7(a1, [3, n7, n8], [n9, n10, 5]), lk(5) = C7(a1, [4, n9, n10], [n11, n12, 6]), lk(6) = C7(a1, [1, n3, n4], [n12, n11, 5]), where ni ∈ Vm for 1 ≤ i ≤ 12. Now considering lk(1), we see that n1 ∈{4, 5, 7}.Case 1. If n1 = 4, then successively, we see that n2 = 5, n3 = 7, and n4 = 8; now considering lk(5) and lk(1), we get two quadrangular faces which share more than one vertex, which is not allowed. So, n1 ≠ 4.Case 2. If n1 = 5, then successively, we get n2 = 4, n3 = 7, n4 = 8, n12 = 9, and n11 = 10. Now completing lk(1), lk(6), lk(5), lk(2), lk(3), lk(4), lk(7), and lk(10), we see that lk(a2) = C6(7, 8, 11, 10, 9, 12) or lk(a2) = C6(7, 8, 9, 10, 11, 12). If lk(a2) = C6(7, 8, 9, 10, 11, 12), then lk(8) is a cycle of length 5, a contradiction. If lk(a2) = C6(7, 8, 11, 10, 9, 12), then completing successively, we get M ≅ K1(2,12)[36: 32, 4, 3, 4] by the map i↦vi, aj↦uj, 1 ≤ i ≤ 12, 1 ≤ j ≤ 2.Case 3. If n1 = 7, then we get (n2, n3) ∈ {(8, 3), (8, 4), (8, 9)}. For (n2, n3) = (8, 3), n4 = 4 and lk(1) = C7(a1, [2, 7, 8], [3, 4, 6]). Now considering successively lk(3) and lk(1), we see two quadrangular faces which share more than one vertex, which is not allowed. Hence, (n2, n3) ≠ (8, 3).For (n2, n3) = (8, 4), completing successively lk(1), lk(4), lk(5), lk(6), lk(3), lk(2), lk(11), and lk(8), we see that lk(a2) = C6(7, 8, 9, 10, 11, 12) or lk(a2) = C6(7, 8, 9, 12, 11, 10).If lk(a2) = C6(7, 8, 9, 10, 11, 12), then lk(7) is a cycle of length 5, a contradiction.If lk(a2) = C6(7, 8, 9, 12, 11, 10), completing successively, we get M ≅ K1(2,12)-[36: 32, 4, 3, 4] via 1↦v9, 2↦v10, 3↦v11, 4↦v8, 5↦v7, 6↦v12, 7↦v5, 8↦v6, 9↦v1, 10↦v4, 11↦v3, 12↦v2, a1↦u2, a2↦u1.If (n2, n3) = (8, 9), n4 = 10. This implies lk(1) = C7(a1, [2, 7, 8], [9, 10, 6]) and lk(6) = C7(a1, [1, 9, 10], [n12, n11, 5]) for (n11, n12) ∈ {(3, 2), (11, 4), (11, 7), (11, 12)}. A small calculation shows no map exists for (n11, n12) ∈ {(11, 4), (11, 7), (11, 12)}. For (n11, n12) = (3, 2), completing successively, we get M ≅ K1(2,12)[36: 32, 4, 3, 4] via 1↦v3, 2↦v4, 3↦v5, 4↦v6, 5↦v1, 6↦v2, 7↦v7, 8↦v12, 9↦v11, 10↦v10, 11↦v9, 12↦v8, a1↦u1, a2↦u2. This completes computation of the DSEM for ≤15. This gives the following result.


#### 4.3.2. Result

Theorem 3.
There exists a unique DSEM of type [36: 32, 4, 3, 4] on the surfaces of Euler characteristic 0 for ≤15 vertices. This is K1(2,12)[36: 32, 4, 3, 4] on Klein bottle, given in Figure 5.


### 4.4. Computation and Classification for Type33,42:32,4,3,41

Consider the following DSEM of type33,42:32,4,3,41, shown in Figure 6, on torus denoted by T14,833,42:32,4,3,41.Figure 6
Doubly semiequivelar map on torus of typeT1(4,8) [33, 42: 32, 4, 3, 4]1.


#### 4.4.1. Computation

LetM be a map of the type 33,42:32,4,3,41 with the vertex set V. Let V33 and V32,4,3,4 denote the sets of vertices with face-sequence types (33, 42) and (32, 4, 3, 4), respectively. It is easy to see that 2V33,42=V32,4,3,4 and V32,4,3,4 is multiple of 4. Therefore, for |V| ≤ 15, we let V=a1,a2,…,aV33,42,1,2,…,2V33,42, where V33,42≤4. Assume that lk(a1) = C7(2, 3, [4, 5, a2, 6, 1]). Then, lk(a2) = C7(7, 8, [5, 4, a1, 1, 6]). This implies lk(2) = C7(a1, [1, x1, x2], [n1, n2, 3]) or lk(2) = C7(a1, [3, x1, x2], [n1, n2, 1]), for x1,x2∈V33,42 and n1,n2∈V32,4,3,4. In the first case of lk(2), considering lk(1), we see three quadrangular faces incident at 1, which is not allowed. On the other hand, when lk(2) = C7(a1, [3, x1, x2], [n1, n2, 1]), we get x1 = a3, x2 = a4, and (n1, n2) ∈ {(5, 8), (7, 8), (8, 5), (8, 7)}.For (n1, n2) = (7, 8), considering lk(1) and lk(2), we see that lk(a3) = C7(1, 8, [3, 2, a4, n3, 6]) or lk(a3) = C7(1, 6, [3, 2, a4, n3, 8]), but for both the cases of lk(a3), we get no suitable value for n3 in V32,4,3,4. So, (n1, n2) ≠ (7, 8).For (n1, n2) = (8, 5), completing lk(2) and lk(1) and proceeding, as in previous case, we see that lk(3) cannot be completed.For (n1, n2) = (8, 7), considering lk(2) and lk(1), we see that lk(7) cannot be completed.For (n1, n2) = (5, 8), successively, we get lk(2) = C7(a1, [3, a3, a4], [5, 8, 1]), lk(5) = C7(a4, [4, a1, a2], [8, 1, 2]), and lk(1) = C7(a3, [6, a2, a1], [2, 5, 8]). Then, lk(a3) = C7(1, 8, [3, 2, a4, 7, 6]) or lk(a3) = C7(1, 6, [3, 2, a4, 7, 8]).When lk(a3) = C7(1, 8, [3, 2, a4, 7, 6]), completing lk(a4), we see that lk(7) cannot be completed.When lk(a3) = C7(1, 6, [3, 2, a4, 7, 8]), completing successively, we get M≅T14,833,42:32,4,3,41 by the map i↦vi, aj↦uj, 1 ≤ i ≤ 8, 1 ≤ j ≤ 4. Thus, the computation is completed for ≤15 vertices. This leads to the following result.


#### 4.4.2. Result

Theorem 4.
There exists a unique DSEM of type33,42:32,4,3,41 with number of vertices ≤15. This is T14,833,42:32,4,3,41 on torus, shown in Figure 6.


### 4.5. Computation and Classification for Type33,42:32,4,3,42

Consider the following DSEM of type33,42:32,4,3,42, shown in Figure 7, on Klein bottle denoted by K16,633,42:32,4,3,42.Figure 7
Doubly semiequivelar map of typeK16,633,42:32,4,3,41 on Klein bottle.


#### 4.5.1. Computation

LetM be a map of the type 33,42:32,4,3,42 with the vertex set V. Let V33,42 and V32,4,3,4 denote the sets of vertices with face-sequence types (33, 42) and (32, 4, 3, 4), respectively. Then, we see easily that V33,42=V32,4,3,4=2k for k∈ℕ. Thus, for V=V33,42+V32,4,3,4≤15, we let V=a1,a2,…,aV33,42,1,2,…,V33,42, where V33,42=2k for k ≤ 3. Assume that, without loss of generality, lk(a1) = C7(a3, 3, [4, 5, a2, 1, 2]). This implies lk(a2) = C7(x1, n1, [5, 4, a1, 2, 1]) or lk(a2) = C7(x1, n1, [1, 2, a1, 4, 5]) for x1∈V33,42 and n1∈V32,4,3,4.If lk(a2) = C7(x1, n1, [5, 4, a1, 2, 1]), then a small calculation shows that no such map exists for the given V. On the other hand, if lk(a2) = C7(x1, n1, [1, 2, a1, 4, 5]), then we have x1 = a4, n1 = 6. This implies lk(2) = C7(a3, [a1, a2, 1], [x2, x3, n2]), where (x2, x3, n2) ∈ {(a5, a4, 5), (a5, a4, 6)}. If (x2, x3, n2) = (a5, a4, 5), then lk(a3) = C7(a1, 2, [5, n3, a6, n4, 3]). Now considering lk(5), we see three quadrangular faces incident at 5, which is not allowed. If (x2, x3, n2) = (a5, a4, 6), lk(2) = C7(a3, [a1, a2, 1], [a5, a4, 6]). This implies lk(a3) = C7(a1, 2, [6, n3, a6, n4, 3]), where (n3, n4) ∈ {(1, 5), (4, 1), (4, 5)}. In case (n3, n4) = (4, 1) and (4, 5), we see, respectively, lk(1) and lk(4) cannot be completed. If n3,n4=1,5,M≅K16,633,42:32,4,3,42 by the map i↦vi, ai↦ui, 1 ≤ i ≤ 6. Thus, the computation is completed. Then, we obtain the following result.


#### 4.5.2. Result

Theorem 5.
There exists a unique DSEM of type33,42:32,4,3,42 with number of vertices ≤15. This is K16,633,42:32,4,3,42 on Klein bottle, shown in Figure 7.


### 4.6. Computation and Classification for Type33,42:441

Consider the DSEMs of type33,42:441, shown in Figure 8, on torus and Klein bottle denoted by Tin,2n33,42:441, for i ∈ {1, …, 6}, and Kin,2n33,42:441, for i ∈ {1, …, 3}, respectively.Figure 8
Doubly semiequivelar maps of type33,42:441 on torus and Klein bottle. (a) K1(3,6) [33, 42: 44]1. (b) T1(3,6) [33, 42: 44]1. (c) T2(3,6) [33, 42: 44]1. (d) K2(4,8) [33, 42: 44]1. (e) T3(4,8) [33, 42: 44]1. (f) T4(4,8) [33, 42: 44]1. (g) T5(5,10) [33, 42: 44]1. (h) T6(5,10) [33, 42: 44]1. (i) K4(5,10) [33, 42: 44]1. (j) T7(5,10) [33, 42: 44]1. (k) K3(5,10) [33, 42: 44]1.
(a)
(b)
(c)
(d)
(e)
(f)
(g)
(h)
(i)
(j)
(k)Claim 4.
For the maps above, we have the following:(a)
T13,633,42:441≇T23,633,42:441.(b)
T34,833,42:441≇T44,833,42:441.(c)
T55,1033,42:441≇T65,1033,42:441≇T75,1033,42:441.(d)
K35,1033,42:441≇K45,10-33,42:441.Proof.
The proof follows by considering the following polynomials:pEGT13,633,42:441=x9−21x7−24x6+72x5+72x4−99x3−54x2+54xpEGT23,633,42:441=x9−21x7−18x6+54x5pEGT34,833,42:441=x12−28x10−16x9+212x8+88x7−684x6−48x5+912x4−272x3−240x2+96xpEGT44,833,42:441=x12−28x10−24x9+212x8+280x7−524x6−976x5+80x4+880x3+528x2+96xpEGK35,1033,42:441=x15−5x13−40x12+385x11+790x10−1100x9−3620x8+55x7+6200x6+3305x5−3500x4−3265x3−190x2+270x−24pEGK45,1033,42:441=x15−35x13−24x12+401x11+434x10−1832x9−2468x8+3123x7+5232x6−939x5−3716x4−1261x3+30x2+30xpEGT55,1033,42:441=x15−35x13−20x12+425x11+294x10−2500x9−1520x8+7855x7+3060x6−12919x5−1100x4+8815x3−2250x2+150xpEGT65,1033,42:441=x15−35x13−20x12+395x11+344x10−1790x9−1960x8+3150x7+3920x6−2059x5−3000x4+235x3+750x2+150xpEGT75,1033,42:441=x15−35x13−30x12+395x11+594x10−1495x9−3360x8+175x7+3990x6+2166x5


#### 4.6.1. Computation

LetM be a map of the type 33,42:441 with the vertex set V. Let V44 and V33,42 denote the sets of vertices with face-sequence types (44) and (33, 42), respectively. Then, counting the number of quadrangular faces in terms of V33,42 and V44 we see easily that V33,42=2V44. Thus, for |V| ≤ 15, we let V=a1,a2,…,aV44,1,2,…,2V44 such that V44≤5. Assume that, without loss of generality, lka1=C8a3,1,2,3,a2,4,5,6. This implies lk(a2) = C8(a1, 2, 3, n1, x1, n2, 4, 5) for x1∈V44 and n1,n2∈V33,42. Observe that x1 ∈ {a3, a4}.Case 1. When x1 = a3, then (n1, n2) ∈ {(1, 6), (6, 1)}. If (n1, n2) = (6, 1), then lk(a2) = C8(a1, 2, 3, 6, a3, 1, 4, 5) and lk(a3) = C8(a2, 3, 6, 5, a1, 2, 1, 4). This implies lk(1) = C7(n3, n4, [2, a1, a3, a2, 4]). It is easy to see that (n3, n4) ∈ {(3, 6), (5, 6), (6, 3), (6, 5)}. But for these values of (n3, n4), we see easily that no map exists.On the other hand, if (n1, n2) = (1, 6), then lk(a2) = C8(a1, 2, 3, 1, a3, 6, 4, 5), lk(a3) = C8(a2, 3, 1, 2, a1, 5, 6, 4). This implies lk(1) = C7(n3, n4, [2, a1, a3, a2, 3]), for (n3, n4) ∈ {(4, 5), (4, 6), (5, 4), (5, 6), (6, 4), (6, 5)}.Observe that (5, 6) ≅ (4, 5) by the map (1, 3, 2) (4, 5, 6) (a1, a3, a2), (6, 4) ≅ (4, 5) by the map (1, 2, 3) (4, 6, 5) (a1, a2, a3), and (6, 5) ≅ (4, 6) by the map (2, 3) (4, 5) (a1, a2). Thus, we search for (n3, n4) ∈ {(4, 5), (4, 6), (5, 4), (5, 6)}. Now doing computation for these cases, we see the following:Ifn3,n4=4,5,M≅K13,633,42:441 by the map i↦v6, aj↦uj, 1 ≤ i ≤ 6, 1 ≤ j ≤ 3.Ifn3,n4=4,6,M≅T13,633,42:441 by the map i↦v6, aj↦uj, 1 ≤ i ≤ 6, 1 ≤ j ≤ 3.Ifn3,n4=5,4,M≅T23,633,42:441 by the map i↦v6, aj↦uj, 1 ≤ i ≤ 6, 1 ≤ j ≤ 3.Case 2. When x1 = a4, then considering lk(a3) = C8(a1, 2, 1, n3, x2, n4, 6, 5), we get x2 ∈ {a4, a5}.Subcase 2.1. If x2 = a4, then (n3, n4) ∈ {(7, 8), (8, 7)}. If (n3, n4) = (8, 7), (n5, n6) ∈ {(4, 5), (5, 4), (5, 6), (6, 5), (6, 7), (7, 3), (7, 6)}. If (n5, n6) = (4, 5), then considering successively lk(1), lk(5), and lk(8), we see that deg(4) > 5, a contradiction. If (n5, n6) = (5, 4), then considering successively lk(1), lk(5), lk(2), and lk(6), we get lk(7) of length 5, a contradiction. Proceeding similarly for the rest of the cases of (n5, n6), we see easily that no map exists. On the other hand, if (n3, n4) = (7, 8), then lk(1) = C7(n5, n6, [2, a1, a3, a4, 7]), where (n5, n6) ∈ {(4, 5), (4, 8), (5, 4), (5, 6), (6, 5), (6, 8), (8, 4), (8, 6)}. But (5, 4) ≅ (4, 8) by the map (1, 3) (4, 6) (a2, a3), (5, 6) ≅ (4, 5) by the map (1, 7) (2, 3) (4, 5) (6, 8) (a1, a2) (a3, a4), (6, 8) ≅ (4, 5) by the map (1, 2, 3, 7) (4, 8, 6, 5) (a1, a2, a4, a3), (8, 4) ≅ (4, 5) by the map (1, 3) (2, 7) (4, 6) (5, 8) (a1, a4) (a2, a3), and (8, 6) ≅ (6, 5) by the map (1, 7) (2, 3) (4, 5) (6, 8) (a1, a2) (a3, a4). So, we search for (n5, n6) ∈ {(4, 5), (4, 8), (6, 5)}. Now doing computation for these cases, we see the following:Ifn5,n6=4,5,M≅K24,833,42:441 by the map i↦v6, aj↦uj, 1 ≤ i ≤ 8, 1 ≤ j ≤ 4.Ifn5,n6=4,8,M≅T34,833,42:441 by the map i↦v6, aj↦uj, 1 ≤ i ≤ 8, 1 ≤ j ≤ 4.Ifn5,n6=6,5,M≅T44,833,42:441 by the map i↦v6, aj↦uj, 1 ≤ i ≤ 8, 1 ≤ j ≤ 4.Subcase 2.2. When x2 = a5, then considering lk(a4), we see that (n5, n6) ∈ {(9, 10), (10, 9)}. If (n5, n6) = (10, 9), (n7, n8) ∈ {(3, 7), (4, 5), (4, 8), (5, 4), (5, 6), (6, 5), (6, 10), (7, 10), (10, 6), (10, 7)}. But a small calculation shows that no map exists for these values of (n7, n8). If $(n5, n6) = (9,10)$, then we get lk(1) = C7(n7, n8, [2, a1, a3, a5, 9]). Then, up to isomorphism, we get (n7, n8) ∈ {(3, 7), (4, 5), (5, 4), (6, 5)}. Doing computation for these cases, we see the following:Ifn7,n8=3,7,M≅K35,1033,42:441 by the map i↦vi, aj↦uj, 1 ≤ i ≤ 10, 1 ≤ j ≤ 5.Ifn7,n8=4,5,M≅K45,1033,42:441 by the map i↦vi, aj↦uj, 1 ≤ i ≤ 10, 1 ≤ j ≤ 5.Ifn7,n8=4,8,M≅T55,1033,42:441 by the map i↦vi, aj↦uj, 1 ≤ i ≤ 10, 1 ≤ j ≤ 5.Ifn7,n8=5,4,M≅T65,1033,42:441 by the map i↦vi, aj↦uj, 1 ≤ i ≤ 10, 1 ≤ j ≤ 5.Ifn7,n8=6,5,M≅T75,1033,42:441 by the map i↦vi, aj↦uj, 1 ≤ i ≤ 10, 1 ≤ j ≤ 5. This completes the computation and we get the following results.


#### 4.6.2. Results

Lemma 3.
LetM be a DSEM of type 33,42:441 with number of vertices ≤15. Then, M is isomorphic to one of the following: T13,633,42:441, T23,633,42:441, K13,633,42:441, T34,833,42:441, T44,833,42:441, K24,833,42:441, T55,1033,42:441, T65,1033,42:441, T75,1033,42:441, K35,1033,42:441, and K45,1033,42:441, as shown in Figure 8.Combining Lemma 3 together with Claim4, we get the following theorem.Theorem 6.
There are exactly 11 non-isomorphic DSEMs of type33,42:441 on the surfaces of Euler characteristic 0 with ≤15 vertices. Out of these, 7 are on torus and remaining 4 are on Klein bottle.


### 4.7. Computation and Classification for Type33,42:442

Consider the following DSEMs of type33,42:442, shown in Figure 9, on torus and Klein bottle denoted by Ti6,633,42:442, for i = 1, 2, and K16,633,42:442, respectively.Figure 9
Doubly semiequivelar maps of type33,42:442 on torus and Klein bottle. (a) K1(6,6) [33, 42: 44]2. (b) T1(6,6) [33, 42: 44]2. (c) T2(6,6) [33, 42: 44]2.
(a)
(b)
(c)Claim 5.
T
1
6,6
3
3
,
4
2
:
4
4
2
≇
T
2
6,6
3
3
,
4
2
:
4
4
2.Proof.
See the following polynomials:pEGT16,633,42:442=x12−27x10−20x9+207x8+168x7−610x6−288x5+723x4−136x3−171x2+84x−11pEGT26,633,42:442=x12−27x10−20x9+201x8+192x7−532x6−552x5+492x4+560x3−84x2−192x−44


#### 4.7.1. Computation

LetM be a map of the type 33,42:442 with the vertex set V. Let V44 and V33,42 denote the sets of vertices with face-sequence types (44) and (33, 42), respectively. Then, it is easy to see that M has the number of quadrangular faces 3V44/2 or V33,42+V44/2. This implies V44=V33,42=2k for k∈ℕ. Therefore, for |V| ≤ 15, we let V=a1,a2,…,aV44,1,2,…,V44, where V44=2k for k ≤ 3. Without loss of generality, we may assume lk(1) = C7(3, 4, [5, a1, a2, a3, 2]). Then, lk(a2) = C8(a3, 2, 1, 5, a1, a6, a5, a4), lk(a1) = C8(a2, 1, 5, n1, x1, x2, a6, a5) for n1 ∈ Vm and x1, x2 ∈ Vl. Observe that (n1, x1, x2) ∈{(2, a3, a4), (3, a4, a3), (6, a4, a3)}.Case 1. When (n1, x1, x2) = (3, a4, a3), then considering successively lk(a1), lk(a4), lk(a5), lk(3), and lk(4), we see that lk(5) cannot be completed. So, (n1, x1, x2) ≠ (2, a4, a3).Case 2. When (n1, x1, x2) = (6, a4, a3), then lk(a1) = C8(a2, 1, 5, 6, a4, a3, a6, a5) and lk(a4) = C8(a1, 5, 6, n2, a5, a2, a3, a6) for n2 ∈{3, 4, 7}. If n2 = 3, then considering successively lk(a4), lk(a5), lk(a6), lk(a3), lk(3), and lk(2), we see lk(4) cannot be completed.Ifn2 = 4, then considering successively lk(a4), lk(a5), lk(a6), lk(a3), and lk(4), as in previous case, we see that lk(5) cannot be completed.Ifn2 = 7, then considering successively lk(a4), lk(a5), and lk(a6), we see that lk(a3) cannot be completed. So, (n1, x1, x2) ≠ (6, a4, a3).Case 3. If (n1, x1, x2) = (2, a3, a4), then successively completing lk(a1), lk(a3), lk(2), and lk(5), we get lk(4) = C7(1, 5, [6, x5, x4, x3, 3]) for (x3, x4, x5) ∈ {(a5, a4, a6), (a6, a4, a5), (a4, a5, a6), (a6, a5, a4), (a4, a6, a5), (a5, a6, a4)}. Note that (a6, a5, a4) ≅ (a5, a4, a6) by the map (1, 5, 2) (3, 4, 6) (a1, a3, a2) (a4, a5, a6), (a4, a6, a5) ≅ (a5, a4, a6) by the map (1, 2, 5) (3, 6, 4) (a1, a2, a3) (a4, a6, a5), and (a5, a6, a4) ≅ (a4, a5, a6) by the map (2, 5) (3, 4) (a1, a3) (a4, a6).Thus, we do computation for (x3, x4, x5) ∈ {(a5, a4, a6), (a6, a4, a5), (a4, a5, a6)}. This gives the following:Ifx3,x4,x5=a5,a4,a6,M≅K16,633,42:442 by the map i↦vi, ai↦ui, 1 ≤ i ≤ 6.Ifx3,x4,x5=a6,a4,a5,M≅T16,633,42:442 by the map i↦vi, ai↦ui, 1 ≤ i ≤ 6.Ifx3,x4,x5=a4,a5,a6,M≅T26,633,42:442 by the map i↦vi, ai↦ui, 1 ≤ i ≤ 6.Thus, the computation is completed and we get the following results.


#### 4.7.2. Results

Lemma 4.
LetM be a DSEM of type 33,42:442 with number of vertices ≤15. Then, M is isomorphic to T16,633,42:442, T26,633,42:442 or K16,633,42:442, as shown in Figure 9.Combining Lemma4 together with Claim 5, we get the following theorem.Theorem 7.
There are exactly 3 non-isomorphic DSEMs of type33,42:442 on the surfaces of Euler characteristic 0 with ≤15 vertices. Out of these, two are on torus and the remaining one is on Klein bottle.


## 4.1. Computation and Classification for Type36:33,421

Consider the following DSEMs of type36:33,421, in Figure 3, on torus and Klein bottle denoted by Tin,2n36:33,421, for i ∈{1, …, 8}, and Kin,2n36:33,421, for i ∈{1, …, 6}, respectively.Figure 3
Doubly semiequivelar maps on torus and Klein bottle of type36:33,421. (a) T1(3, 6) [36: 33, 42]1. (b) T2(3, 6) [36: 33, 42]1. (c) K1(3, 6) [36: 33, 42]1. (d) K2(3,6) [36: 33, 42]1. (e) T3(4,8) [36: 33, 42]1. (f) T4(4,8) [36: 33, 42]1. (g) K3(4,8) [36: 33, 42]1. (h) K4(4,8) [36: 33, 42]1. (i) K5(4,8) [36: 33, 42]1. (j) T5(4,8) [36: 33, 42]1. (k) T6(5,10) [36: 33, 42]1. (l) T7(5,10) [36: 33, 42]1. (m) K6(5,10) [36: 33, 42]1. (n) K7(5,10) [36: 33, 42]1. (o) T8(5,10) [36: 33, 42]1.
(a)
(b)
(c)
(d)
(e)
(f)
(g)
(h)
(i)
(j)
(k)
(l)
(m)
(n)
(o)Claim 1.
For the maps above, we have the following:(a)
T13,636:33,421≇T23,636:33,421.(b)
T34,836:33,421≇T44,836:33,421≇T54,836:33,421.(c)
K34,836:33,421≇K44,836:33,421≇K54,836:33,421.(d)
T65,1036:33,421≇T75,1036:33,421≇T85,1036:33,421.(e)
K65,1036:33,421≇K75,1036:33,421.Proof.
Letp(EG(M)) denote the characteristic polynomial of adjacency matrix associated with the edge graph of M. Then, the proof follows from the following polynomials:pEGT13,636:33,421=x9−24x7−42x6+63x5+138x4−72x3−144x2+48x+32pEGT23,636:33,421=x9−24x7−36x6+45x5+48x4−21x3−18x2+3x+2pEGT34,836:33,421=x12−32x10−40x9+254x8+440x7−628x6−1400x5+105x4+1000x3+300x2pEGT44,836:33,421=x12−32x10−32x9+254x8+224x7−932x6−448x5+1673x4+96x3−1156x2+160x+192pEGT54,836:33,421=x12−32x10−48x9+254x8+656x7−292x6−2352x5−2167x4+624x3+2044x2+1120x+192pEGK34,836:33,421=x12−31x10−39x9+227x8+377x7−561x6−1129x5+416x4+1283x3+92x2−492x−144pEGK44,836:33,421=x12−32x10−40x9+254x8+440x7−644x6−1400x5+457x4+1640x3+156x2−640x−192pEGK54,836:33,421=x12−32x10−48x9+258x8+640x7−364x6−220x5−1635x4+496x3+684x2−32x−64pEGT65,1036:33,421=x15−40x13−40x12+515x11+754x10−282x9−4940x8+6790x7+13430x6−668x5−15340x4+975x3+5490x2+1755x+162pEGT75,1036:33,421=x15−38x13−51x12+462x11+1033x10−1049x9−5533x8−4681x7+2905x6+6351x5+2282x4−1046x3−680x2+32x+48pEGT85,1036:33,421=x15−40x13−60x12+485x11+1374x10−985x9−7910x8−9955x7−1010x6+7623x5+7030x4+2820x3+570x2+55x+2pEGK65,1036:33,421=x15−39x13−58x12+462x11+1309x10−916x9−7455x8−9096x7−203x6+6562x5+3147x4−909x3−761x2−97x−3pEGK75,1036:33,421=x15−40x13−48x12+497x11+1010x10−1973x9−6234x8−111x7+12010x6+9531x5−3294x4−7264x3−3410x2−637x−38Claim 2.
K
1
3,6
3
6
:
3
3
,
4
2
1
≇
K
2
3,6
3
6
:
3
3
,
4
2
1
.Proof.
Note thatpEGK13,636:33,421=pEGK23,636:33,421=x9−24x7−38x6+51x5+78x4−44x3−24x2, but the maps are non-isomorphic. To see this, we use geometric argument as follows: define a basis {a, b} at any vertex vi (for 1 ≤ i ≤ 6), where a and b are minimal nontrivial loops (i.e., nontrivial cycle with minimum number of vertices); now, if we consider K13,636:33,421, then at each vi, we get a and b with length 3 (for example, at v1, a = C3(v1,v6,u2) and b = C3(v1,v2,v5)) while in K23,636:33,421, at each vi, we get a of length 3 and b of length 4 (for example, at v1, we see that a = C3(v4,v1,u1) or a = C3(v4,v1,u2) and b = C4(v1,v2,v6,v4)). Hence, K13,636:33,421≇K23,636:33,421.


### 4.1.1. Computation

LetM be a DSEM of type 36:33,421 with the vertex set V. Let V36 and V33,42 denote the sets of vertices with face-sequence types (36) and (33, 42), respectively. Then, we see that the number of triangular faces in M is 4V36 or 2V33,42. This implies 2V36=V33,42. Thus, for V=V33,42+V33,42≤15, we let V=a1,a2,…,aV36,1,2,…,2|V36|, where V36|≤5. Without loss of generality, we may assume lk(a1) = C6(a2, 1, 2, a3, 3, 4). This implies lk(a2) = C6(a1, 1, n1, x1, n2, 4), lk(1) = C7(a1, a2, [n1, n3, n4, n5, 2]), lk2=C7a1,a3, n6,n7,n5,n4,1, lka3=C6a1,3, n8,x2,n6,2, lk3=C7a1,a3,n8,n9,n10,n11,4, and lk4=C7a1,a2,n2,n12,n11,n10,3 for some x1,x2∈V36 and n1,n2,…,n12∈V33,42.Now considering lk(a2), we see that n1≠2 or 3 (for n1 = 2, the set {2, 1, a2} forms triangular face in lk(a1) but not in lk(a2); for n1 = 3, we get deg(3) > 5). Similarly, we see that n2∉{2, 3}. From these observations, we have (n1, x1, n2) ∈{(5, a3, 6), (5, a4, 6)}.Case 1. if (n1, x1, n2) = (5, a3, 6), then lk(a3) = C6(a1, 3, 6, a2, 5, 2) or lka3=C6a1,3,5,a2,6,2.Whenlka3=C6a1,3,5,a2,6,2, considering lk(2), we have n4,n5,n7∈4,3,5,5,3,4. If n4,n5,n7=5,3,4, then lk(1) is a cycle of length 5, a contradiction. On the other hand, if n4,n5,n7=4,3,5, then lk2=C7a1,a3,6,5,3,4,1,lk1=C7a1,a2,5,6,4,3,2; completing successively, we get lk4=C7a1,a2,6,5,1,2,3, lk3=C7a1,a3,5,6,2,1,4, lk5=C7a2,a3,3,2,6,4,1, and lk6=C7a2,a3,2,3,5,1,4. Then, we get M≅K23,636:33,421 by the map i↦vi, aj↦uj, 1 ≤ i ≤ 6, 1 ≤ j ≤ 3.When lk(a3) = C6(a1, 3, 6, a2, 5, 2), considering lk(2), we get (n4, n5, n7) ∈{(4, 3, 6), (6, 3, 4), (3, 4, 6), (6, 4, 3), (3, 6, 4), (4, 6, 3)}. Observe that (3, 4, 6) ≅ (6, 3, 4) by the map (1, 5, 2) (3, 4, 6) (a1, a2, a3), (3, 6, 4) ≅ (6, 4, 3) by the map (1, 5) (4, 6) (a1, a3), and (4, 6, 3) ≅ (6, 3, 4) by the map (1, 2, 5) (3, 6, 4) (a1, a3, a2). So, we need to search only for (n4, n5, n7) ∈ {(4, 3, 6), (6, 3, 4), (6, 4, 3)}.In case (n4, n5, n7) = (4, 3, 6), completing successively, we get lk2=C7a1,a3,5,6,3,4,1, lk3=C7a1,a3,6,5,2,1,4, lk4=C7a1,a2,6,5,1,2,3, lk1=C7a1,a2,5,6,4,3,2, lk5=C7a2,a3,2,3,6,4,1, lk6=C7a2,a3,3,2,5,1,4. This gives M≅T13,636:33,421 by the map i↦vi, aj↦uj, 1 ≤ i ≤ 6, 1 ≤ j ≤ 3.Proceeding similarly as above, for (n4, n5, n7) = (6, 3, 4), we get M≅K13,636:33,421 by the map i↦vi, aj↦uj, where 1 ≤ i ≤ 6, 1 ≤ j ≤ 3.For (n4, n5, n7) = (6, 4, 3), M≅T23,636:33,421, by the map i↦vi, aj↦uj, for 1 ≤ i ≤ 6, 1 ≤ j ≤ 3.Case 2. For (n1, x1, n2) = (5, a4, 6), considering lk(a3), we get x2 ∈{a4, a5}.Subcase 2.1. If x2 = a4, then lk(a3) = C6(a1, 2, 7, a4, 8, 3). This implies lk(a4) = C6(a2, 5, 7, a3, 8, 6) or lk(a4) = C6(a2, 5, 8, a3, 7, 6).When lk(a4) = C6(a2, 5, 7, a3, 8, 6), then, up to isomorphism, we see that (n13, n14, n15) ∈ {(2, 1, 5), (5, 1, 2), (1, 2, 7), (7, 2, 1), (7, 5, 1)}. Now doing computation for these cases, we see the following:If (n13, n14, n15) = (2, 1, 5), M≅K34,836:33,421 by i↦vi, aj↦uj, 1 ≤ i ≤ 8, 1 ≤ j ≤ 4.Ifn13,n14,n15=5,1,2,M≅T34,836:33,421 by i↦vi, aj↦uj, 1 ≤ i ≤ 8, 1 ≤ j ≤ 4.Ifn13,n14,n15=1,2,7,M≅T44,836:33,421 by i↦vi, aj↦uj, 1 ≤ i ≤ 8, 1 ≤ j ≤ 4.Ifn13,n14,n15=7,2,1,M≅K44,836:33,421 by i↦vi, aj↦uj, 1 ≤ i ≤ 8, 1 ≤ j ≤ 4.Ifn13,n14,n15=7,5,1,M≅T54,836:33,421 by i↦vi, aj↦uj, 1 ≤ i ≤ 8, 1 ≤ j ≤ 4.On the other hand when lk(a4) = C6(a2, 5, 8, a3, 7, 6), we get (n13, n14, n15) ∈{(2, 1, 5), (1, 5, 8), (3, 8, 5), (5, 1, 2), (5, 8, 3), (8, 5, 1)}. If (n13, n14, n15) = (2, 1, 5) and (1, 5, 8), then lk(7) is a cycle of length 5 and 6 respectively, which is not possible. If (n13, n14, n15) = (3, 8, 5) and (5, 1, 2), then we see easily that lk(7) and lk(8) cannot be completed, respectively. If (n13, n14, n15) = (5, 8, 3), then completing lk(7) we get lk(1) of length 5, again a contradiction. If n13,n14,n15=8,5,1,M≅K54,836:33,421 by i↦vi, aj↦uj, 1 ≤ i ≤ 8, 1 ≤ j ≤ 4.Subcase 2.2. When x2 = a5, successively, we get lk(a3) = C6(a1, 2, 7, a5, 8, 3) and lk(a4) = C6(a2, 6, 9, a5, 10, 5). This implies lk(a5) = C6(a4, 9, 7, a3, 8, 10) or lk(a5) = C6(a4, 9, 8, a3, 7, 10).In case lk(a5) = C6(a4, 9, 7, a3, 8, 10), considering lk(1), we get (n3, n4, n5) ∈ {(3, 4, 6), (3, 8, 10), (4, 3, 8), (4, 6, 9), (6, 4, 3), (6, 9, 7), (7, 9, 6), (8, 3, 4), (9, 6, 4), (10, 8, 3)}. But a small calculation shows that no map exists for these cases, except for (n3, n4, n5) = (6, 4, 3). For (n3, n4, n5) = (6, 4, 3), we get M≅K65,1036:33,421 by the map i↦vi, aj↦uj, 1 ≤ i ≤ 10, 1 ≤ j ≤ 5.On the other hand, when lk(a5) = C6(a4, 9, 8, a3, 7, 10), considering lk(1), up to isomorphism, we get (n3, n4, n5) ∈ {(3, 4, 6), (3, 8, 9), (4, 3, 8), (6, 4, 3)}. Now doing computation for these cases, we see the following:If (n3, n4, n5) = (3, 4, 6), M≅K75,1036:33,421 by i↦vi, aj↦uj, 1 ≤ i ≤ 10 and 1 ≤ j ≤ 5.If (n3, n4, n5) = (3, 8, 9), M≅T65,1036:33,421 by i↦vi, aj↦uj, 1 ≤ i ≤ 10 and 1 ≤ j ≤ 5.If (n3, n4, n5) = (4, 3, 8), M≅T75,1036:33,421 by i↦vi, aj↦uj, 1 ≤ i ≤ 10 and 1 ≤ j ≤ 5.If (n3, n4, n5) = (6, 4, 3), M≅T85,1036:33,421 by i↦vi, aj↦uj, 1 ≤ i ≤ 10 and 1 ≤ j ≤ 5. This completes computation for the number of vertices ≤15 and we obtain the following results.


### 4.1.2. Results

Lemma 1.
LetM be a DSEM of type 36:33,421 with number of vertices ≤15. Then, M is isomorphic to one of the following: T13,636:33,421, T23,636:33,421, K13,636:33,421, K23,636:33,421, T34,836:33,421, T44,836:33,421, T54,836:33,421, K34,836:33,421, K44,836:33,421, K54,836:33,421, T65,1036:33,421, T75,1036:33,421, T85,1036:33,421, K65,1036:33,421, and K75,1036:33,421, as shown in Figure 3.Combining Lemma1 together with Claims 1 and 2, we get the following theorem.Theorem 1.
There are exactly 15 DSEMs of type36:33,421 on the surfaces of Euler characteristic 0 with number of vertices ≤15. Out of these, 8 are on the torus and remaining 7 are on the Klein bottle.


## 4.1.1. Computation

LetM be a DSEM of type 36:33,421 with the vertex set V. Let V36 and V33,42 denote the sets of vertices with face-sequence types (36) and (33, 42), respectively. Then, we see that the number of triangular faces in M is 4V36 or 2V33,42. This implies 2V36=V33,42. Thus, for V=V33,42+V33,42≤15, we let V=a1,a2,…,aV36,1,2,…,2|V36|, where V36|≤5. Without loss of generality, we may assume lk(a1) = C6(a2, 1, 2, a3, 3, 4). This implies lk(a2) = C6(a1, 1, n1, x1, n2, 4), lk(1) = C7(a1, a2, [n1, n3, n4, n5, 2]), lk2=C7a1,a3, n6,n7,n5,n4,1, lka3=C6a1,3, n8,x2,n6,2, lk3=C7a1,a3,n8,n9,n10,n11,4, and lk4=C7a1,a2,n2,n12,n11,n10,3 for some x1,x2∈V36 and n1,n2,…,n12∈V33,42.Now considering lk(a2), we see that n1≠2 or 3 (for n1 = 2, the set {2, 1, a2} forms triangular face in lk(a1) but not in lk(a2); for n1 = 3, we get deg(3) > 5). Similarly, we see that n2∉{2, 3}. From these observations, we have (n1, x1, n2) ∈{(5, a3, 6), (5, a4, 6)}.Case 1. if (n1, x1, n2) = (5, a3, 6), then lk(a3) = C6(a1, 3, 6, a2, 5, 2) or lka3=C6a1,3,5,a2,6,2.Whenlka3=C6a1,3,5,a2,6,2, considering lk(2), we have n4,n5,n7∈4,3,5,5,3,4. If n4,n5,n7=5,3,4, then lk(1) is a cycle of length 5, a contradiction. On the other hand, if n4,n5,n7=4,3,5, then lk2=C7a1,a3,6,5,3,4,1,lk1=C7a1,a2,5,6,4,3,2; completing successively, we get lk4=C7a1,a2,6,5,1,2,3, lk3=C7a1,a3,5,6,2,1,4, lk5=C7a2,a3,3,2,6,4,1, and lk6=C7a2,a3,2,3,5,1,4. Then, we get M≅K23,636:33,421 by the map i↦vi, aj↦uj, 1 ≤ i ≤ 6, 1 ≤ j ≤ 3.When lk(a3) = C6(a1, 3, 6, a2, 5, 2), considering lk(2), we get (n4, n5, n7) ∈{(4, 3, 6), (6, 3, 4), (3, 4, 6), (6, 4, 3), (3, 6, 4), (4, 6, 3)}. Observe that (3, 4, 6) ≅ (6, 3, 4) by the map (1, 5, 2) (3, 4, 6) (a1, a2, a3), (3, 6, 4) ≅ (6, 4, 3) by the map (1, 5) (4, 6) (a1, a3), and (4, 6, 3) ≅ (6, 3, 4) by the map (1, 2, 5) (3, 6, 4) (a1, a3, a2). So, we need to search only for (n4, n5, n7) ∈ {(4, 3, 6), (6, 3, 4), (6, 4, 3)}.In case (n4, n5, n7) = (4, 3, 6), completing successively, we get lk2=C7a1,a3,5,6,3,4,1, lk3=C7a1,a3,6,5,2,1,4, lk4=C7a1,a2,6,5,1,2,3, lk1=C7a1,a2,5,6,4,3,2, lk5=C7a2,a3,2,3,6,4,1, lk6=C7a2,a3,3,2,5,1,4. This gives M≅T13,636:33,421 by the map i↦vi, aj↦uj, 1 ≤ i ≤ 6, 1 ≤ j ≤ 3.Proceeding similarly as above, for (n4, n5, n7) = (6, 3, 4), we get M≅K13,636:33,421 by the map i↦vi, aj↦uj, where 1 ≤ i ≤ 6, 1 ≤ j ≤ 3.For (n4, n5, n7) = (6, 4, 3), M≅T23,636:33,421, by the map i↦vi, aj↦uj, for 1 ≤ i ≤ 6, 1 ≤ j ≤ 3.Case 2. For (n1, x1, n2) = (5, a4, 6), considering lk(a3), we get x2 ∈{a4, a5}.Subcase 2.1. If x2 = a4, then lk(a3) = C6(a1, 2, 7, a4, 8, 3). This implies lk(a4) = C6(a2, 5, 7, a3, 8, 6) or lk(a4) = C6(a2, 5, 8, a3, 7, 6).When lk(a4) = C6(a2, 5, 7, a3, 8, 6), then, up to isomorphism, we see that (n13, n14, n15) ∈ {(2, 1, 5), (5, 1, 2), (1, 2, 7), (7, 2, 1), (7, 5, 1)}. Now doing computation for these cases, we see the following:If (n13, n14, n15) = (2, 1, 5), M≅K34,836:33,421 by i↦vi, aj↦uj, 1 ≤ i ≤ 8, 1 ≤ j ≤ 4.Ifn13,n14,n15=5,1,2,M≅T34,836:33,421 by i↦vi, aj↦uj, 1 ≤ i ≤ 8, 1 ≤ j ≤ 4.Ifn13,n14,n15=1,2,7,M≅T44,836:33,421 by i↦vi, aj↦uj, 1 ≤ i ≤ 8, 1 ≤ j ≤ 4.Ifn13,n14,n15=7,2,1,M≅K44,836:33,421 by i↦vi, aj↦uj, 1 ≤ i ≤ 8, 1 ≤ j ≤ 4.Ifn13,n14,n15=7,5,1,M≅T54,836:33,421 by i↦vi, aj↦uj, 1 ≤ i ≤ 8, 1 ≤ j ≤ 4.On the other hand when lk(a4) = C6(a2, 5, 8, a3, 7, 6), we get (n13, n14, n15) ∈{(2, 1, 5), (1, 5, 8), (3, 8, 5), (5, 1, 2), (5, 8, 3), (8, 5, 1)}. If (n13, n14, n15) = (2, 1, 5) and (1, 5, 8), then lk(7) is a cycle of length 5 and 6 respectively, which is not possible. If (n13, n14, n15) = (3, 8, 5) and (5, 1, 2), then we see easily that lk(7) and lk(8) cannot be completed, respectively. If (n13, n14, n15) = (5, 8, 3), then completing lk(7) we get lk(1) of length 5, again a contradiction. If n13,n14,n15=8,5,1,M≅K54,836:33,421 by i↦vi, aj↦uj, 1 ≤ i ≤ 8, 1 ≤ j ≤ 4.Subcase 2.2. When x2 = a5, successively, we get lk(a3) = C6(a1, 2, 7, a5, 8, 3) and lk(a4) = C6(a2, 6, 9, a5, 10, 5). This implies lk(a5) = C6(a4, 9, 7, a3, 8, 10) or lk(a5) = C6(a4, 9, 8, a3, 7, 10).In case lk(a5) = C6(a4, 9, 7, a3, 8, 10), considering lk(1), we get (n3, n4, n5) ∈ {(3, 4, 6), (3, 8, 10), (4, 3, 8), (4, 6, 9), (6, 4, 3), (6, 9, 7), (7, 9, 6), (8, 3, 4), (9, 6, 4), (10, 8, 3)}. But a small calculation shows that no map exists for these cases, except for (n3, n4, n5) = (6, 4, 3). For (n3, n4, n5) = (6, 4, 3), we get M≅K65,1036:33,421 by the map i↦vi, aj↦uj, 1 ≤ i ≤ 10, 1 ≤ j ≤ 5.On the other hand, when lk(a5) = C6(a4, 9, 8, a3, 7, 10), considering lk(1), up to isomorphism, we get (n3, n4, n5) ∈ {(3, 4, 6), (3, 8, 9), (4, 3, 8), (6, 4, 3)}. Now doing computation for these cases, we see the following:If (n3, n4, n5) = (3, 4, 6), M≅K75,1036:33,421 by i↦vi, aj↦uj, 1 ≤ i ≤ 10 and 1 ≤ j ≤ 5.If (n3, n4, n5) = (3, 8, 9), M≅T65,1036:33,421 by i↦vi, aj↦uj, 1 ≤ i ≤ 10 and 1 ≤ j ≤ 5.If (n3, n4, n5) = (4, 3, 8), M≅T75,1036:33,421 by i↦vi, aj↦uj, 1 ≤ i ≤ 10 and 1 ≤ j ≤ 5.If (n3, n4, n5) = (6, 4, 3), M≅T85,1036:33,421 by i↦vi, aj↦uj, 1 ≤ i ≤ 10 and 1 ≤ j ≤ 5. This completes computation for the number of vertices ≤15 and we obtain the following results.


## 4.1.2. Results

Lemma 1.
LetM be a DSEM of type 36:33,421 with number of vertices ≤15. Then, M is isomorphic to one of the following: T13,636:33,421, T23,636:33,421, K13,636:33,421, K23,636:33,421, T34,836:33,421, T44,836:33,421, T54,836:33,421, K34,836:33,421, K44,836:33,421, K54,836:33,421, T65,1036:33,421, T75,1036:33,421, T85,1036:33,421, K65,1036:33,421, and K75,1036:33,421, as shown in Figure 3.Combining Lemma1 together with Claims 1 and 2, we get the following theorem.Theorem 1.
There are exactly 15 DSEMs of type36:33,421 on the surfaces of Euler characteristic 0 with number of vertices ≤15. Out of these, 8 are on the torus and remaining 7 are on the Klein bottle.


## 4.2. Computation and Classification for Type36:33,422

Consider the following DSEMs of type36:33,422, shown in Figure 4, on torus and Klein bottle denoted by Ti6,636:33,422, for i = 1, 2, and K16,636:33,422, respectively.Figure 4
Doubly semiequivelar maps on torus and Klein bottle of type36:33,422. (a) K1(6,6) [36: 33, 42]2. (b) T1(6,6) [36: 33, 42]2. (c) T2(6,6) [36: 33, 42]2.
(a)
(b)
(c)Claim 3.
T
1
6,6
3
6
:
3
3
,
4
2
2
≇
T
2
6,6
3
6
:
3
3
,
4
2
2
.Proof.
The proof follows from the following polynomials:pEGT16,636:33,422=x12−33x10−44x9+258x8+432x7−682x6−1032x5+957x4+560x3−789x2+276x−32pEGT26,636:33,422=x12−33x10−44x9+252x8+456x7−568x6−1296x5+348x4+1328x3+108x2−432x−128


### 4.2.1. Computation

LetM be a map of the type 36:33,422 with the vertex set V. Let V36 and V33,42 denote the sets of vertices with face-sequence types (36) and (33, 42), respectively. Observe that M has the number of edges = 5V36+V33,42/2, number of triangular faces = 3V36, and number of quadrangular faces = V33,42/2. Now by the Euler characteristic equation, we get V36+V33,42−5V36+V33,42/2+3V36+V33,42/2=0. This implies V36=V33,42. Also, considering the number of quadrangular faces, it is evident that the cardinality of both the sets should be positive even integer. Thus, for |V| ≤ 15, we let V=a1,a2,…,aV36,1,2,…,V36, where V36=2k for k ≤ 3.Without loss of generality, assume lk(a1) = C6(a2, a3, a4, a5, 1, 2). This implies lk(1) = C7(a1, a5, [n1, n2, n3, n4, 2]), lk(2) = C7(a2, a1, [1, n3, n4, n6, n5]) and lk(a2) = C6(a1, a3, x1, x2, 3, 2) for some n1, …, n6 ∈ Vm and x1, x2 ∈ Vl. It is easy to see that (x1, x2) ∈{(a5, a4), (a6, a5)}.Case 1. When (x1, x2) = (a5, a4), then successively we get lk(a2) = C6(a1, a3, a5, a4, 3, 2), lk(a5) = C6(a1, a4, a2, a3, 4, 1), lk(a3) = C6(a4, a1, a2, a5, 4, 5) and lk(a4) = C6(a3, a1, a5, a2, 3, 5). Now considering lk(1), we see that (n2, n3, n4) has no value for the V so that the links of remaining vertices can be completed. So, (x1, x2) ≠ (a5, a4).Case 2. When (x1, x2) = (a6, a5), then successively we get lk(a2) = C6(a1, a3, a6, a5, 3, 2), lk(a5) = C6(a1, a4, a6, a2, 3, 1), lk(a4) = C6(a3, a1, a5, a6, 5, 4), lk(a3) = C6(a4, a1, a2, a6, 6, 4), lk(a6) = C6(a3, a2, a5, a4, 5, 6). Considering lk(1), it is easy to see that (n2, n3, n4) ∈{(4, 5, 6), (4, 6, 5), (5, 4, 6), (5, 6, 4), (6, 4, 5), (6, 5, 4)}. Observe that (4, 5, 6) ≅ (6, 4, 5) by the map (1, 3, 2) (5, 6, 4) (a1, a5, a2) (a3, a4, a6), (4, 5, 6) ≅ (5, 6, 4) by the map (1, 2, 3) (4, 6, 5) (a1, a2, a5) (a3, a6, a4), and (5, 4, 6) ≅ (6, 5, 4) by the map (1, 3) (4, 6) (a1, a2) (a4, a6). So, we search for (n2, n3, n4) ∈ {(4, 5, 6), (4, 6, 5), (5, 4, 6)}. Now doing computation for these cases, we see the following:Ifn2,n3,n4=4,5,6,M≅K16,636:33,422 by the map i↦vi, ai↦ui, 1 ≤ i ≤ 6.Ifn2,n3,n4=4,6,5,M≅T16,636:33,422 by the map i↦vi, ai↦ui, 1 ≤ i ≤ 6.Ifn2,n3,n4=5,4,6,M≅T26,636:33,422 by the map i↦vi, ai↦ui, 1 ≤ i ≤ 6.This completes computation for ≤15 vertices. From this we get following results.


### 4.2.2. Results

Lemma 2.
LetM be a DSEM of type 36:33,422 on the surfaces of Euler characteristic 0 with ≤15 vertices. Then, M is isomorphic to one of K16,636:33,422, T16,636:33,422, and T16,636:33,422, given in Figure 4.Combining Lemma2 with Claim 3, we get the following theorem.Theorem 2.
There are exactly 3 non-isomorphic DSEMs of type36:33,422 with number of vertices ≤15. Out of these, 2 are on torus and the remaining one is on Klein bottle.


## 4.2.1. Computation

LetM be a map of the type 36:33,422 with the vertex set V. Let V36 and V33,42 denote the sets of vertices with face-sequence types (36) and (33, 42), respectively. Observe that M has the number of edges = 5V36+V33,42/2, number of triangular faces = 3V36, and number of quadrangular faces = V33,42/2. Now by the Euler characteristic equation, we get V36+V33,42−5V36+V33,42/2+3V36+V33,42/2=0. This implies V36=V33,42. Also, considering the number of quadrangular faces, it is evident that the cardinality of both the sets should be positive even integer. Thus, for |V| ≤ 15, we let V=a1,a2,…,aV36,1,2,…,V36, where V36=2k for k ≤ 3.Without loss of generality, assume lk(a1) = C6(a2, a3, a4, a5, 1, 2). This implies lk(1) = C7(a1, a5, [n1, n2, n3, n4, 2]), lk(2) = C7(a2, a1, [1, n3, n4, n6, n5]) and lk(a2) = C6(a1, a3, x1, x2, 3, 2) for some n1, …, n6 ∈ Vm and x1, x2 ∈ Vl. It is easy to see that (x1, x2) ∈{(a5, a4), (a6, a5)}.Case 1. When (x1, x2) = (a5, a4), then successively we get lk(a2) = C6(a1, a3, a5, a4, 3, 2), lk(a5) = C6(a1, a4, a2, a3, 4, 1), lk(a3) = C6(a4, a1, a2, a5, 4, 5) and lk(a4) = C6(a3, a1, a5, a2, 3, 5). Now considering lk(1), we see that (n2, n3, n4) has no value for the V so that the links of remaining vertices can be completed. So, (x1, x2) ≠ (a5, a4).Case 2. When (x1, x2) = (a6, a5), then successively we get lk(a2) = C6(a1, a3, a6, a5, 3, 2), lk(a5) = C6(a1, a4, a6, a2, 3, 1), lk(a4) = C6(a3, a1, a5, a6, 5, 4), lk(a3) = C6(a4, a1, a2, a6, 6, 4), lk(a6) = C6(a3, a2, a5, a4, 5, 6). Considering lk(1), it is easy to see that (n2, n3, n4) ∈{(4, 5, 6), (4, 6, 5), (5, 4, 6), (5, 6, 4), (6, 4, 5), (6, 5, 4)}. Observe that (4, 5, 6) ≅ (6, 4, 5) by the map (1, 3, 2) (5, 6, 4) (a1, a5, a2) (a3, a4, a6), (4, 5, 6) ≅ (5, 6, 4) by the map (1, 2, 3) (4, 6, 5) (a1, a2, a5) (a3, a6, a4), and (5, 4, 6) ≅ (6, 5, 4) by the map (1, 3) (4, 6) (a1, a2) (a4, a6). So, we search for (n2, n3, n4) ∈ {(4, 5, 6), (4, 6, 5), (5, 4, 6)}. Now doing computation for these cases, we see the following:Ifn2,n3,n4=4,5,6,M≅K16,636:33,422 by the map i↦vi, ai↦ui, 1 ≤ i ≤ 6.Ifn2,n3,n4=4,6,5,M≅T16,636:33,422 by the map i↦vi, ai↦ui, 1 ≤ i ≤ 6.Ifn2,n3,n4=5,4,6,M≅T26,636:33,422 by the map i↦vi, ai↦ui, 1 ≤ i ≤ 6.This completes computation for ≤15 vertices. From this we get following results.


## 4.2.2. Results

Lemma 2.
LetM be a DSEM of type 36:33,422 on the surfaces of Euler characteristic 0 with ≤15 vertices. Then, M is isomorphic to one of K16,636:33,422, T16,636:33,422, and T16,636:33,422, given in Figure 4.Combining Lemma2 with Claim 3, we get the following theorem.Theorem 2.
There are exactly 3 non-isomorphic DSEMs of type36:33,422 with number of vertices ≤15. Out of these, 2 are on torus and the remaining one is on Klein bottle.


## 4.3. Computation and Classification for Type [36: 32, 4, 3, 4]

Consider the following DSEM of type [36: 32, 4, 3, 4], given in Figure 5 on Klein bottle, denoted by K1(2,12)[36: 32, 4, 3, 4].Figure 5
Doubly semiequivelar maps on Klein bottle of typeK1(2,12) [36: 32, 4, 3, 4].


### 4.3.1. Computation

LetM be a map of the type [36: 32, 4, 3, 4] with the vertex set V. Let V36 and V32,4,3,4 denote the sets of vertices with face-sequence types (36) and (32, 4, 3, 4), respectively. It is easy to see that 6V36=V32,4,3,4. Thus, for |V| ≤ 15, we let V=a1,a2,…,aV36,1,2,…,6V36, where V36≤2. Without loss of generality, we assume lk(a1) = C6(1, 2, 3, 4, 5, 6). Then, successively, we have lk(1) = C7(a1, [2, n1, n2], [n3, n4, 6]), lk(2) = C7(a1, [1, n2, n1], [n5, n6, 3]), lk(3) = C7(a1, [2, n5, n6], [n7, n8, 4]), lk(4) = C7(a1, [3, n7, n8], [n9, n10, 5]), lk(5) = C7(a1, [4, n9, n10], [n11, n12, 6]), lk(6) = C7(a1, [1, n3, n4], [n12, n11, 5]), where ni ∈ Vm for 1 ≤ i ≤ 12. Now considering lk(1), we see that n1 ∈{4, 5, 7}.Case 1. If n1 = 4, then successively, we see that n2 = 5, n3 = 7, and n4 = 8; now considering lk(5) and lk(1), we get two quadrangular faces which share more than one vertex, which is not allowed. So, n1 ≠ 4.Case 2. If n1 = 5, then successively, we get n2 = 4, n3 = 7, n4 = 8, n12 = 9, and n11 = 10. Now completing lk(1), lk(6), lk(5), lk(2), lk(3), lk(4), lk(7), and lk(10), we see that lk(a2) = C6(7, 8, 11, 10, 9, 12) or lk(a2) = C6(7, 8, 9, 10, 11, 12). If lk(a2) = C6(7, 8, 9, 10, 11, 12), then lk(8) is a cycle of length 5, a contradiction. If lk(a2) = C6(7, 8, 11, 10, 9, 12), then completing successively, we get M ≅ K1(2,12)[36: 32, 4, 3, 4] by the map i↦vi, aj↦uj, 1 ≤ i ≤ 12, 1 ≤ j ≤ 2.Case 3. If n1 = 7, then we get (n2, n3) ∈ {(8, 3), (8, 4), (8, 9)}. For (n2, n3) = (8, 3), n4 = 4 and lk(1) = C7(a1, [2, 7, 8], [3, 4, 6]). Now considering successively lk(3) and lk(1), we see two quadrangular faces which share more than one vertex, which is not allowed. Hence, (n2, n3) ≠ (8, 3).For (n2, n3) = (8, 4), completing successively lk(1), lk(4), lk(5), lk(6), lk(3), lk(2), lk(11), and lk(8), we see that lk(a2) = C6(7, 8, 9, 10, 11, 12) or lk(a2) = C6(7, 8, 9, 12, 11, 10).If lk(a2) = C6(7, 8, 9, 10, 11, 12), then lk(7) is a cycle of length 5, a contradiction.If lk(a2) = C6(7, 8, 9, 12, 11, 10), completing successively, we get M ≅ K1(2,12)-[36: 32, 4, 3, 4] via 1↦v9, 2↦v10, 3↦v11, 4↦v8, 5↦v7, 6↦v12, 7↦v5, 8↦v6, 9↦v1, 10↦v4, 11↦v3, 12↦v2, a1↦u2, a2↦u1.If (n2, n3) = (8, 9), n4 = 10. This implies lk(1) = C7(a1, [2, 7, 8], [9, 10, 6]) and lk(6) = C7(a1, [1, 9, 10], [n12, n11, 5]) for (n11, n12) ∈ {(3, 2), (11, 4), (11, 7), (11, 12)}. A small calculation shows no map exists for (n11, n12) ∈ {(11, 4), (11, 7), (11, 12)}. For (n11, n12) = (3, 2), completing successively, we get M ≅ K1(2,12)[36: 32, 4, 3, 4] via 1↦v3, 2↦v4, 3↦v5, 4↦v6, 5↦v1, 6↦v2, 7↦v7, 8↦v12, 9↦v11, 10↦v10, 11↦v9, 12↦v8, a1↦u1, a2↦u2. This completes computation of the DSEM for ≤15. This gives the following result.


### 4.3.2. Result

Theorem 3.
There exists a unique DSEM of type [36: 32, 4, 3, 4] on the surfaces of Euler characteristic 0 for ≤15 vertices. This is K1(2,12)[36: 32, 4, 3, 4] on Klein bottle, given in Figure 5.


## 4.3.1. Computation

LetM be a map of the type [36: 32, 4, 3, 4] with the vertex set V. Let V36 and V32,4,3,4 denote the sets of vertices with face-sequence types (36) and (32, 4, 3, 4), respectively. It is easy to see that 6V36=V32,4,3,4. Thus, for |V| ≤ 15, we let V=a1,a2,…,aV36,1,2,…,6V36, where V36≤2. Without loss of generality, we assume lk(a1) = C6(1, 2, 3, 4, 5, 6). Then, successively, we have lk(1) = C7(a1, [2, n1, n2], [n3, n4, 6]), lk(2) = C7(a1, [1, n2, n1], [n5, n6, 3]), lk(3) = C7(a1, [2, n5, n6], [n7, n8, 4]), lk(4) = C7(a1, [3, n7, n8], [n9, n10, 5]), lk(5) = C7(a1, [4, n9, n10], [n11, n12, 6]), lk(6) = C7(a1, [1, n3, n4], [n12, n11, 5]), where ni ∈ Vm for 1 ≤ i ≤ 12. Now considering lk(1), we see that n1 ∈{4, 5, 7}.Case 1. If n1 = 4, then successively, we see that n2 = 5, n3 = 7, and n4 = 8; now considering lk(5) and lk(1), we get two quadrangular faces which share more than one vertex, which is not allowed. So, n1 ≠ 4.Case 2. If n1 = 5, then successively, we get n2 = 4, n3 = 7, n4 = 8, n12 = 9, and n11 = 10. Now completing lk(1), lk(6), lk(5), lk(2), lk(3), lk(4), lk(7), and lk(10), we see that lk(a2) = C6(7, 8, 11, 10, 9, 12) or lk(a2) = C6(7, 8, 9, 10, 11, 12). If lk(a2) = C6(7, 8, 9, 10, 11, 12), then lk(8) is a cycle of length 5, a contradiction. If lk(a2) = C6(7, 8, 11, 10, 9, 12), then completing successively, we get M ≅ K1(2,12)[36: 32, 4, 3, 4] by the map i↦vi, aj↦uj, 1 ≤ i ≤ 12, 1 ≤ j ≤ 2.Case 3. If n1 = 7, then we get (n2, n3) ∈ {(8, 3), (8, 4), (8, 9)}. For (n2, n3) = (8, 3), n4 = 4 and lk(1) = C7(a1, [2, 7, 8], [3, 4, 6]). Now considering successively lk(3) and lk(1), we see two quadrangular faces which share more than one vertex, which is not allowed. Hence, (n2, n3) ≠ (8, 3).For (n2, n3) = (8, 4), completing successively lk(1), lk(4), lk(5), lk(6), lk(3), lk(2), lk(11), and lk(8), we see that lk(a2) = C6(7, 8, 9, 10, 11, 12) or lk(a2) = C6(7, 8, 9, 12, 11, 10).If lk(a2) = C6(7, 8, 9, 10, 11, 12), then lk(7) is a cycle of length 5, a contradiction.If lk(a2) = C6(7, 8, 9, 12, 11, 10), completing successively, we get M ≅ K1(2,12)-[36: 32, 4, 3, 4] via 1↦v9, 2↦v10, 3↦v11, 4↦v8, 5↦v7, 6↦v12, 7↦v5, 8↦v6, 9↦v1, 10↦v4, 11↦v3, 12↦v2, a1↦u2, a2↦u1.If (n2, n3) = (8, 9), n4 = 10. This implies lk(1) = C7(a1, [2, 7, 8], [9, 10, 6]) and lk(6) = C7(a1, [1, 9, 10], [n12, n11, 5]) for (n11, n12) ∈ {(3, 2), (11, 4), (11, 7), (11, 12)}. A small calculation shows no map exists for (n11, n12) ∈ {(11, 4), (11, 7), (11, 12)}. For (n11, n12) = (3, 2), completing successively, we get M ≅ K1(2,12)[36: 32, 4, 3, 4] via 1↦v3, 2↦v4, 3↦v5, 4↦v6, 5↦v1, 6↦v2, 7↦v7, 8↦v12, 9↦v11, 10↦v10, 11↦v9, 12↦v8, a1↦u1, a2↦u2. This completes computation of the DSEM for ≤15. This gives the following result.


## 4.3.2. Result

Theorem 3.
There exists a unique DSEM of type [36: 32, 4, 3, 4] on the surfaces of Euler characteristic 0 for ≤15 vertices. This is K1(2,12)[36: 32, 4, 3, 4] on Klein bottle, given in Figure 5.


## 4.4. Computation and Classification for Type33,42:32,4,3,41

Consider the following DSEM of type33,42:32,4,3,41, shown in Figure 6, on torus denoted by T14,833,42:32,4,3,41.Figure 6
Doubly semiequivelar map on torus of typeT1(4,8) [33, 42: 32, 4, 3, 4]1.


### 4.4.1. Computation

LetM be a map of the type 33,42:32,4,3,41 with the vertex set V. Let V33 and V32,4,3,4 denote the sets of vertices with face-sequence types (33, 42) and (32, 4, 3, 4), respectively. It is easy to see that 2V33,42=V32,4,3,4 and V32,4,3,4 is multiple of 4. Therefore, for |V| ≤ 15, we let V=a1,a2,…,aV33,42,1,2,…,2V33,42, where V33,42≤4. Assume that lk(a1) = C7(2, 3, [4, 5, a2, 6, 1]). Then, lk(a2) = C7(7, 8, [5, 4, a1, 1, 6]). This implies lk(2) = C7(a1, [1, x1, x2], [n1, n2, 3]) or lk(2) = C7(a1, [3, x1, x2], [n1, n2, 1]), for x1,x2∈V33,42 and n1,n2∈V32,4,3,4. In the first case of lk(2), considering lk(1), we see three quadrangular faces incident at 1, which is not allowed. On the other hand, when lk(2) = C7(a1, [3, x1, x2], [n1, n2, 1]), we get x1 = a3, x2 = a4, and (n1, n2) ∈ {(5, 8), (7, 8), (8, 5), (8, 7)}.For (n1, n2) = (7, 8), considering lk(1) and lk(2), we see that lk(a3) = C7(1, 8, [3, 2, a4, n3, 6]) or lk(a3) = C7(1, 6, [3, 2, a4, n3, 8]), but for both the cases of lk(a3), we get no suitable value for n3 in V32,4,3,4. So, (n1, n2) ≠ (7, 8).For (n1, n2) = (8, 5), completing lk(2) and lk(1) and proceeding, as in previous case, we see that lk(3) cannot be completed.For (n1, n2) = (8, 7), considering lk(2) and lk(1), we see that lk(7) cannot be completed.For (n1, n2) = (5, 8), successively, we get lk(2) = C7(a1, [3, a3, a4], [5, 8, 1]), lk(5) = C7(a4, [4, a1, a2], [8, 1, 2]), and lk(1) = C7(a3, [6, a2, a1], [2, 5, 8]). Then, lk(a3) = C7(1, 8, [3, 2, a4, 7, 6]) or lk(a3) = C7(1, 6, [3, 2, a4, 7, 8]).When lk(a3) = C7(1, 8, [3, 2, a4, 7, 6]), completing lk(a4), we see that lk(7) cannot be completed.When lk(a3) = C7(1, 6, [3, 2, a4, 7, 8]), completing successively, we get M≅T14,833,42:32,4,3,41 by the map i↦vi, aj↦uj, 1 ≤ i ≤ 8, 1 ≤ j ≤ 4. Thus, the computation is completed for ≤15 vertices. This leads to the following result.


### 4.4.2. Result

Theorem 4.
There exists a unique DSEM of type33,42:32,4,3,41 with number of vertices ≤15. This is T14,833,42:32,4,3,41 on torus, shown in Figure 6.


## 4.4.1. Computation

LetM be a map of the type 33,42:32,4,3,41 with the vertex set V. Let V33 and V32,4,3,4 denote the sets of vertices with face-sequence types (33, 42) and (32, 4, 3, 4), respectively. It is easy to see that 2V33,42=V32,4,3,4 and V32,4,3,4 is multiple of 4. Therefore, for |V| ≤ 15, we let V=a1,a2,…,aV33,42,1,2,…,2V33,42, where V33,42≤4. Assume that lk(a1) = C7(2, 3, [4, 5, a2, 6, 1]). Then, lk(a2) = C7(7, 8, [5, 4, a1, 1, 6]). This implies lk(2) = C7(a1, [1, x1, x2], [n1, n2, 3]) or lk(2) = C7(a1, [3, x1, x2], [n1, n2, 1]), for x1,x2∈V33,42 and n1,n2∈V32,4,3,4. In the first case of lk(2), considering lk(1), we see three quadrangular faces incident at 1, which is not allowed. On the other hand, when lk(2) = C7(a1, [3, x1, x2], [n1, n2, 1]), we get x1 = a3, x2 = a4, and (n1, n2) ∈ {(5, 8), (7, 8), (8, 5), (8, 7)}.For (n1, n2) = (7, 8), considering lk(1) and lk(2), we see that lk(a3) = C7(1, 8, [3, 2, a4, n3, 6]) or lk(a3) = C7(1, 6, [3, 2, a4, n3, 8]), but for both the cases of lk(a3), we get no suitable value for n3 in V32,4,3,4. So, (n1, n2) ≠ (7, 8).For (n1, n2) = (8, 5), completing lk(2) and lk(1) and proceeding, as in previous case, we see that lk(3) cannot be completed.For (n1, n2) = (8, 7), considering lk(2) and lk(1), we see that lk(7) cannot be completed.For (n1, n2) = (5, 8), successively, we get lk(2) = C7(a1, [3, a3, a4], [5, 8, 1]), lk(5) = C7(a4, [4, a1, a2], [8, 1, 2]), and lk(1) = C7(a3, [6, a2, a1], [2, 5, 8]). Then, lk(a3) = C7(1, 8, [3, 2, a4, 7, 6]) or lk(a3) = C7(1, 6, [3, 2, a4, 7, 8]).When lk(a3) = C7(1, 8, [3, 2, a4, 7, 6]), completing lk(a4), we see that lk(7) cannot be completed.When lk(a3) = C7(1, 6, [3, 2, a4, 7, 8]), completing successively, we get M≅T14,833,42:32,4,3,41 by the map i↦vi, aj↦uj, 1 ≤ i ≤ 8, 1 ≤ j ≤ 4. Thus, the computation is completed for ≤15 vertices. This leads to the following result.


## 4.4.2. Result

Theorem 4.
There exists a unique DSEM of type33,42:32,4,3,41 with number of vertices ≤15. This is T14,833,42:32,4,3,41 on torus, shown in Figure 6.


## 4.5. Computation and Classification for Type33,42:32,4,3,42

Consider the following DSEM of type33,42:32,4,3,42, shown in Figure 7, on Klein bottle denoted by K16,633,42:32,4,3,42.Figure 7
Doubly semiequivelar map of typeK16,633,42:32,4,3,41 on Klein bottle.


### 4.5.1. Computation

LetM be a map of the type 33,42:32,4,3,42 with the vertex set V. Let V33,42 and V32,4,3,4 denote the sets of vertices with face-sequence types (33, 42) and (32, 4, 3, 4), respectively. Then, we see easily that V33,42=V32,4,3,4=2k for k∈ℕ. Thus, for V=V33,42+V32,4,3,4≤15, we let V=a1,a2,…,aV33,42,1,2,…,V33,42, where V33,42=2k for k ≤ 3. Assume that, without loss of generality, lk(a1) = C7(a3, 3, [4, 5, a2, 1, 2]). This implies lk(a2) = C7(x1, n1, [5, 4, a1, 2, 1]) or lk(a2) = C7(x1, n1, [1, 2, a1, 4, 5]) for x1∈V33,42 and n1∈V32,4,3,4.If lk(a2) = C7(x1, n1, [5, 4, a1, 2, 1]), then a small calculation shows that no such map exists for the given V. On the other hand, if lk(a2) = C7(x1, n1, [1, 2, a1, 4, 5]), then we have x1 = a4, n1 = 6. This implies lk(2) = C7(a3, [a1, a2, 1], [x2, x3, n2]), where (x2, x3, n2) ∈ {(a5, a4, 5), (a5, a4, 6)}. If (x2, x3, n2) = (a5, a4, 5), then lk(a3) = C7(a1, 2, [5, n3, a6, n4, 3]). Now considering lk(5), we see three quadrangular faces incident at 5, which is not allowed. If (x2, x3, n2) = (a5, a4, 6), lk(2) = C7(a3, [a1, a2, 1], [a5, a4, 6]). This implies lk(a3) = C7(a1, 2, [6, n3, a6, n4, 3]), where (n3, n4) ∈ {(1, 5), (4, 1), (4, 5)}. In case (n3, n4) = (4, 1) and (4, 5), we see, respectively, lk(1) and lk(4) cannot be completed. If n3,n4=1,5,M≅K16,633,42:32,4,3,42 by the map i↦vi, ai↦ui, 1 ≤ i ≤ 6. Thus, the computation is completed. Then, we obtain the following result.


### 4.5.2. Result

Theorem 5.
There exists a unique DSEM of type33,42:32,4,3,42 with number of vertices ≤15. This is K16,633,42:32,4,3,42 on Klein bottle, shown in Figure 7.


## 4.5.1. Computation

LetM be a map of the type 33,42:32,4,3,42 with the vertex set V. Let V33,42 and V32,4,3,4 denote the sets of vertices with face-sequence types (33, 42) and (32, 4, 3, 4), respectively. Then, we see easily that V33,42=V32,4,3,4=2k for k∈ℕ. Thus, for V=V33,42+V32,4,3,4≤15, we let V=a1,a2,…,aV33,42,1,2,…,V33,42, where V33,42=2k for k ≤ 3. Assume that, without loss of generality, lk(a1) = C7(a3, 3, [4, 5, a2, 1, 2]). This implies lk(a2) = C7(x1, n1, [5, 4, a1, 2, 1]) or lk(a2) = C7(x1, n1, [1, 2, a1, 4, 5]) for x1∈V33,42 and n1∈V32,4,3,4.If lk(a2) = C7(x1, n1, [5, 4, a1, 2, 1]), then a small calculation shows that no such map exists for the given V. On the other hand, if lk(a2) = C7(x1, n1, [1, 2, a1, 4, 5]), then we have x1 = a4, n1 = 6. This implies lk(2) = C7(a3, [a1, a2, 1], [x2, x3, n2]), where (x2, x3, n2) ∈ {(a5, a4, 5), (a5, a4, 6)}. If (x2, x3, n2) = (a5, a4, 5), then lk(a3) = C7(a1, 2, [5, n3, a6, n4, 3]). Now considering lk(5), we see three quadrangular faces incident at 5, which is not allowed. If (x2, x3, n2) = (a5, a4, 6), lk(2) = C7(a3, [a1, a2, 1], [a5, a4, 6]). This implies lk(a3) = C7(a1, 2, [6, n3, a6, n4, 3]), where (n3, n4) ∈ {(1, 5), (4, 1), (4, 5)}. In case (n3, n4) = (4, 1) and (4, 5), we see, respectively, lk(1) and lk(4) cannot be completed. If n3,n4=1,5,M≅K16,633,42:32,4,3,42 by the map i↦vi, ai↦ui, 1 ≤ i ≤ 6. Thus, the computation is completed. Then, we obtain the following result.


## 4.5.2. Result

Theorem 5.
There exists a unique DSEM of type33,42:32,4,3,42 with number of vertices ≤15. This is K16,633,42:32,4,3,42 on Klein bottle, shown in Figure 7.


## 4.6. Computation and Classification for Type33,42:441

Consider the DSEMs of type33,42:441, shown in Figure 8, on torus and Klein bottle denoted by Tin,2n33,42:441, for i ∈ {1, …, 6}, and Kin,2n33,42:441, for i ∈ {1, …, 3}, respectively.Figure 8
Doubly semiequivelar maps of type33,42:441 on torus and Klein bottle. (a) K1(3,6) [33, 42: 44]1. (b) T1(3,6) [33, 42: 44]1. (c) T2(3,6) [33, 42: 44]1. (d) K2(4,8) [33, 42: 44]1. (e) T3(4,8) [33, 42: 44]1. (f) T4(4,8) [33, 42: 44]1. (g) T5(5,10) [33, 42: 44]1. (h) T6(5,10) [33, 42: 44]1. (i) K4(5,10) [33, 42: 44]1. (j) T7(5,10) [33, 42: 44]1. (k) K3(5,10) [33, 42: 44]1.
(a)
(b)
(c)
(d)
(e)
(f)
(g)
(h)
(i)
(j)
(k)Claim 4.
For the maps above, we have the following:(a)
T13,633,42:441≇T23,633,42:441.(b)
T34,833,42:441≇T44,833,42:441.(c)
T55,1033,42:441≇T65,1033,42:441≇T75,1033,42:441.(d)
K35,1033,42:441≇K45,10-33,42:441.Proof.
The proof follows by considering the following polynomials:pEGT13,633,42:441=x9−21x7−24x6+72x5+72x4−99x3−54x2+54xpEGT23,633,42:441=x9−21x7−18x6+54x5pEGT34,833,42:441=x12−28x10−16x9+212x8+88x7−684x6−48x5+912x4−272x3−240x2+96xpEGT44,833,42:441=x12−28x10−24x9+212x8+280x7−524x6−976x5+80x4+880x3+528x2+96xpEGK35,1033,42:441=x15−5x13−40x12+385x11+790x10−1100x9−3620x8+55x7+6200x6+3305x5−3500x4−3265x3−190x2+270x−24pEGK45,1033,42:441=x15−35x13−24x12+401x11+434x10−1832x9−2468x8+3123x7+5232x6−939x5−3716x4−1261x3+30x2+30xpEGT55,1033,42:441=x15−35x13−20x12+425x11+294x10−2500x9−1520x8+7855x7+3060x6−12919x5−1100x4+8815x3−2250x2+150xpEGT65,1033,42:441=x15−35x13−20x12+395x11+344x10−1790x9−1960x8+3150x7+3920x6−2059x5−3000x4+235x3+750x2+150xpEGT75,1033,42:441=x15−35x13−30x12+395x11+594x10−1495x9−3360x8+175x7+3990x6+2166x5


### 4.6.1. Computation

LetM be a map of the type 33,42:441 with the vertex set V. Let V44 and V33,42 denote the sets of vertices with face-sequence types (44) and (33, 42), respectively. Then, counting the number of quadrangular faces in terms of V33,42 and V44 we see easily that V33,42=2V44. Thus, for |V| ≤ 15, we let V=a1,a2,…,aV44,1,2,…,2V44 such that V44≤5. Assume that, without loss of generality, lka1=C8a3,1,2,3,a2,4,5,6. This implies lk(a2) = C8(a1, 2, 3, n1, x1, n2, 4, 5) for x1∈V44 and n1,n2∈V33,42. Observe that x1 ∈ {a3, a4}.Case 1. When x1 = a3, then (n1, n2) ∈ {(1, 6), (6, 1)}. If (n1, n2) = (6, 1), then lk(a2) = C8(a1, 2, 3, 6, a3, 1, 4, 5) and lk(a3) = C8(a2, 3, 6, 5, a1, 2, 1, 4). This implies lk(1) = C7(n3, n4, [2, a1, a3, a2, 4]). It is easy to see that (n3, n4) ∈ {(3, 6), (5, 6), (6, 3), (6, 5)}. But for these values of (n3, n4), we see easily that no map exists.On the other hand, if (n1, n2) = (1, 6), then lk(a2) = C8(a1, 2, 3, 1, a3, 6, 4, 5), lk(a3) = C8(a2, 3, 1, 2, a1, 5, 6, 4). This implies lk(1) = C7(n3, n4, [2, a1, a3, a2, 3]), for (n3, n4) ∈ {(4, 5), (4, 6), (5, 4), (5, 6), (6, 4), (6, 5)}.Observe that (5, 6) ≅ (4, 5) by the map (1, 3, 2) (4, 5, 6) (a1, a3, a2), (6, 4) ≅ (4, 5) by the map (1, 2, 3) (4, 6, 5) (a1, a2, a3), and (6, 5) ≅ (4, 6) by the map (2, 3) (4, 5) (a1, a2). Thus, we search for (n3, n4) ∈ {(4, 5), (4, 6), (5, 4), (5, 6)}. Now doing computation for these cases, we see the following:Ifn3,n4=4,5,M≅K13,633,42:441 by the map i↦v6, aj↦uj, 1 ≤ i ≤ 6, 1 ≤ j ≤ 3.Ifn3,n4=4,6,M≅T13,633,42:441 by the map i↦v6, aj↦uj, 1 ≤ i ≤ 6, 1 ≤ j ≤ 3.Ifn3,n4=5,4,M≅T23,633,42:441 by the map i↦v6, aj↦uj, 1 ≤ i ≤ 6, 1 ≤ j ≤ 3.Case 2. When x1 = a4, then considering lk(a3) = C8(a1, 2, 1, n3, x2, n4, 6, 5), we get x2 ∈ {a4, a5}.Subcase 2.1. If x2 = a4, then (n3, n4) ∈ {(7, 8), (8, 7)}. If (n3, n4) = (8, 7), (n5, n6) ∈ {(4, 5), (5, 4), (5, 6), (6, 5), (6, 7), (7, 3), (7, 6)}. If (n5, n6) = (4, 5), then considering successively lk(1), lk(5), and lk(8), we see that deg(4) > 5, a contradiction. If (n5, n6) = (5, 4), then considering successively lk(1), lk(5), lk(2), and lk(6), we get lk(7) of length 5, a contradiction. Proceeding similarly for the rest of the cases of (n5, n6), we see easily that no map exists. On the other hand, if (n3, n4) = (7, 8), then lk(1) = C7(n5, n6, [2, a1, a3, a4, 7]), where (n5, n6) ∈ {(4, 5), (4, 8), (5, 4), (5, 6), (6, 5), (6, 8), (8, 4), (8, 6)}. But (5, 4) ≅ (4, 8) by the map (1, 3) (4, 6) (a2, a3), (5, 6) ≅ (4, 5) by the map (1, 7) (2, 3) (4, 5) (6, 8) (a1, a2) (a3, a4), (6, 8) ≅ (4, 5) by the map (1, 2, 3, 7) (4, 8, 6, 5) (a1, a2, a4, a3), (8, 4) ≅ (4, 5) by the map (1, 3) (2, 7) (4, 6) (5, 8) (a1, a4) (a2, a3), and (8, 6) ≅ (6, 5) by the map (1, 7) (2, 3) (4, 5) (6, 8) (a1, a2) (a3, a4). So, we search for (n5, n6) ∈ {(4, 5), (4, 8), (6, 5)}. Now doing computation for these cases, we see the following:Ifn5,n6=4,5,M≅K24,833,42:441 by the map i↦v6, aj↦uj, 1 ≤ i ≤ 8, 1 ≤ j ≤ 4.Ifn5,n6=4,8,M≅T34,833,42:441 by the map i↦v6, aj↦uj, 1 ≤ i ≤ 8, 1 ≤ j ≤ 4.Ifn5,n6=6,5,M≅T44,833,42:441 by the map i↦v6, aj↦uj, 1 ≤ i ≤ 8, 1 ≤ j ≤ 4.Subcase 2.2. When x2 = a5, then considering lk(a4), we see that (n5, n6) ∈ {(9, 10), (10, 9)}. If (n5, n6) = (10, 9), (n7, n8) ∈ {(3, 7), (4, 5), (4, 8), (5, 4), (5, 6), (6, 5), (6, 10), (7, 10), (10, 6), (10, 7)}. But a small calculation shows that no map exists for these values of (n7, n8). If $(n5, n6) = (9,10)$, then we get lk(1) = C7(n7, n8, [2, a1, a3, a5, 9]). Then, up to isomorphism, we get (n7, n8) ∈ {(3, 7), (4, 5), (5, 4), (6, 5)}. Doing computation for these cases, we see the following:Ifn7,n8=3,7,M≅K35,1033,42:441 by the map i↦vi, aj↦uj, 1 ≤ i ≤ 10, 1 ≤ j ≤ 5.Ifn7,n8=4,5,M≅K45,1033,42:441 by the map i↦vi, aj↦uj, 1 ≤ i ≤ 10, 1 ≤ j ≤ 5.Ifn7,n8=4,8,M≅T55,1033,42:441 by the map i↦vi, aj↦uj, 1 ≤ i ≤ 10, 1 ≤ j ≤ 5.Ifn7,n8=5,4,M≅T65,1033,42:441 by the map i↦vi, aj↦uj, 1 ≤ i ≤ 10, 1 ≤ j ≤ 5.Ifn7,n8=6,5,M≅T75,1033,42:441 by the map i↦vi, aj↦uj, 1 ≤ i ≤ 10, 1 ≤ j ≤ 5. This completes the computation and we get the following results.


### 4.6.2. Results

Lemma 3.
LetM be a DSEM of type 33,42:441 with number of vertices ≤15. Then, M is isomorphic to one of the following: T13,633,42:441, T23,633,42:441, K13,633,42:441, T34,833,42:441, T44,833,42:441, K24,833,42:441, T55,1033,42:441, T65,1033,42:441, T75,1033,42:441, K35,1033,42:441, and K45,1033,42:441, as shown in Figure 8.Combining Lemma 3 together with Claim4, we get the following theorem.Theorem 6.
There are exactly 11 non-isomorphic DSEMs of type33,42:441 on the surfaces of Euler characteristic 0 with ≤15 vertices. Out of these, 7 are on torus and remaining 4 are on Klein bottle.


## 4.6.1. Computation

LetM be a map of the type 33,42:441 with the vertex set V. Let V44 and V33,42 denote the sets of vertices with face-sequence types (44) and (33, 42), respectively. Then, counting the number of quadrangular faces in terms of V33,42 and V44 we see easily that V33,42=2V44. Thus, for |V| ≤ 15, we let V=a1,a2,…,aV44,1,2,…,2V44 such that V44≤5. Assume that, without loss of generality, lka1=C8a3,1,2,3,a2,4,5,6. This implies lk(a2) = C8(a1, 2, 3, n1, x1, n2, 4, 5) for x1∈V44 and n1,n2∈V33,42. Observe that x1 ∈ {a3, a4}.Case 1. When x1 = a3, then (n1, n2) ∈ {(1, 6), (6, 1)}. If (n1, n2) = (6, 1), then lk(a2) = C8(a1, 2, 3, 6, a3, 1, 4, 5) and lk(a3) = C8(a2, 3, 6, 5, a1, 2, 1, 4). This implies lk(1) = C7(n3, n4, [2, a1, a3, a2, 4]). It is easy to see that (n3, n4) ∈ {(3, 6), (5, 6), (6, 3), (6, 5)}. But for these values of (n3, n4), we see easily that no map exists.On the other hand, if (n1, n2) = (1, 6), then lk(a2) = C8(a1, 2, 3, 1, a3, 6, 4, 5), lk(a3) = C8(a2, 3, 1, 2, a1, 5, 6, 4). This implies lk(1) = C7(n3, n4, [2, a1, a3, a2, 3]), for (n3, n4) ∈ {(4, 5), (4, 6), (5, 4), (5, 6), (6, 4), (6, 5)}.Observe that (5, 6) ≅ (4, 5) by the map (1, 3, 2) (4, 5, 6) (a1, a3, a2), (6, 4) ≅ (4, 5) by the map (1, 2, 3) (4, 6, 5) (a1, a2, a3), and (6, 5) ≅ (4, 6) by the map (2, 3) (4, 5) (a1, a2). Thus, we search for (n3, n4) ∈ {(4, 5), (4, 6), (5, 4), (5, 6)}. Now doing computation for these cases, we see the following:Ifn3,n4=4,5,M≅K13,633,42:441 by the map i↦v6, aj↦uj, 1 ≤ i ≤ 6, 1 ≤ j ≤ 3.Ifn3,n4=4,6,M≅T13,633,42:441 by the map i↦v6, aj↦uj, 1 ≤ i ≤ 6, 1 ≤ j ≤ 3.Ifn3,n4=5,4,M≅T23,633,42:441 by the map i↦v6, aj↦uj, 1 ≤ i ≤ 6, 1 ≤ j ≤ 3.Case 2. When x1 = a4, then considering lk(a3) = C8(a1, 2, 1, n3, x2, n4, 6, 5), we get x2 ∈ {a4, a5}.Subcase 2.1. If x2 = a4, then (n3, n4) ∈ {(7, 8), (8, 7)}. If (n3, n4) = (8, 7), (n5, n6) ∈ {(4, 5), (5, 4), (5, 6), (6, 5), (6, 7), (7, 3), (7, 6)}. If (n5, n6) = (4, 5), then considering successively lk(1), lk(5), and lk(8), we see that deg(4) > 5, a contradiction. If (n5, n6) = (5, 4), then considering successively lk(1), lk(5), lk(2), and lk(6), we get lk(7) of length 5, a contradiction. Proceeding similarly for the rest of the cases of (n5, n6), we see easily that no map exists. On the other hand, if (n3, n4) = (7, 8), then lk(1) = C7(n5, n6, [2, a1, a3, a4, 7]), where (n5, n6) ∈ {(4, 5), (4, 8), (5, 4), (5, 6), (6, 5), (6, 8), (8, 4), (8, 6)}. But (5, 4) ≅ (4, 8) by the map (1, 3) (4, 6) (a2, a3), (5, 6) ≅ (4, 5) by the map (1, 7) (2, 3) (4, 5) (6, 8) (a1, a2) (a3, a4), (6, 8) ≅ (4, 5) by the map (1, 2, 3, 7) (4, 8, 6, 5) (a1, a2, a4, a3), (8, 4) ≅ (4, 5) by the map (1, 3) (2, 7) (4, 6) (5, 8) (a1, a4) (a2, a3), and (8, 6) ≅ (6, 5) by the map (1, 7) (2, 3) (4, 5) (6, 8) (a1, a2) (a3, a4). So, we search for (n5, n6) ∈ {(4, 5), (4, 8), (6, 5)}. Now doing computation for these cases, we see the following:Ifn5,n6=4,5,M≅K24,833,42:441 by the map i↦v6, aj↦uj, 1 ≤ i ≤ 8, 1 ≤ j ≤ 4.Ifn5,n6=4,8,M≅T34,833,42:441 by the map i↦v6, aj↦uj, 1 ≤ i ≤ 8, 1 ≤ j ≤ 4.Ifn5,n6=6,5,M≅T44,833,42:441 by the map i↦v6, aj↦uj, 1 ≤ i ≤ 8, 1 ≤ j ≤ 4.Subcase 2.2. When x2 = a5, then considering lk(a4), we see that (n5, n6) ∈ {(9, 10), (10, 9)}. If (n5, n6) = (10, 9), (n7, n8) ∈ {(3, 7), (4, 5), (4, 8), (5, 4), (5, 6), (6, 5), (6, 10), (7, 10), (10, 6), (10, 7)}. But a small calculation shows that no map exists for these values of (n7, n8). If $(n5, n6) = (9,10)$, then we get lk(1) = C7(n7, n8, [2, a1, a3, a5, 9]). Then, up to isomorphism, we get (n7, n8) ∈ {(3, 7), (4, 5), (5, 4), (6, 5)}. Doing computation for these cases, we see the following:Ifn7,n8=3,7,M≅K35,1033,42:441 by the map i↦vi, aj↦uj, 1 ≤ i ≤ 10, 1 ≤ j ≤ 5.Ifn7,n8=4,5,M≅K45,1033,42:441 by the map i↦vi, aj↦uj, 1 ≤ i ≤ 10, 1 ≤ j ≤ 5.Ifn7,n8=4,8,M≅T55,1033,42:441 by the map i↦vi, aj↦uj, 1 ≤ i ≤ 10, 1 ≤ j ≤ 5.Ifn7,n8=5,4,M≅T65,1033,42:441 by the map i↦vi, aj↦uj, 1 ≤ i ≤ 10, 1 ≤ j ≤ 5.Ifn7,n8=6,5,M≅T75,1033,42:441 by the map i↦vi, aj↦uj, 1 ≤ i ≤ 10, 1 ≤ j ≤ 5. This completes the computation and we get the following results.


## 4.6.2. Results

Lemma 3.
LetM be a DSEM of type 33,42:441 with number of vertices ≤15. Then, M is isomorphic to one of the following: T13,633,42:441, T23,633,42:441, K13,633,42:441, T34,833,42:441, T44,833,42:441, K24,833,42:441, T55,1033,42:441, T65,1033,42:441, T75,1033,42:441, K35,1033,42:441, and K45,1033,42:441, as shown in Figure 8.Combining Lemma 3 together with Claim4, we get the following theorem.Theorem 6.
There are exactly 11 non-isomorphic DSEMs of type33,42:441 on the surfaces of Euler characteristic 0 with ≤15 vertices. Out of these, 7 are on torus and remaining 4 are on Klein bottle.


## 4.7. Computation and Classification for Type33,42:442

Consider the following DSEMs of type33,42:442, shown in Figure 9, on torus and Klein bottle denoted by Ti6,633,42:442, for i = 1, 2, and K16,633,42:442, respectively.Figure 9
Doubly semiequivelar maps of type33,42:442 on torus and Klein bottle. (a) K1(6,6) [33, 42: 44]2. (b) T1(6,6) [33, 42: 44]2. (c) T2(6,6) [33, 42: 44]2.
(a)
(b)
(c)Claim 5.
T
1
6,6
3
3
,
4
2
:
4
4
2
≇
T
2
6,6
3
3
,
4
2
:
4
4
2.Proof.
See the following polynomials:pEGT16,633,42:442=x12−27x10−20x9+207x8+168x7−610x6−288x5+723x4−136x3−171x2+84x−11pEGT26,633,42:442=x12−27x10−20x9+201x8+192x7−532x6−552x5+492x4+560x3−84x2−192x−44


### 4.7.1. Computation

LetM be a map of the type 33,42:442 with the vertex set V. Let V44 and V33,42 denote the sets of vertices with face-sequence types (44) and (33, 42), respectively. Then, it is easy to see that M has the number of quadrangular faces 3V44/2 or V33,42+V44/2. This implies V44=V33,42=2k for k∈ℕ. Therefore, for |V| ≤ 15, we let V=a1,a2,…,aV44,1,2,…,V44, where V44=2k for k ≤ 3. Without loss of generality, we may assume lk(1) = C7(3, 4, [5, a1, a2, a3, 2]). Then, lk(a2) = C8(a3, 2, 1, 5, a1, a6, a5, a4), lk(a1) = C8(a2, 1, 5, n1, x1, x2, a6, a5) for n1 ∈ Vm and x1, x2 ∈ Vl. Observe that (n1, x1, x2) ∈{(2, a3, a4), (3, a4, a3), (6, a4, a3)}.Case 1. When (n1, x1, x2) = (3, a4, a3), then considering successively lk(a1), lk(a4), lk(a5), lk(3), and lk(4), we see that lk(5) cannot be completed. So, (n1, x1, x2) ≠ (2, a4, a3).Case 2. When (n1, x1, x2) = (6, a4, a3), then lk(a1) = C8(a2, 1, 5, 6, a4, a3, a6, a5) and lk(a4) = C8(a1, 5, 6, n2, a5, a2, a3, a6) for n2 ∈{3, 4, 7}. If n2 = 3, then considering successively lk(a4), lk(a5), lk(a6), lk(a3), lk(3), and lk(2), we see lk(4) cannot be completed.Ifn2 = 4, then considering successively lk(a4), lk(a5), lk(a6), lk(a3), and lk(4), as in previous case, we see that lk(5) cannot be completed.Ifn2 = 7, then considering successively lk(a4), lk(a5), and lk(a6), we see that lk(a3) cannot be completed. So, (n1, x1, x2) ≠ (6, a4, a3).Case 3. If (n1, x1, x2) = (2, a3, a4), then successively completing lk(a1), lk(a3), lk(2), and lk(5), we get lk(4) = C7(1, 5, [6, x5, x4, x3, 3]) for (x3, x4, x5) ∈ {(a5, a4, a6), (a6, a4, a5), (a4, a5, a6), (a6, a5, a4), (a4, a6, a5), (a5, a6, a4)}. Note that (a6, a5, a4) ≅ (a5, a4, a6) by the map (1, 5, 2) (3, 4, 6) (a1, a3, a2) (a4, a5, a6), (a4, a6, a5) ≅ (a5, a4, a6) by the map (1, 2, 5) (3, 6, 4) (a1, a2, a3) (a4, a6, a5), and (a5, a6, a4) ≅ (a4, a5, a6) by the map (2, 5) (3, 4) (a1, a3) (a4, a6).Thus, we do computation for (x3, x4, x5) ∈ {(a5, a4, a6), (a6, a4, a5), (a4, a5, a6)}. This gives the following:Ifx3,x4,x5=a5,a4,a6,M≅K16,633,42:442 by the map i↦vi, ai↦ui, 1 ≤ i ≤ 6.Ifx3,x4,x5=a6,a4,a5,M≅T16,633,42:442 by the map i↦vi, ai↦ui, 1 ≤ i ≤ 6.Ifx3,x4,x5=a4,a5,a6,M≅T26,633,42:442 by the map i↦vi, ai↦ui, 1 ≤ i ≤ 6.Thus, the computation is completed and we get the following results.


### 4.7.2. Results

Lemma 4.
LetM be a DSEM of type 33,42:442 with number of vertices ≤15. Then, M is isomorphic to T16,633,42:442, T26,633,42:442 or K16,633,42:442, as shown in Figure 9.Combining Lemma4 together with Claim 5, we get the following theorem.Theorem 7.
There are exactly 3 non-isomorphic DSEMs of type33,42:442 on the surfaces of Euler characteristic 0 with ≤15 vertices. Out of these, two are on torus and the remaining one is on Klein bottle.


## 4.7.1. Computation

LetM be a map of the type 33,42:442 with the vertex set V. Let V44 and V33,42 denote the sets of vertices with face-sequence types (44) and (33, 42), respectively. Then, it is easy to see that M has the number of quadrangular faces 3V44/2 or V33,42+V44/2. This implies V44=V33,42=2k for k∈ℕ. Therefore, for |V| ≤ 15, we let V=a1,a2,…,aV44,1,2,…,V44, where V44=2k for k ≤ 3. Without loss of generality, we may assume lk(1) = C7(3, 4, [5, a1, a2, a3, 2]). Then, lk(a2) = C8(a3, 2, 1, 5, a1, a6, a5, a4), lk(a1) = C8(a2, 1, 5, n1, x1, x2, a6, a5) for n1 ∈ Vm and x1, x2 ∈ Vl. Observe that (n1, x1, x2) ∈{(2, a3, a4), (3, a4, a3), (6, a4, a3)}.Case 1. When (n1, x1, x2) = (3, a4, a3), then considering successively lk(a1), lk(a4), lk(a5), lk(3), and lk(4), we see that lk(5) cannot be completed. So, (n1, x1, x2) ≠ (2, a4, a3).Case 2. When (n1, x1, x2) = (6, a4, a3), then lk(a1) = C8(a2, 1, 5, 6, a4, a3, a6, a5) and lk(a4) = C8(a1, 5, 6, n2, a5, a2, a3, a6) for n2 ∈{3, 4, 7}. If n2 = 3, then considering successively lk(a4), lk(a5), lk(a6), lk(a3), lk(3), and lk(2), we see lk(4) cannot be completed.Ifn2 = 4, then considering successively lk(a4), lk(a5), lk(a6), lk(a3), and lk(4), as in previous case, we see that lk(5) cannot be completed.Ifn2 = 7, then considering successively lk(a4), lk(a5), and lk(a6), we see that lk(a3) cannot be completed. So, (n1, x1, x2) ≠ (6, a4, a3).Case 3. If (n1, x1, x2) = (2, a3, a4), then successively completing lk(a1), lk(a3), lk(2), and lk(5), we get lk(4) = C7(1, 5, [6, x5, x4, x3, 3]) for (x3, x4, x5) ∈ {(a5, a4, a6), (a6, a4, a5), (a4, a5, a6), (a6, a5, a4), (a4, a6, a5), (a5, a6, a4)}. Note that (a6, a5, a4) ≅ (a5, a4, a6) by the map (1, 5, 2) (3, 4, 6) (a1, a3, a2) (a4, a5, a6), (a4, a6, a5) ≅ (a5, a4, a6) by the map (1, 2, 5) (3, 6, 4) (a1, a2, a3) (a4, a6, a5), and (a5, a6, a4) ≅ (a4, a5, a6) by the map (2, 5) (3, 4) (a1, a3) (a4, a6).Thus, we do computation for (x3, x4, x5) ∈ {(a5, a4, a6), (a6, a4, a5), (a4, a5, a6)}. This gives the following:Ifx3,x4,x5=a5,a4,a6,M≅K16,633,42:442 by the map i↦vi, ai↦ui, 1 ≤ i ≤ 6.Ifx3,x4,x5=a6,a4,a5,M≅T16,633,42:442 by the map i↦vi, ai↦ui, 1 ≤ i ≤ 6.Ifx3,x4,x5=a4,a5,a6,M≅T26,633,42:442 by the map i↦vi, ai↦ui, 1 ≤ i ≤ 6.Thus, the computation is completed and we get the following results.


## 4.7.2. Results

Lemma 4.
LetM be a DSEM of type 33,42:442 with number of vertices ≤15. Then, M is isomorphic to T16,633,42:442, T26,633,42:442 or K16,633,42:442, as shown in Figure 9.Combining Lemma4 together with Claim 5, we get the following theorem.Theorem 7.
There are exactly 3 non-isomorphic DSEMs of type33,42:442 on the surfaces of Euler characteristic 0 with ≤15 vertices. Out of these, two are on torus and the remaining one is on Klein bottle.


## 5. Summary

From Theorems1–7, we get the following.Theorem 8.
There are at least 35 non-isomorphic DSEMs on the surfaces of Euler characteristic 0 with ≤15 vertices. Out of these, 20 are on the torus and remaining 15 are on the Klein bottle.A tabular form of the results obtained here is presented in Table1.


## 6. Discussion

In [16], the authors constructed infinite series of semiequivelar maps on torus and Klein bottle from equivelar maps by using elementary map operations: truncation and subdivision (these operations do not affect the symmetry of a map). Here, we present infinite series of the seven type doubly semiequivelar maps for torus, and one can explore similarly for Klein bottle.Infinite series of DSEMs of types36:33,421, 33,42:441, 33,42:442, 33,42:32,4,3,42, and [36: 33, 42] are constructed from infinite series of semiequivelar map of type [44] by subdividing the quadrangular faces as shown in Figures 10–14, respectively. Infinite series of DSEM of type [36: 32, 4, 3, 4] is obtained from an infinite series of semiequivelar map of type [63] by subdividing the hexagonal faces (by introducing a new vertex and joining it to the six vertices of the face by an edge) as shown in Figure 15.Figure 10
DSEM of type-36:33,421:n≥3.Figure 11
DSEM of type-33,42:441:n≥3.Figure 12
DSEM of type-33,42:442:n≥3.Figure 13
DSEM of type-33,42:32,4,3,42:n≥2.Figure 14
DSEM of type-36:33,422:n≥0.Figure 15
DSEM of type-36:32,4,3,4:n≥2.Although, we present infinite series of DSEM of type [33, 42: 32, 4, 3, 4] (see Figure 16), we do not know whether this DSEM can be obtained from any semiequivelar map by the above elementary map operations. This observation leads to the following question.Figure 16
DSEM of type-33,42:32,4,3,41:n≥3.Question 1.
Can we obtain every doubly semiequivelar map (corresponding to the 2-uniform tilings) on torus and Klein bottle from semiequivelar maps (corresponding to the Archimedean tilings) by applying finite sequence of map operations on the same surface?


### 6.1. Infinite Series of DSEMs on Torus

If we study group structures associated to the maps, we see that DSEMs, obtained here, on torus are 2-uniform. For example, in case of type36:33,421, we see that the groups G1 = v1,v3,v5,v4,v2,v6u1,u3,u2, G2 = v1,v2,v5v3,v6,v4u1,u3,u2,v1,v3v2,v4v5,v6u2,u3, G3 = v1,v2,v7,v5v3,v8,v6,v4u1,u3,u4,u2,v1,v4v2,v6v3,v5v7,v8u1,u2u3,u4, G4 = v1,v5,v7,v2v3,v4,v6,v8u1,u2,u4,u3,v1,v8v2,v6v3,v5v4,v7u1,u4u2,u3, G5 = v1,v5,v7,v2v3,v4,v6,v8u1,u2,u4,u3,v1,v4v2,v3v5,v6v7,v8, G6 = v1,v2,v7,v10,v5v3,v8,v9,v6,v4u1,u3,u5,u4,u2,v1,v9v2,v8v3,v7v4,v10v5,v6u1,u5u2,u4, G7 = v1,v7,v5,v2,v10v3,v9,v4,v8,v6u1,u5,u2,u3,u4,v1,v3v2,v4v5,v8v6,v7v9,v10u2,u3u4,u5, and G8 = v1,v10,v2,v5,v7v3,v6,v8,v4,v9u1,u4,u3,u2,u5,v1,v4v2,v3v5,v6v7,v8v9,v10 act on the maps T13,636:33,421, T23,636:33,421, T34,836:33,421, T44,836:33,421, T54,836:33,421, T65,1036:33,421, T75,1036:33,421, and T85,1036:33,421, respectively, such that under the action, the maps have two orbits of vertices. Similarly, we can easily find a group for the DSEMs of types 36:33,422,33,42:32,4,3,41,33,42:441, and 33,42:442 on torus, under which the maps have two orbits of vertices. However, this fact does not hold for the DSEMs on Klein bottle. For example, if we let K13,636:33,421, we get no automorphism which sends v1 to v2. This can be seen as follows: suppose there is f∈AutK13,636:33,421 such that f(v1) = v2. Then, considering lk(v1) and lk(v2), we see, either f(v5) = v5 or f(v5) = v1. In the first case, when f(v5) = v5, we get f(u2) = u3, f(u1) = u1, f(v2) = v1, f(v3) = v6, f(v6) = v3, and f(v4) = v4. Now if we see lk(u1), we get a contradiction of the facts f(u1) = u1 and f(v3) = v6, as v6 ∉ lk(u1). So, f(v5) ≠ v5. Similarly, we see that f(v5) ≠ v1. Combining these, we see that f(v1) ≠ v2. This shows that K13,636:33,421 is not 2-uniform. This observation leads to ask the following question.Question 2.
Are the doubly semiequivelar maps (corresponding to the 2-uniform tilings) on torus 2-uniform?


## 6.1. Infinite Series of DSEMs on Torus

If we study group structures associated to the maps, we see that DSEMs, obtained here, on torus are 2-uniform. For example, in case of type36:33,421, we see that the groups G1 = v1,v3,v5,v4,v2,v6u1,u3,u2, G2 = v1,v2,v5v3,v6,v4u1,u3,u2,v1,v3v2,v4v5,v6u2,u3, G3 = v1,v2,v7,v5v3,v8,v6,v4u1,u3,u4,u2,v1,v4v2,v6v3,v5v7,v8u1,u2u3,u4, G4 = v1,v5,v7,v2v3,v4,v6,v8u1,u2,u4,u3,v1,v8v2,v6v3,v5v4,v7u1,u4u2,u3, G5 = v1,v5,v7,v2v3,v4,v6,v8u1,u2,u4,u3,v1,v4v2,v3v5,v6v7,v8, G6 = v1,v2,v7,v10,v5v3,v8,v9,v6,v4u1,u3,u5,u4,u2,v1,v9v2,v8v3,v7v4,v10v5,v6u1,u5u2,u4, G7 = v1,v7,v5,v2,v10v3,v9,v4,v8,v6u1,u5,u2,u3,u4,v1,v3v2,v4v5,v8v6,v7v9,v10u2,u3u4,u5, and G8 = v1,v10,v2,v5,v7v3,v6,v8,v4,v9u1,u4,u3,u2,u5,v1,v4v2,v3v5,v6v7,v8v9,v10 act on the maps T13,636:33,421, T23,636:33,421, T34,836:33,421, T44,836:33,421, T54,836:33,421, T65,1036:33,421, T75,1036:33,421, and T85,1036:33,421, respectively, such that under the action, the maps have two orbits of vertices. Similarly, we can easily find a group for the DSEMs of types 36:33,422,33,42:32,4,3,41,33,42:441, and 33,42:442 on torus, under which the maps have two orbits of vertices. However, this fact does not hold for the DSEMs on Klein bottle. For example, if we let K13,636:33,421, we get no automorphism which sends v1 to v2. This can be seen as follows: suppose there is f∈AutK13,636:33,421 such that f(v1) = v2. Then, considering lk(v1) and lk(v2), we see, either f(v5) = v5 or f(v5) = v1. In the first case, when f(v5) = v5, we get f(u2) = u3, f(u1) = u1, f(v2) = v1, f(v3) = v6, f(v6) = v3, and f(v4) = v4. Now if we see lk(u1), we get a contradiction of the facts f(u1) = u1 and f(v3) = v6, as v6 ∉ lk(u1). So, f(v5) ≠ v5. Similarly, we see that f(v5) ≠ v1. Combining these, we see that f(v1) ≠ v2. This shows that K13,636:33,421 is not 2-uniform. This observation leads to ask the following question.Question 2.
Are the doubly semiequivelar maps (corresponding to the 2-uniform tilings) on torus 2-uniform?


## 7. Conclusions

In this article, the notion of doubly semiequivelar maps (DSEMs) has been introduced for the first time. A methodology has been presented to enumerate doubly semiequivelar maps on torus and Klein bottle corresponding to the 2-uniform tilings36:33,421, 36:33,422, 36:32,4,3,4, 33,42:32,4,3,41, 33,42:32,4,3,42, 33,42:441, and 33,42:442. The methodology has been demonstrated to enumerate the DSEMs on at most 15 vertices. The enumeration provides at least 35 non-isomorphic DSEMs on the surfaces of Euler characteristic zero, and out of these, 20 are on torus and remaining 15 are on Klein bottle. Further, infinite series of these types DSEMs have been constructed. We know that a study of maps becomes more significant when certain symmetry involves; in view of this, the notion of 2-uniform maps (parallel to the notion of vertex-transitive maps for equivelar or semiequivelar maps) has been introduced. During computation, it has been found that all the maps obtained on torus are 2-uniform, which does not hold in case of DSEMs on Klein bottle. This motivates us to explore the fact whether all the DSEMs on torus are 2-uniform. In the literature, vertex-transitive maps have been studied extensively. It would be interesting to study 2-uniform maps not only for torus and Klein bottle but also for other close surfaces and to explore the analog notions of vertex-transitive maps for 2-uniform maps.



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