Pentagonal Subdivision

We develop the theory of simple pentagonal subdivision of quadrilateral tilings, on orientable as well as non-orientable surfaces. Then we apply the theory to answer questions related to pentagonal tilings of surfaces, especially those related to pentagonal or double pentagonal subdivisions.


Introduction
Pentagonal subdivision appeared in [2], as a way of producing tilings of oriented surface by pentagons. When the process is applied to platonic solids, we get three families of edge-to-edge tilings of sphere by congruent pentagons. We proved in [2] that these are the only tilings of sphere by congruent pentagons with edge length combination a 2 b 2 c. When some among a, b, c become equal, the reduced form of the tilings also appeared in [1] and [3]. A natural combinatorial question is the criterion for a pentagonal tiling to be a pentagonal subdivision.
Pentagonal subdivision is also related to the problem of distribution of vertices of degree > 3 (we call high degree vertices, which reflects negative curvature) in a pentagonal tiling of sphere. The problem is motivated by the fact that most vertices in a pentagonal tiling of sphere have degree 3. In [4], we studied the extreme case of few high degree vertices, which means high concentration of high degree. Specifically, we proved that the tiling cannot have only one high degree vertex. Moreover, if the number is two, then the tiling is the explicit earth map tiling. At the other extreme, we are interested in the case that each tile has exactly one high degree vertex, which means evenly distributed high degree. Any pentagonal subdivision of a triangular tiling in which all vertices have degree > 3 is such an example. We speculated that the converse is also true.
In [2], we also introduced double pentagonal subdivision, and proved that tilings of sphere by congruent pentagons with edge length combination a 3 bc are two pentagonal subdivisions of platonic solids. A natural related question is the criterion for a tiling to admit double pentagonal subdivision.
We answer these questionsin this paper, specifically through Theorems 10, 11, and 13. In our pursue of answers, we discover that underlying these questions is the simple pentagonal subdivision that turns some quadrilateral tilings into pentagonal tilings. In Section 2, we develop this theory over any surface. We find criteria for a quadrilateral tiling to admit simple pentagonal subdivision. For example, Proposition 4 says that on orientable surfaces, the condition is exactly the bipartite property. We further study degenerate quadrilaterals appearing in tilings admitting simple pentagonal subdivisions. Then we get very explicit criteria for some surfaces. For example, Theorem 5 says that a quadrilateral tiling of sphere admits simple pentagonal subdivision if and only if a tile is either non-degenerate, or is the specific degenerate tile Q 13 .
In Section 3, we apply the theory in Section 2 to pentagonal tilings. In addition to answering our original questions, we also answer questions related to simple pentagonal subdivision. Theorems 8 and 9 give criterion for a pentagonal tiling to come from simple pentagonal subdivision. As part of the discussion on double pentagonal subdivision, we also introduce quadrilateral subdivision in Section 3.3. Then in Theorem 12, we give criterion for a quadrilateral tiling to come from quadrilateral subdivision.
Finally, we fix the exact context of this paper. The surfaces are compact and connected. By tiling, we mean a graph embedded in the surface, so that the surface is divided into open regions (i.e., tiles) homeomorphic to the open 2-disk. This means that tilings are edge-to-edge, the degree of each vertex is at least 3, and each tile has at least 3 edges.

Subdivisible Quadrilateral Tiling
Given a quadrilateral tiling, we would like to divide each tile into half, by connecting the middle points of an opposite pair of edges, such that the middle point of each edge is used exactly once. Then each quadrilateral tile is divided into two pentagonal tiles, and we get a pentagonal tiling. See Figure 1, where the subdivision of quadrilaterals are indicated by dotted lines. We call the pentagonal tiling the simple pentagonal subdivision of the quadrilateral tiling. Not all quadrilateral tilings can admit simple pentagonal subdivision. Definition. A quadrilateral tiling of a closed and connected surface is (pentagonally) subdivisible if it admits a simple pentagonal subdivision.
We allow tiles to be degenerate. Here is the definition.

Definition.
A polygon is non-degenerate if its boundary is a simple closed curve. Otherwise the polygon is degenerate.
Non-degenerate means no identification of vertices, and degenerate means that some vertices or even edges are identified. The identification of edges must be in pairs in order to get a surface. Figure 2 shows that the identified edge pair can be opposing or twisted, with the only exception that adjacent opposing edges cannot be identified because it produces a vertex of degree 1.   Figure 3 gives a subdivisible tiling T 4 of 4-gon, a subdivisible tiling T 2 of 2gon, and a subdivisible tiling of torus T 2 . By regarding the outside of T 4 as one quadrilateral, we see that the cube tiling of sphere S 2 is subdivisible. If we identify antipidal points on the boundary of 2-gon, then we get a subdivisible tiling of projective space P 2 , with two degenerate tiles. Figure 3: Subdivisible tilings of 4-gon, 2-gon, and torus. Figure 4 shows three operations for simple pentagonal subdivision. The first is dual subdivision. This means that, for each tile, we switch the connection of middle points of one opposite pair of edges to the other opposite pair of edges. The second is refinement, given by 3 × 3 grid. We may also use 5 × 5 grid, and so on. The refinement can be used to turn degenerate tiles into non-degenerate tiles. The third is connected sum of simple pentagonal subdivisions of two surfaces along non-degenerate tiles. Specifically, we delete one non-degenerate tile from each surface, and then glue along the boundaries of the deleted tiles in twisted way. Proposition 1. Any closed surface has a subdivisible tiling in which all tiles are non-degenerate.

Criterion for Subdivisibility
Proof. Figure 3 gives subdivisible tilings of S 2 , P 2 , T 2 . The tiles for S 2 , T 2 are non-degenerate. We may apply refinement to get a subdivisible tiling of P 2 consisting of non-degenerate tiles only. Then we may use connected sum to construct subdivisible tilings of the other surfaces with only non-degenerate tiles.
A simple pentagonal subdivision is equivalent to assigning an orthogonal direction at the middle of each edge, such that two adjacent edges at a vertex have compatible assignments. See the first of Figure 5. We call the assignment subdivision of edge. All the compatible subdivisions of all edges at a vertex give an orientation of the vertex. The dual subdivision simply  The second of Figure 5 shows that two vertices at the end of any edge should have opposite orientations. This characterises subdivisible tilings. Proof. We only need to argue that an orientation assignment satisfying the proposition induces simple pentagonal subdivision. First, the opposite orientations at the two ends of an edge determine the subdivision of the edge. Then in a quadrilateral in the first of Figure 6, the opposite orientation property implies that subdivisions of the four edges induce a subdivision of the quadrilateral. We note that the process only makes use of the orientation inside the tile, and therefore includes the possibility that the quadrilateral tile is degenerate, such as the second of Figure 6, in which a pair of vertices of a quadrilateral tile are identified. For a quadrilateral tiling of a surface S, we may start with an orientation o v at a vertex v. We connect to any other vertex w by a path v = v 0 − v 1 − · · · − v k = w (each "− ′′ is one edge), and move o v to an orientation o w along the path. Then we assign o w to w for even k, and assign −o w for odd k. The problem is what happens when w = v, or the path is a cycle. This requires the assignment to be compatible with the orientation homomorphism π 1 S → {±1}. If S = kT 2 is an orientable surface, then the fundamental group π 1 S is generated by fundamental cycles a 1 , b 1 , . . . , a k , b k , subject to the relation a 1 b 1 a −1 If S = kP 2 is a nonorientable surface, then π 1 S is generated by fundamental cycles a 1 , . . . , a k , subject to the relation a 2 1 · · · a 2 k = 1. In case S is the sphere, the fundamental group is trivial, and we do not need to be concerned with fundamental cycles. Proof. For the orientation assignment to be compatible with the orientation homomorphism π 1 S → {±1}, we must have the compatibility for generators. For orientable surface, the orientation homomorphism takes all fundamental cycles a i , b i to 1. This means even number of edges in the first part. For non-orientable surface, the orientation homomorphism takes all fundamental cycles a i to −1. This means odd number of edges in the second part.
Conversely, suppose we have the compatibility for fundamental cycles. Then we have the compatibility for combinations of fundamental cycles. Now any cycle is homotopic to a combination of fundamental cycles. The homotopy can be reduced to the replacements of some consecutive edges of a tile by the remaining edges of the same tile. For a quadrilateral tile, this means i edges replaced by 4 − i edges. Such replacement preserves the parity of the number of edges in a cycle. Therefore we have the compatibility for all cycles.
The interpretation of subdivisibility in terms of the parity of number of edges in cycles implies the following.

Degeneracy of Tile
The criteria in Section 2.1 are qualitative. We may get more specific criteria by studying how degenerate a tile can be in a subdivisible quadrilateral tiling. Figure 7 shows all quadrilateral tiles. Some quadrilaterals are not labelled because we will show that they cannot be tiles in subdivisible tilings. The first row consists of the non-degenerate quadrilateral Q, and degenerate quadrilateral tiles obtained by identifying vertices. We label the four vertices of Q by 1, 2, 3, 4, and the vertices are not identified. We use the same labels for the vertices of all quadrilaterals. For example, Q 12 means that the vertices 1, 2 are identified, and Q 12,34 means that 1, 2 are identified and 3, 4 are also identified.
We need to pay attention to how the tiles embed into surfaces. This means taking the union of a quadrilateral tile with disk neighborhood of the identified vertex. Different configuration of the disk neighborhood may give different embedding. This happens when 1, 2, 3 are identified, and there are two possible embeddings Q 123 , Q 132 . Similar phenomenon happens when 1, 2, 3, 4 are identified.
The second row consists of degenerate quadrilateral tiles obtained by identifying one pair of edges. The identification of edges also induces identification of vertices. The first is a cylinder. In the first of Figure 8, we indicate the opposite orientations at vertices 1 and 2 implied by Proposition 2. In Figure 7: Quadrilateral tiles.
the second picture, we glue the two edges to get a cylinder, and find that the two orientations are still opposite at the identified vertex. Therefore the quadrilateral cannot be a tile in subdivisible tiling, and should be dismissed. The second of the second row is a Möbius band, and can be dismissed for the same reason. The third quadrilateral is dismissed because adjacent opposing pair of edges are identified. Such identification is forbidden by Figure 2.  Figure 9 describes the fourth quadrilateral R in the second row. In the first picture, we divide R (by dashed line) into two triangles. In the second picture, we glue the two triangles in different order (single arrow edges are glued first). The result is the Möbius band in the third picture. The fourth picture is the boundary of tile, which is a 2-gon. We also indicate the subdivision on boundary.
The fifth quadrilateral in the second row is further degeneration of R by identifying the vertex 4 with the vertex 1, 2, 3 (already identified in R). Again there are two ways R 1 , R 2 for such tile to embed into surfaces.
In the third row of Figure 7, we have identification of two pairs of edges. This already produces closed surfaces T 2 , 2P 2 , P 2 , S 2 , P 2 , 2P 2 . Therefore we get specific surfaces tiled by only one quadrilateral. The first, second and third can be dismissed for the same reason as the first and second of the second row. The fourth and fifth can be dismissed because adjacent opposing pair of edges are identified. Only the last K (for Klein bottle) remains, and Figure 10 explains that this tiling of Klein bottle is subdivisible. Figure 10: Degenerate quadrilateral K.
In the first of Figure 10, we use dotted lines to divide the quadrilateral into two parts. We denote the halves of edges by a 0 , a 1 , b 0 , b 1 . Then the upper half of the quadrilateral is the pentagon in the second of Figure 10, and the lower half of the quadrilateral is the pentagon in the fourth of Figure  10. Both pentagons are obtained by identifying a pair of edges in twisted way, and both are Möbius bands. The boundaries of the pentagons consist of three edges a 1 , b 0 , x, as in the third of Figure 10. Glueing the two pentagons along the common boundary gives the Klein bottle 2P 2 . There are three vertices. The vertex between a 1 , b 0 has degree 4. The vertices between a 1 , x and between b 0 , x have degree 3.
Only labeled quadrilaterals in Figure 7 are suitable for subdivisible tilings. The total number is 13.

Orientable Degenerate Tile
By Proposition 2, only Q, Q 13 , Q 13,24 can be tiles in subdivisible tilings of orientable surfaces. The first of Figure 11 is Q 13 , and the second is a disk neighborhood of the identified vertex 1, 3. The unionQ 13 of the two means that corners labeled by 1 and 3 in the quadrilateral are the two fans in the disk neighborhood. To help understanding the picture, we imagine walking around the quadrilateral in the direction 1 → 2 → 3 → 4 → 1. We go into the vertex 1 at i 1 (i for into) and come out of 1 at o 1 (o for out). We use similar notations around the other identified vertices. We also note that, to satisfy Proposition 2, the direction i 1 → o 1 around the circle in the second picture must be the same as the direction i 3 → o 3 around the circle.
The union of the first and second of Figure 11 can be more easily interpreted as glueing the third and fourth pictures along thick lines, so that the labels i * and o * match. This is a cylinder, or the sphere with two disks removed. The boundary of the cylinder consists of two circles, illustrated by the fifth picture. At the top of two circles, we also indicate the corners to fit into the second picture. Each boundary circle is the boundary of the subdivisible tiling T 2 in Figure 3. Therefore the union Q 13 ∪ T 2 ∪ T 2 is a subdivisible tiling of sphere by one Q 13 and twelve non-degnerate tiles Q. By connected sum construction, we find that Q 13 can appear in subdivisible tilings of any surface.
The boundary ofQ 13,24 can be filled by a single copy of Q 13,24 . Therefore we get a subdivisible tiling Q 13,24 ∪ Q 13,24 of torus, consisting of exactly two Q 13,24 and no other tiles (and exactly two vertices, both of degree 4). On the other hand, we can also use T 4 in Figure 3 to get a subdivisible tiling Q 13,24 ∪ T 4 of torus. Then we may use connected sum to show that Q 13,24 can appear in subdivisible tilings of closed surfaces of the form T 2 #S, where S can be any closed surface.
Closed surfaces that are not of the form T 2 #S are S 2 , P 2 , 2P 2 . If Q 13,24 appears in a subdivisible tiling of any such surface, then the surface must containQ 13,24 = T 2 − D 2 . Since this is not the case for the three surfaces, we conclude that closed surfaces in which Q 13,24 can appear in subdivisible tilings are kT 2 (k ≥ 1) and kP 2 (k ≥ 3).
We may specialise Proposition 3 to the three orientable tiles and get the following.
Theorem 5. A quadrilateral tiling of sphere is subdivisible if and only if its tiles are Q or Q 13 .
The fundamental group of sphere is trivial, and there is no need to verify the fundamental cycle condition.
Theorem 6. A quadrilateral tiling of torus is subdivisible if and only if its tiles are Q, Q 13 , Q 13,24 , and one of the following holds: 1. All tiles are Q or Q 13 , and the two fundamental cycles are represented by cycles with even number of edges.
If Q 13,24 is a tile, then the two fundamental cycles are represented by the cycles 1 − 2 − 3 and 2 − 3 − 4. Both have even number of edges.

Non-orientable Degenerate Tile
If a subdivisible tiling contains tiles other than Q, Q 13 , Q 13,24 , then the surface is not orientable. The detailed analysis about which tile can appear on what surface is more complicated, but still follows the method for the orientable case.
The first of Figure 13 is the degenerate quadrilateral Q 12 , and the second is a disk neighborhood of the identified vertex 1, 2. Note that by Proposition 2, along the boundary of the disk neighborhood, i 1 → o 1 and i 2 → o 2 are in opposite directions. The unionQ 12 of the two is the glueing of the third and fourth pictures along thick lines. The boundary ofQ 12 is one circle given by the fifth picture. ThereforeQ 12 = S−D 2 for a non-orientable closed surface S, with χ(S) = χ(Q 12 ) + χ(D 2 ) − χ(S 1 ) = (2 − 2) + 1 − 0 = 1. This implies S = P 2 , andQ 12 = P 2 − D 2 is a Möbius band. We may determine the other degenerate tiles in similar way. When three vertices 1, 2, 3 are identified, the neighborhood of the identified vertex has two possible configurations 123 and 132. See the second and third of Figure  14. If we glue the first and second pictures, then we get two boundary circles with χ(S) = χ(Q 123 ) + 2 = (2 − 3) + 2 = 1. ThereforeQ 123 = P 2 − 2D 2 is the punctured Möbius band. If we glue the first and third pictures, then we get one boundary circle When all four vertices 1, 2, 3, 4 are identified, the neighborhood configuration is determined by the shape of the cycle connecting four vertices. The cycle 1234 has the shape , and 1243 has the shape ⋊ ⋉. This gives two degenerate tiles Q 1234 and Q 1243 . For further degeneracy of R, the identification of 4 and 1, 2, 3 can be normal or twisted, which gives R 1 and R 2 . Figure 14: Tile Boundary Union with Nbhd Min Surface Table 1: Degenerate tiles and the surfaces they embed into. Table 1 gives the complete list of degenerate tiles and their boundaries. The notations for boundaries are given in Figure 15. The vertices decorated with • or • are the identified vertices of the original quadrilateral, and already have degree ≥ 3. The undecorated vertices are the original unidentified vertices, and have degree 2. We omit the subdivisions on the boundary edges.
The fourth column in Table 1 is the minimal surfaces on which the tiles may appear in some subdivisible tiling. Since the boundaries 2 * and 4 * can be filled in by subdivisible tilings T 2 and T 4 in Figure 3, for tiles that are not R 1 and K, we find the surface S min in the fourth column of Table 1, such that the tile can appear in a subdivisible tiling of S min . By connected sum construction, the surface can also appear in a subdivisible tiling of S min #S for any other closed surface S. For example, Q 1234 can be a tile in subdivisible tilings of all non-orientable surfaces except P 2 . We already know K must be the single tile of a subdivisible tiling of 2P 2 . Therefore K cannot be a tile on any other surface.
It remains to consider R 1 . We can certainly glue two R 1 along their common boundary to find that 2P 2 has a subdivisible tiling consisting of exactly two R 1 . In order to do the connected sum operation, so that R 1 appears in subdivisible tilings of 2P 2 #S, we may insert a copy of T 2 between two 1 1 . See Figure 16. On the other hand, if R 1 does not appear in a subdivisible tiling on P 2 , then each component of the boundary of R 1 must be filled by a disk. In other words, there is a subdivisible tiling of the disk with 1 1 as the boundary. On the other hand, we note that the boundary of any tile in Table 1 consists of either 2 or 4 edges. Therefore the boundary of any tiling with these tiles must consist of even number of edges. This implies that a single 1 1 cannot be a boundary. We conclude that the surfaces on which R 1 can appear are kP 2 for k ≥ 2.
By Table 1, only Q, Q 12 , Q 13 , Q 123 , R can appear in a subdivisible tiling of P 2 . We may specialise Proposition 3 to the projective space, which has only one fundamental cycle.
Theorem 7. A quadrilateral tiling of projective space is subdivisible if and only if its tiles are Q, Q 13 , Q 12 , Q 123 , R, and one of the following holds: 1. All tiles are Q or Q 13 , and the fundamental cycles is represented by a cycle with odd number of edges.
2. One of Q 12 , Q 123 , R is a tile.
If one of Q 12 , Q 123 , R is a tile, then the loop at the identified vertex represents the fundamental cycle and is a cycle with only one edge.

Pentagonal Subdivision
By subdividing labeled quadrilaterals in Figure 7 into halves in either way, we find three types of pentagonal tiles in simple pentagonal subdivisions. See Figure 17.
The pentagon P 1 is non-degenerate, and P 2 , P 3 are degenerate. The big • (unlike the small • used in Section 2) indicates vertices in the original quadrilateral tiling. If we include the neighborhood disk at the •-vertex, then P 2 is a Möbius band. Of course P 3 is already a Möbius band. The number "3" indicates new vertices at the middle of edges of quadrilaterals. These vertices must have degree 3. The following lists how the pentagons are obtained by subdividing a quadrilateral.
3. P 3 : R, R 1 , R 2 , K. We answer the question when a pentagonal tiling is the result of simple pentagonal subdivision. Then we apply the conclusion to the pentagonal subdivision and double pentagonal subdivision introduced in [2].

Criterion for Simple Pentagonal Subdivision
The only pentagon in a simple pentagonal subdivision on an orientable surface is P 1 . The converse is also true. 2. Each tile has exactly two •-vertices.

The unlabelled vertices have degree 3.
Proof. The three conditions imply that every tile is P 1 . There is a unique edge in P 1 such that both ends are unlabelled. We indicate the edge by dotted line. We start with a tile labeled 1 , in Figure 18. The dotted edge is shared with another tile 2 . By the conditions of theorem, the •-vertices of 2 must be as indicated in Figure 18. The union of 1 and 2 is a quadrilateral.
The two ends of the dotted edge are degree 3 vertices shared by 1 and 2 . Outside one such vertex is a tile 3 . The •-vertices of 3 are already given as vertices of 1 and 2 . Using 3 in place of 1 , we get another tile 4 similar to 2 , and the union of 3 and 4 is another quadrilateral.
For each dotted edge, we get a quadrilateral by taking the union of two tiles on two sides of the edge. These quadrilaterals may be degenerate, but would not have adjacent opposing edges identified as in Figure 2. Therefore the quadrilaterals form a tiling. Moreover, Figure 18 shows that the pentagonal tiling is the simple pentagonal subdivision of the quadrilateral tiling. We note that the proof (of sufficiency) actually does not require the surface to be orientable. Therefore the sufficiency part of Theorem 8 is also valid on non-orientable surfaces. The following is the more general result taking into account of non-orientable tiles.
Theorem 9. A pentagonal tiling of a closed surface is the simple pentagonal subdivision of a quadrilateral tiling if and only if it is possible to label some vertices by •, such that the unlabelled vertices have degree 3, and the tiles are P 1 , P 2 , P 3 in Figure 17.
In the theorem, we require P 2 to embed into surface as a Möbius band.
Proof. In each of P 1 , P 2 , P 3 , there is only one edge with both ends unlabelled. Like the proof of Theorem 8, we indicate the edge by dotted line. Every pentagon has one dotted line, and every dotted line is shared by two pentagons. By taking the union of two pentagons on two sides of dotted line, we get a quadrilateral. For example, Figure 19 shows that P 1 ∪ P 2 = Q 12 and P 2 ∪ P 2 = Q 12,34 (note smaller • and • used for identified vertices in Section 2). By further identifying vertices (for example, one •-vertex of P 1 identified with the •-vertex of P 2 ), P 1 ∪ P 2 = Q 12 may be further reduced to Q 123 or Q 132 . Similar reductions happen to other unions. Then the same argument as the proof of Theorem 8 shows that the quadrilaterals form a tiling of the surface (in particular, the forbidden identification in Figure 2 does not happen), and the pentagonal tiling is the simple pentagonal subdivision of the quadrilateral tiling. P 2 P 1 Q 12 P 2 P 2 Q 12,34 Figure 19: Union of two pentagons is one quadrilateral.

Criterion for Pentagonal Subdivision
Pentagonal subdivision is introduced for any tiling of oriented surface in [2]. Specifically, let T be a tiling of an oriented closed surface, with possibly degenerate tiles. We divide each edge e in T into three segments. If e is one edge of a tile t, we may use the orientation to label the two dividing points on e as the first and the second, from the viewpoint of t. Note that e is shared by t and another tile t ′ . The labelling of e from the viewpoint of t ′ is different from the viewpoint of t. Then we fix a center point of t, and connect to the first dividing points of edges of t. We get the pentagonal subdivision T (5) by doing this for all the pairs (t, e) in the tiling. See Figure 20. We remark that a tiling T and its dual T * (using opposite of the orientation for T ) induce the same pentagonal subdivision, with •and •-vertices exchanged.
Proof. If we do not distinguish • and •, then we may apply Theorem 8 to find that the pentagonal tiling is a simple pentagonal subdivision of a quadrilateral tiling. Then we consider the quadrilateral tiles around a •-vertex, as in Figure 21. By the assumption on pentagonal tiles, we have simple pentagonal subdivision of the neighborhood compatible with the orientation at the •-vertex. Then we combine the pentagons around the •-vertex to form a polygon (enclosed by the dashed lines). The vertices of the polygon are •-vertices, and the edges of the polygon are • − 1 · · · 2 − •. We have one polygon for each •-vertex, and the polygons form a tiling of the surface. The pentagonal tiling is the pentagonal subdivision of this tiling. If we start with a tiling in which all vertices have degree 3, then each tile in the pentagonal subdivision has one •-vertex, and the other four vertices (including •-vertex) have degree 3. Dually, if we start with a triangular tiling, then each tile in the pentagonal subdivision has one •-vertex, and the other four vertices have degree 3. The converse is also true.
Theorem 11. Suppose it is possible to label some vertices in a tiling of closed surface with non-degenerate pentagons by •, such that each tile has exactly one •-vertex, and other four vertices have degree 3, then the surface is orientable, and the tiling is obtained by pentagonal subdivision.
We already know three vertices of 4 shared with 1 , 2 that are not •vertices. By the assumption, among the two remaining vertices of 4 , one is • and the other has degree 3. Without loss of generality, we assume the •-vertex as indicated. Then there is unique way to assign the •-vertex shared by 1 , 4 , such that • and • are not adjacent in 1 and 4 . We note that 1 , 4 have the same orientation following • − 3 − • − 3 − 3 direction. The •-vertex of 4 is also the unique •-vertex of 6 . Therefore the other four vertices of 6 have degree 3. Then we already know four non •-vertices for 5 . This determines the •-vertex of 5 as indicated. We can assign the •-vertex shared by 2 , 5 , 6 the same as before. Then 1 , 2 , 4 , 5 , 6 have the same orientation following We started from the central (first) •-vertex, made assumption on the second •-vertex, and derived the third •-vertex. Since the position of the third •-vertex with respect to the first •-vertex is the same as the position of the second •-vertex with respect to the first •-vertex, and is also the same as the position of the first •-vertex with respect to the second •-vertex, the argument can continue in two directions and eventually cover the whole tiling. In particular, we introduce compatible orientations of tiles following • − 3 − • − 3 − 3 direction. This makes the surface orientable. Moreover, we also assign all the •-vertices. Then we may apply Theorem 10 to show that the pentagonal tiling is a pentagonal subdivision.

Criterion for Double Pentagonal Subdivision
Double pentagonal subdivision is also introduced for any tiling of oriented surface in [2]. This is actually a combination of two subdivisions. First we apply quadrilateral subdivision given by Figure 23. Then we apply simple pentagonal subdivision to the quadrilateral subdivision.
The quadrilateral subdivision T (4) takes any tiling T , select a middle point of each edge, select a center point of each tile, and then connect the center point of any tile to the middle point of each boundary edge of the tile. It is the union of T with its dual T * . Unlike pentagonal subdivision, we do not need orientation to construct quadrilateral subdivision. There are two types of tiles in a quadrilateral subdivision. The first is the quadrilateral Q given by the first of Figure 24. It is non-degenerate, with a •-vertex (from the original tiling T ), a •-vertex (from the center of a tile in T ), and two vertices of degree 4, such that the •-vertex and •-vertex are not adjacent. The second quadrilateral Q ′ is degenerate, obtained by identifying the two degree 4 vertices in the first of Figure 24 in twisted way, and therefore gives a Möbius band. This happens when two adjacent twisted edges of a tile in T are identified as in the second of Figure 24 (also see Figure  2). We denote the quadrilateral by Q ′ . This forces two middle points labeled by small • to be identified, and the identified vertex has degree 4. Together with the two nearby quadrilaterals 1 and 2 , the three quadrilaterals form the Möbius band in the third of Figure 24. We note that more identifications along the boundary of the Möbius band may happen when vertices or edges of 1 and 2 are identified. Proof. Figure 25 shows three consecutive tiles 1 , 2 , 3 around a •-vertex. We have the •-vertices of the three tiles, and indicate the non-labeled degree 4 vertices of 2 by small •. We also indicate four quadrilateral tiles around each small • vertex of degree 4.
The tile 2 is Q or Q ′ . If 2 is the non-degenerate Q, then there is no identification among four edges x, y, u, v (although •-vertices of 1 , 2 , 3 may be identified). If 2 is the degenerate Q ′ , we need to argue that we are in the situation described by the second and third of Figure 24. Since the identification of the two small • is twisted, and the identified vertex has degree 4, we must have 2 identified with 2' , such that x and u are identified. This also implies that y is identified with v, and 3 identified with 3' . Then the edge u + v is identified with x + y, and we get the situation described by the second and third of Figure 24.
The union of the quadrilateral tiles around the initial •-vertex is a polygon. The polygon may be degenerate, but the discussion above shows that it cannot have the forbidden degeneracy in Figure 2. Moreover, by comparing with the nearby •-vertex shared by 2' and 3' , we find the edge of the polygon around initial •-vertex is locally shared with a polygon around another •-vertex (which might be the initial polygon by edge identification). We find that all the polygons form a tiling, and the quadrilateral tiling is the quadrilateral subdivision of this tiling. Proof. Given a tiling T of a surface S, the fundamental cycles of S can be represented by cycles C of T . Then the fundamental cycles of S are represented by the same cycles C in the quadrilateral subdivision T (4), except that every edge has been subdivided into two edges. In particular, the cycles of T (4) representing fundamental cycles have even numbers of edges. By Proposition 3, the quadrilateral tiling T (4) is subdivisible if and only if the surface S is orientable.