9. Basic polyhedra

The next world is polyhedral or P-world [15]. The first problem is the derivation of basic polyhedra: 4-regular graphs without digons. This problem is solved by Kirkman [12], where the polyhedra are obtained by introducing a new triangular face in the link diagrams, in order to eliminate all digons. For this, it is necessary that the link diagram contains at most three digons, and that all of them belong to the same face. In such a face we inscribe a triangle, with the vertices belonging to the face edges (e.g. in their midpoints), and each digon must contain a vertex of the triangle. In the following table are given in Dowker notation all the diagrams of links for 3 £ n £ 9 satisfying this necessary condition, and the list of (n+3)-polyhedra derived from them:

Table 6

 n=3 3 4 6 2 6* n=5 212 6 8| 2 10 4 8* n=6 312 4 8 10 12 2 6 9* 6* 6 8| 10 12| 2 4 9* n=7 21112' 4 8 10 12 2 14 6 10*, 10** .2 6 8| 10 12 14 2 4 10**, 10*** n=8 31112 4 10 12 14 2 16 8 6 11* 21212' 4 10 12 14| 8 2 16 6 11** 21212''' 8 10 14| 2 16 4 6 12 11* .3 6 8| 12 14 16|10 2 4 11*, 11** .21 6 8| 10 14 12 16 2 4 11*, 11** .2.20 6 8 14 12 4 16 2 10 11*, 11*** 8* 6 8 10 12 14 16 2 4 11*, 11** 3#212 11** n=9 31212 4 12 10 16 14 2 18 6 8 12D 21312' 4 10 12 14 18 2 16 6 8 12D 2111112 4 10 12 14 2 18 16 8 6 12A, 12B, 12F 2111112''' 4 12 10 16 18 2 8 6 14 12B, 12F 21,21,21 8 12 16| 2 18 4 10 6 14 12G .4 6 8| 12 14 16 18 2 4 10 12E .31 6 8| 10 14 16 18 2 4 12 12J, 12L .22 6 8| 16 14 12 18 2 4 10 12E .211 6 8| 12 14 18| 16 2 4 10 12B, 12H, 12I, 12J, 12K .3.20 8 10 12| 14 2 16 18 6 4 12D .21.2' 4 8 14 12 2 16 18 10 6 12B, 12F, 12H 2:2:2 8 12 16| 2 14 4 18 6 10 12C .(2,2) 10 12| 14 18| 6 16 8 2 4 12I 8*2 8 10 12| 6 14 16 18 2 4 12B, 12F, 12G, 12H, 12I 8*20 6 8 10 16 14 18 4 2 12 12F, 12I, 12K 9* 6 16 14 12 4 2 18 10 8 12D, 12H, 12L 212#1#3 12E 6*#3 12J

The same results are obtaned by Kirkman [12] for n £ 8. For n = 9 two of the links from which 12-polyhedra could be derived are omitted by Kirkman, but even this uncomplete list is sufficient for the derivation of all the polyhedra for n = 12. Namely, the missing polyhedron 12E is the only polyhedron that could be derived from the projection of link .22, denoted by Kirkman as 9Bn, so the uncomplete result of Kirkman (11 from 12 polyhedrons obtained for n = 12) could be just an omittion in the process of derivation. The other more probable reason for this omittion is that the polyhedron 12E is the only 2-connected graph, and all the others are 3-connected, so their list obtained Kirkman coincides with the enumeration [38]. The complete list of the 12-polyhedrons is obtained by A. Caudron [15] by composing hyperbolic tangles, so we use this list and the notation.

If we generalize Kirkman method, introducing not only new triangular, but also p-gonal faces (p > 3) in the link diagrams in order to eliminate digons, we could derive this way all polyhedrons with 6 £ n £ 12 vertices from the links belongong to the L-world or from their direct products. In this case we may describe the derivation by the corresponding partitions: 6* = 3+3, 8* = 4+4, 9* = 6*+3 = 3+3+3, 10* = 5+5, 10** = 4+3+3, 10*** = 4+3+3, 11* = 8*+3 = 4+3+3, 11* = 8*+3 = 4+3+3, 11** = 8*+3 = 4+3+3, 11*** = 3#3+5, 12A = 6+6, 12B = 3#3+3+3, 12C = 8*+4 = 4+4+4, 12D = 9*+3 = 3+3+3+3, 12E = 6+3+3, 12F = 5+3+4, 12G = 5+3+4, 12H = 9*+3 = 3+3+3+3, 12I=5+3+4, 12J= 3#3+3+3, 12K=3#3+3+3, 12L = 9*+3 = 3+3+3+3. For the larger values of n, the completeness of such derivation is an opened question.