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Semi-equivelar maps of Euler characteristics -2 with few vertices

Profile image of Debashis BhowmikDebashis Bhowmik

2019, arXiv: Geometric Topology

Abstract

We enumerate and classify all the semi equivelar maps on the surface of $ \chi=-2 $ with up to 12 vertices. We also determine which of these are vertex-transitive and which are not.

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References (81)

  1. 4,9,11,3), (4,10,9,3), (4,10,11,3), (4,11,9,3), (4,11,10,3) }, (4, 9, 10, 6) ≈ { (9,3,10,6),
  2. ≈ { (6,3,5,9), (6,3,5,10), (6,3,5,11), (4,10,5,3), (4,11,5,3) }, (4,9,10,11) ≈ { (9,3,10,11),
  3. ≈ { (9,6,10,3), (9,6,11,3), (10,6,9,3), (10,6,11,3), (11,6,9,3), (11,6,10,3), (6,9,11,4), (6,10,9,4), (6,10, 11,4), (6,11,9,4), (6,11,10,4)}, (6,9,10,5) ≈ { (9,10,6,3), (9,11,6,3), (10,9,6,
  4. 9,6,11,5), (10,6,9,5), (10,6,11,5), (11,6,9,5), (11,6,10,5) }, (9,6,10,11) ≈ { (9,6, 11,10), (10,6,9,11), (10,6,11,9), (11,6,9,10), (11,6,10,9), (9,10, 11,4), (9,11,10,4), (10,9, 11,4), (10,11, 9,4), (11,9,10,4), (11,10,9,4) }, (9,10,5,6) ≈ { (9,11,5,6), (10,9,5,6), (10,11,5,6), (11,9,5,6), (11,10,5,6)}, (9,10,5,11) ≈ { (9,11,5,10), (10,9,5,11), (10,11,5, 9), (11,9,5,10), (11,10,5,9)
  5. Now, (a, b, c, d) can not be a combination of 3,4,5,6, so for any (a, b, c, d) ∈ { (4,3,5,6), (4,3,6,5), (4,5,3,6), (4,5,6,3), (4,6,3,5), (4,6,5,3), (5,4,3,6), (5,4,6,3), (5,6,3,4), (5,6,4,3), (6,4, 5,3) } the SEM will not complete.
  6. 6 appear in 1 st 4-gon, so it can not appear in another 4-gon and also not appear in last 4-gon as 4,6 appear in 1 st 4-gon, so it can not appear in another 4-gon.
  7. When (a, b, c, d) = (6, 4, 9, 10), then face sequence will not follow in lk(6).
  8. Case 69. When (a, b, c, d) = (6, 9, 4, 10), then lk(6) = C 8 (1, 9, [5, 4, 0], [7, a 1 , 2]). Six 4-gon are [0, 4, 5, 6], [0, 1, 8, 7], [1, 9, 4, 10], [2, 6, 7, b 1 ], [2, 3, b 2 , b 3 ], [4, b 4 , b 5 , 6 ], where b 1 , b 2 , b 3 , b 4 , b 5 , b 6 ∈ V . Now b 1 = 3 as if b 1 = 3 then 6 will appear two times in lk(3) and 3 will not appear in last 4-gon as [3,4] is an edge of two 3-gon, which implies at 3 has one 4-gon which is not possible. Case 70. When (a, b, c, d) = (6, 9, 5, 3), then lk(6) = C 8 (1, 9, [5, 4, 0], [7, 10, 2]), lk(3) = C 8 (0, 4, [5, 9, 1], [8, a 1 , 2]), where a 1 ∈ V which implies 4 will appear two times in lk(5). Case 71. When (a, b, c, d) = (6, 9, 5, 10), then lk(6) = C 8 (1, 9, [5, 4, 0], [7, a) 1 , 2]), where
  9. ∈ V . Six 4-gon are [0, 4, 5, 6], [0, 1, 8, 7], [1, 9, 5, 10], [2, 6, 7, a 1 ], [2, 3, a 2 , a 3 ], [4, a 4 , a 5 , a 6 ], where a 2 , a 3 , a 4 , a 5 , a 6 ∈ V . 3 will not appear will not appear in 4 th 4-gon as 2,3 will not appear in two 4-gon an 3 will not appear in last 4-gon as [3,4] is an adjacent edge of two 3-gon. Case 72. When (a, b, c, d) = (6, 9, 10, 3), then lk(6) = C 8 (1, 9, [5, 4, 0], [7, a 1 , 2]) and lk(3) = C 8 (0, 4, [10, 9, 1], [8, a 2 , 2]), where a 1 , a 2 ∈ V . Now six 4-gon are [0, 4, 5, 6], [0, 1, 8, 7], [1, 3, 10, 9], [2, 6, 7, a 1 ], [2, 3, 8, a 2 ], [4, a 3 , a 4 , a 5 ], where a 3 , a 4 , a 5 ∈ V . Now 5 will not ap- pear in 4 th 4-gon as 5,6 will not appear in two 4-gon and 5 will also not appear in last 4-gon as 4,5 will not appear in two 4-gon, so 5 will appear in 5 th 4-gon i.e. a 2 = 5. Therefore 11 will appear in 4 th and last 4-gon, therefore a 1 = 11 and 9,10 will appear in last 4-gon, which contradict as 9,10 can not appear in two 4-gon.
  10. B = { (5,7,6,8), (5,7,6,10), (5,7,6,11). (5,8,7,6), (5,8,7,10), (5,8,7,11), (5,10,6,7), (5,10,6,8), (5,10,6,11), (5,10,7,6), (5,10,7,11), (5,10,8,6), (5,10,8,11), (5,10,11,6), (5, 10,11,7), (5,10,11,
  11. in lk(6), which is a contradiction. Therefore for (5, 8, 6, 10), (5, 8, 6, 11), SEM will not complete. If (a 1 , a 2 , a 3 , a 4 ) = (5, 8, 10, b), where b ∈ {6, 7} then lk(10) = C 8 (5, 8, [3, 2, b], [11, 1, 4]). But we have [4, 5] and [5, 8] form an edge in two 4-gon, which implies face sequence is not followed in lk(5). Therefore for (5, 8, 10, 6), (5, 8, 10, 7), SEM will not complete. For (a 1 , a 2 , a 3 , a 4 ) = (a 1 , 8, 11, a 4 ), [8, 1] and [8, 3] form an edge in two 4-gon, which im- plies face sequence will not followed in lk(8). Therefore, for (5,8,11,6), (5,8,11,7), (6,8,11,5), (6,8,11,7), (10,8,11,5), (10,8,11,7), SEM will not complete. If (a 1 , a 2 , a 3 , a 4 ) = (5, 11, 6, 7), then lk(11) = C 8 (8, 6, [3, 9, 5], [10, 4, 1]) ⇒ lk(6) = C 8 (8, 11, [3, 2, 7], [0, 4, 5]) ⇒ lk(5) = C 8 (8, 10, [11, 3, 9], [4, 0, 6]), which implies 4 will occur two times in lk(9). If (a 1 , a 2 , a 3 , a 4 ) = (5, 11, 6, 8), then lk(8) = C 8 (11, 5, [6, 3, 2], [7, 0, 1]) and we have [1, 11] and [5, 11] form an edge of two 4-gon, which implies face sequence will not followed in lk(11). If (a 1 , a 2 , a 3 , a 4 ) = (5, 11, 7, 10), then lk(11) = C 8 (8, 7, [3, 9, 5], [10, 4, 1]) and we have [1, 8] and [7, 8] form an edge of two 4-gon, which implies face sequence will not followed in lk(8). If (a 1 , a 2 , a 3 , a 4 ) = (5, 11, 10, 7), then lk(11) = C 8 (8, b, [5, 9, 3], [10, 4, 1]) ⇒ lk(10) = C 8 (5, b, [7, 2, 3], [11, 1, 4]). From lk(11) we have b ∈ {6, 7} and from lk(10) we have b ∈ {6, 8, 9}, which implies b = 6 and we have [6, 8] and [1, 8] form an edge of two 4-gon, which implies face sequence will not followed in lk(8). If (a 1 , a 2 , a 3 , a 4 ) = (5, 11, 10, 8), then lk(11) = C 8 (8, b 1 , [5, 9, 3], [10, 4, 1]). Now [5, b 1 ] form an edge of a 4-gon and b 1 can not be 9, therefore b 1 = 6. And lk(10) = C 8 (5, b 2 , [8, 2, 3], [11, 1, 4]), where [8, b 2 ] form an edge of a 4-gon and b 2 can not be 2, therefore b 2 = 7. These implies lk(8) = C 8 (11, 6, [2, 3, 10], [7, 0,
  12. ⇒ lk(2) = C 8 (0, 1, [9, 7, 6], [8, 10, 3]) ⇒ lk(
  13. ⇒ lk(9) = C 8 (1, 4, [3, 11, 5], [7, 6,
  14. ⇒ lk(5) = C 8 (10, 7, [9, 3, 11], [6, 0, 4]). This is the SEM KO 1[(3 3 ,4,3,4)] , given in figure does not form an edge, therefore b 1 = 9. b 1 = 11 as [10,11] is an edge in a 4-gon. If (a 1 , a 2 , a 3 , a 4 ) = (8, 10, 6, 11), then lk(10) = C 8 (5, 6, [3, 9, 8], [11, 1, 4]), which implies 6 will appear two times in lk(5). For (a 1 , a 2 , a 3 , a 4 ) = (8, 11, 5, a 4 ), lk(5) = C 8 (10, 11, [3, 2, a 4 ], [6, 0, 4]). We have 4,10,11 occur in same 4-gon, therefore 11 will occur two times in lk(10). Therefore (8,11,5,7), (8,11,5,10) are not possible. If (a 1 , a 2 , a 3 , a 4 ) = (8, 11, 6, 10), then lk(6) = C 8 (7, 11, [3, 2, 10], [5, 4, 0]), which implies 10 will occur two times in lk(5). (a 1 , a 2 , a 3 , a 4 ) = (8, 11, 6, 7) then lk(6) will not possible. If (a 1 , a 2 , a 3 , a 4 ) = (8, 11, 7, 5), then lk(7) = C 8 (6, 11, [3, 2, 5], [8, 1, 0]) ⇒ lk(11) = C 8 (7, 6, [10, 4, 1], [8, 9, 3]) ⇒ lk(5) = C 8 (8, 10, [4, 0, 6], [2, 3, 7]) ⇒ lk(8) = C 8 (5, 10, [9, 3, 11], [1, 0,
  15. ⇒ lk(10) = C 8 (8, 5, [4, 1, 11], [6, 2, 9]) ⇒ lk(2) = C 8 (0, 1, [9, 10, 6], [5, 7,
  16. ⇒ lk(
  17. = C 8 (1, 4, [3, 11, 8], [10, 6, 2]) ⇒ lk(6) = C 8 (11, 7, [0, 4, 5], [2, 9, 10]). This map isomor- phic to KO 1[(3 3 ,4,3,4)] , given in figure 6, under the map (0, 4, 1)(2, 3, 9)(5, 11, 7)(6, 10, 8).
  18. If (a 1 , a 2 , a 3 , a 4 ) = (8, 11, 10, a 4 ), then 8 will appear two times in lk(11). Therefore (8,11,10,5), (8,11,10,6) are not possible. If (a 1 , a 2 , a 3 , a 4 ) = (8, 10, 7, 5), then lk(10) = C 8 (5, 7, [3, 9, 8], [11, 1, 4]). We have [4,5], [5,7] are edges of 4-gon, therefore face sequence is not followed in lk(5). If (a 1 , a 2 , a 3 , a 4 ) = (10, 5, 6, 7) then C(0, 4, 5, 3, 2, 7) ∈ lk(6) which is a contradiction. If (a 1 , a 2 , a 3 , a 4 ) = (10, 5, 6, 8), then lk(6) = C 8 (7, b 1 , [8, 2, 3], [5, 4, 0]). Since [0,7] is an edge in a 4-gon, therefore [8, b 1 ] will form an edge in a 4-gon, which implies b 1 = 1, which is a contradiction as 1,7 appear in same 4-gon.
  19. If (a 1 , a 2 , a 3 , a 4 ) = (10, 5, 7, 11), then lk(7) = C 8 (6, 5, [3, 2, 11], [8, 1, 0]). which implies 5 will appear two times in lk(6). If (a 1 , a 2 , a 3 , a 4 ) = (10, 5, 8, 6), then lk(8) = C 8 (11, 5, [3, 2, 6], [7, 0, 1]), which implies 6 will appear two times in lk(7). If (a 1 , a 2 , a 3 , a 4 ) = (10, 5, 8, 11), then lk(5) = C 8 (b 1 , 8, [3, 9, 10], [4, 0, 6]). We have [3,8] is an edge of a 4-gon, therefore [6, b 1 ] will form an edge in a 4-gon, which implies face [7,10] are edges of 4-gon, therefore face sequence is not followed in lk(7). If (a 1 , a 2 , a 3 , a 4 ) = (11, 6, 10, 8), then lk(6) = C 8 (7, 10, [3, 9, 11], [5, 4, 0]), which implies lk(10) is not possible. If (a 1 , a 2 , a 3 , a 4 ) = (11, 7, 5, 8), then lk(7) = C 8 (6, 5, [3, 9, 11], [8, 1, 0]), which implies 5 will appear two times in lk(6). For (a 1 , a 2 , a 3 , a 4 ) = (11, 7, 6, a 4 ), face sequence will not followed in lk(6). Therefore (11,7,6,8), (11,7,6,10) are not possible. If (a 1 , a 2 , a 3 , a 4 ) = (11, 7, 8, 5), then lk(7) = C 8 (6, b 1 , [11, 9, 3], [8, 1, 0]). Since [0,6] is an edge in a 4-gon, therefore [11, b 1 ] will form an edge in a 4-gon, which implies b 1 ∈ {4, 10}. We see that for each b 1 , 6, b 1 will appear in same 4-gon, which is a contradiction. If (a 1 , a 2 , a 3 , a 4 ) = (11, 7, 8, 6), then lk(7) = C 8 (6, b 1 , [11, 9, 3], [8, 1, 0]). Since [0,6] is an edge in a 4-gon, therefore [11, b 1 ] will form an edge in a 4-gon, which implies b 1 = 10. Then lk(
  20. = C 8 (0, 1, [9, 5, 10], [6, 8, 3]). This map is isomorphic to KO 1[(3 3 ,4,3,4)] , given in figure 6, under the map (0, 2, 1, 9, 4, 3)(5, 10, 11)
  21. a 4 , 2, 3]). Therefore a 4 = 6 is not possible and if a 4 = 5, then from lk(6) we will see that [6,11] will form an edge in a 4-gon, which is not possible. Therefore (11,7,10,5), (11,7,10,6) are not possible. If (a 1 , a 2 , a 3 , a 4 ) = (10, 8, 6, 7), then lk(8) = C 8 (11, 6, [3, 9, 10], [7, 0, 1]) ⇒ lk(6) = C 8 (8, 11, [5, 4, 0], [7, 2,
  22. ⇒ lk(7) = C 8 (b 1 , 10, [8, 1, 0], [6, 3, 2]). Since [8,10] is an edge in a 4- gon, therefore [2, b 1 ] eill form an edge in a 4-gon, which implies b 1 ∈ {5, 9, 11}. b 1 can not be 2 and b 1 = 11 as [10,11] is an edge in a 4-gon. Therefore b 1 = 5, which im- plies lk(5) = C 8 (7, 10, [4, 0, 6], [11, 9, 2]) ⇒ lk(10) = C 8 (5, 7, [8, 3, 9], [11, 1, 4]) ⇒ lk(11) = C 8 (8, 6, [5, 2, 9], [10, 4, 1]) ⇒ lk(2) = C 8 (0, 1, [9, 11, 5], [7, 6, 3]) ⇒ lk(9) = C 8 (1, 4, [3, 8, 10], [11, 5, 2]) ⇒ lk(
  23. = C 8 (10, 5, [2, 3, 6], [0, 1, 8]). This map isomorphic to KO 2[(3 3 ,4,3,4)] , given in figure 6, under the map (0, 3, 4, 9, 1, 2)(5, 7, 6)(8, 10, 11). Non-orientable cases: We see that for any (a 1 , a 2 , a 3 , a 4 ), a 5 , a 6 will appear in same 4-gon, therefore the following cases are not possible: (5,7,6.8), (5,8,7,6), (5,10,11,6), (5,10,6,11), (6,8,7,5), (6,10,11,5), (6,11,5,10), (7,6,5,8), (7,6,8,5), (7,10,11,8), (7,11,8,10), (8,6,5,7), (8,10,11,7), (8,5,7,6), (10,6,5,11), (10,8,7,11), (10,7,11,8), (11,6,5,10), (11,8,7,10), (11,8,10,7), (11,5,10,
  24. = C 8 (10, b 3 , [8, 3, 9], [6, 0, 4]). From here we see that [8, b 3 ] will form an edge in a 4-gon, which implies b 3 = 7, which implies [6, 7] form an edge in a 4-gon, which is not possible. If b 1 = 11 then lk(5) = C 8 (10, 11, [9, 3, 8], [6, 0, 4]) which implies lk(11) is not possible. Therefore for (5, 8, 7, 10), (5, 8, 7, 11) SEM will not exist. If (a 1 , a 2 , a 3 , a 4 ) = (5, 10, 7, 6) then lk(7) = C 8 (10, b 1 , [8, 1, 0], [6, 2, 3]). Since [8, b 1 ] form an edge in a 4-gon, therefore b 1 = 9 which implies face sequence is not followed in lk(9). If (a 1 , a 2 , a 3 , a 4 ) = (5, 10, 11, b 1 ) then 5 will occur two times in lk(10). Therefore for (5, 10, 11, 7), (5, 10, 11, 8) SEM is not possible. If (a 1 , a 2 , a 3 , a 4 ) = (5, 11, 7, 6) then lk(11) = C 8 (8, 7, [3, 9, 5], [10, 4, 1]) which implies face sequence is not followed in lk(8). For (a 1 , a 2 , a 3 , a 4 ) = (5, 11, 8, b 1 ), face sequence is not followed in lk(11). Therefore for (5, 11, 8, 6), (5, 11, 8, 10), SEM will not possible. If (a 1 , a 2 , a 3 , a 4 ) = (5, 10, 6, 7) then lk(10) = C 8 (6, b 1 ), [11, 1, 4], [5, 9, 3]. From here we see that [11, b 1 ] is an adjacent edge of two 3-gon which implies b 1 = V . Therefore in this case SEM will not exist. If (a 1 , a 2 , a 3 , a 4 ) = (5, 10, 6, 8) then lk(10) = C 8 (b 1 , 6, [3, 9, 5], [4, 1, 11]). We see that [11, b 1 ] form an edge in a 4-gon, therefore b 1 = 7. Now lk(
  25. ⇒ lk(
  26. ⇒ lk(11) = C 8 (5, 8, [1, 4, 10], [7, 2,
  27. ⇒ lk(2) = C 8 (0, 1, [9, 11, 7], [8, 6,
  28. ⇒ lk(9) = C 8 (1, 4, [3, 0, 5], [11, 7, 2]). This map is same as the map KN O 2[(3 3 ,4,3,4)] , given in figure 7.
  29. If (a 1 , a 2 , a 3 , a 4 ) = (5, 10, 7, 11) then lk(10) = C 8 (7, b 1 , [11, 1, 4], [5, 9, 3]). We see that [11, b [ 1]] form an edge in a 4-gon, therefore b 1 ∈ {2, 7} which is not possible. If (a 1 , a 2 , a 3 , a 4 ) = (5, 10, 6, p) then lk(10) = C 8 (8, b 1 , [11, 1, 4], [5, 9, 3]). We see that [11, b 1 ] form an edge in a 4-gon which implies b 1 ∈ {2, 7, 9}. But b 1 = 2, 9 and for b 1 = 7 face sequence will not follow in lk(8). Therefore for (5, 10, 8, 6), (5, 10, 8, 11), SEM is not possible. If (a 1 , a 2 , a 3 , a 4 ) = (6, 8, 7, b 1 ) then lk(8) = C 8 (11, b 2 , [6, 9, 3], [7, b 1 , 2]). As we have [1, 11] is an edge in a 4-gon, therefore [6, b 2 ] will form an edge in a 4-gon, which implies b 2 ∈ {0, 5}. But b 2 = 0 and for b 2 = 5 face sequence will not followed in lk(5). Therefore (a 1 , a 2 , a 3 , a 4 ) = (6,8,7,10), (6,8,7,11). If (a 1 , a 2 , a 3 , a 4 ) = (6, 10, 11, b 1 ) then lk(10) = C 8 (b 2 , 5, [4, 1, 11], [3, 9, 6]). As [4, 5] is an edge in a 4-gon, therefore [6, b 2 ] form an edge in a 4-gon which implies b 2 ∈ {0, 5}, which is not possible. Therefore (6, 10, 11, 7) and (6, 10, 11, 8) are not possible. If (a 1 , a 2 , a 3 , a 4 ) = (6, 7, 5, 10) then lk(7) = C 8 (5, b 1 , [8, 1, 0], [6, 9, 3]). As [3, 5] is an edge in a 4-gon, therefore [8, b 1 ] will form an edge in a 4-gon, which implies b 1 = 11. Therefore lk(
  30. b 1 ] is an adjacent edge of a 3-gon and a 4-gon, therefore b 1 = 2 and then face sequence will not follow in lk(2). If (a 1 , a 2 , a 3 , a 4 ) = (11, 8, 6, 10) then lk(8) = C 8 (b 1 , 6, [3, 9, 11], [1, 0, 7]) which implies [7, b 1 ] is an adjacent edge of a 3-gon and a 4-gon, therefore b 1 ∈ {2, 5}. But for b 1 = 2, face sequence will not follow in lk(2) and for b 1 = 5 then face sequence will not follow in lk(6). Similarly (a 1 , a 2 , a 3 , a 4 ) = (11, 8, 10, 5) is not possible. If (a 1 , a 2 , a 3 , a 4 ) = (11, 8, 7, a 4 ) then face sequence will not follow in lk(8), Therefore (a 1 , a 2 , a 3 , a 4 ) = (11, 8, 7, 5), (11, 8, 7, 6) are not possible. If (a 1 , a 2 , a 3 , a 4 ) = (11, 8, 10, 6) then lk(8) = C 8 (b 1 , 10, [3, 9, 11], [1, 0, 7]) which implies [7, b 1 ] is an adjacent edge of a 3-gon and a 4-gon, therefore b 1 = 5. Now lk(10) = C 8 (8, 5, [4, 1, 11], [6, 2, 3]) and lk(11) = C 8 (6, b 2 , [9, 3, 8], [1, 4, 10]). From here we see that [9, b 2 ] is an adjacent edge of a 3-gon and a 4-gon, therefore b 2 = 7. Now lk(
  31. = C 8 (11, 7, [0, 4, 5], [2, 3, 10]) ⇒ lk(9) = C 8 (1, 4, [3, 8, 11], [7, 5, 2]) ⇒ lk(7) = C 8 (6, 11, [9, 2, 5], [8, 1,
  32. ⇒ lk(5) = C 8 (8, 10, [4, 0, 6], [2, 9, 7]) ⇒ lk(
  33. = C 8 (0, 1, [9, 7, 5], [6, 10, 3]). This map isomorphic to the map KN O 1[(3 3 ,4,3,4)] , given in figure 7, under the map (0, 2, 1, 9, 4,
  34. 5, 11)(6, 8)(7, 10). If (a 1 , a 2 , a 3 , a 4 ) = (11, 5, 10, a 4 ) then face sequence will not follow in lk(5). Therefore (a 1 , a 2 , a 3 , a 4 ) = (11, 5, 10, 8), (11, 5, 10, 7) are not possible. If (a 1 , a 2 , a 3 , a 4 ) = (11, 7, 5, 10) then after completing lk(7) we see that C(3, 2, 10, 4, 0, 6,
  35. When (a, b, c, d) = (4, 3, 6, 9) then lk(3) = C 8 (0, 4, [1, 9, 6], [a 1 , b 1 , 2]). Now if a 1 , b 1 / ∈ {10, 11} then vertices in lk(10) i.e. V (lk(10)) = {2, 4, 5, 6, 7, 8, 9, u} = V (lk(11)), where u = 11 for V (lk(10)) and u = 10 for V (lk(11)), then V (lk(
  36. = {0, 1, 3, 4, 5, 7, 9, 10, 11, a 1 }. If a 1 ∈ V \ {2, 8} then number of vertex in lk(6) i.e. |v lk(6) | = 9 and other- wise |v lk(6) | = 10, both cases are not possible. Therefore at least one of a 1 , b 1 is 10. Now if a 1 = 10 then b 1 = 10 and as 4 ∈ V (lk(
  37. V (lk(11)) i.e. 10, 11 ∈ V (lk(4)), therefore either [2, 10] is an adjacent edge of two 4-gon or 10 occur two times in lk(4)
  38. As [4,11] is not an edge in a 4-gon, therefore [4,10] will form an edge in a 4-gon, therefore lk(4) = C 8 (11, 1, [10, b 2 , 3], [0, 6, 5]), where b 2 ∈ {7, 8, } and b 1 = 10. Now since [4,11] is not an edge in a 4-gon, therefore [8,11] will form an edge in a 4-gon and hence b 2 = 7. Now we see that [5,11] and [8,11] will form an edge in two distinct 4-gon. Now lk(
  39. or lk(4) = C 8 (11, 1, [8, c 1 , 3], [0, 6, 5]). Now remaining three 4-gons are [3, 4, b 1 , b 2 ], [2, 11, b 3 , b 4 ], [3, 11, b 5 , b 6 ]. If lk(4) = C 8 (8, 1, [11, c 1 , 3], [0, 6, 5]) then one of b 1 , b 2 is 11 which is not possible. If lk(4) = C 8 (11, 1, [8, c 1 , 3], [0, 6, 5]), since [1,11] is not an edge in a 4-gon, therefore [5,11] and [10,11] are make an edge in two distinct 4-gon. Now b 3 = 10 then (2,10) will appear in two 4-gon. Therefore b 5 = 10 and b 3 = 5. Since (5,6) and (4,6) can not appear in two 4-gon, therefore b 6 = 6. (2,9) can not appear in two 4-gon, there- fore b 2 = 9 and therefore b 4 = 7. Therefore lk(4) = C 8 (11, 1, [8, 9, 3], [0, 6, 5]) ⇒ lk(
  40. = C 8 (0, 2, [11, 10, 6], [9, 8, 4]) ⇒ lk(2) = C 8 (0, 3, [11, 5, 7], [9, 10, 1]) ⇒ lk(9) = C 8 (7, 6, [3, 4, 8], [10, 1, 2]) ⇒ lk(
  41. ⇒ lk(
  42. ⇒ lk(7) = C 8 (6, 9, [2, 11, 5], [8, 1,
  43. ⇒ lk(5) = C 8 (8, 10, [6, 0, 4], [11, 2,
  44. ⇒ lk(11) = C 8 (1, 4, [5, 7, 2], [3, 6, 10]). This map is identical to the map KN O 3[(3 3 ,4,3,4)] , given in figure 7.
  45. Case 100. When (a, b, c, d) = (9, 10, 11, 6) then lk(6) = C 8 (8, 1, [11, a 1 , 7], [0, 4, 5]). Now remaining three incomplete 4-gons are [3, 4, b 1 , b 2 ], [2, 11, b 3 , b 4 ] and [3, 11, b 5 , b 6 ].
  46. Since (4,6) can not appear in two 4-gon, therefore 6 will be one of b 3 , b 4 , b 5 , b 6 and we have [6,11] is an edge, therefore [6,11] form an edge in a 4-gon and then [10,11] will not form an edge in a 4-gon, which implies one of b 1 , b 2 must be 10. Since (2,9) and (9,10) can not appear in two 4-gon, therefore one of b 5 , b 6 must be 9. Now from lk(6) we see that a 1 = 2. Therefore lk(
  47. = C 8 (10, 1, [9, a 1 , 3], [0, 6, 5]) then a 1 = 7, 8 as then (10,11) will occur in two 4-gon, therefore a 1 = 11. Now two incomplete 4-gons are [2, 10, b 1 , b 2 ] and [3, 10, b 3 , b 4 ]. Now since (3,11) will not appear in two 4-gon, therefore one of b 1 , b 2 is 11. If remaining one of b 1 , b 2 is 5 then (7,8) will occur in two 4-gon, therefore one of b 3 , b 4 is 5. From lk(4) we see that [1,10] is not an edge in a 4-gon, therefore [5,10] will form an edge in a 4-gon and then [8,10] will form an edge in a 4-gon, therefore b 1 = 8, b 2 = 11, b 3 = 5 and b 4 =
  48. ⇒ lk(11) = C 8 (6, 7, [3, 4, 9], [8, 10, 2]) ⇒ lk(7) = C 8 (6, 11, [3, 10, 5], [8, 1,
  49. ⇒ lk(8) = C 8 (5, 9, [11, 2, 10], [1, 0,
  50. ⇒ lk(9) = C 8 (5, 8, [11, 3, 4], [1, 2,
  51. ⇒ lk(5) = C 8 (9, 8, [7, 3, 10], [4, 0, 6]) ⇒ lk(6) = C 8 (7, 11, [2, 1, 9], [5, 4, 0]). This map isomorphic to the map KN O 3[(3 3 ,4,3,4)] , given in figure 7, under the map (0, 3, 2)(1, 4, 11, 7, 9, 5, 10). Subcase 101.2. If lk(
  52. = C 8 (6, a 2 , [10, 4, 3], [8, 1, 0]). Since [0,6] is an edge in a 4-gon, therefore [6, a 2 ] will not form an edge in a 4-gon which implies a 2 = 11. Now lk(5) = C 8 (9, a 3 , [a 4 , a 5 , a 6 ], [6, 0, 4])
  53. which implies lk(6) = C 8 (7, 11, [a 7 , a 8 , a 6 ], [5, 4, 0]). From lk(6) we see that a 6 ∈ {2, 9}, but if a 6 = 9 then 9 will occur two times in lk(5), therefore a 6 = 2 and then a 7 = 9, a 8 = 1. Now from lk(5) we see that [2,5] form an edge in a 4-gon, therefore b 2 = 5 and then b 1 = 10 and b 3 = 9, b 4 = 8 which implies a 4 = 10, a 5 = 11. Now since [4,9] is not an [1,10] is not an edge in a 4-gon, therefore [5,10] will form an edge in a 4-gon which implies b 2 = 5 and then b 3 = 11 and b 5 = 7. Now lk(10) = C 8 (4, 1, [9, 7, 3], [2, 11, 5]) ⇒ lk(2) = C 8 (0, 3, [10, 5, 11], [6, 9, 1]) ⇒ lk(3) = C 8 (0, 2, [10, 9, 7], [11, 8, 4]) ⇒ lk(6) = C 8 (7, 11, [2, 1, 9], [5, 4, 0]) ⇒ lk(
  54. = C 8 (6, 11, [3, 10, 9], [8, 1, 0]) ⇒ lk(11) = C 8 (6, 7, [3, 4, 8], [5, 10,
  55. ⇒ lk(
  56. = C 8 (9, 5, [11, 3, 4], [1, 0, 7]) ⇒ lk(5) = C 8 (8, 9, [6, 0, 4], [10, 2, 11]) ⇒ lk(9) = C 8 (8, 5, [6, 2, 1], [10, 3, 7]). This map isomorphic to the map KN O 3[(3 3 ,4,3,4)] , given in figure 7, under the map (0, 2)(4, 11, 6, 7, 9, 8, 10).
  57. Subcase 103.1. If lk(
  58. = C 8 (7, 11, [2, 1, 9], [5, 4, 0]) then lk(2) = C 8 (0, 3, [10, b 3 , 11], [6, 9, 1]). Now remaining three incomplete 4-gons are [3, 4, b 1 , b 2 ], [2, 10, b 3 , 11] and [3, 10, b 4 , b 5 ]. Therefore one of b 1 , b 2 is 11. Now either [5,10] will form an edge in a 4-gon or not. If [5,10] form an edge in a 4-gon then lk(5) = C 8 (1, 8, [4, 0, 6], [9, a 1 , 10]) which implies (9,10) occur in same 4-gon but not form an edge which make contradiction. If [5,10] will not form an edge in a 4-gon, then one of b 1 , b 2 is 5 and [5,8] will form an edge in a 4-gon, which is not possible. Subcase 103.2. If lk(
  59. = C 8 (7, 11, [9, 1, 2], [5, 4, 0]) then either [5,10] form an edge in a 4- gon or not. If [5,10] is not an edge in a 4-gon then lk(5) = C 8 (10, 1, [8, 11, 2], [6, 0, 4]). Now remaining two incomplete 4-gons are [3, 4, b 1 , b 2 ] and [3, 11, b 3 , b 4 ] and then 10 will occur in both 4-gon which is not possible as (3,10) can not appear in two 4-gon. Therefore [5,10] will form an edge in a 4-gon and then lk(5) = C 8 (1, 8, [4, 0, 6], [2, 11, 10]) which implies lk(
  60. = C 8 (0, 3, [11, 10, 5], [6, 9, 1]). Now remaining two incomplete 4-gons are [3, 4, b 1 , b 2 ] and [3, 11, b 3 , b 4 ].
  61. As [10,11] is an edge in a 4-gon, therefore one of b 1 , b 2 is 10. Since [9,10] is not an edge in a 4-gon, therefore one of b 3 , b 4 is 9. Since [1,8] is an edge in a 4-gon, therefore from lk(5) we get [4,8] will not form an edge in a 4-gon which im- plies one of b 3 , b 4 is 8 and then one of b 1 , b 2 is 7. Since [0,7] is an edge in a 4-gon, therefore from lk(6) we see that [9,11] will form an edge in a 4-gon i.e. b 3 = 9 and b 4 = 8. Now lk(
  62. = C 8 (0, 2, [11, 9, 8], [7, 10, 4]) ⇒ lk(4) = C 8 (8, 9, [10, 7, 3], [0, 6, 5]) ⇒ lk(10) = C 8 (9, 1, [5, 2, 11], [7, 3, 4]) ⇒ lk(7) = C 8 (6, 11, [10, 4, 3], [8, 1, 0]) ⇒ lk(9) = C 8 (10, 4, [8, 3, 11], [6, 2, 1]). This map isomorphic to the map KN O 1[(3 3 ,4,3,4)] , given in figure 7, under the map (0, 10)(1, 2, 8, 9, 3, 5)(6, 11, 7).
  63. Case 104. When (a, b, c, d) = (9, 10, 5, 6) then remaining three incomplete 4-gons are [3, 4, b 1 , b 2 ], [2, 11, b 3 , b 4 ] and [3, 11, b 5 , b 6 ]. Now lk(6) = C 8 (1, 8, [c 1 , c 2 , 7], [0, 4, 5]). If b 5 , b 6 ∈ {6, 7} then one of b 3 , b 4 must be 9 or 10 which is not possible. Therefore b 3 , b 4 ∈ {6, 7}. Now lk(2) = C 8 (0, 3, [11, b 3 , b 4 ], [9, 10, 1]), from this and lk(6) we see that b 4 = 6, therefore b 4 = 7, b 3 = 6 and then c 1 = 11, c 2 = 2. Since [1,8] is an edge in a 4-gon, therefore [8,11] will not form an edge in a 4-gon, therefore one og b 1 , b 2 is 8. Now lk(
  64. = C 8 (c 3 , 9, [2, 11, 6], [0, 1, 8]) where c 3 ∈ {3, 4}, which implies one of b 5 , b 6 is 9 and then remaining one of b 1 , b 2 is 10 and therefore remaining one of b 5 , b 6 is 5. Now c 3 = 3 as then face sequence will not followed in lk(3), therefore c 3 = 4. There- fore lk(
  65. ⇒ lk(3) = C 8 (0, 2, [11, 5, 9], [10, 8, 4]) ⇒ lk(10) = C 8 (5, 11, [8, 4, 3], [9, 2, 1]) ⇒ lk(11) = C 8 (10, 8, [6, 7, 2], [3, 9, 5]). This map is isomorphic to the map KO 2[(3 3 ,4,3,4)] , given in figure 6, under the map (0, 5, 3, 10, 2, 1, 6, 9, 8)(4, 11, 7).
  66. Case 105. When (a, b, c, d) = (9, 10, 6, 11) then remaining three incomplete 4-gons are [3, 4, b 1 , b 2 ], [2, 11, b 3 , b 4 ] and [3, 11, b 5 , b 6 ]. (10,11) can not appear in two 4-gon, there- fore one of b 1 , b 2 is 10, and (2,9), (9,10) can not appear in two 4-gon, which implies one of b 5 , b 6 is 9. Since (4,6) will not appear in two 4-gon, therefore (6,11) will appear in same 4-gon and we see that [6,11] is an edge, therefore [6,11] will form an edge in a 4- gon. Now lk(6) = C 8 (10, 1, [11, a 1 , 7], [0, 4, 5]) which implies [6,7] form an edge in a 4-gon, therefore a 1 = 2 and b 3 = 6, b 4 = 7. Since (4,5) can not appear in two 4-gon, there- References
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