1. Embeddings with all triangles faces Dan Archdeacon The University of Vermont

2. Common goal: Embed a simple graph such that every face is a triangle Why? Minimizes the genus of the embedding Examples include n = 0,3,4,7 (mod 12) in the Map Color Theorem, giving triangular embeddings of K n on orientable surfaces Similar nonorientable embeddings exist for n = 0,1 (mod 3), except n = 3,4,7.

3. Change the goal: Instead of “every face is a triangle”, try “every triangle is a face” Trouble: In an embedding every edge is on at most 2 faces, but now every edge should be on n-2 triangles Solution: Replace K n with a multigraph K n (n-2)/2 Note: n must be even

4. The Main Result Theorem: For all even n, there exists an embedding of K n (n-2)/2 such that every triangle is a face. Moreover, if n ≥ 6 the surface can be chosen orientable or nonorientable. Note: The orientable genus is (n + 3)(n - 2)(n - 4)/24 The nonorientable genus is (n + 3)(n - 2)(n - 4)/12 (a handle is worth two crosscaps)

5. Proof: use induction on n in steps of size 2 Start of the induction: For n = 4 use the tetrahedron The inductive step: We use the existing embedding of K n (n-2)/2 and an extender T n to construct the embedding of K n+2 n/2. The key idea: Create 2 embeddings with vertices labeled 1,…,n+2 such that every triangle appears as a labeled face. Equivalently, if we identify all vertices with the same label, construct an embedding of K n+2 n/2 on a pinched surface with every triangle as a face. Yet to do: Describe how to identify vertices with the same label maintaining a surface embedding. Equivalently, describe how to modify our pinched embedding to remove the number of sheets at a pinch point.

6. The extender T n triangulating a sphere Add 2 new vertices with the following properties: Call them N,S for the north and south pole of a sphere. Connect N,S with n/2 meridians separating the sphere into n/2 sectors The remaining vertices all lie on the equator and are labeled 1,…,n such that every pair appears as an equatorial edge Every triangle with both N and S appears incident with a meridian Every triangle with only one of N or S appears using an equitorial edge This is possible because the last edge in the long path and the independent edges form a matching M, and the remaining edges in the long path form an Eulerian cycle in K n – M Equivalently: T n is a triangulation of a sphere with n pinch points whose faces are all triangles involving at least one of N,S Example: T 6

7. Identifying vertices (removing sheets) To glue u 1 to u 2 and v 1 to v 2 first create digons along edges u 1 v 1, u 2 v 2 : Then add a handle between these digons: Contract the cylinder laterally to identify the boundaries of the digons

8. Star Trek: The Game (106 th episode of “The Next Generation”)

9. Effect on pinch points when adding a handle The construction described topologically as adding a handle merges two sheets into one, similar to the representation of the complex function f(z)=z 2 around the origin

10. Adding a handle as a permutation of sides of edges Consider the sides of the edges incident with vertices a & b. Two kinds of relations: 1) corners at a face, 2) left-right side of a given edge. The orbits correspond to the sheets at these vertices 10,10b and 11,11b are the two sides of two edges Modify them so that 10,11b and 10b,11 are the two sides of two edges Net result is that orbits merge at both a and b, reducing the number of sheets AT THE START: a = (10, 30b 30, 40 40b, 10b) (11, 41 41b, 31b 31, 11b) (12, 42 42b, 32b 32, 12b) (13, 43 43b, 33b 33, 13b) (14, 44 44b, 34b 34, 14b) b = (10, 20 20b, 50 50b, 10b) (11b, 51 51b, 21b 21, 11) (12b, 52 52b, 22b 22, 12) (13b, 53 53b, 23b 23, 13) (14b, 54 54b, 24b 24, 14) AFTER ADDING A HANDLE WITH IDENTIFICATION: 10 10b & 11 11b  10 11b & 10b 11 a = (10, 30b 30, 40 40b, 10b 11, 41 41b, 31b 31, 11b) (12, 42 42b, 32b 32, 12b) (13, 43 43b, 33b 33, 13b) (14, 44 44b, 34b 34, 14b) b = (10, 20 20b, 50 50b, 10b 11, 21 21b, 51b 51, 11b) (12b, 52 52b, 22b 22, 12) (13b, 53 53b, 23b 23, 13) (14b, 54 54b, 24b 24, 14)

11. Combinatorial view as GEMs (Graph Encoded Manifolds) This model encodes maps as operations on properly 3-edge- colored cubic graphs Vertices of the GEM are flags of the embedding Permuted by three fixed-point-involutions: rotations, lateral along edges, and transversal across edges Embedding is orientable iff the GEM is bipartite Changes are only on the transversally colored edges Different viewpoints can yield different insights as well as double-checks

12. Summarizing the result of adding a handle Number of vertices decreases by 2 Number of edges remains the same Number of faces remains the same  The Euler characteristic decreases by 2 (so Euler genus increases by 2) If we deal with surfaces some of these operations connect different components. The labeled faces remain the same Equivalently, the number of sheets decreases by 2, one each at u,v Repeat n 2 /4 times using the bold equatorial edges of T n and equivalently labeled edges in K n (n-2)/2

13. A variation: adding an antihandle As before create digons along edges u 1 v 1, u 2 v 2, then add a handle between these digons using opposite orientations. The resulting surface is always nonorientable Similar to adding a handle or antihandle to the sphere forming the torus or Klein bottle respectively Save this step until last to yield a nonorientable surface

14. Conclusion: can identify all vertices with the same label (equivalently eliminate all pinch points), creating both orientable and nonorientable embeddings with every triangle appearing as a face The End

15. The End ?

16. Embeddings with all quadrilaterals faces Dan Archdeacon The University of Vermont

17. The Main Result Theorem: For all n, m = (n – 2)(n – 3)/2, there exists an embedding of K n m on a nonorientable surface such that every quadrilateral is a face Conjecture: The surface can be orientable, provided that n = 0,2,5,7 (mod 8)

18. Proof: use induction on n (steps of size 1) Start of the induction: For n = 4 use the tetrahedron in the projective plane, called the hemicube The inductive step: We use the existing embedding of K n (n-2)(n-3)/2 and n(n-1)(n-2)/6 copies of K 4 in the projective plane to construct the embedding of K n+1 (n-1)(n-2)/2. Each of the projective K 4 ’s use the vertex n+1 The key idea: Create 1+n(n-1)(n-2)/6 embeddings with vertices labeled 1,…,n+1 such that every quadrilateral appears as a labeled face. Equivalently, if we identify all vertices with the same label, construct an embedding of K n+1 (n-1)(n-2)/2 on a pinched surface with every quadrilateral as a face. Yet to do: Describe how to identify vertices with the same label maintaining the desired property. Equivalently, describe how to modify our embedding to remove the number of sheets at a pinch point.

19. Identifying vertices: adding a crosscap Let v be adjacent to u 1 and u 2. As before form digons as shown on the upper right, this time both incident with v Identify e 1 ’ with e 2 ’ and e 1 ’’ with e 2 ’’. The result is shown on the lower right, where the shaded disk is removed and its boundary points identified Reduces v by 1, e and f remain the same, decreasing the Euler characteristic (increasing Euler genus)by 1 with the new surface nonorientable. The faces are still all quadrilaterals Equivalent to a P3 where the two incident edges lie on different pinch sheets. Construction reduces the number of sheets at u by 1.

20. Helaman Ferguson Born August 11, 1940 Professor Brigham Young University “I find that sculpture is a very powerful way to convey mathematics, and mathematics is a very powerful design language for sculpture.” Torus with Cross-cap

21. Iterate the vertex identifications We next do a sequence of the operations “adding a handle” and “adding a crosscap” to complete the vertex identifications Equivalent to slowly reducing the total number of sheets at all vertices Must be done with care to ensure the appropriate constructions can be applied. I’ll spare you the details, but they’re not bad

22. Conclusion: can identify all vertices with the same label (equivalently eliminate all pinch points), creating nonorientable embeddings with every quadrilateral appearing as a face The End

23. The End ?

24. Embeddings with all pentagons faces Dan Archdeacon The University of Vermont

25. Change the goal again: try “every pentagon is a face” Easy calculations: The graph is now K n m with m = (n - 2)(n - 3)(n - 4)/2. There is no restriction on the parity of n The surface is nonorientable with genus 2 – n + 3n(n – 1)(n – 2)(n – 3) (n – 4)/20

26. The Main Result Theorem: For all n, m = (n – 2)(n – 3) (n – 4)/2, there exists an embedding of K n m on a nonorientable surface such that every pentagon is a face Conjecture: The surface can be orientable for even n Proof: Again by induction on n

27. Start of the induction Start of the induction: For n = 5 use the following (very hard to find) net

28. The inductive step Again uses the embedding K n m to construct the embedding for order n + 1 This time we need to use n(n – 1)(n – 2) (n – 3)/24 copies of the embedding for K 5 Use a sequence of “adding handles” and “adding crosscaps” to identify vertices/remove pinch points, with caution about where they apply

29. Conclusion: can identify all vertices with the same label (equivalently eliminate all pinch points), creating nonorientable embeddings with every pentagon appearing as a face The End

30. The End ?

31. Embeddings with all k-gons faces Dan Archdeacon The University of Vermont

32. Change the goal yet again: try “every k-gon is a face” Conjecture: For all (n,k) ≠ (4,3) there is a nonorientable embedding of a complete multigraph K n m such that every k-gon is a face. The same holds for orientable embeddings provided the parity condition on the Euler characteristic is satisfied

33. The difficulty: starting the induction The case n = k: Perhaps the most interesting case is embedding complete multigraphs such that every Hamiltonian cycle is a face. These are unknown for k ≥ 6. They correspond to the start of an induction for all k-gons faces. Theorem: If there exists an embedding of a complete multigraph of order k with all Hamiltonian cycles faces, then there is an embedding a complete multigraph of order n with all k- cycles faces for all n ≥ k.

34. Further variations: Require every triangle appears r ≥ 2 times (n even) Every quadrilateral, or pentagon r ≥ 2 times Every triangle and quadrilateral a face (n even) Every triangle, pentagon, and quadrilateral a face (n even)

35. Further variations: ✔ Require every triangle appears r ≥ 2 times (n even) Every quadrilateral, or pentagon r ≥ 2 times Every triangle and quadrilateral a face (n even) Every triangle, pentagon, and quadrilateral a face (n even)

36. Further variations: ✔ Require every triangle appears r ≥ 2 times (n even) ✔ Every quadrilateral, or pentagon r ≥ 2 times Every triangle and quadrilateral a face (n even) Every triangle, pentagon, and quadrilateral a face (n even)

37. Further variations: ✔ Require every triangle appears r ≥ 2 times (n even) ✔ Every quadrilateral, or pentagon r ≥ 2 times ✔ Every triangle and quadrilateral a face (n even) Every triangle, pentagon, and quadrilateral a face (n even)

38. Further variations: ✔ Require every triangle appears r ≥ 2 times (n even) ✔ Every quadrilateral, or pentagon r ≥ 2 times ✔ Every triangle and quadrilateral a face (n even) ✔ Every triangle, pentagon, and quadrilateral a face (n even)

39. The End ?

40. Embeddings with directed k-gons faces Dan Archdeacon The University of Vermont

41. Change the goal again: try “every directed k-cycle is a face” Surface must be orientable to talk about the faces as directed cycles Easy calculations: The order n must be even The graph is now K n m with m = (n - 2)! / (n – k)!

42. The Main Result Theorem: For all even n, k ε {3,4,5}, m = (n – 2)!/(n – k)!, there exists an embedding of K n m on an orientable surface such that every directed k-gon is a face Proof: Use a double cover of the nonorientable embedding with undirected k-gons faces

43. All directed k-gons for larger k? Most probably exist (n even) Theorem: They exist for n,k even Good area for future research when k is odd

44. The End?

45. This time for real Any questions?