1. Sep./6/2007 Kazushi AHARA, and Keita SAKUGAWA (Meiji University) Subdivision HYPLANE and K=-1 surfaces with symmetry

2. Sep./6/2007 Hyplane is a polyhedron such that (a) Vertex curvature is negative and constant, and (b) vertices are ‘configured uniformly.’ Hyplane is a polyhedral analogue of K=-1(negative constant curvature) surface in R 3. What’s hyplane?

3. Sep./6/2007 Let v be an internal vertex of a polyhedron. Then K(v) = 2π - ∑ v A (figure) Vertex curvature K(v) A: face with v

4. Sep./6/2007 Theorem If a triangle ABC is on a polyhedron and it bounds a triangle region on polyhedron, then ∠ A+ ∠ B+ ∠ C = π+ ∑ K(v) (figure) Gauss theorem on a polyhedron v ∈ ΔABC

5. Sep./6/2007 Let (a,b,c) be a triad of positive integers such that (a) 1/a + 1/b + 1/c < 1/2 (b) If a is odd then b=c. If b is odd then c= a. If c is odd then a=b. Then we can consider a hyperbolic tessellation of triangles with angles ( 2π/a, 2π/b, 2π/c ). Hyplane on triangle tessellation

6. Sep./6/2007 (a, b, c)=(4, 6, 14)

7. Sep./6/2007 Let a triangle ABC be such that ∠ A= bcπ/(ab+bc+ca) ∠ B= caπ/(ab+bc+ca) ∠ C= abπ/(ab+bc+ca) And make a polyhedron P such that (1) all faces are congruent to ΔABC, (2) Any two faces side by side are symmetric, and (3) there are a faces meeting together at each vertex corresponding to A, (and similarly for B and C.) Hyplane on triangle tessellation

8. Sep./6/2007 ΔABC is a triangle with angles 6π/20, 7π/20, 7π/20, (54 degree, 63 degree, 63 degree) Software ‘hyplane’ is on this model. (a,b,c)=(6,6,7) case

9. Sep./6/2007 If the faces of the polyhedron are congruent to each other, we may consider that the second condition (b) vertices are ‘configured uniformly’ is satisfied. (Yes, I think so.) So we may consider a hyplane as an polyhedral analogue of a surface of the negative constant curvature. Hyplane and K=-1 surface

10. Sep./6/2007 When we have a hyperbolic tessellation by triangles of the same size and figure, we can construct a hyplane. Here is an acyclic example: tessellation of rhombi with angles (6π/11,4π/11). Acyclic hyplane

11. Sep./6/2007

12. Examples of K=-1 surfaces (1) Surface of revolution 1 (pseudo sphere, revolution of tratrix, tractoid)

13. Sep./6/2007 Examples of K=-1 surfaces (2) Surface of revolution 2 (Hyperboloid type)

14. Sep./6/2007 Examples of K=-1 surfaces (3) Surface of revolution 3 (conic type)

15. Sep./6/2007 Examples of K=-1 surfaces (4) Kuen surface

16. Sep./6/2007 Examples of K=-1 surfaces (5) Dini surface

17. Sep./6/2007 Hyplane (polyhedron) corresponds to a triangle tessellation on the hyperbolic plane. If the sum of internal angles of the triangle in the tessellation get smaller, the area get larger, the vertex curvature of the hyplane get larger, and it gets difficult to figure up the hyplane model. (Mission) To obtain a K=-1 surface from hyplane, make ‘subdivision’ of hyplane. Subdivision of hyplane

18. Sep./6/2007 We need to preserve the conditions on subdivision: (a) vertex curvature is negative and constant. (b) vertices are ‘configured uniformly.’ So we need to ‘rescale’ the size of all faces. From Gauss-Bonnet theorem, the total of vertex curvature must be constant, so the vertex curvature of each vertex must be near 0 (, since the number of vertices get large.) Difficulty for subdivision

19. Sep./6/2007 We first consider subdivision into 4 triangles (figure) but the middle point on the edge must not the midpoint of the edge. To satisfy (a), we need to rescale all small faces in the subdivision. Rescaling

20. Sep./6/2007 We consider the following condition. That is, in the figure below, the same alphabets mean congruent, and all small triangles are isosceles. (figure) Assumption

21. Sep./6/2007 Let A i and B i be as in the figure below. A 4 and (π-A 1 )/2 is determined directly from the new vertex curvature. About other angles, we can determine them easily. Equations

22. Sep./6/2007 But ….. The area of triangles are not the same. There are no solution for satisfying (b) vertices are ‘configured uniformly’ in any subdivision (in any meaning.) So we use the above solution for subdivision.

23. Sep./6/2007 Surfaces of revolution and hyplane Hyperboloid type

24. Sep./6/2007 Surfaces of revolution and hyplane Hyperboloid type

25. Sep./6/2007 conic type and pseudo-sphere There exists a hyplane model for conic type, but pseudo-sphere.

26. Sep./6/2007 Surface with C 3 symmetry There are NOT known K=-1 surface with C 3 (=cyclic group of order 3) symmetry (other than surfaces of revolution.) But there are some examples of hyplane with C 3 symmetry. We call them ‘omusubi’ hyplane.

27. Sep./6/2007 Omusubi type (1)

28. Sep./6/2007 Omusubi type (2)

29. Sep./6/2007 Surface with A 4 symmetry There are NOT known K=-1 surface with A 4 (=Alternating Group of lengthe 4) symmetry. But there are some examples of hyplane with A 4 symmetry. We call them ‘chimaki’ hyplane.

30. Sep./6/2007 Chimaki type

31. Sep./6/2007 Symmetry axis and surface If K=-1 surface S has a symmetry (of order more than 2) of rotation and let X be the axis of the symmetry. Then X never intersect with S. Because if they intersect, then the intersection point must be umbilical and hence has positive curvature.

32. Sep./6/2007 Punctured examples (1)

33. Sep./6/2007 Punctured examples (1)

34. Sep./6/2007 Punctured examples (1)

35. Sep./6/2007 Singular type If there exists a smooth K=-1 surface with such (C3 or A4) symmetry, each singular point on the symmetry axis never be cusp shape. Because on such K=-1 surface, there exists a tessellation of triangles and in the tessellation viewpoint, the point is not singular.

36. Sep./6/2007 Problem! Find a good coordinate on these ‘surfaces.’ To get a K=-1 surface, we need a specified coordinate (, where the second fundamental form is reduced) and a solution of sine-Gordon equation ω uv =sin ω with a certain boundary condition.