Sep./6/2007

Kazushi AHARA, and Keita SAKUGAWA

(Meiji University)

Subdivision HYPLANE and

K=-1 surfaces with symmetry

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 R3.

What’s hyplane?

Sep./6/2007

Let v be an internal vertex of a polyhedron. Then

K(v) = 2π - ∑ vA

(figure)

Vertex curvature K(v)

A: face with v

Sep./6/2007

Theorem

If a triangle ABC is on a polyhedron and it bounds a t

riangle region on polyhedron, then

∠A+ ∠B+ ∠C = π+ ∑ K(v)

(figure)

Gauss theorem on a polyhedron

v∈ΔABC

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 od

d then a=b.

Then we can consider a hyperbolic tessellation of triangl

es with angles (2π/a, 2π/b, 2π/c ).

Hyplane on triangle tessellation

Sep./6/2007

(a, b, c)=(4, 6, 14)

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

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

Sep./6/2007

If the faces of the polyhedron are congruent to each oth

er , 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 th

e negative constant curvature.

Hyplane and K=-1 surface

Sep./6/2007

When we have a hyperbolic tessellation by triangles of t

he same size and figure, we can construct a hyplan

e. Here is an acyclic example: tessellation of rhombi

with angles (6π/11,4π/11).

Acyclic hyplane

Sep./6/2007

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Examples of K=-1 surfaces (1)

Surface of revolution 1 (pseudo sphere, revolution of tra

trix, tractoid)

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Examples of K=-1 surfaces (2)

Surface of revolution 2 (Hyperboloid type)

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Examples of K=-1 surfaces (3)

Surface of revolution 3 (conic type)

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Examples of K=-1 surfaces (4)

Kuen surface

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Examples of K=-1 surfaces (5)

Dini surface

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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

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

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

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

Sep./6/2007

Let Ai and Bi be as in the figure below. A4 and

(π-A1)/2 is determined directly from the new

vertex curvature. About other

angles, we can determine

them easily.

Equations

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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.

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Surfaces of revolution and hyplane

Hyperboloid type

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Surfaces of revolution and hyplane

Hyperboloid type

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conic type and pseudo-sphere

There exists a hyplane model for conic type , b

ut pseudo-sphere.

Sep./6/2007

Surface with C3 symmetry

There are NOT known K=-1 surface with C3(=cyclic group

of order 3) symmetry (other than surfaces of revoluti

on.) But there are some examples of hyplane with C

3 symmetry. We call them ‘omusubi’ hyplane.

Sep./6/2007

Omusubi type (1)

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Omusubi type (2)

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Surface with A4 symmetry

There are NOT known K=-1 surface with A4(=Alternating

Group of lengthe 4) symmetry. But there are some e

xamples of hyplane with A4 symmetry. We call them

‘chimaki’ hyplane.

Sep./6/2007

Chimaki type

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.

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Punctured examples (1)

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Punctured examples (1)

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Punctured examples (1)

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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.

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) a

nd a solution of sine-Gordon equation

　 ωuv=sin ω

with a certain boundary condition.