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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
Sep./6/2007
Examples of K=-1 surfaces (1)
Surface of revolution 1 (pseudo sphere, revolution of tra
trix, tractoid)
Sep./6/2007
Examples of K=-1 surfaces (2)
Surface of revolution 2 (Hyperboloid type)
Sep./6/2007
Examples of K=-1 surfaces (3)
Surface of revolution 3 (conic type)
Sep./6/2007
Examples of K=-1 surfaces (4)
Kuen surface
Sep./6/2007
Examples of K=-1 surfaces (5)
Dini surface
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
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
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.
Sep./6/2007
Surfaces of revolution and hyplane
Hyperboloid type
Sep./6/2007
Surfaces of revolution and hyplane
Hyperboloid type
Sep./6/2007
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)
Sep./6/2007
Omusubi type (2)
Sep./6/2007
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.
Sep./6/2007
Punctured examples (1)
Sep./6/2007
Punctured examples (1)
Sep./6/2007
Punctured examples (1)
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.
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.