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Polyhedra of positive genus

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Tim Poston

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Sep 22, 1997, 3:00:00 AM9/22/97
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Nick Halloway wrote:
>
> Is it possible to embed in 3-space a
> finite triangulation of a (many-holed) torus, that has the same number of
> triangles meeting at each vertex, with each face an equilateral triangle?

A triangulated surface without boundary has 3/2 times as many
edges as triangles, and if there are n triangles at each
vertex it has 3/n vertices. So the Euler relation

v - e + f = \chi

becomes

(3/n) f - (3/2) f + f = \chi

(3/n - 1/2) f = \chi

With n = 6, this forces \chi = 0.
With smaller values of n, it makes \chi positive,
hence we have a sphere or projective plane.
Multiple holes need at least n = 7,
but
(3/7 - 1/2) f = \chi
f = -14 \chi
with n = 8 you have
f = -8 \chi
and so on.

So you are very topologically limited,
before you even start looking at
constraints like embedding or equilaterality.
But when you do...
If n > 6 , try to construct a meeting of n equilateral
triangles that all lie on one side of a plane.
(Your embedded object must have many such corners,
wherever it can sit on a table.)

I would be most surprised by examples.

Tim

__________________________________________________________________________
Tim Poston Centre for Information-enhanced Medicine, Nat. Univ.
Singapore
Only the paranoid call paranoia survival.

Webber, Ronda

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Sep 23, 1997, 3:00:00 AM9/23/97
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For anyone who is curious, a description of Will's toroids can be found
in the Science News, Vol. 148, Dec. 1995.
If you have questions for him, email web...@math.washington.edu

Ronda Webber
Sehome High School
rowe...@shs.bham.wednet.edu
----------
From: Heidi Burgiel
To: geometry...@forum.swarthmore.edu
Subject: Re: Polyhedra of positive genus
Date: Tuesday, September 23, 1997 11:48AM

Hi!

My colleague Will Webber described in his thesis (University of
Washington, 1994) several toroidal manifolds with, I believe,
congruent nearly-equilateral faces. (I know they were
nearly-equilateral. There may have been more than one (but fewer than
10) congruence class. I can look it up if there is interest.)

He was unable to find a toroidal manifold with equilateral faces, but
from his data it seems likely that you can get arbitrarily close to
equilateral faces.

Another interesting feature of his tori is that they're folded from a
single sheet of paper -- so the angle deficit is the same at each
vertex (except possibly, but not likely, the faces at the "seams") is
the same.

I believe some of his work was described in Science Magazine.

Heidi B.

Heidi Burgiel

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Sep 23, 1997, 3:00:00 AM9/23/97
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Walter Whiteley

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Sep 23, 1997, 3:00:00 AM9/23/97
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I am curious about the observation that equilateral
triangles are the discrete equivalent of uniform curvature.
My own limited experience would be that uniform
angle deficit (or surplus) is the discrete equivalent of
uniform curvature. Perhpas I misunderstand something.

Walter Whiteley
York University

Walter Whiteley

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Sep 23, 1997, 3:00:00 AM9/23/97
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Two comments:
there is a general theorem that 'almost all' realizations in 3-spacce
of any topological type of closed triangulated 2-manifold
[Fogelsanger's Theorem.]
In fact the result is much stronger - applying to
all simplicial minimal homology d-surfaces realized in
d+1 space.
The 'almost all' phrase applies to an open dense subset
of the space of all realizations. While Connelly's flaxibe;
sphere shows that the rare counterexamples exist, and this
can be grafted onto any genus in a trivial way - it does not
occur as an embedded manifold. (Interestingly - the conjecture
is that all flexible triangulated 2-surfaces have constant volume
and constant Dehn Invariant).

IN general, as you increase the genus of a triangulated
torus - you are MORE likely to be rigid (you have extra
constraints beyond the minimum required).

Second comment. The book of Bonnie Stewart Adventures among the Toroids
(published by the author: B.M. Steward, 4494 Wausau Road,
Okemos Michegan, 48864, 1970)
has a number of interesting examples of tori created with triangles and
squares.

Just scannning that book, I do NOT see any closed surfaces
with only equilateral triangles, beyond the spheres and
trees made by joining such spheres across a common face.

Walter Whiteley
York University
Toronto Ontario

John Conway

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Sep 23, 1997, 3:00:00 AM9/23/97
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On Tue, 23 Sep 1997, Walter Whiteley wrote:

> Two comments:
> there is a general theorem that 'almost all' realizations in 3-spacce
> of any topological type of closed triangulated 2-manifold
> [Fogelsanger's Theorem.]

This sentence did not contain a verb, and I suspect you
didn't finish it.

> In fact the result is much stronger - applying to
> all simplicial minimal homology d-surfaces realized in
> d+1 space.
> The 'almost all' phrase applies to an open dense subset
> of the space of all realizations. While Connelly's flaxibe;
> sphere shows that the rare counterexamples exist, and this
> can be grafted onto any genus in a trivial way - it does not
> occur as an embedded manifold.

It really would be nice if you looked over your sentences
after writing them, to see if by any chance they parse! I presume
"flaxibe" was intended to be "flexible"? Connelly's flexible sphere
DOES occur as an embedded manifold. I expecvt you're thinking
of the much earlier flexible octahedra of Bricard.

(Interestingly - the conjecture
> is that all flexible triangulated 2-surfaces have constant volume
> and constant Dehn Invariant).

My thesis advisor, Harold Davenport, was asked (in the fifties or
earlier) by an engineer whether there could be a "bellows" (ie.,
a flexible polyhedron of non-constant volume). As far as I know,
this is the earliest entrance of the problem into mathematics. The
earliest documented place (probably a few years later) that I
know of is the sheet of geometry problems that my former colleague
Hallard Croft used to circulate in the early sixties.

When I first met Connelly (shortly after his discovery), I
asked him whether the volume changed, and he said it almost certainly
did, but then went away and came back after doing the calculation
to say that it didn't.

The constant volume conjecture has recently been proved,
the original idea coming from a Russian mathematician. The
appropriate paper (I don't have a reference) has four authors.


>
> IN general, as you increase the genus of a triangulated
> torus - you are MORE likely to be rigid (you have extra
> constraints beyond the minimum required).
>
> Second comment. The book of Bonnie Stewart Adventures among the Toroids
> (published by the author: B.M. Steward, 4494 Wausau Road,
> Okemos Michegan, 48864, 1970)
> has a number of interesting examples of tori created with triangles and
> squares.
>
> Just scannning that book, I do NOT see any closed surfaces
> with only equilateral triangles, beyond the spheres and
> trees made by joining such spheres across a common face.
>
> Walter Whiteley
> York University
> Toronto Ontario

Martin Gardner asked for the smallest number of equal equilateral
triangles that can be used to make an embedded torus, giving some
answer that I improved to 36 (if I've just counted them correctly
from the model I have here). I'm sure 36 is smallest, and uniquely so.

John Conway

Heidi Burgiel

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Sep 23, 1997, 3:00:00 AM9/23/97
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Hi!

I found the reference. The article on Webber's work was in Science
News, not Science. I believe there are patterns for tori folded from
a plane included with the article. (Allow at least 2 hours to do
this.)

Science News December 23 - 30, 1995, p. 432

Heidi Burgiel

John Conway

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Sep 25, 1997, 3:00:00 AM9/25/97
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On Tue, 23 Sep 1997, Nick Halloway wrote:

>
>
> On Tue, 23 Sep 1997, John Conway wrote:
>
> > On Tue, 23 Sep 1997, Walter Whiteley wrote:
> >
> > > Two comments:
> > > there is a general theorem that 'almost all' realizations in 3-spacce
> > > of any topological type of closed triangulated 2-manifold
> > > [Fogelsanger's Theorem.]
> >
> > This sentence did not contain a verb, and I suspect you
> > didn't finish it.
>

> That OK, I can without verbs ... Looking up this on the web it says
^^^
There's a VERB there! You must be slipping.

> that almost all realizations have first order/static rigidity. Not
> quite sure what that means.

I expect this is what used to be called "infinitesimal rigidity".
I can best exp[lain what it means by saying that if something ISN'T
infinitesimally rigid, you can move its vertices by epsilon-sized
amounts in such a way that its edge-lengths change by o(epsilon)
amounts, all faces remaining planar.

> I suppose this doesn't have the same # triangles meeting at each vertex?
> If it does do you have the triangulation, please?

It doesn't, but I'll give the triangulation anyway. I'll do so
in the form of a net:


____ ____
\ \ / /
___\___\/___/___
\ / /\ \ /
___\/ /__\ \/___
\ \ /\**/\ / /
\___\/__\/__\/___/
/\ /\
/ \____/ \
\ /\ /\ /
\/ \/ \/


A____________A
\ /\ /\ /\
B\/__\/__\/__\B
\ /\ /\ /
\/ \/ \/

Roll up the bottom piece by identifying AB with AB so that you
see six faces of an octahedron which has three flaps affixed to the
sides of the lower one of its two missing faces. Then identify
the boundary of the other ("top") missing face with the hole
indicated by ** in the upper piece, whose outer edges should be
glued up until the 6-edge boundary that remains can be identified
with the 6-edge zigzag that joins the leftmost B to the rightmost one
in the above figure. I have not drawn edges to separate triangles
that lie in the same plane.

John Conway

Heidi Burgiel

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Sep 25, 1997, 3:00:00 AM9/25/97
to

Professor Conway,

Thank you for the pattern for that lovely torus! I have just built
it, and it is wonderfully simple. Do you know if it occurs in nature?

For the advantage of those who don't have Polydrons, I will attempt to
describe it.

Start with an octahedron with one face up and one face down. It will
have six "side" faces. Fit three more octahedra face to face with
three of these side faces (equally spaced, of course). Looking down
on the object, you should see three square-pyramidal shapes pointing
up and away from the central octahedron.

Place tetrahedra face-to-face with the other three side faces of the
central octaedron, and pack the spaces between the outer octahedra and
the tetrahedra with six more tetrahedra. Remove the central
octahedron (I guess you have to drill it out, or partially dissasemble
your model at this point) and you have a torus! It rests nicely on
the exposed faces of the first three tetrahedra, with the three
pyramidal octahedron peaks pointing up and out.

Heidi

Andrew Hume

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Sep 25, 1997, 3:00:00 AM9/25/97
to

there are lots more in a charming book called `Adventures
Among the Toroids' by Bonnie Stewart (self-published).

i have built many of these, mostly with polydrons,
and have a couple sitting on a file drawer at work.
(my office has its own supply of polydrons;
most visitors end up playing with them.)

--

Andrew Hume
and...@research.att.com +1 973-360-8651

Webber, Ronda

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Sep 25, 1997, 3:00:00 AM9/25/97
to

Heidi Has it mostly right. In my thesis I describe several ways of
producing Monohedral Idemvalent Toroidal Polyhedra. Monohedral- meaning
all the faces are congruent, Idemvalent- meaning "same valence" for each
vertex. More precisely, I was able to produce genus 1 toroids with 6
congruent triangles at each vertex. None of them used Equilateral
triangles, but some were remarkably close
The first method is an adaptation to a construction by Aluoglu and
Giese in the 50's. They used monohedral octahedra to produce tori with
six congruent faces at each vertex. Some of the vertices had an angle
sum of more than 360 degrees, some had less than 360. The average of
course was 360. My contribution was to show how to produce ones that
were combinatorally the same but geometrically different. In fact, I
found some examples that were knotted. The number of different
geometric varieties is infinite.
The second method was the one that Heidi was remembering. As Heidi
pointed out, they were folded from apiece of paper. First, tile the
paper with congruent triangles then fold along the edges. The angle sum
at every vertex is 360, even at the seam! In the thesis I describe ways
of doing this folding to produce tori with different types of symmetry.
Some have the symmetry of a prism, some an antiprism, some dihedral, and
some have only cyclic symmetry. Each different fold pattern requires
different triangles to get the "seams" to come together exactly.
Usually there is a 1 parameter family of triangles that work, but
sometimes there is only 1 particular triangle.
As for equilateral triangles--There is one example that I describe that
uses isosceles triangles that have legs of length .99995 with base
length 1. Not equilateral, but mighty close.
It is interesting to look at the models and see how 6 triangles at a
vertex can look like they have positive curvature, and when folded
differently look like they have negative curvature.

William Webber
Whatcom Community College


Typed by Ronda Webber--Blame the typos on me :)
----------
From: Nick Halloway
To: Walter Whiteley
Cc: geometry...@forum.swarthmore.edu


Subject: Re: Polyhedra of positive genus

Date: Tuesday, September 23, 1997 11:25AM

On Tue, 23 Sep 1997, Walter Whiteley wrote:

> I am curious about the observation that equilateral
> triangles are the discrete equivalent of uniform curvature.
> My own limited experience would be that uniform
> angle deficit (or surplus) is the discrete equivalent of
> uniform curvature. Perhpas I misunderstand something.

If you have equilateral triangles and the same # triangles meeting at
each vertex then the surface has to have the same angle surplus at
each vertex. If the triangular faces were not uniform in area you could
make
a surface with the same angle surplus at each vertex, but with
curvature concentrated in some places. If all the triangular faces have
the same area but are not equilateral then you might get a surface which
has more curvature in one direction than in another. That would have
uniform Gaussian curvature but I wonder if such a surface would close?
Or, I wonder if one can show that using equilateral triangles, with the
same curvature in all n directions around a vertex, that the surface
couldn't close -- perhaps forming a torus in 3D requires more
curvature in some directions than in others.

Epimenides The Cretan

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Sep 25, 1997, 3:00:00 AM9/25/97
to

John:

On 23 Sep you wrote:

>The constant volume conjecture has recently been proved,
>the original idea coming from a Russian mathematician. The
>appropriate paper (I don't have a reference) has four authors.

How about THREE only?

B. Connelly - I. Sabitov - A. Walz: The Bellows Conjecture.
Contributions to Algebra and Geometry 38:1(1997) 1-10

More info at:
http://math.cornell.edu/~connelly/

cretAN


John Conway

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Sep 26, 1997, 3:00:00 AM9/26/97
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On Thu, 25 Sep 1997, Heidi Burgiel wrote:

> Professor Conway,
>
> Thank you for the pattern for that lovely torus! I have just built
> it, and it is wonderfully simple. Do you know if it occurs in nature?

You're welcome! I doubt if it exists anywhere except in the collections
of those who have read my message about it (which includes me, of course!).

I've also found what I'm sure are the minimal examples of non-spherical
polyhedra with all faces regular. There are in fact just two such, each
with some square and some triangular faces, which differ in only a trivial
way. ("minimal" here refers to number of faces)

I can't recall if this group was where there was some discussion of the
Csaszar and Szilassi polyhedra? The Csaszar polyhedron is a toroidal
(or genus 1) polyhedron with just 7 vertices, and the Szilassi one is
an embedding of its topological dual, with just 7 (non-convex, hexagonal)
faces.

The minimal numbers of vertices for polyhedra of genus respectively
0,1,2,3,4 are 4,7,8,10,11, and after that the answers are unknown. It's
possible to stick two Csaszar polyhedra together (in two distinct ways)
so as to get 8-vertexers of genus 2, and I have models of these. I
have the papers describing the things of genus 3 and 4,, and intend to
make them up soon. The genus 5 problem is very interesting.

John Conway

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