From: Dinogeorge@aol.com
Date: July 10, 2007 6:47:36 PM PDT
To: polyhedron@lists.mathconsult.ch, george.olshevsky@gmail.com
Cc: Dinogeorge@aol.com
Subject: Re: [Polyhedron] Uniform helical polyhedra

In a message dated 7/10/2007 5:40:09 P.M. Pacific Daylight Time,  
grunbaum@math.washington.edu writes:

Your  illustrations are very nice. It is only a pity that you excluded   
coplanar adjacent faces -- allowing them increases the number of   
kinds by three only.

Yes, three kinds of towers: those whose faces are all squares, those whose  
faces lie in adjoining vertical strips of triangles, and those whose faces  lie 
in both kinds of vertical strips alternating.

But coplanarity of adjacent faces also excludes a collection  of infinite 
uniform "polyhedra" derived from the uniform tilings,  such as any trivalent 
tiling raised into a prism minus all the prisms  of some of its faces (we might 
call these "polyhedra" Swiss cheese  slices). Attached is a picture of a portion 
of the dodecagonal-square Swiss  cheese slice. You could stack an infinite 
number of Swiss cheese  slices to obtain more square-faced infinite "polyhedra," 
except  that the squares again would lie in coplanar vertical strips.

And excluding coplanar adjoining faces eliminates plane tilings  considered 
as infinite polyhedra, as well as those pesky "rippled"  tilings: tilings of 
triangles and/or squares whose edges fall into sets of  parallel apeirogons that 
may then be folded by any arbitrary angle  between 0 and pi/2 along those 
apeirogons.

I have the feeling that none of what I'm doing these days with infinite  
polyhedra is new, but finding the prior references has been rather a  problem for 
me. All such citations concerning these kinds of figures will  be most 
appreciated. But, I don't mind duplicating other work on my own, since  at the very 
least it's a learning experience.

************************************** See what's free at http://www.aol.com.
¨_______________________________________________
Polyhedron mailing list (private)
Contributions to mailto:Polyhedron@lists.mathconsult.ch
Web interface: http://www.mathconsult.ch/lists/cgi/listinfo/polyhedron
¨

From: Branko Grünbaum <grunbaum@math.washington.edu>
Date: July 10, 2007 5:39:43 PM PDT
To: Dinogeorge@aol.com
Cc: george.olshevsky@gmail.com, polyhedron@lists.mathconsult.ch
Subject: Re: [Polyhedron] Uniform helical polyhedra

Hi George,

Your illustrations are very nice. It is only a pity that you excluded coplanar adjacent faces -- allowing them increases the number of kinds by three only.

On the other hand, you may be interested to know that the topic has been considered in the past. Attached please find examples of the six kinds, as illustrated in the book

A. Wachman, M. Burt and M. Kleinmann, Infinite Polyhedra. Technion – Israel Institute of Technology, Faculty of Architecture and Town Planning. Haifa. First ed. 1974. Second (unchanged) ed. 2005.

¨
Helical columns of triangles are also discussed by

E. A. Lord, A. L. Mackay and S. Ranganathan, New Geometries for New Materials. Cambridge Univ. Press 2006.

I have discussed uniform columns (including selfintersecting ones) in my courses on polyhedra in 1998 and 2005.  By coincidence, I am about midway in the writing of a paper surveying uniform columns.  Naturally, my illustrations are nowhere as nice as yours -- but I have to stick to black & white, and to what I can do in Mathematica and similar programs.

Regards,

Branko

On Jul 10, 2007, at 3:33 PM, Dinogeorge@aol.com wrote:

We’re all familiar with finite uniform polytopes. What are not so familiar
are the infinite uniform polytopes, for which I’d like to propose the following
 set of defining properties:

Polytopeness
[1] Faces are planar polygons embedded in Euclidean  three-space
[2] Faces are bounded polygons with a finite number of  edges
[3] Exactly two faces incident with every edge
[4] Faces incident at  an edge not coplanar
[5] No coincident edges or faces
[6] Coincident  vertices OK (equivalent to compound vertex figures with no
coincident  elements)
[7] Faces form a single edge-connected set (so compound infinite  polyhedra
are excluded for now)

Simpleness (simplicity)
[8] Faces do not intersect (later, faces might  be allowed to have a finite
number of other intersecting faces, to permit  infinite uniform star-polyhedra;
but we wouldn’t want to consider an endless  collection of monster
star-polyhedra that fill three-space densely)

Infiniteness (mandatory for this collection)
[9] Countably infinitely  many faces

Uniformity
[10] All faces are regular polygons
[11] All vertex  figures are transitive on the symmetry group (“corners all
alike”; regarding  infinite polyhedra with coincident vertices, for such a
polyhedron—and I have  yet to discover any—to be uniform, each vertex figure
separately must be  transitive on the symmetry group)

In addition to mining the 28 uniform honeycombs (and their various
derivatives and extensions) and an as-yet-unknown number of scaliform honeycombs  for
face sets that fulfill the above properties, we have two kinds of infinitely
tall polyhedral towers and at least one class of what I call helical polyhedra
with equit faces. To make such a helical polyhedron, among whose symmetries
screw dislocations are prominent, consider a portion of the equit tiling
bounded  between two sawtooth edge-apeirogons (see attached picture). Make it at
least  six equits wide (in general, at least 2n for n>2). Score the edges
between  all the equits in that set, then bend the collection along the scored edges
 until the bounding sawtooth edge-apeirogons intermesh. We can glue the
equit-sheet along the edge-apeirogons to form an infinite tower bounded by the
equits. (In the attached picture, for example, glue the apeirogons so that the
colored strips become continuous.) One kind of tower may be formed by gluing
an  outer equit on one side to the outer equit on the other side in the same
equit  row (see attached picture for the case n=5); a different kind of tower
will be  formed by gluing an outer equit to an equit from a different row—say
the first  or second row up, in general the kth row up. (In attached picture,
k=1. The  sawtooth edge-apeirogons will be horizontally aligned when k is even,
antialigned when k is odd.) As long as k is not too large but is >0, this
assembly method will create a uniform infinite polyhedron with six equits
meeting the same way at every corner (with “top” and “bottom” polygons of the
tower “at infinity,” the polyhedra will be squeezable-floppy for n>3, but
there will be a position for which the whole figure is uniform). The larger n,
the more opportunities there are for valid values of k. The cases n=3, k=1 and
n=4, k=1 yield the helical towers in the attached pictures. The first one
Stella4D constructed as an endless twisted string of tetrahedra (portions of
which occur, for example, in some nets of the regular hexacosichoron). The
second was built from the first by fitting in a couple of helical rows of
irregular tetrahedra and using Stella’s “make faces regular” feature; end  effects
offscreen (because the Stella polyhedron is after all not infinite) may  make
the tower ever so slightly out of shape, even though the faces are all  equits
(it is a true deltahedron). Note that each triangle of a helical tower  belongs
to the three girdles that pass through it. We use the k, n values for  the
rows most nearly perpendicular to the symmetry axis. Helical towers are  chiral,
of course.

We could attempt building this kind of helical tower using the square
tiling, but although it can be done in the abstract, in Euclidean three-space  the
ribbons of squares are constrained to join up end to end—no jumping from one
ribbon up or down to the next one anywhere, as this would bend the squares out
of their planes. To build the helical tower from squares, we must go to
four-space and bend the ribbons around and out of their three-space; we will  then
have constructed a uniform skew square helical tower, and even a uniform
skew square helical torus (by bringing the ends of the tower together instead of
letting them extend to infinity—we may have to enter five-space to do this,
though I think four-space might just be enough: certainly the uniform helical
toroidal skew polyhedra with equit faces exist in four-space). Back to
three-space: The infinite non-helical tower (k=0) of squares created by joining  an
infinite number of n-gonal prisms top to bottom is excluded from our
collection because adjacent faces must be coplanar.

No uniform tiling of the plane that has trivalent vertices can possibly be
folded into a tower polyhedron in this manner; all the putative faces are
constrained to be coplanar. So four or more faces per vertex are required. The
square tiling, which is tetravalent, is eliminated from consideration by the  “
square ribbons constraint” and “no adjacent coplanar faces” constraints
described above. The elongated equit tiling, in which ribbons of squares are
intercalated between ribbons of equits, is pentavalent and will fold up into the
uniform tower polyhedra formed by joining n-gonal prisms to n-gonal anitprisms
alternately base to base (k=0; a picture of the n=5 elongated tower is
attached), but helical towers are again prevented by the “square ribbon  constraint.
” Rolling a tower up the other way, with vertical rows of squares in  between
vertical rows of equits, is excluded by the “no adjacent coplanar  faces”
constraint. (But in four-space, we can fold a suitable m×n finite patch of  the
elongated equit tiling around and end-to-end to make a toroidal uniform skew
polyhedron.) Uniform tilings with polygons that have five or more sides cannot
be rolled up into uniform towers, for the same reason that the mantles of
isogonal prismatic polyhedra may have only three or four sides: There are too
many vertices relative to the axis of symmetry of the tower if the polygons
have  more than four sides. This kind of symmetry consideration also kills the
last  remaining possibility, the snub square tiling, which has pairs of equits
adjoining in two perpendicular directions. The only way they could coexist in
a  uniform tower would be to be inclined at the same angle to the axis, but
this  would position the accompanying square faces asymmetrically relative to
the same  axis.

I believe this shows that the only possible simple uniform helical towers
are those with six equits at each vertex, as described in the preceding.
Incidentally, the duals of the helical equit towers are represented in real life  by
none other than various kinds of carbon nanotube  molecules.

************************************** See what's free at http://www.aol.com.
<Helicalapeirohedron.gif>
<Helicalapeirohedron4.gif>
<N3K1helicalpattern.gif>
<Pentagonalelongtower.gif>
<Pentagonaltower.gif>
_______________________________________________
Polyhedron mailing list (private)
Contributions to mailto:Polyhedron@lists.mathconsult.ch
Web interface: http://www.mathconsult.ch/lists/cgi/listinfo/polyhedron

_______________________________________________
Polyhedron mailing list (private)
Contributions to mailto:Polyhedron@lists.mathconsult.ch
Web interface: http://www.mathconsult.ch/lists/cgi/listinfo/polyhedron
¨

From: Dinogeorge@aol.com
Date: July 10, 2007 3:33:51 PM PDT
To: polyhedron@lists.mathconsult.ch
Cc: Dinogeorge@aol.com, george.olshevsky@gmail.com
Subject: [Polyhedron] Uniform helical polyhedra

We’re all familiar with finite uniform polytopes. What are not so familiar  
are the infinite uniform polytopes, for which I’d like to propose the following 
 set of defining properties:

Polytopeness
[1] Faces are planar polygons embedded in Euclidean  three-space
[2] Faces are bounded polygons with a finite number of  edges
[3] Exactly two faces incident with every edge
[4] Faces incident at  an edge not coplanar
[5] No coincident edges or faces
[6] Coincident  vertices OK (equivalent to compound vertex figures with no 
coincident  elements)
[7] Faces form a single edge-connected set (so compound infinite  polyhedra 
are excluded for now)

Simpleness (simplicity)
[8] Faces do not intersect (later, faces might  be allowed to have a finite 
number of other intersecting faces, to permit  infinite uniform star-polyhedra; 
but we wouldn’t want to consider an endless  collection of monster 
star-polyhedra that fill three-space densely)

Infiniteness (mandatory for this collection)
[9] Countably infinitely  many faces

Uniformity
[10] All faces are regular polygons
[11] All vertex  figures are transitive on the symmetry group (“corners all 
alike”; regarding  infinite polyhedra with coincident vertices, for such a 
polyhedron—and I have  yet to discover any—to be uniform, each vertex figure 
separately must be  transitive on the symmetry group)

In addition to mining the 28 uniform honeycombs (and their various  
derivatives and extensions) and an as-yet-unknown number of scaliform honeycombs  for 
face sets that fulfill the above properties, we have two kinds of infinitely  
tall polyhedral towers and at least one class of what I call helical polyhedra  
with equit faces. To make such a helical polyhedron, among whose symmetries  
screw dislocations are prominent, consider a portion of the equit tiling 
bounded  between two sawtooth edge-apeirogons (see attached picture). Make it at 
least  six equits wide (in general, at least 2n for n>2). Score the edges 
between  all the equits in that set, then bend the collection along the scored edges 
 until the bounding sawtooth edge-apeirogons intermesh. We can glue the  
equit-sheet along the edge-apeirogons to form an infinite tower bounded by the  
equits. (In the attached picture, for example, glue the apeirogons so that the  
colored strips become continuous.) One kind of tower may be formed by gluing 
an  outer equit on one side to the outer equit on the other side in the same 
equit  row (see attached picture for the case n=5); a different kind of tower 
will be  formed by gluing an outer equit to an equit from a different row—say 
the first  or second row up, in general the kth row up. (In attached picture, 
k=1. The  sawtooth edge-apeirogons will be horizontally aligned when k is even,  
antialigned when k is odd.) As long as k is not too large but is >0, this  
assembly method will create a uniform infinite polyhedron with six equits  
meeting the same way at every corner (with “top” and “bottom” polygons of the  
tower “at infinity,” the polyhedra will be squeezable-floppy for n>3, but  
there will be a position for which the whole figure is uniform). The larger n,  
the more opportunities there are for valid values of k. The cases n=3, k=1 and  
n=4, k=1 yield the helical towers in the attached pictures. The first one  
Stella4D constructed as an endless twisted string of tetrahedra (portions of  
which occur, for example, in some nets of the regular hexacosichoron). The  
second was built from the first by fitting in a couple of helical rows of  
irregular tetrahedra and using Stella’s “make faces regular” feature; end  effects 
offscreen (because the Stella polyhedron is after all not infinite) may  make 
the tower ever so slightly out of shape, even though the faces are all  equits 
(it is a true deltahedron). Note that each triangle of a helical tower  belongs 
to the three girdles that pass through it. We use the k, n values for  the 
rows most nearly perpendicular to the symmetry axis. Helical towers are  chiral, 
of course.

We could attempt building this kind of helical tower using the square  
tiling, but although it can be done in the abstract, in Euclidean three-space  the 
ribbons of squares are constrained to join up end to end—no jumping from one  
ribbon up or down to the next one anywhere, as this would bend the squares out  
of their planes. To build the helical tower from squares, we must go to  
four-space and bend the ribbons around and out of their three-space; we will  then 
have constructed a uniform skew square helical tower, and even a uniform  
skew square helical torus (by bringing the ends of the tower together instead of  
letting them extend to infinity—we may have to enter five-space to do this,  
though I think four-space might just be enough: certainly the uniform helical  
toroidal skew polyhedra with equit faces exist in four-space). Back to  
three-space: The infinite non-helical tower (k=0) of squares created by joining  an 
infinite number of n-gonal prisms top to bottom is excluded from our  
collection because adjacent faces must be coplanar.

No uniform tiling of the plane that has trivalent vertices can possibly be  
folded into a tower polyhedron in this manner; all the putative faces are  
constrained to be coplanar. So four or more faces per vertex are required. The  
square tiling, which is tetravalent, is eliminated from consideration by the  “
square ribbons constraint” and “no adjacent coplanar faces” constraints  
described above. The elongated equit tiling, in which ribbons of squares are  
intercalated between ribbons of equits, is pentavalent and will fold up into the  
uniform tower polyhedra formed by joining n-gonal prisms to n-gonal anitprisms  
alternately base to base (k=0; a picture of the n=5 elongated tower is  
attached), but helical towers are again prevented by the “square ribbon  constraint.
” Rolling a tower up the other way, with vertical rows of squares in  between 
vertical rows of equits, is excluded by the “no adjacent coplanar  faces”
constraint. (But in four-space, we can fold a suitable m×n finite patch of  the 
elongated equit tiling around and end-to-end to make a toroidal uniform skew  
polyhedron.) Uniform tilings with polygons that have five or more sides cannot  
be rolled up into uniform towers, for the same reason that the mantles of  
isogonal prismatic polyhedra may have only three or four sides: There are too  
many vertices relative to the axis of symmetry of the tower if the polygons 
have  more than four sides. This kind of symmetry consideration also kills the 
last  remaining possibility, the snub square tiling, which has pairs of equits  
adjoining in two perpendicular directions. The only way they could coexist in 
a  uniform tower would be to be inclined at the same angle to the axis, but 
this  would position the accompanying square faces asymmetrically relative to 
the same  axis.

I believe this shows that the only possible simple uniform helical towers  
are those with six equits at each vertex, as described in the preceding.  
Incidentally, the duals of the helical equit towers are represented in real life  by 
none other than various kinds of carbon nanotube  molecules.

************************************** See what's free at http://www.aol.com.
¨¨¨¨¨_______________________________________________
Polyhedron mailing list (private)
Contributions to mailto:Polyhedron@lists.mathconsult.ch
Web interface: http://www.mathconsult.ch/lists/cgi/listinfo/polyhedron