Envelope Polyhedra

TL;DR

This paper introduces envelope polyhedra, a new class of regular polyhedra with unique arrangements of polygons and dihedral angles, leading to novel finite and infinite multiply connected networks with complex topologies.

Contribution

It defines envelope polyhedra, expanding the classification of regular polyhedra by allowing variable dihedral angles and complex topologies, including infinite networks.

Findings

Introduces envelope polyhedra with variable dihedral angles.

Describes finite polyhedra with toroidal topology.

Presents infinite multiply connected polygon networks.

Abstract

This paper presents an additional class of regular polyhedra--envelope polyhedra--made of regular polygons, where the arrangement of polygons (creating a single surface) around each vertex is identical; but dihedral angles between faces need not be identical, and some of the dihedral angles are 0 degrees (i.e., some polygons are placed back to back). For example, squares, 6 around a point, is produced by deleting the triangles from the rhombicuboctahedron, creating a hollow polyhedron of genus 7 with triangular holes connecting 18 interior and 18 exterior square faces. An empty cube missing its top and bottom faces becomes an envelope polyhedron, squares, 4 around a point, with a toroidal topology. This definition leads to many interesting finite and infinite multiply connected regular polygon networks, including one infinite network with squares, 14 around a point, and another with triangles, 18 around a point. These are introduced just over 50 years after my related paper on infinite spongelike pseudopolyhedra in American Mathematical Monthly (Gott, 1967).

Tables

Table 1. Table 1: Nomenclature follows that of WBK: For example squares–6 around a point appears as 4 6 superscript 4 6 4^{6} . This has a Schläfli symbol {4,6}. The symbols ⋄ ⋄ \diamond and ∘ \circ in that box indicate that both finite and infinite envelope polyhedra of this type exist; ∙ ∙ \bullet indicates that an infinite pseudopolyhedron of this type exists. N 𝑁 N represents all integers N ≥ 3 𝑁 3 N\geq 3 .

$N^2$ $\diamond$	$N^3$	$N^4$	$N^5$	$N^6$	$N^7$	$N^8$	$N^9$	$N^{10}$	$N^{11}$
${12}^2$ $\diamond$	${12}^3$	${12}^4$ $\circ$	${12}^5$	${12}^6$	${12}^7$	${12}^8$	${12}^9$	${12}^{10}$	${12}^{11}$
${11}^2$ $\diamond$	${11}^3$	${11}^4$	${11}^5$	${11}^6$	${11}^7$	${11}^8$	${11}^9$	${11}^{10}$	${11}^{11}$
${10}^2$ $\diamond$	${10}^3$	${10}^4$ $\diamond$	${10}^5$	${10}^6$	${10}^7$	${10}^8$	${10}^9$	${10}^{10}$	${10}^{11}$
$9^2$ $\diamond$	$9^3$	$9^4$	$9^5$	$9^6$	$9^7$	$9^8$	$9^9$	$9^{10}$	$9^{11}$
$8^2$ $\diamond$	$8^3$	$8^4$ $\diamond \circ$	$8^5$	$8^6$	$8^7$	$8^8$	$8^9$	$8^{10}$	$8^{11}$
$7^2$ $\diamond$	$7^3$	$7^4$	$7^5$	$7^6$	$7^7$	$7^8$	$7^9$	$7^{10}$	$7^{11}$
$6^2$ $\diamond$	$6^3$ $\bullet$	$6^4$ $\diamond \bullet$	$6^5$	$6^6$ $\bullet$	$6^7$	$6^8\circ$	$6^9$	$6^{10}$	$6^{11}$
$5^2$ $\diamond$	$5^3$ $+$	$5^4$	$5^5$ $\bullet$	$5^6$	$5^7$	$5^8$	$5^9$	$5^{10}$	$5^{11}$
$4^2$ $\diamond$	$4^3$ $+$	$4^4$ $\diamond \bullet \circ$	$4^5$ $\bullet$	$4^6$ $\diamond \bullet \circ$	$4^7$	$4^8$	$4^9$	$4^{10}$ $\circ$	$4^{11}$
$3^2$ $\diamond$	$3^3$ $+$	$3^4$ $+$	$3^5$ $+$	$3^6$ $\diamond \bullet \circ$	$3^7$ $\bullet$	$3^8$ $\diamond \bullet$	$3^9$ $\bullet$	$3^{10}$ $\bullet \circ$	$3^{11}$

Full text

geometry, polyhedra

1 Introduction—Pseudopolyhedra

My work on envelope polyhedra grows directly out of my earlier work on pseudopolyhedra, which I will describe first. This was my high school science fair project which won 1st Place in mathematics at the (May, 1965) National Science Fair International (now the Intel International Science and Engineering Fair). A picture of this project appears in my book The Cosmic Web (2016), along with a description. These were infinite spongelike polyhedra whose polygons were all regular, whose vertices were congruent, and two polygons always shared only one edge. All had a sum of polygon angles around a vertex $>360^{\circ}$, and corresponded to surfaces with negative curvature.

Positively curved surfaces like the sphere can be approximated by regular polyhedra where the sum of angles at a vertex is $<360^{\circ}$. The cube is a very rough approximation to a sphere. A cube is made up of squares meeting 3 around a point. At the corner of a cube, three square faces meet at a vertex, and each square has a $90^{\circ}$ angle at its corner, making the sum of the angles around a point at the corner $3\times 90^{\circ}$, or $270^{\circ}$. This is $90^{\circ}$ less than $360^{\circ}$. All 5 regular Platonic polyhedra: tetrahedron (triangles–3 around a point), octahedron (triangles–4 around a point), icosahedron (triangles–5 around a point), cube (squares–3 around a point), and dodecahedron (pentagons–3 around a point) have a sum of angles around a vertex of $<360^{\circ}$. A plane can be tiled by squares–4 around a point to make a checkerboard pattern, where the sum of angles around each point is therefore $4\times 90^{\circ}$, or $360^{\circ}$ degrees—this is a surface of zero curvature. Johannes Kepler recognized that the three long-known planar networks, triangles–6 around a point, squares–4 around a point, and hexagons–3 around a point, were also regular polyhedra but with an infinite number of faces. (Kepler in addition allowed regular star polygons crossing through each other to count, creating regular stellated polyhedra. With slightly more lenient rules you can find additional interesting structures.)

In early 1965, found 7 regular spongelike polygon networks having a sum of angles around a point $>360^{\circ}$, with an infinite number of faces and an infinite number of holes: triangles–8 around a point, triangles–10 around a point, squares–5 around a point, squares–6 around a point, pentagons–5 around a point, hexagons–4 around a point, and hexagons–6 around a point. I called these pseudopolyhedrons (Fig. 1), after the pseudosphere which is a surface of constant negative curvature encountered in the non-Euclidean geometry of Nikolai Lobachevsky and Janos Bolyai.

After I got to Harvard, math instructor Tom Banchoff, later famous for his professional friendship with Salvador Dali, encouraged me to submit a paper on pseudopolyhedra to the American Mathematical Monthly, which I did. The referee’s report was quite positive, but noted that three of my polygon networks had been discovered before. The reference was to a paper, which I had never heard of, by H. S. M. Coxeter (1937). It described how the first of these figures—squares–6 around a point—was found in 1926 by John Petrie, who also discovered hexagons–4 around a point (the one I found first when I was 18). Petrie gets credit for discovering this entire class of figures. Coxeter himself discovered hexagons–6 around a point. Petrie and Coxeter did this work in 1926 when both were 19. In addition to demanding that the configuration of polygons around each vertex be identical (as I did), their criteria for regularity also demanded that the dihedral angles between all adjacent pairs of faces also be equal and the vertex have rotational symmetry. With those conditions they were able to prove that the three examples they found were the only regular figures of this type. They called them regular skew polyhedra. I was happy to add the Coxeter/Petrie reference. My paper was still publishable, the referee said, because I had discovered four new pseudopolyhedrons. I still required the configuration of polygons around each vertex to be identical, but allowed the dihedral angles between adjacent faces to vary. Some had some dihedral angles of $180^{\circ}$, for example. I rediscovered all three structures discovered by Petrie and Coxeter as well as finding four new ones allowed by my more lenient rules. My paper appeared in print in 1967. It was my first published scientific paper. My pseudopolyhedrons were, along with the Coxeter/Petrie regular skew polyhedra, were included in the Penguin Dictionary of Curious and Interesting Geometry (Wells, 1991). In my paper I used an anglicized plural form, pseudopolyhedrons, thinking this would be modern, but the Latin plural has remained healthy over the past half century, so I will surrender and here refer to them in the plural as pseudopolyhedra. These are today sometimes also called infinite polyhedra, spongelike polyhedra, or infinite skew polyhedra. These have an infinite number of faces and therefore belong to the set of apeirohedra (along with the regular planar networks and cylindrical networks).

When Siobhan Roberts wrote her definitive biography of Coxeter, King of Infinite Space, in 2006, I was happy to contribute my story of the astronomical applications these figures later had in understanding the distribution of galaxies in space. In the early 1980’s there were two schools of thought about how galaxies were clustered in space. The American school, headed by Jim Peebles, maintained that there was a hierarchical pattern of clusters of galaxies floating in a low density void, like isolated meatballs in a low density soup. The Soviet school, headed by Yakov Zeldovich, maintained that galaxies formed on a giant honeycomb punctuated by isolated voids. I realized neither model was consistent with the new theory of inflation, which showed that fluctuations in density in the early universe were produced by random quantum fluctuations. In this case, the regions of above-average and below-average density should have the same topology. This could occur with a spongelike topology, which divided space into two equivalent parts. I knew this because it occurred in some of my pseudopolyhedra (triangles–10 around a point, pentagons-5 around a point, squares–6 around a point, hexagons–4 around a point, hexagons–6 around a point). We showed (Gott, Melott, and Dickinson 1986) that the spongelike initial conditions required by inflation would grow under the influence of gravity into a spongelike structure of galaxy clusters connected by filaments of galaxies, with low density voids connected by tunnels, a structure now known as the cosmic web and verified by many surveys. I tell the story of this discovery in my book The Cosmic Web (Gott, 2016). I gave an invited lecture on this at the Royal Institution which can be seen on YouTube.

Additional regular pseudopolyhedra have been discovered by crystallographer A. F. Wells: including triangles–7 around a point; triangles–9 around a point; and triangles–12 around a point. These are illustrated in Wells’s 1969 paper and his 1977 book, Three Dimensional Nets and Polyhedra. Wells, like me, did not demand equal dihedral angles between adjacent faces. All are spongelike with an infinite number of faces and an infinite number of holes. Melinda Green rediscovered my pentagons–5 around a point, and has illustrated many pseudopolyhedra (see references). Avraham Wachmann, Michael Burt, and Menachem Kleinman (abbreviated WBK) have discovered many semi-regular spongelike polyhedra, composed of polygons of more than one kind, for example, two squares and two hexagons around each point. (But they failed to find pentagons–5 around a point.) These are to the Petrie/Coxeter/Gott/Wells pseudopolyhedra as the Archimedean polyhedra are to the 5 classic Platonic polyhedra and are illustrated in their 1974 book Infinite Polyhedra. WBK also allow different dihedral angles between faces as I and Wells did, and in addition allowed networks to contain pairs of mirror vertices, ones where the arrangement of polygons was congruent only under mirror reflection. Envelope polyhedra containing such mirror vertices will be discussed in the second half of the Appendix.

2 Finding Envelope Polyhedra—Dihedra

I thought of envelope polyhedra in 1991, while visiting Aspen, Colorado to attend a seminar on cosmology, to talk about my two-moving cosmic string solution in general relativity which allowed time travel to the past. I was thinking about the classic regular polyhedra as approximations to a sphere. For this reason, there are polyhedral maps of the Earth. Perhaps the most famous and successful is the Gnomonic Cahill Butterfly map. It maps the Earth onto a regular octahedron. Then one unfolds the 8 triangular faces in a butterfly pattern sitting on a plane. It shows relatively low distortion but has a number of “interruptions.” Another successful polyhedron map was invented by the famous architect Buckminster Fuller, the inventor of the geodesic dome. Fuller mapped the Earth onto a 20-sided icosahedron. He unfolded the 20 triangular faces to sit on the plane. I remembered that there was also a conformal projection invented by Emile Guyou in 1887, which maps the two hemispheres of the Earth onto two squares sitting side by side. If one folds the two squares together as one folds a billfold closed, and seals them one would create an envelope with the Western Hemisphere mapped on the front and the Eastern Hemisphere on the back. This envelope has two square faces taped together along their edges back to back. I realized this is also a polyhedron. It has 2 square faces (the front and back of the envelope), 4 edges (which form the edges of the envelope) and 4 vertices (which form the four corners of the envelope. This obeys Euler’s rules for convex polyhedra, namely that the number of faces minus the number of edges plus the number of vertices equals 2: $F-E+V=2$. This envelope has $F=2$, $E=4$, $V=4$, so $2-4+4=2$. It also obeys Descartes’ rule that the sum of the angle deficits in a convex polyhedron must be $720^{\circ}$. Only 2 faces come together at a vertex: the front of the envelope and the back. This is a polyhedron we would designate as squares–2 around a point. This has a Schläfli symbol {4,2} and in the WBK nomenclature would be designated $4^{2}$. Imagine an ant tethered to a vertex (one of the four corners) with a tiny string. It stays at a constant distance from the vertex as it circles it. An ant crawling around this vertex would traverse an angle of $90^{\circ}$ on the front square, then go over an edge and start crawling on the back square through another $90^{\circ}$. So the total angle at the vertex is $90^{\circ}+90^{\circ}=180^{\circ}$. This is $180^{\circ}$ less than we would get circling a point on a plane which is $360^{\circ}$, giving an angle deficit for this vertex of $180^{\circ}$. There are 4 such vertices or corners, and so the total angle deficit is $4\times 180^{\circ}=720^{\circ}$, just as Descartes would have figured. This is of course a polyhedron with zero volume, which is why the ancients did not count it. In general, one has envelope polyhedra which are $N$-gons–2 around a point, for all $N\geq 3$. Each has 2 Faces: triangles–2 around a point, squares–2 around a point, pentagons–2 around a point, hexagons–2 around a point, and so forth.

Years later, I found that this was just a rediscovery on my part of dihedra which have already been accepted as polyhedra for some time. (Coxeter, 1937) mentioned dihedra, for example (but did not include them in his lists of regular polyhedra). They all have the topology of a sphere. (One can therefore make a conformal map of the Earth on two hexagons: one covering the northern hemisphere, one covering the southern hemisphere. This answers in the affirmative the gamer’s perennial question: can the sphere be tessellated with identical geodesic hexagons? Yes, with two. Each has six geodesic sides, six $60^{\circ}$ geodesic arcs, along the equator.) There are an infinite number of envelope polyhedra with 2 faces, $N$ edges, and $N$ vertices. These all satisfy the $F-E+V=2$ rule for convex polyhedra, in the most transparent way possible. They also satisfy the Coxeter-Petrie condition that all dihedral angles be equal (in this case $0^{\circ}$) and that the vertex figure should have rotational symmetry (in this case $n=2$). These are not new. But they are just a subset of the larger class of envelope polyhedra that are the subject of this paper. From my work on pseudopolyhedra I already knew in 1991 that if one allowed dihedral angles of $0^{\circ}$, there would be many new envelope polyhedra of zero and negative curvature, both finite and infinite with $360^{\circ}$ around a vertex and $>360^{\circ}$ around a vertex. And this would make for many additional interesting structures. I started adding these to my list of envelope polyhedra. If dihedra can have dihedral angles between their two faces of $0^{\circ}$, then this should be allowed for dihedral angles in general.

3 A Wealth of Envelope Polyhedra

I was used to polyhedra approximating surfaces that divided space into two regions: the inside and the outside in the case of finite polyhedra, into two regions in the case of the plane tessellations: above the plane and below the plane, and into two spongelike interlocking regions in the case of the pseudopolyhedra. Envelope polyhedra, do not divide space into two parts. They have some dihedral angles (angles between two faces) that are $0^{\circ}$. But they do represent a surface an ant could crawl over. I don’t allow the surface to cross itself (I am not considering starred polyhedra). A regular envelope polyhedron has faces that are regular polygons, and the arrangement of polygons touching each vertex must be identical. Some of the angles between adjacent faces in envelope polyhedra will be $0^{\circ}$—those faces will be back to back as in an envelope. This leads to a wealth of forms.

4 Finite Envelope Polyhedra with Holes

See Fig. 2, squares–6 around a point. This and following photographs are stereo pairs with the left-eye view on the left and the right-eye view on the right. Place your nose on the page and the left eye view will be in front of your left eye and the right eye view will be in front of your right eye. You will see a blurry 3D view, slowly back away from the page and the fused central image will come into clear focus, with side images to the left and right. These may also be viewed with a standard stereograph viewer. The models in Figs. 2–5 are made from Polydron plastic polygons (from polydron.com), ignore their slightly serrated edges which allow them to be hinged together to make polyhedra.

Squares–6 around a point, as illustrated in Fig. 2 starts with a rhombicuboctahedron with 18 square faces and 8 triangular faces. Remove the triangular faces to create triangular holes into the interior. This leaves an envelope polyhedron that is hollow with 18 exterior square faces and 18 interior square faces. The polygonal faces thus partially envelop a hollow void. Thus, the name envelope polyhedron seems appropriate here too. Each vertex connects 3 exterior and 3 interior squares. An ant crawling around a vertex would visit all 6 of these squares. [The pseudorhombicuboctahedron, a Johnson polyhedron, where the top in Fig. 2 is twisted relative to the bottom by $45^{\circ}$ provides an alternate way to start the construction and gives an alternate form (2)]. All vertices are congruent with a sum of angles of $6\times 90^{\circ}>360^{\circ}$ around each, so this approximates a negatively curved surface. It has 36 faces (18 exterior plus 18 interior faces), $6\times 4=24$ interior edges, $6\times 4=24$ exterior edges, and $8\times 3=24$ edges on the 8 triangular holes connecting the interior and exterior for a total of 72 edges, as well as $8\times 3=24$ vertices.

Note that just as there are pairs of faces (exterior and interior) that are back to back, there are also pairs of edges that are back to back. But there are no vertices back to back. Vertices connect the exterior and interior faces. Cut one triangular hole in the rhombicuboctahedron and it looks like a bowl with a small mouth which can be distorted into a disk and fattened into a sphere. Each of the additional 7 triangular holes you cut creates a doughnut hole and creates a handle on the sphere. So the genus of this figure (number of doughnut holes) is 7. It obeys the rule $F-E+V=2\times(1-\mbox{genus})=36-72+24=-12$. This gives it a genus of 7 and a topology equal to that of a sphere with 7 handles. It has negative curvature and so could be considered a finite envelope pseudopolyhedron. Note that the faces around each vertex must create a single surface around each vertex that an ant tethered to the vertex could visit. Thus two cubes touching at a point would not be considered squares–6 around a point because an ant tethered to the vertex would circle the vertex visiting 3 squares on one cube and complete her circuit and return to where she started visiting only those 3 squares and never visit the other cube. Likewise, a cube with a square fin attached at an edge would not count as part of a squares–5 around a point structure because although one ant tethered at the vertex would visit 5 outside square faces, another ant tethered on the inside of the cube would visit only 3 squares, again creating multiple surfaces with different angle deficits at a single vertex. We are not allowing such multiple surfaces. The envelope polyhedra is one continuous two-dimensional surface.

Here is another interesting example: octagons–4 around a point. Get an empty cubical cardboard box. Cut off each of its corners with a saw. This will make a truncated cube. Each of the 6 square faces of the cube will have its corners cut off, becoming an octagon. Cutting the corners of the cubical box will create 8 triangular holes where the 8 corners of the box used to be. This has 12 octagonal faces (6 inside, 6 outside), 48 edges (12 outside edges, 12 inside edges, and 24 edges on the 8 triangular holes connecting the inside and outside), and 24 vertices (3 on each of the triangular holes). $F-E+V=12-48+24=-12$. That’s $2\times(1-\mbox{genus})$ as predicted. This has a genus of 7. It has the same number of triangular openings as the squares–6 around a point envelope polyhedron in Fig. 2 so it also has a genus of 7. Some dihedral angles between adjacent octagons are $90^{\circ}$ (when both are either outside or both inside) and some are $0^{\circ}$ (when one is inside and one is outside).

An ant tethered to one of the vertices will circle it by crawling over two outside-facing octagons and two inside-facing octagons as it goes around the outside and then through the triangular hole to visit the inside, making 4 octagons around each vertex. The interior angle in an octagon is $135^{\circ}$. So the total angle the ant traverses circling the vertex is $4\times 135^{\circ}=540^{\circ}>360^{\circ}$. This is a negatively curved surface with the curvature all concentrated at the vertices. It even looks like a saddle-shaped surface. One can imagine an ant sitting on one of the vertices and draping its little hind legs over each side, one inside and one outside, like he was riding a horse. This negatively curved finite polyhedron, like squares–6 around a point in Fig. 2, is multiply connected. But all of my original multiply connected pseudopolyhedra were infinite.

Hexagons–4 around a point. Take an octahedron shaped box and saw off its corners. You will be left with a truncated octahedron shape made of hexagons with 6 square holes. See Fig. 3. An ant circling a vertex will traverse two hexagons on the exterior, before visiting another two hexagons on the interior to make hexagons–4 around a point.

Take a triangular prism and remove the top and bottom triangles, to create the envelope polyhedron squares–4 around a point. Take a regular octahedron and remove two opposite faces, to create the envelope polyhedron triangles–6 around a point. Take a cube and remove two opposite faces to create the envelope polyhedron squares–4 around a point. These have the topology of a doughnut, and $360^{\circ}$ and zero curvature at the vertices. Take an icosahedron box and remove four triangles to create the envelope polyhedron triangles–8 around a point. These are illustrated in Fig. 4.

5 Infinite Envelope Polyhedra

Squares–10 around a point. The hollow rhombicuboctahedron envelope polyhedron shown in Fig. 2 fits perfectly in an imaginary cube. Stack cubes like this to fill space with the rhombicuboctahedra glued back to back at a single square (red in Fig. 2) in a repeating pattern. Each vertex connects two of these rhombicuboctahedra, and as an ant circles the vertex, it travels on an exterior square face of the first, then three interior faces of the first, then another exterior face of the first, an exterior face of the second, three interior faces of the second, and another exterior face of the second—giving 10 squares around the vertex.

Octagons–8 around a point. Take an infinite number of the truncated cubical cardboard boxes (octagons–4 around a point) discussed above and fill all of space with them by stacking them like cubical boxes in a warehouse. We will be gluing the boxes together. Two adjacent boxes will be glued together on their outside octagons. These will disappear from the surface. All that will be left are the inside octagons of each cube. Four boxes (numbered 1, 2, 3, 4) will fit together at their edges. A vertex will connect 8 octagons. The vertex is at the end of an edge that connects two inside octagons of cube 1. Cubes 2, 3, and 4 also come together at this vertex. An ant tethered to the vertex will circle the vertex by first traversing the two inside octagons of cube 1, then it will cross an edge of a triangular hole, to enter cube 2 and traverse two of its inside octagons, then enter cube 3 traversing two of its interior octagons, before entering cube 4 and traversing two of its interior octagons as she returns to the place she started. This makes a complicated saddle shaped surface that goes up and down, up and down, up and down, and up and down. It is a saddle with a four-fold symmetry, like a +. It is a saddle a horse could sit on, hanging its legs down, one each into boxes 1, 2, 3, 4. It is cradled by the four triangular holes. The angle around each vertex is $8\times 135^{\circ}=1080^{\circ}$. This is an infinite polyhedron: it has an infinite number of faces, an infinite number of cubical cells with triangular holes. A fly could fly through the whole structure, visiting any cell he wanted.

(Note: Pellicer and Schulte (2009) and Schulte and Weiss (2016) considered skeletal polyhedra. These had skeletal faces consisting only of edges, based on the idea of Grünbaum (1994) that one could consider polyhedra consisting of polygons with edges only without membranes spanning them to form a surface. These form additional polyhedra, having skew polygons, and even infinite helical polygons. In these polyhedra skeletal polygons come together two at an edge. But Pellicer and Shulte (2009) also consider regular polygonal complexes where more than two skeletal faces meet at an edge. If we removed all the faces from octagons–8 around a point, we would have 4 skeletal octagons meeting at an edge. But to be clear, we are not considering such skeletal structures here—envelope polyhedra all have surfaces. In octagons–8 around a point we have 8 octagons and 8 edges meeting at a point. In the surface, two octagons always share an edge. Edges can be placed back to back, of course, in the structure, just as polygons can be placed back to back. These constitute, in the surface, separate polygons and separate edges which an ant would encounter circling the vertex.)

Triangles–18 around a point. (1) Start with the pseudopolyhedron triangles–10 around a point shown in Fig. 1, then remove triangles in the planes of triangles that are connected to other triangles in the same plane (with dihedral angles of $180^{\circ}$) along each of their three edges. This creates triangular holes in the planes of triangles. Each vertex used to have 10 triangles around it on one convoluted surface; take one triangle away and that leaves 9 triangles, but the hole allows the ant to traverse to the opposite side of the plane and visit the 9 opposite sides of these triangles to make a total of 18 around a point. The angle around each vertex is $18\times 60^{\circ}=1080^{\circ}$. There are left- and right-handed versions of this.

Squares–12 around a point. Make a single layer of triangular prisms that cover a plane. Delete the triangular tops and bottoms of the prisms, leaving a triangular grid of cells. Six triangular prisms meet at a vertex, and each has two interior squares that touch the vertex, giving 12 squares around a point.

Squares–14 around a point. Take the previous structure and add east-west fins of back-to-back squares above this single layer of cells in a dashed pattern. See Fig. 5. This way each vertex has the 12 squares around a point that it had before plus the two squares in the fin. These fins are then connected to another plane of squares–12 around a point triangular cell layer above it. Layers of fins and cells alternate vertically forever. Each of the 14 squares has a vertex angle of $90^{\circ}$, so the total angle around a vertex is $(14\times 90^{\circ})=1260^{\circ}$, a trick for an Olympic snowboarder. That is $3\nicefrac{{1}}{{2}}$ rotations. That gives it an angle excess of $900^{\circ}$ (above $360^{\circ}$), the most found for any polyhedron so far.

There are many other envelope polyhedra. The Appendix gives a list of those I have found so far and a summary of how these additional ones are constructed. In general they are found by: (a) deleting faces from regular polyhedra to create polygonal holes (as illustrated by some of the examples in Fig. 4); (b) deleting polygons from semi-regular polyhedra containing faces of different types (like the triangular prism in Fig. 4) and the Archimedean polyhedra to create polygonal holes; (c) deleting polygons from regular and semi-regular planar networks and cylindrical networks; (d) deleting polygons from regular pseudopolyhedra like those shown in Fig. 1, and from semi-regular pseudopolyhedra like those shown in WBK, and (e) Creating planar networks of open-ended prisms with fins (like Fig. 5) and without fins, based on regular and semi-regular (Archimedean) planar networks.

6 Summary

Polyhedra fall into three groups depending on whether the sum of face angles around a vertex are $<360^{\circ}$, $=360^{\circ}$, or $>360^{\circ}$ (i.e., whether the curvature at each vertex is positive, zero, or negative. Dihedral angles do not all have to be identical. But the arrangement of polygons around each vertex must be identical. Envelope Polyhedra, which we introduce here, just have some dihedral angles which are $0^{\circ}$. Using the nomenclature of WBK, where $N$-gons–$M$ around a point with Schläfli symbol $\{N,M\}$ are designated $N^{M}$, we find the following structures so far:

[TABLE]

[TABLE]

[TABLE]

These are summarized in Table 1.

A number of these have several geometrical forms as described in the Appendix. Open symbols refer to envelope polyhedra which have some dihedral angles equal to $0^{\circ}$: the symbol $\diamond$ indicates a finite number of sides and refers to polyhedra with a finite number of faces, $\circ$ is a circle and refers to envelope polyhedra with an infinite number of faces. Closed symbols refer to Polyhedra: $+$ (which are finite), and $\bullet$ which are either infinite Planar or Cylindrical Networks, or infinite Pseudopolyhedra which have no dihedral angles equal to $0^{\circ}$.

7 Envelope Polyhedra with Mirror Vertices

Below, and in the Appendix, we also list additional infinite structures ($3^{12}$, $4^{6}$, $4^{8}$, $4^{10}$, $6^{6}$) with mirror vertices. All the previous polyhedra we have discussed have vertices that are identical. But one may obtain a considerable number of additional structures by allowing vertices and their mirror images to be considered as congruent. One example starts with the pseudopolyhedron hexagons–4 around a point shown in Fig. 1. Now remove the same set of two opposite faces on each truncated octahedron box. You get a repeating pattern of hexagons–6 around a point: $6^{6}$, but with pairs of mirror vertices whose vertex figures are not identical (except under mirror reflection) which is shown in Fig. 6. Because it includes the mirror image vertices, this structure $6^{6}$ is not included in Table 1 which includes only structures with vertices that are identical.

Also starting with squares–6 around a point in Fig. 1, on can delete horizontal squares that are tops and bottoms of cubes open in the east-west direction to get squares–10 around a point, which also includes mirror vertices (with mirror symmetric vertex figures).

Many additional such structures with mirror vertex figures can be obtained by deleting polygons from structures in WBK. I should mention that WBK consider mirror image vertices as congruent and many of their semi-regular pseudopolyhedra have vertices that are congruent only under mirror reflection. In plane geometry we are used to considering a triangle and its mirror image as congruent, but in three dimensions they are actually identical since they can be superimposed by rotation in the third dimension. Chemists know that in 3D, glucose comes in distinct mirror image right-handed (D-glucose = dextrose) and left-handed (L-glucose) forms. Dextrose can be digested by the body to produce energy while L-glucose cannot. In 3D solid Euclidean geometry mirror image vertices cannot be superimposed. So some might regard those with mirror image vertices as less regular. Therefore I have separated the envelope polyhedra into two classes, those whose vertices have identical vertex figures, listed in the previous sections and the first part of the Appendix, and those containing mirror image vertex figures listed in this section and in a separate section of the Appendix.

No doubt there are more envelope polyhedra yet to be discovered. I leave that as a challenge for interested readers—to add to the list. Surprisingly, pseudopolyhedra eventually had an application in astrophysics. Will any of these envelope polyhedra eventually find applications, in chemistry, biology, art, or architecture perhaps? Time will tell.

I thank Robert J. Vanderbei for 3D photography of the models, and Wesley N. Colley for help in preparation of the manuscript.

Appendix A Additional Envelope Polyhedra

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Triangles–6 around a point. (1) Start with an $N$-gon antiprism ($N\geq 3$) and remove the top and bottom $N$-gons. This leaves a band of triangles connecting them, like the corrugated circumference of a drum without the top and bottom. Each vertex links 3 outside triangles with 3 interior triangles. These have $2N$ faces with the topology of a doughnut. The $N=5$ case is an icosahedron with its top and bottom caps of 5 triangles each deleted, leaving a ring of 10 exterior triangles and 10 interior triangles. (2) A straight filmstrip of triangles that is infinitely long. They alternate between triangles with points up and points down. Each vertex links 3 triangles on one side with 3 triangles on the other side. These vertices have zero curvature and $360^{\circ}$ around a vertex. (3) Such an infinite filmstrip may be twisted to form a helix, there are left- and right-handed versions of these.

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Triangles–8 around a point. (1) Start with a hollow snub cube, one of the Archimedean polyhedra where 4 triangles and one square meet at each vertex, and delete the square faces. This leaves 32 exterior triangular faces, 32 interior triangular faces, and 6 square holes connecting the interior and exterior. There are left and right handed versions of this. (2) Start with a hollow snub dodecahedron, another Archimedean polyhedron, and delete the pentagonal faces. This leaves 80 exterior triangular faces, 80 interior triangular faces, and 12 pentagonal holes connecting the interior and exterior. This has the topology of a sphere with 11 handles. There are both left and right handed versions of this.

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Triangles–12 around a point. (1) Start with squares–8 around a point, a single plane layer of cubical boxes with open ends. This is a checkerboard of cubical boxes with open tops and bottoms. The vertical sides of the boxes are oriented either east-west or north-south. All the boxes have vertical edges. Distort this by leaning all the vertical edges in the northwest direction until they are at $45^{\circ}$ to the vertical, like the diagonal of a vertical octahedron. The boxes now have rhombus sides that can be tiled by two equilateral triangles. Each cube is now leaning, with 4 rhombus interior sides made up of two equilateral triangles each. The rhombus sides of each open-ended box have angles of $120^{\circ}$ (where two triangles meet) and $60^{\circ}$ where there is one triangle. An ant circling a vertex on top of the layer would cover 2 interior triangles in the southeast box, 3 interior triangles in the northeast box, 4 interior triangles in the northwest box, and 3 interior triangles in the southwest box before returning to where it started, making 12 triangles around a point. Flip the layer of boxes over to see that the vertices on the bottom of the layer are identical to those on the top. The total angle around the vertex is $720^{\circ}$, just as in the squares–8 around a point structure we started with. Both are topologically equivalent. Both structures are flexible.

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Triangles–14 around a point. (1) Start with the Gott (1967) pseudopolyhedron triangles–10 around a point, and pull out for consideration two adjacent planes of triangles and the tunnels that connect them. Each vertex in the plane of triangles will have a triangular hole in it where the column would have been to connect it to the next plane of triangles in the triangles–10 around a point configuration. The two planes of triangles, with their triangular holes are connected to each other by tunnels that are antiprisms (octahedrons with two missing [top and bottom] opposite faces). For example an ant circling each vertex on the top plane would traverse 1 triangle on the bottom of that plane of triangles, 3 triangles that are part of the exterior of a tunnel, 3 more triangles on the bottom of the top plane, then through the open triangular hole onto the top of the top plane, where it would traverse 3 triangles, on the top of the plane, 3 triangles in the interior of the tunnel, 1 more triangle on the top of the plane, and it is back where it started. That gives 14 triangles around a point. There are left and right handed versions of this. (2) Start with Wells (1977) pseudopolyhedron $3^{8}$ made of snub cubes connected at their missing square faces in a cubic array. Left and right handed snub cubes alternate like black and white cubes in a 3D chessboard. Left handed ones join to right handed ones at the vertices of the missing squares. Now delete from the array of triangles those triangles on left-handed snub cubes that touch three different missing squares. That leaves left-handed snub cubes with 8 triangular holes. Each vertex lies on the corner of a square tunnel connecting a left and right handed snub cube. It is missing one triangle out of 8 giving its ant a triangular hole to allow it to visit both sides of the 7 remaining triangles to give a surface with 14 triangles around a point. There are left and right-handed versions of this structure (depending on whether one takes the triangles out of the left-handed, or right-handed snub cubes. But within each case the vertices are all identical—either all left-handed or all right-handed.)

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Squares–4 around a point. (1) Start with an $N$-gon prism (for $N>2$) with an $N$-gon top, $N$ square sides, and an $N$-gon bottom, then remove the $N$-gon top and bottom. This leaves an envelope polyhedron with 2N faces, N outside square faces and N inside square faces. There are an infinite number of these which all have the topology of a doughnut-like the sides of a drum with the top and bottom missing. Each vertex joins two outside squares and two inside squares. Thus, the polyhedron has an angle deficit of $0^{\circ}$. For $N=4$, this is a cube with the top and bottom missing. (2) There is also an arrangement where one has two strips of squares back to back extending in a straight line like a filmstrip. This has the topology of an infinite cylinder. It has zero curvature. There are also arrangements where the squares line up back to back in a zig-zag patterns, like a filmstrip with folds in it back and forth.

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Squares–6 around a point. (1) Start with just the top plane of Gott’s (1967) pseudopolyhedron squares–5 around a point ($4^{5}$). This is a single checkerboard plane punctuated by holes so that three squares on the top surface of the plane surround each vertex, while three more squares surround it on the bottom surface of the plane. As the ant circles the vertex it visits 6 squares giving $4^{6}$—or the holes can be in a staggered arrangement as in the front plane of squares in $4^{5}$ in WBK at the bottom of page 2, or a plane of squares in $4^{5}$ in WBK at top right of page 16. (2) A single honeycomb layer of hexagonal prisms tiling a planar layer, with the tops and bottoms removed. An ant circles a vertex visiting the two interior squares of each of three hexagonal cells making squares—6 around a point. This is like Fig. 5 without the fins and with hexagonal cells instead. (3) Squares and octagons ($4\times 8^{2}$) tile the plane, to use the nomenclature of WBK. Convert them into a layer of octagonal and square prisms (cubes) without their tops and bottoms. Each vertex is surrounded by two interior squares of the cube and two interior squares of each of two octagonal prisms: 6 squares around a point. (4) Triangles and dodecagons ($3\times 12^{2}$) tile the plane. Similarly, make them into triangular and dodecagonal prisms without their tops and bottoms. This makes a single layer with 6 squares around a point.

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Squares–8 around a point. (1) Start with the semi-regular pseudopolyhedron shown by Wachman, Burt, and Kleinmann (WBK) on page 9 at the top. It looks like triangular prisms joined by cubes to form a surface with 4 squares and one triangle around each vertex (designated $4^{4}\times 3$). Simply remove the triangles, to make an infinite envelope polyhedron with squares–8 around a point. With triangles gone there are 4 squares around each point, with two sides each which the ant traverses circling the vertex. (2) On the same page at the bottom are hexagonal prisms joined by cubes in a similar manner ($4^{4}\times 6$), remove the hexagons to get squares–8 around a point. (3) On page 15, is a multilayer structure ($4^{5}$). Instead of having two planes of squares connected by tunnels as in (Gott 1967), this has planes of squares connected by tunnels going down and columns going up to adjoining punctuated planes of squares. Remove selected squares on the planes of squares (those touching at their corners 2 columns and 2 tunnels) to create an envelope polyhedron squares–8 around a point. (4) A similar $4^{5}$ structure on page 16 has a different arrangement of columns and tunnels. Remove selected squares (on the planes of squares, those touching at their corners 2 columns and 1 tunnel on one plane and 1 column and 2 tunnels on the next and repeat) to create an envelope polyhedron squares–8 around a point. (5) Make a single layer of cubes that cover a plane. From the top they look like a checkerboard. Delete the tops and bottoms of the cubes, to leave a checkerboard shaped grid of cells. Four cubes join at a vertex, so an ant circling the vertex, traverses two interior squares of cube 1, then two interior squares of cube 2, then two interior squares of cube 3, and finally two interior squares of cube 4 giving 8 squares around a point. (6) Take the structure $4^{4}\times 6$ on page 40 of WBK. Delete the hexagons. Equivalently, hexagons, triangles and squares tile the plane, remove the hexagons and triangles and connect such planes with triangular prisms (tunnels and columns) with their tops and bottoms off. This gives squares–8 around a point. (7) There is a biplane version of the previous structure where the two planes are connected by triangular tunnels. (8) Take the filmstrip, triangles–6 around a point (2, listed above) and build triangular prisms on top of it. Now delete the triangular faces. This leaves an infinite row of interlaced triangular prisms without their tops and bottoms, giving squares–8 around a point. (9) Take the structure $4^{5}$ on page 50 of WBK, Delete the squares in the planes of squares whose four corners touch two columns and two tunnels. (10) Take the structure $3\times 4^{4}$ on page 71 of WBK and remove the triangles to make squares–8 around a point. This structure has hollow shells like that shown in Fig. 2, missing their red squares joined by cubes missing two opposite squares attaching them where their two missing red squares used to be. (11) Start with the structure $3\times 4^{4}$ on page 78 of WBK, remove the triangles to produce envelope polyhedron $4^{8}$. An ant at each vertex navigates both sides of 4 squares as it circles the vertex. This has a “diamond” structure with a hexagonal ring of squares circling each “carbon” bond as an axis. (12) Start with the Wells (1977) structure $4^{5}$ on page 87 of WBK and delete squares which are at the center of a flat cross shaped pattern. These squares have dihedral angles on all their four edges equal to $180^{\circ}$. (13) Start with the Wells (1977) structure $4^{5}$ on page 88 of WBK and delete the squares that have 4 dihedral angles of $215^{\circ}16^{\prime}$ at their edges, as indicated in WBK (square COB in their diagram). These deleted squares sit on the surfaces of truncated octahedrons, the squares that remain lie on surfaces of hexagonal prisms. (14) Squares and octagons ($4\times 8^{2}$) tile the plane, to use the nomenclature of WBK. Make them into octagonal prisms and square prisms (cubes) without their tops and bottoms. Add fins (as in Fig. 5) to the tops of the octagonal prism sides where two octagonal prisms meet. Then repeat vertically to make alternate layers of prisms and fins. Gives 8 squares around a point. (15) Triangles and dodecagons ($3\times 12^{2}$) also tile the plane. Similarly, make them into a single planar layer of triangular and dodecagonal prisms without their tops and bottoms. Add fins (as in Fig. 5) to each dodecagonal prism side that attaches to another dodecagonal prism. This likewise gives 8 squares around a point. (16) Triangles, squares, and hexagons tile the plane ($3\times 4\times 6\times 4$), turn these into prisms without their tops and bottoms to make a single planar layer. An ant circling a vertex would visit 2 interior squares of each of 4 prisms, giving 8 squares around a point. (17) Triangles and hexagons tile the plane $(3\times 6)^{2}$. Turn them into prisms without their tops and bottoms to make a single planar layer. An ant circling a vertex would visit 2 interior squares of each of 4 prisms, giving 8 squares around a point.

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Squares–10 around a point. (1) Start with the structure $3\times 4^{5}$ on page 99 of WBK and delete the triangles. This takes envelope polyhedra squares–6 around a point (as shown in Fig. 2) and replaces the yellow squares with cubes without ends linking the envelope polyhedron to similar copies of itself. The envelope polyhedron squares–6 around a point shown in Fig. 2 fits in an imaginary cube, the added cubes without ends link it along the imaginary cube’s 12 diagonals to similar envelope polyhedra in nearby cubes. Circling a vertex the ant will traverse an outside red square of Fig. 2, two outside squares of a cube without ends, it will then go through the triangular hole, visit the two inside squares of that same cube without ends, then the inside red square, the inside two squares of another cube without ends, before coming out of the triangular tunnel and traversing the corresponding two outside squares of that cube without ends before returning to where it started: 10 squares around the vertex. (2) Start with a single layer of cubes tiling a plane, remove the tops and bottoms of the cubes, then add fins to every other east-west square (similar to Fig. 5). Then repeat vertically to create alternating layers of planes of open cubes and fins. An ant crawling around each vertex will visit two interior squares of each of 4 cubes joining each other, plus the two sides of one fin, giving squares–10 around a point. (3) Triangles and hexagons tile the plane ($3^{4}\times 6$). Turn these into prisms without their tops and bottoms to make a single layer. Each vertex is surrounded by 5 prisms, each with two interior faces the ant must visit making it squares–10 around a point. (4) Triangles and squares tile the plane ($3^{3}\times 4^{2}$). Turn these into prisms without tops and bottoms to make a single layer. Each vertex is surrounded by 5 prisms, each with two interior faces the ant must visit so it is also squares–10 around a point. (5) Triangles and squares tile the plane ($3^{2}\times 4\times 3\times 4$). Turn these into prisms without their tops and bottoms. Similarly, this makes squares–10 around a point.

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Squares–12 around a point. (1) Triangles and squares tile the plane ($3^{3}\times 4^{2}$). Turn these into prisms without tops and bottoms to make a single layer. Add fins to squares joining two open cubes. Then repeat vertically to make alternating layers of prisms and fins. An ant will traverse two sides of a fin, then interior pairs of faces of 5 prisms as it circles each vertex, giving squares–12 around a point. (2) Triangles and squares tile the plane ($3^{2}\times 4\times 3\times 4$). Turn these into prisms without their tops and bottoms. Add fins above the squares connecting two triangular prisms. Again the ant traverses two sides of the square fin, then interior pairs of faces of 5 prisms as it circles each vertex: squares–12 around a point.

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Hexagons–4 around a point. (1) Start with a hollow tetrahedron and cut its corners off. Cutting the corners off each triangular face creates 4 hexagonal exterior faces and 4 hexagonal interior faces, with 4 triangular holes connecting the interior and exterior. Each vertex connects 2 exterior and 2 interior hexagons. (2) Start with a hollow icosahedron and cut its corners off. This creates 20 exterior hexagonal faces, 20 interior hexagonal faces and 12 pentagonal holes connecting the interior and exterior. (3) Start with a plane tessellation of hexagons. It is possible to remove every third hexagon in such a way that hexagonal holes are created in the plane of hexagons such that every vertex borders one of the hexagonal holes. These holes connect to the other side of the plane, so an ant circling the vertex will visit 2 hexagons on the top of the plane, go through the hole and traverse 2 hexagons on the bottom of the plane before returning to where it started.

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Hexagons–8 around a point. Tessellate the entire three dimensional space with octahedrons and tetrahedrons. This tessellates space into cells with all triangular faces. Now cut the corners off each of the triangular faces turning each of them into regular hexagons. Each edge in the original tessellation is truncated to 1/3 its former length as the corners of the triangular faces are cut off. At each end of one of these truncated original edges is a vertex of the envelope polyhedron. Each original edge is bordered by two octahedral and two tetrahedral volumes, so four back to back hexagons join at a truncated original edge, and intersect at the vertex at the end of that edge. As an ant circles this vertex he will traverse 2 interior hexagons of an octahedral volume, then 2 interior hexagons of a tetrahedral volume, then 2 interior hexagons of an octahedral volume, then 2 interior hexagons of a tetrahedral volume–8 hexagons around a point. This envelope polyhedron is reminiscent of octagons–8 around a point which was based on a tessellation of space by cubes.

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Octagons—4 around a point. (1) Start with the regular skew polyhedron squares–6 around a point, and cut the corners off all the squares, making all of them into octagons. Where two squares formerly met at an edge, two octagons from one side of the surface would meet at a shortened edge, and each vertex at the end of each of these shortened edges will now connect these two octagons with two more from the other side of the surface of the original regular skew polyhedron. (2) Start with a plane tessellation of squares, truncate all the squares to create octagons and leave square holes in the plane. Vertices connect 2 octagons on the top side of the plane with 2 octagons on the bottom side.

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Decagons–4 around a point. Start with a hollow dodecahedron and cut off the corners. This turns each pentagonal face into a decagon with 10 sides. The cut off corners become 20 triangular holes, connecting 12 interior decagons with 12 exterior decagons. This has the topology of a sphere with 19 handles, the most complicated multiply-connected topology of any finite envelope polyhedra.

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Dodecagons–4 around a point. (1) Start with the regular skew polyhedron, hexagons–4 around a point. Cut off the corners of all the hexagons turning them into dodecagons (with 12 sides each). Where two hexagons met on an edge, the edge will now be shortened, and a new vertex will be created at each end point of each shortened original edge. This vertex will now connect two dodecagons from one side of the original regular skew polyhedron with two dodecagons from the other side. That makes dodecagons–4 around a point. (2) Do the same operation starting with the regular skew polyhedron hexagons–6 around a point. (3) Start with the semi-regular pseudopolyhedron $3^{3}\times 12^{2}$ on page 30 of WBK and remove the triangles. This creates an envelope polyhedron dodecagons–4 around a point where a single plane of decagons is punctuated by triangular holes. (Dodecagons and triangles tessellate a plane with one triangle and two dodecagons around each vertex, once the triangles are holes, there are two dodecagons with top and bottom faces left around each vertex to give 4 around each vertex).

Appendix B Additional Envelope Polyhedra with Mirror Vertices

Below, in the WBK nomenclature are additional envelope polyhedra containing mirror image vertices. Page references are from WBK.

$\mbox{\bf{3}}^{\mbox{\scriptsize{\bf{12}}}}$: (1) From $3^{6}\times 6$ page 33, by deleting the hexagons. (2) From $6^{6}\times 6$ page 34, by deleting the hexagons. (3) Start with Wells’s $3^{7}$ shown on page 85 of WBK at bottom right. These are icosahedrons connected to each other by octahedral tunnels in a structure where the icosahedrons resemble carbon atoms in a diamond structure and the four octahedral tunnels originating from each resemble the carbon bonds in the diamond structure. Remove four triangles from each icosahedron, which are opposite the four octahedral tunnels. This creates a $3^{12}$ envelope polyhedron with mirror image vertices where there are three exterior and three interior octahedral tunnel triangles, and three exterior and three interior icosahedral triangles around each vertex with a triangular hole connecting the interior and exterior triangles around each vertex. (4) From $3^{6}\times 4$ on page 95 delete the squares. This leaves snub cubes missing their square faces linked by open-ended octagonal tunnels.

$\mbox{\bf{4}}^{\mbox{\scriptsize{\bf{6}}}}$: (1) From $4^{4}$ page VIII, 1, by deleting a square from each vertex of an infinitely tall cylinder with $2n$ sides ($n\geq 2$) to make a series of holes. There are several ways to do this; see the punctuated planes squares–6 around a point (1) described in the previous section of the Appendix. (2) From $4^{3}\times 8$ page 3 bottom right, by deleting the octagons. (3) From $4^{3}\times 12$ page 10, by deleting the dodecagons. (4) from $4^{3}\times 6$ page 11, by deleting the hexagons. (5) From $3^{3}\times 4^{3}$ page 13, by deleting the triangles to make a ladder made of squares with holes in the side rails and cubic rungs with open ends. This is a vertical stack of cubes with alternately open east-west ends and open north-south ends. (6) From $4^{3}\times 8$ page 19, by deleting the octagons. (7) From $4^{3}\times 12$ on page 41, by deleting the dodecagons. (8) From $3^{3}\times 4^{3}$ page 48, by deleting the triangles. (9) From $3^{3}\times 4^{3}$ on page 61 delete the triangles. (10) Tessellate the plane with dodecagons, squares and hexagons, build dodecagonal prisms, cubes and hexagonal prisms on these. Delete their tops and bottoms. (11) From $4^{3}\times 6$ on page 72 delete the hexagons. (12) From $4^{3}\times 6$ on page 86 delete the hexagons. (13) From $4^{3}\times 8$ on page 89 delete the octagons. (14) From $4^{3}\times 8$ on page 98 delete the octagons. (15) From $(4\times 8)^{2}$ on page 97, deleted the octagons and add a square at each vertex connecting the remaining two squares to make a bent L shaped pattern of 3 squares (where the two legs of the L are bent in opposite directions at the edges of the squares by $45^{\circ}$ each). This gets traversed twice by an ant circling the vertex to give $4^{8}$. The structure consists of octagonal prisms, missing their octagonal tops and bottoms, pasted together at right angles at square sides—octagonal rings of squares meeting at right angles in a three-dimensional structure. (This can also be constructed from squares–10 around a point mentioned in the main body of this paper by deletion of the appropriate squares).

$\mbox{\bf{4}}^{\mbox{\bf{\scriptsize{8}}}}$: (1) From $4^{5}$ page 17, by deleting squares. Bottom right picture shows some squares face on; these represent towers of open cubes seen from the top, eliminate squares from these towers seen edge on in the bottom right picture, so as to leave the squares seen face on in the bottom right picture as fins connecting the other 2 by 2 boxlike structures. (2, 3, 4, & 5) From $4^{5}$ on page 20, delete the horizontal squares in the right hand picture. Or from $4^{5}$ on page 20, instead delete alternate squares in the filmstrips of squares connecting octagonal columns of squares. There are two ways to do this. Or delete the squares on the vertical sides of half of the open cubes connecting the octagonal columns. (6) From $4^{4}\times 6$ on page 37, by deleting the hexagons. (7) From $4^{4}\times 6$ on page 43, by deleting the hexagons. (8, 9, 10, & 11) From $4^{5}$ on page 46, delete the squares seen face-on in the upper right hand picture. Or from $4^{5}$ on page 46 instead delete alternate squares in the filmstrips of squares connecting hexagonal and octagonal columns of squares. There are two ways to do this. Or delete squares on the vertical sides of the open cubes connecting the hexagonal columns in the bottom right figure. (12) From $4^{4}\times 8$ on page 49, by deleting the octagons. (13) From $4^{4}\times 8$ on page 55 delete the octagons. (14) From $4^{4}\times 8$ on page 56 delete the octagons. (15) From $4^{4}\times 6$ on page 57 delete the hexagons. (16) From $4^{4}\times$ 6 on page 58 delete the hexagons. (17) From $4^{4}\times 12$ on page 59 delete the dodecagons. (18) From $3^{3}\times 4^{4}$ on page 60 at left delete the triangles, leaving two parallel layers of checkerboard pattern of just the white squares connected by filmstrips of squares. (19) From $3^{3}\times 4^{4}$ on page 62 delete the triangles. (20, 21, 22) Attach square fins (one per vertex) to the single layer of $4^{6}$ (10) described above to attach to layers above to make a vertical stack of layers. There are three ways to do this, as three double-sided squares meet at a vertex in a single layer, and so we have three different places to attach a fin. (23) Octagons and squares tile the plane. Turn them into prisms and make a planar layer of octagonal prisms and cubes without their tops and bottoms. Add a series of parallel square fins above opposite sides of each cube. Repeat so that these parallel fins connect to a similar layer above with mirror image vertices.

$\mbox{\bf{4}}^{\mbox{\scriptsize{\bf{10}}}}$: (1) From $4^{6}$ on page 45, delete every other square forming the hexagonal rings. (2) From $4^{6}$ on page 53, delete horizontal squares from North-South ties in the lower left figure. (3) Delete similar squares from $4^{6}$ on page 54 at upper right. (4) Delete horizontal squares on North-South tunnels from Petrie $4^{6}$ on page 67, upper right figure. (5) Take squares–8 around a point (8) described above and add fins connecting it to other copies of squares–8 around a point. This can also be obtained from Fig. 5 by deleting the 12 squares at the back of Fig. 5, leaving one row of open ended triangular prisms with fins sticking up. This structure repeats vertically. It may also be viewed as two vertical checkerboards punctuated with holes [squares–6 around a point (1) described above] connected by horizontal east-west zig-zag filmstrips [squares–4 around a point] giving $4^{10}$. The holes in the squares–6 around a point punctured checkerboards jog back and forth half a square from layer to layer. (6) Start with squares–10 around a point (2). This has vertical checkerboards punctuated by square holes, connected by horizontal sequences of squares which are sides of cubes. Now between checkerboards rotate by $90^{\circ}$ to produce vertical sequences of squares that are sides of cubes. We now have punctuated checkerboards connected by alternating horizontal and vertical sequences of squares that are sides of cubes.

$\mbox{\bf{4}}^{\mbox{\scriptsize{\bf{12}}}}$: (1 & 2) Start with squares–10 around a point (3) and add fins to the top of the single layer of open ended triangular and hexagonal prisms tiling a plane. There are two ways to do this. Add vertical fins to alternate vertical square sides of the hexagonal prisms, or add vertical fins to square sides between two triangle prisms where the fin is in the same plane as sides of the two hexagonal prisms that it touches at vertices. Now repeat vertically to create alternate layers of open prisms and fins. These of course have mirror vertices. (3, 4) Start with the plane tessellation $3^{3}\times 4^{2}$ and construct prisms over it. Now take the tops and bottoms off these prisms. These are cubes and triangular prisms. Add square fins to this to connect to the next vertical layer of open prisms. There are two ways to do this that produce mirror vertices. Place the fins in parallel above alternate squares joining a cube and a triangular prism, or place the fins instead above squares connecting triangular prisms

$\mbox{\bf{4}}^{\mbox{\scriptsize{\bf{14}}}}$: In Fig. 5 the repeated fins and rows of squares repeated vertically produce vertical checkerboards punctuated with holes [squares–6 around a point (1)] connected by horizontal zig-zag filmstrips of squares. Now rotate the set of 12 squares at the back of the figure by $90^{\circ}$ in a vertical plane, so that the zig-zag filmstrip of squares stands vertically. We now have a series of punctuated checkerboards connected to adjacent checkerboards by horizontal rows of zig-zag filmstrips of squares–4 around a point on one side and vertical rows of zig-zag filmstrips of squares–4 around a point on the other side. This gives 4 + 6 + 4 = 14 squares around a point: $4^{14}$.

Interestingly, squares, which have 4 sides, seem to produce the greatest variety of structures in 3D, just as carbon, which has four bonds with adjacent atoms, produces the richest chemistry.