Combinatorially twoorbit convex polytopes
Abstract.
Any convex polytope whose combinatorial automorphism group has two orbits on the flags is isomorphic to one whose group of Euclidean symmetries has two orbits on the flags (equivalently, to one whose automorphism group and symmetry group coincide). Hence, a combinatorially twoorbit convex polytope is isomorphic to one of a known finite list, all of which are 3dimensional: the cuboctahedron, icosidodecahedron, rhombic dodecahedron, or rhombic triacontahedron. The same is true of combinatorially twoorbit normal facetoface tilings by convex polytopes.
Key words and phrases:
Twoorbit, convex polytopes, tilings, halfregular, quasiregular2010 Mathematics Subject Classification:
Primary 52B15; Secondary 51M20, 51F15, 52C22twoorbit.bib
1. Introduction
Regular polytopes are those whose symmetry groups act transitively on their flags (see Section 2 for definitions; throughout this article, “polytope” means convex polytope.) We say that a polytope whose symmetry group has orbits on the flags is a orbit polytope, so the regular polytopes are the oneorbit polytopes. The oneorbit polytopes in the plane (the regular polygons) and in 3space (the Platonic solids) have been known for millenia; the six oneorbit 4polytopes and the three oneorbit polytopes for every have been known since the 19th century. In [matteo2014two], the author found all the twoorbit polytopes. These exist only in the plane and in 3space. In the plane, there are two infinite families, one consisting of the irregular isogonal polygons, and the other consisting of the irregular isotoxal polygons. Here, isogonal means that the symmetry group acts transitively on the vertices, and isotoxal means that the symmetry group acts transitively on the edges. In 3space, there are only four: the two quasiregular polyhedra, namely the cuboctahedron and the icosidodecahedron, and their duals, the rhombic dodecahedron and the rhombic triacontahedron.
A polytope is combinatorially orbit if its automorphism group has orbits on the flags. In general, a polytope has more combinatorial automorphisms of its face lattice than it has Euclidean symmetries. Hence, if the symmetry group has flag orbits and the automorphism goup has flag orbits, then ; in fact . Furthermore, not every polytope can be realized such that every automorphism is also a Euclidean isometry; [bokowski1984combinatorial] constructs a combinatorially 84orbit 4polytope which is not isomorphic to any polytope whose symmetry group is equal to the automorphism group . However, it is proved in [McMThesis, Theorem 3A1] that a polytope is combinatorially oneorbit if and only if it is isomorphic to a (geometrically) oneorbit polytope. In this paper, we show that every combinatorially twoorbit polytope is isomorphic to a (geometrically) twoorbit polytope. The converse is not quite true, since any gon is isomorphic to a twoorbit polytope, yet is not combinatorially twoorbit.
In Section 5 we show, similarly, that combinatorially twoorbit normal facetoface tilings by convex polytopes are isomorphic to twoorbit tilings. It seems that the corresponding question for oneorbit tilings remains open, with a finite list of possible exceptions. We summarize the results in these theorems.
Theorem 1.
Any combinatorially twoorbit convex polytope is isomorphic to a (geometrically) twoorbit convex polytope. Hence, if is a combinatorially twoorbit convex polytope, then and is isomorphic to one of the cuboctahedron, the icosidodecahedron, the rhombic dodecahedron, or the rhombic triacontahedron.
In light of the fact that, for , all twoorbit convex polytopes are combinatorially twoorbit, and that both conditions are vacuous for , we can say that a convex polytope with is combinatorially twoorbit if and only if it is isomorphic to a twoorbit convex polytope.
Theorem 2.
A locally finite, combinatorially twoorbit tiling by convex polytopes need not be isomorphic to a twoorbit tiling by convex polytopes. However, locally finite, combinatorially twoorbit tilings of by convex polytopes only occur for or .
Terms related to tilings (such as “normal”) are defined at the beginning of Section 5.
Theorem 3.
Any combinatorially twoorbit, normal tiling by convex polytopes is isomorphic to a twoorbit tiling by convex polytopes. Hence, if is a combinatorially twoorbit normal tiling of by convex polytopes, then either

and is isomorphic to one of the trihexagonal tiling or the rhombille tiling, or

and is isomorphic to one of the tetrahedraloctahedral honeycomb or the rhombic dodecahedral honeycomb.
2. Preliminaries
We briefly review the terminology used. See [ARP, grunbaum1967convex, coxeter1973regular] for details.
2.1. Basic terminology for polytopes
A convex polytope is the convex hull of finitely many points in . Throughout this article, “polytope,” unqualified, means “convex polytope.” The dimension of a polytope is the dimension of its affine hull; a polytope of dimension is called a polytope, and the faces of with dimension are its faces. The 0faces are called vertices, 1faces are called edges, faces are called ridges, and faces are called facets. In addition to these proper faces, we admit two improper faces, namely a face (the empty face) and a face (which is itself). With the inclusion of these improper faces, the faces of ordered by inclusion form a lattice, the face lattice of , denoted . A flag of is a maximal chain (linearly ordered subset) in . For any flag , an adjacent flag is one which differs from in exactly one face. The flags are adjacent if they differ in only the face. Every flag has a unique adjacent flag for , denoted (this is due to the “diamond condition” on polytopes.) Two faces are said to be incident if one contains the other. A section of , for incident faces , is the portion of the face lattice consisting of all the faces containing and contained in , and is denoted . So , inheriting the order. Every such section can be realized as the face lattice of a convex polytope, and we will often identify convex polytopes with their face lattices.
For a convex polytope , the symmetries of are the Euclidean isometries which carry onto itself, and form a group denoted . The automorphisms of are inclusionpreserving bijections from the face lattice of to itself, and form a group denoted . Each symmetry of also acts on the faces of in an inclusionpreserving manner, so we can identify with a subgroup of . A polytope is said to be fully transitive if its symmetry group acts transitively on its faces for every . It is combinatorially fully transitive if its automorphism group acts transitively on the faces of each dimension. In this case we may instead say that is fully transitive.
2.2. Class
Let and be a flag of a combinatorially twoorbit polytope . If the adjacent flag is in the same orbit as if and only if , then we say is in class [hubard2010two, hubard2009monodromy]. It is not hard to see that this class is welldefined; see [hubard2010two] for proofs of this and the following remarks. The automorphism group is fully transitive if and only if . We cannot have , because then would be combinatorially regular. The only other case is that , and then acts transitively on all faces with , but has two orbits on the faces for the unique not in .
Definition.
An (abstract) polytope is intransitive if does not act transitively on the faces, but acts transitively on the faces for all .
We shall see that all twoorbit polytopes are intransitive for some .
2.3. Modified Schläfli symbol
The Schläfli symbol of a polytope is a standard concept; see e.g. [ARP, 11],[mcmullen1967combinatorially], or [coxeter1973regular]. For a regular polytope , it is a list of numbers, , where is the order of the automorphism , where is an involution which carries a base flag to its adjacent flag . For convex polytopes, it is equivalent to say that for every incident pair of an face and an face , the section is a gon. This is the meaning we will focus on.
For the purposes of the article, we will use a modified Schläfli symbol. It is like the standard symbol for a polytope , but possibly with some positions replaced by a stack of two distinct numbers, . Wherever a single number appears, it means (as usual) that every section is a gon. If two numbers appear, it means that all such sections are either gons or gons. If is a twoorbit polytope with such a symbol, then the orbit of a flag is determined by whether is a gon or a gon. If it is a gon, and is of class , then the corresponding section of is a gon precisely when . In order for the section to have a different size, must differ from in either the face or the face—but by definition it differs in exactly the face. We conclude that or (or both) are not in .
Beware that you cannot read off the symbols for sections from the symbol for , as you can with a standard Schläfli symbol, without additional information. For instance, in the type discussed below, the facets are of type (the rhombic dodecahedron) and the vertex figures are of type (the cuboctahedron). However, in the tetrahedraloctahedral tiling of type , the vertex figures are cuboctahedra , but the facets alternate between two types, tetrahedra and octahedra .
Those polytopes with standard Schläfli symbols (with just one number in each position), so that the size of every section depends only on , are called equivelar. Equivelar convex polytopes are combinatorially regular [ARP, Theorem 1B9]. On the other hand, in a combinatorially twoorbit polytope, obviously the sections for a given can have at most two sizes. So every combinatorially twoorbit convex polytope has a modified Schläfli symbol, with at least one stack of two numbers appearing.
2.4. Results on combinatorially twoorbit polytopes
For a polytope and , a chain of type is a chain of faces in with an face for each , and no others. A chain of cotype is a chain in with an face for each , and no others.
Lemma 1.
If is in class and , then acts transitively on chains of cotype .
Proof.
Let and be two chains of cotype . Each of these may be extended to two flags of which, being adjacent, are in different flag orbits. Thus, we extend to a flag and to a flag such that both are in the same orbit; then the automorphism carrying to also takes to . ∎
Corollary 1.
If is in class and , then has a modified Schläfli symbol whose entry is singlevalued except possibly at and .
Proof.
By Lemma 1, acts transitively on the sections for each rank unless or . ∎
Recall that if all entries of the Schläfli symbol are singlevalued, then is combinatorially regular. But by the Corollary, if two distinct ranks are missing from , then all the entries would be singlevalued unless , so that coincides with . This also shows that no three ranks can be missing from .
Lemma 2.
If a polytope is in class and , then the entries (if ) and (if ) are even.
Proof.
If , then consider any section with incident faces of the indicated ranks. This is a polygon whose edges correspond to faces of . A walk along the edges of this polygon can be extended to a sequence of adjacent flags of , alternately adjacent and adjacent. The flags change orbits whenever the face is changed. But changing faces (corresponding to vertices of the polygon) will not change the orbit, since and do not differ by 3. Thus the polygon has evenly many sides. Hence , the th entry in the Schläfli symbol for (which is singlevalued by Corollary 1) is even.
Similarly, if , then any section is a polygon whose vertices correspond to faces of . A walk along the edges of this polygon corresponds to a sequence of adjacent flags of , alternately adjacent or adjacent, with the adjacent flags in different orbits. Hence the polygon again has evenly many sides, so is even. ∎
Corollary 2.
If a polytope is in class and , then or .
Proof.
If and , then both the entries and appear in the Schläfli symbol. But this contradicts Euler’s theorem; a polyhedral section would have the symbol with two even entries, which is impossible for convex polytopes [grunbaum1967convex, §13.1]. ∎
Continuing the preceding remarks, we conclude that the only way two distinct ranks can be missing from , where is in class , is if omits both 0 and and , i.e. must be a 4polytope in class . We will postpone considering this special case until Section 4. Otherwise, and any twoorbit polytope of type must be either vertexintransitive or facetintransitive. Since vertexintransitive polytopes are the duals of the facetintransitive polytopes, we will deal with the latter in Section 3.
3. Combinatorially facetintransitive twoorbit polytopes
Suppose is a combinatorially twoorbit polytope which is facetintransitive, i.e. it is in class where . It follows that is what is called an alternating semiregular polytope in [monson2012semiregular]. Fix a flag , the base flag. Then for each , there is an automorphism such that . There is no automorphism carrying to , which is in the other orbit. However, the flag , reached by changing the facet of , then changing the ridge, then flipping facets again, is in the same orbit as , so there is an automorphism carrying to . This automorphism is referred to as in [hubard2010two].
Lemma 3.
The automorphisms and generate the whole automorphism group of , so .
Proof.
Write , and say the facetadjacent flag has the facet . First we show that the given generators suffice to carry the flag to each of its adjacent flags for .
Let . Since fixes and , it must also fix . Hence, it fixes all faces of except for its face ; so .
On the other hand, cannot fix . Since , the image of the adjacent flag must be adjacent to , i.e. . But the automorphism which carries to must carry to .
Thus, the given generators carry to each of its adjacent flags except for .
Now let be any automorphism of . The automorphism is the unique one carrying to . By exhibiting an automorphism in carrying to , we show that the arbitrary element lies in this subgroup.
By the flagconnectedness property of polytopes, there is a sequence of adjacent flags . For each there is some such that the flag is adjacent to the preceding flag . Suppose and we have written either or for some .
If , then is or , respectively.
If , then is or , respectively.
If , then is or , respectively.
Thus we continue until we have written or for some . Since is in the same orbit as and is not, we must in fact have and . ∎
By Corollary 1 with , will have a modified Schläfli symbol of the form , where , since cannot be equivelar. Figure 1 shows the Coxeter diagram for these generators, modified by labeling the nodes with the corresponding generator. Such a diagram is dubbed a “tailtriangle diagram” in [monson2012semiregular], making a “tailtriangle group.” Note that must be even, by Lemma 2.
Since the generators of satisfy all the Coxeter relations implied by the diagram, is a quotient of the Coxeter group associated with the diagram. However, in principle the generators of might satisfy additional relations. We shall show that, in fact, there are no additional relations in , so that is exactly the Coxeter group associated with the diagram in Figure 1. Since is finite, we can then have recourse to the classification of finite Coxeter groups.
Lemma 4.
The automorphism group is a Coxeter group, with Coxeter diagram as in Figure 1.
The proof is a modification of that of [McMThesis, Theorem 3A1], that a combinatorially regular convex polytope is isomorphic to a regular one. The method is also in Coxeter’s proof [coxeter1973regular, §5.3] that the Coxeter relations fully define the group generated by reflections in the walls of the fundamental region described by the diagram. The essence is that any relation in the group (i.e. a word in the generators representing the identity) can be represented as a loop in the boundary of the polytope ; contracting this loop to a point gives a guide to reducing the word, using the given relations, until it is empty. This shows that every relation in the group is a consequence of the Coxeter relations. The following proof is modeled on, and sometimes verbatim from, [coxeter1973regular, §5.3].
Proof.
We associate flags of with chambers of a “barycentric subdivision” of the boundary of . Each flag is associated to the simplex whose vertices are “barycenters” of each proper face of . These barycenters can be any preassigned points in the relative interior of each face of . So the vertices of the simplex for are the vertex , the midpoint (say) of the edge , and so on up an interior point of the facet . Each face of this simplex corresponds to a subchain of . A facet of the simplex is a simplex involving the centers of all but one of the proper faces in . Say the missing face is . Then the facet, called the th wall, forms the boundary between the simplex for and the simplex for the adjacent flag . We identify each flag with its corresponding chamber in the boundary of .
The union of the chambers and constitute a “fundamental region” for , since every flag is the image of one of these. For , the th wall of and the th wall of are contiguous, and we will call their union the th wall of . The th wall of is called the th wall of , and the th wall of is called the th wall of ( is just a symbol distinct from .) The th walls of and are in the interior of and are not walls of . Thus, has walls labeled .
Say the vertex of lying in the relative interior of a face of is . Recall that is the facet in ; say the vertex in is . Then contains the vertices , plus , but is on the “edge” from to and is not a vertex of , so that has vertices and is again a simplex, with vertices . See Figure 2. (Some facets may be skew, rather than linear.) In the left figure, where , the th wall is labeled . In the right figure, where , the face is the th wall; the face is the 2nd wall; the face is the 1st wall; and the face is the 0th wall.
Now for , the chambers for and are adjacent, and their union is called “region .” We pass through the th wall of region into region (for ), where denotes . Each automorphism carries faces to other faces, and the two orbits of faces are carried only to themselves. Although does not actually map points to other points, if we consider a vertex of as representing the face in whose relative interior it lies, it makes sense to say that each vertex of is carried only to the unique vertex of the same type in each region .
To a word , where , we associate a path from to region passing through the th wall of , then the th wall of region , and so on. (By a path we mean a continuous curve which avoids any face of .)
If the word represents the identity, we must show that the relation is a consequence of the Coxeter relations inherent in Figure 1. These relations are , where for all , and otherwise is the label on the edge from to , or 2 if there is no edge. If , the path associated to is a closed path back to . Consider what happens to the expression as the closed path is gradually shrunk until it lies wholly within region . Whenever the path goes from one region to another and then immediately returns, this detour may be eliminated by canceling a repeated in the expression, in accordance with the relation . The only other kind of change that can occur during the shrinking process is when the path momentarily crosses a face .
If is the intersection of the th and th walls of one region, so that it does not contains vertices of the types opposite the th and th walls, then does not contain vertices of those types in any region that contains it. So the walls containing alternate between th walls and th walls, and is contained in regions.
This change will replace by (or vice versa) in accordance with the relation . The shrinkage of the path thus corresponds to an algebraic reduction of the expression by means of the Coxeter relations. Since the boundary of is topologically a sphere, and simply connected if , we can shrink the path to a point. It follows that every relation in is a Coxeter relation. ∎
We can now complete the proof of
Theorem 4.
Any combinatorially twoorbit facetintransitive convex polytope is isomorphic to a twoorbit convex polytope.
Since has finitely many flags, we know that is a finite Coxeter group. Consulting the list of finite Coxeter groups, we see that must be 2, since no loops appear in diagrams of finite Coxeter groups. Furthermore, the only diagram with four or more nodes that branches as in Figure 1 is , where every edge has the label 3. But we must have , since is not equivelar. Hence the diagram must not have a “tail”: we must have , and the only admissible diagrams of finite Coxeter groups are those in Figure 3.
We know that both of these Coxeter groups occur as the automorphism group of a twoorbit facetintransitive polyhedron: for the cuboctahedron, and for the icosidodecahedron. The next lemma will show that the isomorphism type of a 2orbit facetintransitive polytope is determined by its automorphism group (as a Coxeter system), so these are the only possibilities.
For the purposes of the Lemma, we will fix a canonical form of the Coxeter group presentation, as encoded in the diagram of Figure 1 or the Schläfli symbol , such that . A flag will be said to be an appropriate base flag if the generators corresponding to , defined as in Lemma 3, satisfy . We prove the Lemma generally, rather than restricting to the two presentations in Figure 3, so that it also applies to tilings, or indeed, any abstract polytopes.
Lemma 5.
Proof.
If is an isomorphism, let be an appropriate base flag for . Then the generators of corresponding to the base flag must satisfy the same relations that the generators of corresponding to do, so that the groups have the same presentation.
Conversely, suppose and are combinatorially twoorbit facetintransitive polytopes with the same presentation. For , let be an appropriate base flag of and define the generators of with respect to as in Lemma 3. Since and have the same presentation, the map carrying and extends to a group isomorphism . Then the bijection of the sets of flags taking and , for all , determines the required isomorphism between the lattices and . ∎
4. Exceptional possibilities in
We now return to the exceptional possibilities left open for combinatorially twoorbit 4polytopes (see the end of Section 2.) Recall that such a polytope is in class , so it is combinatorially fully transitive. For any flag , the 1adjacent flag and 2adjacent flag are in the same orbit as , while the 0adjacent flag and 3adjacent flag are not. By 2facetransitivity, all the 2faces have the same number of sides, . All the edges are in the same number of facets, . By Lemma 2 with and , and are even. Since is not equivelar, the Schläfli symbol has the form where and are even.
Each facet and vertex figure of has at most two combinatorial flag orbits, by the action of the automorphism group of restricted to these sections. Since is facettransitive, each facet must have both gons and gons as vertex figures. Since is vertextransitive, each vertex figure must have both gons and gons as faces. Thus the facets and vertex figures are not combinatorially regular: they are combinatorially twoorbit 3polytopes. By the preceding proof, the facets and vertex figures are isomorphic to one of the four twoorbit polyhedra. Since all 2faces are the same, and by the necessary compatibility of the vertex figures with the facets, the two possibilities are:

A polytope whose facets are isomorphic to the rhombic dodecahedron, and whose vertex figures are isomorphic to cuboctahedra; the modified Schläfli symbol is , and

A polytope whose facets are isomorphic to the rhombic triacontahedron, and whose vertex figures are isomorphic to icosidodecahedra; the modified Schläfli symbol is .
In each case, the polytope would be combinatorially selfdual. However, we demonstrate that such polytopes cannot exist.
Suppose that has the first combinatorial type above, . Consider the angle at a vertex in a 2face containing . That is, in the affine hull , which is a plane, we take the interior angle of the quadrilateral at . The sum of all these angles at the 4 vertices of is . So, if we take the sum of all such angles in the whole polytope —i.e. the sum of the angle for every incident pair of vertex and 2face in —the sum is , where is the number of 2faces of . Since every vertex is in 24 2faces (the number of edges of the cuboctahedron), and each 2face has 4 vertices, (where is the number of vertices of ).
On the other hand, let be any vertex of and consider the sum of the angles in each 2face incident to . Each 2face lies in exactly two facets: one where is in 4 edges, and one where is in 3 edges. (Correspondingly, each edge of the vertex figure, the cuboctahedron, is in one square and one triangle.) We may partition the 24 2faces at into 6 sets of 4, each set consisting of the 2faces of a facet containing wherein has valence 4. The sum of the angles of within these four 2faces must be less than (the difference from is the angular deficiency or defect.) Hence the sum of the angles at in all the 2faces containing is less than , and the sum of the angles of all incident pairs of vertices and 2faces is therefore less than .
But this contradicts the earlier conclusion that the sum is exactly . Therefore, no such polytope can exist.
The same argument rules out the possibility of a polytope of the second type, . Each vertex is in 60 2faces (the number of edges of the icosidodecahedron), and each 2face has 4 vertices, so we have , and the sum of the angles over all incident pairs of vertex and 2face is .
On the other hand, the 2faces at each vertex can be partitioned into 12 sets of 5, each set consisting of the 2faces of a particular facet containing wherein has valence 5. The sum of the angles at in all these 2faces is less than , so the sum of all the angles of in the 60 2faces containing is less than .
Thus we have , a contradiction, so no such polytope can exist.
With these possibilities disposed of, we have proved Theorem 1.
5. Tilings
In this section, we deal with combinatorially twoorbit tilings of Euclidean space . All the tilings we consider are by convex polytopes, and are facetoface, which means that the intersection of any two tiles is a face of each (possibly the empty face). A tiling is locally finite if every bounded set meets only finitely many tiles.
Definition.
An LFC tiling is a facetoface locally finite tiling of by convex polytopes.
An LFC tiling of is an abstract polytope of rank . A tiling is normal if it satisfies three conditions:

Every tile is a topological ball.

The intersection of every two tiles is connected (or empty).

The tiles are uniformly bounded. That is, there are positive numbers and such that every tile contains a ball of radius and is contained in a ball of radius .
Any convex tiling automatically satisfies properties 1 and 2. So when we require a tiling to be normal, it is equivalent to require the tile sizes to be bounded. Every normal tiling is locally finite.
In Section 2.4, Lemmas 1 and 2 and Corollary 1 apply also to combinatorially 2orbit LFC tilings. Corollary 2 holds, but requires a modified proof:
Corollary 2.
If a rank LFC tiling is in class and , then or .
Proof.
Suppose . Then there is an incident pair of faces and . If is a proper section, then it is isomorphic to a convex polytope, with symbol with two even entries (by Lemma 2). It is impossible for convex polytopes to have such a symbol [grunbaum1967convex, §13.1].
On the other hand, if is all of , then and , i.e. and , so we have an edgeintransitive planar tiling. By Corollary 1, has type with singlevalued entries. Hence is a regular tiling, a contradiction. ∎
Lemmas 3, 4, and 5 in Section 3 also apply to LFC tilings, but since the automorphism group of a tiling is not finite, we get no corresponding short list of potential diagrams. If we could conclude that the automorphism group was of socalled “affine type,” then the analog to Theorem 4 would follow.
Theorem 4A4 of McMullen’s thesis [McMThesis] says, for , a rank convex polytope with combinatorially regular vertex figures and combinatorially regular facets is combinatorially regular. The proof works equally well for LFC tilings; we sketch it here.
Lemma 6.
A rank LFC tiling , , whose vertex figures and facets are all combinatorially regular is itself combinatorially regular.
Proof.
Each vertex figure is combinatorially regular, hence equivelar, so for any , every incident pair of face and face containing gives a polygonal section of the same size, say . Each edge figure, being contained in a vertex figure, is also equivelar, so for any , the size of each section of incident faces containing a given edge is also constant. Since any two vertices of may be linked by a chain of vertices and edges, the Schläfli entries with are welldefined on all of .
Similarly, facechains of facets and ridges show that is welldefined for , and facechains of vertices and facets cover the remaining case when and . ∎
Theorem 5.
All combinatorially 2orbit LFC tilings are of or .
Proof.
A combinatorially twoorbit tiling has facets and vertex figures which are either combinatorially regular or combinatorially twoorbit. If we are tiling , and , then by Theorem 1 the facets and vertex figures are actually combinatorially regular, so the tiling is combinatorially regular. Thus .
Of course, LFC tilings of and are trivial, and no combinatorially twoorbit ones exist. The remaining cases are tilings of or . ∎
5.1. Planar tilings
Planar tilings are the only case, in light of Lemma 6, where a combinatorially twoorbit tiling can have combinatorially regular tiles and vertex figures. Indeed, any planar tiling has combinatorially regular tiles and vertex figures, since all polygons are combinatorially regular.
Theorem 6.
There are infinitely many (isomorphism classes of) combinatorially twoorbit LFC tilings of the plane.
To see this, first we show that there are combinatorially regular tilings by convex gons, with three tiles at each vertex, for every . This is a consequence of result 4.7.1 of Tilings and Patterns [grunbaum1986tilings, 194]. We paraphrase the result, taking advantage of result 4.1.1 [grunbaum1986tilings] that homeomorphisms preserving a tiling are equivalent to combinatorial automorphisms, and of convexification [grunbaum1986tilings, 202].
Lemma 7 ([grunbaum1986tilings, \nopp4.7.1]).
There exists a combinatorially regular LFC tiling of type , for positive integers , if and only if . Such a tiling can be normal only if equality holds.
Since for every , there is a combinatorially regular tiling for every such . From this tiling, we can form a combinatorially twoorbit tiling by “truncating” at each vertex to the midpoints of its incident edges, analogously to the formation of the cuboctahedron from the cube, of the icosidodecahedron from the dodecahedron, or of the trihexagonal tiling from the regular hexagonal tiling. Each edge of is reduced to its midpoint. The midpoints of the three edges incident to a vertex become the vertices of a triangular tile. The midpoints of the edges of a gonal tile in become the vertices of a smaller gonal tile. For instance, with , this is a “triheptagonal” tiling. Each vertex of this new tiling (formerly an edge midpoint) is in four tiles: two triangles (the vertex figures of the endpoints of the former edge), and two gons. Thus the tiling can be described , a notation that gives, in cyclic order, the number of sides of each tile incident to a vertex of the tiling.
However, none of these examples are normal for . We proceed to show
Theorem 7.
Every normal combinatorially twoorbit planar tiling is isomorphic to one of the (geometrically) twoorbit planar tilings: either the trihexagonal tiling or its dual, the rhombille tiling.
By Corollary , a combinatorially twoorbit tiling of is either facetintransitive, in which case acts transitively on its vertices, or vertexintransitive, in which case acts transitively on its facets (tiles).
In the former case, we apply
Lemma 8 ([grunbaum1986tilings, \nopp3.5.4]).
If every vertex of a normal tiling has valence , and is incident with tiles which have adjacents, then
Each vertex is incident to evenly many tiles (by Lemma 2 with ), which alternate orbits. With 6 tiles at each vertex, the only solution is when all tiles are triangles, ; but this is the regular tiling by triangles. So we consider 4 tiles at each vertex. If none of the tiles are triangles, the only solution is four squares, ; but this is the regular tiling by squares. So we must have . The only solution is , which is the trihexagonal tiling. This tiling has two triangles and two hexagons alternating at each vertex.
On the other hand, if acts transitively on facets, we apply
Lemma 9 ([grunbaum1986tilings, \nopp3.5.1]).
If every tile of a normal tiling has vertices, and these vertices have valences , then
Every facet has evenly many sides (by Lemma 2 with ), and the valence of each vertex alternates. Clearly, this has the same solutions as before. In a notation giving, in cyclic order, the valence of each vertex adjacent to a tile, we have , the regular tiling by hexagons; , the regular tiling by squares; and , the rhombille tiling. The latter has rhombus tiles, with three tiles meeting at the obtuse corners, and 6 tiles meeting at the acute corners.
5.2. Tilings of
A tiling of has rank 4. Hence, by Lemma 6, if it has combinatorially regular facets and vertex figures, it must be combinatorially regular. So a combinatorially twoorbit tiling of must have some tiles or some vertex figures from the list of twoorbit polyhedra. By Corollary , the class of must be (tileintransitive), (vertexintransitive), or (fully transitive). We consider these cases.
5.2.1. Tileintransitive
In this case, is in class and the automorphism group is transitive on the vertices, edges, and 2faces of the tiling. There are two different tile orbits, and each tile must be combinatorially regular (since the orbit of a flag is determined entirely by which type of tile it includes). Thus, all the vertices have isomorphic vertex figures, which must be a twoorbit polyhedron; since there are two types of tile, the vertex figure must be facetintransitive, i.e. the cuboctahedron or the icosidodecahedron.
With the cuboctahedron as vertex figure, each vertex is 3valent in some tiles, and 4valent in others. The only regular polyhedron with 4valent vertices is the octahedron; the only regular polyhedron with 3valent vertices and triangular faces (to match the octahedron) is the tetrahedron. But the tiling built from tetrahedra and octahedra in this manner is the tetrahedraloctahedral honeycomb, , one of the (geometrically) twoorbit tilings.
With the icosidodecahedron as vertex figure, each vertex is 3valent in some tiles, and 5valent in others. The only regular polyhedron with 5valent vertices is the icosahedron, and the other tiles must be tetrahedra. Such a tetrahedralicosahedral tiling has type . Indeed, a tiling can be built up in such a way, in hyperbolic space; it is known as the alternated order5 cubic honeycomb. It can be carved out of a tiling by cubes, with 5 around each edges, , which is a regular tiling of hyperbolic space. Inscribe a tetrahedron in each cube, so that tetrahedra in adjacent cubes alternate direction. The shape left around a vertex which is not part of the tetrahedron is an icosahedron (there are 20 cubes around each vertex in .)
We show by contradiction that there is no normal tiling of of this type. Suppose is a normal tetrahedralicosahedral tiling. Divide each icosahedron of into twenty pyramids (over each of its 2faces), and adjoin each of these pyramids to the tetrahedron with which it shares the 2face. Thus we have partitioned into tiles, one for each tetrahedron in , consisting of the tetrahedron with a pyramidal cap added to each of its 2faces. These tiles need not be convex, but are isomorphic to cubes: Each tile has six neighboring tiles, with each of which it shares two triangular faces of its pyramidal caps; we treat each such pair of triangular faces as a single “skew” 4gonal face. Thus we get a tiling topologically isomorphic to the order5 cubic tiling . If we start with a normal tiling, this one will be, also. Say each tile contains a ball of radius and is contained in a ball of radius . Then the number of tiles in a ball of radius is at most .
Next consider a growing sequence of patches of the tiling . (For our purposes, a patch can be defined as any finite set of tiles of whose union is homeomorphic to a ball.) Begin with a vertex of the tiling, designated . Let consist of all the tiles containing , consist of all the tiles with nonempty intersection with the union of , and so on, so consists of all the tiles with nonempty intersection with the union of .
Let us call a tile of which has any 2face on the boundary of a tile if it has 2faces on the boundary of . By induction, we see that all tiles on the boundary are 1tiles, 2tiles, or 3tiles; that every edge on the boundary of is contained in either 1 or 2 tiles of ; and that every vertex on the boundary of is contained in either 1, 2, or 5 tiles of .

A tile not in which contains a 2face in becomes a 1tile of . Each of its four exposed edges is in two tiles of . See the leftmost example in Figure 4.

A tile not in which contains only an edge in becomes a 2tile of . Six of its exposed edges are in two tiles of , while one exposed edge is only in this tile. See the middle example in Figure 4.

A tile not in which contains only a vertex in becomes a 3tile of . Six of its exposed edges are in two tiles of , while three exposed edges are only in this tile. See the rightmost example in Figure 4.
Let be the number of 1tiles in , be the number of 2tiles, and be the number of 3tiles. The 1tiles in are those tiles which share a 2face with some tile of , so we have
The 2tiles in are added above edges on the boundary of . One such 2tile is added above each edge contained in two tiles of , and two such 2tiles are added above each edge contained in a unique tile of . Counting the number of edges of each type in the boundary of yields
The 3tiles in are added above vertices on the boundary of . A vertex contained in five tiles of is also contained in five 1tiles of , added above the five incident 2faces in the boundary of , and in five 2tiles of added above the five incident edges in the boundary of , so we need to add five more 3tiles to include all 20 of the incident tiles. A vertex contained in two tiles of is incident to eight 3tiles of , and a vertex in a unique tile of is incident to ten 3tiles of . Hence
Thus we have an equation for the number of tiles on the boundary of each patch , beginning with the patch consisting of the 20 tiles incident to :
This matrix is diagonalizable, making it straightforward to solve for the total number of tiles in the patch :
This is exponential in ; the number of tiles increases by a factor of roughly 30 in each successive patch. Now consider a ball centered at with radius . This ball contains the patch , but the number of tiles in the ball is at most . An exponential function of cannot remain bounded by a cubic function of , so there must be some such that , a contradiction.
Therefore, no normal tiling of has type , even allowing nonconvex tiles. So no tetrahedralicosahedral tiling of type can be normal either. On the other hand, there seems to be no obstruction to constructing (nonnormal) LFC tilings of these types.
5.2.2. Vertexintransitive
In this case, the orbit of a flag is determined by the vertex it contains. So the vertex figures are combinatorially regular. The tiles are twoorbit vertexintransitive polyhedra, i.e. the rhombic dodecahedron or rhombic triacontahedron.
With the rhombic dodecahedron, a vertex which is incident to 4 edges in a given tile has a vertex figure with a square face; hence the vertex figure is a cube. Hence each edge incident to the vertex is in 3 tiles. A vertex which is incident to 3 edges in a given tile has a vertex figure with triangular faces. Every edge of the tiling is incident to one vertex of each type, hence is in 3 tiles, so the second type of vertex figure must be a tetrahedron. Rhombic dodecahedra put together in this way form the rhombic dodecahedral honeycomb , one of the (geometrically) twoorbit tilings.
With the rhombic triacontahedron as tile, any vertex which is incident to five edges in a given tile has a pentagon in its vertex figure; hence its vertex figure is a combinatorially regular dodecahedron. Every edge is incident to one vertex of this type, so every edge is in three tiles. Thus the other vertices, which are incident to three edges in each tile, have tetrahedra for vertex figures. This potential tiling has type and is dual to the tetrahedralicosahedral tiling discussed above. Like that one, this tiling can be realized in hyperbolic space, with a twoorbit symmetry group. For any normal tiling there is a dual tiling which is also normal (but the tiles of which are not necessarily convex). Hence, if a normal tiling of type existed, we could find a normal tiling of type , which we know to be impossible.
5.2.3. Class
These is the same class discussed in Section 4, and the same considerations establish that we have either a cuboctahedron vertex figure with rhombic dodecahedra as tiles, type , or an icosidodecahedron vertex figure with rhombic triacontahedra as tiles, type . (We note that the cuboctahedron and icosidodecahedron are nontiles, meaning there cannot be any tiling of using only tiles isomorphic to these; see [schulte1985existence].)
Perhaps these types can be realized as LFC tilings. However, there is no such normal tiling. Essentially the same proof as in Section 4 applies, along with the Normality Lemma [schattschneider1997tilings, 45], which says that in a normal tiling, the ratio of (the number of tiles that meet the boundary of a spherical patch of the tiling) to (the number of tiles in the patch) goes to zero as the radius of the patch grows. The two methods of counting internal angles of 2faces in Section 4 hold for all the faces in the interior of a given patch. Discrepancies occur only at tiles on the boundary, where a vertex is not surrounded by all the 2faces incident to it in the tiling. Taking the limit as the patch grows, the discrepancies go to zero and the inequality remains. The details are too tedious to include here.
With these ruled out, we have established
Theorem 8.
Every normal combinatorially twoorbit tiling of is isomorphic to one of the (geometrically) twoorbit tilings: either the tetrahedraloctahedral honeycomb or its dual, the rhombic dodecahedral honeycomb.
6. Open Questions
Question.
Is a combinatorially regular LFC tiling of , , necessarily isomorphic to a regular tiling of ? (Except for , this says that any combinatorially regular tiling is isomorphic to the tiling by cubes.)
The answer is probably no, but the author does not know a counterexample.
Question.
Is a combinatorially regular normal LFC tiling of necessarily isomorphic to a regular tiling of ?
The answer is probably yes, but the author knows a proof only for the cases .
Question.
Are there combinatorially twoorbit LFC tilings of not isomorphic to any twoorbit tiling?
Any such tiling would have one of the previously discussed types , , , or . The author believes that nonnormal tilings of these types probably do exist.
For results in these directions, as well as other open questions of this type, see [schulte2010combinatorial].