---
format:
  html:
    html-math-method: katex
---


Appendix: Partial Cayley Graphs of Coxeter Diagrams
===================================================

This post is an appendix to [another post]() discussing the basics of Coxeter diagrams. It focuses on transforming path-like swap diagrams into proper $A_n$ Coxeter diagrams, which correspond to symmetric groups. This post focuses on the graphs made by the cosets made by removing a single generator (Coxeter diagram vertex).

For finite diagrams whose order is not prohibitively big, I will provide an embedding as a permutation group by labelling each generator in the Coxeter diagram. Since each generator is the product of disjoint swaps, I will also show their swap diagrams, as well as interactions via the edges.


Platonic Symmetry
-----------------

The symmetric group $S_n$ also happens to describe the symmetries of an $(n-1)$-dimensional simplex. The 3-simplex is simply a tetrahedron and has symmetry group $S_4$, also called $T_h$. We know that $S_4$ can be encoded by the diagram $A_3$. The string {3, 3} can be read across the edges of $A_3$, denoting the order of certain symmetries.

This happens to coincide with another description of the tetrahedron: its [Schläfli symbol](https://en.wikipedia.org/wiki/Schl%C3%A4fli_symbol). It describes triangles (the first 3) which meet in triples (the second 3) at a vertex. It may also be interpreted as the symmetry of the 2-dimensional components (faces) and the vertex-centered symmetry. [The Wikipedia article](https://en.wikipedia.org/wiki/Tetrahedron) on the tetrahedron presents both of these objects in its information column.

::: {}
![]()

From the Wikipedia article on the tetrahedron. The Conway notation, Schläfli symbol, Coxeter diagram, and symmetry groups are shown.
:::

The image above shows more sophisticated diagrams alongside $A_3$, which I will not attempt describing (mostly because I don't completely understand them myself). Other Platonic solids and their higher-dimensional analogues have different Schläfli symbols, and correspond to different Coxeter diagrams.


### $B_3$: Octahedral Group

Adding an order-4 product into the mix makes things a lot more interesting. The cube ({4, 3}) and octahedron ({3, 4}) share a symmetry group, $O_h$, which corresponds to the Coxeter diagram $B_3$.

:::: {layout-ncol="3"}
::: {}
![]()
:::

::: {}
![]()
:::

::: {}
![]()
:::
::::

The third graph is entirely path-like, similar to the ones created by removing an endpoint from the $A_n$ diagrams. In the same vein, the graph for $B_3$ resembles the graph for $A_3$ made by removing the center vertex, albeit with two extra vertices. The center diagram is the first to have a hexagon created by removing a single vertex. Going across left to right, the order suggested by each index is

::: {}
- $8 \cdot |A_2| = 8 \cdot 6 = 48$
- $12 \cdot |A_1 A_1| = 12 \cdot 4 = 48$
- $6 \cdot |B_2| = 6 \cdot 8 = 48$
    - $B_2$ describes the symmetry of a square, i.e., $Dih_4$, the dihedral group of order 8
:::

Each diagram suggests the same order, which is good. A simple embedding which obeys the edge condition assigns $(1 ~ 2)(3 ~ 4)$ to *a*, $(2 ~ 3)$ to *b*, and $(1 ~ 2)$ to *c*. Then $ab = (1 ~ 2 ~ 4 ~ 3)$ and $bc = (1 ~ 3 ~ 2)$, with *ac* obviously having order 2.

There's a problem though. These generate a group of order 24 (actually, the rotational symmetries of a cube, $O \cong S_4 \cong T_h$). The group we want is $O_h \cong S_4 \times \mathbb{Z}_2$. If we want to embed a group of order 48 in a symmetric group, we need one for which 48 divides its order. $|S_6| = 720$ divides 48, and indeed, a quick fix is just to multiply each generator by $(5 ~ 6)$.

These two embeddings generate different (proper) Cayley graphs. The one for O has 24 vertices and is nonplanar. On the other hand, the one for $O_h$ is planar, and is the skeleton of the [truncated cuboctahedron](https://en.wikipedia.org/wiki/Truncated_cuboctahedron), a figure containing octagons, hexagons, and squares. This is exactly what is suggested by the orders of the products in the Coxeter diagram. Note also that the cuboctahedron is the rectification of the cube and octahedron, i.e., is midway between them with respect to the dual operation.

In either case, the products of adjacent generators (the permutations on the edges) are the same. When made into a Cayley graph, these products generate the [rhombicuboctahedron](https://en.wikipedia.org/wiki/Rhombicuboctahedron), which is another shape midway between the cube and octahedron. Since all of these generators are in $S_4$, it only has half the number of vertices as the truncated cuboctahedron.

:::: {layout-ncol="3"}
::: {.column width="32%"}
![]()

"Bad" embedding
:::

::: {.column width="32%"}
![]()

Good embedding
:::

::: {.column width="32%"}
![]()

Edge-generated
:::
::::

![]()


### $H_3$: Icosahedral Group

Continuing with groups based on 3D shapes, the dodecahedron ({5, 3}) and icosahedron ({3, 5}) also share symmetry groups. It is known as $I_h$ and corresponds to Coxeter diagram $H_3$.

:::: {layout-ncol="3"}
::: {.column width="32%"}
![]()
:::

::: {.column width="32%"}
![]()
:::

::: {.column width="32%"}
![]()
:::
::::

Two of these graphs are similar to the cube/octahedron graphs. The other contains a decagon, corresponding to the order 5 product between *a* and *b*. We have the orders:

::: {}
- $20 \cdot |A_2| = 20 \cdot 6 = 120$
    - Graph resembles an extension of $B_3 / A_1 A_1$, as hexagons joining blocks of squares
- $30 \cdot |A_1 A_1| = 30 \cdot 4 = 120$
- $12 \cdot |H_2| = 12 \cdot 10 = 120$
    - Graph resembles an extension of $B_3 / A_2$, as a single square joining paths
:::

The order 120 is the same as the order of $S_5$, which corresponds to diagram $A_4$. However, these are not the same group, since $I_h \cong \text{Alt}_5 \times \mathbb{Z}_2 \ncong S_5$. A naive (though slightly less obvious) embedding, found similarly to $B_3$'s, incorrectly assigns the following:

::: {}
- $a = (1 ~ 2)(3 ~ 4)$
- $b = (2 ~ 3)(4 ~ 5)$
- $c = (2 ~ 3)(1 ~ 4)$
:::

This is certainly wrong, since all these permutations are within $S_5$. Actually, they are all even permutations and in fact generate $I \cong \text{Alt}_5$, with order 60. Yet again, multiplying $(6 ~ 7)$ to each vertex boosts the order to 120 and gives a proper embedding of $I_h$.

Similarly, the first, incorrect embedding gives a nonplanar Cayley graph. Correspondingly, the second one gives a planar graph, the skeleton of the [truncated icosidodecahedron](https://en.wikipedia.org/wiki/Truncated_icosidodecahedron). It consists of decagons, hexagons, and squares, just like those which appear in the graphs above. Again, note that icosidodecahedron is the rectification of the dodecahedron and icosahedron. In this case, the edges generate the [rhombicosidodecahedronal graph](https://en.wikipedia.org/wiki/Rhombicosidodecahedron).

:::: {layout-ncol="3"}
::: {.column width="32%"}
![]()

"Bad" embedding
:::

::: {.column width="32%"}
![]()

Good embedding
:::

::: {.column width="32%"}
![]()

Edge-generated
:::
::::

![]()

It is remarkable that the truncations of the rectifications (also called omnitruncation) have skeleta that are the same as the Cayley graphs generated by their respective Platonic solids' Coxeter diagrams. In a way, these figures describe their own symmetry. Notably, both of these figures belong to a class of polyhedra known as [zonohedra](https://en.wikipedia.org/wiki/Zonohedron).


### $B_4$: Hyperoctahedral Group

Up a dimension from the cube and octahedron lie their 4D counterparts, the tesseract ({4, 3, 3}, interpreted as three cubes ({4, 3}) around an edge) and 16-cell ({3, 3, 4}). They correspond to Coxeter diagram $B_4$.

:::: {layout-ncol="2"}
::: {.column width="49%"}
![]()
:::

::: {.column width="49%"}
![]()
:::
::::

Three of these graphs are *also* similar to those encountered one dimension lower. The remaining graph is best understood in three dimensions, befitting the 4D symmetries it encodes. It appears to have similar regions to the [omnitruncated tesseract](https://en.wikipedia.org/wiki/Runcinated_tesseracts#Omnitruncated_tesseract), featuring both the truncated octahedron and hexagonal prism cells. We have the orders:

::: {}
- $16 \cdot |A_3| = 16 \cdot 24 = 384$
    - Graph resembles an extension of $B_3 / A_2$, as squares connecting paths
- $32 \cdot |A_1 A_2| = 32 \cdot 2 \cdot 6 = 384$
- $24 \cdot |B_2 A_1| = 24 \cdot 8 \cdot 2 = 384$
    - Graph resembles an extension of $B_3 / A_1 A_1$, as hexagons joining blocks of squares
- $8 \cdot |B_3| = 8 \cdot 48 = 384$
    - Graph resembles an extension of $B_3 / B_2$, a simple path
:::

The order of the group, 384, suggests that it needs to be embedded in at least $S_8$ since $384 ~ \vert ~ 8! ~ ( = 40320)$. Indeed, such an embedding exists, found by computer search rather than by hand:

::: {}
- $a = (1 ~ 3)$
- $b = (1 ~ 2)(3 ~ 4)(5 ~ 6)(7 ~ 8)$
- $c = (1 ~ 3)(2 ~ 6)(4 ~ 5)(7 ~ 8)$
- $d = (1 ~ 3)(2 ~ 4)(5 ~ 7)(6 ~ 8)$
:::

Notably, this embedding takes advantage of the possibility of the product of an order 2 and an order 4 element having order 4. A similar computer search yielded an insufficient embedding in $S_8$, with order 192:

::: {}
- $a = (1 ~ 2)(3 ~ 4)(5 ~ 6)(7 ~ 8)$
- $b = (1 ~ 3)(2 ~ 5)(4 ~ 7)(6 ~ 8)$
- $c = (1 ~ 2)(3 ~ 4)(5 ~ 7)(6 ~ 8)$
- $d = (1 ~ 2)(3 ~ 5)(4 ~ 6)(7 ~ 8)$
:::

The latter embedding *cannot* be "fixed" by going to $S_{10}$, either by multiplying one or all elements by $(9 ~ 10)$. Only *a* is a viable choice for the former since its products with the rest of the generators have an order divisible by 2. Quickly "running" the generators shows that the order of the group is unchanged by this maneuver. Much of the structure permutations ensures that nonadjacent vertices still have order-2 products.

![]()

I won't try to identify either of these generating sets' Cayley graphs since they will in all likelihood correspond to a 4D object's skeleton and because it is impractical to try comparing graphs of this size. In fact, *H* appears to not be isomorphic to a subgroup of $W(B_4)$. The latter has at least 2 subgroups of order 192: one generated by the edges in the above embedding, and one containing only the even permutations. These are distinct from one another, since the number of elements of a particular order is different. The latter subgroup is closer to *H*, matching the number of elements of each order, but the even permutations have commutator subgroup has order 96 while *H* has a commutator subgroup of order 48.


Other Finite Diagrams
---------------------

Higher-dimensional Platonic solids are hardly the limits of what these diagrams can encode. The following three diagrams also give rise to finite graphs.

### $D_4$

$D_4$ is the first Coxeter diagram with a branch. Like $B_4$ before it, it is corresponds to the symmetries of a 4D object. We only really have two choices in which vertex to remove, which generate the following diagrams

:::: {layout-ncol="2"}
::: {.column width="49%"}
![]()
:::

::: {.column width="49%"}
![]()
:::
::::

While we could also remove *c* or *d*, this would just produce a graph identical to the one on the left, just with different labelling. The right diagram is rather interesting, as it can be described geometrically as two cubes attached to a triple of coaxial hexagons. Both of these cases give the order

::: {}
- $8 \cdot |A_3| = 8 \cdot 24 = 192$
- $24 \cdot |A_1 A_1 A_1| = 24 \cdot 8 = 192$
:::

The order of 192 requires at minimum, $S_8$, since $192 ~ \vert ~ 40320$. Fortunately, a computer search yields a correct embedding immediately:

::: {}
- $b = (1 ~ 2)(3 ~ 4)(5 ~ 6)(7 ~ 8)$
- $a = (1 ~ 8)(2 ~ 3)(4 ~ 7)(5 ~ 6)$
- $c = (1 ~ 5)(2 ~ 7)(3 ~ 4)(6 ~ 8)$
- $d = (1 ~ 8)(2 ~ 4)(3 ~ 7)(5 ~ 6)$
:::

::: {}
![]()
:::

In fact, the group generated by this diagram is isomorphic to the even permutation subgroup of $W(B_4)$. This can be verified by selecting order 2 elements from the latter which obey the laws in $D_4$. *H*, and the edge-generated subgroup, on the other hand do not satisfy this diagram.


### $D_5$

D-type diagrams continue by elongating one of the paths. The next diagram, $D_5$ has really only four distinct graphs, of which I will show only two:

:::: {layout-ncol="2"}
::: {.column width="49%"}
![]()
:::

::: {.column width="49%"}
![]()
:::
::::

First, note how the graph to the right is generated by removing the vertex *e*, but the left and right sides of the diagram are asymmetrical. This is because *d* and *e* are equivalent with respect to the branch, and shows us that removing either vertex *d* or *e* results in the same graph. The order suggested by each is

::: {}
- $10 \cdot |D_4| = 10 \cdot 192 = 1920$
- $16 \cdot |A_4| = 16 \cdot 120 = 1920$
:::

If we were to remove vertex b, we'd end up with the diagram $A_1 A_3$, which has order 48. The diagram would have 1920 / 48 = 40 vertices, which would be fairly difficult to render. Removing vertex c would be even worse, since the resulting diagram, $A_2 A_1 A_1$, has order 24, and would require 80 vertices, twice as many.

Finding an embedding at this point is difficult. The order of 1920 also divides $40320 = |S_8|$, but computer search has failed to find an embedding in up to $S_{11}$.


### $E_6$

If one of the shorter paths of $D_5$ is extended, then we end up with the diagram $E_6$. I will only show one of its graphs.

![]()

Similarly to one of the graphs for $D_5$, this graph goes in with *a* and comes out with *e*, which are again symmetric with respect to the branch. I particularly like how on this graph, most of the squares are structured so that they can bridge to the *ae* commuting square in the middle.

The order of this group is $27 \cdot 1920 = 51840$, which is starting to be incomprehensible, if it hasn't been already. The graphs not shown would have the following number of vertices, which is precisely why I won't draw them out:

::: {}
- Removing *f*: $[E_6 : A_5] = 51840 / 720 = 72$ vertices
- Removing *b* or *d*: $[E_6 : A_1 A_4] = 51840 / 240 = 216$ vertices
- Removing *c*: $[E_6 : A_2 A_2 A_1] = 51840 / 72 = 720$ vertices
:::

Going by order alone, $E_6$ should embed in $S_9$ ($51840 ~ \vert ~ 9!$), but since $S_11$ was too small for its subgroup $D_5$, this too optimistic. I don't know what the minimum degree is required to embed $S_6$, but finding it directly it is beyond my computational power.

The E diagrams continue with $E_7$ and $E_8$. Each of the three corresponds to the symmetries of semi-regular higher-dimensional objects, whose significance and structure I can't even begin to comprehend. Their size alone makes continuing onward by hand a fool's errand, and I won't be attempting to draw them out right just now.


Infinite (Affine) Diagrams
--------------------------

Not every Coxeter diagram results in a closed graph. Instead, they may proliferate vertices forever. They are termed either "affine" or "hyperbolic", depending respectively on whether the diagrams join up with themselves or seem to require more and more room as the algorithm advances. This is also related to the collection of fundamental domains and roots that the diagram describes.

Since hyperbolic graphs are difficult to draw, I'll be restricting myself to the affine diagrams. Unlike hyperbolic diagrams, which are unnamed, affine ones are typically named by altering finite diagrams.

### $\widetilde A_2$

The Coxeter diagram associated to the triangle graph $K_3$ is called $\widetilde A_2$. This graph is also the line graph of $\bigstar_3$ so it might make sense to assign to the vertices the generators $(1 ~ 2)$, $(2 ~ 3)$, and $(1 ~ 3)$, which we know to generate $S_3$. However, $S_3$ already has a diagram, $A_2$, which is clearly a subdiagram of $\widetilde A_2$, so the new group must be larger.

![]()

Attempting to make a graph by following the generators results in an infinite tiling by hexagons. There are three distinct types of hexagon -- *ab*, *ac*, and *bc* -- since each pair of elements has a product of order 3. Removing a pair of vertices from the diagram would get rid of the initial edge (*a* in this case), and the "limiting case" where all three vertices are removed is just the true hexagonal tiling.


### $\widetilde G_2$

The hexagonal tiling has a Schläfli symbol of {6, 3}, and is dual to the triangular tiling. As a Coxeter diagram, this symbol matches the Coxeter diagram $\widetilde G_2$.

![]()

The graph generated by removing a vertex is another infinite tiling, in this case the [truncated trihexagonal tiling](https://en.wikipedia.org/wiki/Truncated_trihexagonal_tiling). Once again, this tiling is the truncation of the rectification of the symmetry it typically describes. Each of the three figures corresponds to a unique pair of products: dodecagons to *ab*, squares to *ac*, and hexagons to *bc*.


### $\widetilde C_2$

The only remaining regular 2D tiling is the square tiling ({4, 4}), whose Coxeter diagram is named $\widetilde C_2$.

![]()

The tiling generated by this diagram is known as the [truncated square tiling](https://en.wikipedia.org/wiki/Truncated_square_tiling). Mirroring the other cases, it is also the truncated *rectified* square tiling, since rectifying the square tiling merely rotates it by 45°. In this tiling, while the squares distinctly correspond to the product *ac*, the octagons are either order-4 product, *ab* or *bc*.


### $\widetilde A_3$

The above diagrams are the only rank-2 affine diagrams. The simplest diagram of rank 3 is $\widetilde A_3$, which appears similar to 4-cyclic graph. Similar to how $\widetilde A_2$'s graph is the tiling of $A_2$'s, its Cayley graph is the honeycomb of $A_3$'s.

:::: {layout-ncol="2"}
::: {.column width="49%"}
![]()
:::

::: {.column width="49%"}
![]()
:::
::::

The left diagram shows the initial branches, while the left image shows the four distinct cells which form the honeycomb. From left to right, the diagrams contain only *abd*, *abc*, *bcd*, and *acd*. The squares always represent commuting pairs on opposite ends of the diagram: *ac* and *bd*.

I am not certain in general whether $\widetilde A_n$ generates the honeycomb formed by $A_n$, but this tessellation should always exist, since the solids formed by the generators of $A_n$ are always permutohedra.


Closing
-------

The diagrams I have studied here are only the smaller ones which I can either visualize or compute. I might have gotten carried away with studying the groups themselves in the 4D case, since there appears to be so much contention. Despite this, I examined only half of the available Platonic solids in 4D, missing out on the 24-cell ({3, 4, 3}, $F_4$), 120-cell ({5, 3, 3}, $H_4$), and 600-cell ({3, 3, 5}, $H_4$). If 4D symmetries are hard to understand, then things can only get worse in higher dimensions.

One of the things I'm curious about is the minimal degree of symmetric group needed to embed the finite diagrams (known as $\mu$). While $S_6$ is minimal for the octahedral group, it doesn't appear to be big enough for the icosahedral one. Certain groups obey $\mu(G \times H) = \mu(G) + \mu(H)$ (typically abelian ones), but I'm not sure whether that's the case here.

Coset and embedding diagrams made with GeoGebra. Cayley graph images made with NetworkX (GraphViz).


### Additional Links

- [Point groups in three dimensions](https://en.wikipedia.org/wiki/Point_groups_in_three_dimensions) (Wikipedia)
- [Point groups in four dimensions](https://en.wikipedia.org/wiki/Point_groups_in_four_dimensions) (Wikipedia)
- [Finding the minimal n such that a given finite group G is a subgroup of Sn](https://math.stackexchange.com/questions/1597347/finding-the-minimal-n-such-that-a-given-finite-group-g-is-a-subgroup-of-s-n) (Mathematics Stack Exchange)
- [Omnitruncated](https://community.wolfram.com/groups/-/m/t/774393), by Clayton Shonkwiler
