Is there a nice way to directly read off the whole face lattice of the polytope from its Coxeter diagram equipped with the black markings?

I’m sure there is, because everything in this subject is maximally nice… and it’s also very well-understood. However, I don’t know how to to it!

I’ll give a clue that may help someone figure this out. I’ve been taking a deliberately lowbrow approach in this series, but there’s a deeper way to understand these labelled Coxeter diagrams which I might or might not get around to explaining in my posts. It goes like this:

A Coxeter diagram like

can be seen as nothing but a quick way of writing down a group presentation. From the viewpoint, the letters don’t mean anything: they’re just names of the generators. The relations say the generators always square to one:

V² = E² = F² = 1

and then there are the interesting relations, which are described by the diagram:

(VE)⁴ = 1

(EF)³ = 1

(VF)² = 1

where last has a 2 in the exponent because there’s no edge joining V and F–that’s just one of the rules.

Let’s call this group G. In this outlook, a labelled Coxeter diagram is a way of naming a subgroup H. The vertices of the corresponding polyhedron (or higher-dimensional polytope) are the points of G/H.

I think the way I have it set up, the subgroup H is generated by the white vertices. So, for example, if we take this labelling:

we get a subgroup H that’s isomorphic to

But this diagram describes the symmetry group of an equilateral triangle. You can see this directly from the presentation encoded by the diagram E—3—F, namely:

E² = F² = 1

(EF)³ = 1

So in this case H is the symmetry group of an equilateral triangle, with 6 elements. So, since G has 48 elements, G/H has 48/6 = 8 elements, which sounds right for the vertices of the cube.

But it’s not just that we’re getting the right number! If you look at our operation E that changes which edge of the cube a chosen triangle touches:

and the operation F that changes which face of the cube that triangle lies on:

you’ll see they both fix the same vertex of the cube: the vertex closest to us! And indeed, H is the group of all symmetries of the cube that fix that vertex—the so-called ‘stabilizer’ of the vertex. So, by some famous abstract nonsense, G/H can be seen as the set of vertices of the cube.

The same idea works for all other more interesting examples. Thinking more about this stuff should eventually lead us to a nice answer to your puzzle.

Ok, I understand that the black dots tell us where the vertices of those “intermediate” polytopes are. But what about the edges and the faces of the intermediate polytopes? While those are determined by the vertices, I still wonder: do they also correspond to certain sublattices of the face lattice of the cube? (For example, every edge of the truncated cube corresponds to either an edge of the cube or a diamond in the face lattice of a cube, i.e. two edges meeting in a vertex and spanning a face.)

Is there a nice way to directly read off the whole face lattice of the polytope from its Coxeter diagram equipped with the black markings?

Yes, if we based our diagrams on the octahedron instead of the cube they wouldn’t change at all!

Okay, well, sure, we’d write them backwards… but that doesn’t really matter: they’ve got enough information in them that we can write them forwards, backwards, or whatever and they still name the same polyhedron.
We should think of them as abstract graphs with edges labelled by numbers and vertices labelled black or white.

Yes, these diagrams are a lot of fun, and the fun is just beginning.

There’s a tiny typo in the text. The last time you mention the diagram for the octahedron, you repeat the diagram for the truncated octahedron, i.e. you give it as “o—4—•—3—•” rather than “o—4—o—3—•”.

To answer the puzzle as to what the diagrams would look like if they were based on the Coxeter diagram for the octahedron instead of that of the cube, I guess you can just write all the same diagrams backwards.

The Coxeter diagram for the octahedron is:

V—3—E—4—F

So you get:

Cube: o—3—o—4—•

The cube has a vertex for each of the 8 faces of the octahedron.

Truncated cube: o—3—•—4—•

The truncated cube has a vertex for every edge-face flag of the octahedron, so three distinct vertices for each face of the octahedron for a total of 24.

Cuboctahedron: o—3—•—4—o

The cuboctahedron has a vertex for each of the 12 edges of the octahedron.

Truncated octahedron: •—3—•—4—o

The truncated octahedron has a vertex for every vertex-edge flag of the octahedron, so two distinct vertices for each edge for a total of 24.

Octahedron: •—3—o—4—o

And the octahedron is just itself.