Symmetries of a 4-demicube

Question

I do not really understand, what is the symmetry group of a 4-demicube.

I've read that the 4-demicube coincides with the 16-cell. In what sense? I can see that the corresponding graph (i.e. taking only vertices and edges) of the 4-demicube and the 16-cell coincide. Do they coincide also as polytopes?

The symmetry group of the 16-cell is the hyperoctahedral group (i.e. Coxeter group of type B), while the symmetry group of the 4-demicube is supposed to be the Coxeter group of type D. Why is that? Is it because although they coincide as graphs, they do not coincide as polytopes? Or do they actually also coincide as polytopes, but only by a chance and actually of the 16 cells in the 4-demicube, we have 8 and 8 that arised in a different way, so we do not want to mix them for some reason?

In the Wikipedia article 16-cell, they write

There is a lower symmetry form of the 16-cell, called a demitesseract or 4-demicube, ...

What exactly means this "lower symmetry form"?

As you can see, I am mostly confused because I do not know the exact meaning of all the terms like "polytope", "16-cell", "demicube", "symmetry" and so on that is probably always different in different contexts, so I will be happy also for any reference that would make this clear.

EDIT: I guess I found somewhat satisfying answer to my question or, better to say, a confirmation of what I thought is the answer. In arXiv:1403.2125, there is written:

Hence the “two-orbit” 4-demicube is actually a regular 4-crosspolytope with artificially restricted symmetries, essentially by coloring the facets depending whether they were formed inside a facet, or at a missing vertex, of the 4-cube.

Nevertheless, I am still interested to find some book, with some clear definition, what precisely is meant by a 4-demicube. I mean, what kind of mathematical object is that? Because its apparntly not "just a polytope", but a polytope with some coloring of facets, or a polytope with some prescribed symmetry group (which sounds a bit silly).

Note that there is the book The Symmetries of things that contains a section about those n-demicubes (calling them hemicubes) and even a sentence about 4-demicube. But what they write does not seem to me clear at all

... the four-dimensional one is an orthoplex, but with only half its symmetry as is better revealed by the generalized Schläfli symbols.

I guess if I read the whole book and understood for example those Schläfli symbols, then I would maybe understand it, so its probably partially my fault that it is not clear to me, what they mean. Nevertheless, they definitelly do not give a proper definition of what an n-demicube really is.

By the way, since all the Wikipedia articles are written in such a mysterious way regarding the symmetries as well, I think, it would be nice to edit them, so I invite any expert to do so. Or I can do it also myself if I have some proper reference.

Answer 1

The Wikipedia pages on these topics are a little bit heavy on jargon (it took me quite a while to get used to it, and I am still not very comfortable there), so I will try my best to use as little of that as possible.

First of all, the notations $I_n$, $A_n$, $B_n$, $D_n$, $E_n$, $F_n$ and $H_n$ denote certain matrix groups in Euclidean space (sets of orthogonal matrices, closed under matrix multiplication and matrix inverse). Namely, these groups are the ones generates by reflections (I will say more about this below). If you write "a polytope from the $B_4$-family", then this is initially only a statement about how the polytope was generated, and less a statement about how its symmetry group looks like (but these notions are not independent).

Let me first talk about, how a polytope is generated from such a group. You might want to look up Wythoff construction for more.

• Choose a reflection group (or actually any matrix group if you want). For simplicity, lets take $I_3$, which is the symmetry group of the triangle. This is a reflection group, i.e. there exists a set of hyper-planes (that is, lines in 2D), so that every element of the group can be written as a composition of reflection on these hyper-planes. You might want to check that all symmetries of a triangle can be written in such a way. This is clear for the reflection symmetries. Every rotational symmetry of the triangle is the composition of two reflections.

• Now, choose an arbitrary point $v\in\Bbb R^2$ (most often you do not want to choose the oirigin though). This point is visualized as the red point in the images below.

• Generate the set of all reflection of that point on all the hyper-planes in your reflection group. Also generate all reflections of reflections, and all reflections of reflections of reflections, and so on. That is, we consider the set $\{Tv\in\Bbb R^2\mid T\in I_3\}$. In the image below, these reflections are the black points.

• Finally, take the convex hull of these points to obtain a polytope (this is also called orbit polytope, because the vertices of that polytope are the orbit of $v$ under a group).

Now, above image shows what polytope is obtained for the same group $I_3$ with different choices of the point $v$. One immediate observation is that all these polytopes have all the elements of $I_3$ as symmetries (reflections at certain axes and $120^\circ$ rotations). So, the triangle on the left, as well as the "truncated triangle" in the middle and the hexagon on the right are invariant under all tranformations in $I_3$. But what stands out is, that the right polygon has in fact much more symmetries than the other two (e.g. rotations by $60^\circ$). These additional symmetries cannot be written as compositions of elements in $I_3$.

That is our core observation: a polytope generated by some matrix group $G$ has all the symmetries in $G$, but might have more than these. Obviously, the hexagon could have been obtained in a different way, starting from the reflection group $I_6$ (the actual symmetry group of the hexagon). The reflections of $I_6$ are shown in the image below on the right.

$\qquad\qquad\qquad$

So there are several ways to generate a hexagon by a reflection group. You might consider it as a $I_6$-polytope, or as a $I_3$-polytope. So, a polytope from the $I_3$-family will have all the symmetries in $I_3$, but might have more! The symmetry group of a $I_3$-polytope contains $I_3$, but might be larger.

And this is exactly what happens in the case of the 4-demicube. It just so happens, that when you start from the $D_4$-reflection group, and you generate a polytope in the above way, you then obtain the same polytope (and this means exactly the same polytope, the exact same vertices, faces etc.) as when you start from $B_4$ (for an appropriately chosen point $v$). And so there are just two ways to obtain the 16-cell, one via $D_4$ and one via $B_4$. Considered as plain polytopes without any further colors etc., the 4-demicube and the 16-cell are indistinguishable, and both have the symmetry group $B_4$ (which contains $D_4$ as a subgroup).

The construction of the $n$-demicube is quite simple: the edge-graph of the $n$-cube is bipartite. Choose one of the partition classes and delete these vertices from the cube. The convex hull of the remaining vertices is the $n$-demicube. You might try to visualize this in 3D. There you will observe that the 3-demicube is the tetrahedron. So you have two ways to generate the tetrahedron: as a $A_3$-polytope, and as a $D_3$-polytope. However, in 3D these two groups have the same number of elements. This coincidence does not happen in higher dimensions. And as far as I know, in $n\ge 5$ dimensions, the $n$-demicube is not just a polytope from one of the other classes.

Answer 2

Just as the tetrahedron can be vertex inscribed into a cube, in fact it is the vertex alternated faceting of the cube, the hexadecachoron (16-cell) too can be vertex inscribed into the tesseract. And again it is nothing but its vertex alternated faceting.

The same holds for any demi-hypercube: it is nothing but the vertex alternated faceting of the hypercube - for any higher dimension. In fact the above ones are just the concrete examples of dimensions 3 and 4.

Due to that vertex alternation the number of vertices of the demi-hypercube always is just half of the number of vertices of the containing hypercube itself.

Obviously any square by this vertex alternation gets replaced by one of its diagonals. Accordingly the edge length of the vertex inscribed demi-hypercube always will be $\sqrt{2}$ times as large as the edge length of the containing hypercube. And furthermore the number of squares of the hypercube coincides with the number of edges of the demi-hypercube.

That the hexadecachoron happens to be dual to the tesseract is a mere coincidence of dimension 4. This will not be the case beyond. And obviously it neither is so within dimension 3.

--- rk

Edit/Addition: For usual the symmetry of the demicube is just have of that of the (hyper)cube because of the to be applied vertex alternation. But in 4D this is different: both the alternated facet cubes and the additionally introduced vertex figures here are isomorphic regular tetrahedra. Thus, this alternation result here has an additional color interchanging symmetry, then resulting in the full symmetry again.

So eg. in 5D the according demihypercube will have 2 different facets: the alternated tesseractic facets (thus the above 16-cells) plus the thereby additionally introduced vertex figures, ie. 5-cells. Thence there is no color interchanging additional symmetry, or in other words $[B_5:D_5]=2$. - This is what occurs in all diemensions above in a quite similar manner.

Sure in 3D there also is just a single facet type in that alternated figure (the tetrahedron is indeed a regular polyhedron), but note what is going on there: The vertex alternations of the cube faces (squares) are their diagonal lines only, ie. those are becoming subdimensional here and thus do not count as further facets. Here only the vertex figures of half the former vertices (the alternates) come in, thus providing the only faces, the triangles.

--- rk

Answer 3

Let's use here the typewriter friendly inline notation of Coxeter-Dynkin graphs. Then demi-hypercubic $D_4$ symmetry is denoted as o3o3o *b3o, i.e. the tridental graph, as the virtual node *b is just the revisit of the b-th real node (from the left). And the hypercubic $B_4$ symmetry is denoted as o3o3o4o. Those symbols just encrypt the submultiplicatives of $\pi$ (in radians) of the angles between according mirror types, which in turn are being denoted by the various nodes.

Next, by Wythoff's kaleidoscopical construction the various Mirror symmetrical uniform polytopes can be encoded by the various ringings of the nodes of those Coxeter-Dynkin graphs. Thereby any ringed node itself represents a single edge class, in fact the off-ness of the generating point (vertex) from the according mirror. The size of that edge type can be thought to vary, but for uniform polytopes it is pre-defined to be unity. Such ringed nodes (of unit size) usually are represented in this typewriter friendly notation by the symbol x. On the other hand, several nodes might remain un-ringed (i.e. still o). This represents occasions where the generating point of that construction just happens to be on the specific mirror (off-ness size zero).

Now it happens that any mirror symmetric uniform polytope, which can be encoded wrt. the $D_4$ symmetry, also can be given wrt. the $B_4$ symmetry:

x3o3o *b3o = x3o3o4o = hexadecachoron
o3x3o *b3o = o3x3o4o = icositetrachoron
x3x3o *b3o = x3x3o4o = truncated hexadecachoron
x3o3x *b3o = o3o3x4o = rectified tesseract
x3x3x *b3o = o3x3x4o = tesseracti-hexadecachoron
x3o3x *b3x = x3o3x4o = rectified icositetrachoron
x3x3x *b3x = x3x3x4o = truncated icositetrachoron

Thus any possible decoration of the tridental graph gets reflected into a decoration of the linear graph, there leaving out the last node wrt. ringings. Thus all polytopes with full $D_4$ symmetry necessarily have their vertices on that mirror type, which is represented by the last node in the $B_4$ symmetry.

Therefore e.g. the hypercube (tesseract, o3o3o4x) itself cannot be given wrt. $D_4$ symmetry in kaleidoscopical construction (i.e. with a single generating point / vertex).

--- rk