
This polytope goes by various names. Most people call it the root polytope of \(\mathrm{E}_8\). Jonathan Bowers called it the
apparently following the principle that an absurdly complex shape needs an absurdly complicated name. But it was discovered by Thorold Gosset in his 1900 paper classifying semiregular polytopes, and he called it the 8ic semiregular figure, since it's the only semiregular polytope in 8dimensions... and the highestdimensional semiregular polytope that isn't regular!
Remember:
(Uniform and semiregular polytopes are the same in 3 dimensions, since uniform and semiregular and regular polytopes are the same in 2 dimensions, but the differences become noticeable in higher dimensions.)
Coxeter later called 'Gosset's 8ic semiregular figure' the 4_{21} polytope, as part of a systematic naming scheme, since it's part of a little series of semiregular polytopes built from the E_{n} Coxeter groups.
Now, a honeycomb is the higherdimensional analogue of a tiling of the plane, and from the viewpoint of Coxeter groups, it makes a lot of sense to treat honeycombs in Euclidean and hyperbolic space as generalized polytopes. If we do that, the 4_{21} polytope is not the end of the line. There's also a honeycomb in 8dimensional Euclidean space whose symmetry group is the Coxeter group \(\mathrm{E}_9\), and a honeycomb in 9dimensional hyperbolic space whose symmetry group is the Coxeter group \(\mathrm{E}_{10}\)!
Though you might not have noticed, since I didn't use the word 'honeycomb', I've secretly been alluding to both these guys in Part 1, Part 2 and Part 3 of this series:
This honeycomb is made of 8dimensional simplexes and 8dimensional orthoplexes. Remember, a nsimplex is the regular polytope in \(n\) dimensions with \(n+1\) equidistant vertices: it's like a generalized tetrahedron. The northoplex is the regular polytope in \(n\) dimensions with vertices that look like this in a suitable coordinate system: $$\begin{array}{c} (\pm 1 , 0 ,\dots, 0), \\ (0, \pm 1, \dots, 0 ), \\ ..\dots\dots\dots, \\ (0, 0, \dots, \pm 1) \end{array} $$ It's like a generalized octahedron. So, you can think of the 5_{21} honeycomb as an 8dimensional relative of the tetrahedraloctahedral honeycomb in 3 dimensions:
I would like to keep exploring these structures and how they're related to integral octonions... but we should start small, and look at the dischiliahectohexacontamyriaheptachiliadiacosioctacontazetton.
Here are some fun facts about this 8dimensional polytope:
Now, when you hear something has 17280 7dimensional simplexes as facets, you should want to run away and scream! But overcoming this impulse, and actually understanding this sort of thing, has considerable rewards. So, let's give it a try.
While girding myself for a task like this, I generally find it helpful to eat a 7dimensional snack. Luckily it's easy to find one here in Singapore:
Okay, I'm ready now. First, remember that the vertices of our polytope are these points:
That gives \(112 + 128 = 240\) vertices.
Next, what about the edges? In fact each vertex is connected by an edge to each of its 56 nearest neighbors, but each edge connects two vertices, so our polytope has a total of $$ \frac{ 240 \times 56}{2} = 120 \times 56 = 5600 + 1120 = 6720 $$
edges.Why does each vertex have 56 nearest neighbors? We saw this last time, but it will be good to remember, since we'll need the ideas for our next calculation too. The idea is to take the 240 vertices and notice how they lie on 5 hyperplanes:
If you take your favorite vertex to be the one where the sum of all coordinates is 4:
$$ (\frac{1}{2}, \frac{1}{2}, \frac{1}{2}, \frac{1}{2}, \frac{1}{2}, \frac{1}{2}, \frac{1}{2}, \frac{1}{2}) $$then you can show its nearest neighbors are the 56 where the sum of all coordinates is 2:
$$ (1, 1, 0, 0, 0, 0, 0, 0 ) $$ $$ (\frac{1}{2}, \frac{1}{2}, \frac{1}{2}, \frac{1}{2}, \frac{1}{2}, \frac{1}{2}, \frac{1}{2}, \frac{1}{2}) $$and permutations of these. You can easily check that the distance of our favorite guy to any of these is \(\sqrt{2}\), just as its distance to the origin is \(\sqrt{2}\).
Next, why does our polytope have facets shaped like 7orthoplexes?
To see this, let's think about how we get one of these topdimensional faces. We get it by taking a plane far from origin and slowly moving it straight toward the origin until it hits our polytope. Sometimes when we do this we're unlucky and the plane hits a single vertex, or 2, or 3... but sometimes it hits 8 or more, and then it has really hit a topdimensional face, or facet! If it hits exactly 8 vertices at the same time, that facet is a 7simplex.
Let's take a hyperplane defined by some equation like
$$ \ell (x) = c $$where \(\ell : \mathbb{R}^8 \to \mathbb{R}\) is a linear functional. We'll start with \(c\) big and reduce it until our hyperplane first hits a vertex of the polytope.
What linear functional should we use?
One obvious choice is the sum of all 8 coordinates:
$$ \ell(x) = x_1 + \cdots + x_8 $$But we know already this won't work. If we use this, our hyperplane first hits the polytope when \(\ell(x) = 2\), and then it hits just one point, our favorite point:
$$ (\frac{1}{2}, \frac{1}{2}, \frac{1}{2}, \frac{1}{2}, \frac{1}{2}, \frac{1}{2}, \frac{1}{2}, \frac{1}{2}) $$So, we need to try something else. Another obvious choice is
$$ \ell(x) = x_1 $$If we use this, our hyperplane first hits the polytope when \(\ell(x) = 1\), and then it hits a whole bunch of points, namely
$$ (1, \pm 1, 0, 0, 0, 0, 0, 0) $$ $$ (1, 0, \pm 1, 0, 0, 0, 0, 0) $$ $$ (1, 0, 0, \pm 1, 0, 0, 0, 0) $$ $$ (1, 0, 0, 0, \pm 1, 0, 0, 0) $$ $$ (1, 0, 0, 0, 0, \pm 1, 0, 0) $$ $$ (1, 0, 0, 0, 0, 0, \pm 1, 0) $$ $$ (1, 0, 0, 0, 0, 0, 0, \pm 1) $$Hey, these are the vertices of a 7dimensional orthoplex! So that's how we get the orthoplex facets.
Puzzle 1. Show that there are 2160 7orthoplex facets.
It may helpful to note that $$2160 = 240 \times 9$$
Next, let's try to find the 7simplex facets.
To find these, we can try the same trick we used to find the 7orthoplex facets, just with a different linear functional \(\ell\).
For example, we could try letting \(\ell(x)\) be the sum of the first two coordinates. Then the hyperplane \(\ell (x) = c\) first hits our polytope when \(c\) decreases to \(2\), and it hits it in a single point. We could let \(\ell(x)\) be the sum of the first three coordinates, but then we get a triangle. Or we could let \(\ell(x)\) be the sum of the first four coordinates, but then we get another 7orthoplex.
So we have to try something weirder, and this point I got a lot of help from Greg Egan. He used the hyperplane
$$ \ell(x) = 3 $$where
$$ \ell(x) = 2x_1 +x_2 + x_3 + x_4 + x_5 $$and he showed this hyperplane contains these 8 vertices:
$$(\frac{1}{2}, \frac{1}{2}, \frac{1}{2}, \frac{1}{2}, \frac{1}{2}, \frac{1}{2}, \frac{1}{2}, \frac{1}{2} )$$ $$(\frac{1}{2}, \frac{1}{2}, \frac{1}{2}, \frac{1}{2}, \frac{1}{2}, \frac{1}{2}, \frac{1}{2}, \frac{1}{2} )$$ $$(\frac{1}{2}, \frac{1}{2}, \frac{1}{2}, \frac{1}{2}, \frac{1}{2}, \frac{1}{2}, \frac{1}{2}, \frac{1}{2} )$$ $$(\frac{1}{2}, \frac{1}{2}, \frac{1}{2}, \frac{1}{2}, \frac{1}{2}, \frac{1}{2}, \frac{1}{2}, \frac{1}{2} )$$ $$(1,1,0,0,0,0,0,0)$$ $$(1,0,1,0,0,0,0,0)$$ $$(1,0,0,1,0,0,0,0)$$ $$(1,0,0,0,1,0,0,0)$$These are all at distance \(\sqrt{2}\) from each other, so they're the vertices of a 7simplex!
Puzzle 2. Show there are 17280 7simplex facets.
It may be helpful to note that \(17280 = 240 \times 72\).
Puzzle 1. Show that the \(\mathrm{E}_8\) root polytope has 2160 7orthoplex facets.
Answer. Here is Greg Egan's answer:
Moving on to the 7orthoplexes...
John already gave the prototypical normal \((1,0,0,0,0,0,0,0)\) and listed the fourteen vertices with which it has a dot product of 1. We can put the 1 in any coordinate, and we can negate it, giving a total of 16 facets.
For normals of the form $$\left(\pm \frac{1}{2},\pm \frac{1}{2},\pm \frac{1}{2},\pm \frac{1}{2},0,0,0,0\right)$$ and permutations of the coordinates there are \(2^4 \cdot \binom{8}{4} = 1120\) possibilities. For example, $$\left(\frac{1}{2},\frac{1}{2},\frac{1}{2},\frac{1}{2},0,0,0,0\right)$$ has a dot product of 1 with the following 14 vertices:
$$\begin{array}{l} \left( \frac{1}{2} , \frac{1}{2} , \frac{1}{2} , \frac{1}{2} , \frac{1}{2} , \frac{1}{2} , \frac{1}{2} , \frac{1}{2} \right)\\ \left( \frac{1}{2} , \frac{1}{2} , \frac{1}{2} , \frac{1}{2} , \frac{1}{2} , \frac{1}{2} , \frac{1}{2} , \frac{1}{2} \right)\\ \left( \frac{1}{2} , \frac{1}{2} , \frac{1}{2} , \frac{1}{2} , \frac{1}{2} , \frac{1}{2} , \frac{1}{2} , \frac{1}{2} \right)\\ \left( \frac{1}{2} , \frac{1}{2} , \frac{1}{2} , \frac{1}{2} , \frac{1}{2} , \frac{1}{2} , \frac{1}{2} , \frac{1}{2} \right)\\ \left( \frac{1}{2} , \frac{1}{2} , \frac{1}{2} , \frac{1}{2} , \frac{1}{2} , \frac{1}{2} , \frac{1}{2} , \frac{1}{2} \right)\\ \left( \frac{1}{2} , \frac{1}{2} , \frac{1}{2} , \frac{1}{2} , \frac{1}{2} , \frac{1}{2} , \frac{1}{2} , \frac{1}{2} \right)\\ \left( \frac{1}{2} , \frac{1}{2} , \frac{1}{2} , \frac{1}{2} , \frac{1}{2} , \frac{1}{2} , \frac{1}{2} , \frac{1}{2} \right)\\ \left( \frac{1}{2} , \frac{1}{2} , \frac{1}{2} , \frac{1}{2} , \frac{1}{2} , \frac{1}{2} , \frac{1}{2} , \frac{1}{2} \right)\\ \left( 1 , 1 , 0 , 0 , 0 , 0 , 0 , 0 \right)\\ \left( 1 , 0 , 1 , 0 , 0 , 0 , 0 , 0 \right)\\ \left( 1 , 0 , 0 , 1 , 0 , 0 , 0 , 0 \right)\\ \left( 0 , 1 , 1 , 0 , 0 , 0 , 0 , 0 \right)\\ \left( 0 , 1 , 0 , 1 , 0 , 0 , 0 , 0 \right)\\ \left( 0 , 0 , 1 , 1 , 0 , 0 , 0 , 0 \right)\\ \end{array}$$
For normals of the form $$\left(\pm \frac{3}{4},\pm \frac{1}{4},\pm \frac{1}{4},\pm \frac{1}{4},\pm \frac{1}{4},\pm \frac{1}{4},\pm \frac{1}{4},\pm \frac{1}{4}\right)$$ with an odd number of minus signs and permutations of the coordinates there are \(8 \cdot \left(\binom{8}{1}+\binom{8}{3}+\binom{8}{5}+\binom{8}{7}\right) = 1024\) possibilities. For example, $$\left(\frac{3}{4},\frac{1}{4},\frac{1}{4},\frac{1}{4},\frac{1}{4},\frac{1}{4},\frac{1}{4},\frac{1}{4}\right)$$ has a dot product of 1 with the following 14 vertices:
$$\begin{array}{l} \left( \frac{1}{2} , \frac{1}{2} , \frac{1}{2} , \frac{1}{2} , \frac{1}{2} , \frac{1}{2} , \frac{1}{2} , \frac{1}{2} \right)\\ \left( \frac{1}{2} , \frac{1}{2} , \frac{1}{2} , \frac{1}{2} , \frac{1}{2} , \frac{1}{2} , \frac{1}{2} , \frac{1}{2} \right)\\ \left( \frac{1}{2} , \frac{1}{2} , \frac{1}{2} , \frac{1}{2} , \frac{1}{2} , \frac{1}{2} , \frac{1}{2} , \frac{1}{2} \right)\\ \left( \frac{1}{2} , \frac{1}{2} , \frac{1}{2} , \frac{1}{2} , \frac{1}{2} , \frac{1}{2} , \frac{1}{2} , \frac{1}{2} \right)\\ \left( \frac{1}{2} , \frac{1}{2} , \frac{1}{2} , \frac{1}{2} , \frac{1}{2} , \frac{1}{2} , \frac{1}{2} , \frac{1}{2} \right)\\ \left( \frac{1}{2} , \frac{1}{2} , \frac{1}{2} , \frac{1}{2} , \frac{1}{2} , \frac{1}{2} , \frac{1}{2} , \frac{1}{2} \right)\\ \left( \frac{1}{2} , \frac{1}{2} , \frac{1}{2} , \frac{1}{2} , \frac{1}{2} , \frac{1}{2} , \frac{1}{2} , \frac{1}{2} \right)\\ \left( 1 , 1 , 0 , 0 , 0 , 0 , 0 , 0 \right)\\ \left( 1 , 0 , 1 , 0 , 0 , 0 , 0 , 0 \right)\\ \left( 1 , 0 , 0 , 1 , 0 , 0 , 0 , 0 \right)\\ \left( 1 , 0 , 0 , 0 , 1 , 0 , 0 , 0 \right)\\ \left( 1 , 0 , 0 , 0 , 0 , 1 , 0 , 0 \right)\\ \left( 1 , 0 , 0 , 0 , 0 , 0 , 1 , 0 \right)\\ \left( 1 , 0 , 0 , 0 , 0 , 0 , 0 , 1 \right)\\ \end{array}$$
That gives a total of 16+1120+1024=2160 facets.
I suppose you could argue purely from symmetry that these sets of 14 vertices form 7orthoplexes, given that John's example clearly is a 7orthoplex.
Puzzle 2. Show that the \(\mathrm{E}_8\) root polytope has 17280 7simplex facets.
Answer. Here is Greg Egan's answer:
You can also read comments on the nCategory Café, and make your own comments or ask questions there!From the normals \((\pm 2,\pm 1,\pm 1,\pm 1,\pm 1,0,0,0)\) and permutations of their coordinates, there are \(2^5 \cdot 8 \cdot \binom{7}{3} =8960\) facets that are 7simplices.
From the normals \((\pm 1, \pm 1, \pm 1, \pm 1, \pm 1, \pm 1, \pm 1, \pm 1)\) with an odd number of minus signs there are \(\binom{8}{1}+\binom{8}{3}+\binom{8}{5}+\binom{8}{7}=128\) facets that are 7simplices.
From the normals \((\pm \frac{3}{2}, \pm \frac{3}{2}, \pm \frac{3}{2}, \pm \frac{1}{2},\pm \frac{1}{2},\pm \frac{1}{2},\pm \frac{1}{2},\pm \frac{1}{2} )\) with an odd number of minus signs and permutations of their coordinates, there are \(\binom{8}{3} \cdot \left(\binom{8}{1}+\binom{8}{3}+\binom{8}{5}+\binom{8}{7}\right)=7168\) facets that are 7simplices.
From the normals \((\pm \frac{5}{2}, \pm \frac{1}{2}, \pm \frac{1}{2}, \pm \frac{1}{2},\pm \frac{1}{2},\pm \frac{1}{2},\pm \frac{1}{2},\pm \frac{1}{2} )\) with an even number of minus signs and permutations of their coordinates, there are \(8 \cdot \left(\binom{8}{0}+\binom{8}{2}+\binom{8}{4}+\binom{8}{6}+\binom{8}{8}\right)=1024\) facets that are 7simplices.
The total of these counts is 17,280. This isn't a very rigorous enumeration, since I just found it by trial and error starting from \((2,1,1,1,1,0,0,0)\) and looking for other vectors with the same squared norm of 8.
I suppose I ought to exhibit the 8 vertices of the 7simplex for one example of each kind of normal vector. I already did that for \((2,1,1,1,1,0,0,0)\), so here are examples for the other three kinds.
In each case, the dot product of the normal vector with the vertices is 3.
For \((1,1,1,1,1,1,1,1)\), the 8 vertices of the 7simplex are:
$$ \begin{array}{l} \left(\frac{1}{2} , \frac{1}{2} , \frac{1}{2} , \frac{1}{2} , \frac{1}{2} , \frac{1}{2} , \frac{1}{2} , \frac{1}{2} \right)\\ \left( \frac{1}{2} , \frac{1}{2} , \frac{1}{2} , \frac{1}{2} , \frac{1}{2} , \frac{1}{2} , \frac{1}{2} , \frac{1}{2} \right)\\ \left( \frac{1}{2} , \frac{1}{2} , \frac{1}{2} , \frac{1}{2} , \frac{1}{2} , \frac{1}{2} , \frac{1}{2} , \frac{1}{2} \right)\\ \left( \frac{1}{2} , \frac{1}{2} , \frac{1}{2} , \frac{1}{2} , \frac{1}{2} , \frac{1}{2} , \frac{1}{2} , \frac{1}{2} \right)\\ \left( \frac{1}{2} , \frac{1}{2} , \frac{1}{2} , \frac{1}{2} , \frac{1}{2} , \frac{1}{2} , \frac{1}{2} , \frac{1}{2} \right)\\ \left( \frac{1}{2} , \frac{1}{2} , \frac{1}{2} , \frac{1}{2} , \frac{1}{2} , \frac{1}{2} , \frac{1}{2} , \frac{1}{2} \right)\\ \left( \frac{1}{2} , \frac{1}{2} , \frac{1}{2} , \frac{1}{2} , \frac{1}{2} , \frac{1}{2} , \frac{1}{2} , \frac{1}{2} \right)\\ \left( \frac{1}{2} , \frac{1}{2} , \frac{1}{2} , \frac{1}{2} , \frac{1}{2} , \frac{1}{2} , \frac{1}{2} , \frac{1}{2} \right) \end{array} $$
For \(\left(\frac{3}{2},\frac{3}{2},\frac{3}{2},\frac{1}{2},\frac{1}{2},\frac{1}{2},\frac{1}{2},\frac{1}{2}\right)\), the 8 vertices of the 7simplex are:
$$ \begin{array}{l} \left( \frac{1}{2} , \frac{1}{2} , \frac{1}{2} , \frac{1}{2} , \frac{1}{2} , \frac{1}{2} , \frac{1}{2} , \frac{1}{2} \right)\\ \left( \frac{1}{2} , \frac{1}{2} , \frac{1}{2} , \frac{1}{2} , \frac{1}{2} , \frac{1}{2} , \frac{1}{2} , \frac{1}{2} \right)\\ \left( \frac{1}{2} , \frac{1}{2} , \frac{1}{2} , \frac{1}{2} , \frac{1}{2} , \frac{1}{2} , \frac{1}{2} , \frac{1}{2} \right)\\ \left( \frac{1}{2} , \frac{1}{2} , \frac{1}{2} , \frac{1}{2} , \frac{1}{2} , \frac{1}{2} , \frac{1}{2} , \frac{1}{2} \right)\\ \left( \frac{1}{2} , \frac{1}{2} , \frac{1}{2} , \frac{1}{2} , \frac{1}{2} , \frac{1}{2} , \frac{1}{2} , \frac{1}{2} \right)\\ \left( 1 , 1 , 0 , 0 , 0 , 0 , 0 , 0 \right)\\ \left( 1 , 0 , 1 , 0 , 0 , 0 , 0 , 0 \right)\\ \left( 0 , 1 , 1 , 0 , 0 , 0 , 0 , 0 \right) \end{array} $$
For \(\left(\frac{5}{2},\frac{1}{2},\frac{1}{2},\frac{1}{2},\frac{1}{2},\frac{1}{2},\frac{1}{2},\frac{1}{2}\right)\), the 8 vertices of the 7simplex are:
$$ \begin{array}{l} \left(\frac{1}{2} , \frac{1}{2} , \frac{1}{2} , \frac{1}{2} , \frac{1}{2} , \frac{1}{2} , \frac{1}{2} , \frac{1}{2} \right)\\ \left( 1 , 1 , 0 , 0 , 0 , 0 , 0 , 0 \right)\\ \left( 1 , 0 , 1 , 0 , 0 , 0 , 0 , 0 \right)\\ \left( 1 , 0 , 0 , 1 , 0 , 0 , 0 , 0 \right)\\ \left( 1 , 0 , 0 , 0 , 1 , 0 , 0 , 0 \right)\\ \left( 1 , 0 , 0 , 0 , 0 , 1 , 0 , 0 \right)\\ \left( 1 , 0 , 0 , 0 , 0 , 0 , 1 , 0 \right)\\ \left( 1 , 0 , 0 , 0 , 0 , 0 , 0 , 1 \right)\\ \end{array} $$
