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August 3, 2013

Integral Octonions (Part 4)

Posted by John Baez

This mini-series will take a break soon, since I’ll be wandering around China from August 6th to 20th. But before I go, I’d like to say a bit about the 8-dimensional polytope whose vertices are the 240 integral octonions with smallest norm (apart from zero)—or if you prefer, the 240 root vectors of E 8\mathrm{E}_8.

This polytope goes by various names. Most people call it the root polytope of E 8\mathrm{E}_8. Jonathan Bowers called it the dischiliahectohexacontamyriaheptachiliadiacosioctacontazetton, 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 8-ic semiregular figure, since it’s the only semiregular polytope in 8-dimensions… and the highest-dimensional semiregular polytope that isn’t regular!

Remember:

  • A regular polytope is one whose symmetry group acts transitively on its flags, where a flag is a vertex lying on an edge lying on a face lying on… etc.
  • A uniform polytope is one whose symmetry group acts transitively on its vertices and whose top-dimensional faces are all uniform.
  • A semiregular polytope is a uniform polytope whose top-dimensional faces are all regular.

(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 8-ic semiregular figure’ the 421 polytope, as part of a systematic naming scheme, since it’s part of a little series of semiregular polytopes built from the En Coxeter groups.

Now, a honeycomb is the higher-dimensional 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 421 polytope is not the end of the line. There’s also a honeycomb in 8-dimensional Euclidean space whose symmetry group is the Coxeter group E 9\mathrm{E}_9, and a honeycomb in 9-dimensional hyperbolic space whose symmetry group is the Coxeter group E 10\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:

  • The 8-dimensional honeycomb, which Coxeter called 521 honeycomb, is a way of packing 8d Euclidean space with regular polytopes such that there’s one vertex at each point of the E 8\mathrm{E}_8 lattice. The honeycomb itself has the Coxeter group E 9\mathrm{E}_9 as symmetries. This is the ‘affine’ version of E 8\mathrm{E}_8, so it includes the translations preserving the E 8\mathrm{E}_8 lattice, but also a bunch of reflections.

    This honeycomb is made of 8-dimensional simplexes and 8-dimensional orthoplexes. Remember, a n-simplex is the regular polytope in nn dimensions with n+1n+1 equidistant vertices: it’s like a generalized tetrahedron. The n-orthoplex is the regular polytope in nn dimensions with vertices that look like this in a suitable coordinate system: (±1,0,,0),(0,±1,,0),,(0,0,,±1) (\pm 1 , 0 ,\dots, 0), \quad (0, \pm 1, \dots, 0 ), \quad \dots, \quad (0, 0, \dots, \pm 1) It’s like a generalized octahedron. So, you can think of the 521 honeycomb as an 8-dimensional relative of the tetrahedral-octahedral honeycomb in 3 dimensions:



  • The 9-dimensional honeycomb, which Coxeter called the 621 honeycomb, is a way of packing 9-dimensional hyperbolic space with regular polytopes that has the Coxeter group E10 as symmetries. This honeycomb is made of 9-dimensional simplexes and 9-dimensional orthoplexes.

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 8-dimensional polytope:

  • It has 240 vertices.
  • It has 6720 edges.
  • It has 19440 facets, where a facet is a top-dimensional face—so in this case, a 7-dimensional face.
  • It has 17280 facets that are 7-simplexes.
  • All 2160 of the other facets are 7-orthoplexes.

Now, when you hear something has 17280 7-dimensional 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 7-dimensional 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:

4×(82)=1124 \times \binom{8}{2} = 112 vectors like this: (±1,±1,0,0,0,0,0,0)(\pm 1,\pm 1,0,0,0,0,0,0) where we take any combination of signs and any permutation of coordinates, and

2 7=1282^7 = 128 vector like this: (±12,±12,±12,±12,±12,±12,±12,±12) (\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},\pm \frac{1}{2}) where we take an even number of minus signs.

That gives 112+128=240112 + 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

240×562=120×56=5600+1120=6720 \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:

  • There is 11 vertex where the sum of the coordinates is 44: (12,12,12,12,12,12,12,12) (\frac{1}{2}, \frac{1}{2}, \frac{1}{2}, \frac{1}{2}, \frac{1}{2}, \frac{1}{2}, \frac{1}{2}, \frac{1}{2})
  • There are 5656 vertices where the sum of the coordinates is 22. There are 2828 like this: (1,1,0,0,0,0,0,0) (1, 1, 0, 0, 0, 0, 0, 0 ) and all permutations of these numbers, and there are 2828 like this: (12,12,12,12,12,12,12,12) (\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 all permutations of these.
  • There are 126126 roots where the sum of the coordinates is 00. There are 5656 like this: (1,1,0,0,0,0,0,0) (1, -1, 0, 0, 0, 0, 0, 0 ) and all permutations of these numbers. And there are 7070 like this: (12,12,12,12,12,12,12,12) (\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 all permutations of these.
  • There are 5656 roots where the sum of the coordinates is 2-2. These are just the negatives of those where the sum is 1, namely (1,1,0,0,0,0,0,0) (-1, -1, 0, 0, 0, 0, 0, 0 ) and all its permutations, and (12,12,12,12,12,12,12,12) (\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 all its permutations.
  • There is 11 root where the sum of the coordinates is 4-4, namely (12,12,12,12,12,12,12,12) (-\frac{1}{2}, -\frac{1}{2}, -\frac{1}{2}, -\frac{1}{2}, -\frac{1}{2}, -\frac{1}{2}, -\frac{1}{2}, -\frac{1}{2})

If you take your favorite vertex to be the one where the sum of all coordinates is 4:

(12,12,12,12,12,12,12,12) (\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) (1, 1, 0, 0, 0, 0, 0, 0 ) (12,12,12,12,12,12,12,12) (\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 2\sqrt{2}, just as its distance to the origin is 2\sqrt{2}.

Next, why does our polytope have facets shaped like 7-orthoplexes?

To see this, let’s think about how we get one of these top-dimensional 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 top-dimensional face, or facet! If it hits exactly 8 vertices at the same time, that facet is a 7-simplex.

Let’s take a hyperplane defined by some equation like

(x)=c \ell (x) = c

where : 8\ell : \mathbb{R}^8 \to \mathbb{R} is a linear functional. We’ll start with cc 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:

(x)=x 1++x 8 \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 (x)=2\ell(x) = 2, and then it hits just one point, our favorite point:

(12,12,12,12,12,12,12,12) (\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

(x)=x 1 \ell(x) = x_1

If we use this, our hyperplane first hits the polytope when (x)=1\ell(x) = 1, and then it hits a whole bunch of points, namely

(1,±1,0,0,0,0,0,0) (1, \pm 1, 0, 0, 0, 0, 0, 0) (1,0,±1,0,0,0,0,0) (1, 0, \pm 1, 0, 0, 0, 0, 0) (1,0,0,±1,0,0,0,0) (1, 0, 0, \pm 1, 0, 0, 0, 0) (1,0,0,0,±1,0,0,0) (1, 0, 0, 0, \pm 1, 0, 0, 0) (1,0,0,0,0,±1,0,0) (1, 0, 0, 0, 0, \pm 1, 0, 0) (1,0,0,0,0,0,±1,0) (1, 0, 0, 0, 0, 0, \pm 1, 0) (1,0,0,0,0,0,0,±1) (1, 0, 0, 0, 0, 0, 0, \pm 1)

Hey, these are the vertices of a 7-dimensional orthoplex! So that’s how we get the orthoplex facets.

Problem 1. Show that there are 2160 7-orthoplex facets.

I haven’t done this yet. It may helpful to note that 2160=240×92160 = 240 \times 9.

Next, let’s try to find the 7-simplex facets.

To find these, we can try the same trick we used to find the 7-orthoplex facets, just with a different linear functional \ell.

For example, we could try letting (x)\ell(x) be the sum of the first two coordinates. Then the hyperplane (x)=c \ell (x) = c first hits our polytope when cc decreases to 22, and it hits it in a single point. We could let (x)\ell(x) be the sum of the first three coordinates, but then we get a triangle. Or we could let (x)\ell(x) be the sum of the first four coordinates, but then we get another 7-orthoplex.

So we have to try something weirder. At this point I started making calculational mistakes, which I’ve corrected thanks to Greg Egan’s comments below. He used the hyperplane

(x)=3 \ell(x) = 3

where

(x)=2x 1+x 2+x 3+x 4+x 5 \ell(x) = 2x_1 +x_2 + x_3 + x_4 + x_5

This hyperplane contains the 8 vertices

(12,12,12,12,12,12,12,12)(\frac{1}{2}, \frac{1}{2}, \frac{1}{2}, \frac{1}{2}, \frac{1}{2}, \frac{1}{2}, \frac{1}{2}, \frac{1}{2} )

(12,12,12,12,12,12,12,12)(\frac{1}{2}, \frac{1}{2}, \frac{1}{2}, \frac{1}{2}, \frac{1}{2}, \frac{1}{2}, -\frac{1}{2}, -\frac{1}{2} )

(12,12,12,12,12,12,12,12)(\frac{1}{2}, \frac{1}{2}, \frac{1}{2}, \frac{1}{2}, \frac{1}{2}, -\frac{1}{2}, \frac{1}{2}, -\frac{1}{2} )

(12,12,12,12,12,12,12,12)(\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)(1,1,0,0,0,0,0,0)

(1,0,1,0,0,0,0,0)(1,0,1,0,0,0,0,0)

(1,0,0,1,0,0,0,0)(1,0,0,1,0,0,0,0)

(1,0,0,0,1,0,0,0)(1,0,0,0,1,0,0,0)

These are all at distance 2\sqrt{2} from each other, so they’re the vertices of a 7-simplex.

Problem 2. Show there are 17280 7-simplex facets.

I haven’t done this either. It may be helpful to note that 17280=240×7217280 = 240 \times 72.

So there’s more work for me to do—or you, if you choose. For these deeper problems, it’s probably good to use more high-powered math, like the theory of Coxeter groups, though a brute force approach should still succeed.

Posted at August 3, 2013 11:11 AM UTC

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Re: Integral Octonions (Part 4)

As far as I can see, the linear equation 2x 1x 2=32x_1-x_2=3 is only satisfied by a single vertex, (1,1,0,0,0,0,0,0)(1,-1,0,0,0,0,0,0). The other points you give in your 6-simplex all have x 1=1x_1=1 and x 2=0x_2=0.

Unrelated to that, I think you’ve mixed two normalisation conventions in this post: you initially describe vertices that are twice as large as the version you work with later in the post.

Posted by: Greg Egan on August 4, 2013 6:26 AM | Permalink | Reply to this

Re: Integral Octonions (Part 4)

Whoops! I meant that

(x)=2x 1x 2x 3x 4x 4x 5x 6x 7 \ell(x) = 2x_1 - x_2 - x_3 - x_4 - x_4 - x_5 - x_6 - x_7

equals 3 at the seven vertices I listed:

(1,1,0,0,0,0,0,0) (1, -1, 0, 0, 0, 0, 0, 0) (1,0,1,0,0,0,0,0) (1, 0, -1, 0, 0, 0, 0, 0) (1,0,0,1,0,0,0,0) (1, 0, 0, -1, 0, 0, 0, 0) (1,0,0,0,1,0,0,0) (1, 0, 0, 0, -1, 0, 0, 0) (1,0,0,0,0,1,0,0) (1, 0, 0, 0, 0, -1, 0, 0) (1,0,0,0,0,0,1,0) (1, 0, 0, 0, 0, 0, -1, 0) (1,0,0,0,0,0,0,1) (1, 0, 0, 0, 0, 0, 0, -1)

But for some strange reason I wrote

(x)=2x 1x 2 \ell(x) = 2x_1 - x_2

and this mistake infected my later calculations. So, the specific way I planned to ‘tilt the hyperplane’ doesn’t work. However, if the vertices I listed are seven of the eight vertices of some 7-simplex face, there must be some way to tilt the hyperplane until it hits the eighth. I felt sure we could use

(12,12,12,12,12,12,12,12)(\frac{1}{2}, \frac{1}{2}, \frac{1}{2}, \frac{1}{2}, \frac{1}{2}, \frac{1}{2}, \frac{1}{2}, \frac{1}{2})

as the eighth, and I still feel confident, but I should check this.

In general, there’s always the danger that when you tilt a hyperplane touching seven vertices of the E 8\mathrm{E}_8 root polytope until it touches an eighth, it winds up touching 14: the vertices of a 7-orthoplex.

There’s also the danger that a plane containing eight equidistant vertices of the polytope actually cuts through the polytope, instead of just touching a face.

These dangers can be eliminated by simple calculations of the sort I outlined… but only if you don’t screw up.

Posted by: John Baez on August 4, 2013 11:17 AM | Permalink | Reply to this

Re: Integral Octonions (Part 4)

Ah, okay! That makes a lot more sense now!

Unfortunately, though, if you follow through your original calculation with (2,1,1,1,1,1,1,1)+α(1,1,1,1,1,1,1,1)(2,-1,-1,-1,-1,-1,-1,-1)+\alpha(1,1,1,1,1,1,1,1) as the normal and then require that its dot product with (12,12,12,12,12,12,12,12)\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) is equal to 3, you get α=1\alpha=1 and a normal of (3,0,0,0,0,0,0,0)(3,0,0,0,0,0,0,0), which is just the normal to your 7-orthoplex example.

Posted by: Greg Egan on August 4, 2013 11:43 AM | Permalink | Reply to this

Re: Integral Octonions (Part 4)

I think I can get one 7-simplex from (2,1,1,1,1,0,0,0)v=3(2,1,1,1,1,0,0,0)\cdot v=3, which is satisfied by the vertices:

(12,12,12,12,12,12,12,12)(\frac{1}{2}, \frac{1}{2}, \frac{1}{2}, \frac{1}{2}, \frac{1}{2}, \frac{1}{2}, \frac{1}{2}, \frac{1}{2} )

(12,12,12,12,12,12,12,12)(\frac{1}{2}, \frac{1}{2}, \frac{1}{2}, \frac{1}{2}, \frac{1}{2}, \frac{1}{2}, -\frac{1}{2}, -\frac{1}{2} )

(12,12,12,12,12,12,12,12)(\frac{1}{2}, \frac{1}{2}, \frac{1}{2}, \frac{1}{2}, \frac{1}{2}, -\frac{1}{2}, \frac{1}{2}, -\frac{1}{2} )

(12,12,12,12,12,12,12,12)(\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)(1,1,0,0,0,0,0,0)

(1,0,1,0,0,0,0,0)(1,0,1,0,0,0,0,0)

(1,0,0,1,0,0,0,0)(1,0,0,1,0,0,0,0)

(1,0,0,0,1,0,0,0)(1,0,0,0,1,0,0,0)

Posted by: Greg Egan on August 4, 2013 7:10 AM | Permalink | Reply to this

Re: Integral Octonions (Part 4)

From the normals (±2,±1,±1,±1,±1,0,0,0)(\pm 2,\pm 1,\pm 1,\pm 1,\pm 1,0,0,0) and permutations of their coordinates, there are 2 58(73)=89602^5 \cdot 8 \cdot \binom{7}{3} =8960 facets that are 7-simplices.

From the normals (±1,±1,±1,±1,±1,±1,±1,±1)(\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 (81)+(83)+(85)+(87)=128\binom{8}{1}+\binom{8}{3}+\binom{8}{5}+\binom{8}{7}=128 facets that are 7-simplices.

From the normals (±32,±32,±32,±12,±12,±12,±12,±12)(\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 (83)((81)+(83)+(85)+(87))=7168\binom{8}{3} \cdot \left(\binom{8}{1}+\binom{8}{3}+\binom{8}{5}+\binom{8}{7}\right)=7168 facets that are 7-simplices.

From the normals (±52,±12,±12,±12,±12,±12,±12,±12)(\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((80)+(82)+(84)+(86)+(88))=10248 \cdot \left(\binom{8}{0}+\binom{8}{2}+\binom{8}{4}+\binom{8}{6}+\binom{8}{8}\right)=1024 facets that are 7-simplices.

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)(2,1,1,1,1,0,0,0) and looking for other vectors with the same squared norm of 8.

Posted by: Greg Egan on August 4, 2013 9:33 AM | Permalink | Reply to this

Re: Integral Octonions (Part 4)

Wow, great! You’re so much better at calculations than I am. It’s sort of absurd for me to be studying the E 8\mathrm{E}_8 root polytope by means of calculations, when my forte is reading stuff and making connections between different things people say. But I need to try, to get some hands-on feel for this structure… even if I lose a few fingers in the process.

I do have a thought about how to rigorize, or at least pretty up, this type of calculation. You almost said it. We can try to characterize the facets of the E 8\mathrm{E}_8 root polytope more invariantly as planes orthogonal to certain vectors in the E 8\mathrm{E}_8 lattice. We seem to be getting:

  • The 7-orthoplex facets lie in planes of the form nx=2n \cdot x = 2 where nn is a vector in the E 8\mathrm{E}_8 lattice with nn=4n \cdot n = 4.
  • The 7-simplex facets lie in planes of the form nx=3n \cdot x = 3 where nn is a vector in the E 8\mathrm{E}_8 lattice with nn=8n \cdot n = 8.

There could theoretically be more facets, but at least we could count all the vectors in the E 8\mathrm{E}_8 lattice of a given small norm, and check that there are 2160 of norm squared 4 and 17280 of norm squared 8.

In fact people who study theta functions of lattices know how to calculate the number of vectors in the E 8\mathrm{E}_8 lattice of any given norm squared. Let’s see if I can find it by googling the numbers 240, 2160 and 17280.

Hmm, most of the top hits are about the E 8\mathrm{E}_8 root polytope. But there’s a promising one, a paper by Conway and Queen called ‘Computing the character table of a Lie group’, that contains the line:

This was done by computer since these sums contain respectively 240, 2160, 17280 terms.

(I see this on http://www.google.com.sg but not http://www.google.com, in case anyone cares.)

I also see something about the 5 215_{21} honeycomb in 8 dimensions:

Each vertex of the 5 215_{21} honeycomb is surrounded by 2160 8-orthoplexes and 17280 8-simplices.

But since this honeycomb has vertices at the E 8\mathrm{E}_8 lattice points, we should expect some nice relationship like this.

But there I go again: looking stuff up when a more athletic mathematician would just compute it.

Posted by: John Baez on August 4, 2013 11:58 AM | Permalink | Reply to this

Re: Integral Octonions (Part 4)

John wrote:

we could count all the vectors in the E8 lattice of a given small norm, and check that there are 2160 of norm squared 4 and 17280 of norm squared 8.

This seems to be almost true! With a computer calculation, I count 2160 lattice vectors of norm squared 4, but 17520=17280+240 lattice vectors of norm squared 8. For example, the vector (2,2,0,0,0,0,0,0)(2,2,0,0,0,0,0,0) has norm squared of 8 but it does not have a dot product of 3 with any vertices.

But of course (2,2,0,0,0,0,0,0)(2,2,0,0,0,0,0,0) is just double one of the vertices, and we wouldn’t expect a vector like that to be a facet normal. If we exclude those 240 doubled-vertices, the remaining 17280 are exactly the normals to the 7-simplices.

And though I did a brute-force computer calculation just to be absolutely sure, those combinatorial formulas I gave that counted the facet normals more or less amount to a proof of the same result, given that the properties of the lattice vectors are simple enough to check pretty easily that they cover all the possibilities.

Posted by: Greg Egan on August 4, 2013 2:11 PM | Permalink | Reply to this

Re: Integral Octonions (Part 4)

Greg wrote:

With a computer calculation, I count 2160 lattice vectors of norm squared 4, but 17520=17280+240 lattice vectors of norm squared 8.

Oh, okay: the 17280 and the 240 live in different orbits of the rotation/reflection symmetry group of the E 8\mathrm{E}_8 lattice, even though they have the same norm.

Table 3 at the end of that paper by Conway and Queen lists the sizes of various orbits—but they’re indexed by a notation that doesn’t make the norms of the vectors apparent to the untutored eye, so I didn’t bother to mention this. Now I’m a bit more motivated. The sizes of the orbits are:

1, 240, 2160, 5720, 240, 17280, 30240, 60480,…

So I imagine there’s 1 vector of norm squared 0, 240 of norm squared 2, 2160 of norm squared 4, 5720 of norm squared 6, 240+17280 of norm squared 8, and so on. The table goes on much further.

Posted by: John Baez on August 5, 2013 10:10 AM | Permalink | Reply to this

Re: Integral Octonions (Part 4)

I suppose I ought to exhibit the 8 vertices of the 7-simplex for one example of each kind of normal vector. I already did that for (2,1,1,1,1,0,0,0)(2,1,1,1,1,0,0,0) a few comments back, 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)(1,-1,1,1,1,1,1,1), the 8 vertices of the 7-simplex are:

(12,12,12,12,12,12,12,12) (12,12,12,12,12,12,12,12) (12,12,12,12,12,12,12,12) (12,12,12,12,12,12,12,12) (12,12,12,12,12,12,12,12) (12,12,12,12,12,12,12,12) (12,12,12,12,12,12,12,12) (12,12,12,12,12,12,12,12) \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 (32,32,32,12,12,12,12,12)\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 7-simplex are:

(12,12,12,12,12,12,12,12) (12,12,12,12,12,12,12,12) (12,12,12,12,12,12,12,12) (12,12,12,12,12,12,12,12) (12,12,12,12,12,12,12,12) (1,1,0,0,0,0,0,0) (1,0,1,0,0,0,0,0) (0,1,1,0,0,0,0,0) \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 (52,12,12,12,12,12,12,12)\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 7-simplex are:

(12,12,12,12,12,12,12,12) (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) (1,0,0,0,0,1,0,0) (1,0,0,0,0,0,1,0) (1,0,0,0,0,0,0,1) \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}

Posted by: Greg Egan on August 4, 2013 10:11 AM | Permalink | Reply to this

Re: Integral Octonions (Part 4)

Moving on to the 7-orthoplexes …

John already gave the prototypical normal (1,0,0,0,0,0,0,0)(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 (±12,±12,±12,±12,0,0,0,0)\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(84)=11202^4 \cdot \binom{8}{4} = 1120 possibilities. For example, (12,12,12,12,0,0,0,0)\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:

(12,12,12,12,12,12,12,12) (12,12,12,12,12,12,12,12) (12,12,12,12,12,12,12,12) (12,12,12,12,12,12,12,12) (12,12,12,12,12,12,12,12) (12,12,12,12,12,12,12,12) (12,12,12,12,12,12,12,12) (12,12,12,12,12,12,12,12) (1,1,0,0,0,0,0,0) (1,0,1,0,0,0,0,0) (1,0,0,1,0,0,0,0) (0,1,1,0,0,0,0,0) (0,1,0,1,0,0,0,0) (0,0,1,1,0,0,0,0) \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 (±34,±14,±14,±14,±14,±14,±14,±14)\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((81)+(83)+(85)+(87))=10248 \cdot \left(\binom{8}{1}+\binom{8}{3}+\binom{8}{5}+\binom{8}{7}\right) = 1024 possibilities. For example, (34,14,14,14,14,14,14,14)\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:

(12,12,12,12,12,12,12,12) (12,12,12,12,12,12,12,12) (12,12,12,12,12,12,12,12) (12,12,12,12,12,12,12,12) (12,12,12,12,12,12,12,12) (12,12,12,12,12,12,12,12) (12,12,12,12,12,12,12,12) (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) (1,0,0,0,0,1,0,0) (1,0,0,0,0,0,1,0) (1,0,0,0,0,0,0,1) \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 7-orthoplexes, given that John’s example clearly is a 7-orthoplex.

Posted by: Greg Egan on August 4, 2013 11:30 AM | Permalink | Reply to this

Re: Integral Octonions (Part 4)

With respect to the Integral Octonions there are 7 distinct E8 lattices (they are distinct as Octonion Integral Domains) each with its own distinct 240-vertex E8 root vector polytope.

The distinctions are useful when you build a 24-dim Leech Lattice from 8-dim E8 lattices as described in the work of Geoffrey Dixon and in the paper by Robert A. Wilson at www.maths.qmul.ac.uk/~raw/pubs_files/octoLeech1rev.pdf

You can see the differences when you look at the nearest neighbor vertices to the origin (In your coordinates, they are of square norm 2, smaller than your square norm 4 that gives 2160 next-to-nearest neighbor vertices to the origin).

Here is an example of two of the 7 sets of 240, using my favorite coordinate basis { 1, i, j, k, e, ie, je, ke } normalized to square norm 1 and looking at the 16 vertices ±1, ±i, ±j, ±k, ±e, ±ie, ±je, ±ke of the form ( +/-1, 0,0,0,0,0,0,0) and permutations and the {8|4} = (choose 4 of 8) = 70 x 16 = 1120 of the form (1/2)( +/-1, +/-1, +/-1, +/-1, 0,0,0,0) and permutations (where depending on position the 1 is 1 or i or j or k or e or ie or je or ke)

you see that you get 7 sets of 240 = 112 + 128 = 16 ( 1 + 6 + 8 ) that close under Octonion Multiplication as needed for Integral Domain structure.

Here are two of the 7 written explicitly in my favorite coordinates:

E8 lattice A with 15 sets of 16 vertices = 240 vertices:

±1, ±i, ±j, ±k, ±e, ±ie, ±je, ±ke

(±1 ±i ±k ±e)/2

(±j ±ie ±je ±ke)/2

(±1 ±e ±je ±j)/2

(±i ±k ±ie ±ke)/2

(±1 ±j ±ke ±k)/2

(±i ±e ±ie ±je)/2

(±1 ±k ±ie ±je)/2

(±i ±j ±e ±ie)/2

(±1 ±je ±i ±ke)/2

(±j ±k ±e ±ie)/2

(±1 ±ke ±e ±ie)/2

(±i ±j ±k ±je)/2

(±1 ±ie ±j ±i)/2

(±k ±e ±je ±ke)/2

E8 lattice B with 15 sets of 16 vertices = 240 vertices:

±1, ±i, ±j, ±k, ±e, ±ie, ±je, ±ke

(±1 ±k ±ke ±ie)/2

(±i ±j ±e ±je)/2

(±1 ±ie ±i ±e)/2

(±j ±k ±je ±ke)/2

(±1 ±e ±j ±ke)/2

(±i ±k ±ie ±je)/2

(±1 ±ke ±je ±i)/2

(±j ±k ±e ±ie)/2

(±1 ±i ±k ±j)/2

(±e ±ie ±je ±ke)/2

(±1 ±j ±ie ±je)/2

(±i ±k ±e ±ke)/2

(±1 ±je ±e ±k)/2

(±i ±j ±ie ±ke)/2

Note that, for example, the 64 points

(±1 ±e ±je ±j)/2

(±i ±k ±ie ±ke)/2

(±1 ±k ±ie ±je)/2

(±i ±j ±e ±ie)/2

are in A but not in B

and

the 64 points

(±1 ±ie ±i ±e)/2

(±j ±k ±je ±ke)/2

(±1 ±i ±k ±j)/2

(±e ±ie ±je ±ke)/2

are in B but not in A

Tony Smith

Posted by: Tony Smith on August 5, 2013 11:18 AM | Permalink | Reply to this

Re: Integral Octonions (Part 4)

Hi, Tony! Despite the title of this thread I haven’t really gotten around to describing the integral octonions yet, just some things you can do with them and how they look at a lattice, ignoring the multiplication. If and when I get serious, I’ll want to talk more about these 7 ways of extending the integer linear combinations of 1,i,j,k,e,ie,je,ke1,i,j,k,e,i e,j e,k e to a copy of the Cayley integral octonions.

Back here I asked Mike Rios if there’s a way to embed the Leech lattice in the exceptional Jordan algebra so that it’s closed under the Jordan product. His work gives various hints that it’s possible, but I believe studying this question requires thinking about the 7 forms of the Cayley integral octonions and what happens when you multiply a Cayley integral octonion of one form by one of another form. I haven’t gotten around to doing this.

Posted by: John Baez on August 8, 2013 11:14 AM | Permalink | Reply to this

Re: Integral Octonions (Part 4)

Sometime I would like to explain a more systematic way to calculate the number of vertices, edges, triangles, tetahedra, 4-simplexes, … , 7-simplexes and 7-orthoplexes in the E 8\mathrm{E}_8 root polytope.

The basic idea was explained in week187 of This Week’s Finds. For any Dynkin diagram there’s a polynomial in one variable qq which lets you count many things. A ratio of such polynomials, evaluated at a particular prime power qq, gives the number of points in a flag variety G/PG/P where GG is a semisimple algebraic group over the field with qq elements and PP is a parabolic subgroup. But if we set q=1q = 1, the same sort of ratio lets us answer questions like the ones we have here. This is part of the circle of ideas concerning ‘the field with one element’.

To get the idea to work, I need to know this polynomial for E 8\mathrm{E}_8 and all the Dynkin diagrams contained in that. I know what they are for A n\mathrm{A}_n and D n\mathrm{D}_n (see the article). So, I just need to know it for E 8,E 7\mathrm{E}_8, \mathrm{E}_7 and E 6\mathrm{E}_6. The information to figure this out is on page 11 of this paper:

• J. A. de Azcarraga, J. M. Izquierdo and J. C. Perez Bueno, An introduction to some novel applications of Lie algebra cohomology and physics.

For example, I believe the polynomial for E 8\mathrm{E}_8 is

[2][8][12][14][18][20][24][30] [2] [8] [12] [14] [18] [20] [24] [30]

where

[n]=1+q+q 2++q n1 [n] = 1 + q + q^2 + \cdots + q^{n-1}

For E 7\mathrm{E}_7 it should be

[2][6][8][10][12][14][18] [2] [6] [8] [10] [12] [14] [18]

and for E 6\mathrm{E}_6 it should be

[2][5][6][8][9][12] [2] [5] [6] [8] [9] [12]

In particular, the number of complete flags for the E 8\mathrm{E}_8 root polytope should be

2×8×12×14×18×20×24×30=696729600 2 \times 8 \times 12 \times 14 \times 18 \times 20 \times 24 \times 30 = 696729600

Good—this is right!! A complete flag, in this case, is a vertex, edge, triangle, tetahedron, 4-simplex, … , 7-simplex and 7-orthoplex, all incident to each other. Since the symmetry group of the E 8\mathrm{E}_8 acts freely and transitively, this symmetry group should have 696729600 elements—and it does!

Posted by: John Baez on August 5, 2013 12:29 PM | Permalink | Reply to this

Re: Integral Octonions (Part 4)

Personally, I wouldn’t be able to mention the tetra-/octrahedral honeycomb without referring to the relevant Escher print ;-)

http://www.wikipaintings.org/en/m-c-escher/flat-worms

Re: John Baez on August 5, 2013 12:29 PM:

Would that be the Hilbert polynomial ?

Posted by: Isabel Pirsic on August 5, 2013 2:36 PM | Permalink | Reply to this

Re: Integral Octonions (Part 4)

Isabel wrote:

Would that be the Hilbert polynomial?

Is that what people call it? In week186 I explained several equivalent ways to associate a polynomial to a Dynkin diagram. Here are five:

1) the coefficient of q iq^i in this polynomial is the number of ii-cells in the Bruhat decomposition of the flag variety G/BG/B. Here BB is the Borel subgroup of GG, and the “Bruhat decomposition” is a standard way of writing G/BG/B as disjoint union of ii-cells, that is, copies of F iF^i where FF is our field and ii is a natural number.

2) if the coefficient of q iq^i in this polynomial is kk, the (2i)(2i)th homology group of G/BG/B defined over the complex numbers is k.\mathbb{Z}^k.

3) the value of this polynomial at qq a prime power is the cardinality of G/BG/B defined over the field 𝔽 q\mathbb{F}_q.

4) the coefficient of q iq^i in this polynomial is the number of Coxeter group elements of length ii. Here the “length” of any element in the Coxeter group is its length as a word when we write it as product of the generating reflections.

5) the coefficient of q iq^i in this polynomial is the number of top-dimensional simplices of distance ii from a chosen top-dimensional simplex in the Coxeter complex. Here we measure “distance” between top-dimensional simplices in the hopefully obvious way, based on how many walls you need to cross to get from one to the other.

Posted by: John Baez on August 5, 2013 3:47 PM | Permalink | Reply to this

Re: Integral Octonions (Part 4)

Thanks for the answer !

Looking through my notes, I find that I was actually thinking of the Ehrhart polynomial ( http://en.wikipedia.org/wiki/Ehrhart_polynomial , informally described as a generalization of Pick’s theorem ), which may apparently only be related to the Hilbert polynomial in special cases.

Furthermore, both may actually have nothing to do with the one you present … – I have to confess that to me both are mostly buzz words that I remembered from colleagues’ talks about counting in a lattice polytope context and wondered if this was the same thing. So, I apologize for not being able to resolve this any further … ; if anyone can, I’d be interested, though.

Posted by: Isabel Pirsic on August 5, 2013 4:34 PM | Permalink | Reply to this

Re: Integral Octonions (Part 4)

By the way, there’s been a long debate over the plural of ‘simplex’: simplexes, or simplices?

I think Conway may have invented the parallel word ‘orthoplex’ to tilt the terms of that debate. Simplices sounds acceptable to me… but ‘orthoplices’ sounds merely comic.

Posted by: John Baez on August 5, 2013 5:00 PM | Permalink | Reply to this

Re: Integral Octonions (Part 4)

Regarding etymology, I would say “orthoplices” = “orthant complices” is on a weaker footing than either “simplices” or “orthoplexes” = “orthant complexes” (even though “complices” does exist as an archaic (English) plural) – analogously, I don’t think there are any, e.g., “chain complices” either (sounds vaguely nefarious, doesn’t it?).

In fact, the plural of the latin root is, well, complex: for the noun “complexus” it would be “complexūs”, or, if you consider the participle of “complector”, where it comes from, any of complexī/-ae/-a. So I’d think “complexes” and “orthoplexes” are a good way to go.

As for “simplex”, things are, well, simpler, since “simplicēs” is the established Latin plural, unless you would consider “simplicia” (which to me sounds very baroque, but not entirely possible).

A propos: simplex is supposed to be derived from semel + plico , i.e., once-folded … German humanists may here be reminded of Winckelmann’s “Edle Einfalt, stille Größe”. Whereas complexus is nothing but “woven together”.

The method of using the -es affix is of course also admissible. I seem to recall from (German) linguistics that using a native plural paradigm rather than that of the original language reflects a greater degree of acceptance as part of the vocabulary - with ‘elitist’ usage acting as counterforce.

Perhaps one day we might even more commonly say “matrixes” ?

Posted by: Isabel Pirsic on August 5, 2013 6:05 PM | Permalink | Reply to this

Re: Integral Octonions (Part 4)

This is an unrelated question. What is the significance of the numbers that are the coordinates of the vertices of all of the regular polytopes when mapped onto a Cartesian space of the normed division algebras, centered at the origin, with one vertex at “1”? Are there interesting mathematical properties or relations of these numbers?

What is the significance of the complex numbers that are the coordinates of the vertices of the regular polygons (equilateral triangle, square, regular pentagon, etc.), when mapped onto the complex Argand plane, centered at the origin, with one vertex at 1 + 0i?

What is the significance of the quaternions that are the coordinates of the vertices of the regular polychora (4D simplex, 24-cell, 120-cell, etc.), when mapped onto a 4D Cartesian coordinate system of the quaternions, centered at the origin, with one vertex at 1 + 0i + 0j +0k?

What is the significance of the octonions that are the coordinates of the vertices of the regular 8D polytopes (8D simplex, etc.), when mapped onto a 8D Cartesian coordinate system of the octonions, centered at the origin, with one vertex at 1 + 0e1 + 0e2 +0e3+ …0e7?

Posted by: Jeffery Winkler on August 13, 2013 9:10 PM | Permalink | Reply to this

Re: Integral Octonions (Part 4)

Jeffrey wrote:

What is the significance of the numbers that are the coordinates of the vertices of all of the regular polytopes when mapped onto a Cartesian space of the normed division algebras, centered at the origin, with one vertex at “1”? Are there interesting mathematical properties or relations of these numbers?

There’s a lot to say about this, but the short answer is that these numbers are connected to algebra and number theory in a lot of interesting ways.

What is the significance of the complex numbers that are the coordinates of the vertices of the regular polygons (equilateral triangle, square, regular pentagon, etc.), when mapped onto the complex Argand plane, centered at the origin, with one vertex at 1 + 0i?

This is the most well-known case.

When you set it up correctly, the vertices of the regular nn-gon can be described by some very nice complex numbers: precisely the nnth roots of unity!

exp(2πik/n)k=1,2,,n \exp(2 \pi i k / n) \qquad k = 1,2,\dots, n

For any nn, when you take all these numbers and add, subtract, multiply and divide them to your heart’s content you get an algebraic number field, a subfield of the complex numbers. An algebraic number field generated by the nnth roots of unity is called the cyclotomic field (ζ n)\mathbb{Q}(\zeta_n). Gauss, Galois and others noticed that the question of which regular nn-gons you can construct with ruler and compass is closely connected to properties of the symmetry group of this field (its so-called ‘Galois group’).

In certain very special cases, like n=3n = 3 and n=4n = 4, if you only add, subtract, and multiply these numbers to your heart’s content, you get the ring of integers in the aforementioned algebraic number field. For n=3n = 3 these are called the Eisenstein integers:

For n=4n = 4 they’re called the Gaussian integers:

A lot of this stuff generalizes to the quaternionic and octonionic cases, and for that I suggest starting with my review of Conway and Smith’s book. To dig deeper, read their book! It’s a lot of fun.

However, instead of focusing on the real numbers that are coordinates of vertices of the highly symmetrical polytopes in dimensions 1, 2, 4, and 8, it seems better to focus directly on the real numbers, complex numbers, quaternions or octonions that are these vertices. And that’s what Conway and Smith do.

Posted by: John Baez on August 14, 2013 3:33 PM | Permalink | Reply to this

Re: Integral Octonions (Part 4)

I remember you discussing these issues in your wonderful talk about the number 24

http://math.ucr.edu/home/baez/numbers/24.pdf

The mysteries of the number 24 are also discussed in the following interesting paper.

http://arxiv.org/pdf/1308.5233.pdf

What do you think of this paper about the Dedekind Eta Function and K3 surfaces?

Posted by: Jeffery Winkler on August 27, 2013 9:17 PM | Permalink | Reply to this

Re: Integral Octonions (Part 4)

It looks like a very interesting paper, but it jumps a bit too quick from the stuff I already know to the stuff I’ve never heard about. The jump happens here, for example:

Thus, the question which instantly emerges is whether it is possible to find physical systems whose partition functions are exactly these elegant products. Remarkably, at this list did the authors of [9] arrive when considering the counting of electrically charged, 1/2-BPS states in the N=4N = 4 supersymmetric CHL orbifolds of the heterotic string on the six-torus. What is particularly fascinating for our present purposes is that upon string duality the situation is equivalent to the compactication of type IIB string theory on the product of a two-torus with a K3 surface of a specic type, viz., one which admits certain Nikulin involutions.

I understand most of the individual words here except for “CHL” (which means “Chaudhuri-Hockney-Lykken”, not that this helps me any). However, I could use a few pages to help me understand why “these elegant products” might arise this way.

(By the way, I have trouble believing that John McKay was involved in writing the phrase “remarkably, at this list did the authors arrive”.)

Posted by: John Baez on August 29, 2013 1:37 PM | Permalink | Reply to this

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