December 25, 2021

The Binary Octahedral Group

Posted by John Baez

It’s been pretty quiet around the $n$-Café lately! I’ve been plenty busy myself: Lisa and I just drove back from DC to Riverside with stops at Roanoke, Nashville, Hot Springs, Okahoma City, Santa Rosa (a small town in New Mexico), Gallup, and Flagstaff. A lot of great places! Hot Springs claims to have the world’s shortest street, but I’m curious what the contenders are. Tomorrow I’m supposed to talk with James Dolan about hyperelliptic curves. And I’m finally writing a paper about the number 24.

But for now, here’s a little Christmas fun with Platonic solids and their symmetries. For more details, see:

All the exciting animations in my post here were created by Greg. And if you click on any of the images in my post here, you’ll learn more.

The complex numbers together with infinity form a sphere called the Riemann sphere. The 6 simplest numbers on this sphere lie at points we could call the north pole, the south pole, the east pole, the west pole, the front pole and the back pole. They’re the corners of an octahedron!

On the Earth, we all know where the north pole and south pole are. I’d say the “front pole” is where the prime meridian meets the equator at 0°N 0°E. It’s called Null Island, but there’s no island there—just a buoy. Here it is:

Where’s the back pole, the east pole and the west pole? I’ll leave two of these as puzzles, but I discovered that in Singapore I’m fairly close to the east pole:

If you think of the octahedron’s corners as the quaternions $\pm i, \pm j, \pm k,$ you can look for unit quaternions $q$ such that whenever $x$ is one of these corners, so is $q x q^{-1}$. There are 48 of these! They form a group called the binary octahedral group.

By how we set it up, the binary octahedral group acts as rotational symmetries of the octahedron: any transformation sending $x$ to $q x q^{-1}$ is a rotation. But this group is a double cover of the octahedron’s rotational symmetry group! That is, pairs of elements of the binary octahedral group describe the same rotation of the octahedron.

If we go back and think of the Earth’s 6 poles as points $0, \pm 1,\pm i, \infty$ on the Riemann sphere instead of $\pm i, \pm j, \pm k$, we can think of the binary octahedral group as a subgroup of $\mathrm{SL}(2,\mathbb{C}$), since this acts as conformal transformations of the Riemann sphere!

If we do this, the binary octahedral group is actually a subgroup of $\mathrm{SU}(2)$, the double cover of the rotation group—which is isomorphic to the group of unit quaternions. So it all hangs together.

It’s fun to actualy see the unit quaternions in the binary octahedral group. First we have 8 that form a group on their own, called the quaternion group:

$\pm 1, \pm i , \pm j , \pm k$

These are the vertices of the 4d analogue of an octahedron, called a cross-polytope. It looks like this:

Then we have 16 that form the corners of a hypercube (the 4d analogue of a cube, also called a tesseract or 4-cube):

$\displaystyle{ \frac{\pm 1 \pm i \pm j \pm k}{2} }$

They look like this:

These don’t form a group, but if we take them together with the 8 previous ones we get a 24-element subgroup of the unit quaternions called the binary tetrahedral group. These 24 elements are also the vertices of the 24-cell, which is the one regular polytope in 4 dimensions that doesn’t have a 3d analogue. It looks like this:

This shape is called the 24-cell not because it has 24 vertices, but because it also has 24 faces, which happen to be regular octahedra. You can see one if you slice the 24-cell like this:

The slices here have real part 1, ½, 0, -½, and -1 respectively. Note that the slices with real part ±½ contain the vertices of a hypercube, while the rest contain the vertices of a cross-polytope.

And here’s another great way to think way about binary tetrahedral group. We’ve seen that if you take every other vertex of a cube you get the vertices of a regular tetrahedron. Similarly, if you take every other vertex of a 4d hypercube you get a 4d cross-polytope. So, you can take the vertices of a 4d hypercube and partition them into the vertices of two cross-polytopes. As a result, the 24 elements of the binary tetrahedral group can be partitioned into three cross-polytopes! Greg Egan shows how it looks:

So far we’ve accounted for half the quaternions in the binary octahedral group! Here are the other 24:

$\displaystyle{ \frac{\pm 1 \pm i}{\sqrt{2}}, \frac{\pm 1 \pm j}{\sqrt{2}}, \frac{\pm 1 \pm k}{\sqrt{2}}, }$

$\displaystyle{ \frac{\pm i \pm j}{\sqrt{2}}, \frac{\pm j \pm k}{\sqrt{2}}, \frac{\pm k \pm i}{\sqrt{2}} }$

These form the vertices of another 24-cell! So the binary octahedral group can be built by taking the vertices of two separate 24-cells. And in fact, these two 24-cells are ‘dual’ to each other: the vertices of each one hover right above the centers of the faces of the other!

The first 24 quaternions, those in the binary tetrahedral group, give rotations that preserve each one of the two tetrahedra that you can fit around an octahedron—or in a cube:

while the second 24 switch these tetrahedra.

Greg has nicely animated the 48 elements of the binary octahedral group here:

He’s colored them according to the rotations of the octahedron they represent. Remember: in the binary octahedral group, two elements $q$ and $-q$ describe each rotational symmetry of the octahedron. And here they are:

Black: the 2 elements

$\pm 1$

act as the identity.

Blue: the 12 elements

$\displaystyle{ \frac{\pm 1 \pm i}{\sqrt{2}}, \frac{\pm 1 \pm j}{\sqrt{2}}, \frac{\pm 1 \pm k}{\sqrt{2}} }$

describe ±90° rotations around the octahedron’s 3 axes.

Red: the 16 elements

$\displaystyle{ \frac{\pm 1 \pm i \pm j \pm k}{2} }$

describe 120° rotations of the octahedron’s 8 triangles.

Yellow: The 6 elements

$\pm i , \pm j , \pm k$

describe 180° rotations around the octahedron’s 3 axes.

Cyan: the 12 elements

$\displaystyle{ \frac{\pm i \pm j}{\sqrt{2}}, \frac{\pm j \pm k}{\sqrt{2}}, \frac{\pm k \pm i}{\sqrt{2}} }$

describe 180° rotations of the octahedron’s 6 opposite pairs of edges.

Finally, here’s another fun fact about the binary octahedral group that follows from what we’ve already seen. Note that because

• we can partition the 48 vertices of the binary octahedral group into two 24-cells

and

• we can partition the 24 vertices of the 24-cell into three cross-polytopes

it follows that we can partition the 48 vertices of the binary octahedral group into six cross-polytopes!

I don’t know the deep meaning of this fact. I know that the vertices of the 24-cell correspond to the 24 roots of the Lie algebra $\mathfrak{so}(8).$ I know that the the famous ‘triality’ symmetry of $\mathfrak{so}(8)$ permutes the three cross-polytopes in the 24-cell, which are in some rather sneaky way related to the three 8-dimensional irreducible representations of $\mathfrak{so}(8).$ I also know that if we take the two 24-cells in the binary octahedral group, and expand one by a factor of $\sqrt{2},$ so the vertices of other lie exactly at the center of its faces, we get the 48 roots of the Lie algebra $\mathfrak{f}_4.$ But I don’t know how to extend this story to get a nice story about the six cross-polytopes in the binary octahedral group.

All I know is that if you pick a quaternion group sitting in the binary octahedral group, it will have 6 cosets, and these will be six cross-polytopes.

Posted at December 25, 2021 3:04 AM UTC

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Re: The Binary Octahedral Group

On Azimuth Allen Knutson is helping me with the problem near the end of this article—namely, what does triality have to do with the 6 cross-polytopes inside the binary octahedral group?

Posted by: John Baez on December 25, 2021 9:03 PM | Permalink | Reply to this

Re: The Binary Octahedral Group

And I’m finally writing a paper about the number 24.

That’s the happy news I didn’t know I wanted to hear!

Posted by: Blake Stacey on December 25, 2021 9:22 PM | Permalink | Reply to this

Re: The Binary Octahedral Group

Thanks! I now think I see how a certain big picture fits together, so this will not be an enormous sprawling list of ways the number 24 shows up in math—it’ll be more of a coherent story.

Posted by: John Baez on December 26, 2021 1:14 AM | Permalink | Reply to this

24

Here’s 12

https://math.mit.edu/~poonen/papers/lattice12.pdf

and 24$\times$24

https://en.wikipedia.org/wiki/Topological$\underline{\;}$modular$\underline{\;}$forms

and recent work on central charge 24

https://arxiv.org/abs/2112.12291

“We associate with a generalised deep hole of the Leech lattice vertex operator algebra a generalised hole diagram. We show that this Dynkin diagram determines the generalised deep hole up to conjugacy and that there are exactly 70 such diagrams.” Presumably this is related to 1^2 + 2^2 + … + 24^2 = 70^2, but I didn’t see that in the paper.

Posted by: Allen Knutson on December 26, 2021 4:54 AM | Permalink | Reply to this

Re: The Binary Octahedral Group

The most beautiful elliptic curve

$Y^2 = X^3 - X$

in the known universe has automorphism group $Sl_2(F_3)$ of order 24 at the prime two…

Posted by: jackjohnson on December 26, 2021 1:13 PM | Permalink | Reply to this

Re: The Binary Octahedral Group

See this old guest post of mine for the next-level numerology of relating that 24 with a 27 that is one member of an Arnold trinity.

Posted by: Blake Stacey on December 27, 2021 2:00 AM | Permalink | Reply to this

Mozibur

Plato is famous for saying that the Platonic solids were responsible for the physical world in his book the Timeaus.

Of course today we pooh-pooh that idea and point to our U(1) x SU(2) x SU(3) gauge theory.

But it occured to me that their ought to be polytopes associated with, that is the polytope associated with the root lattice of u(1) x su(2) x su(3) by simply taking their convex hull.

Do these show any signs of Platonic regularity?

Posted by: Mozibur Ullah on December 26, 2021 9:55 PM | Permalink | Reply to this

Re: The Binary Octahedral Group

There are successive lifts of the octahedral group up the Whitehead tower of $O(3)$ through the binary octahedral group to the corresponding Platonic 2-group.

I’d be interested to know of a natural action for this 2-group.

Posted by: David Corfield on December 27, 2021 2:32 PM | Permalink | Reply to this

Re: The Binary Octahedral Group

Yes, there should be a ton of fascinating math in this direction! (I wish they’d fix the typo in the title on the abstract.)

A small step in this direction: since the group $O(n)$ has at least two nontrivial double covers, the pin groups $Pin_+(n)$ and $Pin_-(n)$, the symmetric groups $S_n \subset O(n)$ get two double covers, which are usually nonisomorphic, sometimes called $2 \cdot S_n^+$ and $2 \cdot S_n^-$.

But the rotational symmetry group of the octahedron is $S_4$, so it gets two different double covers! Our friend the binary octahedral group is $2 \cdot S_4^-$. But $2 \cdot S_4^+$ is also interesting: it’s $GL(2,\mathbb{F}_3)$, the group of invertible $2 \times 2$ matrices with entries in the field with 3 elements! And it’s the symmetry group of the most symmetrical Riemann surface of genus 2: the Bolza curve. Not surprisingly, this curve is a branched double cover of the Riemann sphere with branch points located at the vertices of an octahedron.

Posted by: John Baez on December 27, 2021 5:55 PM | Permalink | Reply to this

24

I’m really looking forward to reading about 24.👏 I’ll possibly only understand the first sentence and yet still no doubt find it utterly fascinating. Which is often the case on this forum where I get to treasure a layman’s sense of the majesty of the material, even where most of the details elude me.

Posted by: Bertie on December 28, 2021 9:04 AM | Permalink | Reply to this

Mozibur

A minor gripe:

Is there a nicer name for the 4d analogue of an octahedron other than a cross polytope? This name gives nothing away about its relationship to the octahedron and I’m a believer in names reflecting properties.

After all, the 4d analogue of the cube is called the 4-cube or hypercube. And these names nod to its cube like property.

Maybe octotope or 4-octon or 4-octoid!

Posted by: Mozibur Ullah on December 28, 2021 1:12 PM | Permalink | Reply to this

Re: Mozibur

‘Cross-polytope’ is actually a name for the generalization of the octahedron to any dimension. Conway has suggested ‘orthoplex’, in analogy to ‘simplex’, and I like this a lot. Another name is ‘hyperoctahedron’, in analogy to ‘hypercube’. I think that might help for people who only know a little math.

For the particular case of the 4d cross-polytope people often say ‘16-cell’, since it has 16 faces. In this terminology the 4d convex regular polytopes are called the 5-cell, the 8-cell, the 16-cell, the 24-cell, the 120-cell and the 600-cell.

Posted by: John Baez on December 28, 2021 7:34 PM | Permalink | Reply to this

Re: Mozibur

One argument in favour of calling it a hyperoctahedron is that its symmetry group is, as far as I know, universally referred to as the hyperoctahedral group.

Posted by: lambda on December 30, 2021 11:36 PM | Permalink | Reply to this

Re: Mozibur

I’m fond of the name crosspolytope myself. To me, that name reflects more about the structure of the polytope (which is the convex hull of the $n$-dimensional “cross” shape $\bigcup_{i=1}^n \{(0,...,0,x_i,0,...,0) : -1 \le x_i \le 1 \}$) than anything reminiscent of “octahedron”, whose name is tied to the $n=3$ case. In my subfield of high-dimensional convex geometry, “crosspolytope” is firmly established. (Though I like “orthoplex”, too.)

I rather dislike “hyperoctahedron”, partly because when $n \neq 3$ there’s no 8 involved, but also because in my field nobody ever refers to “hypercubes” or “hyperspheres” either. We just say “cube” or “sphere”, whatever the dimension, and I tend to associate “hyper-” with people who (as John suggests) know little enough mathematics that they think of 4 or more dimensions as exotic. (Not to disparage such people, but I don’t want to make myself look like one in front of colleagues!)

On the other hand, we do talk about “hyperplanes” in my field, so we’re not any more logical or consistent in our terminology on the whole than anyone else.

Posted by: Mark Meckes on December 31, 2021 3:43 PM | Permalink | Reply to this

Re: Mozibur

I mainly use the name ‘hyperoctahedron’ on Twitter, only for the 4d cross-polytope — along with names like ‘hypercube’, ‘hyperdodecahedron’ for the 120-cell, and ‘hypericosahedron’ for the 600-cell. The idea is that people who know nothing of math will get some vague idea of what I’m talking about (some sort of super-duper versions of the shapes they know), while experts can guess what I really mean and smile knowingly, or smirk condescendingly.

Posted by: John Baez on December 31, 2021 10:41 PM | Permalink | Reply to this

Mozibur

Ah, now that you mention it, I have come across orthoplex before! I agree that it’s great suggestion and much better than the other alternative that you mention - hyperoctohedron - it’s too long: I’m also a believer in names being as short as possible but no shorter.

Posted by: Mozibur Ullah on December 30, 2021 9:08 PM | Permalink | Reply to this

Re: The Binary Octahedral Group

Ha! Gallup is (unless some conspiracy has hidden a recent change from me) the exact opposite of a great place, as I can safely and vividly attest as a native of and formerly frequent visitor to the place.

The surrounding natural areas are a different story.

Later this month I hope to start back on KS stuff FWIW.

Posted by: Steve Huntsman on January 8, 2022 12:20 AM | Permalink | Reply to this

Re: The Binary Octahedral Group

Gallup is the nearest big Anglo town to the Navajo Nation, so a lot of Navajo, Hopi, and Zuni artists sell their creations there. That’s why Lisa and I got interested in the first place… but soon she went to a store called Weaving in Beauty, wound up taking lessons there on Navajo-style weaving, and the rest is history: now we go there once a year, usually right before Christmas.

It’s definitely a poor, hard-bitten town. A lot of folks on the nearby reservations go there to do laundry because they don’t have their own facilities. And this has caused problems lately:

On Thursday April 30, 2020, on his last full day in office Gallup City Mayor Jackie McKinney sent a letter to New Mexico Governor Michelle Lujan Grisham requesting that she invoke the state’s Riot Control Act to restrict travel in and out of Gallup in an effort to “mitigate the uninhibited spread of COVID-19 in that city.” The following morning (May 1) Governor Grisham issued a statement and executive order 2020-027 which included the directive that:

“All roadways providing access to the City of Gallup shall be closed and only Gallup residents, those who work in Gallup, and members of the media shall be permitted entry into the municipality [….]”

Gallup is a border town to both the Navajo Nation and the Zuni Pueblo. Every weekend tens of thousands of people from both Navajo and Zuni travel to Gallup to purchase groceries, do laundry and haul water, among other essentials. On the first weekend of every month, because so many people on the reservation live on fixed incomes, travel into Gallup is even more critical as funds grow short and food supplies low. The Navajo Nation is a food desert, almost 26,000 square miles, nearly 200,000 people and only about a dozen full service grocery stores. The Navajo Nation does not have the capacity to sell groceries to all of its residents. Travel to the border towns is essential, especially on the first weekend of the month. And with almost no notice, the mayor of Gallup and the governor of New Mexico closed access to a critical border town that provides essential services, medical care and resources to a large portion of the Navajo Nation and Zuni Pueblo.

The Gallup Police Department, New Mexico State Police and the National Guard were deployed to set up checkpoints and block access into the city.

And so on.

Posted by: John Baez on January 9, 2022 1:42 AM | Permalink | Reply to this

Re: The Binary Octahedral Group

Hot Springs lacks any strong claim for the shortest street, without pretty contrived caveats. At 98 feet, it is simply outclassed by rivals, such as Ebenezer Place, Wick, at under 7 feet.

Hot Springs seems to have a much stronger claim for the world’s shortest St Paddy’s Day parade, held along said ‘shortest’ street 😃😃

Posted by: bertie on January 8, 2022 4:51 AM | Permalink | Reply to this

Re: The Binary Octahedral Group

Apparently everything is bigger in America… even the shortest street!

You’re right: Ebenezer Place in Wick, Scotland is definitely shorter.

Posted by: John Baez on January 9, 2022 1:47 AM | Permalink | Reply to this

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