## From the Octahedron to E8

Here’s a fun challenge for people confined due to coronavirus.

The E8 lattice is a thing of beauty, taking full advantage of the magic properties of the number 8. The octahedron has 8 sides. Wouldn’t it be cool if you could build the E8 lattice from the humble octahedron?

David Harden thinks he’s found a way! But I haven’t carefully checked that it works: it takes some calculations. Can you check it?

• David L. Harden, What other lattices are obtainable from this noncommutative ring?, MathOverflow, 14 March 2020.

Let me dive in and explain it.

You start with the octahedron. You take the double cover of its group of rotational symmetries and think of this as a group of unit quaternions. You get 48 very special quaternions this way. You then take integer linear combination of these quaternions: call the set of these $R.$ It turns out that every element of $R$ is of the form

$(a + b\sqrt{2}) + (c + d\sqrt{2})i + (e + f\sqrt{2})j + (g + h\sqrt{2})k$

where $a,b,c,d,e,f,g,h$ are rational.

Not every quaternion of this form is in $R.$ But that’s okay: what we have is enough to let us think of $R$ as sitting inside $\mathbb{Q}[\sqrt{2}]^4,$ which is an 8-dimensional vector space over the rational numbers.

To see this lattice as E8, we use a special inner product on $\mathbb{Q}[\sqrt{2}]^4.$ If

$v = a + bi + cj + dk \in \mathbb{Q}[\sqrt{2}]^4$

and

$v' = a' + b'i + c'j + d'k \in \mathbb{Q}[\sqrt{2}]^4$

then their usual quaternion inner product is

$\langle v, v' \rangle_{\mathbb{H}} = aa' + bb' + cc' + dd' = p + q \sqrt{2}$

for some rational numbers $p$ and $q.$ We then define the inner product on $\mathbb{Q}[\sqrt{2}]^4$ by

$\langle v, v' \rangle = 2(p + q)$

So, we have a lattice $R$ in an 8-dimensional rational vector space with an inner product… and David claims that it’s a copy of the E8 lattice!

To prove this, it’s enough to show that

1) the inner product of any two vectors in $R$ is an integer,

and that either

2a) the inner product of any vector in $R$ with itself is even, and if $v_1, \dots , v_8$ is any list of generators for $R$ then the determinant of the 8 × 8 matrix $\langle v_i, v_j \rangle$ is 1

or

2b) there are no vectors in $R$ of length 1, and 240 vectors in $R$ whose inner product with themselves is 2.

Either of these characterizes E8. David has actually checked 1) and both 2a) and 2b), but some of his calculations take work, so I hope some of you can check them again.

If we start with the octahedron and take the double cover of its group of rotational symmetries, you get something called the binary octahedral group, which has 48 elements. I described it here:

• John Baez, The binary octahedral group, Azimuth, 29 August 2019.

I even described how to think of its elements as unit quaternions. We get 8 like this:

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

and 16 like this:

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

These 24 are the vertices of a wonderful shape called the 24-cell, drawn here by Greg Egan:

The remaining 24 elements of the binary octahedral group form the vertices of a second 24-cell! Here they are:

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

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

Starting from the 48 elements of the binary octahedral group, we can easily get ahold of 8 generators of the lattice $R.$ David Harden chose these:

$\begin{array}{ccl} v_{1} &=& \frac{1}{2}+\frac{i}{2}+\frac{j}{2}+\frac{k}{2} \\ v_{2} &=& \frac{1}{\sqrt{2}}+\frac{i}{\sqrt{2}} \\ v_{3} &=&i \\ v_{4} &=&\frac{i}{\sqrt{2}}+\frac{j}{\sqrt{2}} \\ v_{5} &=&j \\ v_{6} &=&\frac{j}{\sqrt{2}}+\frac{k}{\sqrt{2}} \\ v_{7} &=&k \\ v_{8} &=&k\sqrt{2} \end{array}$

I think it’s pretty easy to see that these generate the lattice $R.$ I also think it’s pretty easy to check 1). For this we need to check that the inner product of any two of these vectors $v_1, \dots, v_8$ is an integer. But we have to use the correct inner product: the one described above!

So, for example, the ordinary quaternion inner product of $v_7$ and $v_8$ is

$\langle v_7, v_8 \rangle_{\mathbb{H}} = \langle k, \sqrt{2}k \rangle_{\mathbb{H}} = \sqrt{2}$

so

$\langle v_7, v_8 \rangle = 2$

Or:

$\langle v_8, v_8 \rangle_{\mathbb{H}} = \langle \sqrt{2}k, \sqrt{2}k \rangle_{\mathbb{H}} = 2$

so

$\langle v_8, v_8 \rangle = 4$

After doing a bunch of these, I’m convinced that all the inner products are integers. So the hard part is checking 2a) or 2b).

Can you do it? The matrix of inner products $\langle v_i , v_j \rangle$ needed for checking 2a) is called the Gram matrix, and David Harden has computed it, but you could compute it yourself and check that its determinant is 1.

By the way, all this is similar to the construction of the E8 lattice from the icosahedron, explained in Conway and Sloane’s book and later here:

• John Baez, From the icosahedron to E8.

There you start with the icosahedron. You take the double cover of its group of rotational symmetries, which is called the binary icosahedral group, and think of this as a group of unit quaternions. You then take integer linear combination of these quaternions: call the set of these $I.$ It turns out that every element of $I$ is of the form

$(a + b\sqrt{5}) + (c + d\sqrt{5})i + (e + f\sqrt{5})j + (g + h\sqrt{5})k$

where $a,b,c,d,e,f,g,h$ are rational.

Not every quaternion of this form is in $I.$ But that’s okay: this is enough to let us think of $I$ as sitting inside $\mathbb{Q}[\sqrt{5}]^4,$ which is an 8-dimensional vector space over the rational numbers. And since the rationals sit in the reals, you can think of $I$ as a lattice in an 8-dimensional real vector space.

To see this lattice as $\mathrm{E}_8$, we put this norm on $\mathbb{Q}[\sqrt{5}]^4:$

$\|(a + b\sqrt{5}) + (c + d\sqrt{5})i + (e + f\sqrt{5})j + (g + h\sqrt{5})k \|^2 =$

$a^2 + b^2 + c^2 + d^2 + e^2 + f^2 + g^2 + h^2$

And we get a copy of the E8 lattice this way, since 1) and 2a) hold, and for that matter also 2b).

The lattice $I$ is actually a subring of the quaternions, which Conway and Sloane call the icosians. Harden’s lattice $R$ is also a subring of the quaternions, and he has dubbed it the octians.

It’ll be cute to see an octahedron giving E8, since they are both connected to the number eight! But even if this construction really works, I have no idea what it ‘really means’.

Is this construction new?

### 15 Responses to From the Octahedron to E8

1. pendantry says:

OMG I think my brain just exploded

2. I kind of think that both this and the icosahedron examples are not that interesting. My issue is the contrived inner product. You have a ring which, as an abelian group, is free on eight generators. Great: you can certainly put an inner product on it to make it an $E_8$ lattice. Maybe you can even arrange your inner product to look somewhat reasonable — say, to look good on an index-2 sublattice $D_8 \subset E_8$ or something. But these norms on $\mathbb{Q}[\sqrt{2}]$ and $\mathbb{Q}[\sqrt{5}]$ in which you declare that $\sqrt{2}$ or $\sqrt{5}$ has length $1$ seem completely arbitrary to me. They don’t respect the ring structure. So what distinguishes them?

• John Baez says:

I don’t know what’s good about these norms. $\mathbb{Z}[\sqrt{2}]$ and $\mathbb{Z}[\sqrt{5}]$ have completely multiplicative norms, but in a different sense of ‘norm’—and worse, those norms aren’t closely related to the ones I’m using here.

So, I wouldn’t try to convince a skeptic like you that these constructions of the E8 lattice are interesting. But Conway and Sloane found the icosahedral example interesting enough to write about, and it seems worth noting that the most straightforward way of adapting it to the octahedron works just as well.

In my paper From the icosahedron to E8 I described two ways to construct the E8 lattice from the icosahedron: the one here, and another connected to the McKay correspondence and du Val’s work on the resolution of Kleinian singularities. I think I could convince you that the latter is interesting. I posed as a puzzle to find a connection between the two constructions. Perhaps a good solution to that puzzle would count as a proof that the construction here is also interesting.

By the way, to get LaTeX to work here you have to follow the instructions that appear in boldface right above the box where you enter your comment. I fixed your LaTeX.

• John Baez says:

Actually, I take it back: the Conway–Sloane construction of E8 from the icosians is interesting, because it’s a special case of a systematic procedure relating the noncrystallographic Coxeter groups H2 H3 and H4 to Dynkin diagrams with twice as many dots in them, namely A4, D6 and E8.

I explained this back in week270:

If you look at my slides you’ll also see an appendix that describes two ways to get the E8 lattice starting from the dodecahedron. This is a nice interaction between the magic powers of the number 5 and those of the number 8. After my talk, Christian Korff from the University of Glasgow showed me a paper that fits this relation into a bigger pattern:

7) Andreas Fring and Christian Korff, Non-crystallographic reduction of Calogero-Moser models, Jour. Phys. A 39 (2006), 1115-1131. Also available as arXiv:hep-th/0509152.

They set up a nice correspondence between some non-crystallographic Coxeter groups and some crystallographic ones:

the H2 Coxeter group and the A4 Coxeter group,
the H3 Coxeter group and the D6 Coxeter group,
the H4 Coxeter group and the E8 Coxeter group.

A Coxeter group is a finite group of linear transformations of Rn that’s generated by reflections. We say such a group is "non-crystallographic" if it’s not the symmetries of any lattice. The ones listed above are closely tied to the number 5:

H2 is the symmetry group of a regular pentagon.
H3 is the symmetry group of a regular dodecahedron.
H4 is the symmetry group of a regular 120-cell.

Note these live in 2d, 3d and 4d space. Only in these dimensions are there regular polytopes with 5-fold rotational symmetry! Their symmetry groups are non-crystallographic, because no lattice can have 5-fold rotational symmetry.

A Coxeter group is "crystallographic", or a "Weyl group", if it is symmetries of a lattice. In particular:

A4 is the symmetry group of a 4-dimensional lattice.
D6 is the symmetry group of a 6-dimensional lattice.
E8 is the symmetry group of an 8-dimensional lattice.

You can see precise descriptions of these lattices in "week65" – they’re pretty simple.

Both crystallographic and noncrystallographic Coxeter groups are described by Coxeter diagrams, as explained back in "week62". The H2, H3 and H4 Coxeter diagrams look like this:

   5
o---o

5
o---o---o

5
o---o---o---o


The A4, D6 and E8 Coxeter diagrams (usually called Dynkin diagrams) have twice as many dots as their smaller partners H2, H3 and H4. In every case, each dot in the diagram corresponds to one of the reflections that generates the Coxeter group. The edges in the diagram describe relations – you can read how in "week62".

All this is well-known stuff. But Fring and Korff investigated something more esoteric. Each dot in the big diagram corresponds to 2 dots in its smaller partner. And if we map each generator of the smaller group to the product of the two corresponding generators in the bigger one, we get a group homomorphism.

In fact, we get an inclusion of the smaller group in the bigger one!

This is just the starting point of Fring and Korff’s work. Their real goal is to show how certain exactly solvable physics problems associated to crystallographic Coxeter groups can be generalized to these three noncrystallographic ones. For this, they must develop more detailed connections than those I’ve described. But I’m already happy just pondering this small piece of their paper.

For example, what does the inclusion of H2 in A4 really look like?

It’s actually quite beautiful. H2 is the symmetry group of a regular pentagon, including rotations and reflections. A4 happens to be the symmetry group of a 4-simplex. If you draw a 4-simplex in the plane, it looks like a pentagram. So, any symmetry of the pentagon gives a symmetry of the 4-simplex. So, we get an inclusion of H2 in A4.

People often say that Penrose tilings arise from lattices in 4d space. Maybe I’m finally starting to understand how! The A4 lattice has a bunch of 4-simplices in it – but when we project these onto the plane correctly, they give pentagrams. I’d be very happy if this were the key.

What about the inclusion of H3 in D6?

Here James Dolan helped me make a guess. H3 is the symmetry group of a regular dodecahedron, including rotations and reflections. D6 consists of all linear transformations of R6 generated by permuting the 6 coordinate axes and switching the signs of an even number of coordinates. But a dodecahedron has 6 "axes" going between opposite pentagons! If we arbitrarily orient all these axes, I believe any rotation or reflection of the dodecahedron gives an element of D6. So, we get an inclusion of H3 in D6.

And finally, what about the inclusion of H4 in E8?

H4 is the symmetry group of the 120-cell, including rotations and reflections. In 8 dimensions, you can get 240 equal-sized balls to touch a central ball of the same size. E8 acts as symmetries of this arrangement. There’s a clever trick for grouping the 240 balls into 120 ordered pairs, which is explained by Fring and Korff and also by Conway’s "icosian" construction of E8 described at the end of my talk on the number 8. Each element of H4 gives a permutation of the 120 faces of the 120-cell – and thanks to that clever trick, this gives a permutation of the 240 balls. This permutation actually comes from an element of E8. So, we get an inclusion of H4 in E8.

• Scott Hotton says:

The lattice ${\bf Z}^5 \subset {\bf R}^5$ has 5-fold symmetry from the cyclic permutation of the standard basis of ${\bf R}^5$. The permutation action is the direct sum of its action on ${\rm span}\{(1,1,1,1,1)\}$ and its orthogonal complement $V={\rm span}\{(1,1,1,1,1)\}^{\perp}$. The intersection of a two dimensional subspace of $V$ with the Voronoi cells of the lattice ${\bf Z}^5 \; \cap \; V$ gives a quasiperiodic tiling of the plane with 5-fold symmetry. This can be generalized to give quasiperiodic tilings of the plane with $n$-fold symmetry. I think these pictures from the web may be examples:

https://geometricolor.files.wordpress.com/2017/09/ballsimprov.jpg?w=640&h=448

3. Just used sympy to check 1) and 2a). I think there is a discrepancy between the inner product you define here in the post and the inner product David defines in his question, and it seems the determinant you indicate checking is different than the determinant David checks. I uploaded my jupyter notebook for people to check if they feel like it: https://nbviewer.jupyter.org/github/jarthurgross/from-the-octahedron-to-e8/blob/master/from-the-octahedron-to-e8.ipynb

• John Baez says:

Could you say how my inner product is different from David Harden’s and how the determinant I indicate checking is different from his? (I felt sure they were the same.)

• Well, for $v_1$ I have $a=c=e=g=1/2$ and $b=d=f=h=0$, so by your definition $\langle v_1,v_1\rangle=4(1/2)^2=1$, but the Gram matrix David lists in his question indicates $\langle v_1, v_1\rangle=2$, which is consistent with what I get when I tried coding up his definition of the inner product. And for the determinants, it sounded to me like you were suggesting taking the determinant of the matrix whose rows were the coefficients of the generators in the $a,\ldots,h$ basis, while David mentions taking the determinant of the Gram matrix. Perhaps I misinterpreted something, though.

• John Baez says:

Thanks! I need to think some more and maybe fix my article.

• John Baez says:

Okay, I fixed the article. I feel like explaining my mistakes in detail, to justify my foolishness, but I’ll just say that I thought Harden’s inner product was more like Conway and Sloane’s than it really is, and if it had been what I thought it was, the determinant of the matrix of components of the vectors $v_i$ would be 1 (or -1) iff the determinant of the Gram matrix $\langle v_i, v_j \rangle$ was 1.

4. Layra says:

If you try to run this construction with the tetrahedron, you run into the problem that the natural embedding of the binary tetrahedral group into the quaternions is the 24 cell, so everything has rational coefficients.

• John Baez says:

You’re right. The binary tetrahedral group gives the first 24-cell I mentioned above, with vertices

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

and

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

So the integral linear combinations of these give a lattice in the quaternions, the Hurwitz integral quaternions. This is also a subring of the quaternions: Harden points out that Wilson calls it the ‘tetrian ring’. In this case we can’t pull the dimension-doubling trick that we can with the octian ring or the icosian ring!

5. J Gregory Moxness says:

I thought it might be interesting to visualize the related Platonic solid 3D structures of the 2160 E8 $2_41$ Witting Polytope from complex projective 4-space. You may find it interesting, if not helpful.

6. Taylor Ogata says:

Stupid question that is bugging… why isn’t the kissing number in 8 dimensions 1136?

16 from plus/minus 1 in all 8 dimensions

Then, 1/2 in any 4 of the dimensions, combination of (8,4) = 70, and the 1/2 choice in any of the 4 dimensions could be plus or minus so for each combination of the 70 there is 16 possibles ones.
Ex: (0, 1/2, 1/2, 0, 1/2, 0, 0, 1/2) or (0, -1/2, 1/2, 0, 1/2, 0, 0, 1/2) or (1/2, 1/2, -1/2, 1/2, 0, 0, 0, 0), etc…

Total: 16 + 70(16) = 1136

All of these points are distance 1 away from origin and distance 1 or more away from each other

Thanks for any insight to error in thought or what is not be considered here.

• John Baez says:

What’s the distance between

(1/2, 1/2, 1/2,1/2, 0, 0, 0, 0)

and

(1/2, 1/2, 1/2, 0, 1/2, 0, 0, 0) ?

You seem to be claiming it’s at least 1.

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