## November 15, 2014

### Integral Octonions (Part 8)

#### Posted by John Baez

This time I’d like to summarize some work I did in the comments last time, egged on by a mysterious entity who goes by the name of ‘Metatron’.

As you probably know, there’s an archangel named Metatron who appears in apocryphal Old Testament texts such as the Second Book of Enoch. These texts rank Metatron second only to YHWH himself. I don’t think the Metatron posting comments here is the same guy. However, it’s a good name for someone interested in lattices and geometry, since there’s a variant of the Cabbalistic Tree of Life called Metatron’s Cube, which looks like this:

This design includes within it the $\mathrm{G}_2$ root system, a 2d projection of a stellated octahedron, and a perspective drawing of a hypercube.

Anyway, there are lattices in 26 and 27 dimensions that play rather tantalizing and mysterious roles in bosonic string theory. Metatron challenged me to find octonionic descriptions of them. I tried.

Given a lattice $L$ in $n$-dimensional Euclidean space, there’s a way to build a lattice $L^{++}$ in $(n+2)$-dimensional Minkowski spacetime. This is called the ‘over-extended’ version of $L$.

If we start with the lattice $\mathrm{E}_8$ in 8 dimensions, this process gives a lattice called $\mathrm{E}_{10}$, which plays an interesting but mysterious role in superstring theory. This shouldn’t come as a complete shock, since superstring theory lives in 10 dimensions, and it can be nicely formulated using octonions, as can the lattice $\mathrm{E}_8$.

If we start with the lattice called $\mathrm{D}_{24}$, this over-extension process gives a lattice $\mathrm{D}_{24}^{++}$. This describes the ‘cosmological billiards’ for the 3d compactification of the theory of gravity arising from bosonic string theory. Again, this shouldn’t come as a complete shock, since bosonic string theory lives in 26 dimensions.

Last time I gave a nice description of $\mathrm{E}_{10}$: it consists of $2 \times 2$ self-adjoint matrices with integral octonions as entries.

It would be nice to get a similar octonionic description of $\mathrm{D}_{24}^{++}$. But it’s actually easier to go up to 27 dimensions, because the space of $3 \times 3$ self-adjoint matrices with octonion entries is 27-dimensional. And indeed, there’s a 27-dimensional lattice waiting to be described with octonions.

You see, for any lattice $L$ in $n$-dimensional Euclidean space, there’s also a way to build a lattice $L^{+++}$ in $(n+3)$-dimensional Minkowski spacetime, called the ‘very extended’ version of $L$.

If we do this to $L = \mathrm{E}_8$ we get an 11-dimensional lattice called $\mathrm{E}_{11}$, which has mysterious connections to M-theory. But if we do it to $\mathrm{D}_{24}$ we get a 27-dimensional lattice sometimes called $\mathrm{K}_{27}$. You can read about both these lattices here:

I’ll prove that $\mathrm{E}_{11}$ has a nice description in terms of integral octonions. I’ll almost do something similar for $\mathrm{K}_{27}$, but in fact I’ll succeed for a lattice containing it, which is twice as dense.

To do these things, I’ll use the explanation of over-extended and very extended lattices given here:

These constructions use a 2-dimensional lattice called $\mathrm{H}$. Let’s get to know this lattice. It’s very simple.

### A 2-dimensional Lorentzian lattice

Up to isometry, there’s a unique even unimodular lattice in Minkowski spacetime whenever its dimension is 2 more than a multiple of 8. The simplest of these is $\mathrm{H}$: it’s the unique even unimodular lattice in 2-dimensional Minkowski spacetime.

There are various ways to coordinatize $\mathrm{H}$. The easiest, I think, is to start with $\mathbb{R}^2$ and give it the metric $g$ with

$g(x,x) = -2 u v$

when $x = (u,v)$. Then, sitting in $\mathbb{R}^2$, the lattice $\mathbb{Z}^2$ is even and unimodular. So, it’s a copy of $\mathrm{H}$.

Let’s get to know it a bit. The coordinates $u$ and $v$ are called lightcone coordinates, since the $u$ and $v$ axes form the lightcone in 2d Minkowski spacetime. In other words, the vectors

$\ell = (1,0), \quad \ell' = (0,1)$

are lightlike, meaning

$g(\ell,\ell) = 0 , \quad g(\ell', \ell') = 0$

Their sum is a timelike vector

$\tau = \ell + \ell' = (1,1)$

since the inner product of $\tau$ with itself is negative; in fact

$g(\tau,\tau) = -2$

Their difference is a spacelike vector

$\sigma = \ell - \ell' = (1,-1)$

since the inner product of $\sigma$ with itself is positive; in fact

$g(\sigma,\sigma) = 2$

Since the vectors $\tau$ and $\sigma$ are orthogonal and have length $\sqrt{2}$ in the metric $g$, we get a square of area $2$ with corners

$0, \tau, \sigma, \tau + \sigma$

that is,

$(0,0),\; (1,1),\; (1,-1), \;(2,0)$

If you draw a picture, you can see by dissection that this square has twice the area of the unit cell

$(0,0),\; (1,0), \; (0,1) , \; (1,1)$

So, the unit cell has area 1, and the lattice is unimodular as claimed. Furthermore, every vector in the lattice has even inner product with itself, so this lattice is even.

### Over-extended lattices

Given a lattice $L$ in Euclidean $\mathbb{R}^n$,

$L^{++} = L \oplus \mathrm{H}$

is a lattice in $(n+2)$-dimensional Minkowski spacetime, also known as $\mathbb{R}^{n+1,1}$. This lattice $L^{++}$ is called the over-extension of $L$.

A direct sum of even lattices is even. A direct sum of unimodular lattices is unimodular. Thus if $L$ is even and unimodular, so is $L^{++}$.

All this is obvious. But here are some deeper facts about even unimodular lattices. First, they only exist in $\mathbb{R}^n$ when $n$ is a multiple of 8. Second, they only exist in $\mathbb{R}^{n+1,1}$ when $n$ is a multiple of 8.

But here’s the really amazing thing. In the Euclidean case there can be lots of different even unimodular lattices in a given dimension. In 8 dimensions there’s just one, up to isometry, called $\mathrm{E}_8$. In 16 dimensions there are two. In 24 dimensions there are 24. In 32 dimensions there are at least 1,160,000,000, and the number continues to explode after that. On the other hand, in the Lorentzian case there’s just one even unimodular lattice in a given dimension, if there are any at all.

More precisely: given two even unimodular lattices in $\mathbb{R}^{n+1,1}$, they are always isomorphic to each other via an isometry: a linear transformation that preserves the metric. We then call them isometric.

Let’s look at some examples. Up to isometry, $\mathrm{E}_8$ is the only even unimodular lattice in 8-dimensional Euclidean space. We can identify it with the lattice of integral octonions, $\mathbf{O} \subseteq \mathbb{O}$, with the inner product

$g(X,X) = 2 X X^*$

$L^{++}$ is usually called $E_{10}$. Up to isometry, this is the unique even unimodular lattice in 10 dimensions. There are lots of ways to describe it, but last time we saw that it’s the lattice of $2 \times 2$ self-adjoint matrices with integral octonions as entries:

$\mathfrak{h}_2(\mathbf{O}) = \left\{ \; \left( \begin{array}{cc} a & X \\ X^* & b \end{array} \right) : \; a,b \in \mathbb{Z}, \; X \in \mathbf{O} \; \right\}$

where the metric comes from $-2$ times the determinant:

$x = \left( \begin{array}{cc} a & X \\ X^* & b \end{array} \right) \;\; \implies \;\; g(x,x) = - \det(x) = 2 X X^* - 2 a b$

We’ll see a fancier formula like this later on.

There are 24 even unimodular lattices in 24-dimensional Euclidean space. One of them is

$\mathrm{E}_8 \oplus \mathrm{E}_8 \oplus \mathrm{E}_8$

Another is $\mathrm{DD}_{24}$ the lattice of vectors in $\mathbb{R}^{24}$ where the components are either all integers or all half-integers and their sum is even. This contains the lattice $\mathrm{D}_{24}$ that I mentioned earlier. It’s twice as dense. I’ll explain it later.

If we take the over-extension of any of these even unimodular lattices in 24-dimensional Euclidean space, we get an even unimodular lattice in 26-dimensional Minkowski spacetime… and all these are isometric! The over-extension process ‘washes out the difference’ between them. In particular,

$\mathrm{DD}_{24}^{++} \cong (\mathrm{E}_8 \oplus \mathrm{E}_8 \oplus \mathrm{E}_8)^{++}$

This is nice because up to a scale factor, $\mathrm{E}_8$ is the lattice of integral octonions. So, there’s a description of $\mathrm{DD}_{24}^{++}$ using three integral octonions! But the story is prettier if we go up an extra dimension.

### Very extended lattices

After the over-extended version $L^{++}$of a lattice $L$ in Euclidean space comes the ‘very extended’ version, called $L^{+++}$. If you ponder the paper by Gaberdiel et al, you can see this is the direct sum of the over-extension $L^{++}$ and a 1-dimensional lattice called $\mathrm{A}_1$. $\mathrm{A}_1$ is just $\mathbb{Z}$ with the metric

$g(x,x) = 2 x^2$

It’s even but not unimodular.

In short, the very extended version of $L$ is

$L^{+++} = L^{++} \oplus \mathrm{A}_1 = L \oplus \mathrm{H} \oplus \mathrm{A}_1$

If $L$ is even, so is $L^{+++}$. But if $L$ is unimodular, this will not be true of $L^{+++}$.

The very extended version of $\mathrm{E}_8$ is called $\mathrm{E}_{11}$. The very extended version of $\mathrm{D}_{24}$ is called $\mathrm{K}_{27}$. This a fascinating thing, and it would be nice to describe it using octonions. But it will be easier to work with a lattice that’s twice as dense: the very extended version of $\mathrm{DD}_{24}$.

Now it’s time to explain this ‘twice as dense’ business.

### Doubling the density of Dn

The $\mathrm{D}_n$ lattice is very simple. Take an $n$-dimensional checkerboard with alternating red and black hypercubes. Put a dot in the middle of each black hypercube. That’s the $\mathrm{D}_n$ lattice!

More precisely, the $\mathrm{D}_n$ lattice consists of all $n$-tuples of integers that sum to an even integer. Requiring that they sum to an even integer picks out the center of every other hypercube in our checkerboard.

Here’s a basis of vectors for the $\mathrm{D}_4$ lattice:

$\begin{array}{cccc} (-1, & -1, & 0, & 0) \\ (1, & -1, & 0, & 0 ) \\ (0, & 1, & -1 & 0 ) \\ (0, & 0, & 1, & -1) \end{array}$

The same pattern works in any dimension: the first vector has two $-1$s followed by a bunch of zeros, but the rest of the vectors are all just shifted versions of the second. I’ve chosen them to be simple roots for $\mathrm{D}_n$, but that’s not so important now.

What really matters is this. The dot product of any basis vector with itself is even, and the dot product of any two different ones is an integer. Thus, the lattice they generate, the $\mathrm{D}_n$ lattice, is even: the dot product of any vector with itself is even.

Also, the determinant of the matrix formed by our basis vectors:

$\left( \begin{array}{rrrr} -1 & -1 & 0 & 0 \\ 1 & -1 & 0 & 0 \\ 0 & 1 & -1 & 0 \\ 0 & 0 & 1 & -1 \end{array} \right)$

is 2. In any dimension, this sort of determinant gives $\pm 2$. So, the volume of the unit cell in $\mathrm{D}_n$ lattice is always 2. So, it’s not unimodular.

To get something unimodular, we can double the density by taking the union of $\mathrm{D}_n$ and a copy of $\mathrm{D}_n$ translated by the vector

$\left(\frac{1}{2}, \dots, \frac{1}{2} \right)$

Let’s call this union $\mathrm{DD}_n$. A lot of people call it $\mathrm{D}_n^+$, but we’re using plus signs for too many other things already. I’m calling it $\mathrm{D}_n$ because it’s doubly dense, and also because it’s the union of two shifted copies of $\mathrm{D}_n$.

$\mathrm{DD}_3$ is the way carbon atoms are arranged in a diamond!

This pattern is called the diamond cubic. It’s beautiful, but it’s not a lattice in the mathematical sense. $\mathrm{DD}_n$ is a lattice only when $n$ is even. Here’s the story:

• In any dimension, the volume of the Voronoi cells of $\mathrm{DD}_n$ is 1, so we can say it’s unimodular.
• In even dimensions, and only those, $\mathrm{DD}_n$ is lattice. After all, only then does the sum of $(\frac{1}{2}, \dots, \frac{1}{2})$ with itself again lie in $\mathrm{DD}_n$.
• In dimensions that are multiples of 4, and only those, $\mathrm{DD}_n$ is an integral lattice, meaning that the dot product of any two vectors in the lattice is an integer. After all, only then is the inner product of $(\frac{1}{2}, \dots, \frac{1}{2})$ with itself an integer.
• In dimensions that are multiples of 8, and only those, $\mathrm{DD}_n$ is an even lattice, meaning that the dot product of any vector with itself is even. After all, only then is the inner product of $(\frac{1}{2}, \dots, \frac{1}{2})$ even.

As I mentioned before, even unimodular lattices are possible in Euclidean space only when the dimension is a multiple of 8. $\mathrm{DD}_8$ is none other than our friend $\mathrm{E}_8$! But what we really need now is $\mathrm{DD}_{24}$, since it’s an even unimodular lattice in 24 dimensions. I’d like to have a nice octonionic description of

$\mathrm{K}_{27} = \mathrm{D}_{24}^{+++}$

but I’ll actually get one for

$\mathrm{DD}_{24}^{+++}$

which is twice as dense. We could use this to get an octonionc description of $\mathrm{K}_{27}$, but I won’t do that.

### Lattices in the exceptional Jordan algebra

Now we are ready to have some fun!

Let $\mathfrak{h}_3(\mathbb{O})$ be the space of $3 \times 3$ self-adjoint octonionic matrices. It’s 27-dimensional since a typical element looks like

$x = \left( \begin{array}{ccc} a & X & Y \\ X^* & b & Z \\ Y^* & Z^* & c \end{array} \right)$

where $a,b,c \in \mathbb{R}, X,Y,Z \in \mathbb{O}$. It’s called the exceptional Jordan algebra. We don’t need to know about Jordan algebras now, but this concept encapsulates the fact that if $x \in \mathfrak{h}_3(\mathbb{O})$, so is $x^2$.

There’s a 2-parameter family of metrics on the exceptional Jordan algebra that are invariant under all Jordan algebra automorphisms. They have

$g(x,x) = \alpha \tr(x^2) + \beta \tr(x)^2$

for $\alpha, \beta \in \mathbb{R}$ with $\alpha \ne 0$. Some are Euclidean and some are Lorentzian.

Sitting inside the exceptional Jordan algebra is the lattice of $3 \times 3$ self-adjoint matrices with integral octonions as entries:

$\mathfrak{h}_3(\mathbf{O}) = \left\{ \; \left( \begin{array}{ccc} a & X & Y \\ X^* & b & Z \\ Y^* & Z^* & c \end{array} \right) :\; a,b,c \in \mathbb{Z}, \; X,Y,Z \in \mathbf{O} \; \right\}$

And here’s the cool part:

Theorem. There is a Lorentzian inner product $g$ on the exceptional Jordan algebra that is invariant under all automorphisms and makes the lattice $\mathfrak{h}_3(\mathbf{O})$ isometric to $\mathrm{DD}_{24}^{+++}$.

Proof. We will prove that the metric

$g(x,x) = \tr(x^2) - \tr(x)^2$

obeys all the conditions of this theorem. From what I’ve already said, it is invariant under all Jordan algebra automorphisms. The challenge is to show that it makes $\mathfrak{h}_3(\mathbf{O})$ isometric to $\mathrm{DD}_{24}^{+++}$. But instead of $\mathrm{DD}_{24}^{+++}$, we can work with $(\mathrm{E}_8 \oplus \mathrm{E}_8 \oplus \mathrm{E}_8)^{+++}$, since we have seen that

$\mathrm{DD}_{24}^{+++} \cong (\mathrm{E}_8 \oplus \mathrm{E}_8 \oplus \mathrm{E}_8)^{+++}$

Let us examine the metric $g$ in more detail. Take any element $x \in \mathfrak{h}_3(\mathbf{O})$:

$x = \left( \begin{array}{ccc} a & X & Y \\ X^* & b & Z \\ Y^* & Z^* & c \end{array} \right)$

where $a,b,c \in \mathbb{R}, X,Y,Z \in \mathbb{O}$. Then

$tr(x^2) = a^2 + b^2 + c^2 + 2(X X^* + Y Y^* + Z Z^*)$

while

$tr(x)^2 = (a + b + c)^2$

Thus

$\begin{array}{ccl} g(x,x) &=& tr(x^2) - tr(x)^2 \\ &=& 2(X X^* + Y Y^* + Z Z^*) - 2(a b + b c + c a) \end{array}$

It follows that with this metric, the diagonal matrices are orthogonal to the off-diagonal matrices. An off-diagonal matrix $x \in \mathfrak{h}_3(\mathbf{O})$ is a triple $(X,Y,Z) \in \mathbf{O}^3$, and has

$g(x,x) = 2(X X^* + Y Y^* + Z Z^*)$

Thanks to the factor of 2, this metric makes the lattice of these off-diagonal matrices isometric to $\mathrm{E}_8 \oplus \mathrm{E}_8 \oplus \mathrm{E}_8$. Since

$(\mathrm{E}_8 \oplus \mathrm{E}_8 \oplus \mathrm{E}_8)^{+++} = \mathrm{E}_8 \oplus \mathrm{E}_8 \oplus \mathrm{E}_8 \oplus \mathrm{H} \oplus \mathrm{A}_1$

it thus suffices to show that the 3-dimensional Lorentzian lattice of diagonal matrices in $\mathfrak{h}_3(\mathbf{O})$ is isometric to

$\mathrm{H} \oplus \mathrm{A}_1$

A diagonal matrix $x \in \mathfrak{h}_3(\mathbf{O})$ is a triple $(a,b,c) \in \mathbb{Z}^3$, and on these triples the inner product $g$ is given by

$g(x,x) = -2(a b + b c + c a)$

If we restrict attention to triples of the form $x = (a,b,0)$, we get a 2-dimensional Lorentzian lattice: a copy of $\mathbb{Z}^2$ with inner product

$g(x,x) = -2a b$

This is just $\mathrm{H}$.

We can use this to show that the lattice of all triples $(a,b,c) \in \mathbb{Z}^3$, with the inner product $g$, is isometric to $\mathrm{H} \oplus \mathrm{A}_1$.

Remember, $\mathrm{A}_1$ is a 1-dimensional lattice generated by a spacelike vector whose norm squared is 2. So, it suffices to show that the lattice $\mathbb{Z}^3$ is generated by vectors of the form $(a,b,0)$ together with a spacelike vector of norm squared 2 that is orthogonal to all those of the form $(a,b,0)$.

To do this, we need to describe the inner product $g$ on $\mathbb{Z}^3$ more explicitly. For this, we can use polarization identity

$g(x,x') = \frac{1}{2}( g(x+x',x+x') - g(x,x) - g(x',x'))$

Remember, if $x = (a,b,c)$ we have

$g(x,x) = -2(a b + b c + c a)$

So, if we also have $x' = (a',b',c')$, the polarization identity gives

$g(x,x') = -(a b'+a' b) - (b c'+ b c') - (c a' + c'a)$

We are looking for a spacelike vector $x' = (a',b',c')$ that is orthogonal to all those of the form $x = (a,b,0)$. For this, it is necessary and sufficient to have

$0 = g((1,0,0),(a',b',c')) = - b' - c'$

and

$0 = g((0,1,0), (a',b',c')) = - a' - c'$

An example is $x' = (1,1,-1)$. This has

$g(x',x') = -2(1 - 1 - 1) = 2$

so it is spacelike, as desired. Even better, it has norm squared two. And even better, this vector $x'$, along with those of the form $(a,b,0)$, generates the lattice $\mathbb{Z}^3$.

So we have shown what we needed: the lattice of all triples $(a,b,c) \in \mathbb{Z}^3$ is generated by those of the form $(a,b,0)$ together with a spacelike vector with norm squared 2 that is orthogonal to all those of the form $(a,b,0)$. $\blacksquare$

This theorem has three nice spinoffs:

Corollary 1. With the same Lorentzian inner product $g$ on the exceptional Jordan algebra, the lattice $\mathrm{DD}_{24}^{++}$ is isometric to the sublattice of $\mathfrak{h}_3(\mathbf{O})$ where a fixed diagonal entry is set equal to zero, e.g.:

$\left\{ \; \left( \begin{array}{ccc} a & X & Y \\ X^* & b & Z \\ Y^* & Z^* & 0 \end{array} \right) : \; a,b \in \mathbb{Z}, \; X,Y,Z \in \mathbf{O} \; \right\}$

Proof. Use the fact that with the metric $g$, the diagonal matrices

$\left\{ \; \left( \begin{array}{ccc} a & 0 & 0 \\ 0 & b & 0 \\ 0 & 0 & 0 \end{array} \right) : \; a,b \in \mathbb{Z} \; \right\}$

form a copy of $\mathrm{H}$, so the matrices above form a copy of

$\mathrm{E}_8 \oplus \mathrm{E}_8 \oplus \mathrm{E}_8 \oplus \mathrm{H} \cong (\mathrm{E}_8 \oplus \mathrm{E}_8 \oplus \mathrm{E}_8)^{++} \cong \mathrm{DD}_{24}^{++} \qquad \qquad \qquad \blacksquare$

Corollary 2. With the same Lorentzian inner product $g$ on the exceptional Jordan algebra, the lattice $\mathrm{E}_{11} = E_8^{+++}$ is isometric to the sublattice of $\mathfrak{h}_3(\mathbf{O})$ where two fixed off-diagonal entries are set equal to zero, e.g.:

$\left\{ \; \left( \begin{array}{ccc} a & X & 0 \\ X^* & b & 0 \\ 0 & 0 & c \end{array} \right) : \; a,b,c \in \mathbb{Z}, \; X\in \mathbf{O} \; \right\}$

Proof. Use the fact that with the metric $g$, the diagonal matrices

$\left\{ \; \left( \begin{array}{ccc} a & 0 & 0 \\ 0 & b & 0 \\ 0 & 0 & c \end{array} \right) : \; a,b \in \mathbb{Z} \; \right\}$

form a copy of $\mathrm{H} \oplus \mathrm{A}_1$, so the matrices above form a copy of

$\mathrm{E}_8 \oplus \mathrm{H} \oplus \mathrm{A}_1 \cong \mathrm{E}_8^{+++} \qquad \qquad \qquad \blacksquare$

Corollary 3. With the same Lorentzian inner product $g$ on the exceptional Jordan algebra, the lattice $\mathrm{E}_{10} = \mathrm{E}_8^{++}$ is isometric to the sublattice of $\mathfrak{h}_3(\mathbf{O})$ where two fixed off-diagonal entries and one diagonal entry are set equal to zero, e.g.:

$\left\{ \; \left( \begin{array}{ccc} a & X & 0 \\ X^* & b & 0 \\ 0 & 0 & 0 \end{array} \right) : \; a,b,c \in \mathbb{Z}, \; X\in \mathbf{O} \; \right\}$

Proof. Use the previous corollary; this is the obvious copy of $\mathrm{E}_8^{++} \cong \mathrm{E}_8 \oplus \mathrm{H}$ inside $\mathrm{E}_8^{+++} \cong \mathrm{E}_8 \oplus \mathrm{H} \oplus \mathrm{A}_1$. $\blacksquare$

Posted at November 15, 2014 6:06 AM UTC

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### Re: Integral Octonions (Part 8)

This is very nice!

I’ve been pondering that metric:

$g(x,x) = \tr(x^2) - \tr(x)^2$

and wondering if there’s some setting where things work as nicely in the $n\times n$ case as they do for the $2\times 2$. I think this metric is Lorentzian for any $n\ge 2$ on the $n\times n$ self-adjoint matrices over any division algebra $\mathbb{K}$, with the identity matrix as a timelike vector, while all self-adjoint pairs of off-diagonal matrix elements are spacelike, as well as the $n-1$ diagonal matrices such as $diag(1,-1,0,0,...)$.

You say that for $n=3$ this metric is invariant under the Jordan algebra automorphisms. But do you know if there’s some kind of systematic representation of $SO_0(m,1)$ on the $n\times n$ matrices (where $m = n-1 + \frac{1}{2}n(n-1) dim \mathbb{K}$)?

Posted by: Greg Egan on November 15, 2014 12:02 PM | Permalink | Reply to this

### Re: Integral Octonions (Part 8)

Interesting! I guess the existence of this representation follows abstractly from what you said: you’ve made $\mathfrak{h}_n(\mathbb{K})$ into a copy of $(m+1)$-dimensional Minkowski spacetime whenever $\mathbb{K}$ is a normed division algebra. But I haven’t seen this before. I’ll think about it.

The space $\mathfrak{h}_4(\mathbb{O})$ exerts a dangerous appeal that so far has never done anyone any good: like $\mathfrak{h}_2(\mathbb{O})$ and $\mathfrak{h}_3(\mathbb{O})$ it is closed under the product

$x \circ y = x y + y x$

but in this case the Jordan identity

$(x \circ y) \circ (x \circ x) = x \circ (y \circ (x \circ x))$

fails to hold, so it’s not a Jordan algebra. This is closely related to the fact that there’s a projective line $\mathbb{O}\mathrm{P}^1$ and a projective plane $\mathbb{O}\mathrm{P}^2$ but no projective 3-space $\mathbb{O}\mathrm{P}^3$.

One could hope that someday someone will figure out something really interesting to do with $\mathfrak{h}_n(\mathbb{O})$ for $n \ge 3$.

Posted by: John Baez on November 15, 2014 4:41 PM | Permalink | Reply to this

### Re: Integral Octonions (Part 8)

Here’s one nice thing about the metric

$g(x,x) = \tr(x^2) - \tr(x)^2$

on the space of $n \times n$ self-adjoint real, complex, quaternionic or octonionic matrices. If you look at it on the diagonal matrices, you get a very pretty version of $n$-dimensional Minkowski spacetime. In the obvious basis of diagonal matrices, this metric takes the form

$g_{i j} = \left\{ \begin{array}{cc} 1 & i \ne j \\ 0 & i = j \end{array} \right.$

For example, when $n = 4$ we get

$\left( \begin{array}{cccc} 0 & 1 & 1 & 1 \\ 1 & 0 & 1 & 1 \\ 1 & 1 & 0 & 1 \\ 1 & 1 & 1 & 0 \end{array} \right)$

which is a strikingly beautiful different coordinatization of the usual Minkowski metric on $\mathbb{R}^{3,1}$. I first learned about this from David Finkelstein. In this setup, all 4 coordinate axes are lightlike; as you move along one of them you trace out the motion of the corner of a tetrahedron that is expanding at the speed of light!

I need to think about what the lattice $\mathbb{Z}^n$ is like with this sort of Lorentzian metric. In the case $n = 3$, my blog article here showed it’s isomorphic to $\mathrm{H} \oplus \mathrm{A}_1$. I think you’ve given the clue to showing that in general we get $\mathrm{H} \oplus \mathrm{A}_{n-2}$.

Posted by: John Baez on November 15, 2014 4:54 PM | Permalink | Reply to this

### Re: Integral Octonions (Part 8)

That is nice!

One tiny nitpick, I think that with the metric:

$g(x,x) = \tr(x^2) - \tr(x)^2$

and the basis $\{diag(1,0,0,...), diag(0,1,0,...),...\}$, the coordinates are:

$g_{i j} = \left\{ \begin{array}{cc} -1 & i \ne j \\ 0 & i = j \end{array} \right.$

So that gives the opposite of the Finkelstein matrix, but that’s OK, it just corresponds to signature $(-+++)$ rather than Finkelstein’s $(+---)$.

I believe you’re right that with this metric, $\mathbb{Z}^n$ is isomorphic to $\mathrm{H} \oplus \mathrm{A}_{n-2}$. Everything you’ve already shown about embedding H in the integer vectors of the form $(a,b,0,0,0,...)$ remains, e.g. it’s generated by lightlike vectors I’ll call $e_1=(1,0,0,0,...)$ and $e_2=(0,1,0,0,...)$, and we get a copy of $\mathrm{A}_{n-2}$ that is orthogonal to those first two vectors, with simple roots:

$\begin{array}{rcl}r_1 & = & (1,1,-1,0,0,...)\\ r_2 & = & (0,0,1,-1,0,0,...) \\ r_3 & = &(0,0,0,1,-1,0,...) \\ & ... & \end{array}$

To see that we really get all of $\mathbb{Z}^n$ from this basis, note that we can get the remaining $n-2$ elements of the standard basis $\{e_i\}$ of $\mathbb{Z}^n$ as:

$e_i = e_1 + e_2 - \sum_{j=1}^{i-2} r_j$

Posted by: Greg Egan on November 16, 2014 12:31 AM | Permalink | Reply to this

### Re: Integral Octonions (Part 8)

Great!—thanks for checking this, and catching that mistake.

I still need to get a better mental image of this sort of Lorentzian lattice. Of course all lattices look like $\mathbb{Z}^n$ with respect to some inner product, so that’s not very helpful. The picture of it as $\mathrm{H} \oplus \mathrm{A}_{n-2}$ is better, but it makes me think that we’ve got the lattice $\mathrm{H}$ sitting in plane containing the time axis and one space axis (I imagine $\mathrm{H}$ as a kind of ‘diamond’ lattice with the time axis pointing up, and the lightcones at 45° from the vertical), and then all the other orthogonal space directions involved in an $\mathrm{A}_{n-2}$ lattice. This is both hard to visualize and somehow misleading, since there’s not really one space coordinate that’s different from all the rest.

I guess the really good picture is something like this. Take a cubical $\mathbb{Z}^n$ lattice and tilt it so that the $(1,1,\dots, 1)$ direction is pointing ‘up’, in the time direction. In the 3d case you can just imagine a cube chopped into smaller cubelets, standing on one corner. Take a horizontal plane—a spacelike slice orthogonal to the $(1,1,1)$ direction—and move it up through the cube. You’ll get a slice of the cube which starts out as an expanding triangle, its corners moving outward at the speed of light. And certain evenly spaced slices will contain a bunch of cubelet corners arranged in an $\mathrm{A}_2$ lattice!

I guess the same trick should work in higher dimensions, but give slices that contain $\mathrm{A}_{n-1}$ lattices.

David Finkelstein had some radical ideas about physics based on a discrete spacetime model with this sort of lattice geometry. It’s not invariant under all Lorentz transformations, of course, but I find it more attractive than the usual spacetime model that cellular automaton fans like: a $\mathbb{Z}^n$ lattice where one coordinate is time and the rest are space. At least here we’ve got $n$ preferred axes that are all the same! And in the $n = 3$ case, I guess we get spacelike slices with the ‘face-centered cubic’ lattice in them, much cooler than a plain old cubic lattice. Not that I believe in this model… but it’s got some panache.

Posted by: John Baez on November 16, 2014 2:18 AM | Permalink | Reply to this

### Re: Integral Octonions (Part 8)

Metatron suggested that the virtue of the metric

$g(x,x) = \tr(x^2) - \tr(x)^2$

on $\mathfrak{h}_3(\mathbb{O})$ was that any projection $x \in \mathfrak{h}_3(\mathbb{O})$ is lightlike. I assented, but this isn’t quite right. It’s still a great idea, though.

A projection $x$ is an element with $x^2 = x$. A projection in $\mathfrak{h}_3(\mathbb{O})$ can have rank 0, 1, 2, or 3, where its rank is $tr(x)$.

We only get $g(x,x) = 0$ for a projection $x$ when it has rank 1. Of course, any multiple of a rank-1 projection will also have $g(x,x) =0$. But not every element with $g(x,x) = 0$ is a scalar multiple of a rank-1 projection.

But the rank-1 projections are very nice: they correspond in a 1-1 way with points in the octonionic projective plane $\mathbb{O}\mathrm{P}^2$. Indeed this is a nice way to define $\mathbb{O}\mathrm{P}^2$.

So, with this metric on $\mathfrak{h}_3(\mathbb{O})$, we are seeing $\mathbb{O}\mathrm{P}^2$ as a subspace of the set of light rays through the origin in 27-dimensional Minkowski spacetime! Each rank-1 projection determines a light ray.

Thanks to your remark, Greg, a lot of this stuff generalizes to $\mathfrak{h}_n(\mathbb{K})$ whenever this is a Jordan algebra: $\mathbb{K} = \mathbb{R}, \mathbb{C}, \mathbb{H}$ for any $n$, and $\mathbb{O}$ at least when $n \le 3$.

Posted by: John Baez on November 15, 2014 5:04 PM | Permalink | Reply to this

### Re: Integral Octonions (Part 8)

Yes, the rank one projections are what I meant, those that satisfy $X^{#}=X\times X=0$. We can find nice examples of these via Dray and Manogue’s algorithm give in math-ph/9910004. I still need to get around to writing an app that spits out the eigenmatrices and eigenvalues for any chosen element of $\mathfrak{h}_3(\mathbb{O})$.

Looking forward, it’s a great time to re-visit the Leech and the 26-dimensional unimodular lattice $\Pi_{25,1}$. Recall that the monster Lie algebra $\hat{M}$ (which the Monster group acts on) is defined as the generalized Kac-Moody algebra whose Dynkin diagram is the Dynkin diagram of $\Pi_{25,1}$, so $\hat{M}$ has a simple root for each point of the Leech lattice $\Lambda$. The group of the automorphisms of the Leech, $Co_0$, can be cast in octonion form via $F_4$ transformations, and the quotient of $Co_0$ by its center, the largest Conway group $Co_1$, in turn sits inside the Monster group. This begs the question: can we cast automorphisms of $\Pi_{25,1}$ using $F_4$ transformations?

Another cool fact: there exists a 27-dimensional unimodular lattice $L$ with no roots, where $\mathrm{Aut}(L)$ has order $2^{12}\cdot 3^3\cdot 5^2\cdot 7^2\cdot 13$, the order of $2\times^3D_4(2)\cdot 3$. There is a 27-dimensional representation $V$ of $2\times^3D_4(2)\cdot 3$ such that $V$ can be given the structure of $\mathfrak{h}_3(\mathbb{O})$ such that $L$ is closed under multiplication. Maybe $F_4$ also plays a role here.

Posted by: Metatron on November 15, 2014 9:12 PM | Permalink | Reply to this

### Re: Integral Octonions (Part 8)

I’m slowly gearing up for an assault on the Leech lattice — thanks for the tips. I’m just now learning some of the really good ways of thinking about the Leech.

Is $\mathrm{Co}_0$ a subgroup of $\mathrm{F}_4$? If so, is there a comprehensible account of this somewhere? This seems like a good step.

Posted by: John Baez on November 16, 2014 4:27 PM | Permalink | Reply to this

### Re: Integral Octonions (Part 8)

These are new results, so all we have so far is R.A. Wilson’s construction of $Co_0=2\cdot Co_1$ generators in terms of $3\times 3$ matrices over $\mathbb{O}$. These matrices are of type $F_4$, as can be checked quickly. As $Co_0$ is the double cover of $Co_1$, and $Co_1$ is a large subgroup of the Monster, a significant portion of the Monster can be generated by $F_4$ transformations. And we know $F_4$ is the isometry group of $\mathbb{OP}^2$, as well as the automorphism group of $\mathfrak{h}_3(\mathbb{O})$. So the bigger question is: can one generate the rest of the Monster with (exceptional Lie group) transformations that act on octonion structures?

Posted by: Metatron on November 16, 2014 6:04 PM | Permalink | Reply to this

### Re: Integral Octonions (Part 8)

Proposition: All rank-1 elements in the exceptional Jordan algebra $\mathfrak{h}_3(\mathbb{O})$ are lightlike under the metric $g(X,X)=\mathrm{tr}(X^2)-\mathrm{tr}X^2$.

Proof: Given $X\in\mathfrak{h}_3(\mathbb{O})$ satisfying $X\times X=0$ (i.e., rank-1), we have $\mathrm{tr}(X\times X)=0$. Expanding, this gives: $\mathrm{tr}(X\times X)=\mathrm{tr}(X^2)-\mathrm{tr}X^2+\frac{3}{2}\mathrm{tr}X^2-\frac{3}{2}\mathrm{tr}(X^2)=0$ $\Rightarrow \mathrm{tr}X^2-\mathrm{tr}(X^2)=0$ $\Rightarrow \mathrm{tr}X^2=\mathrm{tr}(X^2)$ $\Rightarrow g(X,X)=0$. $\blacksquare$

Corollary: All points of the Cayley plane $\mathbb{OP}^2$ are lightlike under the metric $g(X,X)=\mathrm{tr}(X^2)-\mathrm{tr}X^2$.

Posted by: Metatron on November 17, 2014 5:16 PM | Permalink | Reply to this

### Re: Integral Octonions (Part 8)

Nice! The way I saw this fact was to start by noting that my favorite rank-1 projection in $\mathfrak{h}_3(\mathbb{O})$, e.g. the diagonal matrix

$x = \mathrm{diag}(1,0,0)$

obeys

$tr(x^2) - tr(x)^2 = 0$

Then I used the fact that $\mathrm{F}_4$ acts on $\mathfrak{h}_3(\mathbb{O})$ in a way that preserves the trace, and acts transitively on these projections—which are the points of $\mathbb{O}\mathrm{P}^2$, a space well known to have a transitive action of $\mathrm{F}_4$. So, all rank-1 projections have $tr(x^2) - tr(x)^2 = 0$.

Needless to say, my proof doesn’t really save work, except locally, since you need to have done a bunch more work previously. I’m glad to see a more direct argument.

Either way, the upshot is that we can think of $\mathbb{O}\mathrm{P}^2$ as sitting inside the projectivized lightcone in $\mathfrak{h}_3(\mathbb{O})$: that is, the set of light rays. Two points in the lightcone define the same ray if they’re scalar multiples of each other. $\mathbb{O}\mathrm{P}^2$ is 16-dimensional, and the projectivized lightcone is 25-dimensional: 27 minus one for the condition $tr(x^2) - tr(x)^2 = 0$ and one for projectivizing.

In fact, the projectivized lightcone is really just a 25-dimensional sphere: imagine slicing the lightcone with a spacelike plane that doesn’t contain the origin: each ray will hit this plane at one point, and these points form a sphere.

So, we get a nice embedding

$\mathbb{O}\mathrm{P}^2 \subset S^{25}$

There’s another metric on where the rank-2 projections are lightlike:

$g(x,x) = 2 \tr(x^2) - \tr(x)^2$

These correspond naturally to the lines in the octonionic projective plane. But the space of lines is isomorphic to the space of points, by duality. So, we get another way to embed $\mathbb{O}\mathrm{P}^2$ in a projectivized lightcone.

Posted by: John Baez on November 17, 2014 7:18 PM | Permalink | Reply to this

### Re: Integral Octonions (Part 8)

Tony Smith let me quote the following email:

I apologize for sending this by email instead of as blog comment,
but this way you can decide whether it is useful or not,
and you are free to quote any or all of it as a blog comment.

You and Metatron are having an interesting discussion about K27 E11 etc
and the 27-dim Jordan algebra
and an Integral Octonion version of the Leech Lattice and related structures.

In it you and Metatron have cited some interesting papers:
math-ph/9910004 - Dray and Manogue on exceptional Jordan algebra
hep-th/0205068 - Gaberdiel, Olive, and West on very extended Lie algebras
hep-th/l0104081 - West on Closed Bosonic String and K27 which contains E11
and
referred to the work of R. A. Wilson

Have you looked at these papers:

hep-th/9703084 - Barwald, Gebert, Gunayadin, and Nicolai, on
fake monster and true monster being algebras of 26-dim bosonic string states
compactified on torus or Z2 orbifold, respectively

hep-th/0208214 - West, on closed bosonic string vacuum breaking K27 down
to E11 since K27 contains E11 + D16 (note that D16 is part of
E8 = D16 + half-spinor D16)

0805.4451 - Cook and West, on K27 = D24+++  extended to K28

hep-th/0106235 - Englert, Houart, and Taormina, on bosonic strings compactified on
E8 x SO(16) lattice

hep-th/0203098 -  Englert, Houart, and Taormina, on bosonic string in terms of
E8 x SO(16) leading to non-supersymmetric fermionic strings

hep-th/0407055 - Nicolai and Samtleben, on how fermionic degrees of freedom
should be spinors under the maximal compact subgroups of Kac Moody groups as in
the chain

... included in Spin(16) included in K(E9) included in K(E10) included in ...

To work out the relevant spinor representations for K(E10) (and also
for K(E11)) will be no easy task.

As to the work of R. A. Wilson which I guess includes his paper “Octonions
and the Leech Lattice” at

http://www.maths.qmul.ac.uk/~raw/pubs_files/octoLeech1rev.pdf

which overlaps a lot with the work of Geoffrey Dixon:

Construction of the Leech Lattice uses not only the Lattice structure of E8
lattices but also their Octonion Multiplication Integral Domain structure
in terms of which there are 7 independent E8 lattices, plus an 8th (Kirmse’s
mistake) that does not close, so you have to be specific as to which E8 lattices
you are using when you build stuff like the Leech Lattice.

Also, in his paper “Conway’s group and octonions”,
Wilson also has to pay attention to the distinctions among the algebraically
distinct E8 lattices.

Posted by: John Baez on November 17, 2014 7:27 PM | Permalink | Reply to this

### Re: Integral Octonions (Part 8)

John Baez and John Huerta wrote a series of papers about the 3-Psi rule, which explains why string theory works in 10 dimensions, and M-theory works in 11 dimensions.

Hisham Sati wrote the following paper claiming that just as 10D string theory is a limiting case of a 11D M-theory, that 11D M-theory is a limiting case of a more fundamental theory that works in 27 dimensions.

http://arxiv.org/pdf/0807.4899v3.pdf

I was wondering if there is some version of the 3-Psi rule that explains Hisham Sati’s 27D theory. What would be the analog of the 3-Psi rule for Hisham Sati’s 27 D theory?

I then thought this could also relate to what this current article is about.

If the over extension of E8 gives E10 which relates to 10D string theory, and the over extension of D24 gives D24 ++ which relates to 11D M-theory, what would be the possible other lattice would then relate to Hisham Sati’s 27D theory? What would be the analog?

Posted by: Jeffery Winkler on November 18, 2014 11:14 PM | Permalink | Reply to this

### Re: Integral Octonions (Part 8)

Jeffrey wrote:

Hisham Sati wrote the following paper claiming that just as 10D string theory is a limiting case of a 11D M-theory, that 11D M-theory is a limiting case of a more fundamental theory that works in 27 dimensions.

Wow, thanks! Thia adds to the short list of papers I know suggesting that the biggest, best theory of all lives 27 dimensions! The evidence is indirect and inconclusive, and the idea seems to be hotly disputed, but it’s interesting.

I was wondering if there is some version of the 3-Psi rule that explains Hisham Sati’s 27D theory. What would be the analog of the 3-Psi rule for Hisham Sati’s 27 D theory?

That’s an interesting question. However, the 3-$\Psi$’s rule shows up when one proves the action for 11d supergravity is supersymmetric, or alternatively, that the action for 3-branes in 11 dimensions is supersymmetric. As far as I know, most people are guessing that this mysterious 27-dimensional theory is bosonic, not supersymmetric! That would make some sense, using the analogy:

10d superstring theory:11d M-theory::26d bosonic string theory::27d bosonic M-theory

Horowitz and Susskind say:

We have proposed that a bosonic version of M theory exists, which is a 27 dimensional theory with 2-branes and 21-branes.

They don’t write down an action for this theory; their evidence for its existence is circumstantial.

On the other hand, Sati seems to think the hoped-for theory in 27 dimensions might be supersymmetric:

In physics, the lifting of M-theory via the sixteen-dimensional manifold $\mathbb{O}\mathrm{P}^2$ brings us to 27 dimensions. Given a Kaluza-Klein interpretation, this suggests the existence of a theory in 27 dimensions, whose dimensional reduction over $\mathbb{O}\mathrm{P}^2$ leads to M-theory. The higher dimensional theory involves spinors, and it is natural to ask whether or not the theory can be supersymmetric. In one form we propose this as a candidate for the ‘bosonic M-theory’ sought after in [24], from gravitational geometric arguments, and in [45], from matrix model arguments.

It’s a bit odd to say a supersymmetric theory could be a candidate for ‘bosonic M-theory’; maybe he’s saying it’s not really bosonic.

If the over extension of $\mathrm{E}_{8}$ gives $\mathrm{E}_{10}$ which relates to 10D string theory, and the over extension of $\mathrm{D}_{24}$ gives $\mathrm{D}_{24}^{++}$ which relates to 11D M-theory […]

Actually I said that $\mathrm{D}_{24}^{++}$ is related to bosonic string theory in 26 dimensions: it seems to describe the ‘gravitational billiards effect’ for the gravity theory that appears as a limit of bosonic string theory. And this makes some rough sense, since this lattice is 26-dimensional.

what would be the possible other lattice would then relate to Hisham Sati’s 27D theory? What would be the analog?

Well, the obvious candidate lattice in 27 dimensions is the one I described in my theorem here: $\mathrm{D}_{24}^{+++}$. I proved that this is the lattice of $3 \times 3$ self-adjoint integral octonion matrices! So, it’s like an integral version of the exceptional Jordan algebra, consisting of $3 \times 3$ self-adjoint octonion matrices. Some people call this lattice $\mathrm{K}_{27}$.

Posted by: John Baez on November 19, 2014 12:28 AM | Permalink | Reply to this

### Re: Integral Octonions (Part 8)

Adding to the “pro-D27” papers you listed, Smolin actually did write down an action for a candidate D=27 $\mathfrak{h}_3(\mathbb{O})$ matrix model for bosonic M-theory.

Smolin uses the trilinear form $\mathrm{tr}(X\circ (Y\circ Z))$ (invariant under $F_4$) to define his Chern-Simons-like action. Ohwashi, extends Smolin’s model using the cubic form $\mathrm{tr}(X\circ (Y\times Z))$ to get a model with $E_6$ symmetry.

Yuhi Ohwashi, E6 Matrix Model

This brings us to present time. Brezin-Kazakov, Douglas-Shenker and Gross-Migdal back in 1990, worked out a nonperturbative formulation of string theory with 0-dimensional target space (see [1], [2], [3]) with:

$Z=\int{d\phi e^{-S}}$

$S=N\mathrm{tr}(\frac{1}{2}\phi^2-\frac{k}{3}\phi^3)$

Such a model looks like strikingly similar to a matrix model with a cubic form action expanded in terms of trace:

$S_{c}=\frac{1}{3}\mathrm{tr}(X\circ (X\times X))=\frac{1}{3}\mathrm{tr}(X^3)-\frac{1}{2}\mathrm{tr}(X^2)\mathrm{tr}X+\frac{1}{6}\mathrm{tr}X^2$.

Perhaps an $E_{6(-26)}$ invariant $\mathfrak{h}_3(\mathbb{O})$ matrix model would have such an action.

Posted by: Metatron on November 20, 2014 7:56 PM | Permalink | Reply to this

### Re: Integral Octonions (Part 8)

H. Sati’s D=27 proposal includes M-theory in D=11 with 16-dimensional $\mathbb{OP}^2$ fibers. As $\mathbb{OP}^2$ consists of the set of rank-1 elements in the exceptional Jordan algebra $\mathfrak{h}_3(\mathbb{O})$, which is 27-dimensional, it’s natural to suspect there may be a matrix theory formulation for Sati’s construction. In the matrix picture, locally, the total space $M_{27}$ would be locally homeomorphic to $\mathfrak{h}_3(\mathbb{O})$.

Posted by: Metatron on November 19, 2014 4:53 AM | Permalink | Reply to this

### Re: Integral Octonions (Part 8)

I recently read Robert Wilson’s beautiful paper Octonions and the Leech Lattice, which constructs the Leech lattice as a sublattice of the space of triples of a certain set of octonions.

The particular set of octonions Wilson chooses to work with are not Cayley integral octonions, so they are not closed under multiplication. However, I realised that it’s quite easy to tweak his construction to work with (any of the seven sets of) Cayley integral octonions instead. What’s more, the geometrical aspects of his construction can be abstracted away from the octonionic aspects, giving a nice way to produce the Leech lattice (up to a rescaling) as a sublattice of $L_0^3$, where $L_0$ is any lattice isomorphic (up to a rescaling) to $E_8$.

So, suppose $L_0$ is either isomorphic to $E_8$, or to some multiple of $E_8$. For the sake of concreteness, I will illustrate what follows using a specific example, the lattice generated by the rows of this matrix, which we will call $r_1,\dots,r_8$:

$r_i=\left( \begin{array}{cccccccc} -1 & -1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & -1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & -1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 & -1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 & -1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 & -1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 1 & -1 \\ \frac{1}{2} & -\frac{1}{2} & -\frac{1}{2} & -\frac{1}{2} & -\frac{1}{2} & \frac{1}{2} & \frac{1}{2} & \frac{1}{2} \end{array}\right)$

These vectors have squared norms of 2, which is the usual normalisation convention for $E_8$, and they comprise a set of simple roots whose mutual dot products are $-1$ for roots joined by an edge in the Dynkin diagram for $E_8$, and 0 otherwise.

In this lattice, the shortest non-zero vectors have squared norms of 2, and the next-shortest vectors have squared norms of 4, i.e. they are $\sqrt{2}$ larger. For every simple root $r_i$, there are numerous vectors $w_{i,j}$ of squared norm 4 for which $r_i\cdot w_{i,j}=1$. (In fact there are 576 such vectors in every case, but that particular number is not important.) If we were dealing with a differently scaled lattice, then this dot product would be different, of course, so the scale-invariant aspect here is that the angle between $r_i$ and all these $w_{i,j}$ is $arccos{\frac{1}{2\sqrt{2}}}$.

The $w_{i,j}$ split up into pairs that lie precisely opposite each other with respect to $r_i$, with all three vectors lying in the same plane. In fact, because of the particular angle, $arccos{\frac{1}{2\sqrt{2}}}$, between the vectors and the ratio of $\sqrt{2}$ between their lengths, the sum of the two $w_{i,j}$ in each such pair is simply equal to $r_i$.

Now, we have so many choices of these pairs of $w_{i,j}$ that we can find an ordered pair $(w_{i,1}, w_{i,2})$ for every root $r_i$ such that the vectors $w_{i,1}$ all lie at the same angles to each other as the corresponding $r_i$, and so do the second members of each pair, $w_{i,2}$. The mutual dot products between these vectors will be twice as large as those of the $r_i$, but in other respects they just follow the $E_8$ Dynkin diagram as usual.

We will define $L_1$ to be the lattice generated by the eight $w_{i,1}$ and $L_2$ to be the lattice generated by the eight $w_{i,2}$; both these lattices are sublattices of $L_0$ that are isomorphic to a version of $E_8$ that is $\sqrt{2}$ larger than that of $L_0$ itself.

To prove that such sets of vectors really exist, here is an example. (I haven’t calculated how many different choices there are, but the number runs at least into the thousands.)

$w_{i,1}=\left( \begin{array}{cccccccc} 0 & -1 & 1 & 1 & 1 & 0 & 0 & 0 \\ -1 & 0 & -1 & 0 & -1 & 1 & 0 & 0 \\ -\frac{1}{2} & \frac{1}{2} & \frac{1}{2} & -\frac{1}{2} & \frac{1}{2} & -\frac{3}{2} & \frac{1}{2} & \frac{1}{2} \\ 1 & -1 & 0 & 0 & -1 & 0 & 0 & -1 \\ 0 & 1 & -1 & 1 & 1 & 0 & 0 & 0 \\ -\frac{1}{2} & \frac{1}{2} & \frac{3}{2} & -\frac{1}{2} & -\frac{1}{2} & \frac{1}{2} & -\frac{1}{2} & -\frac{1}{2} \\ \frac{1}{2} & -\frac{1}{2} & -\frac{1}{2} & -\frac{1}{2} & \frac{1}{2} & \frac{1}{2} & \frac{3}{2} & \frac{1}{2} \\ 0 & -1 & 0 & -1 & 0 & 0 & -1 & 1 \end{array} \right)$

$w_{i,2}=\left( \begin{array}{cccccccc} -1 & 0 & -1 & -1 & -1 & 0 & 0 & 0 \\ 1 & 1 & 0 & 0 & 1 & -1 & 0 & 0 \\ \frac{1}{2} & -\frac{1}{2} & \frac{1}{2} & -\frac{1}{2} & -\frac{1}{2} & \frac{3}{2} & -\frac{1}{2} & -\frac{1}{2} \\ -1 & 1 & 0 & 1 & 0 & 0 & 0 & 1 \\ 0 & -1 & 1 & -1 & 0 & -1 & 0 & 0 \\ \frac{1}{2} & -\frac{1}{2} & -\frac{3}{2} & \frac{1}{2} & \frac{1}{2} & \frac{1}{2} & -\frac{1}{2} & \frac{1}{2} \\ -\frac{1}{2} & \frac{1}{2} & \frac{1}{2} & \frac{1}{2} & -\frac{1}{2} & -\frac{1}{2} & -\frac{1}{2} & -\frac{3}{2} \\ \frac{1}{2} & \frac{1}{2} & -\frac{1}{2} & \frac{1}{2} & -\frac{1}{2} & \frac{1}{2} & \frac{3}{2} & -\frac{1}{2} \end{array} \right)$

It’s not hard to check that for any $i$, we have $w_{i,1}+w_{i,2} = r_i$, that $w_{i,1}\cdot r_i = w_{i,2}\cdot r_i = 1$, and the individual sets $\{w_{i,1}\}$ and $\{w_{i,2}\}$ generate lattices $L_1$ and $L_2$ that are versions of $E_8$ larger by a factor of $\sqrt{2}$ than the version generated by $\{r_i\}$.

Because of the geometry of the roots of these three lattices as we’ve defined them, we can concisely capture the relationship between the lattices themselves as follows. We define a linear operator $R$ by the requirement that:

$R w_{i,1} = \sqrt{2} r_i, \; \;i=1,\dots,8$

That is, in each plane spanned by $w_{i,1}$ and $r_i$, $R$ rotates by an angle of $arccos{\frac{1}{2\sqrt{2}}}$. But $w_{i,2}$ lies in that same plane, separated from $r_i$ by the same angle, so we have:

$R^2 w_{i,1} = R \sqrt{2} r_i = w_{i,2},\; \; i=1,\dots,8$

The entire lattices generated by these vectors are related in the same fashion. We will write this as:

$T_1 L_0 = L_1, T_2 L_0 = L_2$ where we have now defined: $T_1 = \sqrt{2} R^{-1}, T_2 = \sqrt{2} R$

We also have the nice relation, which follows from $w_{i,1}+w_{i,2}=r_i$, that:

$T_1 + T_2 = I$

We’ve now reached the point where we can make use of Wilson’s construction. Wilson showed (using octonionic methods, and his own particular choice of the lattices I’m calling $L_0$, $L_1$ and $L_2$) that:

Theorem: The lattice $L_L$ consisting of all triples $(a,b,c)$ such that:

$a, b, c \in L_0$ $a+b, a+c, b+c \in L_1$ $a+b+c \in L_2$

is isomorphic to a version of the Leech lattice which is rescaled by $\sqrt{2}$ times the scaling of $L_0$ relative to $E_8$.

For our concrete example where the roots of $L_0$ have squared norm 2, we will get a lattice $L_L$ from the construction that is $\sqrt{2}$ times larger than the Leech lattice as it is normally defined. In the following, I will only talk about the particular scaling where $L_0$ is isomorphic to $E_8$, but it’s trivial to rescale everything in sight to suit any particular application where we want $L_0$ to be larger or smaller.

To prove the claimed result, it suffices to show that $L_L / \sqrt{2}$ is an even, unimodular lattice with no vectors of squared norm 2, since the Leech lattice is the unique 24-dimensional lattice with these properties. An even lattice is one where every vector’s squared norm is an even integer, and a unimodular lattice is one where the determinant of any basis is $\pm 1$. But rather than requiring a basis, we will use an equivalent property: every unimodular lattice is self-dual. The dual of a lattice $L$ is the set of vectors $d$ such that for every $v\in L$, the dot product $d\cdot v$ is an integer. That set is precisely the original lattice if and only if the lattice is unimodular.

To proceed, we start from the well-known fact that $E_8$ itself is unimodular, and so $L_0$ will be self-dual. The dual of $L_1$, because of its different scaling, will be $\frac{1}{2} L_1$, and the dual of $L_2$ will be $\frac{1}{2} L_2$.

If you ponder the definition of a dual lattice, you’ll see that the dual of the intersection of two lattices is the lattice spanned by all the vectors in the individual duals. So, the dual of $L_1 \cap L_2$ is the lattice spanned by $\frac{1}{2} L_1$ and $\frac{1}{2} L_2$, which certainly includes the entire lattice $\frac{1}{2} L_0$, since every one of the $r_i$ that generates $L_0$ can be written as the sum of $w_{i,1}$ in $L_1$ and $w_{i,2}$ in $L_2$. But since $L_1, L_2 \subset L_0$, the span of $\frac{1}{2} L_1$ and $\frac{1}{2} L_2$ can be no larger than $\frac{1}{2} L_0$ and must be precisely equal to it. Taking duals, we then have:

$2 L_0 = L_1 \cap \L_2$

This result lets us see that our criteria for membership of $L_L$ are still satisfied if we change the sign of any element in $(a,b,c)$; say we change $c$ to $-c$. Clearly the first condition, $a, b, -c \in L_0$, is still satisfied. For the second, we now have $a-c = (a+c) - 2c$ that needs to be in $L_1$, but since $c \in L_0$ we have $2c \in L_1$, and hence $a-c \in L_1$. Similarly, $a+b-c = (a+b+c)-2c$ needs to be in $L_2$, and because $2c \in L_2$, it is.

The same result tells us that all vectors of the form:

$(2\lambda, 0, 0), \lambda \in L_0$

and permutations are elements of $L_L$.

Suppose we have some $(x,y,z) \in \mathrm{dual}(L_L)$. Then $(x,y,z)\cdot (2\lambda, 0, 0) = 2\lambda \cdot x$ must be an integer for all $\lambda \in L_0$, so $x \in \mathrm{dual}(2 L_0) = \frac{1}{2} L_0$, and similarly for $y$ and $z$.

Vectors of the form:

$(T_2 \lambda, -\lambda, \lambda), \lambda \in L_0$

and permutations are also elements of $L_L$, since $T_2 \lambda \in L_2$, and $T_2 \lambda - \lambda = -T_1 \lambda \in L_1$. Changing signs and permuting coordinates, we have $(-T_2 \lambda, \lambda, \lambda) + (\lambda, -T_2 \lambda, -\lambda) = (T_1 \lambda, T_1 \lambda, 0) \in L_L$ for all $\lambda \in L_0$, or simply $(\mu, \mu, 0)\in L_L$ for all $\mu \in L_1$.

This means the dot product $(x,y,z)\cdot (\mu,\mu,0)=(x+y)\cdot \mu$ must be an integer, for all $\mu \in L_1$. The dual of $L_1$ is $\frac{1}{2} L_1$, so $x+y \in \frac{1}{2} L_1$, and similarly for $x+z$ and $y+z$.

And we also have $(T_2 \lambda, \lambda, \lambda) + (0, -T_1 \lambda, -T_1 \lambda) = (T_2 \lambda, T_2 \lambda, T_2 \lambda) \in L_L$, or simply $(\nu, \nu, \nu) \in L_L$ for all $\nu \in L_2$. The dot product $(x,y,z)\cdot (\nu,\nu,\nu)=(x+y+z)\cdot \nu$ must be an integer, for all $\nu \in L_2$, and the dual of $L_2$ is $\frac{1}{2} L_2$, so $x+y+z \in \frac{1}{2} L_2$.

What we have shown so far is that:

$\mathrm{dual}(L_L) \subseteq \frac{1}{2} L_L$

or equivalently:

$\mathrm{dual}(L_L/\sqrt{2}) \subseteq L_L/\sqrt{2}$

We want to make this an equality. Suppose $(x,y,z) \in \frac{1}{2} L_L$, so that $x+y, x+z, y+z \in \frac{1}{2} L_1$ and $x+y+z \in \frac{1}{2} L_2$. Then:

$(x,y,z)\cdot (x,y,z) = (x+y)\cdot (x+y) + (x+z)\cdot (x+z) + (y+z)\cdot (y+z) - (x+y+z)\cdot (x+y+z)$

The squared norms of vectors in $L_1$ and $L_2$ are multiples of four, so the squared norms of vectors in $\frac{1}{2} L_1$ and $\frac{1}{2} L_2$ are integers. So we have shown that any $(x,y,z) \in \frac{1}{2} L_L$ has a squared norm that is an integer.

It follows that $L_L/\sqrt{2}$ is an even lattice, with all squared norms even integers. What’s more, any dot product of two elements $p, q \in L_L/\sqrt{2}$ can be written as:

$p \cdot q = \frac{1}{2}\left((p+q)\cdot (p+q) - p \cdot p - q \cdot q\right)$

Since all the squared norms here are even, the dot product must be an integer. Therefore every element of $L_L/\sqrt{2}$ also lies in $\mathrm{dual}(L_L/\sqrt{2})$, and we’ve proved that $L_L/\sqrt{2}$ is self-dual and hence unimodular.

Suppose $L_L/\sqrt{2}$ has a vector of squared norm 2, or equivalently, $L_L$ has a vector of squared norm 4. Recall that for any $(a,b,c) \in L_L$ we have $a, b, c \in L_0$, and the squared norms of non-zero vectors in $L_0$ are all at least 2. If all three vectors were non-zero that would mean a squared norm of at least 6, so at least one vector, say $c$, is zero. Then $a+c=a$ and $b+c=b$ must lie in $L_1$, where the non-zero vectors have squared norms of at least 4, and if two vectors were non-zero that would mean a squared norm of at least 8. Our last chance, then, is a single non-zero vector, which must belong to $L_1 \cap L_2 = 2 L_0$, giving it a squared norm of at least 8. So we’ve shown that $L_L/\sqrt{2}$ has no vector with a squared norm of 2.

Putting together everything we now know about $L_L/\sqrt{2}$, it must be the Leech lattice.$\blacksquare$

Posted by: Greg Egan on November 23, 2014 4:02 PM | Permalink | Reply to this

### Re: Integral Octonions (Part 8)

As a postscript to the construction I just described of the Leech lattice as a sublattice of $L_0^3$ where $L_0$ is isomorphic to some possibly rescaled version of $E_8$ … if we want to set $L_0$ to any of the seven lattices of Cayley integral octonions, which have roots of length 1, and hence are re-scaled from the usual $E_8$ by a factor of $\frac{1}{\sqrt{2}}$, it turns out that we can always choose one of the larger sublattices of $L_0$, say $L_1$, to be equal to a certain well-known version of $E_8$: one where every vector consists solely of either integer coordinates or integers plus $\frac{1}{2}$, and where the sum of the coordinates is an even integer. This is true regardless of which of the seven sets of Cayley integral octonions we choose for $L_0$, because this “well-known version” of $E_8$ is a sublattice of the Kirmse integers, and it is invariant under any permutation of its coordinates, including the coordinate-swaps that yield the various Cayley integers.

The second sublattice, $L_2$, is then found by identifying the rotation $R$ that maps $L_1$ into $\sqrt{2} L_0$ and applying it a second time, to give $L_2 = R^2 L_1$.

Posted by: Greg Egan on November 23, 2014 4:40 PM | Permalink | Reply to this

### Re: Integral Octonions (Part 8)

Wow, this is great! I’m going to promote it to a post of its own, since not everyone may see it down here. But first maybe I’ll post a quick summary of Wilson’s paper.

So here’s one curious and nice thing: if we take $L_0$ in your construction to be the lattice of integral octonions, we get a way to build the Leech lattice with its standard scaling as triples of integral octonions obeying some conditions.

(In this situation the shortest vectors in $L_0$ have length 1, so they’re $1/\sqrt{2}$ as long as the shortest vectors in the standard $E_8$ lattice. But your construction takes the standard $E_8$ lattice and produces a lattice $L_L$ that’s $\sqrt{2}$ times as big as the standard Leech lattice. These effects cancel out.)

So, in particular, we get a way, or actually a bunch of ways, to embed the Leech lattice with its standard scaling inside the off-diagonal part of $\mathfrak{h}_3(\mathbf{O})$.

But we’ve also seen that the Leech lattice rescaled by $1/\sqrt{2}$ embeds inside $\mathfrak{h}_3(\mathbf{O})$, in a way that seems much less pretty (so far).

I’m saying these things mainly to make sure I’m not confused.

Posted by: John Baez on November 23, 2014 9:41 PM | Permalink | Reply to this

### Re: Integral Octonions (Part 8)

John wrote:

But we’ve also seen that the Leech lattice rescaled by $1/\sqrt{2}$ embeds inside $\mathfrak{h}_3(\mathbf{O})$, in a way that seems much less pretty (so far).

Right. (John and I discussed this by email earlier.) That’s where we take the isomorphism between a 26-dimensional sublattice of $\mathfrak{h}_3(\mathbf{O})$ that you construct in the blog post above and a version of $II_{25,1}$ rescaled by $1/\sqrt{2}$ (where by $II_{25,1}$ I mean the unique even unimodular lattice in $\mathbb{R}^{25,1}$ with the usual Minkowksi metric), and find the 24-dimensional sublattice of that defined as the orthogonal complement of $w'$, modulo $w'$, where $w'$ is the element of $\mathfrak{h}_3(\mathbf{O})$ that is the image under the isomorphism and change of scale of the lightlike vector in $II_{25,1}$:

$w=(70,0,1,2,3,4,5,...,23,24)$

That $w$ is famous for yielding the Leech lattice with its usual scaling as a sublattice of $II_{25,1}$ by the same process in the original space. But when we repeat all the steps in $\mathfrak{h}_3(\mathbf{O})$, everything is rescaled by $1/\sqrt{2}$.

Posted by: Greg Egan on November 23, 2014 10:44 PM | Permalink | Reply to this

### Re: Integral Octonions (Part 8)

Ultimately, it would be nice to relate these octonionic lattice constructions to the heterotic string. I found a nice 1986 paper, giving hints along this path:

Ten dimensional Heterotic Strings from Niemeier Lattices

W. Lerche, D. Luest

Essentially, 24-dimensional Euclidean even self-dual lattices that can be decomposed as $\Gamma_{24}=\Gamma_{16}\otimes D_8$ lead to heterotic string theory in D=10. The $E_8\times E_8$ heterotic string comes from the lattice $E_8^3$. The only other supersymmetric heterotic theory lives on $E_8\times D_16$.

The root vectors of $\Gamma_{16}$ give rise to massless gauge bosons of the heterotic string theory and the corresponding gauge group. All such string theories, relevant for four-dimensional physics are based on two conformally invariant two-dimensional field theories: one consists of 26 bosons whose conformal anomaly cancels that of the reparametrization ghost, while the other is a supersymmetric model with 10 bosons and 10 fermions, cancelling the reparametrization and superconformal ghost. Either one can be used as the left-moving or right-moving sector of a closed string theory. The spacetime dimension is determined by how many of the bosons play the role of spacetime coordinates.

A gauge theory based on $\mathfrak{h}_3(\mathbb{O})$, by virtue of its traceless structure, automatically gives 26 bosons. Going to larger matrices, over $\mathbb{O}$, is hindered by the loss of the Jordan algebraic structure, as well as the topological reasons for the non-existence of $\mathbb{OP}^3$. At the end of the day, octonion quantum mechanics is quite restrictive, so it would be elegant to see the bosonic and heterotic string through an octonionic lens.

Posted by: Metatron on November 26, 2014 6:16 PM | Permalink | Reply to this

### Re: Integral Octonions (Part 8)

Thanks! I very much want to formulate (and study) the heterotic string using octonions, because there are so many signs that this is a good idea.

The $\mathrm{E}_8\times \mathrm{E}_8$ heterotic string comes from the lattice $\mathrm{E}_8^3$. The only other supersymmetric heterotic theory lives on $\mathrm{E}_8\times \mathrm{D}_{16}$.

To be quite nitipicky, the other comes from the lattice $\mathrm{E}_8 \times \mathrm{DD}_{16}$, where $\mathrm{DD}_{16}$ is the ‘doubly dense’ version of $\mathrm{D}_{16}$ described starting here. $\mathrm{D}_{16}$ is an even lattice, but not unimodular. $\mathrm{DD}_{16}$ repairs this defect.

I’m being so nitpicky because I made a similar mistake myself.

Posted by: John Baez on November 28, 2014 10:36 PM | Permalink | Reply to this

### Re: Integral Octonions (Part 8)

I need to fix a mistake in this post. I wrote:

There are 24 even unimodular lattices in 24-dimensional Euclidean space. One of them is

$\mathrm{E}_8 \oplus \mathrm{E}_8 \oplus \mathrm{E}_8$

Another is $\mathrm{D}_24$. This is the lattice of vectors in $\mathbb{R}^{24}$ where the components are integers and their sum is even. It’s also the root lattice of the Lie group $\mathrm{Spin}(48)$.

The mistake here is that the lattice $\mathrm{D}_{24}$ is not unimodular: the volume of its unit cell is 2, not 1. To get an even unimodular lattice we need to double the density of $\mathrm{D}_{24}$ and form a lattice I’ll call $\mathrm{DD}_{24}$ — for two reasons: it’s doubly dense, and it’s the union of $\mathrm{D}_{24}$ and a translated copy of $\mathrm{D}_{24}$. People often call this doubled lattice $\mathrm{D}^+_{24}$, but I’m using superscript pluses for too many other things in this post!

This requires changing Corollary 2 to the following:

Corollary 1. With the aforementioned Lorentzian inner product $g$ on the exceptional Jordan algebra, the lattice $\mathrm{DD}_{24}^{++}$ is isometric to the sublattice of $\mathfrak{h}_3(\mathbf{O})$ where a fixed diagonal entry is set equal to zero, e.g.:

$\left\{ \; \left( \begin{array}{ccc} a & X & Y \\ X^* & b & Z \\ Y^* & Z^* & 0 \end{array} \right) : \; a,b \in \mathbb{Z}, \; X,Y,Z \in \mathbf{O} \; \right\}$

I’ll explain how the $\mathrm{DD}_n$ lattice works in a separate comment, and also rewrite the post to include all this stuff.

Posted by: John Baez on November 28, 2014 10:29 PM | Permalink | Reply to this

### Re: Integral Octonions (Part 8)

To get the $\mathrm{D}_n$ lattice, take an $n$-dimensional checkerboard with alternating red and black hypercubes. Put a dot in the middle of each black hypercube. That’s the $\mathrm{D}_n$ lattice!

More precisely the $\mathrm{D}_n$ lattice consists of all $n$-tuples of integers that sum to an even integer: requiring that they sum to an even integer picks out the center of every other hypercube in our checkerboard.

Here’s a basis of vectors for the $\mathrm{D}_4$ lattice:

$\begin{array}{cccc} (-1, & -1, & 0, & 0) \\ (1, & -1, & 0, & 0 ) \\ (0, & 1, & -1 & 0 ) \\ (0, & 0, & 1, & -1) \end{array}$

The same pattern works in any dimension: the first vector has two $-1$s followed by a bunch of zeros, but the rest of the vectors are all just shifted versions of the second. I’ve chosen them to be simple roots for $\mathrm{D}_n$, but that’s not so important now.

What really matters is this. The dot product of any basis vector with itself is even, and the dot product of any two different ones is an integer. Thus, the lattice they generate, the $\mathrm{D}_n$ lattice, is even: the dot product of any vector with itself is even.

Also, the determinant of the matrix formed by our basis vectors:

$\left( \begin{array}{cccc} -1 & -1 & 0 & 0 \\ 1 & -1 & 0 & 0 \\ 0 & 1 & -1 & 0 \\ 0 & 0 & 1 & -1 \end{array} \right)$

is 2. In any dimension, this sort of determinant gives $\pm 2$. So, the volume of the unit cell in $\mathrm{D}_n$ lattice is always 2. So, it’s not unimodular.

To get something unimoduar, we can double the density by taking the union of $\mathrm{D}_n$ and a copy of $\mathrm{D}_n$ translated by the vector

$\left(\frac{1}{2}, \dots, \frac{1}{2} \right)$

Let’s call this union $\mathrm{DD}_n$.

$\mathrm{DD}_3$ is the way carbon atoms are arranged in a diamond!

This pattern is called the diamond cubic. It’s beautiful, but it’s not a lattice in the mathematical sense. Here’s the story:

• In any dimension, the volume of the Voronoi cells of $\mathrm{DD}_n$ is 1, so we can say it’s unimodular.
• In even dimensions, and only those, $\mathrm{DD}_n$ is lattice. After all, only then does the sum of $(\frac{1}{2}, \dots, \frac{1}{2} )$ with itself again lie in $\mathrm{DD}_n$.
• In dimensions that are multiples of 4, and only those, $\mathrm{DD}_n$ is an integral lattice, meaning that the dot product of any two vectors in the lattice is an integer. After all, only then is the inner product of $(\frac{1}{2}, \dots, \frac{1}{2} )$ with itself an integer.
• In dimensions that are multiples of 8, and only those, $\mathrm{DD}_n$ is an even lattice, meaning that the dot product of any vector with itself is even. After all, only then is the inner product of $(\frac{1}{2}, \dots, \frac{1}{2} )$ even.

As I mentioned before, even unimodular lattices are possible in Euclidean space only when the dimension is a multiple of 8. $\mathrm{DD}_8$ is none other than our friend $\mathrm{E}_8$!

Posted by: John Baez on November 28, 2014 11:07 PM | Permalink | Reply to this

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