## November 23, 2014

### Integral Octonions (Part 9)

#### Posted by John Baez

Now let’s talk about building the Leech lattice from integral octonions!

The Leech lattice is a marvelous thing. Here are three ways to define it:

• Take a bunch of balls of radius one in 24 dimensions. Get as many balls to touch a single ball as you possibly can. Then, get as many balls to touch these as you possibly can. Go on like this forever. Unlike in 3 dimensions, you have no real choice about how to do this: the balls will inevitably be centered at points of the Leech lattice.

• Take disks of radius one in the plane and pack them in the densest possible lattice arrangement. Use this pattern to form a layer of balls in 3 dimensions. Stack layers like this on top of each other, pushing them down to get the densest possible lattice packing of 3d balls. Use this pattern to form a layer of balls in 4 dimensions! Stack layers like this on top of each other, pushing them down to get the densest possible lattice packing of 4d balls. Repeat this until you get to 24 dimensions. There will be some choices — but no matter how you make these choices, you’ll inevitably be led to the Leech lattice.

• Up to isometry, the Leech lattice is the unique even unimodular lattice in 24 dimensions whose shortest nonzero vectors have length 2.

Since the Leech lattice lives in 24 dimensions and octonions live in 8, we should try to describe points in the Leech lattice using triples of octonions.

That’s what Geoffrey Dixon and Robert Wilson did:

As a spinoff, this explains why there are

$3 \cdot 240 \cdot (1 + 16 + 16^2) = 196,560$

shortest nonzero vectors in the Leech lattice. 240 is the number of shortest nonzero vectors in the $\mathrm{E}_8$ lattice — or in other words, the number of integral octonions of norm 1.

In a further paper, Wilson used his idea to give an octonionic description of the symmetries of the Leech lattice:

If you take all the rotations and reflections that are symmetries of the Leech lattice, they generate a group called $\mathrm{Co}_0$ with

$8,315,553,613,086,720,000$

elements. This is not a simple group, because it contains the transformation $x \mapsto -x$, which commutes with everything else. But when you mod out by that, you get a simple group called $\mathrm{Co}_1$. This is half as big, with only

$4,157,776,806,543,360,000$

elements. It’s one of the most important sporadic finite simple groups!

All of this sounds quite fun, though also quite terrifying.

Luckily, Greg Egan found a way to simplify Wilson’s ideas in a comment to Part 8. So, let me describe Egan’s construction, and then reprint his proof that it works.

### Summary of Egan’s construction

Egan took Wilson’s construction and showed how to eliminate all mention of the octonions, focusing simply on the $\mathrm{E}_8$ lattice. Here’s one way to state Egan’s result:

Theorem. Let $L_0 \subseteq \mathbb{R}^8$ be the E8 lattice, and let $r_1, \dots, r_8 \in L_0$ be a basis of simple roots. There exists a rotation

$R : \mathbb{R}^8 \to \mathbb{R}^8$

such that:

• The transformation $T_1 = \sqrt{2} R^{-1}$ maps each $r_i$ to a vector $w_{i,1} \in L_0$ whose dot product with $r_i$ is 1.

• The transformation $T_2 = \sqrt{2} R$ maps each $r_i$ to a vector $w_{i,2} \in L_0$ whose dot product with $r_i$ is 1.

For any rotation $R$ with these properties, define lattices

$L_1 = T_1 L_0, \qquad L_2 = T_2 L_0$

Then the lattice consisting of all triples $(a,b,c) \in \mathbb{R}^{24}$ such that:

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

is isometric to a copy of the Leech lattice that has been rescaled by a factor of $\sqrt{2}$.

Proof. See below.     $\blacksquare$

Here’s the picture in my head:

The vertical vector is the root $r_i$. It has length $\sqrt{2}$, as roots of $\mathrm{E}_8$ do. The other two vectors are $w_{i,1}$ and $w_{i,2}$. They have length 2 and their dot product with $r_i$ is 1. And in fact, their sum is $r_i$.

Why? Since they’ve been formed by rescaling $r_i$ and then rotating the result in opposite directions, their sum must be some number times $r_i$. But since this sum has dot product 2 with $r_i$, and so does $r_i$ itself, that number must be 1.

Of course, the neat part is that all 24 vectors — the $r_i$, the $w_{1,i}$ and the $w_{2,i}$ — all live in the same $\mathrm{E}_8$ lattice.

So, if we want, we can take them to be integral octonions! We have to be careful: the integral octonions $\mathbf{O}$ form a lattice that’s a copy of the $\mathrm{E}_8$ lattice rescaled by $1/\sqrt{2}$. But luckily, this cancels the factor of $\sqrt{2}$ in the theorem above. So, Egan’s construction gives a copy of the Leech lattice at its correct scale as a sublattice of $\mathbf{O}^3$!

To understand why his construction works, we can’t do better than read what he wrote. So, in everything that follows — even in the ‘postscript’ at the end – I’ll just be quoting him.

### Constructing the Leech lattice from E8

by Greg Egan

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 $\mathrm{E}_8$.

So, suppose $L_0$ is either isomorphic to $\mathrm{E}_8$, or to some multiple of $\mathrm{E}_8$. 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 $\mathrm{E}_8$, 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 $\mathrm{E}_8$, 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 $\mathrm{E}_8$ 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 $\mathrm{E}_8$ 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.)

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

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

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 $\mathrm{E}_8$ 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:

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

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:

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

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 $\mathrm{E}_8$.

Proof. 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 $\mathrm{E}_8$, 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 $\mathrm{E}_8$ 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:

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

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:

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

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.     $\blacksquare$

### Postscript

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 $\mathrm{E}_8$ … 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 $\mathrm{E}_8$ 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 $\mathrm{E}_8$: 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 $\mathrm{E}_8$ 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 at November 23, 2014 11:03 PM UTC

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### Re: Integral Octonions (Part 9)

Is there a map available showing all the embeddings (and other relationships) between the various lattices that have come up in this series? I’m dizzy!

Posted by: Tim Campion on November 25, 2014 3:29 PM | Permalink | Reply to this

### Re: Integral Octonions (Part 9)

I don’t know of such a map. What kind would you like?

I agree that this subject gets rather intricate and deserves some sort of visual overview.

We get one lattice for each Dynkin diagram, and one inclusion of lattices for each inclusion of Dynkin diagrams, along with other inclusions which are subtler, which we mostly haven’t needed yet.

So, you want to know your ‘finite Dynkin diagrams’ down pat:

These give lattices in Euclidean space. We’ve also been playing around with a couple ‘Lorentzian Dynkin diagrams’, which give lattices in Minkowski space: $\mathrm{E}_{10}$, $\mathrm{E}_{11}$ and $\mathrm{D}_{27}$. The most important of these is $\mathrm{E}_{10}$:

We’ve also looked at a couple of functors, $++$ and $+++$, which send lattices in Euclidean space to lattices in Minkowski space. For example, $\mathrm{E}_8^{++}$ is $\mathrm{E}_{10}$, and $\mathrm{E}_8^{+++}$ is $\mathrm{E}_{11}$.

The really tricky beautiful part I’m trying to understand now (with massive help from Greg) are the relations between various lattices in the octonions. There’s the Kirmse integers, which are not closed under multiplication, and 7 possible ways of reflecting this lattice to get Cayley integers that are closed under multiplication. These 7 are the densest lattices in the octonions that are closed under multiplication.

All 8 lattices mentioned so far are isometric the usual $\mathrm{E}_8$ lattice rescaled by a factor of $1/\sqrt{2}$. Greg’s construction of the Leech lattice involves nothing about octonion multiplication. Thus, we can use any of these 8 lattices as $L_0$ in Greg’s proof. But the lattices he calls $L_1$ and $L_2$ are also interesting.

Things will get tricky and interesting when we bring octonion multiplication into the game. This is what makes the $3 \times 3$ self-adjoint octonion matrices into a Jordan algebra: the exceptional Jordan algebra. We can use Greg’s construction to fit the Leech lattice into the off-diagonal part of the exceptional Jordan algebra… in various ways, depending on the choices we made for $L_0, L_1, L_2$.

Throwing in integer diagonal entries, we get various 27-dimensional lattices in the exceptional Jordan algebra. These lattices are all isometric to the Leech lattice plus $\mathbb{Z}^3$. A nice question is whether we can find a lattice like this that is closed under the Jordan product

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

This would be a kind of “exceptional Leech–Jordan ring”.

But anyway, I agree that it would be good, while scaling these heights, to provide a detailed map for readers stuck on various cliff faces.

Posted by: John Baez on November 25, 2014 7:41 PM | Permalink | Reply to this

### Re: Integral Octonions (Part 9)

John wrote:

A nice question is whether we can find a lattice like this that is closed under the Jordan product

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

This would be a kind of “exceptional Leech–Jordan ring”.

I can make this work if I’m allowed to either rescale the lattice or double the right-hand side of the Jordan product.

If you perform the construction described in the post, with $L_0$ a set of Cayley integral octonions, and $L_1$ the version of $E_8$ given by:

$L_1 = \{v \in \mathbb{Z}^8 \cup (\mathbb{Z}+\frac{1}{2})^8 : \sum_i v_i \in 2 \mathbb{Z}\}$

and then you embed the resulting version of the Leech lattice $L_L$ in $\mathfrak{h}_3(\mathbf{O})$ as the off-diagonal elements, along with a copy of $\mathbb{Z}^3$ for the diagonal elements … this sublattice $\mathbb{Z}^3\oplus L_L \subset \mathfrak{h}_3(\mathbf{O})$ is not closed under the Jordan product:

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

but it is closed under:

$x \circ y = 2(x y + y x)$

Alternatively, you get a lattice closed under the first Jordan product if you simply double the original lattice, i.e. if you use $2(\mathbb{Z}^3\oplus L_L)$.

Posted by: Greg Egan on November 26, 2014 3:41 AM | Permalink | Reply to this

### Re: Integral Octonions (Part 9)

Thanks, Greg! It’s very nice that $\mathbb{Z}^3 \oplus L_L \subset \mathfrak{h}_3(\mathbf{O})$ is closed under the ‘doubled’ Jordan product

$2 (x y + y x )$

I don’t know how concerned to be about that factor of 2.

You’re right of course that the usual identity matrix doesn’t act as a multiplicative identity for the Jordan product

$x y + y x$

Working with real Jordan algebras, people are somewhat flexible about whether they use this product or the ‘halved’ version

$\frac{1}{2} (x y + y x)$

The former is nice because it lets you avoid lots of $1/2$’s in your formulas and it resembles the Lie bracket $x y - y x$. The latter is nice because if you start with self-adjoint matrices with entries in some $\ast$-algebra, the usual identity matrix will act as the identity. Over the real numbers, or any field of characteristic $\ne 2$, you can easily switch between these two products. So, it’s no big deal.

But you need to really worry about this issue when you’re working over the integers, as we are now, or over a field of characteristic 2.

I’ve never seen anyone use the ‘doubled’ Jordan product!

Posted by: John Baez on November 26, 2014 9:55 PM | Permalink | Reply to this

### Re: Integral Octonions (Part 9)

Why this obsession with these 2s? However these things are motivated, I’ve always ignored any reason for positing that the inner shell should not consist of unit elements. Anyway, my Leech rep was scaled so that inner shell lies in unit 24 sphere, the hope being that if a product existed using O triples it would be easier since O multiplication of subset of unit elements is closed.

Posted by: Geoffrey Dixon on November 27, 2014 7:35 PM | Permalink | Reply to this

### Re: Integral Octonions (Part 9)

The 2’s are there not because of ‘obsession’ but because of mathematical facts. Greg Egan has just explained to us a way to find:

a) a sublattice of the $3 \times 3$ self-adjoint matrices with Cayley integral octonions as entries

such that:

b) this lattice is isometric to the direct sum of the Leech lattice (for the off-diagonal entries of our matrix) and $\mathbb{Z}^3$ (for the on-diagonal entries)

and:

c) this lattice is closed under the doubled Jordan product $2(x y + y x)$.

More recently he’s done a computer search and gotten a lot of evidence that you can’t get rid of that factor of 2 in condition c). We’ll try to present a proof at some point.

Furthermore, the Leech lattice in his construction is the usual one, where the shortest vectors have length 2. I’m pretty sure that 2 is unavoidable as well.

Here we are working with Cayley integral octonions with their usual norm, where the shortest one have length 1. So it’s not as if we’re sticking in 2’s just for the thrill of it!

In fact, the factor of 2 in condition c) really annoys me. But if it’s unavoidable, I’ll need to learn to live with it.

Posted by: John Baez on November 28, 2014 4:21 AM | Permalink | Reply to this

### Re: Integral Octonions (Part 9)

Well, the self-adjoint matrices are the focus so the Jordan product is: $X\circ Y = \frac{1}{2}(XY+YX)$. As for $3\times 3$ generators of $Co_0$, that are also unitary of type $F_4$, we pick up $\frac{1}{\sqrt{2}}$ and $\frac{1}{2}$ factors. However, Wilson gave two octonionic constructions of the Leech, one with the traditional norm and another with the more natural octonion $\mathbb{O}^3$ norm $x\overline{x}+y\overline{y}+z\overline{z}$. One would have to choose a construction and build $F_4$ generators of $Co_0$, accordingly.

Posted by: Metatron on December 2, 2014 6:12 PM | Permalink | Reply to this

### Re: Integral Octonions (Part 9)

Metatron wrote:

Well, the self-adjoint matrices are the focus so the Jordan product is: $X\circ Y = \frac{1}{2}(X Y+Y X)$

Note however that we get an isomorphic Jordan algebra by starting with $\mathfrak{h}_3(\mathbb{O})$ and using the Jordan product $X \circ Y = \alpha (X Y + Y X)$ for any nonzero $\alpha \in \mathbb{R}$.

Jordan algebras are just like Lie algebras in this respect: if we scale the Lie bracket (or Jordan product) by any nonzero real number, we get another isomorphic Lie algebra (or Jordan algebra). So, the factor of $\frac{1}{2}$ above is not fundamental: it’s just conveniently chosen to make the usual identity matrix be the identity for the Jordan product.

However, when we start working over the integers instead of working over the real numbers — working with ‘Jordan rings’ like $\mathfrak{h}_3(\mathbf{O})$ instead of Jordan algebras like $\mathfrak{h}_3(\mathbb{O})$ — this result no longer applies. Rescaling the Jordan product in a Jordan ring can give a nonisomorphic Jordan ring.

Posted by: John Baez on December 4, 2014 3:44 PM | Permalink | Reply to this

### Re: Integral Octonions (Part 9)

Given $(2,2,0),(2(e_1+e_2),0,0)\in\Lambda$, map $(2,2,0)\mapsto X\in\mathfrak{h}_3(\mathbb{O})$ and $(2(e_1+e_2),0,0)\mapsto Y\in\mathfrak{h}_3(\mathbb{O})$, such that

$X=\left(\begin{array}{ccc} 0 & 2 & 2 \\ 2 & 0 & 0 \\ 2 & 0 & 0 \end{array}\right)$ and $Y=\left(\begin{array}{ccc} 0 & 2(e_1+e_2) & 0 \\ 2(e_1+e_2)^{\ast} & 0 & 0 \\ 0 & 0 & 0 \end{array}\right)$.

Then $XY+YX=\left(\begin{array}{ccc} 0 & 0 & 0 \\ 0 & 0 & 4(e_1+e_2)^{\ast} \\ 0 & 4(e_1+e_2) & 0 \end{array}\right)$.

Hence, with the usual Jordan product $X\circ Y=\frac{1}{2}(XY+YX)$ we get the nice result $X\circ Y=\left(\begin{array}{ccc} 0 & 0 & 0 \\ 0 & 0 & 2(e_1+e_2)^{\ast} \\ 0 & 2(e_1+e_2) & 0 \end{array}\right)$, corresponding to a norm 4 vector in the Leech lattice.

Posted by: Metatron on December 3, 2014 1:10 AM | Permalink | Reply to this

### Re: Integral Octonions (Part 9)

I just checked what happens if you use Kirmse integers in place of Cayley integers, and exactly the same results hold: you need to double the Jordan product or double the size of the lattice to make it closed.

Actually, I don’t think we’d get a ring even if we found a lattice that was closed under:

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

because we don’t have a multiplicative identity. For the usual identity matrix to work, I think the Jordan product needs to be:

$x \circ y = \frac{x y + y x}{2}$

Posted by: Greg Egan on November 26, 2014 6:09 AM | Permalink | Reply to this

### Re: Integral Octonions (Part 9)

Hi. I’ve been enjoying reading these posts and watching the development of the extensions as you increase the nodes on your quiver. What I have to say here is only tangentially related, but I can’t resist.

Given any quiver, or even symmetrisable generalised Cartan lattice, one has the usual root lattice and Weyl group as in Lie theory. Denote by $(-,-)$ the symmetric bilinear form on the root lattice.

Fixing a reduced expression for the Coxeter element (or orientation of the quiver) yields a non-degenerate (but not symmetric) bilinear form $\langle -,- \rangle$ such that $\langle x,y \rangle + \langle y,x \rangle = (x,y)$.

In this setting one obtains a classification of (certain nice) embeddings of one lattice into another in terms of the non-crossing partitions in the Weyl group.

References include

• Colin Ingalls and Hugh Thomas, Noncrossing partitions and representations of quivers. Compos. Math. 145 (2009), no. 6, 1533–1562.
• Kiyoshi Igusa and Ralf Schiffler, Exceptional sequences and clusters. J. Algebra 323 (2010), no. 8, 2183–2202.

as well as a recent preprint (updated version coming soon!) by myself and Henning Krause.

Posted by: Andrew Hubery on November 27, 2014 9:33 AM | Permalink | Reply to this

### Re: Integral Octonions (Part 9)

Thanks! As Greg and I gradually gain expertise on the $\mathrm{E}_8$ lattice, Leech lattice and related structures, I’m looking around for fun things to do that could make use of this expertise. So, I’ll take a look at those papers sometime.

Posted by: John Baez on November 28, 2014 10:12 PM | Permalink | Reply to this

### Re: Integral Octonions (Part 9)

It changes nothing, but the rep of L24 over O^3 was posted by me in 1995 on hep-th. Wilson’s version is identical, but his work is deeper in that it covers automorphisms. Basically because of this one can continue to safely ignore my earlier version. However, I would like to quibble with the 7 versions of integral octonions idea. There aren’t 7: there are many, all of which can be derived using the octonion XY-product. Those 7 are just examples of this aligned with the 7 imaginary basis units. Anyhum, this stuff is fascinating. I’d been wondering if there is something in laminated lattice theory analogous to parrellizability in the spheres S1, S3, S7 - maybe something to do with fixed points in the automorphism groups that is distinct for L2, L4, L8 (E8). And then if such an analogous property were found, how would it relate to L24? However, I know very little about those automorphism groups, so I will carry on wondering.

As I am an incorrigible maverick and quite possibly a crank, you’ll need to add a grain of salt to anything I post.

Posted by: Geoffrey Dixon on November 26, 2014 6:27 PM | Permalink | Reply to this

### Re: Integral Octonions (Part 9)

Hi, Geoffrey! You wrote:

It changes nothing, but the rep of $\Lambda_{24}$ over $\mathbb{O}^3$ was posted by me in 1995 on hep-th. Wilson’s version is identical, but his work is deeper in that it covers automorphisms.

You’re talking, of course, about these papers:

I don’t know if you’ve looked at the 20 March 2009 revision of

Near the end he writes:

• [Note: After this paper was submitted, I became aware of some work in progress by Geoffrey Dixon [12], which comes much closer to the description of the minimal vectors given above, at the end of Section 3. However, as far as I can see his conjectured formula for the last set of vectors needs modification: in effect, his formula is equivalent to $(\lambda s, \pm \lambda k, \pm(\lambda j)k)$, that is, omitting the (necessary) multiplication by $j$ in the first coordinate.]

Here [12] is this reference:

Unfortunately the link now takes us to a 21 November 2010 revision of your paper which refers to Wilson’s 2009 paper! So, we can’t really tell what you did in the 2005 version of this paper.

I guess when he says that 2005 paper came ‘much closer’ he’s comparing it to your other two papers, which he’d referred to earlier, as follows:

The above construction is deceptively simple. In fact, however, finding the correct definition (beginning of Section 3) was not at all easy. Over the years, many people have noticed the suggestive fact that $196560 = 3 \times 240 \times (1 + 16 + 16 \times 16)$ and tried to build the Leech lattice from triples of integral octonions (see for example [10, 11, 13]), but until now no-one has provided a convincing explanation for this numerology.

Here [10, 11] are your other two papers, while [13] is

• N. Elkies and B. Gross, The exceptional cone and the Leech lattice, Internat. Mat. Res. Notices (1996), 665–698.

Anyway, I haven’t investigated the differences, if any, between your construction and Wilson’s! I apologize for focusing on Wilson’s work; that’s just the paper which I happened to be looking at… and then Greg Egan generalized Wilson’s construction into something I find much easier to understand and remember. If I ever write a review paper about this stuff, I’ll be better about giving credit where credit is due.

I’ll talk about actual math issues in some other comment, after I recover from all this scholarship.

Posted by: John Baez on November 26, 2014 7:30 PM | Permalink | Reply to this

### Re: Integral Octonions (Part 9)

No worries. In my 2nd book I proved my 1995 Leech rep and Wilson’s are the same. In that book I also show that those 7 octonion integral domains that are frequently referenced - and often weirdly treated as though there are no others - are special cases of how one turns E8 into an integral domain using the XY-product (my extension of Cederwall’s X-product). I also expend several pages musing about developing a product on L24, thinking that it might be ternary. Jens Köplinger ran some computer programs, but results were pessimistic. I have insufficient optimism that if a product does exist under which L24 closes that I’d be able to find it. That fruit is too high off the ground. Basing the quest on the Jordan algebra is certainly worth the effort, but over 3 decades I’ve observed lots of enthusiastic effort poured into the exceptional Jordan algebra, and its promise has IMO never been fulfilled. But what you are doing is promising.

Anyhum, I’m mostly an observer now, and at this very moment we are in a cottage overlooking the Bay of Fundy. I should get back to that. During a fossil hunt yesterday we found an ammonite fossil the size of a car tire.

Posted by: Geoffrey Dixon on November 27, 2014 12:23 PM | Permalink | Reply to this

### Re: Integral Octonions (Part 9)

Posted by: Jens Koeplinger on November 30, 2014 2:41 AM | Permalink | Reply to this

### Re: Integral Octonions (Part 9)

By the way, in 1995 when I worked on that stuff I had a couple face-to-face meetings and emails with Gross about the work he’d done with Elkies. We both needed to introduce that invariant element (1+e1+…e7)/2 to get our respective Leech ideas to work.

Posted by: Geoffrey Dixon on November 27, 2014 12:31 PM | Permalink | Reply to this

### Re: Integral Octonions (Part 9)

Having a basis for a lattice can be incredibly useful when you actually want to calculate things. So, here’s a basis for $L_L$, the version of the Leech lattice that I constructed in the blog post:

$\left( \begin{array}{ccc} 2 r_i & 0 & 0 \\ w_{i,1} & w_{i,1} & 0 \\ w_{i,2} & r_i & r_i \end{array} \right)$

Each subscripted vector belongs to $\mathbb{R}^8$, with 8 different vectors for $i=1,...,8$, and the $0$ entries here are to be taken as $8\times 8$ blocks of zeroes. So this is shorthand for a $24\times 24$ matrix whose rows comprise the basis for the lattice $L_L$.

The vectors $r_i$ are roots of the “primary” copy of $E_8$ that is called $L_0$ in the post, and the vectors $w_{i,1}$ and $w_{i,2}$ are roots of two different copies of $E_8$ that are sublattices of $L_0$ and larger by a factor of $\sqrt{2}$ than $L_0$. Those two lattices are called $L_1$ and $L_2$. For each $i$, we have:

$r_i = w_{i,1} + w_{i,2}$

Given this equation, along with the crucial fact, proved in the post, that:

$2 L_0 = L_1 \cap L_2$

it’s pretty easy to check that each of the 24 vectors is an element of $L_L$, defined as the triples $(a,b,c)\in L_0^3$ such that:

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

For the first eight vectors this is obvious, since all the sums are either 0 or $2r_i \in L_1 \cap L_2$.

For the second group of eight vectors, the pairwise sums are either $w_{i,1}$ or $2 w_{i,1}$, which are in $L_1$ by definition. The triple sum, $2 w_{i,1}$, can be rewritten as:

$2 w_{i,1} = 2 r_i - 2 w_{i,2} \in L_2$

For the third group of eight vectors, the pairwise sums are either $2 r_i \in L_1$, or $w_{i,2} + r_i$, which can be rewritten as:

$w_{i,2} + r_i = 2 r_i - w_{i,1} \in L_1$

The triple sum, $2 r_i + w_{i,2}$ is clearly in $L_2$ since $2 r_i \in L_2$.

To show that this is actually a basis, we exploit the fact that we’ve already proven that a suitably scaled version of this lattice is unimodular. The determinant of the basis we’ve given is simply:

$D = (2^8 d) \times ({\sqrt{2}}^8 d) \times d = 2^{12} d^3$

where $d$ is the determinant of any basis for $L_0$. If $L_0$ is $E_8$ at the usual scale, it is a unimodular lattice, with $d=1$. Then the determinant $D$ of this basis for $L_L$ is $2^{12}$, and the lattice that we showed is the Leech lattice with the normal scaling, $L_L/\sqrt{2}$, has a basis with determinant $D/{\sqrt{2}}^{24}=1$.

Posted by: Greg Egan on November 27, 2014 4:47 AM | Permalink | Reply to this

### Re: Integral Octonions (Part 9)

Yesterday I gave a talk on The dodecahedron, the icosahedron and E8 at the Hong Kong Mathematical Society.

Today at breakfast Roger Howe suggested trying to construct the Leech lattice (a 24-dimensional lattice or the Golay code (a linear code consisting of 24-bit strings) starting the tetrahedron, roughly as follows. There are lots of significant problems left to solve, but it’s a cute idea.

The tetrahedron has 12 rotational symmetries, giving

$\mathrm{A}_4 \subseteq \mathrm{SO}(3)$

Using the double cover

$\pi : \mathrm{SU}(2) \to \mathrm{SO}(3)$

we get a 24-element subgroup of $\mathrm{SU}(2).$ This is quite famous:

• Wikipedia, Binary tetrahedral group.

Since $\mathrm{SU}(2)$ is the unit sphere in the quaternions the binary tetrahedral group forms the vertices of a 4d polytope, called the 24-cell:

• Wikipedia, 24-cell.

The 24-cell is a 4-dimensional regular polytope. The vertices of the 24-cell can be broken up into 3 sets of 8, each set being the vertices of another 4-dimensional polytope, which is called the ‘hyperoctahedron’ or ‘cross-polytope’ or ‘4-dimensional orthoplex’ or ‘16-cell’:

• Wikipedia, 16-cell.

If you set up things nicely, these 3 cross-polytopes inside the 24-cell get permuted when we permute the quaternions $i, j,$ and $k$. This is called ‘triality’.

Now the idea is this: since the 3 cross-polytopes inside the 24-cell are a very nice way to think about

$24 = 8 + 8 + 8$

perhaps they can be used to help understand how we build the Leech lattice out of 3 copies of the E8 lattice, or build the Golay code out of 3 copies of the 8-bit Hamming code!

We (in some sense of ‘we’) already know how to build the Leech lattice out of 3 copies of the E8 lattice, or build the Golay code out of 3 copies of the 8-bit Hamming code. It’s called the ‘Turyn construction’. In the case of building the Leech from 3 E8’s, Greg Egan and I described it above.

The question is whether this is at all illuminated by the geometry I just described.

It’s worth noting that Greg and I already did connect triality to the Turyn construction of the Leech lattice. The three E8 lattices used to build the Leech lattice can be thought of as lying in the vector, left-handed spinor, and right-handed spinor representations of $\mathrm{Spin}(8)$. These representations are permuted by the triality automorphisms of $\mathrm{Spin}(8)$. Furthermore, the weight lattice of $\mathrm{Spin}(8)$ is a 4-dimensional lattice generated by the points in the 24-cell, and the weights of the vector, left-handed spinor and right-handed spinor representations lie in the 3 cross-polytopes I mentioned!

So, all this stuff does fit together. I guess the question is whether the Turyn construction makes a bit more sense if we examine it in this larger context.

Posted by: John Baez on May 21, 2017 4:19 AM | Permalink | Reply to this

### Re: Integral Octonions (Part 9)

One potentially relevant thing was pointed out nicely by Tim Silverman in a comment to another post. The weights of the 3 8-dimensional irreps $V, S_+, S_-$ of $\mathfrak{so}(8)$ are the vertices of three cross-polytopes that fit together to form a 24-cell… but the nonzero weights of $\mathfrak{so}(8)$ as a representation of itself, form the vertices another 24-cell.

The second 24-cell is bigger than the first. If we take this bigger 24-cell, and take any one of the cross-polytopes in it, and look at the midpoints of the edges of that cross-polytope, we get the vertices of the smaller 24-cell.

Using this fact, just as we found a copy of the Leech lattice in $V \oplus S_+ \oplus S_-$, we can find a copy of the Leech lattice in $\mathfrak{so}(8)$ itself. Of course $\mathfrak{so}(8)$ is 28-dimensional, but this latter Leech lattice lives in the 24-dimensional subspace spanned by the nonzero weights. (Warning: I’m using $\mathfrak{so}(8) \cong \mathfrak{so}(8)^*$ to think of the weights as vectors in $\mathfrak{so}(8)$.)

I want to know if the action

$\mathfrak{so}(8) \otimes (V \oplus S_+ \oplus S_-) \to V \oplus S_+ \oplus S_-$

restricts to an action of the Leech lattice in $\mathfrak{so}(8)$ on the Leech lattice in $V \oplus S_+ \oplus S_-$.

If so, everything about the 8-dimensional rotation Lie algebra will start seeming very ‘Leechy’.

Posted by: John Baez on May 21, 2017 11:36 AM | Permalink | Reply to this

### Re: Integral Octonions (Part 9)

Roger Howe suggested trying to construct the Leech lattice … [from] … the 3 cross-polytopes inside the 24-cell

Yes, I’ve been wanting to do this for a while now. Among other things it should clarify the relationship between $F_4$ and the Conway groups.

Greg and I already did connect triality to the Turyn construction of the Leech lattice

Is this written up anywhere? I’ve been wanting to do this for a while too, but last time I tried it still didn’t quite work.

Posted by: Tim Silverman on May 23, 2017 6:57 AM | Permalink | Reply to this

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