## July 29, 2013

### Integral Octonions (Part 3)

#### Posted by John Baez

You’ve probably heard about E8. It’s the name of several closely connected structures: a Lie group, a Lie algebra, a Coxeter group and a lattice in 8 dimensions.

The lattice in 8 dimensions is what concerns me most right now, because this lattice underlies the Cayley integers: the ‘integral octonions’ that are the theme of this little series. You can think of the integral octonions as giving the $\mathrm{E}_8$ lattice a multiplication making it into a nonassociative division ring.

But before getting into the multiplicative aspects, it’s good to be thoroughly familiar with the additive and geometrical aspects. So, let’s start with a few calculations with $\mathrm{E}_8$ and get to know it as a lattice in 8-dimensional Euclidean space. Then let’s think a bit about how it’s related to the Lie algebra $\mathrm{E}_8$, and the smaller Lie algebras $\mathrm{E}_7$ and $\mathrm{E}_6$.

All this stuff is fun in its own right, but it will also come in handy later if this series progresses as I’m imagining. Part 1 and Part 2 were very sketchy. But ideally, I’d like to show you the algebra and geometry related to Cayley integers very vividly, and for that we need some details.

### The E8 lattice

Here’s the easiest description of the $\mathrm{E}_8$ lattice. Define a half-integer to be an integer plus $\frac{1}{2}$. The E8 lattice consists of all 8-tuples of real numbers

$(x_1, x_2, x_3, x_4, x_5, x_6, x_7, x_8)$

such that

• the $x_i$ are either all integers or all half-integers, and
• the sum of all the $x_i$ is even.

It’s easy to see this set is closed under addition and subtraction, so it’s a lattice. It’s also easy to see that the dot product of any vector with itself is even, so it’s an even lattice.

This lattice is also unimodular, meaning that the volume of the unit cell is 1. To check, this just write out a list of basis vectors for this lattice as a matrix:

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

The determinant is a huge alternating sum of products, but all but one is zero, namely the product of the diagonal entries of this matrix. And that product is 1. So, the volume of the parallelipiped spanned by these vectors—the ‘unit cell’—is 1.

All this is good, because $\mathrm{E}_8$ is famous for being the smallest even unimodular lattice in a Euclidean space. Such things only exist in dimensions that are multiples of 8.

Next, let’s see that there are 240 vectors in the $\mathrm{E}_8$ lattice having the smallest possible nonzero length:

• First, there are a bunch of vectors like this: $(\pm \frac{1}{2}, \pm \frac{1}{2}, \pm \frac{1}{2}, \pm \frac{1}{2}, \pm \frac{1}{2}, \pm \frac{1}{2}, \pm \frac{1}{2}, \pm \frac{1}{2})$ all of which have length $\sqrt{ \frac{1}{4} + \frac{1}{4} + \frac{1}{4} + \frac{1}{4} + \frac{1}{4} + \frac{1}{4} + \frac{1}{4} + \frac{1}{4} } = \sqrt{2}$ There are $2^8$ vectors where all the components are $\pm \frac{1}{2}$. For half of them the sum of the components is even; for the rest the sum is odd. So, $2^7 = 128$ of these vectors lie in $\mathrm{E}_8$.
• Second, there are a bunch of vectors like this: $(\pm 1, \pm 1, 0, 0 , 0 , 0, 0, 0)$ together with those we get from permuting the components. These too have length $\sqrt{2}$. There are 4 times $\binom{8}{2}$ of these, for a total of $4 \times 28 = 112$.

All other nonzero vectors in this lattice are longer. So, there are

$128 + 112 = 240$

as promised!

These vectors are the roots of the Lie algebra $\mathfrak{e}_8$. The dimension of the Lie algebra itself is 8 more: that is, 248. That’s because the dimension of any simple Lie algebra is the number of roots plus the dimension of its Cartan subalgebra, an arbitrarily chosen maximal abelian subalgebra. For $\mathfrak{e}_8$ this is 8. Indeed, we can think of the $\mathrm{E}_8$ lattice as lying in this 8-dimensional space.

(It’s ultimately wiser to think of it as lying in the dual space, but there’s a canonical way to identify them, so it’s pretty harmless.)

### The Lie algebra E8

To have more fun with the roots of $\mathrm{E}_8$, it helps to know some general facts about graded Lie algebras. Here I don’t mean $\mathbb{Z}/2$-graded Lie algebras, also known as Lie superalgebras. Instead, I mean Lie algebras $\mathfrak{g}$ that have been written as a direct sum of subspaces $\mathfrak{g}(i)$, one for each integer $i$, such that

$[\mathfrak{g}(i),\mathfrak{g}(j)] \subseteq \mathfrak{g}(i+j)$

If only the middle 3 of these subspaces are nonzero, so that

$\mathfrak{g} = \mathfrak{g}(-1) \oplus \mathfrak{g}(0) \oplus \mathfrak{g}(1)$

we say that $\mathfrak{g}$ is ‘3-graded’. Similarly, if only the middle 5 are nonzero, so that

$\mathfrak{g} = \mathfrak{g}(-2) \oplus \mathfrak{g}(-1) \oplus \mathfrak{g}(0) \oplus \mathfrak{g}(1) \oplus \mathfrak{g}(2)$

we say $\mathfrak{g}$ is ‘5-graded’, and so on.

In these situations, some nice things happen. First of all, $\mathfrak{g}(0)$ is always a Lie subalgebra of $\mathfrak{g}$. Second of all, it acts on each other space $\mathfrak{g}(i)$ by means of the bracket. Third of all, if $\mathfrak{g}$ is 3-graded, we can give $\mathfrak{g}(1)$ a product by picking any element $k \in \mathfrak{g}(-1)$ and defining

$x \circ y = [[x,k],y]$

This product automatically satisfies the identities defining a Jordan algebra:

$x \circ y = y \circ x$

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

so 3-graded Lie algebras are a great source of Jordan algebras.

To get gradings of a complex simple Lie algebra $\mathfrak{g}$ we can write $\mathfrak{g}$ as the direct sum of a bunch of 1-dimensional ‘weight spaces’ $\mathfrak{g}_r$, one for each root, together with its Cartan algebra, which deserves to be called $\mathfrak{g}_0$. The great thing about this decomposition is that

$[\mathfrak{g}_r , \mathfrak{g}_{r'} ] \subseteq \mathfrak{g}_{r + r'}$

whenever $r$ and $r'$ are either roots or zero. So, to put a grading on $\mathfrak{g}$, we just need to slice $\mathfrak{t}$ with evenly spaced parallel hyperplanes in such a way that each root, as well as the origin, lies on one of these hyperplanes.

Now let’s get to the point, and look at the case where $\mathfrak{g}$ is the complex form of the Lie algebra $\mathfrak{e}_8$. To put a grading on $\mathfrak{e}_8$, you should imagine its $240$ roots as the vertices of a gleaming 8-dimensional diamond. Imagine yourself as a gem cutter, turning around this diamond, looking for nice ways to slice it. You need to slice it with evenly spaced parallel hyperplanes that go through every vertex, as well as the center of the diamond.

The easiest way to do this is to let each slice go through all the roots where the sum of the coordinates takes a fixed value. There are five cases:

• There is $1$ root where the sum of the coordinates is $4$: $(\frac{1}{2}, \frac{1}{2}, \frac{1}{2}, \frac{1}{2}, \frac{1}{2}, \frac{1}{2}, \frac{1}{2}, \frac{1}{2})$
• There are $56$ roots where the sum of the coordinates is $2$. There are $\binom{8}{2}$ or $28$ like this: $(1, 1, 0, 0, 0, 0, 0, 0 )$ and all permutations of these numbers. And there are $\binom{8}{2}$ or $28$ like this: $(\frac{1}{2}, \frac{1}{2}, \frac{1}{2}, \frac{1}{2}, \frac{1}{2}, \frac{1}{2}, -\frac{1}{2}, -\frac{1}{2})$ and all permutations of these.
• There are $126$ roots where the sum of the coordinates is $0$. There are $2$ times $\binom{8}{2}$ or $56$ like this: $(1, -1, 0, 0, 0, 0, 0, 0 )$ and all permutations of these numbers. And there are $\binom{8}{4}$ like this: $(\frac{1}{2}, \frac{1}{2}, \frac{1}{2}, \frac{1}{2}, -\frac{1}{2}, -\frac{1}{2}, -\frac{1}{2}, -\frac{1}{2})$ and all permutations of these. Note that $\binom{8}{4} = 70$. So, the total is indeed $56 + 70 = 126$.
• There are $56$ roots where the sum of the coordinates is $-2$. These are just the negatives of those where the sum is 1, namely

$(-1, -1, 0, 0, 0, 0, 0, 0 )$ and all its permutations, and $(\frac{1}{2}, \frac{1}{2}, -\frac{1}{2}, -\frac{1}{2}, -\frac{1}{2}, -\frac{1}{2}, -\frac{1}{2}, -\frac{1}{2})$ and all its permutations.
• There is $1$ root where the sum of the coordinates is $-4$, namely $(-\frac{1}{2}, -\frac{1}{2}, -\frac{1}{2}, -\frac{1}{2}, -\frac{1}{2}, -\frac{1}{2}, -\frac{1}{2}, -\frac{1}{2})$

It follows that there is a 5-grading of $\mathfrak{e}_8$:

$\mathfrak{e}_8 = \mathfrak{e}_8(-2) \oplus \mathfrak{e}_8(-1) \oplus \mathfrak{e}_8(0) \oplus \mathfrak{e}_8(1) \oplus \mathfrak{e}_8(2) ,$

where the dimensions of the subspaces work as follows:

$248 = 1 + 56 + 134 + 56 + 1$

Here we must remember to include $\mathfrak{g}_0$ in $\mathfrak{e}_8(0)$, obtaining a Lie subalgebra of dimension

$126 + 8 = 134$

In fact, this $134$-dimensional Lie algebra $\mathfrak{e}_8(0)$ is the direct sum of the Lie algebra $\mathfrak{e}_7$ and the 1-dimensional abelian Lie algebra $\mathfrak{gl}(1)$. This comes as little surprise if you know that the dimension of $\mathfrak{e}_7$ is $133$, but the reason is that if we take all the roots of $\mathfrak{e}_8$ that are orthogonal to a given root, we obtain the roots of $\mathfrak{e}_7$. From this point of view, the 5-grading of $\mathfrak{e}_8$ looks like this:

$\mathfrak{e}_8 = \mathbb{C} \oplus \mathbf{F}^* \oplus (\mathfrak{e}_7 \oplus \mathfrak{gl}(1)) \oplus \mathbf{F} \oplus \mathbb{C} .$

where $\mathbf{F}$ is some 56-dimensional space and $\mathfrak{gl}(1)$ is the 1-dimensional abelian complex Lie algebra.

Recall that $\mathfrak{e}_8(0) = \mathfrak{e}_7 \oplus \mathfrak{gl}(1)$ acts on all the other spaces $\mathfrak{e}_8(i)$. In particular, $\mathbb{C}$ is the 1-dimensional trivial representation of $\mathfrak{e}_7 \oplus \mathfrak{gl}(1)$, while $\mathbf{F}$ is the Freudenthal algebra: a 56-dimensional representation of $\mathfrak{e}_7 \oplus \mathfrak{gl}(1)$, which happens to be the smallest nontrivial representation of $\mathfrak{e}_7$.

There are many more games to play along these lines. For example, we can repeat our ‘gem-cutting’ trick to get a 3-grading of $\mathfrak{e}_7$:

$\mathfrak{e}_7 = \mathfrak{e}_7(-1) \oplus \mathfrak{e}_7(0) \oplus \mathfrak{e}_7(1)$

Now the dimensions work like this:

$133 = 27 + 79 + 27$

Since the dimension of $\mathfrak{e}_6$ is 78, it is not very surprising that $\mathfrak{e}_7(0)$ is the direct sum of $\mathfrak{e}_6$ and a one-dimensional abelian Lie algebra. Since 3-gradings give Jordan algebras, it is also not surprising that $\mathfrak{e}_7(1)$ is a famous 27-dimensional Jordan algebra.

The exceptional Jordan algebra is the space $\mathfrak{h}_3(\mathbb{O})$ of $3 \times 3$ self-adjoint matrices with octonion entries, equipped with the product $a \circ b = \frac{1}{2}(a b + b a)$. This is a Jordan algebra over the real numbers. It is 27-dimensional, since its elements look like this:

$\mathfrak{h}_3(\mathbb{O}) = \{ \left( \begin{array}{ccc} \alpha & z^* & y^* \\ z & \beta & x \\ y & x^* & \gamma \end{array} \right) : \; x,y,z \in \mathbb{O} , \; \alpha , \beta, \gamma \in \mathbb{R} \} .$

Its complexification

$\mathbf{J} = \mathbb{C} \otimes \mathfrak{h}_3(\mathbb{O})$

is none other than $\mathfrak{e}_7(1)$, and it’s a Jordan algebra over the complex numbers. The space $\mathfrak{e}_7(-1)$ is the dual of this, so we have : $\mathfrak{e}_7 = \mathbf{J}^\ast \oplus (\mathfrak{e}_6 \oplus \mathfrak{gl}(1)) \oplus \mathbf{J} .$

Using our facts about graded Lie algebras, this implies that $\mathfrak{e}_6$ acts on $\mathbf{J}$ and $\mathbf{J}^\ast$. In fact, $\mathbf{J}$ and its dual are the smallest nontrivial representations of $\mathfrak{e}_6$. (They are not isomorphic representations, while $\mathbf{F}$ and its dual are isomorphic as representations of $\mathfrak{e}_7$.)

Furthermore, if we use the above inclusion of $\mathfrak{e}_6$ in $\mathfrak{e}_7$, we can take our previous decomposition of $\mathfrak{e}_8$:

$\mathfrak{e}_8 = \mathbb{C} \oplus \mathbf{F} \oplus (\mathfrak{e}_7 \oplus \mathfrak{gl}(1)) \oplus \mathbf{F} \oplus \mathbb{C} .$

and decompose everything in sight as irreducible representations of $\mathfrak{e}_6$. When we do this, the Freudenthal algebra decomposes as

$\mathbf{F} = \mathbb{C} \oplus \mathbf{J}^\ast \oplus \mathbf{J} \oplus \mathbb{C} ,$

with the dimensions working as follows:

$56 = 1 + 27 + 27 + 1 .$

### Conclusion

There’s more to do, but this is a good place to take a break. We’ve seen how $\mathrm{E}_7$, $\mathrm{E}_6$, the Freudenthal algebra and the exceptional Jordan algebra spring forth from repeatedly slicing the 240 roots of $\mathrm{E}_8$, and we’ve started to see the eerie, alien-looking numbers that pervade exceptional mathematics:

• 26, the dimension of the space of $3 \times 3$ traceless hermitian octonionic matrices, or $\mathbb{O}^3 \oplus \mathbb{R}^3$.
• 27 = 26 + 1, the dimension of the space of $3 \times 3$ hermitian octonionic matrices $\mathfrak{h}_3(\mathbb{O}) \cong \mathbb{O}^3 \oplus \mathbb{R}^3$, which is the exceptional Jordan algebra — or alternatively, the complex dimension of the Jordan algebra $\mathbb{C} \otimes \mathfrak{h}_3(\mathbb{O})$, which is the smallest nontrivial complex representation of $\mathfrak{e}_6$.
• 56 = 1 + 27 + 27 + 1, the complex dimension of the Freudenthal algebra $\mathbf{F} \cong \mathbb{C} \oplus \mathbb{J}^* \oplus \mathbb{J} \oplus \mathbb{C}$, the smallest nontrivial complex representation of $\mathfrak{e}_7$.
• 78 = 26 + 26 + 26, the dimension of $\mathfrak{e}_6$.
• 79 = 78 + 1, the dimension of the $\mathfrak{e}_6 \oplus \mathfrak{gl}(1)$, which is the grade-zero part of $\mathfrak{e}_7$ with respect to its 3-grading.
• 133 = 27 + 79 + 27, the dimension of $\mathfrak{e}_7 \cong \mathbf{J}^* \oplus (\mathfrak{e}_6 \oplus \mathfrak{gl}(1)) \oplus \mathbf{J}$.
• 134 = 133 + 1, the dimension of $\mathfrak{e}_6 \oplus \mathfrak{gl}(1)$, which is the grade-zero part of $\mathfrak{e}_8$ with respect to its 5-grading.
• 248 = 1 + 56 + 134 + 56 + 1, the dimension of $\mathfrak{e}_8 \cong \mathbb{C} \oplus \mathbf{F}^* \oplus \mathfrak{e}_7 \oplus \mathbf{F} \oplus \mathbb{C}$, which is the smallest nontrivial complex representation of itself.

All these numbers are visible in the $\mathbf{E}_8$ lattice!

Posted at July 29, 2013 2:59 PM UTC

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### Re: Integral Octonions (Part 3)

If you add up the nonnegative parts of a k-grading, you get a parabolic, whose 0 component is a Levi and whose strictly positive parts add up to the nilpotent radical.

The 3-gradings then correspond to parabolics with abelian radical, which in turn correspond to elements of the center of the simply-connected group (the Levi being the centralizer of the element), up to conjugacy. Of course there’s a stupid 3-grading in which the Levi is the whole Lie algebra (really a 1-grading (really THE 1-grading)).

What’s notable about E8, then (and only true of it, G2, and F4) is that it has no center, so no interesting 3-gradings. E7 has only the one you mention.

Posted by: Allen K. on July 29, 2013 4:03 PM | Permalink | Reply to this

### Re: Integral Octonions (Part 3)

Interesting! I don’t really understand why there’s a 1-1 correspondence between elements of the center of the simply-connected group and conjugacy classes of parabolics with abelian radical, but the rest makes sense.

I believe this particular parabolic $P \subset\mathrm{E}_8$, the one whose Lie algebra is

$\mathfrak{e}_8(-2) \oplus \mathfrak{e}_8(-1) \oplus \mathfrak{e}_8(0),$

is special, because $\mathrm{E}_8/P$ is the lowest-dimensional manifold on which $\mathrm{E}_8$ has a nontrivial action. Its dimension is

$dim(\mathrm{E}_8/P) = dim \mathfrak{e}_8(1) + dim \mathfrak{e}_8(2) = 56 + 1 = 57,$

a number one more often associates with varieties of ketchup than algebraic varieties.

Posted by: John Baez on July 30, 2013 5:11 PM | Permalink | Reply to this

### Re: Integral Octonions (Part 3)

Every such symmetric grading is given by a grading element that acts by $i$ on $\mathfrak{g}(i)$.

A side note: In some literature these symmetric gradings are called $|k|$-gradings

(1)$\mathfrak{g}(-k) \oplus \cdots \oplus \mathfrak{g}(0) \oplus \cdots \oplus \mathfrak{g}(k)$

to distinguish them from nonsymmetrical $k$-gradings

(2)$\mathfrak{g}(0)\oplus\cdots\oplus \mathfrak{g}(k).$
Posted by: Vít Tuček on August 4, 2013 9:57 AM | Permalink | Reply to this

### Re: Integral Octonions (Part 3)

A small point of confusion for me: the 5-grading of $\mathfrak{e}_8$ is not by the sum of coordinates, but rather by the sum of coordinates divided by 2, right? For instance $(\frac{1}{2}, \frac{1}{2}, \frac{1}{2}, \frac{1}{2}, \frac{1}{2}, \frac{1}{2}, \frac{1}{2}, \frac{1}{2}) \in \mathfrak{e}_8(2)$ has coordinates summing to 4.

Posted by: Tim Campion on July 29, 2013 10:38 PM | Permalink | Reply to this

### Re: Integral Octonions (Part 3)

Whoops! You’re right, my conventions were inconsistent. Let’s divide the sum of the coordinates by 2. I’ll fix my blog article. Thanks!

Posted by: John Baez on July 30, 2013 2:17 PM | Permalink | Reply to this

### Re: Integral Octonions (Part 3)

There’s something wrong with your matrix - every row is orthogonal to (+1,-1,+1,-1,+1,-1,+1,-1). The rows aren’t even a basis and the determinant must be zero. I think you need some minus signs above the diagonal.

Posted by: Jeremy Henty on July 30, 2013 9:00 PM | Permalink | Reply to this

### Re: Integral Octonions (Part 3)

Whoops! I guess I’d better use this:

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

I slipped—I was acting like a cyclic permutation of 8 things was even rather than odd! Since it’s odd, the determinant of this matrix is

$1^7 \times \frac{1}{2} - (-1)^7 \times \frac{1}{2} = 1$

as desired, while my original bungled version

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

$1^7 \times \frac{1}{2} - 1^7 \times \frac{1}{2} = 0$

Posted by: John Baez on July 31, 2013 1:40 AM | Permalink | Reply to this

### Re: Integral Octonions (Part 3)

Gradings are fascinating. If you’d like to play with rotating the E8 root system in eight dimensions to see gradings, you might enjoy the Elementary Particle Explorer. (Choose E8, then click and drag in the view.)

Posted by: Garrett on July 31, 2013 12:48 AM | Permalink | Reply to this

### Re: Integral Octonions (Part 3)

That matrix is still not right. The inner product of all of the rows with v := (1,1,1,1,1,1,1,1) is 0 modulo 4. But (1,1,0,0,0,0,0,0) is in E_8 and its inner product with v is 2, so it is not in the lattice generated by the rows.

(Am I the only one here who does their homework?)

Posted by: Jeremy Henty on July 31, 2013 7:47 PM | Permalink | Reply to this

### Re: Integral Octonions (Part 3)

Hmm, I didn’t bother to check that the 8 vectors listed as rows here

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

generate the whole $\mathrm{E}_8$ lattice, because I reasoned as follows. First, they’re in the $\mathrm{E}_8$ lattice. Second, they’re linearly independent, so they generate some lattice $L \subseteq \mathrm{E}_8$. Third, the volume of the unit cell of $L$ is 1, since that’s the determinant of the above matrix. Fourth, the volume of the unit cell of $\mathrm{E}_8$ is also 1, since $\mathrm{E}_8$ is famous for being the first interesting ‘unimodular even lattice’. So, $L$ has to be all of $\mathrm{E}_8$. If, for example, only ‘every other vector’ in $\mathrm{E}_8$ were in $L$, the volume of its unit cell would need to be twice as big.

But clearly I made a mistake somewhere or other. Where is it?

I could of course have just copied down a basis for $\mathrm{E}_8$ from a standard reference, like this:

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

But I didn’t feel like it…

Posted by: John Baez on August 1, 2013 11:32 AM | Permalink | Reply to this

### Re: Integral Octonions (Part 3)

Cool stuff.

Maybe you can write up a part 4 about constructing the Leech lattice from triplets of $E_8$ roots? Generating the automorphism group of the Leech lattice, $Co_0$, using 3x3 unitary matrices over the octonions also gives a nice tie in with $F_4$.

Posted by: Mike Rios on August 1, 2013 2:00 PM | Permalink | Reply to this

### Re: Integral Octonions (Part 3)

That’s great stuff. But it won’t be in part 4. I was planning to talk about your paper in part 5 or 6 or 7… after I’ve explained the integral octonions a bit better.

But I had a question. Suppose I think of $\mathbb{O}^3$ as the elements of the exceptional Jordan algebra $\mathfrak{h}_3(\mathbb{O})$ that vanish down the diagonal. Then any embedding of the Leech lattice $L$ in $\mathbb{O}^3$ gives an embedding

$L \oplus \mathbb{Z}^3 \subset \mathfrak{h}_3(\mathbb{O})$

where we put integers down the diagonal of the $3 \times 3$ matrix and $L$ into the off-diagonal entries.

Is there a way to embed the Leech lattice $L$ in $\mathbb{O}^3$ so that this copy of $L \oplus \mathbb{Z}^3$ is closed under the Jordan algebra multiplication?

The fact that

1) $\mathrm{F}_4$ is the automorphism group of the Jordan algebra $\mathfrak{h}_3(\mathbb{O})$,

2) you’re getting the symmetry group $Co_0$ of the Leech lattice to be a subgroup of $\mathrm{F}_4$, and

3) you’re using integral octonions to embed the Leech lattice in $\mathbb{O}^3$

seems to suggest there’s hope! But you don’t seem to come out and say you’ve got a copy of $L \oplus \mathbb{Z}^3$ in $\mathfrak{h}_3(\mathbb{O})$ that’s closed under the Jordan algebra product! I’d really like the Leech lattice to be related to an ‘integral form’ of the exceptional Jordan algebra in this way.

Posted by: John Baez on August 2, 2013 3:08 AM | Permalink | Reply to this

### Re: Integral Octonions (Part 3)

The determinant isn’t 1, it is 4. There are 8 terms, since for each element in the bottom row you can select that element, the elements on the diagonal to its left and the elements above the diagonal to its right. 8 terms each of value 1/2 = 4.

Alternatively, add the first column to the second, then add the new second column to the third, then add the new third column to fourth etc. At each stage you eliminate a -1 and the bottom of the newest column is 1/2 greater than the bottom of the previous column. Eventually you have an lower triangular matrix with a 4 in the bottom right and a 1 everywhere else on the diagonal.

Posted by: Jeremy Henty on August 1, 2013 10:05 PM | Permalink | Reply to this

### Re: Integral Octonions (Part 3)

Okay, that dissolves my confusion. By the way, for anyone wondering, this comment was meant as a reply to this one.

Posted by: John Baez on August 2, 2013 3:24 AM | Permalink | Reply to this

### Re: Integral Octonions (Part 3)

What is the algebra structure of Freudenthal algebra?

Posted by: Vít Tuček on August 5, 2013 10:08 AM | Permalink | Reply to this

### Re: Integral Octonions (Part 3)

The 56-dimensional representation $F$ of $\mathrm{E}_7$ has an invariant symplectic structure

$\omega : F \times F \to \mathbb{C}$

and trilinear product

$\tau: F \times F \times F \to F$

obeying some identities I find hard to remember, which are the definition of a Freudenthal algebra, also called a Freudenthal triple system. The transformations of $F$ preserving this symplectic structure and trilinear product are exactly the group $\mathrm{E}_7$.

Using the symplectic structure, we can trade the trilinear product for a quadrilinear form, and then the identities I find hard to remember are listed starting on page 12 of this paper:

• Fred W. Helenius, Freudenthal triple systems by root system methods.

A Freudenthal triple system is a special case of a Freudenthal–Kantor triple system, which I’d like to learn more about.

Also see:

• K. Meyberg, Eine Theorie der Freudenthalschen Tripelsysteme, I, II, Ned. Akad. Wetenschap. 71 (1968), 162–190.

• Robert B. Brown, Groups of type $\mathrm{E}_7$, Jour. Reine Angew. Math. 236 (1969), 79–102.

• John R. Faulkner, A construction of Lie algebras from a class of ternary algebras, Trans. Amer. Math. Soc. 155 (1971), 397–408.

Posted by: John Baez on August 5, 2013 3:26 PM | Permalink | Reply to this

### Re: Integral Octonions (Part 3)

Thanks! This is a very interesting direction of research!

Glancing through Helenius’s article it seems that one can define such symplectic form and quartic form for any contact grading. Contact grading is a $|2|$-grading with some more properties such as $\dim\, \mathfrak{g}(2)=1$. (For detail see book by Čap, Slovák: Parabolic geometries.)

If I remember correctly, such grading exists also for $F_4$ with $\mathfrak{g}(0)$ being $\mathfrak{sp}(8)$.

The $|2|$-grading that corresponds to $\mathfrak{so}(9)$ has $dim\, \mathfrak{g}(2) = 7$; figuring out the “octonionic maps” there is on my todo list for far too long.

Posted by: Vít Tuček on August 5, 2013 8:22 PM | Permalink | Reply to this

### Re: Integral Octonions (Part 3)

Actually, such a grading exists for any simple Lie algebra and correspond to the (usually maximal, except $A_n$ case) extra-special parabolic subgroups. You throw away all (usually one) vertices from the Dynkin diagram that are joined to the additional vertex on the extended Dynkin diagram, and this gives you the Dynkin diagram of the Levi subalgebra.

Posted by: Victor Petrov on September 24, 2013 6:29 PM | Permalink | Reply to this

### Re: Integral Octonions (Part 3)

Is $(1/2,/2,1/2,1/2, 0, 0, 0 ,0)$ integral? The sum of all the $x_i$ is 2 which is even. Maybe, you want the sum of squares to be even?

Posted by: L. Vaserstein on September 5, 2013 11:59 AM | Permalink | Reply to this

### Re: Integral Octonions (Part 3)

I don’t know what you mean by ‘integral’. In this post I defined the E8 lattice to consist of all 8-tuples of real numbers

$(x_1, x_2, x_3, x_4, x_5, x_6, x_7, x_8)$

such that

• the $x_i$ are either all integers or all half-integers, and
• the sum of all the $x_i$ is even.

So, the vector

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

isn’t in this lattice.

Posted by: John Baez on September 5, 2013 12:04 PM | Permalink | Reply to this

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