Integral Octonions (Part 3)
Posted by John Baez
You’ve probably heard about E_{8}. 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 8dimensional 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 E_{8} lattice
Here’s the easiest description of the $\mathrm{E}_8$ lattice. Define a halfinteger to be an integer plus $\frac{1}{2}$. The E_{8} lattice consists of all 8tuples 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 halfintegers, 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 8dimensional 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 E_{8}
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 ‘3graded’. 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 ‘5graded’, 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 3graded, 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 3graded 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 1dimensional ‘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 8dimensional 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 5grading 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 1dimensional 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 5grading 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 56dimensional space and $\mathfrak{gl}(1)$ is the 1dimensional 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 1dimensional trivial representation of $\mathfrak{e}_7 \oplus \mathfrak{gl}(1)$, while $\mathbf{F}$ is the Freudenthal algebra: a 56dimensional 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 ‘gemcutting’ trick to get a 3grading 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 onedimensional abelian Lie algebra. Since 3gradings give Jordan algebras, it is also not surprising that $\mathfrak{e}_7(1)$ is a famous 27dimensional Jordan algebra.
The exceptional Jordan algebra is the space $\mathfrak{h}_3(\mathbb{O})$ of $3 \times 3$ selfadjoint 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 27dimensional, 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, alienlooking 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 gradezero part of $\mathfrak{e}_7$ with respect to its 3grading.
 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 gradezero part of $\mathfrak{e}_8$ with respect to its 5grading.
 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!
Re: Integral Octonions (Part 3)
If you add up the nonnegative parts of a kgrading, you get a parabolic, whose 0 component is a Levi and whose strictly positive parts add up to the nilpotent radical.
The 3gradings then correspond to parabolics with abelian radical, which in turn correspond to elements of the center of the simplyconnected group (the Levi being the centralizer of the element), up to conjugacy. Of course there’s a stupid 3grading in which the Levi is the whole Lie algebra (really a 1grading (really THE 1grading)).
What’s notable about E8, then (and only true of it, G2, and F4) is that it has no center, so no interesting 3gradings. E7 has only the one you mention.