Integral Octonions (Part 3)

July 29, 2013

Integral Octonions (Part 3)

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

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:

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:

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:

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

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


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:

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

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© 2013 John Baez