## December 1, 2014

### Integral Octonions (Part 10)

#### Posted by John Baez

The Leech lattice gives the densest packing of spheres in 24 dimensions. The exceptional Jordan algebra, consisting of matrices

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

where $a,b,c$ are real and $X,Y,Z$ are octonions, has dimension $3 + 24 = 27$. They’re both remarkable entities. If the mathematical universe is a harmonious place, they should be connected.

More precisely: we should be able to fit the Leech lattice into the exceptional Jordan algebra in a nice way. And Greg Egan has shown that we can!

In fact, we can fit the Leech lattice into the space of matrices like this

$\left( \begin{array}{ccc} 0 & X & Y \\ X^* & 0 & Z \\ Y^* & Z^* & 0 \end{array} \right)$

where $X, Y, Z$ are integral octonions. We can do it in only finitely many ways — but Egan showed we can do it in at least 244,035,421 ways. Of these, at least 17,280 are compatible with the product on the exceptional Jordan algebra, in a way that I will describe.

Big numbers! Amusingly, Egan found the first number through a quick calculation which I’ll describe here — but the second, prettier number through an exhaustive computer search.

Let me start by stating the results more precisely, but in a way that doesn’t require much background. This subject can be intimidating, but I think you can enjoy it without being an expert.

In quantum mechanics, observable quantities are often described by self-adjoint complex matrices. So, the space $\mathfrak{h}_n(\mathbb{C})$ of $n \times n$ self-adjoint complex matrices is very important. But what kind of thing is it?

It’s not an associative algebra in any interesting way. If we multiply two such matrices, the result won’t usually be self-adjoint. Their symmetrized product

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

will be. But this product $\circ$ is not associative. Instead, it makes $\mathfrak{h}_n(\mathbb{C})$ into a formally real Jordan algebra! This is a real vector space with a bilinear product $\circ$ that is

• commutative,
• power-associative: we can reparenthesize any expression built using one element $x$ and the product $\circ$ without changing the result,
• formally real: a sum of squares can only be zero if each element being squared is zero:
$x_1 \circ x_1 + \cdots + x_n \circ x_n = 0 \; \implies \; x_1, \dots, x_n = 0$

Pascual Jordan invented this concept in 1932. He noticed that these axioms imply the weird identity

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

A vector space with a bilinear product obeying this identity is now called a Jordan algebra. But I find this identity annoying and largely beside the point. To me what really matters are the formally real Jordan algebras, since these have a comprehensible definition and nice properties. Each one gives a projective geometry, which is a kind of logic for quantum theory. And each one gives a convex cone, which consists of the ‘positive’ observables.

In 1934, Jordan teamed up with Eugene Wigner and John von Neumann and published a paper on the foundations of quantum mechanics in which they classified all finite-dimensional formally real Jordan algebras. They’re all direct sums of ‘simple’ ones. The simple ones all belong to four infinite series, with one exception:

1. the space $\mathfrak{h}_n(\mathbb{R})$ of $n \times n$ self-adjoint real matrices with the product $x \circ y = \frac{1}{2}(x y + y x)$
2. the space $\mathfrak{h}_n(\mathbb{C})$ of $n \times n$ self-adjoint complex matrices with the product $x \circ y = \frac{1}{2}(x y + y x)$
3. the space $\mathfrak{h}_n(\mathbb{H})$ of $n \times n$ self-adjoint quaternionic matrices with the product $x \circ y = \frac{1}{2}(x y + y x)$
4. the space $\mathbb{R}^n \oplus \mathbb{R}$ with the product $(v,\alpha) \circ (w, \beta) = (\alpha w + \beta v, \langle v,w\rangle + \alpha \beta)$
5. the space $\mathfrak{h}_3(\mathbb{O})$ of $3 \times 3$ self-adjoint octonionic matrices with the product $x \circ y = \frac{1}{2}(x y + y x)$

Ever since then people have been trying to understand the meaning of this result.

The first three series of Jordan algebras are telling us that we can do quantum mechanics with all three associative normed division algebras. So why do we only use the complex numbers? I’ve argued that this is not true: secretly we use all three.

The fourth series, called spin factors, are more mysterious. They show up in special relativity: we get them by taking a Minkowski spacetime of any dimension and picking out a time axis. Formally real Jordan algebras give convex cones — but for a spin factor this is the ‘future cone’ in Minkowski spacetime. I believe this a clue of some sort: a relation between quantum mechanics and spacetime. And string theory enters, too: the spin factor for 10-dimensional Minkowski spacetime is isomorphic to $\mathfrak{h}_2(\mathbb{O})$.

But the last of the formally real Jordan algebras, $\mathfrak{h}_3(\mathbb{O})$, is even more mysterious. It’s called the exceptional Jordan algebra. It could be the key to profound cosmic secrets… or a red herring that will do nothing but lead us astray. Today we’ll ponder how it might be related to another enigmatic entity: the Leech lattice.

The octonions, $\mathbb{O}$, are an 8-dimensional nonassociative algebra. There’s a concept of conjugation for octonions, a lot like complex conjugation: it switches the sign of all the square roots of -1. The octonions with $X = X^*$ form a copy of the real numbers $\mathbb{R} \subset \mathbb{O}$. Thus, $\mathfrak{h}_3(\mathbb{O})$ consists of matrices like this:

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

where $a,b,c$ are real numbers and $X,Y,Z$ are octonions. So, the exceptional Jordan algebra has dimension $1+1+1+8+8+8 = 27$.

The Leech lattice, on the other hand, lives in 24 dimensions. How could it be related?

Sitting inside the octonions you can find a lattice that’s closed under multiplication and contains the identity, $1 \in \mathbb{O}$. There are many different lattices with this property, but if you also want your lattice to be a rescaled version of the E8 lattice — the best lattice in 8 dimensions, the one that gives the densest lattice packing of spheres in this dimension — there’s essentially just one. That is: there’s a continuum of such lattices, but you can get any of them from any other by applying an automorphism of the octonions!

So, I will choose any one of these special lattices and call it the integral octonions, or $\mathbf{O}$. It contains 240 octonions of length 1, and it’s a version of the $\mathrm{E}_8$ lattice rescaled by a factor of $1/\sqrt{2}$.

Now we can take the exceptional Jordan algebra and get a lattice in it by considering only the $3 \times 3$ self-adjoint matrices with integral octonions as entries! I’ll call this lattice $\mathfrak{h}_3(\mathbf{O})$. So:

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

$\mathfrak{h}_3(\mathbf{O})$ is not closed under the usual Jordan product $\frac{1}{2}(x y + x y)$, because we get half-integers showing up. But it is closed under the doubled Jordan product $x \circ y = x y + y x$, and this operation makes it into a Jordan ring. In other words, it’s an abelian group under addition, and equipped with a bilinear product $x \circ y$ that’s commutative and obeys the annoying identity

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

Even better, it’s formally real, meaning that

$x_1 \circ x_1 + \cdots + x_n \circ x_n = 0 \; \implies \; x_1, \dots, x_n = 0$

So, $\mathfrak{h}_3(\mathbf{O})$ deserves to be called the exceptional Jordan ring.

It would be fun to relate the Leech lattice to the exceptional Jordan ring. But the Leech lattice lives in a 24-dimensional real inner product space. So, there’s some work to do.

Luckily, the exceptional Jordan algebra has a positive definite inner product $h$ that’s invariant under automorphisms of this algebra, determined by the property that

$x = \left( \begin{array}{ccc} a & X & Y \\ X^* & b & Z \\ Y^* & Z^* & c \end{array} \right) \quad \implies \quad h(x,x) = a^2 + b^2 + c^2 + X X^* + Y Y^* + Z Z^*$

With this, the subspace of off-diagonal matrices

$\left\{ \; \left( \begin{array}{ccc} 0 & X & Y \\ X^* & 0 & Z \\ Y^* & Z^* & 0 \end{array} \right) : \; X,Y,Z \in \mathbb{O} \; \right\} \subset \mathfrak{h}_3(\mathbb{O})$

becomes a 24-dimensional inner product space! And sitting inside this is a lattice where the matrix entries are integral octonions:

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

So, we can ask if this lattice is the Leech lattice. Or more precisely: is it isometric to the Leech lattice? Two lattices $L, L'$ in inner product spaces $V, V'$ are isometric if there’s an inner product preserving map $f : V \to V'$ that maps $L$ in a one-to-one and onto way to $L'$.

In other words, does the 24-dimensional lattice I just described look just like the Leech lattice?

The answer is no! The integral octonions are isometric to $\mathrm{E}_8$ lattice rescaled by a factor of $1/\sqrt{2}$, so the above lattice $\Lambda$ is isometric to $\mathrm{E}_8 \oplus \mathrm{E}_8 \oplus \mathrm{E}_8$, rescaled by the same factor. This is completely different from the Leech lattice. For example, the smallest nonzero vectors in the Leech lattice have length 4, while the smallest nonzero vectors in this lattice have length 1. But it’s worse than that: the Leech lattice is just not a direct sum of other lattices.

So here’s a better question: does

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

contain a sublattice isometric to the Leech lattice? And the answer, shown by Greg Egan but probably known to experts, is:

Theorem 1. The lattice

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

contains at least 244,035,421 sublattices isometric to the Leech lattice (but only a finite number of them).

They say beggars can’t be choosers. But here we have an overabundance of riches. So, we should demand some extra properties to pick out ‘nice’ sublattices among this huge pile.

The obvious property is that we’d like our sublattice to be closed under the doubled Jordan product $x y + y x$. But this is impossible, since the Jordan product of two off-diagonal matrices typically has a nonzero diagonal part. So, we should look at a bigger lattice consisting of matrices like this:

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

where the off-diagonal part

$\left( \begin{array}{ccc} 0 & X & Y \\ X^* & 0 & Z \\ Y^* & Z^* & 0 \end{array} \right)$

lies in some sublattice $\Lambda$ isometric to the Leech lattice, while the diagonal entries $a,b,c$ are integers. Such a lattice has a fighting chance to be closed under the doubled Jordan product! It will be a 27-dimensional lattice in the exceptional Jordan algebra.

Puzzle 1. Is there a sublattice

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

isometric to the Leech lattice, such that

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

is closed under the doubled Jordan product $x y + y x$?

I should warn you that my ‘puzzles’ range from easy questions to research projects. I don’t know the answer to this one. I suspect the answer is no!

But if we use a quadrupled Jordan product, the answer becomes yes. More precisely, Greg Egan has shown:

Result 1. The lattice

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

contains at least 17,280 sublattices $\Lambda$ isometric to the Leech lattice such that $\widetilde{\Lambda}$ is closed under the quadrupled Jordan product $2(x y + y x)$. For each of these, $\widetilde{\Lambda}$ is a formally real Jordan ring.

Why am I calling this a ‘result’ instead of a ‘theorem’? The construction of these lattices $\Lambda$ is perfectly rigorous and not so hard to understand: I’ll give it here. So is the proof that they give formally real Jordan rings. But so far, Egan has only counted them using a computer program. It would be nice to make the count easier for humans to comprehend. I think that’s a doable project.

Puzzle 2. Are all 17,280 Jordan rings mentioned in Result 1 isomorphic?

Puzzle 3. Are there more then 17,280 lattices $\Lambda$ that obey the conditions of Result 1?

Puzzle 4. If so, do any of these give Jordan rings $\widetilde{\Lambda}$ that are not isomorphic to the Jordan rings in Result 1?

I have no idea!

### Lots of Leech lattices in $\mathbf{O}^3$

Now let’s see what’s really going on here. The flashy big numbers are not really the main point. Theorem 1 is actually quite easy to understand if you’re willing to trust a few facts that it relies on. It’s about copies of the Leech lattice sitting inside the lattice

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

But with the inner product I specified earlier, this lattice is isometric to $\mathbf{O}^3 \subset \mathbb{O}^3$. So, an equivalent way to state Theorem 1 is as follows:

So, an equivalent way to state Theorem 1 is as follows:

Theorem 1$\prime$. The lattice $\mathbf{O}^3$ contains at least 244,035,421 sublattices isometric to the Leech lattice (but only a finite number of them).

Proof. In Part 9, Greg Egan showed how to find sublattices of $\mathbf{O}^3$ isometric to the Leech lattice; I’ll review that later.

Starting from one sublattice of $\mathbf{O}^3$ isometric to the Leech lattice, we can get a vast number by applying symmetries — that is, isometries — of $\mathbf{O}^3$. How many symmetries does this lattice have?

The $\mathrm{E}_8$ lattice is isometric to $\mathbf{O}$ rescaled by a factor of $\sqrt{2}$. So, its isometry group, called the $\mathrm{E}_8$ Weyl group, is the same as that of $\mathbf{O}$. In Part 5, I explained why the $\mathrm{E}_8$ Weyl group has 696,729,600 elements.

It follows that the isometry group of $\mathbf{O}^3$ has at least

$3! \times 696,729,600^3 = 2,029,289,625,631,919,702,016,000,000$

elements, since we can apply an isometry to each component but also permute them.

Of course, we may not get this many sublattices isometric to the Leech lattice by starting with a particular one and acting by these isometries. After all, it may be preserved by some of these isometries. How many? I don’t know, but this number is at most the size of the isometry group of the Leech lattice: the Conway group $\mathrm{Co}_0$. This has only

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

elements.

If we have a finite group $G$ acting transitively on a finite set $X$, we have $|X| = |G|/|H|$ where $H$ is the stabilizer of any point of $X$. So, if we start with a sublattice of $\mathbf{O}^3$ isometric to the Leech lattice and act on it by isometries of $\mathbf{O}^3$, we get at least

$\frac{ 2,029,289,625,631,919,702,016,000,000 }{8,315,553,613,086,720,000} = 244,035,420.857...$

different Leech lattices in $\mathrm{O}^3$. The weird fractional answer simply means that the Conway group is not a subgroup of the larger group we’re considering here; we are just getting a lower bound on the number of Leech lattices in $\mathbf{O}^3$, so we can round up to the nearest integer.

On the other hand, there can only be finitely many of these Leech lattices, since any one is determined by its shortest vectors, which are of length 2, and there are finitely many vectors of this length in $\mathbf{O}^3$.     $\blacksquare$

The estimates here are crude, and it would be fun, though perhaps not very important, to get a more precise answer.

Puzzle 4. How many copies of the Leech lattice are there in $\mathbb{O}^3$? Equivalently, how many sublattices of $\mathrm{E}_8 \oplus \mathrm{E}_8 \oplus \mathrm{E}_8$ are isometric to the Leech lattice rescaled by a factor of $\sqrt{2}$?

I apologize for talking about these rescalings, but they matter. The scale factor here is the smallest one for which you can stick a rescaled copy of the Leech lattice into $\mathrm{E}_8 \oplus \mathrm{E}_8 \oplus \mathrm{E}_8$. If we make the scale factor larger, we might be able to stick in more rescaled Leech lattices.

And while we’re at it:

Puzzle 5. What is the order of the isometry group of $\mathrm{E}_8 \oplus \mathrm{E}_8 \oplus \mathrm{E}_8$?

### How to get a Leech lattice in $\mathbf{O}^3$

But how do we actually get a Leech lattice in $\mathbf{O}^3$? Egan explained this last time, borrowing and polishing some ideas of Dixon and Wilson. But let me review it, since we’ll need to know a bit about this construction.

Egan mainly explained it in terms of the $\mathrm{E}_8$ lattice, but now let me use the integral octonions $\mathbf{O}$ instead, since we’ll be needing that viewpoint. The $\mathrm{E}_8$ lattice is $\sqrt{2}$ times bigger than $\mathbf{O}$. So, a bunch of numbers will be smaller this time!

Remember that there are 240 integral octonions of length 1. Suppose we find a rotation

$R : \mathbb{O} \to \mathbb{O}$

such that the two transformations

$T_1 = \sqrt{2} R^{-1}, \qquad T_2 = \sqrt{2} R$

map each integral octonion $X$ of length 1 to two integral octonions whose inner product with $X$ is $1/2$.

So, both $T_1$ and $T_2$ rotate and stretch the 240 smallest nonzero integral octonions, which have length 1, and give 240 of the second smallest nonzero integral octonions, which have length $\sqrt{2}$. Moreover, each is being rotated by the same cleverly chosen angle!

Using these transformations we get lattices

$L_1 = T_1 \mathbf{O}, \qquad L_2 = T_2 \mathbf{O}$

and we have:

Theorem 3. The lattice $\Lambda$ consisting of all triples $(X,Y,Z) \in \mathbf{O}^3$ such that:

$X,Y, Z \in \mathbf{O}$ $X + Y, X + Z, Y + Z \in L_1$ $X + Y + Z \in L_2$

is isometric to the Leech lattice.

Proof. This is a restatement of the theorem in Part 9.     $\blacksquare$

Theorem 4. For any lattice $\Lambda \subset \mathbf{O}^3$ as in Theorem 3,

$\widetilde{\Lambda} = \left\{ \; \left( \begin{array}{ccc} a & X & Y \\ X^* & b & Z \\ Y^* & Z^* & c \end{array} \right) : \;\; a,b,c \in \mathbb{Z}, \;\; (X,Y,Z) \in \Lambda \; \right\}$

is closed under the quadrupled Jordan product $2(x y + y x)$. $\widetilde{\Lambda}$ is thus a formally real Jordan ring.

Proof. By our assumptions on the transformations $T_1$ and $T_2$, we have

$L_1, L_2 \subseteq \mathbf{O}$

On the other hand, while proving the theorem in Part 9, Egan proved that

$L_1 \cap L_2 = 2 \mathbf{O}$

Using this it is easy to see that whenever $(X,Y,Z) \in \mathbf{O}^3$, we have $(2X,2Y,2Z) \in \Lambda$. The reason is that the pairwise sums lie in $2 \mathbf{O} \subseteq L_1$ and the sum of all three lies in $2 \mathbf{O} \subseteq L_2$.

If we take any two matrices $x , y \in \mathfrak{h}_3(\mathbb{O})$, we have $x y + y x \in \mathfrak{h}_3(\mathbb{O})$, so the the off-diagonal entries of this matrix give a triple $(X,Y,Z) \in \mathbf{O}^3$. If we then double this matrix, the off-diagonal entries are $(2X,2Y,2Z)$, which lies in $\Lambda$.     $\blacksquare$

As Egan noted, the proof seems almost too easy: in the end, the point is simply that doubling anything in the exceptional Jordan ring $\mathfrak{h}_3(\mathbb{O})$ gives something in $\tilde{\Lambda}$.

Of course, we still need to check that Leech lattices of the sort we’re discussing actually exist. We can do it by finding a rotation $R$ of the octonions such that the transformations

$T_1 = \sqrt{2} R^{-1}, \qquad T_2 = \sqrt{2} R$

map each integral octonion $X$ of length 1 to two integral octonions whose inner product with $X$ is $1/2$. Egan did this last time by actually exhibiting such a rotation $R$. Alternatively, following the ideas of Dixon and Wilson, we can do it using the integral octonion

$U = \frac{1}{2}(1 + e_1 + e_2 + e_3 + e_4 + e_5 + e_6 + e_7)$

Multiplying on the left by $U$ gives a transformation $T_1$ that sends integral octonions to integral octonions. The inner product of $U$ with $1$ is $\frac{1}{2}$, and this implies that $T_1$ does what we want: if $X$ has length 1, then

$\langle U X, X \rangle = \langle U, 1 \rangle = \frac{1}{2}$

since right multiplication by an octonion of length 1 preserves the octonion inner product $\langle A, B \rangle = \mathrm{Re}(A B^*)$. We can then solve for $R$ and the transformation $T_2$, and check that $R$ really is a rotation and $T_2$ does what we want. particular, $T_2$ is just multiplication by

$U^* = \frac{1}{2}(1 - e_1 - e_2 - e_3 - e_4 - e_5 - e_6 - e_7)$

The octonion $U$ looks cute in the standard basis of octonions, but there’s nothing really sacred about it. For example, we could put a minus signs in front of some the $e_i$’s, and everything would still work.

So, there are lots of rotations $R$ that give Leech lattices $\Lambda \subset \mathbf{O}^3$ as in Theorem 3 and Jordan rings $\widetilde{\Lambda} \subset \mathfrak{h}_3(\mathbb{O})$ as in Theorem 4. How many? This is where Egan’s programming virtuosity was really crucial:

Result 1. There are 138,240 rotations $R : \mathbb{O} \to \mathbb{O}$ such that the two transformations $T_1 = \sqrt{2} R^{-1}, T_2 = \sqrt{2} R$ map any integral octonion $X$ of length 1 to two integral octonions whose inner product with $X$ is $1/2$. These give rise to 17,280 distinct Leech lattices $\Lambda \subset \mathbf{O}^3$.

I think with some work we can find a human-readable proof. A clue: the numbers 17,280 and 138,240 featured prominently in Part 5. 17,280 is the number of 7-simplex faces of the $\mathrm{E}_8$ root polytope, while 138,240 is the number of 6-simplex faces that touch both a 7-simplex and a 7-orthoplex!

### Previous parts

Here are the previous parts of this thrilling series:

Posted at December 1, 2014 5:16 AM UTC

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### Re: Integral Octonions (Part 10)

As for Puzzle 5, I feel it should be easy to show that every isometry of the lattice $\mathrm{E}_8 \oplus \mathrm{E}_8 \oplus \mathrm{E}_8$ should be one of the ‘obvious’ ones, namely those in $S_3 \ltimes W(\mathrm{E}_8)^3$. A ‘nonobvious’ isometry would be one that didn’t preserve the direct sum decomposition of this lattice, so it would give a ‘nonobvious’ direct sum decomposition. I feel sure there isn’t any direct sum decomposition of $\mathrm{E}_8 \oplus \mathrm{E}_8 \oplus \mathrm{E}_8$ other than the obvious one. But why? This should be a special case of a much more general result, nothing special to $\mathrm{E}_8$.

Posted by: John Baez on December 5, 2014 4:44 PM | Permalink | Reply to this

### Re: Integral Octonions (Part 10)

This direct sum decomposition reminds me of the spectral decomposition of elements of $\mathfrak{h}_3(\mathbb{O})$. Octonionic quantum mechanics, in a sense, forces one into the Jordan interpretation. Along these lines, it’s not straightforward that all elements of $\mathfrak{h}_3(\mathbb{O})$ have real eigenvalues, due to non-associativity.

Challenge to readers: Prove that all elements of $\mathfrak{h}_3(\mathbb{O})$ have real eigenvalues.

Posted by: Metatron on December 8, 2014 3:40 PM | Permalink | Reply to this

### Re: Integral Octonions (Part 10)

The Frobenius inner product on $\mathfrak{h}_3(\mathbb{O})$ reduces to an inner product, and hence natural norm on $\mathbb{O}^3$, when we use Wilson’s second construction of the octonionic Leech lattice with entries in $B=\frac{1}{2}(1+e_1)A(1+e_1)$, where $A$ is the nonassociative ring of Coxeter-Dickson integral octonions. The stuff in $B$ is not closed under multiplication, however. Are you guys calling elements of $B$ Kirmse integers?

The embedding of the Leech in $\mathbb{O}^3$ and $\mathfrak{h}_3(\mathbb{O})$, surely yields an abundance of riches. If we take Wilson’s construction serious, we can look at it as a state space construction, as $\mathbb{O}^3$ is the state space acted on by the self-adjoint operator algebra $\mathfrak{h}_3(\mathbb{O})$. We can map elements of $\mathbb{O}^3$ to $\mathfrak{h}_3(\mathbb{O})$ via the outer product mapping.

The other technique, mentioned here, is embedding the Leech straight in the off diagonals.

Given a vector in the Leech $\Lambda$, each mapping leads to inequivalent matrices in $\mathfrak{h}_3(\mathbb{O})$. I don’t know which is preferable at this point, as $\mathbb{OP}^2$ is usually constructed with the outer product technique. Yet, for the split octonions, there’s a problem.

Posted by: Metatron on December 5, 2014 5:34 PM | Permalink | Reply to this

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