## November 5, 2014

### Integral Octonions (Part 7)

#### Posted by John Baez

Greg Egan and I have been thinking about Lorentzian lattices, and he just proved something nice about $\mathrm{E}_{10}$: it’s the lattice of $2 \times 2$ self-adjoint matrices with integral octonions as entries!

This might be known already, but I’ve never seen it anywhere. So, let me show you the proof. But first, let me remind you why it matters.

If you missed the earlier parts of this series, you can see polished-up versions on my website:

$\mathrm{E}_8$ is the name of four related things:

• the unique lattice in 8-dimensional Euclidean space that is even (the dot product of any lattice vector with itself is even) and unimodular (the volume of each cell is 1),
• the finite group containing all reflections and rotations that are symmetries of this lattice,
• the simple Lie algebra that has this root lattice, and
• the Lie group that has this Lie algebra.

These things are endlessly absorbing, since they have deep connections to the icosahedron, the topology of 4-manifolds, string theory, and more. $\mathrm{E}_8$ seems to be a wormhole in the fabric of mathematics, a portal that connects the most beautiful, special features of otherwise distant subjects.

For example: since the $\mathrm{E}_8$ lattice lives in 8 dimensions, you can think of the vectors in this lattice as octonions! And if you rescale and rotate it correctly, this lattice becomes closed under octonion multiplication, as well as addition and subtraction. Then we call it the lattice of Cayley integers, or integral octonions:

$\mathbf{O} \subseteq \mathbb{O}$

$\mathbb{O}$ is the octonions, and $\mathbf{O}$ is the integral octonions.

I explained exactly how this works last time, in Part 6.

$\mathrm{E}_{8}$ has a big brother called $\mathrm{E}_{10}$. Again this is the name of four things:

• the unique even unimodular lattice in 10-dimensional Minkowski spacetime,
• the infinite discrete group containing all reflections, rotations and Lorentz transformations that are symmetries of this lattice,
• the infinite-dimensional Lie algebra that has this root lattice, and
• the infinite-dimensional Lie group that has this Lie algebra.

It’s been known for a while that the symmetry group of the $\mathrm{E}_{10}$ lattice has a nice description in terms of integral octonions:

Namely, the even part of this group, the transformations with determinant 1 rather than -1, is $\mathrm{PSL}(2,\mathbf{O})$. Making sense of this group takes bit of work, since the octonions are nonassociative! But the paper above does it.

Recently Greg Egan and I have been thinking about this stuff, and a couple days ago he found something I’d long been hoping for: a description of the $\mathrm{E}_{10}$ lattice itself in terms of integral octonions! It consists of $2 \times 2$ self-adjoint matrices with entries in $\mathbf{O}$.

This is a discrete 10-dimensional analogue of a well-known 4-dimensional fact. Good old 4d Minkowksi spacetime can be seen as consisting of $2 \times 2$ self-adjoint matrices with complex entries. Such matrices look like this:

$T = \left( \begin{array}{cc} t + z & x - i y \\ x + i y & t - z \end{array} \right)$

with $(t,x,y,z) \in \mathbb{R}^4$, and

$-\det (T) = x^2 + y^2 + z^2 - t^2$

is the Minkowski metric. The group $\mathrm{SL}(2,\mathbb{C})$ acts on these $2 \times 2$ self-adjoint matrices via

$T \mapsto g T g^*$

and this action preserves the determinant. So, we get a map from $\mathrm{SL}(2,\mathbb{C})$ to the Lorentz group! The kernel of this is $\pm 1$, so the group

$\mathrm{PSL}(2,\mathbb{C}) = \mathrm{SL}(2,\mathbb{C})/ \{\pm I \}$

is a subgroup of the Lorentz group, and in fact it’s the connected component of the identity:

$\mathrm{PSL}(2,\mathbb{C}) \cong \mathrm{SO}_0(3,1)$

The same sort of thing works for all four normed division algebras, though it gets kind of funky in the noncommutative and nonassociative cases:

$\begin{array}{ccl} \mathrm{PSL}(2,\mathbb{R})& \cong &\mathrm{SO}_0(2,1) \\ \mathrm{PSL}(2,\mathbb{C})& \cong &\mathrm{SO}_0(3,1) \\ \mathrm{PSL}(2,\mathbb{H})& \cong &\mathrm{SO}_0(5,1) \\ \mathrm{PSL}(2,\mathbb{O})& \cong &\mathrm{SO}_0(9,1) . \end{array}$

The same idea should explain why $\mathrm{PSL}(2,\mathbf{O})$ is the even part of the symmetry group of the $\mathrm{E}_{10}$ lattice. But implementing this demands that we see the $\mathrm{E}_{10}$ lattice as consisting of $2 \times 2$ matrices with entries in $\mathbf{O}$, the integral octonions.

We already know that 10-dimensional Minkowski spacetime is isometric to the space of $2 \times 2$ self-adjoint octonionic matrices:

$\mathfrak{h}_2(\mathbb{O}) = \left\{ \; \left( \begin{array}{cc} a & b \\ b^* & c \end{array} \right) \; : \; a,b,c \in \mathbb{O} , \; a = a^*, c=c^* \; \right\}$

with its ‘Minkowski metric’

$- \det \left( \begin{array}{cc} a & b \\ b^* & c \end{array} \right) = |b|^2 - a c \in \mathbb{R} \subseteq \mathbb{O}$

(You see, an octonion that’s its own conjugate can be identified with a real number.)

So, we want:

Theorem. Up to a rescaling, the $\mathrm{E}_{10}$ lattice in 10-dimensional Minkowski spacetime is isometric to the set of $2 \times 2$ self-adjoint matrices with integral octonions as entries: $\mathfrak{h}_2(\mathbf{O}) = \left\{ \; \left( \begin{array}{cc} a & b \\ b^* & c \end{array} \right) \; : \; a,b,c \in \mathbf{O} , \; a = a^*, c=c^* \; \right\}$ viewed as a sublattice of $\mathfrak{h}_2(\mathbb{O})$ with its Minkowski metric.

Proof. Up to isometry, $\mathrm{E}_{10}$ is the unique lattice in 10-dimensional Minkowski spacetime with a basis described by this Dynkin diagram:

meaning that the basis vectors $e_i$ (called ‘simple roots’) have inner products

$e_i \cdot e_i = 2$

and for $i \ne j$,

$e_i \cdot e_j = -1$

when the corresponding dots in the diagram are connected by an edge, but

$e_i \cdot e_j = 0$

when they’re not. So, it suffices to find basis vectors for the lattice $\mathfrak{h}_2(\mathbf{O})$ obeying identical relations up to rescaling, where the inner product is now the one coming from the Minkowski metric on $\mathfrak{h}_2(\mathbb{O})$.

Since the shortest nonzero vectors in $\mathfrak{h}_2(\mathbf{O})$ have length 1, we need to rescale by a factor of $1/\sqrt{2}$. We thus seek 10 basis vectors $v_i$ for the lattice $\mathfrak{h}_2(\mathbf{O})$ obeying

$v_i \cdot v_i = 1$

and for $i \ne j$,

$v_i \cdot v_j = -\frac{1}{2}$

when the corresponding dots in the $\mathrm{E}_{10}$ Dynkin diagram are connected by an edge, but

$v_i \cdot v_j = 0$

otherwise.

Now, just as an octonion that’s its own conjugate can be identified with a real number, an integral octonion that is its own conjugate can be identified with an integer. So, we have

$\mathfrak{h}_2(\mathbb{O}) = \left\{ \; \left( \begin{array}{cc} t + z & v \\ v^* & t-z \end{array} \right) \; : \; v \in \mathbb{O} , \; t, z \in \mathbb{R} \; \right\}$

and

$\mathfrak{h}_2(\mathbf{O}) = \left\{ \; \left( \begin{array}{cc} t + z & v \\ v^* & t-z \end{array} \right) \; : \; v \in \mathbf{O} , \; t+z, t-z \in \mathbb{Z} \; \right\}$

and in this description, the Minkowski metric is given by

$-t^2 + z^2 + |v|^2$

Note that setting $t$ and $z$ to zero gives a way to identify any octonion $v$ with an element of $\mathfrak{h}_2(\mathbb{O})$, namely

$\left( \begin{array}{cc} 0 & v \\ v^* & 0 \end{array} \right)$

This lets us identify any octonionic integer with an element of $\mathfrak{h}_2(\mathbf{O})$. The octonionic integers form a rescaled copy of the $\mathrm{E}_8$ lattice. Since the Dynkin diagram for $\mathrm{E}_8$:

is contained in that for $\mathrm{E}_{10}$, a good first step is to find 8 choices for $v \in \mathbf{O}$ that correspond to rescaled versions of the simple roots for $\mathrm{E}_8$.

To do this, we identify $\mathbb{O}$ with $\mathbb{R}^8$ in such a way that the first component is real and the other seven components are imaginary, and the multiplication table is the one I described last time using this mnemonic:

Given all this, Egan chose these octonions as his rescaled versions of the simple roots for $\mathrm{E}_8$:

$\begin{array} {cccccccccc} v_1 &=& (-\frac{1}{2},& \frac{1}{2}, &-\frac{1}{2},& 0,& -\frac{1}{2},& 0,& 0,& 0) \\ v_2&=&(0,&0,&0,& 0,& 1,& 0,& 0,& 0) \\ v_3&=&(0, &0,& \frac{1}{2},& -\frac{1}{2}, &-\frac{1}{2},& 0,& 0, &-\frac{1}{2}) \\ v_4&=&(0, &0,& 0,& 0, &0, &0, &0,& 1) \\ v_5&=&(0, &0, &0, &\frac{1}{2},& 0,& -\frac{1}{2},& -\frac{1}{2},& -\frac{1}{2}) \\ v_6&=&(0,& 0,& 0,& 0,& 0, &0, &1, &0) \\ v_7&=&(\frac{1}{2},& 0,& -\frac{1}{2},& -\frac{1}{2},& 0,& 0,& -\frac{1}{2},& 0) \\ v_8&=&(0,& 0,& 0, &0, &0, &1,& 0, & 0) \end{array}$

The diligent reader will check that each of these arises by taking a Kirmse integer (a concept explained last time) and swapping its first two coordinates, thus obtaining an element of $\mathbf{O}$. It is easy to see that each of these vectors has norm one:

$v_i \cdot v_i = 1$

But the diligent reader will also use this $\mathrm{E}_8$ Dynkin diagram with its nodes numbered:

and verify that for $i \ne j$ we have

$v_i \cdot v_j = -\frac{1}{2}$

when the corresponding dots in the diagram are connected by an edge, but

$v_i \cdot v_j = 0$

otherwise.

Now we move up to 10 dimensions. We give $\mathbb{R}^{10}$ a Minkowski metric where the first coordinate is timelike:

$(t, z, x_1, \dots, x_8) \cdot (t, z, x_1, \dots, x_8) = - t^2 + z^2 + x_1^2 + \cdots + x_8^2$

This allows us to identify $\mathfrak{h}_2(\mathbb{O})$ and its Minkowski metric with $\mathbb{R}^{10}$ as follows:

$\left( \begin{array}{cc} t + z & x \\ x^* & t-z \end{array} \right) \mapsto (t, z, x_1, \dots, x_8)$

Next, we make the $v_i$ into vectors in $\mathbb{R}^{10}$ by sticking two zero components onto the front of each one. Their inner products are unchanged. Thus, if we can find two more unit spacelike vectors $v_a, v_b \in \mathbb{R}^{10}$ with

$v_a \cdot v_b = -\frac{1}{2}$ $v_b \cdot v_1 = -\frac{1}{2}$

and with all their other inner products with the $v_i$ being zero, we will have a full set of (rescaled) simple roots for $E_{10}$ sitting inside $\mathfrak{h}_2(\mathbf{O})$.

Here are two such vectors:

$\begin{array}{ccccccccccc} v_a &=& (0,& -1, & 0, & 0, & 0, & 0,& 0,& 0,& 0, & 0) \\ v_b &=& (\frac{1}{2} , & \frac{1}{2} , & 0, & -1, & 0, & 0, & 0,& 0, & 0, & 0) \end{array}$

It is obvious that $v_a$ is orthogonal to $v_1,\dots,v_8$, and not too hard to check that $v_b$ is orthogonal to $v_2,\dots,v_8$ and has $v_b \cdot v_1 = -1/2$.

So, now we have rescaled copies of the simple roots for $\mathrm{E}_{10}$, $v_a, v_b, v_1, ..., v_8$, all contained in $\mathfrak{h}_2(\mathbf{O})$. To complete the proof, it suffices to check that they are a basis for the lattice $\mathfrak{h}_2(\mathbf{O})$. To be precise: integer linear combinations of these vectors give the whole lattice.

One basis for $\mathfrak{h}_2(\mathbf{O})$ consists of the matrices

$\left( \begin{array}{cc} 0 & v_i \\ v_i^* & 0 \end{array} \right)$

for $i=1,\dots,8$, along with two more matrices:

$\left( \begin{array}{cc} 1 & 0 \\ 0 & 0 \end{array} \right)$

and

$\left( \begin{array}{cc} -1 & 0 \\ 0 & 1 \end{array} \right)$

These two matrices correspond to the following vectors in $\mathbb{R}^{10}$, respectively:

$\begin{array}{ccccccccccc} v_c &=& (\frac{1}{2} , &\frac{1}{2} ,& 0,& 0,& 0, & 0,& 0,& 0,& 0,& 0) \\ v_a &=& (0,& -1, &0, &0,& 0,& 0,& 0,& 0,& 0, &0) \end{array}$

So, we now have 10 vectors in $\mathbb{R}^{10}$ that correspond to a basis of the lattice $\mathfrak{h}_2(\mathbf{O})$: $v_c, v_a$, and $v_1, ..., v_8$ corresponding to the first eight matrices. The only difference between this basis and the rescaled $\mathrm{E}_{10}$ roots is that this basis includes $v_c$ rather than $v_b$. But in fact we can write:

$v_b = v_c - 2v_1 - 3v_2 - 4v_3 - 5v_4 - 6v_5 - 4v_6 - 2v_7 - 3v_8$

So the matrices that convert back and forth between the bases $\{v_a, v_c, v_1, \dots, v_8\}$ and $\{v_a, v_b, v_1, ..., v_8\}$ both have integer entries, and we are done.   $\blacksquare$

Of course, a more conceptual proof would be nice. I’m sure that will come in time. But for now I’m just basking in the warm glow of knowing this fact is true. It ties together a lot of things about the octonions, string theory in 10 dimensions, $\mathrm{E}_{10}$, and cosmological billiards in 11-dimensional supergravity!

In the current proof, a lot of work happens at the very last step. Personally, I got mixed up when checking this step. If you want to verify that

$v_b = v_c - 2v_1 - 3v_2 - 4v_3 - 5v_4 - 6v_5 - 4v_6 - 2v_7 - 3v_8$

or in other words

$-v_b = -v_c + 2v_1 + 3v_2 + 4v_3 + 5v_4 + 6v_5 + 4v_6 + 2v_7 + 3v_8$

this should help:

$\begin{array}{ccccccccccc} - v_c&=& (-\frac{1}{2}, & -\frac{1}{2}, & 0, & 0, & 0, & 0, & 0, & 0, & 0, & 0) \\ 2 v_1&=&(0, & 0, & -1, & 1, & -1, & 0, & -1, & 0, & 0, & 0) \\ 3 v_2&=&(0, & 0, & 0, & 0, & 0, & 0, & 3, & 0, & 0, & 0) \\ 4 v_3&=&(0, & 0, & 0, & 0, & 2, & -2, & -2, & 0, & 0, & -2) \\ 5 v_4&=&(0, & 0, & 0, & 0, & 0, & 0, & 0, & 0, & 0, & 5) \\ 6 v_5&=&(0, & 0, & 0, & 0, & 0, & 3, & 0, & -3, & -3, & -3) \\ 4 v_6&=&(0, & 0, & 0, & 0, & 0, & 0, & 0, & 0, & 4, & 0) \\ 2 v_7&=&(0, & 0, & 1, & 0, & -1, & -1, & 0, & 0, & -1, & 0) \\ 3 v_8&=&(0, & 0, & 0, & 0, & 0, & 0, & 0, & 3, & 0, & 0) \\ - & - & - & - &- & - &- & - &- & - &- & - \\ -v_b &=& (-\frac{1}{2}, & -\frac{1}{2}, & 0, & 1, & 0, & 0, & 0, & 0, & 0, & 0) \end{array}$

Just add up all the vectors on top and check that you get the one at the very bottom! Of course you should also check that I wrote down these vectors correctly.

Posted at November 5, 2014 4:07 PM UTC

TrackBack URL for this Entry:   https://golem.ph.utexas.edu/cgi-bin/MT-3.0/dxy-tb.fcgi/2777

### Re: Integral Octonions (Part 7)

Here’s a new version of the key step in the proof that makes all the arithmetic go away.

$E_8$ with its usual normalisation is famously a unimodular lattice. That means that if we take a set of simple roots $e_i$, and make them the columns of a matrix $E$, we will have:

$det E = \pm 1$

It follows that the Grammian matrix $G$ of dot products between the simple roots:

$G = E^T E$

has determinant 1. Now, $G$ will have only integer entries (specifically, members of $\{-1,0,2\}$), so with a determinant of 1 its inverse $G^{-1}$ must also have only integer entries. Therefore the vector $b$ defined as:

$b = G^{-1} (-1,0,0,0,0,0,0,0)$

will have only integer entries. Why does that matter? Well, if we’ve ordered our simple roots in the same way as the Dynkin diagram with numbered nodes that John has drawn in the article, what this is telling us is that the integer-linear combination of simple roots:

$f = \sum_i b_i e_i$

will have:

$f \cdot e_1 = -1$

and for $i\neq 1$

$f \cdot e_i = 0$

In other words, we have shown that there is a point $f$ in the $E_8$ lattice itself that has the dot products we want in order to start extending the lattice towards the ten-dimensional version $E_{10}$.

Of course our $f$ is an element of $\mathbb{R}^8$, and it uses the wrong normalisation for our present purposes where the lattice vectors have squared norm 1 rather than 2, but we can define:

$g = \sum_i b_i v_i$

where the $v_i$ are now construed as vectors in $\mathbb{R}^{10}$ and have the correct normalisation. The crucial point is that the $b_i$ are integers.

The 10-dimensional $v_i$ form a basis of $\mathfrak{h}_2(\mathbf{O})$, along with:

$\begin{array}{ccccccccccc} v_c &=& (\frac{1}{2} , &\frac{1}{2} ,& 0,& 0,& 0, & 0,& 0,& 0,& 0,& 0) \\ v_a &=& (0,& -1, &0, &0,& 0,& 0,& 0,& 0,& 0, &0) \end{array}$

So we can simply define:

$v_b = v_c + g = v_c + \sum_i b_i v_i$

Now it is plain that $v_b$ is an integer-linear combination of the basis vectors of $\mathfrak{h}_2(\mathbf{O})$ that has the correct dot products with $v_a$ and $v_1,...,v_8$ in order for $v_a, v_b, v_1, .., v_8$ to comprise a set of simple roots whose Dynkin diagram is that of $E_{10}$.

Posted by: Greg Egan on November 6, 2014 3:13 AM | Permalink | Reply to this

### Re: Integral Octonions (Part 7)

Ah, I missed one key point that doesn’t come quite so easily.

It’s crucial that not only does the root $v_b$ have the correct dot products with other roots, it also needs to have the correct norm! In the terms of my previous comment, what we need is that:

$f\cdot f=2$

From the way we have obtained $f$, this is equivalent to saying:

$(-1,0,0,0,0,0,0,0)^T G^{-1} (-1,0,0,0,0,0,0,0) = 2$

or just:

$(G^{-1})_{1,1}=2$

This is in fact the case for the matrix $G$ we get from the dot products of the simple roots of the $E_8$ lattice, but it’s not obvious in the way that the determinant of $G$ being 1 is obvious.

Posted by: Greg Egan on November 6, 2014 3:54 AM | Permalink | Reply to this

### Re: Integral Octonions (Part 7)

OK, here’s a way to complete the minimal-arithmetic version of the proof, albeit with a little more arithmetic than I was hoping for.

One obvious set of simple roots of the $E_8$ lattice is given by the rows of the matrix:

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

Note that the pattern of signs is uniform for the first five rows, but there’s a change for the sixth and seventh that allows the Dynkin diagram to branch at the fifth node.

There’s an obvious candidate for my vector $f$ with respect to these simple roots. I’ll just stick it at the start of the matrix, for the sake of easily seeing what its dot products with the simple roots are:

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

It’s pretty easy to check that the first row of this matrix has a dot product of -1 with the second row, and dot products of zero with all the rest. It’s also clear that its squared norm is 2.

But we already know that there is a unique vector $f$ with these eight dot products, and that it is an integer-linear combination of the simple roots $e_i$, given by:

$f = \sum_i b_i e_i$ $b = G^{-1}(-1,0,0,0,0,0,0,0)$

So what we’ve shown is that, not only is $f$ a point in the $E_8$ lattice, it is a point with squared norm 2, which is what we need to make the revised proof complete.

Posted by: Greg Egan on November 6, 2014 5:40 AM | Permalink | Reply to this

### Re: Integral Octonions (Part 7)

Nice! When I began reading your first comment and saw only the top:

Here’s a new version of the key step in the proof that makes all the arithmetic go away.

I thought “oh no, in the very first comment some smart-aleck wants to show he can easily do a much better proof than Egan’s. No ‘wow, that’s cool!’, no thanks, no nothing.”

Posted by: John Baez on November 6, 2014 9:13 PM | Permalink | Reply to this

### Re: Integral Octonions (Part 7)

Nice work. Another tie in with supergravity is using $J(2,\mathbb{O})$ elements to close a set of Gamma matrices in D<11. (This can be done quite easily with $SO(9)$ generators, as well as the split case which makes use of $SO(5,4)$ split octonion 2x2 generators.) Duff and his team were able to use 4x4 octonion matrices to close a Gamma set in D=11. (I was able to close a different set in D=11, by using the full complex octonions. This gives a 2-time signature.) However, 4x4’s over $\mathbb{O}$ have obvious Jordan algebraic shortcomings.

Lifting to $J(3,\mathbb{O})$, your $E_{10}$ findings, by triality, should fit nicely in D=27. I’m sure there’s a split octonion analog of your construction (both of which embed in $J(3,\mathbb{O}\times\mathbb{C})$ in integral form). Both would be transformed with $F_4$ or $F_{4(4)}$ in $F_4(\mathbb{C})$, as can the Leech.

An octonionic $E_11$ lattice construction should point the way to the proper octonionic parametrization for M-theory.

Posted by: Metatron on November 6, 2014 5:38 PM | Permalink | Reply to this

### Re: Integral Octonions (Part 7)

I find existing octonionic constructions of the Leech lattice hard to understand. The one that I understand better, I have trouble believing. How well do you understand this? Maybe sometime Greg and I should tackle this issue someday, since it involves going up from $2 \times 2$ self-adjoint octonionic matrices to $3 \times 3$’s. Even if existing constructions work fine and can’t be improved, I may need to redo them myself to really understand them.

Posted by: John Baez on November 7, 2014 6:47 PM | Permalink | Reply to this

### Re: Integral Octonions (Part 7)

The existing constructions are not fully satisfactory, from the matrix perspective. A truly Jordan algebraic construction would be handy. Before tackling the Leech, an $E_11$ lattice construction from Hermitian octonionic matrices would be of interest. Have you tried this yet? I’m thinking a 3x3 construction would do the job, with orthogonality of the matrices given by the Frobenius inner product $\langle X, Y \rangle=\mathrm{tr}(X\circ Y^{\ast})$ and using the third real diagonal entry in $J(3,\mathbb{O})$.

Posted by: Metatron on November 8, 2014 9:31 AM | Permalink | Reply to this

### Re: Integral Octonions (Part 7)

Metatron wrote:

The existing constructions are not fully satisfactory, from the matrix perspective.

Can you say why not?

Before tackling the Leech, an $\mathrm{E}_{11}$ lattice construction from Hermitian octonionic matrices would be of interest. Have you tried this yet?

No, we just did $\mathrm{E}_{10}$ last week and I’ve never thought about $\mathrm{E}_{11}$ at all. If you can point me to some clear descriptions of the $\mathrm{E}_{11}$ lattice I can try to cook up an octonionic description along the lines you suggest.

Posted by: John Baez on November 9, 2014 12:04 AM | Permalink | Reply to this

### Re: Integral Octonions (Part 7)

Okay, I think I have a good enough description of $\mathrm{E}_{11}$ to tackle this project: it’s just like $\mathrm{E}_{10}$ but with an extra dot on the longest branch of the Dynkin diagram.

Posted by: John Baez on November 9, 2014 7:06 AM | Permalink | Reply to this

### Re: Integral Octonions (Part 7)

This is also helpful: E11, SL(32) and central charges.

The $K_27$ Kac-Moody algebra also looks interesting. See pg. 18 of hep-th/0104081 and pg. 66 of the thesis you cited.

An off-diagonal $J(3,\mathbb{O})$ representation of the Leech would be more satisfactory.

Posted by: Metatron on November 9, 2014 11:50 AM | Permalink | Reply to this

### Re: Integral Octonions (Part 7)

As motivation for $K_27$, it’s stated on pg. 19 of hep-th/0104081 that, “It is interesting to note that the algebra $K_27$ contains the algebra $E_11$ and one might take this as an indication that the closed bosonic string contains the known superstrings in ten dimensions and in effect M theory. The closed bosonic string on a torus is invariant under the fake monster Lie algebra and it would be interesting to ask if $K_27$ was contained in this algebra.”

Challenge: construct the $K_27$ lattice using $J(3,\mathbb{O})$.

Posted by: Metatron on November 10, 2014 5:37 PM | Permalink | Reply to this

### Re: Integral Octonions (Part 7)

Metatron wrote:

Challenge: construct the $K_27$ lattice using $J(3,\mathbb{O})$.

If you allow me to wallow in the muck of 26 dimensions a bit before soaring to 27, here are some comments.

In a certain classical limit, bosonic string theory reduces to gravity coupled to a massless 2-form field 26-dimensional spacetime. I believe, but I’m not sure, that when we compactify this theory down to 3 spacetime dimensions, we get a theory that exhibits the ‘cosmological billiard’ effect as we approach a spacelike singularity—i.e., the Big Bang. And I believe that in this case the billiard ball bounces around in the 25-dimensional hyperbolic space modulo the Lorentzian Coxeter group $\mathrm{D}_{24}^{++}$, also known as $DE_{26}$.

The relevant lattice lives in 26-dimensional Minkowski spacetime. It’s

$\mathrm{D}_{24}^{++} = \mathrm{D}_24 \oplus \mathrm{H}$

Here $\mathrm{D}_24$ is the lattice of vectors in $\mathbb{R}^{24}$ with integer components that sum to an even integer—a ‘checkerboard’ pattern in 24 dimensions. $\mathrm{H}$ is the 2-dimensional Lorentzian lattice described in another comment.

So, we can clearly stuff $\mathrm{D}_{24}^{++}$ into the 26-dimensional space of traceless $3 \times 3$ octonion matrices. $\mathrm{H}$ fits in the diagonal and $\mathrm{D}_24$ fits in the off-diagonal spots.

But what I’ve said so far gives no indication that this is a really good idea. The off-diagonal part of the traceless $3 \times 3$ integral octonion matrices is a copy of

$\mathrm{E}_8 \oplus \mathrm{E}_8 \oplus \mathrm{E}_8$

which is a different lattice than $\mathrm{D}_{24}$. They’re two of the 24 even unimodular lattices in 24 dimensions.

There is a faint glimmer of hope, though. The Dynkin diagram for $\mathrm{D}_{24}^{++}$ contains the Dynkin diagram for $\mathrm{E}_8$! It looks like a row of 22 dots, together with 2 extra dots. One of these is connected to the third dot in the row, while one is connected to the second-to-last dot in the row. So, the tail end of $\mathrm{D}_{24}^{++}$ looks like a typical $\mathrm{D}_n$ Dynkin diagram, while the head looks like $\mathrm{E}_8$. Hence the name $DE_{26}$.

I believe the so-called $\mathrm{K}_{27}$ Dynkin diagram is this $DE_{26}$ diagram with one extra dot. Is that correct?

Posted by: John Baez on November 11, 2014 7:28 PM | Permalink | Reply to this

### Re: Integral Octonions (Part 7)

Indeed. For the Coxeter construction, we have:

$K_{27}\equiv D_{24}^{+++}$

and

$K_{27}\equiv E_{11}+D_{16}$.

Posted by: Metatron on November 12, 2014 8:27 AM | Permalink | Reply to this

### Re: Integral Octonions (Part 7)

Considerable commentaries in:

• Augustin-Liviu Mare and Matthieu Willems, Topology of the octonionic flag manifold, Münster J. Math. 6 (2013), 485.

Posted by: Sabino Guillermo Echebarria Mendieta on November 7, 2014 11:59 AM | Permalink | Reply to this

### Re: Integral Octonions (Part 7)

Greg and I came up with yet another proof.

Theorem. Up to a rescaling, the $\mathrm{E}_{10}$ lattice in 10-dimensional Minkowski spacetime is isometric to $\mathfrak{h}_2(\mathbf{O}) = \left\{ \; \left( \begin{array}{cc} a & b \\ b^* & c \end{array} \right) : \; b \in \mathbf{O} , \; a,c \in \mathbb{Z} \; \right\}$ the lattice of $2 \times 2$ self-adjoint integral octonion matrices in the space $\mathfrak{h}_2(\mathbb{O})$ with its Minkowski metric.

Proof. This proof relies on two well-known but fairly heavy-duty facts:

• up to isometry, $\mathrm{E}_8$ is the unique even unimodular lattice in the Euclidean space $\mathbb{R}^8$.

• up to isometry, $\mathrm{E}_{10}$ is the unique even unimodular lattice in the Minkowski spacetime $\mathbb{R}^{10}$.

We note that $\mathfrak{h}_2(\mathbb{O})$ is the direct sum of the lattice

$L = \left\{ \; \left( \begin{array}{cc} 0 & b \\ b^* & 0 \end{array} \right) : \; b \in \mathbf{O} \; \right\}$

and the lattice

$M = \left\{ \; \left( \begin{array}{cc} a & 0 \\ 0 & c \end{array} \right) : \; a, c \in \mathbb{Z}\; \right\}$

All vectors in $L$ are orthogonal to those in $M$, so this is an orthogonal direct sum

$\mathfrak{h}_2(\mathbf{O}) = L \oplus M$

$L$ is isometric to the lattice of Cayley integers, so as well-known (and shown in Egan’s original proof) when we rescale it by a factor of $\sqrt{2}$ we get the $\mathrm{E}_8$ lattice in an 8-dimensional Euclidean space. When we rescale $M$ by a factor of $\sqrt{2}$, we get an even unimodular lattice $\mathrm{H}$ in a 2-dimensional Minkowski spacetime.

So, after rescaling by a factor of $\sqrt{2}$, $\mathfrak{h}_2(\mathbf{O})$ is isometric to

$\mathrm{E}_8 \oplus \mathrm{H}$

A direct sum of even lattices is even, and the direct sum of unimodular lattices is unimodular. So, this direct sum is an even unimodular lattice in a 10-dimensional Minkowski spacetime, and thus

$\mathrm{E}_8 \oplus \mathrm{H} \cong \mathrm{E}_{10}$

So, after rescaling, $\mathfrak{h}_2(\mathbf{O})$ is isometric to $\mathrm{E}_{10}$.   $\blacksquare$

Note that we didn’t really need that $\mathrm{E}_8$ was the unique even unimodular lattice in 8d Euclidean space, but we did need the uniqueness for $\mathrm{E}_{10}$.

The advantage of this proof is that it’s short and slick. The advantage of the original proof is that it doesn’t rely on heavy-duty results, and it gives us an explicit set of simple roots for $\mathrm{E}_{10}$ sitting inside $\mathfrak{h}_2(\mathbf{O})$. This will be nice when we try to apply this theorem.

Posted by: John Baez on November 7, 2014 4:16 PM | Permalink | Reply to this

### Re: Integral Octonions (Part 7)

By the way, I’ll point out something that puzzled me for a while in Unimodular lattice: classification. Paraphrasing, they say:

For indefinite lattices, the classification is easy to describe. Write $\mathbb{R}^{m,n}$ for the $m+n$ dimensional vector space $\mathbb{R}^{m+n}$ with the inner product of $(a_1, \dots, a_{m+n})$ and $(b_1, \dots, b_{m+n})$ given by

$a_1 b_1 + \cdots + a_m b_m - a_{m+1} b_{m+1} - \cdots - a_{m+n} b_{m+n}$

This is inner product is indefinite when $m, n > 0$.

In $\mathbb{R}^{m,n}$ where $m, n >0$ there is one odd unimodular lattice up to isomorphism, denoted by $\mathrm{I}_{m,n}$, which is given by all vectors $(a_1, \dots, a_{m+n})$ whose coordinates are all integers.

There are no even unimodular lattices unless $m - n$ is divisible by 8, and in this case when $m , n > 0$ there is a unique example up to isomorphism, denoted by $\mathrm{II}_{m,n}$. This is given by all vectors $(a_1, \dots , a_{m+n})$ such that either all the coordinates $a_i$ are integers or they are all integers plus 1/2, and their sum is even.

Their lattice $\mathrm{II}_{1,1}$ different from the 2-dimensional even unimodular lattice I called $\mathrm{H}$… but isomorphic, as the uniqueness requires.

If we look at all $(z,t)$ such that either both $z,t$ are integers or both are integers plus 1/2, and their sum is even, we get

$\{ (z,t) : \; z,t \in \mathbb{Z}, \; t,z \; even \}$

This has a basis given by $(1,1)$ and $(1/2,-1/2)$.

My lattice

$\mathrm{H} = \{ \sqrt{2} (z,t) : \; t+z , t-z \in \mathbb{Z} \}$

has a basis given by $(1/\sqrt{2}, 1/\sqrt{2})$ and $(1/\sqrt{2}, -1/\sqrt{2})$.

But these two bases are related by a Lorentz transformation!

Posted by: John Baez on November 7, 2014 7:55 PM | Permalink | Reply to this

### Re: Integral Octonions (Part 7)

Okay, Metatron posed a challenge:

Challenge: construct the $\mathrm{K}_{27}$ lattice using $J(3,\mathbb{O})$.

and I think I’ve made some progress. More or less by definition, the $\mathrm{K}_{27}$ lattice is the $\mathrm{D}^{+++}_{24}$ lattice, the ‘very extended’ version of the $\mathrm{D}_{24}$ lattice. The process of ‘very extension’ is described quite concretely here:

The idea is that you start with a lattice $L$ in $n$-dimensional Euclidean space and get a lattice $L^{+++}$ in $(n+3)$-dimensional Minkowski spacetime. To do this we use $II_{1,1}$, which is the unique (up to isometry) even unimodular lattice in 2-dimensional Minkowski spacetime. I gave a couple of coordinatizations of it here, but Gaberdiel et al use the first: $II_{1,1}$ is the lattice of vectors $(z,t)$ where $z,t$ are either both integers or both integers plus $\frac{1}{2}$, and $z + t$ is even.

There are two future-pointing lightlike vectors that generate the lattice $L$. Their sum is timelike vector, say $v$. (These vectors are $(-\frac{1}{2},\frac{1}{2})$ and $(1,1)$, but it doesn’t really matter now.)

Gaberdiel et al define $L^{+++}$ to be the sublattice of

$L \oplus II_{1,1} \oplus II_{1,1}$

consisting of vectors that are orthogonal to the vector $v$ in the second copy of $II_{1,1}$. Since this vector is timelike, $L^{+++}$ is an $(n+3)$-dimensional Lorentzian lattice.

My claim is now this: suppose $L$ is any even unimodular lattice in $n$-dimensional Euclidean space (where $n$ is necessarily a multiple of 8). Then $L^{++} = L \oplus II_{1,1}$ is an even unimodular lattice in $n+2$-dimensional Minkowski spacetime. There is a unique such thing, up to isometry, so up to isometry $L^{++}$ is independent of which even unimodular $n$-dimensional lattice $L$ we chose! The same is true for $L^{+++}$, since $L^{+++}$ depends only on $L^{++}$.

Given this, $\mathrm{D}_{24}^{+++}$ is isometric to

$(\mathrm{E}_8 \oplus \mathrm{E}_8 \oplus \mathrm{E}_8)^{+++}$

since both $\mathrm{D}_{24}$ and $\mathrm{E}_8 \oplus \mathrm{E}_8 \oplus \mathrm{E}_8$ are even unimodular lattices in 24-dimensional Euclidean space.

I believe there is a natural way to identify $(\mathrm{E}_8 \oplus \mathrm{E}_8 \oplus \mathrm{E}_8)^{+++}$ with the lattice $\mathfrak{h}_3(\mathbf{O})$ of $3 \times 3$ self-adjoint matrices with integral octonions as entries. I need to check this, but the point is that the 3 off-diagonal entries $x \in \mathfrak{h}_3(\mathbf{O})$ are integral octonions, which we can identify with points in the $\mathrm{E}_8$ lattice, while the three diagonal entries are integers, which I want to identify with a vector in $s II_{1,1} \oplus II_{1,1}$ that is orthogonal to $v$.

I believe there is a nice Minkowski metric on the exceptional Jordan algebra $\mathfrak{h}_3(\mathbb{O})$, such that the identification of $(\mathrm{E}_8 \oplus \mathrm{E}_8 \oplus \mathrm{E}_8)^{+++}$ with $\mathfrak{h}_3(\mathbf{O})$ becomes an isometry.

Given this, we get an isometry between $\mathrm{D}_{24}^{+++}$ and $\mathfrak{h}_3(\mathbf{O})$.

Posted by: John Baez on November 12, 2014 8:34 PM | Permalink | Reply to this

### Re: Integral Octonions (Part 7)

For this construction, what is the Minkowski metric you have in mind?

Posted by: Metatron on November 13, 2014 12:39 PM | Permalink | Reply to this

### Re: Integral Octonions (Part 7)

That’s the problem, of course.

The ‘reduced structure group’ of the Jordan algebra $\mathfrak{h}_2(\mathbb{O})$ is $SO_0(9,1)$, so that Jordan algebra has a very natural Minkowski metric: the determinant. But there is no Minkowski metric on $\mathfrak{h}_3(\mathbb{O})$ that’s invariant under the whole reduced structure group of this Jordan algebra, which is a certain noncompact real form of $\mathrm{E}_6$. (See the chart on page 9 here), and earlier for more on the concept of structure group.)

However, there’s a 2-parameter family of Minkowski metrics on $\mathfrak{h}_3(\mathbb{O})$ that are invariant under a smaller group: the automorphism group of this Jordan algebra, which is the compact real form of $\mathrm{F}_4$. So, I’m trying to pick one of these such that with this metric, the lattice $\mathfrak{h}_3(\mathbf{O}) \subset \mathfrak{h}_3(\mathbb{O})$ becomes isometric to

$(\mathbf{E}_8 \oplus \mathbf{E}_8 \oplus \mathbf{E}_8)^{+++} \cong D_{24}^{+++}$

Getting the off-diagonal part of $\mathfrak{h}_3(\mathbf{O})$ to be isometric to $\mathbf{E}_8 \oplus \mathbf{E}_8 \oplus \mathbf{E}_8$ is certainly possible. Achieving this reduces the degrees of freedom to just one. So I’ve got a one-parameter family of Minkowski metrics on $\mathfrak{h}_3(\mathbb{O})$ to get the diagonal entries to work out right: we need them to form a lattice isometric to

$H \oplus \{ u \in H : u \perp v \}$

where $H$ and $v$ are as described earlier.

It’s very suspenseful right now, but I’ve got to grade homework. I’ll report back when I get it figured out.

Posted by: John Baez on November 13, 2014 4:20 PM | Permalink | Reply to this

### Re: Integral Octonions (Part 7)

Sounds great. The Frobenius inner product on $J(3,\mathbb{O})=\mathfrak{h}_3(\mathbb{O})$ induces a metric that comes close. However, one doesn’t get any minus signs to split the signature. But if one works within the larger $\mathfrak{h}_3(\mathbb{C}\times\mathbb{O})$, one can have imaginary units on the diagonal, giving $(26,1)$, $(25,2)$ and $(24,3)$ signatures, based on the number of $i$’s, respectively.

As $\mathrm{tr}(X\circ X^{\ast})=\mathrm{tr}(X^2)$ is preserved by automorphisms of $\mathfrak{h}_3(\mathbb{C}\times\mathbb{O})$, this might just do the trick. Yet, this does fall outside $\mathfrak{h}_3(\mathbb{O})$ just a bit. Another real, non-compact form of $F_4$ should be applicable here.

Posted by: Metatron on November 13, 2014 7:00 PM | Permalink | Reply to this

### Re: Integral Octonions (Part 7)

It seems to work!

Let $\mathfrak{h}_3(\mathbb{O})$ be the exceptional Jordan algebra, consisting of $3 \times 3$ self-adjoint octonionic matrices. There is a unique Lorentzian inner product

$g : \mathfrak{h}_3(\mathbb{O}) \times \mathfrak{h}_3(\mathbb{O}) \to \mathbb{R}$

such that

$g(x,x) = tr(x^2) - tr(x)^2$

Let $\mathfrak{h}_3(\mathbf{O}) \subset \mathfrak{h}_3(\mathbb{O})$ be the lattice of $3 \times 3$ self-adjoint matrices having integral octonions as entries.

Theorem. As a lattice in $\mathfrak{h}_3(\mathbb{O})$ with Lorentzian inner product $g$, $\mathfrak{h}_3(\mathbf{O})$ is isometric to the lattice $\mathrm{D}_{24}^{+++}$ in 27-dimensional Minkowski spacetime.

Proof. It’s sufficent to prove $\mathfrak{h}_3(\mathbf{O})$ is isometric to the lattice $(\mathrm{E}_8 \oplus \mathrm{E}_8 \oplus \mathrm{E}_8)^{+++}$, because we have already seen there is an isometry

$\mathrm{D}_{24}^{+++} \cong (\mathrm{E}_8 \oplus \mathrm{E}_8 \oplus \mathrm{E}_8)^{+++}$

Take any element $x \in \mathfrak{h}_3(\mathbf{O})$:

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

where $a,b,c \in \mathbb{R}, X,Y,Z \in \mathbb{O}$. Then

$tr(x^2) =a^2 + b^2 + c^2 + 2(X X^* + Y Y^* + Z Z^*)$

while

$tr(x)^2 = (a + b + c)^2$

Thus

$\begin{array}{ccl} g(x,x) &=& tr(x^2) - tr(x)^2 \\ &=& 2(X X^* + Y Y^* + Z Z^*) - 2(a b + b c + c a) \end{array}$

It follows that the diagonal matrices are orthogonal to the off-diagonal matrices. An off-diagonal matrix $x \in \mathfrak{h}_3(\mathbf{O})$ is a triple $(X,Y,Z) \in \mathbf{O}^3$, and has

$g(x,x) = 2(X X^* + Y Y^* + Z Z^*)$

Thanks to the factor of 2, this makes the lattice of these off-diagonal matrices isometric to $\mathrm{E}_8 \oplus \mathrm{E}_8 \oplus \mathrm{E}_8$. (See Egan’s argument in the blog article for a proof that twice the usual inner product on octonions makes $\mathbf{O}$ into a lattice isometric to the even unimodular lattice $\mathrm{E}_8$.)

Recall how the ‘triple plus’ construction works:

$(\mathrm{E}_8 \oplus \mathrm{E}_8 \oplus \mathrm{E}_8)^{+++} \cong \mathrm{E}_8 \oplus \mathrm{E}_8 \oplus \mathrm{E}_8 \oplus H \oplus \{u \in H : u \perp v \}$

where $H$ is the Lorentzian even unimodular lattice in 2 dimensions and $v$ is the sum of two future-pointing lightlike vectors generating $H$.

It thus suffices to show that the 3-dimensional Lorentzian lattice of diagonal matrices in $\mathfrak{h}_3(\mathbf{O})$ is isometric to

$H \oplus \{u \in H : u \perp v \}$

A diagonal matrix $x \in \mathfrak{h}_3 (\mathbf{O})$ is a triple $(a,b,c) \in \mathbb{Z}^3$, and on these triples the inner product $g$ is given by

$g(x,x) = -2(a b + b c + c a)$

If we restrict attention to triples of the form $x = (a,b,0)$, we get a 2-dimensional Lorentzian lattice: a copy of $\mathbb{Z}^2$ with inner product

$g(x,x) = -2a b$

By results we’ve seen earlier, this lattice is isometric to $\mathrm{H}$. (Or, check directly that it’s even and unimodular.)

This is a big step toward showing that the lattice of all triples $(a,b,c) \in \mathbb{Z}^3$, with the inner product $g$, is isometric to

$\mathrm{H} \oplus \{u \in \mathrm{H} : u \perp v \}$

We can identify $\mathrm{H}$ with $\mathbb{Z}^2$ using the aforementioned inner product on $\mathbb{Z}^2$. We can take $(1,0)$ and $(0,1)$ as two future-pointing light-like vectors generating $\mathbb{Z}^2$, so that $v = (1,1)$ is timelike.

Note that the lattice $\mathrm{H} \oplus \{u \in \mathrm{H} : u \perp v \}$ is generated by the first copy of $\mathrm{H}$ together with a vector $u_0$ in the second copy of $\mathrm{H}$ that is orthogonal to the timelike vector $v$ and has the shortest possible length. So, we can write

$\mathrm{H} \oplus \{u \in \mathrm{H} : u \perp v \} = \mathrm{H} \oplus \langle u_0 \rangle$

In the picture where we identify $\mathrm{H}$ with $\mathbb{Z}^2$, we can take $u_0$ to be $(1,-1)$. Note that

$g(u_0,u_0) = 2$

so $u_0$ has length $\sqrt{2}$.

Indeed, the inner product on $\mathrm{H} \oplus \langle u_0 \rangle$ is fixed by the inner product on the first copy of $H$, the fact that $u_0$ is orthogonal to all vectors in $\mathrm{H}$, and the fact that $u_0$ has length $\sqrt{2}$.

So, it suffices to show the lattice of all triples $(a,b,c) \in \mathbb{Z}^3$ is generated by those of the form $(a,b,0)$ together with a spacelike vector with length $\sqrt{2}$ that is orthogonal to all those of the form $(a,b,0)$.

To work out the inner product $g$ on $\mathbb{Z}^3$ more explicitly, we need to use the polarization identity

$g(x,x') = \frac{1}{2}( g(x+x',x+x') - g(x,x) - g(x',x'))$

Remember, if $x = (a,b,c)$ we have

$g(x,x) = -2(a b + b c + c a)$

So, if we also have $x' = (a',b',c')$, the polarization identity gives

$g(x,x') = -(a b'+a' b) - (b c'+ b c') - (c a' + c'a)$

We are looking for a spacelike vector $x' = (a',b',c')$ that is orthogonal to all those of the form $x = (a,b,0)$. For this, it is necessary and sufficient to have

$0 = g((1,0,0),(a',b',c')) = - b' - c'$

and

$0 = g((0,1,0), (a',b',c')) = - a' - c'$

An example is $x' = (1,1,-1)$. This has

$g(x',x') = -2(1 - 1 - 1) = 2$

so it is spacelike with length $\sqrt{2}$. Even better, this vector $x'$, along with those of the form $(a,b,0)$, generates the lattice $\mathbb{Z}^3$.

So we have shown what we needed to show: the lattice of all triples $(a,b,c) \in \mathbb{Z}^3$ is generated by those of the form $(a,b,0)$ together with a spacelike vector with length $\sqrt{2}$ that is orthogonal to all those of the form $(a,b,0)$. $\blacksquare$

Posted by: John Baez on November 14, 2014 1:40 AM | Permalink | Reply to this

### Re: Integral Octonions (Part 7)

Astonishing and elegant result indeed. And with your choice of inner product, $\mathfrak{h}_3(\mathbb{O})$ elements satisfying $X^2=X$ (i.e. projectors) are automatically lightlike. I’d like to think one can relate the 27-dimensional picture to the 11-dimensional one by a natural compactification via such lightlike elements.

Posted by: Metatron on November 14, 2014 5:58 PM | Permalink | Reply to this

### Re: Integral Octonions (Part 7)

Thanks for the praise! It was nail-bitingly suspenseful. There is no ‘wiggle room’ in choosing a Lorentzian metric on the exceptional Jordan algebra $\mathfrak{h}_3(\mathbb{O})$ if we want it to be invariant under automorphisms of this algebra and make the lattice $\mathfrak{h}_3(\mathbf{O})$ be isometric to $\mathrm{D}_{24}^{+++}$.

I hadn’t thought about the meaning of the Lorentzian metric I used. It’s very cool that all projectors work out to be lightlike. This is also how it works in the usual determinant Lorentzian metric for $\mathfrak{h}_2(\mathbb{R}) \cong \mathbb{R}^{2,1}$, $\mathfrak{h}_2(\mathbb{C})\cong \mathbb{R}^{3,1}$, $\mathfrak{h}_2(\mathbb{H})\cong \mathbb{R}^{5,1}$ and $\mathfrak{h}_2(\mathbb{O})\cong \mathbb{R}^{9,1}$.

So, thanks for pointing this out. I now feel this result is on the right track. (I’m not sure where the track leads, though.)

Posted by: John Baez on November 14, 2014 10:12 PM | Permalink | Reply to this

### Re: Integral Octonions (Part 7)

Hopefully this sheds light on a matrix version of the hypothetical bosonic M-theory. The Lorentzian inner product you invoke is very natural to anyone expanding the Freudenthal cross product for a single element: $X^{#}=X\times X=X^2-\mathrm{tr}(X)X+\frac{1}{2}(\mathrm{tr}(X)^2-\mathrm{tr}(X^2))I$. Vanishing of this product defines rank one element, and the projectors are in this family.

The automorphism group $F_4$ contains generators of the Leech lattice automorphism group so there is more to be explored along these lines as well. The big picture is still murky, however.

Posted by: Metatron on November 15, 2014 1:01 AM | Permalink | Reply to this

### Re: Integral Octonions (Part 7)

John wrote:

It’s very cool that all projectors work out to be lightlike.

Actually it’s not true that every projector is lightlike: only those of trace 1. See this comment for more detail.

But that’s fine. The projectors of trace 1 and 2 in $\mathfrak{h}_3(\mathbb{O})$ correspond to the points and lines in the octonionic projective plane, respectively, and a point $p$ lies on a line $\ell$ iff the projector for $p$ plus the projector for some other point equals the projector for $\ell$.

Posted by: John Baez on November 15, 2014 8:51 PM | Permalink | Reply to this

### Re: Integral Octonions (Part 7)

Would this work well for Garrett Lisi, so that he could put each generation in one of those E8?

Posted by: Daniel Rocha on November 15, 2014 5:43 AM | Permalink | Reply to this

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