Contents

# Contents

This is a record of some calculations done by John Baez, Greg Egan and John Huerta around November 2015.

## 9+1-dimensional geometry

The behavior of spinors depends heavily on the dimension of space, or spacetime, modulo 8. For example, in spacetimes of any dimension that’s 2 more than a multiple of 8 there exist ‘Majorana–Weyl spinors’: spin-1/2 particles that have an intrinsic handedness (that’s the ‘Weyl’ part) and are their own antiparticles (that’s the ‘Majorana’ part).

In 10d Minkowski spacetime something special happens: both kinds of Majorana–Weyl spinors can be described using octonions. Mathematically this originates from the fact that there are two representations of $Spin(9,1)$, called the left-handed and right-handed Majorana–Weyl spinor representations, which can both be identified with $\mathbb{O}^2$, the space of pairs of octonions. This in turn follows from a simpler fact: 10-dimensional Minkowski spacetime can be identified with $\mathfrak{h}_2(\mathbb{O})$, the space of self-adjoint $2 \times 2$ octonionic matrices.

All this is very similar to something that happens in 4 dimensions. 4d Minkowski spacetime can be identified with $\mathfrak{h}_2(\mathbb{C})$, the space of self-adjoint $2 \times 2$ complex matrices, and $Spin(3,1)$ has two representations on $\mathbb{C}^2$, the left-handed and right-handed Weyl spinor representations. All this should be familiar to students of particle physics. The 10d case works essentially the same way: we just replace complex numbers with octonions. Of course the octonions are noncommutative and nonassociative, but it doesn’t really cause a problem here.

For a careful treatment of all this, try Section 2 here:

• John C. Baez, John Huerta, Division algebras and supersymmetry I, in Superstrings, Geometry, Topology, and C${}^\ast$-algebras, eds. R. Doran, G. Friedman and J. Rosenberg, Proc. Symp. Pure Math. 81, AMS, Providence, 2010, pp. 65–80.

Here we will start with four vector spaces, that will turn out to be representations of $Spin(9,1)$:

• $\mathbb{R}$ is the real numbers: the trivial representation of $Spin(9,1)$, which physicists call the scalar representation.

• $S_-$ is $\mathbb{O}^2$, treated as the left-handed Majorana-Weyl spinor representation of $Spin(9,1)$.

• $S_+$ is $\mathbb{O}^2$ treated as right-handed Majorana-Weyl spin representation, the dual of $S_-$.

• $V = \mathfrak{h}_2(\mathbb{O})$ is the space of $2 \times 2$ self-adjoint octonionic matrices: the vector representation of $Spin(9,1)$.

Concretely, we can write $v \in V$ as

$v = \left( \begin{array}{cc} \beta & x \\ x^\ast & \gamma \end{array} \right)$

with $\beta, \gamma \in \mathbb{R}$ and $x \in \mathbb{O}$. Playing a key supporting role in the following algebra is the operation of trace reversal:

$\tilde{v} = v - tr(v)1$

Concretely, this looks like:

$\tilde{v} = \left( \begin{array}{cc} -\gamma & x \\ x^\ast & -\beta \end{array} \right)$

The Minkowski metric on $V$

$g : V \times V \to \mathbb{R}$

is given by

$g(v,w) = \frac{1}{2} Re \, tr(v \tilde{w}) = \frac{1}{2} Re \,tr(\tilde{v} w)$

and we also have

$v \tilde{v} = \tilde{v} v = -det(v) 1$

so

$g(v,v) = -det(v) = x x^\ast - \beta \gamma$

Thus, the metric $g$ has signature $(9,1)$, that is 9 plus signs and 1 minus.

### Invariant bilinear maps

Besides the metric

$g : V \times V \to \mathbb{R}$

there are various other important $Spin(9,1)$-invariant bilinear maps.

We have an invariant bilinear map

$V \times S_- \to S_+$

given by

$(v, s_-) \mapsto \tilde{v} s_-$

and also one

$V \times S_+ \to S_-$

given by

$(v, s_+) \mapsto v s_+$

We also have brackets:

$[-,-]: S_- \times S_- \to V = \mathfrak{h}_2(\mathbb{O})$

given by

$[s_-,t_-] = s_- t_-^\dagger + t_- s_-^\dagger$

and

$[-,-]: S_+ \times S_+ \to V = \mathfrak{h}_2(\mathbb{O})$

given by

$[s_+,t_+] = \widetilde{(s_+ t_+^\dagger + t_+ s_+^\dagger)}$

In both of these formulas, $s_\pm$ and $t_\pm$ are column vectors consisting of pairs of octonions, like this:

$\left( \begin{matrix} x \\ y \end{matrix} \right), \quad x, y \in \mathbb{O}$

so the result of bracketing is a $2 \times 2$ hermitian matrix.

We also have pairings of left-handed with right-handed spinors:

$\langle -,- \rangle : S_+ \times S_- \to \mathbb{R}$

given by:

$\langle s_+, t_- \rangle = Re(s_+^\dagger t_-)$

and

$\langle -,- \rangle : S_- \times S_+ \to \mathbb{R}$

given by:

$\langle s_-, t_+ \rangle = Re(s_-^\dagger t_+)$

## The exceptional Jordan algebra

$\mathfrak{h}_3(\mathbb{O})$ consists of $3 \times 3$ self-adjoint matrices of octonions, thus matrices of the form

$a = \left( \begin{array}{ccc} \alpha & z & y^\ast \\ z^\ast & \beta & x \\ y & x^\ast & \gamma \end{array} \right)$

where $x,y,z \in \mathbb{O}$ and $\alpha, \beta, \gamma \in \mathbb{R}$.

We can also think of $\mathfrak{h}_3(\mathbb{O})$ as consisting of matrices

$a = \left( \begin{array}{cc} r & s^\dagger \\ s & v \end{array} \right)$

where $r \in \mathbb{R}$, $s \in S_-$ and $v \in V$. Concretely, we have

$s = \left( \begin{array}{cc} z^\ast \\ y \end{array} \right)$

and

$v = \left( \begin{array}{cc} \beta & x \\ x^\ast & \gamma \end{array} \right)$

So, we have a chosen isomorphism

$\mathfrak{h}_3(\mathbb{O}) \cong \mathbb{R} \oplus V \oplus S_-$

which is equivariant under $Spin(9,1)$.

### The determinant as a cubic form on $\mathfrak{h}_3(\mathbb{O})$

The determinant of a matrix $a \in \mathfrak{h}_3(\mathbb{O})$ is given by

$det(a) = \alpha \beta \gamma - (\alpha |x|^2 + \beta |y|^2 + \gamma |z|^2) + 2 Re(x y z)$

If we write

$a = \left( \begin{array}{cc} r & s^\dagger \\ s & v \end{array} \right)$

as above, then

$det(a) = r det(v) + Re(s^\dagger \tilde{v} s) = r det(v) + g(v, [s, s])$

where

$\tilde{v} = v - tr(v)1$

is the trace-reversed version of $v$, and

$det(v) = \beta \gamma - x x^\ast$

is the determinant of $v \in \mathfrak{h}_2(\mathbb{O})$.

### The trilinear form on $\mathfrak{h}_3(\mathbb{O})$

There is a unique symmetric trilinear form

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

with

$det(a) = t(a,a,a)$

Explicitly, if we have

$a_i = \left( \begin{array}{ccc} \alpha_i & z_i & y^\ast_i \\ z_i^\ast & \beta_i & x_i \\ y_i & x^\ast_i & \gamma_i \end{array} \right)$

then

$6 t(a_1,a_2,a_3) =$
$\sum_{all \; 6 \; permutations \; of \; \{1,2,3\}} \alpha_i \beta_j \gamma_k + 2 Re(x_i y_j z_k) - 2 \sum_{i \in \{1,2,3\}, with \; \{j,k\} = \{1,2,3\}-i} \left( \alpha_i Re(x_j^\ast x_k) + \beta_i Re(y_j^\ast y_k) + \gamma_i Re(z_j^\ast z_k) \right)$

Alternatively, if we write $a_i$ in terms of scalars, spinors and vectors:

$a_i = \left( \begin{array}{cc} r_i & s^\dagger_i \\ s_i & v_i \end{array} \right)$

then we have

$\begin{array}{rcl} t(a_1,a_2,a_3) & = & (1/3) \sum_{i \in \{1,2,3\}, with \; \{j,k\} = \{1,2,3\}-i } Re(s_j^\dagger \tilde{v}_i s_k) - r_i g(v_j, v_k) \\ & = & (1/3) \sum_{i \in \{1,2,3\}, with \; \{j,k\} = \{1,2,3\}-i } g(v_i, [s_j, s_k]) - r_i g(v_j, v_k) \end{array}$

### The cross product on $\mathfrak{h}_3(\mathbb{O})$

Dualizing the trilinear form

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

we get a symmetric bilinear map called the cross product:

$\times : \mathfrak{h}_3(\mathbb{O}) \times \mathfrak{h}_3(\mathbb{O}) \to \mathfrak{h}_3(\mathbb{O})^\ast$

We can think of $\mathfrak{h}_3(\mathbb{O})^\ast$ as consisting of $3 \times 3$ self-adjoint matrices of octonions, where the pairing

$\langle -,- \rangle : \mathfrak{h}_3(\mathbb{O})^\ast \times \mathfrak{h}_3(\mathbb{O}) \to \mathbb{R}$

is given by

$\langle h', h \rangle = \frac{1}{2} Re\;tr(h' h)$

Furthermore, as representations of $Spin(9,1)$ there is an isomorphism

$\mathfrak{h}_3(\mathbb{O})^\ast \cong \mathbb{R} \oplus V \oplus S_+$

by which any element $(r,v,s_+) \in \mathbb{R} \oplus V \oplus S_+$ corresponds to the matrix

$\left( \begin{array}{cc} r & s_+^\dagger \\ s_+ & \tilde{v} \end{array} \right) \in \mathfrak{h}_3(\mathbb{O})^\ast$

In these terms the cross product is:

$(r_A, v_A, s_A) \times (r_B, v_B, s_B) = (1/3) \big( -2g(v_A, v_B), -r_A v_B - r_B v_A + [s_A , s_B], \tilde{v}_A s_B + \tilde{v}_B s_A \big)$

## The dual of the exceptional Jordan algebra

The dual of the exceptional Jordan algebra, $\mathfrak{h}_3(\mathbb{O})^\ast$, is inequivalent to $\mathfrak{h}_3(\mathbb{O})$ as a representation of $\mathrm{E}_6$. So, we must carefully distinguish between them. Nonetheless, every $\mathrm{E}_6$-invariant structure carried by one is also carried by the other, since there’s an outer automorphism of $\mathrm{E}_6$ that interchanges these two representations.

We shall think of $\mathfrak{h}_3(\mathbb{O})^\ast$ as consisting of $3 \times 3$ self-adjoint matrices of octonions, where the pairing

$\langle -,- \rangle : \mathfrak{h}_3(\mathbb{O})^\ast \times \mathfrak{h}_3(\mathbb{O}) \to \mathbb{R}$

is given by

$\langle h', h \rangle = \frac{1}{2} Re\;tr(h' h)$

As a representation of $Spin(9,1)$ there is an isomorphism

$\mathfrak{h}_3(\mathbb{O})^\ast \cong \mathbb{R} \oplus V \oplus S_+$

under which $(r,v,s_+) \in \mathbb{R} \oplus V \oplus S_+$ corresponds to

$\left( \begin{array}{cc} r & s_+^\dagger \\ s_+ & \tilde{v} \end{array} \right) \in \mathfrak{h}_3(\mathbb{O})^\ast$

In these terms, the pairing

$\langle -,- \rangle : \mathfrak{h}_3(\mathbb{O})^\ast \times \mathfrak{h}_3(\mathbb{O}) \to \mathbb{R}$

is given as follows:

$\begin{array}{rcl} \langle (r', v', s_+') , (r, v, s_-) \rangle & = & \frac{1}{2} Re\;tr\big( \left( \begin{array}{cc} r' & s_+'^\dagger \\ s_+' & \tilde{v}'\end{array} \right) \left( \begin{array}{cc} r & s_-^\dagger \\ s_- & v\end{array} \right) \big)\\ & = & \frac{1}{2} r' r + g(v',v) + \langle s_+', s_- \rangle \end{array}$

### The cubic form on $\mathfrak{h}_3(\mathbb{O})^\ast$

There is a cubic form on $\mathfrak{h}_3(\mathbb{O})^\ast$ given by

$det'(r, s_+, v) = r det(v) + g(v, [s_+, s_+])$

### The trilinear form on $\mathfrak{h}_3(\mathbb{O})^\ast$

There is a unique symmetric trilinear form

$t' : \mathfrak{h}_3(\mathbb{O})^\ast \times \mathfrak{h}_3(\mathbb{O})^\ast \times \mathfrak{h}_3(\mathbb{O})^\ast \to \mathbb{R}$

with

$det'(a) = t'(a,a,a)$

and this has an explicit form that looks very similar to $t$ on $\mathfrak{h}_3(\mathbb{O})$:

$\begin{array}{rcl} t'(a_1,a_2,a_3) & = & (1/3) \sum_{i \in \{1,2,3\}, with \; \{j,k\} = \{1,2,3\}-i } Re(s_j^\dagger v_i s_k) - r_i g(\tilde{v}_j, \tilde{v}_k) \\ & = & (1/3) \sum_{i \in \{1,2,3\}, with \; \{j,k\} = \{1,2,3\}-i } g(v_i, [s_j, s_k]) - r_i g(v_j, v_k) \end{array}$

The second formula here appears identical to that for $t$. However, the bracket operation on right-handed spinors differs from that on left-handed spinors, so strictly speaking it is not the same.

### The cross product on $\mathfrak{h}_3(\mathbb{O})^\ast$

Dualizing the trilinear form

$t' : \mathfrak{h}_3(\mathbb{O})^\ast \times \mathfrak{h}_3(\mathbb{O})^\ast \times \mathfrak{h}_3(\mathbb{O})^\ast \to \mathbb{R}$

we get a symmetric bilinear map, another cross product:

$\times : \mathfrak{h}_3(\mathbb{O})^\ast \times \mathfrak{h}_3(\mathbb{O})^\ast \to \mathfrak{h}_3(\mathbb{O})$

This has an explicit form similar to the original cross product above:

$(r_A, v_A, s_A) \times (r_B, v_B, s_B) = (1/3) \big( -2g(v_A, v_B), -r_A v_B - r_B v_A + [s_A , s_B], v_A s_B + v_B s_A \big)$

## The action of $\mathrm{E}_6$ on $\mathfrak{h}_3(\mathbb{O})$

A certain noncompact real form of $\mathrm{E}_6$, technically $E_{6(26)}$, is the group of collineations of $\mathbb{O}\mathrm{P}^2$, and also the group of determinant-preserving linear transformations of $\mathfrak{h}_3(\mathbb{O})$. We call this group simply $\mathrm{E}_6$.

$\mathrm{E}_6$ is 78-dimensional, so we have

$\begin{array}{ccc} dim(E_8) &=& dim(Spin(9,1)) + dim(S_+) + dim(S_-) + dim(\mathbb{R}) \\ 78 &=& 45 + 16 + 16 + 1 \end{array}$

This suggests that perhaps as vector spaces we have

$e_6 \cong so(9,1) \oplus S_+ \oplus S_- \oplus \mathbb{R}$

This is true, but even better, $Spin(9,1)$ and abelian Lie groups isomorphic to $S_+$, $S_-$ and $\mathbb{R}$ show up as Lie subgroups of $\mathrm{E}_6$, in a way that gives rise to this direct sum decomposition.

1) First, $Spin(9,1)$ is a Lie subgroup of $\mathrm{E}_6$: if we use our identification

$\mathfrak{h}_3(\mathbb{O}) \cong \mathbb{R} \oplus V \oplus S_-$

then each summand is a representation of $Spin(9,1)$, and its action preserves the determinant

$det \left( \begin{array}{cc} r & s_-^\dagger \\ s_- & v \end{array} \right) = r det(v) + g(v, [s_-, s_-])$

2) Any $u_+$ in $S_+$ acts on $(r,v,s_-) \in \mathfrak{h}_3(\mathbb{O})$ by

$\begin{array}{ccl} r & \mapsto & r + g(v, [u_+, u_+]) + 2 \langle u_+,s_-\rangle \\ v & \mapsto & v \\ s_- & \mapsto & s_- + v u_+ \end{array}$

Here the angle brackets denote the dual pairing between $S_+ = S_-^\ast$ and $S_-$. We can check that these formulas define a transformation that preserves the determinant $r det(v) + g(v, [s_-, s_-])$. By adding $v u_+$ to $s_-$, $g(v, [s_-, s_-])$ gains these 3 extra terms:

$g(v, [s_-, v u_+]) = -det(v) \langle u_+,s_- \rangle$
$g(v, [v u_+, s_-]) = -det(v) \langle u_+,s_- \rangle$
$g(v, [v u_+, v u_+]) = -det(v) g(v, [u_+, u_+])$

These cancel out the extra terms in $r det(v)$, so the determinant is unchanged.

3) Similarly, any $u_-$ in $S_-$ acts on $(r,v,s_-) \in \mathfrak{h}_3(\mathbb{O})$ by

$\begin{array}{ccl} r & \mapsto & r \\ v & \mapsto & v + (1/2) r [u_-, u_-] + [s_-, u_-] \\ s_- & \mapsto & s_- + r u_- \end{array}$

4) Any positive real number $t$ acts on $(r,v,s_-) \in \mathfrak{h}_3(\mathbb{O})$ by

$\begin{array}{ccl} r & \mapsto & t^4 r \\ v & \mapsto & t^{-2} v \\ s_- & \mapsto & t s_- \end{array}$

These rescalings clearly preserve the determinant $r det(v) + g(v, [s_-, s_-])$. Note that $t$ can be negative, too, so in fact the multiplicative group $\mathbb{R}^\ast = \mathbb{R} - \{0\}$ appears as a subgroup of $\mathrm{E}_6$.

5) It is possible to implement the three actions described here by transformations on the matrix:

$h = \left( \begin{array}{cc} r & s_-^\dagger \\ s_- & v \end{array} \right)$

that all take the form:

$h \mapsto \frac{1}{2}((g h)g^\dagger + g (h g^\dagger))$

For the action of a right-handed spinor $u_+$, we set:

$g = \left( \begin{array}{cc} 1 & u_+ \\ 0 & 1 \end{array} \right)$

where the 1 in the bottom-right corner of the matrix is a $2 \times 2$ identity matrix. For the action of a left-handed spinor $u_-$, we set:

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

And for the action of a scalar $t$, we set:

$g = \left( \begin{array}{ccc} t^2 & 0 & 0\\ 0 & t^{-1} & 0\\ 0 & 0 & t^{-1} \end{array} \right)$

### Orbits of the $\mathrm{E}_6$ action on $\mathfrak{h}_3(\mathbb{O})$

Any element in $\mathfrak{h}_3(\mathbb{O})$ can be diagonalized by an element of $\mathrm{F}_4 \subseteq \mathrm{E}_6$. So, when computing the orbit of any element, we may assume without loss of generalize that it has the form

$\left( \begin{array}{ccc} \alpha & 0 & 0 \\ 0 & \beta & 0 \\ 0 & 0 & \gamma \end{array} \right)$

with $\alpha, \beta, \gamma \in \mathbb{R}$. Then its determinant is $\alpha \beta \gamma$, and this is $\mathrm{E}_6$-invariant.

We can use transformations in $\mathrm{F}_4$ (or even $O(3) \subseteq \mathrm{F}_4$) to permute $\alpha, \beta,$ and $\gamma$, so their order doesn’t matter.

We can use transformations in $Spin_0(9,1) \subseteq \mathrm{E}_6$ to multiply $\beta$ by any positive constant and divide $\gamma$ by that same constant. Thanks to our ability to permute, the same is true of $\alpha$ and $\beta$, or $\alpha$ and $\gamma$.

Thus, if $\alpha, \beta, \gamma \gt 0$, their product is a complete invariant for the action of $\mathrm{E}_6$. We thus get one $\mathrm{E}_6$ orbit for each value of $\delta \gt 0$:

$+++{}_\delta$: the orbit containing the matrices $diag(\alpha, \beta, \gamma)$ with $\alpha, \beta, \gamma \gt 0$, $\alpha \beta \gamma = \delta$.

We cannot, it seems, use a transformation in $\mathrm{E}_6$ to multiply two of $\alpha, \beta, \gamma$ by $-1$ and leave the third alone. Thus, there is a separate family of $\mathrm{E}_6$ orbits, one for each $\delta \gt 0$:

$+--{}_\delta$: the orbit containing the matrices $diag(\alpha, \beta, \gamma)$ with $\alpha, \beta \lt 0$, $\gamma \gt 0$, $\alpha \beta \gamma = \delta$.

By the same reasoning, there are two more one-parameter families of orbits with $\delta \lt 0$:

$++-{}_\delta$: the orbit containing the matrices $diag(\alpha, \beta, \gamma)$ with $\alpha, \beta \gt 0$, $\gamma \lt 0$, $\alpha \beta \gamma = \delta$.

$+- -{}_\delta$: the orbit containing the matrices $diag(\alpha, \beta, \gamma)$ with $\alpha \ge 0$, $\beta, \gamma \lt 0$, $\alpha \beta \gamma = \delta$.

We’re left with the case $\alpha \beta \gamma = 0$. This case gives 6 orbits:

$++0$: The orbit of the matrix $diag(1,1,0)$.

$+-0$: The orbit of the matrix $diag(1,-1,0)$.

$--0$: The orbit of the matrix $diag(-1,-1,0)$.

$+00$: The orbit of the matrix $diag(1,0,0)$.

$-00$: The orbit of the matrix $diag(-1,0,0)$.

$000$: The orbit of the matrix $diag(0,0,0)$.

So, there are 6 orbits and 4 one-parameter families of orbits where the parameter takes values in an open half-line. We can organize these by rank: every element of $\mathfrak{h}_3(\mathbb{O})$ has a rank, which is the number of nonzero entries in the matrix after it has been diagonalized. This is an $\mathrm{E}_6$-invariant concept.

Rank 3: a rank-3 element $h \in \mathfrak{h}_3(\mathbb{O})$ is one where $h \times h \in \mathfrak{h}_3(\mathbb{O})^\ast$ doesn’t annihilate any nonzero element of $\mathfrak{h}_3(\mathbb{O})$. There are 4 one-parameter families of rank-3 orbits.

For any value of $\delta \gt 0$, there are 2 orbits of matrices with determinant $\delta$:

• $+++{}_\delta$: the orbit containing the matrices $diag(\alpha, \beta, \gamma)$ with $\alpha, \beta, \gamma \gt 0$, $\alpha \beta \gamma = \delta$. This orbit is 26-dimensional. The stabilizer of any point in this orbit is a 52-dimensional group isomorphic to the compact real form of $\mathrm{F}_4$.

• $+--{}_\delta$: the orbit containing the matrices $diag(\alpha, \beta, \gamma)$ with $\alpha \gt 0$, $\beta, \gamma \lt 0$, $\alpha \beta \gamma = \delta$. This orbit is 26-dimensional. The identity component of the stabilizer of any point in this orbit is probably a 52-dimensional group isomorphic to the noncompact real form of $\mathrm{F}_4$ called $F_{4(20)}$, which is diffeomorphic to $Spin(9) \times \mathbb{R}^{16}$.

For any value of $\delta \lt 0$, there are 2 orbits of matrices with determinant $\delta$:

• $++-{}_\delta$: the orbit containing the matrices $diag(\alpha, \beta, \gamma)$ with $\alpha, \beta \gt 0$, $\gamma \lt 0$, $\alpha \beta \gamma = \delta$. This orbit is 26-dimensional. The identity component of the stabilizer of any point in this orbit is probably a 52-dimensional group isomorphic to the noncompact real form of $\mathrm{F}_4$ called $F_{4(20)}$, which is diffeomorphic to $Spin(9) \times \mathbb{R}^{16}$.

• $---{}_\delta$: the orbit containing the matrices $diag(\alpha, \beta, \gamma)$ with $\alpha ,\beta, \gamma \lt 0$, $\alpha \beta \gamma = \delta$. This orbit is 26-dimensional. The stabilizer of any point in this orbit is a 52-dimensional group isomorphic to the compact real form of $\mathrm{F}_4$.

Rank 2: a rank-2 element $h \in \mathfrak{h}_3(\mathbb{O})$ is one where $h \times h \in \mathfrak{h}_3(\mathbb{O})^\ast$ annihilates some nonzero element of $\mathfrak{h}_3(\mathbb{O})$ but $h \times h \ne 0$. There are 3 orbits of rank-2 elements:

• $++0$: The orbit of the matrix $diag(1,1,0)$. This orbit is 26-dimensional. The identity component of the stabilizer of any point in this orbit is a 52-dimensional group isomorphic to $Spin(9) \ltimes S_+$.

• $+-0$: The orbit of the matrix $diag(1,-1,0)$. This orbit is 26-dimensional dimensional. The identity component of the stabilizer of any point in this orbit is a 52-dimensional group isomorphic to $\mathrm{Spin}(8,1) \ltimes S_-$.

• $--0$: The orbit of the matrix $diag(-1,-1,0)$. This orbit is 26-dimensional. The identit component of the stabilizer of any point in this orbit is a 52-dimensional group isomorphic to $Spin(9) \ltimes S_+$.

Rank 1: a rank-1 element $h \in \mathfrak{h}_3(\mathbb{O})$ is one where $h \times h = 0$ but $h \ne 0$. There are 2 orbits of rank-1 elements:

• $+00$: The orbit of the matrix $diag(1,0,0)$. This orbit is 17-dimensional. The identity component of the stabilizer of any point in this orbit is a 61-dimensional group isomorphic to $Spin(9,1) \ltimes S_+$.

• $-00$: The orbit of the matrix $diag(-1,0,0)$. This orbit is 17-dimensional. The identity component of the stabilizer of any point in this orbit is a 61-dimensional group isomorphic to $Spin(9,1) \ltimes S_+$.

Rank 0: The only rank-0 element of $\mathfrak{h}_3(\mathbb{O})$ is zero, so there is just one orbit:

• $000$: The orbit of the matrix $diag(0,0,0)$. This orbit is 0-dimensional. The stabilizer of the unique point in this orbit is all of $\mathrm{E}_6$.

Some of these orbits, or unions of these orbits, have alternate descriptions. Most notably we have the large lightcone:

$\{ a \in \mathfrak{h}_3(\mathbb{O}) : \; det(a) = 0 \} = +\!+0 \; \cup +\!-0 \;\cup \; -\!-0 \; \cup \; +00 \; \cup \; -00 \; \cup\; 000$

and the small lightcone:

$\{ a \in \mathfrak{h}_3(\mathbb{O}) : \; a \times a = 0 \} = +00 \;\cup\; -00 \;\cup\; 000$

We also have the forwards small lightcone:

$\{ a \in \mathfrak{h}_3(\mathbb{O}) : \; a \times a = 0, \; \tr(a) \gt 0 \} = +00$

and the backwards small lightcone:

$\{ a \in \mathfrak{h}_3(\mathbb{O}) : \; a \times a = 0, \; \tr(a) \lt 0 \} = -00$

The forwards small lightcone is diffeomorphic to $\mathbb{O}\mathrm{P}^2 \times \mathbb{R}^+$, and so is the backwards small lightcone.

Points in the small lightcone can be explicitly described using the identification $\mathfrak{h}_3(\mathbb{O}) \cong \mathbb{R} \oplus V \oplus S_-$ as follows. It is the union of these two sets:

• (1a): Points of the form $a = (r, \frac{1}{2r} [s,s], s)$ where we can choose $s$ freely, including the origin, and choose any $r \ne 0$. This set is 17-dimensional. It does not include points with $r = 0$.

• (1b): Points of the form $a = (0, v_{LL}, 0)$ where $v_{LL}$ is a nonzero lightlike vector. This set is 9-dimensional. We can think of its elements as the limit points of points in set (1a) where $r \to 0$ and $s \to 0$ together.

In this parametrization we get the forwards small lightcone from (1a) with $r \gt 0$ and from (1b) with $v_{LL}$ on the forwards lightcone in $\mathbb{R}^{9,1}$.

Here is another parametrization of the small lightcone. Again, it is the union of two sets:

• (2a): Points of the form $a = (\frac{1}{2} tr([n_v,n_v]) / tr(v_{LL}), v_{LL}, n_v)$ where $v_{LL}$ is a nonzero lightlike vector and $n_v$ belongs to the 8-dimensional kernel of $\tilde{v}_{LL}$. This set is 17-dimensional. It does not include points with $v_{LL} = 0$.

• (2b): Points of the form $a = (r, 0, 0)$ where $r \ne 0$. This set is 1-dimensional. We can think of its elements as the limit points of points in set (2a) where $v_{LL} \to 0$ and $n_v \to 0$ together.

In this parametrization we get the forwards small lightcone from (2a) with $v_{LL}$ on the forwards lightcone in $\mathbb{R}^{9,1}$, and from (2b) with $r \gt 0$.

#### Stabilizer computation: +00 orbit

If we look at the stabilizer of the rank-1 element $h = diag(1,0,0)$, with pieces $(r,v,s_-)=(1,0,0)$, this will be fixed by:

1) any element $g$ of $Spin(9,1)$, since $v=0$

2) any element $u_+$ of $S_+$, since that action is given by:

• $r \mapsto r + g(v, [u_+,u_+]) + 2\langle u_+,s_- \rangle$, which takes $r = 1$ to $r = 1$ since $v=0$ and $s_- = 0$,

• $v \mapsto v$ which clearly preserves $v = 0$,

and

• $s_- \mapsto s_- + v u_+$ which takes $s_- = 0$ to $s_- = 0$ since $v = 0$.

These interact nicely so that the semidirect product $Spin(9,1) \ltimes S_+$ stabilizes $diag(1,0,0)$. By dimension counting, this is the whole stabilizer, since

$dim (Spin(9,1) \ltimes S_+) = 45 + 16 = 61$

and the orbit of $diag(1,0,0)$ is 17-dimensional, and $61 + 17 = 78$.

#### Stabilizer computation: ++- orbit

The stabilizer of the rank-3 element $h = diag(-1,1,1)$, with pieces $(r,v,s_-)=(-1,diag(1,1),0)$ is a 52-dimensional group. Page 67 of this book:

describes all 3 real forms of $\mathrm{F}_4$, and I believe this can be used to prove the stabilizer of $h$ is $F_{4(20)}$, the real form whose Killing form has signature 20, meaning that it’s positive definite on a 36-dimensional subspace and negative definite on a 16-dimensional subspace (or the other way around if you use the other sign convention). Among the real forms of $\mathrm{F}_4$, this one is characterized by having $Spin(9)$ as its maximal compact subgroup, so it is diffeomorphic to $Spin(9) \times \mathbb{R}^{16}$.

It is easy to see that $Spin(9)$ appears as a subgroup of the stabilizer of $h$, so there is only a bit left to prove here.

## The action of $\mathrm{E}_6$ on $\mathfrak{h}_3(\mathbb{O})^\ast$

The action of $\mathrm{E}_6$ on $\mathfrak{h}_3(\mathbb{O})^\ast$ is very much like its action on $\mathfrak{h}_3(\mathbb{O})$, but with the roles of $S_+$ and $S_-$ reversed.

Explicitly, we act on the matrices:

$h' = \left( \begin{array}{cc} r' & s_+'^\dagger \\ s_+' & \tilde{v}' \end{array} \right)$

with the transformation:

$h' \mapsto \frac{1}{2}((g' h')g'^\dagger + g' (h' g'^\dagger))$

where:

$g' = (g^{-1})^\dagger$

Acting this way on $h'$ while acting with $g$ on $h$ preserves the pairing:

$\langle h',h \rangle = \frac{1}{2} Re\;tr(h'h)$

The orbits of $\mathrm{E}_6$ on $\mathfrak{h}_3(\mathbb{O})^\ast$ are classified almost exactly as for $\mathfrak{h}_3(\mathbb{O})$. The differences are all routine: for example, in set (2a), where we had demanded that $n_v$ belong to the 8-dimensional null space of $\tilde{v}_{LL}$, we should now use the kernel of $v_{LL}$. This is the appropriate action of $V$ on $S_+$. We also need to switch the roles of $S_+$ and $S_-$. For example, since a point in the forwards small lightcone in $\mathfrak{h}_3(\mathbb{O})$, i.e. the $+00$ orbit in $\mathfrak{h}_3(\mathbb{O})$, is stabilized by a group conjugate to $Spin(9,1) \ltimes S_+$, so a ‘null momentum vector’, i.e. a point in the $+00$ orbit in $\mathfrak{h}_3(\mathbb{O})^\ast$, is stabilized by a group conjugate to $Spin(9,1) \ltimes S_-$.

### The $\mathrm{E}_6$ action on $\mathbb{O}\mathrm{P}^2$

We can think of the octonionic projective plane $\mathbb{O}\mathrm{P}^2$ as the projectivization of the small lightcone

$\{ a \in \mathfrak{h}_3(\mathbb{O}) : \; a \times a = 0 \} = +00 \;\cup\; -00 \;\cup\; 000$

or in other words, $+00 \cup -00$ modulo the $\mathbb{R}^\ast$ action given by rescaling. Since the small lightcone is 17-dimensional this makes $\mathbb{O}\mathrm{P}^2$ 16-dimensional, as it must be.

$\mathrm{E}_6$ acts transitively on the small lightcone and thus on $\mathbb{O}\mathrm{P}^2$. The stabilizer of a point in the forward small lightcone is $Spin(9,1) \ltimes S_+$. Thus, the stabilizer of a point in $\mathbb{O}\mathrm{P}^2$ is the 62-dimensional group $(Spin(9,1) \times \mathbb{R}^\ast) \ltimes S_+$.

$\mathrm{E}_6$ also acts transitively on the space of lines in $\mathbb{O}\mathrm{P}^2$. The space of lines is diffeomorphic to $\mathbb{O}\mathrm{P}^2$, but with a different action of $\mathrm{E}_6$. The stabilizer of any is the 62-dimensional group $(Spin(9,1) \times \mathbb{R}^\ast) \ltimes S_-$.

The group $\mathrm{E}_6$ also acts transitively on the space of antiflags: pairs consisting of a point and line where the line does not contain the point. The space of antiflags is a 32-dimensional manifold, so the stabilizer of any antiflag must be a group of dimension $78 - 32 = 46$. This group is isomorphic to $Spin(9,1) \times \mathbb{R}^\ast$.

The last fact is easiest to see if we treat $\mathbb{O}\mathrm{P}^2$ as the union of $\mathbb{O}^2$ and the ‘line at infinity’, a copy of $\mathbb{O} P^1$. The subgroup stabilizing the line at infinity is $(Spin(9,1) \times \mathbb{R}^\ast) \ltimes S_-$, so this group acts on the complement, $\mathbb{O}^2$. If we identify $\mathbb{O}^2$ with $S_-$ the action is easy to understand. $Spin(9,1)$ acts on $S_-$ via its spinor representation, $\mathbb{R}^\ast$ acts on $S_-$ via dilations, and the additive group $S_-$ acts on $S_-$ by translations. The subgroup of $(Spin(9,1) \times \mathbb{R}^\ast) \ltimes S_-$ that also fixes the origin in $S_-$ is thus $Spin(9,1) \times \mathbb{R}^\ast$.

### Lines in $\mathbb{O}\mathrm{P}^2$

We can identify the space of lines in $\mathbb{O}\mathrm{P}^2$ either with:

• the projectivization of the small lightcone in $\mathfrak{h}_3(\mathbb{O})^\ast$, or
• equivalence classes of rank-2 elements in $\mathfrak{h}_3(\mathbb{O})$.

With the first identification, a point $p$ in $\mathbb{O}\mathrm{P}^2$, viewed as a 1-dimensional subspace contained in the small lightcone of $\mathfrak{h}_3(\mathbb{O})$, is incident on a line $L$ in $\mathbb{O}\mathrm{P}^2$, viewed as a 1-dimensional subspace contained in the small lightcone of $\mathfrak{h}_3(\mathbb{O})^\ast$, iff

$L_0(p_0) = 0$

for any $p_0 \in p$ and $L_0 \in L$.

The second identification is a bit more complicated. Given any rank-2 element $v$ of $\mathfrak{h}_3(\mathbb{O})$, if we translate the small lightcone by $v$, it will intersect the untranslated small lightcone in an 8-dimensional set. The span of this 8-dimensional set will be a 10-dimensional subspace $V(v)$ of $\mathfrak{h}_3(\mathbb{O})$ containing elements of ranks 0, 1 and 2, including $v$.

$V(v) = span (\{ p | p\,\text{ and }\, p-v\, \text{ are both on the small lightcone}\})$

The intersection of the translated and untranslated small lightcones will be precisely the intersections of the ordinary Minkowski light cones centred at the origin and at $v$ within $V(v)$ (where the conformal structure on $V(v)$ is derived from the trilinear form on $\mathfrak{h}_3(\mathbb{O})$). We can view all the rank-2 elements in $V(v)$ as belonging to an equivalence class with $v$, since the same construction performed with any of them will yield the same subspace.

If we identify this equivalence class of rank-2 elements with a line in $\mathbb{O}\mathrm{P}^2$, then the 1-dimensional spaces of rank-1 elements of $V(v)$, which lie on the Minkowski lightcone, correspond to points that are incident on the line.

We can also identify elements $L$ of the projectivized small lightcone in $\mathfrak{h}_3(\mathbb{O})^\ast$ with 10-dimensional Minkowski subspaces of $\mathfrak{h}_3(\mathbb{O})$, by defining:

$V(L) = span(\{p | p\, \text{ is on the small lightcone and }\, L_0(p) = 0\})$

where as before $L_0 \in L$.

We can connect the two ways of thinking about lines by noting that, for rank-2 elements $v_2$:

$v_2 \in V(L)\,\text{ iff }\,v_2 \times v_2 \in L$

This also gives us a new way of describing the equivalence relationship between rank-2 elements that describe the same line:

$v_2 ~ v_2'\,\text{ iff }\, v_2 \times v_2\,\text{ is a multiple of }\,v_2' \times v_2'$

We’ve seen that the choice of an antiflag in $\mathbb{O}\mathrm{P}^2$ (a point, and a line not containing that point) gives rise to a choice of a 1-dimensional subspace $\mathbb{R}$ of elements with ranks 0 and 1 in $\mathfrak{h}_3(\mathbb{O})$, and a 10-dimensional subspace $V$ of elements with ranks 0, 1 and 2 in $\mathfrak{h}_3(\mathbb{O})$, which intersect only at the origin.

Additionally, this choice singles out a 16-dimensional subspace $S$ of elements with ranks 0 and 2 only, defined by:

$S = \{s | t(v,v,s) = 0 \forall v \in V\,\text{ and }\, s \times r = 0 \forall r \in \mathbb{R}\}$

In fact, if we pick any elements of maximum rank in each subspace, $v_2 \in V$ and $r_1 \in \mathbb{R}$, we can obtain $S$ from the same conditions applied to that single choice of elements:

$S = \{s | t(v_2,v_2,s) = 0 \,\text{ and }\, s \times r_1 = 0 \}$

The first condition means that $s$ belongs to the tangent space to the orbit of $v_2$ at $v_2$, and the second condition means that $s$ belongs to the tangent space to the orbit of $r_1$ at $r_1$, where we treat these tangent spaces as subspaces of $\mathfrak{h}_3(\mathbb{O})$. So we can write:

$S = T_{v_2}(orbit(v_2)) \cap T_{r_1}(orbit(r_1))$

The rank-2 elements of $S$ all belong to the orbit $+-0$.

We can describe the spaces $\mathbb{R}$ and $V$ as:

$\mathbb{R} = cl\{r | T_{r}(orbit(r)) = T_{r_1}(orbit(r_1)) \}$
$V = cl\{v | T_{v}(orbit(v)) = T_{v_2}(orbit(v_2)) \}$

## The kernel of the cross product map

For any vector $p \in \mathfrak{h}_3(\mathbb{O})^\ast$ there is a cross product map

$p\times : \mathfrak{h}_3(\mathbb{O})^\ast \to \mathfrak{h}_3(\mathbb{O})$

mapping any vector $v$ to $p \times v$. The dimension of the kernel of this map depends on which orbit $p$ lies in:

• If $p$ has rank 3 then the kernel of this map is 0-dimensional.

• If $p$ has rank 2 then the kernel of this map is 9-dimensional. In this case the point $p$ lies in a copy of $\mathbb{R}^{9,1}$ inside $\mathfrak{h}_3(\mathbb{O})^\ast$, and as such it is either a timelike vector (if $p \in ++0$ or $--0$) or a spacelike vector (if $p \in +-0$). The subgroup of $Spin(9,1)$ stabilizing this vector is thus either $Spin(9)$ or $Spin(8,1)$. In either case it acts on $ker p \!\times$, giving a representation isomorphic to the 9-dimensional ‘vector’ representation of this group.

• If $p$ has rank 1 then the kernel of this map is 17-dimensional. In this case the kernel can be identified with the tangent space of the small lightcone at $p$. The point $p$ lies in a copy of $\mathbb{R}^{9,1}$ inside $\mathfrak{h}_3(\mathbb{O})^\ast$, and as such it is a lightlike vector. The subgroup of $Spin(9,1)$ stabilizing this vector is isomorphic to $Spin(8) \ltimes \mathbb{R}^8$, and it acts on $ker p\!\times$, giving a representation isomorphic to the direct sum of the 1-dimensional ‘scalar’ representation of $Spin(8)$ on $\mathbb{R}$ and the 16-dimensional ‘left-handed plus right-handed spinor’ representation of $Spin(8)$ on $\mathbb{O} \oplus \mathbb{O}$.

• If $p$ has rank 0 then the kernel of this map is 27-dimensional.

Last revised on December 22, 2020 at 15:10:36. See the history of this page for a list of all contributions to it.