This is a record of some calculations done by John Baez, Greg Egan and John Huerta around November 2015.
The behavior of spinors depends heavily on the dimension of space, or spacetime, modulo 8. For example, in spacetimes of any dimension thatβs 2 more than a multiple of 8 there exist βMajoranaβWeyl spinorsβ: spin-1/2 particles that have an intrinsic handedness (thatβs the βWeylβ part) and are their own antiparticles (thatβs the βMajoranaβ part).
In 10d Minkowski spacetime something special happens: both kinds of MajoranaβWeyl 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 MajoranaβWeyl 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 doesnβt really cause a problem here.
For a careful treatment of all this, try Section 2 here:
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
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:
Concretely, this looks like:
The Minkowski metric on $V$
is given by
and we also have
so
Thus, the metric $g$ has signature $(9,1)$, that is 9 plus signs and 1 minus.
Besides the metric
there are various other important $Spin(9,1)$-invariant bilinear maps.
We have an invariant bilinear map
given by
and also one
given by
We also have brackets:
given by
and
given by
In both of these formulas, $s_\pm$ and $t_\pm$ are column vectors consisting of pairs of octonions, like this:
so the result of bracketing is a $2 \times 2$ hermitian matrix.
We also have pairings of left-handed with right-handed spinors:
given by:
and
given by:
$\mathfrak{h}_3(\mathbb{O})$ consists of $3 \times 3$ self-adjoint matrices of octonions, thus matrices of the form
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
where $r \in \mathbb{R}$, $s \in S_-$ and $v \in V$. Concretely, we have
and
So, we have a chosen isomorphism
which is equivariant under $Spin(9,1)$.
The determinant of a matrix $a \in \mathfrak{h}_3(\mathbb{O})$ is given by
If we write
as above, then
where
is the trace-reversed version of $v$, and
is the determinant of $v \in \mathfrak{h}_2(\mathbb{O})$.
There is a unique symmetric trilinear form
with
Explicitly, if we have
then
Alternatively, if we write $a_i$ in terms of scalars, spinors and vectors:
then we have
Dualizing the trilinear form
we get a symmetric bilinear map called the cross product:
We can think of $\mathfrak{h}_3(\mathbb{O})^\ast$ as consisting of $3 \times 3$ self-adjoint matrices of octonions, where the pairing
is given by
Furthermore, as representations of $Spin(9,1)$ there is an isomorphism
by which any element $(r,v,s_+) \in \mathbb{R} \oplus V \oplus S_+$ corresponds to the matrix
In these terms the cross product is:
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 thereβs 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
is given by
As a representation of $Spin(9,1)$ there is an isomorphism
under which $(r,v,s_+) \in \mathbb{R} \oplus V \oplus S_+$ corresponds to
In these terms, the pairing
is given as follows:
There is a cubic form on $\mathfrak{h}_3(\mathbb{O})^\ast$ given by
There is a unique symmetric trilinear form
with
and this has an explicit form that looks very similar to $t$ on $\mathfrak{h}_3(\mathbb{O})$:
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.
Dualizing the trilinear form
we get a symmetric bilinear map, another cross product:
This has an explicit form similar to the original cross product above:
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
This suggests that perhaps as vector spaces we have
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
then each summand is a representation of $Spin(9,1)$, and its action preserves the determinant
2) Any $u_+$ in $S_+$ acts on $(r,v,s_-) \in \mathfrak{h}_3(\mathbb{O})$ by
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:
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
4) Any positive real number $t$ acts on $(r,v,s_-) \in \mathfrak{h}_3(\mathbb{O})$ by
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:
that all take the form:
For the action of a right-handed spinor $u_+$, we set:
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:
And for the action of a scalar $t$, we set:
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
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 doesnβt 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 $\alpha \ge 0$, $\beta, \gamma \lt 0$, $\alpha \beta \gamma = \delta$.
Weβre 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 $h \times h \in \mathfrak{h}_3(\mathbb{O})^\ast$ doesnβt 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:
Some of these orbits, or unions of these orbits, have alternate descriptions. Most notably we have the large lightcone:
and the small lightcone:
We also have the forwards small lightcone:
and the backwards small lightcone:
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 $r \gt 0$.
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
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
and the orbit of $diag(1,0,0)$ is 17-dimensional, and $61 + 17 = 78$.
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 itβs 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 $\mathfrak{h}_3(\mathbb{O})^\ast$ 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:
with the transformation:
where:
Acting this way on $h'$ while acting with $g$ on $h$ preserves the pairing:
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 βnull momentum vectorβ, i.e. a point in the $+00$ orbit in $\mathfrak{h}_3(\mathbb{O})^\ast$, is stabilized by a group conjugate to $Spin(9,1) \ltimes S_-$.
We can think of the octonionic projective plane $\mathbb{O}\mathrm{P}^2$ as the projectivization of the small lightcone
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 βline at infinityβ, 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 $Spin(9,1) \times \mathbb{R}^\ast$.
We can identify the space of lines in $\mathbb{O}\mathrm{P}^2$ either with:
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
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 $v$.
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:
where as before $L_0 \in L$.
We can connect the two ways of thinking about lines by noting that, for rank-2 elements $v_2$:
This also gives us a new way of describing the equivalence relationship between rank-2 elements that describe the same line:
Weβve 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:
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:
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:
The rank-2 elements of $S$ all belong to the orbit $+-0$.
We can describe the spaces $\mathbb{R}$ and $V$ as:
For any vector $p \in \mathfrak{h}_3(\mathbb{O})^\ast$ there is a cross product map
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 βvectorβ 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 βscalarβ representation of $Spin(8)$ on $\mathbb{R}$ and the 16-dimensional βleft-handed plus right-handed spinorβ 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 20:10:36. See the history of this page for a list of all contributions to it.