# On the twisted octonionic eigenvalue problem and some sextics hypersurfaces
related to the Cartan cubic

Roland Abuaf 111Rectorat de Paris, 47 rue des Écoles, 75005 Paris.
<EMAIL_ADDRESS>
###### Abstract

We revisit the octonionic eigenvalue problem from a geometric perspective. In
particular, we study a tautological sheaf defined on a sextic related to this
problem, the Ogievetskiî-Dray-Manogue sextic. We then define and study a
twisted version of the octonionic eigenvalue problem. A new sextic arises in
this setting and we study the corresponding tautological sheaf supported on
it. This twisted version of the octonionic eigenvalue problem is eminently
more symmetric than the original one, as reflected by the last result we prove
in this paper : the automorphism group of the twisted octonionic eigenvalue
problem, though not isomorphic to $\displaystyle\mathrm{E}_{6}$, acts
prehomogeneously on the exceptional Jordan algebra
$\displaystyle\mathrm{J}_{3}(\mathbb{O})$. This is in sharp contrast with the
fact that the generic orbit for the action of the automorphism group of the
classical octonionic eigenvalue problem has (at least) codimension
$\displaystyle 6$ in $\displaystyle\mathrm{J}_{3}(\mathbb{O})$.

###### Contents

  1. 1 Introduction
  2. 2 The eigenvalue problem for $\displaystyle\mathrm{J}_{3}(\mathbb{O})$ and the Ogievetskiî-Dray-Manogue sextic
    1. 2.1 The octonionic eigenvalue problem
    2. 2.2 The geometry of the Ogievetskiî-Dray-Manogue sextic
    3. 2.3 A tautological rank $\displaystyle 4$ sheaf on the Ogievetskiî-Dray-Manogue sextic
  3. 3 The geometry of the twisted eigenvalue problem for $\displaystyle\mathrm{J}_{3}(\mathbb{O})$
    1. 3.1 The twisted octonionic eigenvalue problem
    2. 3.2 A quasi-tautological rank $\displaystyle 4$ sheaf on the Cartan cubic
    3. 3.3 On the automorphism group of twisted eigenvalue problem
  4. A On the dimension the Lie-algebra $\displaystyle\mathfrak{aut}(\SS)$

##  1 Introduction

One starting point of our paper is the claim made by Bhargava and Ho that
there exists a << less familiar to most people>> (as they state it) rank
$\displaystyle 8$ tautological vector bundle on the (complexified) octonionic
projective plane $\displaystyle\mathbb{OP}^{2}\subset\mathbb{P}^{26}$ (see
Theorem $\displaystyle 5.27$ of [BH16]). Following their definition 5.21, one
should believe that this vector bundle is endowed with a global faithful right
action of the (complexified) octonions $\displaystyle\mathbb{O}$. Let us
recall the well-known construction of such a tautological bundle in the case
of an associative composition algebra instead of $\displaystyle\mathbb{O}$.

We denote by $\displaystyle\mathbb{A}$ one of the (complexified) algebras
$\displaystyle\mathbb{R}\otimes_{\mathbb{R}}\mathbb{C}$,
$\displaystyle\mathbb{C}\otimes_{\mathbb{R}}\mathbb{C}$,
$\displaystyle\mathbb{H}\otimes\mathbb{C}$ and we let
$\displaystyle\mathrm{J}_{3}(\mathbb{A})$ be the space of hermitian matrices
with coefficients in $\displaystyle\mathbb{A}$:

$\displaystyle\mathrm{J}_{3}(\mathbb{A})=\left\\{\begin{pmatrix}\lambda_{1}&c&\overline{b}\\\
\overline{c}&\lambda_{2}&a\\\ b&\overline{a}&\lambda_{3}\end{pmatrix},\
a,b,c\in\mathbb{A},\ \lambda_{i}\in\mathbb{C}\right\\}.$

For any $\displaystyle A\in\mathrm{J}_{3}(\mathbb{A})$, we denote by
$\displaystyle\mathrm{det}_{\mathbb{A}}(A)$ the number:

$\begin{split}\mathrm{det}_{\mathbb{A}}(A)&=\lambda_{1}\lambda_{2}\lambda_{3}+c(ab)+\big{(}(\overline{b})(\overline{a})\big{)}(\overline{c})-\lambda_{2}b\overline{b}-\lambda_{1}a\overline{a}-\lambda_{3}c\overline{c}\\\
&=\lambda_{1}\lambda_{2}\lambda_{3}+2\mathrm{Re}(cab)-\lambda_{2}|b|^{2}-\lambda_{1}|a|^{2}-\lambda_{3}|c|^{2},\end{split}$

where $\displaystyle\mathrm{Re}(a)$ is the real part of $\displaystyle
a\in\mathbb{A}$ and $\displaystyle|a|^{2}$ is the square of its norm. This
expression corresponds to the Sarrus expansion for a formal determinant of the
$\displaystyle 3\times 3$ matrix $\displaystyle A$. We notice the remarkable
that $\displaystyle\mathrm{det}_{\mathbb{A}}(A)\in\mathbb{C}$, for all
$\displaystyle A\in\mathrm{J}_{3}(\mathbb{A})$. Finally, we denote by
$\displaystyle\mathscr{C}_{\mathbb{A}}$ the cubic hypersurface in
$\displaystyle\mathbb{P}(\mathrm{J}_{3}(\mathbb{A}))$ given by the equation
$\displaystyle\mathrm{det}_{\mathbb{A}}=0$ and $\displaystyle X_{\mathbb{A}}$
the singular locus of this hypersurface. The case by case analysis is well-
known ([Rob88] for instance):

  * •

if $\displaystyle\mathbb{A}=\mathbb{R}\otimes\mathbb{C}$, then $\displaystyle
X_{\mathbb{A}}=v_{2}(\mathbb{P}^{2})\subset\mathbb{P}^{5}$ is the Veronese
embedding of $\displaystyle\mathbb{P}^{2}$ and
$\displaystyle\mathscr{C}_{\mathbb{A}}$ is the determinantal cubic for
$\displaystyle 3*3$ symmetric matrices.

  * •

if $\displaystyle\mathbb{A}=\mathbb{C}\otimes\mathbb{C}$, then $\displaystyle
X_{\mathbb{A}}=\mathbb{P}^{2}\times\mathbb{P}^{2}\subset\mathbb{P}^{8}$ is the
Segre embedding of $\displaystyle\mathbb{P}^{2}\times\mathbb{P}^{2}$ and
$\displaystyle\mathscr{C}_{\mathbb{A}}$ is the determinantal cubic for
$\displaystyle 3*3$ general matrices.

  * •

if $\displaystyle\mathbb{A}=\mathbb{H}\otimes\mathbb{C}$, then $\displaystyle
X_{\mathbb{A}}=\mathrm{Gr}(2,6)\subset\mathbb{P}^{14}$ is the Plücker
embedding of $\displaystyle\mathrm{Gr}(2,6)$ and
$\displaystyle\mathscr{C}_{\mathbb{A}}$ is the Pfaffian cubic for
$\displaystyle 6*6$ skew-symmetric matrices.

In each case above, one can provide an easy ad-hoc construction of a rank
$\displaystyle n_{\mathbb{A}}=\dim_{\mathbb{C}}\mathbb{A}$ tautological vector
bundle on $\displaystyle X_{\mathbb{A}}$. We shall rather give a uniform
construction of this vector bundle which endows it with a global faithful
right action of $\displaystyle\mathbb{A}$. For all $\displaystyle
A=\begin{pmatrix}\lambda_{1}&c&\overline{b}\\\ \overline{c}&\lambda_{2}&a\\\
b&\overline{a}&\lambda_{3}\end{pmatrix}\in\mathrm{J}_{3}(\mathbb{A})$, we
denote by $\displaystyle\mathrm{Com}(A)$, the element of
$\displaystyle\mathrm{J}_{3}(\mathbb{A})$ defined by:

$\displaystyle\mathrm{Com}(A)=\begin{pmatrix}\lambda_{2}\lambda_{3}-a\overline{a}&(\overline{b})(\overline{a})-\lambda_{3}c&ca-\lambda_{2}\overline{b}\\\
ab-\lambda_{3}\overline{c}&\lambda_{1}\lambda_{3}-b\overline{b}&(\overline{c})(\overline{b})-\lambda_{1}a\\\
(\overline{a})(\overline{c})-\lambda_{2}b&bc-\lambda_{1}\overline{a}&\lambda_{1}\lambda_{2}-c\overline{c}\end{pmatrix}.$

Owing to the associativity of $\displaystyle\mathbb{A}$ and to fundamental
relation $\displaystyle\mathrm{Re}(cab)=\mathrm{Re}(abc)=\mathrm{Re}(bca)$,
for any $\displaystyle c,b,a\in\mathbb{A}$, we get, for any $\displaystyle
A\in\mathrm{J}_{3}(\mathbb{A})$, the matrix factorization:

$\mathrm{Com}(A)\times
A=A\times\mathrm{Com}(A)=\mathrm{det}_{\mathbb{A}}(A).I_{3},$ (1)

where $\displaystyle I_{3}$ is the identity $\displaystyle 3\times 3$ matrix.
Furthermore, there is a natural embedding of algebras:

$\displaystyle L:\mathbb{A}\xrightarrow{\hskip
22.76228pt}\mathscr{M}_{n_{\mathbb{A}},n_{\mathbb{A}}}(\mathbb{C}),$

sending an element to its left multiplication table and which induces an
embedding of algebras:

$\displaystyle L:\mathrm{J}_{3}(\mathbb{A})\xrightarrow{\hskip
22.76228pt}\mathscr{M}_{3n_{\mathbb{A}},3n_{\mathbb{A}}}(\mathbb{C}).$

We fix $\displaystyle e_{1},\ldots,e_{n_{\mathbb{A}}}$ a basis of
$\displaystyle\mathbb{A}$ over $\displaystyle\mathbb{C}$ and consider the ring
$\displaystyle\mathbb{C}[x_{1},\ldots,x_{3n_{\mathbb{A}}+3}]$. We put
$\displaystyle c=x_{1}e_{1}+\ldots+x_{n}e_{n_{\mathbb{A}}}$, $\displaystyle
b=x_{n_{\mathbb{A}}+1}e_{1}+\ldots+x_{2n_{\mathbb{A}}}e_{n_{\mathbb{A}}}$,
$\displaystyle
a=x_{2n_{\mathbb{A}}+1}e_{1}+\ldots+x_{3n_{\mathbb{A}}}e_{n_{\mathbb{A}}}$,
$\displaystyle\lambda_{1}=x_{3n_{\mathbb{A}}+1}e_{1}$,
$\displaystyle\lambda_{2}=x_{3n_{\mathbb{A}}+2}e_{1}$ and
$\displaystyle\lambda_{3}=x_{3n_{\mathbb{A}}+3}e_{1}$. Consider the
$\displaystyle 3n_{\mathbb{A}}\times 3n_{\mathbb{A}}$ matrix with entries in
$\displaystyle\mathbb{C}[x_{1},\ldots,x_{3n_{\mathbb{A}}+3}]$ described in
$\displaystyle n_{\mathbb{A}}\times n_{\mathbb{A}}$ blocks by:

$\displaystyle
M_{\mathbb{A}}:=L(A)=\begin{pmatrix}\lambda_{1}.I_{n_{\mathbb{A}}}&L_{c}&{}^{t}L_{b}\\\
{}^{t}L_{c}&\lambda_{2}.I_{n_{\mathbb{A}}}&L_{a}\\\
L_{b}&{}^{t}L_{a}&\lambda_{3}.I_{n_{\mathbb{A}}}\end{pmatrix}.$

One immediately notices that $\displaystyle M_{\mathbb{A}}$ is a symmetric
$\displaystyle 3n_{\mathbb{A}}\times 3n_{\mathbb{A}}$ matrix with entries in
the ring $\displaystyle\mathbb{C}[x_{1},\ldots,x_{3n_{\mathbb{A}}+3}]$. We
also consider the matrix:

$\displaystyle\mathrm{Com}(M_{\mathbb{A}}):=L(\mathrm{Com}(A))=\begin{pmatrix}L_{\lambda_{2}\lambda_{3}-a\overline{a}}&L_{(\overline{b})(\overline{a})-\lambda_{3}c}&L_{ca-\lambda_{2}\overline{b}}\\\
L_{ab-\lambda_{3}\overline{c}}&L_{\lambda_{1}\lambda_{3}-b\overline{b}}&L_{(\overline{c})(\overline{b})-\lambda_{1}a}\\\
L_{(\overline{a})(\overline{c})-\lambda_{2}b}&L_{bc-\lambda_{1}\overline{a}}&L_{\lambda_{1}\lambda_{2}-c\overline{c}}\end{pmatrix}.$

The matrix $\displaystyle\mathrm{Com}(M_{\mathbb{A}})$ is also a symmetric
$\displaystyle 3n_{\mathbb{A}}\times 3n_{\mathbb{A}}$ symmetric matrix with
coefficients in $\displaystyle\mathbb{C}[x_{1},\ldots,x_{3n_{\mathbb{A}}+3}]$.
The matrix factorizations in equation (1) and the fact that $\displaystyle L$
is a morphism of algebras show that we have a matrix factorizations in
$\displaystyle\mathscr{M}_{3n_{\mathbb{A}},3n_{\mathbb{A}}}(\mathbb{C}[x_{1},\ldots,x_{3n_{\mathbb{A}}+3}]$:

$\mathrm{Com}(M_{\mathbb{A}})\times
M_{\mathbb{A}}=M_{\mathbb{A}}\times\mathrm{Com}(M_{\mathbb{A}})=\mathrm{Det}_{\mathbb{A}}.I_{3n_{\mathbb{A}}},$
(2)

where $\displaystyle\mathrm{Det}_{\mathbb{A}}$ is the cubic polynomial in
$\displaystyle\mathbb{C}[x_{1},\ldots,x_{3n_{\mathbb{A}}+3}]$ corresponding to
$\displaystyle\mathrm{det}_{\mathbb{A}}$. We deduce from Equation (2) that the
degree $\displaystyle 3n_{\mathbb{A}}$ polynomial
$\displaystyle\mathrm{det}_{\mathscr{M}_{3n_{\mathbb{A}},3n_{\mathbb{A}}}}(M_{\mathbb{A}})$
and the cubic polynomial $\displaystyle\mathrm{Det}_{\mathbb{A}}$ define set-
theoretically the same hypersurface in
$\displaystyle\mathbb{P}^{3n_{\mathbb{A}}+2}$ (that is the locus where
$\displaystyle M_{\mathbb{A}}$ is not full-rank). On the other hand, it is
known that $\displaystyle\mathrm{Det}_{\mathbb{A}}$ is irreducible (see the
appendix of [Rob88]), so that we get the relation:

$\mathrm{det}_{\mathscr{M}_{3n_{\mathbb{A}},3n_{\mathbb{A}}}}(M_{\mathbb{A}})=(\mathrm{Det}_{\mathbb{A}})^{n_{\mathbb{A}}}.$
(3)

Let us denote by $\displaystyle E_{\mathbb{A}}$ the cokernel of:

$\displaystyle\mathbb{A}^{\oplus
3}\otimes_{\mathbb{C}}\mathscr{O}_{\mathbb{P}(\mathrm{J}_{3}(\mathbb{A}))}(-1)\stackrel{{\scriptstyle
M_{\mathbb{A}}}}{{\xrightarrow{\hskip 22.76228pt}}}\mathbb{A}^{\oplus
3}\otimes_{\mathbb{C}}\mathscr{O}_{\mathbb{P}(\mathrm{J}_{3}(\mathbb{A}))}$

The sheaf $\displaystyle E_{\mathbb{A}}$ is supported on
$\displaystyle\mathscr{C}_{\mathbb{A}}$, as can be read from equation 2. It is
generically of rank $\displaystyle n_{\mathbb{A}}$ as follows from equation 3.
Since $\displaystyle\mathbb{A}$ is associative, we have:

$\displaystyle M_{\mathbb{A}}\times\begin{pmatrix}x\times a\\\ y\times a\\\
z\times a\end{pmatrix}=\left(M_{\mathbb{A}}\times\begin{pmatrix}x\\\ y\\\
z\end{pmatrix}\right)\times a,$

for all $\displaystyle\begin{pmatrix}x\\\ y\\\
z\end{pmatrix}\in\mathbb{A}^{\oplus 3}$ and all $\displaystyle
a\in\mathbb{A}$. This implies that $\displaystyle E_{\mathbb{A}}$ is induced
with a global faithful right action of $\displaystyle\mathbb{A}$. The sheaf
$\displaystyle E_{\mathbb{A}}$ is a tautological sheaf on
$\displaystyle\mathscr{C}_{\mathbb{A}}$. Indeed, for any $\displaystyle A$ in
the smooth locus of $\displaystyle\mathscr{C}_{\mathbb{A}}$, the stalk
$\displaystyle E_{\mathbb{A}}|_{A}$ is the cokernel of $\displaystyle A$. In
order to transport this sheaf on $\displaystyle X_{\mathbb{A}}$, we recall the
conormal construction. We denote by $\displaystyle N_{X_{\mathbb{A}}}$ the
normal bundle of $\displaystyle X_{\mathbb{A}}$ in
$\displaystyle\mathbb{P}(\mathrm{J}_{3}(\mathbb{A}))$. We have the conormal
diagram:

$\displaystyle\textstyle{\ignorespaces\ignorespaces\ignorespaces\ignorespaces\mathbb{P}(N_{X_{\mathbb{A}}}^{*}(1))\subset\mathbb{P}(\mathrm{J}_{3}(\mathbb{A}))\times\mathbb{P}(\mathrm{J}_{3}(\mathbb{A})^{*})\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\displaystyle\scriptstyle{p_{\mathbb{A}}}$$\displaystyle\scriptstyle{q_{\mathbb{A}}}$$\displaystyle\textstyle{X_{\mathbb{A}}\subset\mathbb{P}(\mathrm{J}_{3}(\mathbb{A}))}$$\displaystyle\textstyle{X_{\mathbb{A}}^{*}\subset\mathbb{P}(\mathrm{J}_{3}(\mathbb{A})^{*})}$

where $\displaystyle
X_{A}^{*}\subset\mathbb{P}(\mathrm{J}_{3}(\mathbb{A})^{*})$ is the projective
dual of $\displaystyle X_{\mathbb{A}}$. It happens that $\displaystyle
X_{A}^{*}\simeq\mathscr{C}_{\mathbb{A}}$ ([Rob88]) under an identification
$\displaystyle\mathbb{P}(\mathrm{J}_{3}(\mathbb{A}))\simeq\mathbb{P}(\mathrm{J}_{3}(\mathbb{A})^{*})$
and that $\displaystyle
E_{\mathbb{A}}=(q_{\mathbb{A}})_{*}\big{(}p_{\mathbb{A}}^{*}F_{\mathbb{A}}\big{)}$,
where $\displaystyle F_{\mathbb{A}}$ is a rank $\displaystyle n_{\mathbb{A}}$
vector bundle on $\displaystyle X_{\mathbb{A}}$. The bundle $\displaystyle
F_{\mathbb{A}}$ is the tautological bundle we were looking for. Indeed, for
any $\displaystyle A\in X_{A}$, the stalk $\displaystyle F_{\mathbb{A}}|_{A}$
is the orthogonal in $\displaystyle(\mathbb{A}^{\oplus 3})^{*}$ of the kernel
of $\displaystyle A$ (which is a $\displaystyle 2n_{\mathbb{A}}$-dimensional
complex vector sub-space of $\displaystyle\mathbb{A}^{\oplus 3}$ as all
matrices $\displaystyle A\in X_{\mathbb{A}}$ have rank $\displaystyle
n_{\mathbb{A}}$). For the convenience of the reader, we describe explicitly
$\displaystyle F_{\mathbb{A}}$ for each $\displaystyle\mathbb{A}$:

  * •

if $\displaystyle\mathbb{A}=\mathbb{R}\otimes\mathbb{C}$, then $\displaystyle
F_{\mathbb{A}}=\mathscr{O}_{\mathbb{P}^{2}}(1)$,

  * •

if $\displaystyle\mathbb{A}=\mathbb{C}\otimes\mathbb{C}$, then $\displaystyle
F_{\mathbb{A}}=\mathscr{O}_{\mathbb{P}^{2}\times\mathbb{P}^{2}}(1,0)\oplus\mathscr{O}_{\mathbb{P}^{2}\times\mathbb{P}^{2}}(0,1)$,

  * •

if $\displaystyle\mathbb{A}=\mathbb{H}\otimes\mathbb{C}$, then $\displaystyle
F_{\mathbb{A}}=(\mathscr{R}^{*})^{\oplus 2}$, where $\displaystyle\mathscr{R}$
is the tautological bundle on $\displaystyle\mathrm{Gr}(2,6)$.

In the case of the octonions, the determinant
$\displaystyle\mathrm{det}_{\mathbb{O}}(A)$ still makes sense for any
$\displaystyle A\in\mathrm{J}_{3}(\mathbb{O})$ and the equation
$\displaystyle\mathrm{det}_{\mathbb{O}}=0$ defines a cubic in
$\displaystyle\mathbb{P}(\mathrm{J}_{3}(\mathbb{O}))$ known as the Cartan
cubic. Its singular locus is the octonionic projective plane
$\displaystyle\mathbb{OP}^{2}\subset\mathbb{P}(\mathrm{J}_{3}(\mathbb{O}))$, a
projective homogeneous space under $\displaystyle\mathrm{E}_{6}$ (see
[Rob88]). There are however many key steps where the above construction of
$\displaystyle F_{\mathbb{A}}$ fails for $\displaystyle\mathbb{A}=\mathbb{O}$.
First of all, the matrix factorization:

$\displaystyle\mathrm{Com}(A)\times
A=A\times\mathrm{Com}(A)=\mathrm{det}_{\mathbb{O}}(A).I_{3},$

does not hold for all $\displaystyle A\in J_{3}(\mathbb{O})$, due to the lack
of associativity of $\displaystyle\mathbb{O}$. Furthermore, the embedding
$\displaystyle L:\mathbb{O}\xrightarrow{\hskip
22.76228pt}\mathscr{M}_{8,8}(\mathbb{C})$ is not an algebra embedding (because
$\displaystyle\mathbb{O}$ is not associative). Hence, even if for all
$\displaystyle A\in\mathrm{J}_{3}(\mathbb{O})$ we could find $\displaystyle
B_{A}\in\mathrm{J}_{3}(\mathbb{O})$ such that $\displaystyle B_{A}\times
A=A\times B_{A}=\mathrm{det}_{\mathbb{O}}(A).I_{3}$, there wouldn’t
necessarily be any corresponding matrix factorization involving $\displaystyle
L(A)$.

Finally, the very notion of a locally free
$\displaystyle\mathbb{O}\otimes\mathscr{O}_{X}$-module on a scheme
$\displaystyle X$ is a notoriously challenging one. While alternative modules
over $\displaystyle\mathbb{O}$ have been studied by many authors (see for
instance [Jac54]), there are serious difficulties to sheafify these modules.
Indeed, let $\displaystyle X$ be a scheme over $\displaystyle\mathbb{C}$ and
$\displaystyle F$ be a sheaf of locally free (right)
$\displaystyle\mathbb{O}\otimes\mathscr{O}_{X}$-modules of rank $\displaystyle
n$ over $\displaystyle\mathbb{O}\otimes\mathscr{O}_{X}$. Let $\displaystyle
U,V$ trivializing opens in $\displaystyle X$. The transition map
$\displaystyle t_{U,V}$ gives a right $\displaystyle\mathbb{O}$-linear
isomorphism:

$\begin{split}t_{U,V}\ :&\ \mathbb{O}\otimes\mathbb{C}^{n}\times\left\\{U\cap
V\right\\}\stackrel{{\scriptstyle\sim}}{{\xrightarrow{\hskip
22.76228pt}}}\mathbb{O}\otimes\mathbb{C}^{n}\times\left\\{U\cap V\right\\}\\\
&\ (a\otimes v,x)\xrightarrow{\hskip 22.76228pt}(t_{U,V}(x)(a\otimes
v),x).\end{split}$

The algebra $\displaystyle\mathbb{O}$ being non-associative, the only right
$\displaystyle\mathbb{O}$-linear endomorphisms of $\displaystyle\mathbb{O}$
are given by multiplication by pure (complexified) real numbers in
$\displaystyle\mathbb{O}$. We thus find that the transition map $\displaystyle
t_{U,V}$ is of the form:

$\begin{split}t_{U,V}\ :&\ \mathbb{O}\otimes\mathbb{C}^{n}\times\left\\{U\cap
V\right\\}\stackrel{{\scriptstyle\sim}}{{\xrightarrow{\hskip
22.76228pt}}}\mathbb{O}\otimes\mathbb{C}^{n}\times\left\\{U\cap V\right\\}\\\
&\ (a\otimes v,x)\xrightarrow{\hskip 22.76228pt}(a\otimes
r_{U,V}(x)(v),x),\end{split}$

for some cocyle $\displaystyle r\in H^{1}(X,\mathrm{GL}_{n})$. As a
consequence $\displaystyle F=\mathbb{O}\otimes E$ where $\displaystyle E$ is a
vector bundle on $\displaystyle X$, and the global $\displaystyle\mathbb{O}$
structure on $\displaystyle F$ is trivial. This is in sharp contrast with the
fact that the vector bundle $\displaystyle F_{\mathbb{A}}$ described above is
a locally free
$\displaystyle\mathbb{A}\otimes\mathscr{O}_{X_{\mathbb{A}}}$-module of rank
$\displaystyle n_{\mathbb{A}}$ (over
$\displaystyle\mathscr{O}_{X_{\mathbb{A}}}$) whose global right
$\displaystyle\mathbb{A}$-structure is not trivial.

The above analysis suggests that the proof we give that the cokernel of the
matrix $\displaystyle M_{\mathbb{A}}$ is a tautological rank $\displaystyle
n_{\mathbb{A}}$ bundle on $\displaystyle\mathscr{C}_{\mathbb{A}}$ endowed with
a right global $\displaystyle\mathbb{A}$ action can not be adapted to the case
$\displaystyle\mathbb{A}=\mathbb{O}$. But it doesn’t actually show that the
cokernel of $\displaystyle M_{\mathbb{O}}$ is the not, by some miracle, the
bundle which existence is claimed by Bhargava and Ho. As a matter of fact, it
happens that this cokernel doesn’t even live on
$\displaystyle\mathscr{C}_{\mathbb{O}}$ and our first main result describes
some of its geometric and representation theoretic properties. Let
$\displaystyle c=x_{1}e_{1}+\ldots+x_{8}e_{8}$, $\displaystyle
b=x_{9}e_{1}+\ldots+x_{16}e_{n}$, $\displaystyle
a=x_{17}e_{1}+\ldots+x_{24}e_{n}$, $\displaystyle\lambda_{1}=x_{25}e_{1}$,
$\displaystyle\lambda_{2}=x_{26}e_{1}$ and
$\displaystyle\lambda_{3}=x_{27}e_{1}$. Consider the $\displaystyle 24\times
24$ matrix with entries in $\displaystyle\mathbb{C}[x_{1},\ldots,x_{27}]$
given in $\displaystyle 8\times 8$ blocks by:

$\displaystyle
M_{\mathbb{O}}=\begin{pmatrix}\lambda_{1}.I_{8}&L_{c}&{}^{t}L_{b}\\\
{}^{t}L_{c}&\lambda_{2}.I_{8}&L_{a}\\\
L_{b}&{}^{t}L_{a}&\lambda_{1}.I_{8}\end{pmatrix}.$

Let $\displaystyle\mathrm{S}_{\mathrm{ODM}}$ be the sextic polynomial in
$\displaystyle\mathbb{C}[x_{1},\ldots,x_{27}]$ defined by:

$\displaystyle\mathrm{S}_{\mathrm{ODM}}=\mathrm{Det}_{\mathbb{O}}^{2}-4\phi(c,b,a)\mathrm{Det}_{\mathbb{O}}-\Big{|}[c,b,a]\Big{|}^{2},$

where
$\displaystyle\phi(c,b,a)=\frac{1}{2}\mathrm{Re}((c\overline{b}-\overline{b}c)a)$,
$\displaystyle[c,b,a]=(cb)a-c(ba)$ and
$\displaystyle\mathrm{Det}_{\mathbb{O}}$ is the cubic polynomial corresponding
to $\displaystyle\mathrm{det}_{\mathbb{O}}$.

###### Theorem 1

  1. 1.

The vanishing locus of $\displaystyle\mathrm{S}_{\mathrm{ODM}}$ is an
irreducible sextic (equally denoted by
$\displaystyle\mathrm{S}_{\mathrm{ODM}}$) in
$\displaystyle\mathbb{P}(\mathrm{J}_{3}(\mathbb{O}))$ which singular locus has
codimension $\displaystyle 4$ in
$\displaystyle\mathbb{P}(\mathrm{J}_{3}(\mathbb{O}))$. Furthermore, up to a
finite quotient, we have:

$\displaystyle\mathrm{Aut}^{0}(\mathrm{S}_{\mathrm{ODM}})=\mathrm{SL}_{3}(\mathbb{C})\times\mathrm{SO}_{7}(\mathbb{C}).$

  2. 2.

The cokernel of:

$\displaystyle\mathbb{O}^{\oplus
3}\otimes_{\mathbb{C}}\mathscr{O}_{\mathbb{P}(\mathrm{J}_{3}(\mathbb{O}))}(-1)\stackrel{{\scriptstyle
M_{\mathbb{O}}}}{{\xrightarrow{\hskip 22.76228pt}}}\mathbb{O}^{\oplus
3}\otimes_{\mathbb{C}}\mathscr{O}_{\mathbb{P}(\mathrm{J}_{3}(\mathbb{O}))}$

is supported on $\displaystyle\mathrm{S}_{\mathrm{ODM}}$. It has generically
rank $\displaystyle 4$ and is
$\displaystyle\mathrm{SL}_{3}(\mathbb{C})\times\mathrm{SO}_{7}(\mathbb{C})$-equivariant.

The sextic in Theorem 1 has been first found out in [Ogi81] and was then
rediscovered later in [DM98] with computer aid. We call it the Ogievetskiî-
Dray-Manogue sextic and denote it $\displaystyle\mathrm{S}_{\mathrm{ODM}}$.
Theorem 1 will be proved in section in the setting of the octonionic
eigenvalue problem. As far as I know, the objects introduced in Theorem 1 have
not been yet investigated from the algebraic geometry perspective and they
certainly deserve to be studied in details. They seem however not directly
related to the geometry of the Cartan cubic and the tautological bundles that
might exist on it. Our second main result relates to this more classical
object. We denote by $\displaystyle N_{\mathbb{O}}$ the $\displaystyle
24\times 24$ matrix with entries in
$\displaystyle\mathbb{C}[x_{1},\ldots,x_{27}]$ given in $\displaystyle 8\times
8$ blocks by:

$\displaystyle
N_{\mathbb{O}}=\begin{pmatrix}\lambda_{1}.I_{8}&R_{c}&{}^{t}L_{b}\\\
{}^{t}R_{c}&\lambda_{2}.I_{8}&L_{a}\\\
L_{b}&{}^{t}L_{a}&\lambda_{1}.I_{8}\end{pmatrix},$

where $\displaystyle R_{c}$ is the matrix of right multiplication by
$\displaystyle c\in\mathbb{O}$.

###### Theorem 2

  1. 1.

Let $\displaystyle E_{\mathbb{O}}$ be the cokernel of

$\displaystyle\mathbb{O}^{\oplus
3}\otimes_{\mathbb{C}}\mathscr{O}_{\mathbb{P}(\mathrm{J}_{3}(\mathbb{O}))}(-1)\stackrel{{\scriptstyle
N_{\mathbb{O}}}}{{\xrightarrow{\hskip 22.76228pt}}}\mathbb{O}^{\oplus
3}\otimes_{\mathbb{C}}\mathscr{O}_{\mathbb{P}(\mathrm{J}_{3}(\mathbb{O}))}.$

The sheaf $\displaystyle E_{\mathbb{O}}$ is supported on a degree
$\displaystyle 9$ reducible hypersurface in
$\displaystyle\mathbb{P}(\mathrm{J}_{3}(\mathbb{O}))$, which is the union of
hypersurface projectively isomorphic to
$\displaystyle\mathscr{C}_{\mathbb{O}}$ (the Cartan cubic) and a sextic
hypersurface, which we denote $\displaystyle\mathfrak{S}$, and whose equation
is:

$\begin{split}\mathfrak{S}&=\left(\lambda_{1}|a|^{2}+\lambda_{2}|b|^{2}+\lambda_{3}|c|^{2}-\lambda_{1}\lambda_{2}\lambda_{3}\right)^{2}\\\
&\hskip
14.22636pt-4\left(\lambda_{1}|a|^{2}+\lambda_{2}|b|^{2}+\lambda_{3}|c|^{2}-\lambda_{1}\lambda_{2}\lambda_{3}\right)\mathrm{Re}(c)\mathrm{Re}(ab)\\\
&\hskip
14.22636pt+4\left(|b|^{2}|a|^{2}\mathrm{Re}(c)^{2}+|c|^{2}\mathrm{Re}(ab)^{2}-|a|^{2}|b|^{2}|c|^{2}\right),\end{split}$

where $\displaystyle|z|^{2}=z\overline{z}$ and
$\displaystyle\mathrm{Re}(z)=\frac{z+\overline{z}}{2}$.

  2. 2.

The cubic hypersurface is
$\displaystyle\langle\mathrm{Spin}_{7}(\mathbb{C}),\mathrm{SL}_{3}(\mathbb{C})\rangle$-invariant,
where
$\displaystyle\langle\mathrm{Spin}_{7}(\mathbb{C}),\mathrm{SL}_{3}(\mathbb{C})\rangle$
is the subgroup of $\displaystyle\mathrm{SL}_{24}$ generated by
$\displaystyle\mathrm{Spin}_{7}(\mathbb{C})$ and
$\displaystyle\mathrm{SL}_{3}(\mathbb{C})$. The restriction of $\displaystyle
E_{\mathbb{O}}$ to the cubic has generically rank $\displaystyle 4$ and is
$\displaystyle\langle\mathrm{Spin}_{7}(\mathbb{C}),\mathrm{SL}_{3}(\mathbb{C})\rangle$-equivariant.

  3. 3.

The hypersurface $\displaystyle\mathfrak{S}$ is equally
$\displaystyle\langle\mathrm{Spin}_{7}(\mathbb{C}),\mathrm{SL}_{3}(\mathbb{C})\rangle$-invariant.
The restriction of $\displaystyle E_{\mathbb{O}}$ to
$\displaystyle\mathfrak{S}$ has generically rank $\displaystyle 2$ and is
$\displaystyle\langle\mathrm{Spin}_{7}(\mathbb{C}),\mathrm{SL}_{3}(\mathbb{C})\rangle$-equivariant.

It should be noted that the explicit equation of $\displaystyle\mathfrak{S}$
has been first found by Jonathan Hauenstein using the Bertini software. The
embedding of $\displaystyle\mathrm{Spin}_{7}$ in
$\displaystyle\mathrm{SL}_{24}$ we consider does not factor through
$\displaystyle\mathrm{E}_{6}$. As a matter of fact, we haven’t been able to
identify the subgroup of $\displaystyle\mathrm{SL}_{24}$ generated by
$\displaystyle\mathrm{Spin}_{7}(\mathbb{C})$ and
$\displaystyle\mathrm{SL}_{3}(\mathbb{C})$. We can however show the:

###### Theorem 3

Let $\displaystyle G$ be the subgroup of $\displaystyle\mathrm{SL}_{24}$
generated by $\displaystyle\mathrm{Spin}_{7}(\mathbb{C})$ and
$\displaystyle\mathrm{SL}_{3}(\mathbb{C})$. The action of
$\displaystyle\mathbb{C}^{*}\times G$ on
$\displaystyle\mathrm{J}_{3}(\mathbb{O})$ is prehomogeneous.

Theorem 3 is slightly surprising as the subgroup $\displaystyle G$ of
$\displaystyle\mathrm{SL}_{24}$ generated by
$\displaystyle\mathrm{Spin}_{7}(\mathbb{C})$ and
$\displaystyle\mathrm{SL}_{3}(\mathbb{C})$ is not
$\displaystyle\mathrm{E}_{6}$ (and is even not a subgroup of
$\displaystyle\mathrm{E}_{6}$). We will prove this fact in details in section
3.3, but less us have a glimpse at the argument. Any element of
$\displaystyle\mathrm{J}_{3}(\mathbb{O})$ can be diagonalized under the action
of $\displaystyle\mathrm{E}_{6}$. Hence, any vector of
$\displaystyle\mathrm{J}_{3}(\mathbb{O})$ can be transformed, under the action
of $\displaystyle\mathrm{E}_{6}$, to a vector whose corresponding
$\displaystyle 24*24$ matrix has rank $\displaystyle 0,8,16$ or $\displaystyle
24$. On the other hand, the action of $\displaystyle G$ on
$\displaystyle\mathrm{J}_{3}(\mathbb{O})$ preserves the rank. Since there are
vectors in $\displaystyle\mathrm{J}_{3}(\mathbb{O})$ whose corresponding
$\displaystyle 24*24$ matrix have rank $\displaystyle 22$, we conclude that
$\displaystyle G$ is not $\displaystyle\mathrm{E}_{6}$. The prehomogeneous
space $\displaystyle(\mathrm{J}_{3}(\mathbb{O}),\mathbb{C}^{*}\times G)$ does
not appear in the list of Kimura-Sato [KS77]. This is not surprising as it is
most likely not irreducible as a prohomogeneous space. It thus provides an
interesting example of a non-irreducible prehomogeneous vector space which
does not split as a sum of lower dimensional prehomogeneous vector spaces.

We prove Theorems 2 and 3 in the second section of this paper. There, we
introduce a variant of the octonionic eigenvalue problem, which we call the
twisted octonionic eigenvalue problem. We deduce the aforementioned results
from both algebraic and geometric features of this twisted problem. We note
that the geometry of the twisted octonionic eigenvalue problem differs
strongly from that of the corresponding untwisted problem. Indeed, one easily
notices that the action of $\displaystyle\mathrm{GL}_{3}\times\mathrm{SO}_{7}$
on $\displaystyle\mathrm{J}_{3}(\mathbb{O})$ is far from being homogeneous
(the generic orbit has codimension at least $\displaystyle 6$ in
$\displaystyle\mathrm{J}_{3}(\mathbb{O})$). Theorem 3 thus shows that the
twisted problem is eminently more symmetric than the untwisted problem.

The sheaf $\displaystyle E_{\mathbb{O}}$ appearing in Theorem 2 may be
considered as a quasi-tautological sheaf on
$\displaystyle\mathscr{C}_{\mathbb{O}}$ as it is the cokernel of a matrix
whose tautological interpretation is clear. However, the description of
$\displaystyle E_{\mathbb{O}}|_{A}$ for generic $\displaystyle
A\in\mathscr{C}_{\mathbb{O}}$ is not completely obvious. Furthermore, the
restriction of $\displaystyle E_{\mathbb{O}}$ to the smooth locus of
$\displaystyle\mathscr{C}_{\mathbb{O}}$ is not a vector bundle, so that
$\displaystyle E_{\mathbb{O}}$ is not the Fourier-Mukai transform, by the
conormal diagram, of a vector bundle on $\displaystyle\mathbb{OP}^{2}$. One
can however expect that it is the Fourier-Mukai transform of a rank
$\displaystyle 4$ coherent sheaf on $\displaystyle\mathbb{OP}^{2}$ which could
be conceived as quasi-tautological sheaf on $\displaystyle\mathbb{OP}^{2}$.
The tautological interpretation of this sheaf is again not completely
transparent. Further investigations should reveal many more fascinating
features associated to the twisted octonionic eigenvalue problem. In a
forthcoming paper with Jonathan Hauenstein [AH], we will we study in more
details the geometry and singularities of $\displaystyle\SS$ and
$\displaystyle\mathfrak{S}$.

Acknowledgements : I am very grateful to Jonathan Hauenstein for the many
discussions we have had on the geometry of $\displaystyle\SS$ and
$\displaystyle\mathfrak{S}$ and especially for providing me with the explicit
equation of $\displaystyle\mathfrak{S}$ using the software Bertini. Many
thanks to Oleg Ogievetskiî for sharing with me an English translation of his
paper [Ogi81]. I would also like to acknowledge discussions I have had with
Robert Bryant and I thank him for the clean argument he communicated to me
showing that the subgroup of $\displaystyle\mathrm{E}_{6}$ generated by
$\displaystyle\mathrm{Spin_{7}}$ and $\displaystyle\mathrm{SL}_{3}$ is
$\displaystyle\mathrm{E}_{6}$ itself.

##  2 The eigenvalue problem for $\displaystyle\mathrm{J}_{3}(\mathbb{O})$ and
the Ogievetskiî-Dray-Manogue sextic

###  2.1 The octonionic eigenvalue problem

Let $\displaystyle\mathbb{A}$ be $\displaystyle\mathbb{C}$,
$\displaystyle\mathbb{C}\otimes_{\mathbb{R}}\mathbb{C}$,
$\displaystyle\mathbb{H}$ or $\displaystyle\mathbb{O}$ (the algebra
$\displaystyle\mathbb{H}$ and $\displaystyle\mathbb{O}$ being the complexified
quaternions and octonions). The left (complex) eigenvalue problem for
$\displaystyle\mathrm{J}_{3}(\mathbb{A})$ is the following. Given
$\displaystyle A=\begin{pmatrix}\lambda_{1}&c&\overline{b}\\\
\overline{c}&\lambda_{2}&a\\\
b&\overline{a}&\lambda_{3}\end{pmatrix}\in\mathrm{J}_{3}(\mathbb{A})$, what
are the $\displaystyle\mu\in\mathbb{C}$ such that there exists a non zero
vector $\displaystyle\begin{pmatrix}x\\\ y\\\ z\end{pmatrix}\in\mathbb{A}^{3}$
with $\displaystyle A\begin{pmatrix}x\\\ y\\\
z\end{pmatrix}=\mu.\begin{pmatrix}x\\\ y\\\ z\end{pmatrix}$? So that there is
no confusion due to the non-commutativity (and possibly non associativity) of
$\displaystyle\mathbb{A}$, the equality $\displaystyle A\begin{pmatrix}x\\\
y\\\ z\end{pmatrix}=\mu.\begin{pmatrix}x\\\ y\\\ z\end{pmatrix}$ is meant to
be:

$\left\\{\begin{split}\lambda_{1}x+cy+\overline{a}_{2}z&\ =\mu x\\\
\overline{c}x+\lambda_{2}y+az&\ =\mu y\\\ bx+\overline{a}y+\lambda_{2}z&\ =\mu
z.\\\ \end{split}\right.$

Let us say that $\displaystyle A\in\mathrm{J}_{3}(\mathbb{A})$ is singular or
degenerate if $\displaystyle 0$ is an eigenvalue of $\displaystyle A$. In case
$\displaystyle\mathbb{A}$ is associative, the following proposition is well-
known. We recall a proof for the convenience of the reader.

###### Proposition 2.1.1

Let $\displaystyle\mathbb{A}$ be an associative composition algebra which is
finite dimensional over $\displaystyle\mathbb{C}$. Let
$\displaystyle\begin{pmatrix}\lambda_{1}&c&\overline{b}\\\
\overline{c}&\lambda_{2}&a\\\
b&\overline{a}&\lambda_{3}\end{pmatrix}\in\mathrm{J}_{3}(\mathbb{A})$ and
$\displaystyle\mu\in\mathbb{C}$. Then, there exists a non zero vector
$\displaystyle\begin{pmatrix}x\\\ y\\\ z\end{pmatrix}\in\mathbb{A}^{3}$ with
$\displaystyle A\begin{pmatrix}x\\\ y\\\
z\end{pmatrix}=\mu.\begin{pmatrix}x\\\ y\\\ z\end{pmatrix}$ if and only if
$\displaystyle\mathrm{det}_{\mathbb{A}}(A-\mu.I_{3})=0$.

Proof :
$\displaystyle\blacktriangleright\;$ If $\displaystyle A=0$ the result is
obvious. We now assume that $\displaystyle A\neq 0$, for instance
$\displaystyle c\neq 0$ (the non-vanishing of any other coefficient in
$\displaystyle A$ is dealt with exactly in the same way). Assume that
$\displaystyle\mathrm{det}_{\mathbb{A}}(A-\mu.I_{3})=0$. If
$\displaystyle\mathrm{Com}(A-\mu.I_{3})\neq 0$ then, the equation:

$\displaystyle(A-\mu.I_{3})\times\mathrm{Com}(A-\mu.I_{3})=\mathrm{det}_{\mathbb{A}}(A-\mu.I_{3}).I_{3}=0$

shows that any non-zero column vector of
$\displaystyle\mathrm{Com}(A-\mu.I_{3})$ is an eigenvector for $\displaystyle
A$ with respect to $\displaystyle\mu$. If
$\displaystyle\mathrm{Com}(A-\mu.I_{3})=0$, then one easily checks that
$\displaystyle\begin{pmatrix}\lambda_{2}-\mu\\\ -\overline{c}\\\
0\end{pmatrix}$ is a non-zero eigenvector for $\displaystyle A$ with respect
to $\displaystyle\mu$.

Assume on the other side that there exists $\displaystyle\begin{pmatrix}x\\\
y\\\ z\end{pmatrix}\neq 0$ such that
$\displaystyle(A-\mu.I_{3})\begin{pmatrix}x\\\ y\\\ z\end{pmatrix}=0$. Let
$\displaystyle B=\begin{pmatrix}x&x&x\\\ y&y&y\\\ z&z&z\end{pmatrix}$. We have
$\displaystyle(A-\mu.I_{3})\times B=0$ and since $\displaystyle\mathbb{A}$ is
associative, we also have
$\displaystyle\mathrm{Com}(A-\mu.I_{3})\times(\mathbb{A}-\mu.I_{3})\times
B=0$. But
$\displaystyle\mathrm{Com}(A-\mu.I_{3})\times(A-\mu.I_{3})=\mathrm{det}_{\mathbb{A}}(A-\mu.I_{3}).I_{3}$,
so that we get:

$\displaystyle\mathrm{det}_{\mathbb{A}}(A-\mu.I_{3}).B=0.$

The matrix $\displaystyle B$ is non-zero by hypothesis and
$\displaystyle\mathrm{det}_{\mathbb{A}}(A-\mu.I_{3})\in\mathbb{C}$ imply that
$\displaystyle\mathrm{det}_{\mathbb{A}}(A-\mu.I_{3})=0$.
$\displaystyle\blacktriangleleft$

The algebra $\displaystyle\mathbb{O}$ being not associative, the above result
does not apply in this case. Nevertheless, Ogievetskiî found out a
characteristic equation that determines the complex eigenvalues of
$\displaystyle A\in\mathrm{J}_{3}(\mathbb{O})$:

######  Theorem 2.1.2 (see [Ogi81, DM98])

  1. 1.

Let $\displaystyle A=\begin{pmatrix}\lambda_{1}&c&\overline{b}\\\
\overline{c}&\lambda_{2}&a\\\
b&\overline{a}&\lambda_{3}\end{pmatrix}\in\mathrm{J}_{3}(\mathbb{O})$. The
matrix $\displaystyle A$ is singular if and only if the $\displaystyle 24*24$
matrix $\displaystyle\begin{pmatrix}\lambda_{1}.I_{8}&L_{c}&{}^{t}L_{b}\\\
{}^{t}L_{c}&\lambda_{2}.I_{8}&L_{a}\\\
L_{b}&{}^{t}L_{a}&\lambda_{3}.I_{8}\end{pmatrix}$ is singular.

  2. 2.

The degeneracy locus of the matrix
$\displaystyle\begin{pmatrix}\lambda_{1}.I_{8}&L_{c}&{}^{t}L_{b}\\\
{}^{t}L_{c}&\lambda_{2}.I_{8}&L_{a}\\\
L_{b}&{}^{t}L_{a}&\lambda_{3}.I_{8}\end{pmatrix}$ is given by the equation:

$\displaystyle(\mathrm{det}_{\mathbb{O}}(A))^{2}-4\phi(c,b,a)\mathrm{det}_{\mathbb{O}}(A)-\Big{|}[c,b,a]\Big{|}^{2}=0.$

As we will build upon it when dealing with the twisted version of the
octonionic eigenvalue problem, we recall the main steps of the proof from
[Ogi81]:

Proof :
$\displaystyle\blacktriangleright\;$ The first assertion in the Theorem is
obvious and follows from the definitions. We prove the second assertion. For
$\displaystyle a,b,c\in\mathbb{O}$ and
$\displaystyle\lambda_{1},\lambda_{2},\lambda_{3}\in\mathbb{C}$ such that
$\displaystyle|a|,|b|,|c|,\lambda_{1},\lambda_{2},\lambda_{3}\neq 0$ and
$\displaystyle|c|^{2}\lambda_{1}-\dfrac{|c|^{4}}{\lambda_{2}}\neq 0$, we
define the following $\displaystyle 24*24$ matrices in block form:

$\displaystyle R_{1}=\begin{pmatrix}\frac{1}{|c|^{2}}L_{c}&0_{8,8}&0_{8,8}\\\
0_{8,8}&I_{8}&0_{8,8}\\\
0_{8,8}&0_{8,8}&\frac{1}{|a|^{2}}{}^{t}L_{a}\end{pmatrix}\ \ \textrm{and}\ \
R_{2}=\begin{pmatrix}I_{8}&\frac{|c|^{2}}{\lambda_{2}}&0_{8,8}\\\
0_{8,8}&I_{8}&0_{8,8}\\\
\frac{1}{\delta}B&\frac{|a|^{2}}{\lambda_{2}}I_{8}&I_{8}\end{pmatrix},$

where $\displaystyle
B=L_{a}L_{b}L_{c}-\dfrac{|a|^{2}|c|^{2}}{\lambda_{2}}.I_{8}$,
$\displaystyle\delta=|c|^{2}\lambda_{1}-\dfrac{|c|^{4}}{\lambda_{2}}$,
$\displaystyle 0_{8,8}$ is the zero $\displaystyle 8*8$ matrix and
$\displaystyle I_{8}$ the identity matrix of size $\displaystyle 8$. One
easily checks that:

$\displaystyle\begin{pmatrix}\lambda_{1}.I_{8}&L_{c}&{}^{t}L_{b}\\\
{}^{t}L_{c}&\lambda_{2}.I_{8}&L_{a}\\\
L_{b}&{}^{t}L_{a}&\lambda_{3}.I_{8}\end{pmatrix}=R_{1}R_{2}\begin{pmatrix}\delta.I_{8}&0_{8,8}&0_{8,8}\\\
0_{8,8}&\lambda_{2}.I_{8}&0_{8,8}\\\
0_{8,8}&0_{8,8}&\nu.I_{8}-\dfrac{1}{\delta}B{}^{t}B\end{pmatrix}{}^{t}R_{2}{}^{t}R_{1},$

where $\displaystyle\nu=\lambda_{3}|a|^{2}-\frac{|a|^{4}}{\lambda_{2}}$. Since
$\displaystyle\mathrm{det}(R_{1})=\dfrac{1}{|a|^{8}|c|^{8}}$ and
$\displaystyle\mathrm{det}(R_{2})=1$, we get:

$\begin{split}\mathrm{det}\begin{pmatrix}\lambda_{1}.I_{8}&L_{c}&{}^{t}L_{b}\\\
{}^{t}L_{c}&\lambda_{2}.I_{8}&L_{a}\\\
L_{b}&{}^{t}L_{a}&\lambda_{3}.I_{8}\end{pmatrix}&=\dfrac{1}{|a|^{16}|c|^{16}}\mathrm{det}\left(\delta\nu\lambda_{2}.I_{8}-\lambda_{2}.B{}^{t}B\right)\\\
&=\mathrm{det}\left((\lambda_{1}\lambda_{2}\lambda_{3}-\lambda_{1}|a|^{2}-\lambda_{2}|b|^{2}-\lambda_{3}|c|^{2}).I_{8}+(L_{a}L_{b}L_{c}+{}^{t}L_{c}{}^{t}L_{b}{}^{t}L_{a})\right)\\\
&=\mathrm{det}(-(2\mathrm{Re}(cab)-\mathrm{det}_{\mathbb{O}}(A)).I_{8}+(L_{a}L_{b}L_{c}+{}^{t}L_{c}{}^{t}L_{b}{}^{t}L_{a})),\end{split}$

where $\displaystyle A=\begin{pmatrix}\lambda_{1}&c&\overline{b}\\\
\overline{c}&\lambda_{2}&a\\\ b&\overline{a}&\lambda_{3}\end{pmatrix}.$

The expression
$\displaystyle\mathrm{det}(-(2\mathrm{Re}(cab)-\mathrm{det}_{\mathbb{O}}(A)).I_{8}+(L_{a}L_{b}L_{c}+{}^{t}L_{c}{}^{t}L_{b}{}^{t}L_{a}))$
has no denominators so that the above equality holds for all $\displaystyle
a,b,c\in\mathbb{O}$ and
$\displaystyle\lambda_{1},\lambda_{2},\lambda_{3}\in\mathbb{C}$. We deduce the
following:

The determinant of
$\displaystyle\begin{pmatrix}\lambda_{1}.I_{8}&L_{c}&{}^{t}L_{b}\\\
{}^{t}L_{c}&\lambda_{2}.I_{8}&L_{a}\\\
L_{b}&{}^{t}L_{a}&\lambda_{3}.I_{8}\end{pmatrix}$ vanishes if and only if
$\displaystyle 2\mathrm{Re}(cab)-\mathrm{det}_{\mathbb{O}}(A)$ is a (complex)
eigenvalue of
$\displaystyle(L_{a}L_{b}L_{c}+{}^{t}L_{c}{}^{t}L_{b}{}^{t}L_{a})$. We shall
now determine the (complex) eigenvalues of
$\displaystyle(L_{a}L_{b}L_{c}+{}^{t}L_{c}{}^{t}L_{b}{}^{t}L_{a})$. This means
we are looking for the $\displaystyle\kappa\in\mathbb{C}$ such that:

$\exists x\in\mathbb{O},\ x\neq 0\ \
a.(b.(c.x))+\overline{c}.(\overline{b}.(\overline{a}.x))=\kappa.x$ (4)

Let us consider a four dimensional quaternionic subalgebra of
$\displaystyle\mathbb{O}$ which contains $\displaystyle a$ and $\displaystyle
c$ (its existence is granted by a Theorem of Hurwitz). Let $\displaystyle
e\in\mathbb{O}$ be a vector orthogonal to $\displaystyle M$ such that
$\displaystyle\mathbb{O}=M\oplus M.e$. We write $\displaystyle
b=b_{0}+b_{1}.e$ and $\displaystyle x=x_{0}+x_{1}.e$, with $\displaystyle
b_{0},b_{1},x_{0},x_{1}\in M$. Using the relations:

$\displaystyle u.(v.e)=(v.u).e,\ (u.e).(v.e)=-\overline{v}.u,\
u.e=e.\overline{u}\ \textrm{and}\ (u.e).v=(u.\overline{v}).e,$

valid for all $\displaystyle(u,v)\in M^{2}$, we find that equation 4 is
equivalent to the system:

$\left\\{\begin{split}2\mathrm{Re}(c.a.b_{0}).x_{0}+(\overline{c}.a-a.\overline{c}).\overline{x_{1}}.b_{1}&=\kappa.x_{0}\\\
(\overline{a}.c-c.\overline{a}).x_{0}.\overline{b_{1}}+2\mathrm{Re}(c.b_{0}.a).\overline{x_{1}}&=\kappa.\overline{x_{1}}.\end{split}\right.$

This system can be written in matrix form:

$\displaystyle\begin{pmatrix}(2\mathrm{Re}(c.a.b_{0})-\kappa).I_{4}&({}^{t}L_{c}L_{a}-L_{a}{}^{t}L_{c})R_{b_{1}}\\\
({}^{t}L_{a}L_{c}-L_{c}{}^{t}L_{a}){}^{t}R_{b_{1}}&(2\mathrm{Re}(c.b_{0}.a)-\kappa).I_{4}\end{pmatrix}\times\begin{pmatrix}x_{0}\\\
\overline{x_{1}}\end{pmatrix}=\begin{pmatrix}0_{4,1}\\\ 0_{4,1}\end{pmatrix}.$

The existence of such an $\displaystyle\begin{pmatrix}x_{0}\\\
\overline{x_{1}}\end{pmatrix}\neq 0$ is then equivalent to:

$\displaystyle\mathrm{det}\begin{pmatrix}(2\mathrm{Re}(c.a.b_{0})-\kappa).I_{4}&({}^{t}L_{c}L_{a}-L_{a}{}^{t}L_{c})R_{b_{1}}\\\
({}^{t}L_{a}L_{c}-L_{c}{}^{t}L_{a}){}^{t}R_{b_{1}}&(2\mathrm{Re}(c.b_{0}.a)-\kappa).I_{4}\end{pmatrix}=0.$

The bottom right entry of the above block-matrix commutes with every matrix,
so that the block-computation of the determinant gives:

$\begin{split}\mathrm{det}&\begin{pmatrix}(2\mathrm{Re}(c.a.b_{0})-\kappa).I_{4}&({}^{t}L_{c}L_{a}-L_{a}{}^{t}L_{c})R_{b_{1}}\\\
({}^{t}L_{a}L_{c}-L_{c}{}^{t}L_{a}){}^{t}R_{b_{1}}&(2\mathrm{Re}(c.b_{0}.a)-\kappa).I_{4}\end{pmatrix}\\\
&=\mathrm{det}((2\mathrm{Re}(c.a.b_{0})-\kappa)(2\mathrm{Re}(c.b_{0}.a)-\kappa).I_{4}-(({}^{t}L_{c}L_{a}-L_{a}{}^{t}L_{c})R_{b_{1}})\times(({}^{t}L_{a}L_{c}-L_{c}{}^{t}L_{a}){}^{t}R_{b_{1}}))\\\
&=\mathrm{det}(\left((2\mathrm{Re}(c.a.b_{0})-\kappa)(2\mathrm{Re}(c.b_{0}.a)-\kappa)-|\overline{c}.a-a.\overline{c}|^{2}.|b_{1}|^{2}\right).I_{4}).\end{split}$

We deduce that:

$\displaystyle\kappa$ is an eigenvalue of
$\displaystyle(L_{a}L_{b}L_{c}+{}^{t}L_{c}{}^{t}L_{b}{}^{t}L_{a})\
\Leftrightarrow\
(2\mathrm{Re}(c.a.b_{0})-\kappa)(2\mathrm{Re}(c.b_{0}.a)-\kappa)-|\overline{c}.a-a.\overline{c}|^{2}.|b_{1}|^{2}=0$.

We note that $\displaystyle b=b_{0}+b_{1}.e$ and $\displaystyle e\perp M$,
hence:

$\displaystyle\mathrm{Re}(c.a.b_{0})=\mathrm{Re}(c.a.b),\
\textrm{Re}(c.b_{0}.a)=\mathrm{Re}(c.b.a)\ \textrm{and}\
|\overline{c}.a-a.\overline{c}|^{2}.|b_{1}|^{2}=|[a,b_{1}.e,c]|^{2}=|[c,b,a]|^{2}.$

We furthermore recall that $\displaystyle
2\mathrm{Re}(c.a.b)-\mathrm{det}_{\mathbb{O}}(A)$ is an eigenvalue of
$\displaystyle(L_{a}L_{b}L_{c}+{}^{t}L_{c}{}^{t}L_{b}{}^{t}L_{a})$ if and only
if the matrix
$\displaystyle\begin{pmatrix}\lambda_{1}.I_{8}&L_{c}&{}^{t}L_{b}\\\
{}^{t}L_{c}&\lambda_{2}.I_{8}&L_{a}\\\
L_{b}&{}^{t}L_{a}&\lambda_{3}.I_{8}\end{pmatrix}$ is singular. We finally find
that the degeneracy locus of the matrix
$\displaystyle\begin{pmatrix}\lambda_{1}.I_{8}&L_{c}&{}^{t}L_{b}\\\
{}^{t}L_{c}&\lambda_{2}.I_{8}&L_{a}\\\
L_{b}&{}^{t}L_{a}&\lambda_{3}.I_{8}\end{pmatrix}$ is defined by the equation:

$\begin{split}&\
(2\mathrm{Re}(c.a.b)-\kappa)(2\mathrm{Re}(c.b.a)-\kappa)-|[c,b,a]|^{2}=0\\\
\Leftrightarrow\ &\
\mathrm{Det}_{\mathbb{O}}(2\mathrm{Re}(c.a.b)-2\mathrm{Re}(c.b.a)+\mathrm{Det}_{\mathbb{O}})-|[c,b,a]|^{2}=0\\\
\Leftrightarrow\ &\ \mathrm{Det}_{\mathbb{O}}^{2}+2\mathrm{Re}(c(ab-
ba))\mathrm{Det}_{\mathbb{O}}-|[c,b,a]|^{2}=0\\\ \Leftrightarrow\ &\
\mathrm{Det}_{\mathbb{O}}^{2}+4\phi(a,\overline{b},c)\mathrm{Det}_{\mathbb{O}}-|[c,b,a]|^{2}=0\\\
\Leftrightarrow\ &\
\mathrm{Det}_{\mathbb{O}}^{2}-4\phi(a,b,c)\mathrm{Det}_{\mathbb{O}}-|[c,b,a]|^{2}=0.\\\
\end{split}$

$\displaystyle\blacktriangleleft$

###  2.2 The geometry of the Ogievetskiî-Dray-Manogue sextic

We will now describe elementary geometric features of the sextic
$\displaystyle\SS\subset\mathbb{P}^{26}$ defined by the equation:

$\displaystyle(\mathrm{det}_{\mathbb{O}}(A))^{2}-4\phi(c,b,a)\mathrm{det}_{\mathbb{O}}(A)-\Big{|}[c,b,a]\Big{|}^{2}=0.$

Some of them have been obtained with the help of Macaulay2 ([GS]):

###### Theorem 2.2.1

  1. 1.

The sextic $\displaystyle\SS\subset\mathbb{P}^{26}$ is irreducible and its
singular locus is of codimension $\displaystyle 4$ in
$\displaystyle\mathbb{P}^{26}$.

  2. 2.

The neutral component the automorphism group of $\displaystyle\SS$ is
$\displaystyle\mathrm{SO_{7}}\times\mathrm{SL}_{3}$.

  3. 3.

The map:

$\displaystyle\mathrm{Gr}(6,\mathrm{J}_{3}(\mathbb{O}))/\\!\\!/\mathrm{SO_{7}}\times\mathrm{SL}_{3}\xrightarrow{\hskip
22.76228pt}\mathrm{S}^{6}\mathbb{C}^{6}/\\!\\!/\mathrm{GL}_{6}$

which sends a six dimensional subspace $\displaystyle
L\subset\mathrm{J}_{3}(\mathbb{O})$ to the sextic $\displaystyle\SS\cap L$ is
generically étale onto its image.

Proof :
$\displaystyle\blacktriangleright\;$ The codimension of the singular locus of
$\displaystyle\SS\subset\mathbb{P}^{26}$ is obtained computationally with the
help of Macaulay2. The only <<challenge>> is to find a way to write explicitly
the equation $\displaystyle\SS$ as, to the best of my knowledge, there is no
obvious package to simulate the octonion algebra in Macaulay2. We refer to the
first appendix for a description of our algorithm. The irreducibility of
$\displaystyle\SS$ follows as it is then a normal hypersurface in
$\displaystyle\mathbb{P}^{26}$.

We now prove the second and third assertion. We first describe the action of
$\displaystyle\mathrm{SO_{7}}\times\mathrm{SL}_{3}$ on
$\displaystyle\mathrm{J}_{3}(\mathbb{O})$. Let $\displaystyle
T_{1}\in\mathrm{SO_{7}}$ and $\displaystyle
A=\begin{pmatrix}\lambda_{1}&c&\overline{b}\\\ \overline{c}&\lambda_{2}&a\\\
b&\overline{a}&\lambda_{3}\end{pmatrix}\in\mathrm{J}_{3}(\mathbb{O})$, we put:

$\displaystyle
T_{1}.A=\begin{pmatrix}\lambda_{1}&T_{1}(c)&(\overline{T_{1}(b)})\\\
(\overline{T_{1}(c)})&\lambda_{2}&T_{1}(a)\\\
T_{1}(b)&(\overline{T_{1}(a)})&\lambda_{3}\end{pmatrix}\in\mathrm{J}_{3}(\mathbb{O}),$

where $\displaystyle\mathrm{SO_{7}}\subset\mathrm{SO_{8}}$ is the isotropy
group of $\displaystyle 1\in\mathbb{O}$. On the other hand, for $\displaystyle
C\in\mathrm{SL_{3}}$ and $\displaystyle
A=\begin{pmatrix}\lambda_{1}&c&\overline{b}\\\ \overline{c}&\lambda_{2}&a\\\
b&\overline{a}&\lambda_{3}\end{pmatrix}\in\mathrm{J}_{3}(\mathbb{O})$, we put:

$\displaystyle C.A={}^{t}C\times A\times C\in\mathrm{J}_{3}(\mathbb{O}),$

where $\displaystyle\times$ denotes here the product of $\displaystyle 3*3$
matrices (there is no associativity issues since all coefficients of
$\displaystyle C$ are complex and thus commute with all coefficients of
$\displaystyle A$). Since $\displaystyle T_{1}(x)=x$, for all $\displaystyle
T_{1}\in\mathrm{SO_{7}}$ and for all $\displaystyle x\in\mathbb{O}$ with zero
imaginary part, we deduce that:

$\displaystyle C.(T_{1}.A)=T_{1}.(C.A),$

for all $\displaystyle T_{1}\in\mathrm{SO_{7}},\ C\in\mathrm{SL_{3}}$ and
$\displaystyle A\in\mathrm{J}_{3}(\mathbb{O})$. Furthermore, we notice that
$\displaystyle\mathrm{SO_{7}}\cap\mathrm{SL_{3}}=\\{Id\\}$ as subgroups of
$\displaystyle\mathrm{SL}_{27}$. As a consequence, we find that the subgroup
of $\displaystyle\mathrm{SL}_{27}$ generated by $\displaystyle\mathrm{SO_{7}}$
and $\displaystyle\mathrm{SL_{3}}$ is
$\displaystyle\mathrm{SO_{7}}\times\mathrm{SL_{3}}$. Let $\displaystyle
A\in\mathrm{J}_{3}(\mathbb{O})$ be a singular matrix. By definition, there
exists $\displaystyle\begin{pmatrix}x\\\ y\\\
z\end{pmatrix}\in\mathbb{O}^{\oplus 3}$, with
$\displaystyle\begin{pmatrix}x\\\ y\\\ z\end{pmatrix}\neq
0_{\mathbb{O}^{\oplus 3}}$ such that:

$\displaystyle A\times\begin{pmatrix}x\\\ y\\\
z\end{pmatrix}=0_{\mathbb{O}^{\oplus 3}}.$

For any $\displaystyle T_{1}\in\mathrm{SO_{7}}$, the Triality principle (see
section 2 of [DeS01] for instance) insures the existence of $\displaystyle
T_{2}\in\mathrm{SO}_{8}$ such that:

$\displaystyle\forall(u,v)\in\mathbb{O}^{2},\ T_{2}(u.v)=T_{1}(u).T_{2}(v).$

One then easily computes that:

$\displaystyle(T_{1}.A)\times\begin{pmatrix}T_{2}(x)\\\ T_{2}(y)\\\
T_{2}(z)\end{pmatrix}=\begin{pmatrix}T_{2}&&\\\ &T_{2}&\\\
&&T_{2}\end{pmatrix}\times\left(A\times\begin{pmatrix}x\\\ y\\\
z\end{pmatrix}\right)=0_{\mathbb{O}^{\oplus 3}}.$

We have $\displaystyle\begin{pmatrix}T_{2}(x)\\\ T_{2}(y)\\\
T_{2}(z)\end{pmatrix}\neq 0_{\mathbb{O}^{\oplus 3}}$, hence $\displaystyle
T_{1}.A$ is also a singular matrix. Furthermore, we obviously note that for
all $\displaystyle C\in\mathrm{SL_{3}}$, we have:

$\displaystyle(C.A)\times\left(C^{-1}\times\begin{pmatrix}x\\\ y\\\
z\end{pmatrix}\right)=0_{\mathbb{O}^{\oplus 3}},$

which shows that $\displaystyle C.A$ is also a singular element of
$\displaystyle\mathrm{J}_{3}(\mathbb{O})$. The variety of singular matrices in
$\displaystyle\mathrm{J}_{3}(\mathbb{O})$ is thus preserved by the action of
$\displaystyle\mathrm{SO_{7}}\times\mathrm{SL_{3}}$. By Theorem 2.1.2, we
deduce that $\displaystyle\mathrm{SO_{7}}\times\mathrm{SL_{3}}$ is a subgroup
of the neutral component of $\displaystyle\mathrm{Aut}(\SS)$.

In order to prove that
$\displaystyle\mathrm{Aut}^{0}(\SS)=\mathrm{SO_{7}}\times\mathrm{SL_{3}}$ (up
to a finite quotient), we show, using Macaulay2, that the Lie algebra of
$\displaystyle\mathrm{Aut}^{0}(\SS)$ has dimension less or equal to
$\displaystyle 29$. Consider the map:

$\begin{split}\varphi\,\,\,:\,\,\,&M_{6\times
27}/\\!\\!/\mathrm{Aut}^{0}(\SS)\xrightarrow{\hskip
22.76228pt}S^{6}\mathbb{C}^{6}\\\ &\left[(m)_{i,j}\right]\xrightarrow{\hskip
22.76228pt}\SS(m_{1,1}.z_{1}+\cdots+m_{1,6}.z_{6},\cdots,m_{27,1}.z_{1}+\cdots+m_{27,6}.z_{6}),\\\
\end{split}$

where $\displaystyle z_{1},\cdots,z_{6}$ is a basis for
$\displaystyle(\mathbb{C}^{6})^{*}$. Let us denote by
$\displaystyle\mathfrak{aut}(\SS)$ the Lie algebra of
$\displaystyle\mathrm{Aut}^{0}(\SS)$. The differential of $\displaystyle\phi$
at a point $\displaystyle\left[(m)_{i,j}\right]\in M_{6\times 27}$ is given by
:

$\begin{split}d\varphi_{\left[m_{i,j}\right]}\,\,\,:\,\,\,&M_{6\times
27}/\mathfrak{aut}(\SS)\xrightarrow{\hskip 22.76228pt}S^{6}\mathbb{C}^{6}\\\
&\left[q_{i,j}\right]\xrightarrow{\hskip
22.76228pt}\sum_{i=1}^{27}\left(\sum_{j=1}^{6}q_{i,j}.z_{j}\right)\times\dfrac{\SS}{\partial
x_{i}}(m_{1,1}.z_{1}+\cdots+m_{1,6}.z_{6},\cdots,m_{27,1}.z_{1}+\cdots+m_{27,6}.z_{6}),\\\
\end{split}$

We compute with the help of Macaulay2 that the rank of $\displaystyle
d\varphi$ at a generic point in $\displaystyle
M_{6\times\mathrm{J}_{3}(\mathbb{O})}/\mathfrak{aut}(\SS)$ is bigger or equal
to $\displaystyle 133$ (see the appendix A for the details of the algorithm).
We deduce that:

$\displaystyle\dim
M_{6\times\mathrm{J}_{3}(\mathbb{O})}/\mathfrak{aut}(\SS)\geq 133,$

which implies that $\displaystyle\dim\mathfrak{aut}(\SS)\leq 29$. Since
$\displaystyle\mathrm{SO_{7}}\times\mathrm{SL_{3}}\subset\mathrm{Aut}^{0}(\SS)$
and $\displaystyle\dim\mathrm{SO_{7}}\times\mathrm{SL_{3}}=29$, we find that:

$\displaystyle\mathrm{Aut}^{0}(\SS)=\mathrm{SO_{7}}\times\mathrm{SL_{3}}.$

We can now infer that the map:

$\displaystyle\mathrm{Gr}(6,\mathrm{J}_{3}(\mathbb{O}))/\\!\\!/\mathrm{SO_{7}}\times\mathrm{SL}_{3}\xrightarrow{\hskip
22.76228pt}\mathrm{S}^{6}\mathbb{C}^{6}/\\!\\!/\mathrm{GL}_{6}$

is generically étale onto its image. Indeed, the rank of $\displaystyle d\phi$
being generically $\displaystyle 133$ and $\displaystyle\dim
M_{6\times\mathrm{J}_{3}(\mathbb{O})}/\\!\\!/\mathrm{SO_{7}}\times\mathrm{SL}_{3}=162-29=133$,
we find that $\displaystyle\phi$ is generically étale onto its image. We
finally notice that $\displaystyle\phi$ is
$\displaystyle\mathrm{GL}_{6}$-equivariant, so that the map induces by
$\displaystyle\varphi$ on the quotients by $\displaystyle\mathrm{GL}_{6}$:

$\displaystyle\big{(}M_{6\times\mathrm{J}_{3}(\mathbb{O})}/\\!\\!/\mathrm{SO_{7}}\times\mathrm{SL}_{3}\big{)}/\\!\\!/\mathrm{GL}_{6}=\mathrm{Gr}(6,\mathrm{J}_{3}(\mathbb{O}))/\\!\\!/\mathrm{SO_{7}}\times\mathrm{SL}_{3}\xrightarrow{\hskip
22.76228pt}\mathrm{S}^{6}\mathbb{C}^{6}/\\!\\!/\mathrm{GL}_{6}$

is equally generically étale onto its image. $\displaystyle\blacktriangleleft$

###### Remark 2.2.2

The generic isotropy group for the action of $\displaystyle\mathrm{SO_{7}}$ on
$\displaystyle\mathbb{C}^{\oplus 3}\otimes\mathbb{O}$ is
$\displaystyle\mathrm{SO_{4}}$. The generic isotropy group for the action of
$\displaystyle\mathrm{SL}_{3}$ on $\displaystyle M_{3}(\mathbb{C})$ by
transpose-conjugation is $\displaystyle\mathrm{O}(3)$. We deduce that the
generic isotropy group for the action of
$\displaystyle\mathrm{SO_{7}}\times\mathrm{SL}_{3}$ on
$\displaystyle\mathrm{J}_{3}(\mathbb{O})$ contains
$\displaystyle\mathrm{SO}_{4}\times\mathrm{O}(3)$, which has dimension
$\displaystyle 9$. As a consequence, the triplet
$\displaystyle(\mathrm{J}_{3}(\mathbb{O}),\mathrm{SO_{7}}\times\mathrm{SL}_{3}\times\mathbb{C}^{*},\SS)$
is not prehomogeneous and the generic orbit of
$\displaystyle\mathrm{SO_{7}}\times\mathrm{SL}_{3}\times\mathbb{C}^{*}$ in
$\displaystyle\mathrm{J}_{3}(\mathbb{O})$ has codimension at least
$\displaystyle 27-(21+8+1-(3+6))=6$.

###  2.3 A tautological rank $\displaystyle 4$ sheaf on the Ogievetskiî-Dray-
Manogue sextic

The main result of this section is related to the cokernel of
$\displaystyle\begin{pmatrix}\lambda_{1}.I_{8}&L_{c}&{}^{t}L_{b}\\\
{}^{t}L_{c}&\lambda_{2}.I_{8}&L_{a}\\\
L_{b}&{}^{t}L_{a}&\lambda_{3}.I_{8}\end{pmatrix}$. In the following, we work
over the polynomial ring $\displaystyle R=\mathbb{C}[x_{1},\ldots,x_{27}]$. We
let $\displaystyle c=x_{1}e_{1}+\ldots+x_{8}e_{8}$, $\displaystyle
b=x_{9}e_{1}+\ldots+x_{16}e_{n}$, $\displaystyle
a=x_{17}e_{1}+\ldots+x_{24}e_{n}$, $\displaystyle\lambda_{1}=x_{25}e_{1}$,
$\displaystyle\lambda_{2}=x_{26}e_{1}$ and
$\displaystyle\lambda_{3}=x_{27}e_{1}$, where $\displaystyle
e_{1},\ldots,e_{8}$ is the canonical basis of $\displaystyle\mathbb{O}$.

###### Theorem 2.3.1

Let
$\displaystyle\mathrm{M_{\mathbb{O}}}=\begin{pmatrix}\lambda_{1}.I_{8}&L_{c}&{}^{t}L_{b}\\\
{}^{t}L_{c}&\lambda_{2}.I_{8}&L_{a}\\\
L_{b}&{}^{t}L_{a}&\lambda_{3}.I_{8}\end{pmatrix}$ in
$\displaystyle\mathscr{M}_{24,24}(R)$ and let $\displaystyle\mathrm{E_{ODM}}$
be the cokernel of:

$\displaystyle\mathbb{C}^{24}\otimes\mathscr{O}_{\mathbb{P}^{26}}(-1)\stackrel{{\scriptstyle\mathrm{M_{\mathbb{O}}}}}{{\xrightarrow{\hskip
22.76228pt}}}\mathbb{C}^{24}\otimes\mathscr{O}_{\mathbb{P}^{26}}.$

The sheaf $\displaystyle\mathrm{E_{ODM}}$ is a
$\displaystyle\mathrm{SO}_{7}\times\mathrm{SL}_{3}$-equivariant sheaf
supported on $\displaystyle\SS$ and has generically rank $\displaystyle 4$ on
$\displaystyle\SS$.

Proof :
$\displaystyle\blacktriangleright\;$ Theorem 2.1.2 insures that the degeneracy
of $\displaystyle\mathrm{M_{\mathbb{O}}}$ is $\displaystyle\SS$, so that
$\displaystyle\mathrm{E_{ODM}}$ is supported on $\displaystyle\SS$. Note that
the determinant of $\displaystyle\mathrm{M_{\mathbb{O}}}$ is of degree
$\displaystyle 24=4\times\deg(\SS)$. In particular, we have
$\displaystyle\mathrm{rank}(\mathrm{E_{ODM}})=4$. Let us now prove that
$\displaystyle\mathrm{M_{\mathbb{O}}}$ is
$\displaystyle\mathrm{SO}_{7}\times\mathrm{SL}_{3}$-equivariant. The
$\displaystyle\mathrm{SL}_{3}$-equivariance is obvious given the definition of
the $\displaystyle\mathrm{SL}_{3}$ action. We are left to prove the
$\displaystyle\mathrm{SO}_{7}$-equivariance. Let $\displaystyle
T_{1}\in\mathrm{SO}_{7}$, the triality principle insures that there exists
$\displaystyle T_{2}\in\mathrm{SO}_{8}$ such that:

$\displaystyle\forall\ (u,v)\in\mathbb{O}^{2},\ T_{2}(uv)=T_{1}(u)T_{2}(v).$

It easily follows that for any $\displaystyle z\in\mathbb{O}$, we have:

$\displaystyle L_{T_{1}(z)}=T_{2}L_{z}T_{2}^{-1}.$

As a consequence, for any $\displaystyle A\in\mathrm{J}_{3}(\mathbb{O})$ and
any $\displaystyle T_{1}\in\mathrm{SO}_{7}$ as above, we have:

$\displaystyle T_{1}.\mathrm{M_{\mathbb{O}}}=\begin{pmatrix}T_{2}&&\\\
&T_{2}&\\\
&&T_{2}\end{pmatrix}\mathrm{M_{\mathbb{O}}}\begin{pmatrix}T_{2}^{-1}&&\\\
&T_{2}^{-1}&\\\ &&T_{2}^{-1}\end{pmatrix}.$

This proves that $\displaystyle\mathrm{M_{\mathbb{O}}}$ is also
$\displaystyle\mathrm{SO}_{7}$-equivariant. We can conclude that its cokernel,
$\displaystyle\mathrm{E_{ODM}}$ is
$\displaystyle\mathrm{SL}_{3}\times\mathrm{SO}_{7}$-equivariant.
$\displaystyle\blacktriangleleft$

###### Remark 2.3.2

The complete description of the higher degeneracy loci of
$\displaystyle\mathrm{M_{\mathbb{O}}}$ seems to be an interesting question. It
seems however to be also a difficult one, as the action of
$\displaystyle\mathrm{SO}_{7}\times\mathrm{SL}_{3}$ on
$\displaystyle\mathbb{P}^{26}$ has infinitely many orbits (see remark 2.2.2).

##  3 The geometry of the twisted eigenvalue problem for
$\displaystyle\mathrm{J}_{3}(\mathbb{O})$

The motivation for introducing the twisted eigenvalue problem is the lack of
<<prehomogeneity>> of the space
$\displaystyle\big{(}\mathbb{P}(\mathrm{J}_{3}(\mathbb{O})),\mathrm{SO}_{7}\times\mathrm{SL}_{3}\times\mathbb{C}^{*},\SS\big{)}$.
This lack of prehomogeneity is partly explained by the relatively big
codimension of the subgroup of $\displaystyle\mathrm{E}_{6}$ generated by
$\displaystyle\mathrm{SO}_{7}$ and $\displaystyle\mathrm{SL}_{3}$. On the
other hand, it is well-known (see chapter $\displaystyle 14$ of [Har90]) that
$\displaystyle\mathrm{E}_{6}$ is generated by $\displaystyle\mathrm{Spin}_{8}$
and $\displaystyle\mathrm{SL}_{3}$. Thus, one could hope for an eigenvalue
problem having a larger group of symmetries. As it happens, the symmetry group
of the twisted octonionic eigenvalue problem contains the subgroup of
$\displaystyle\mathrm{SL}_{27}$ generated by $\displaystyle\mathrm{Spin}_{7}$
and $\displaystyle\mathrm{SL}_{3}$ 222The action of
$\displaystyle\mathrm{Spin}_{7}$ on $\displaystyle\mathrm{J}_{3}(\mathbb{O})$
that we exhibit is not the standard one and does not factor through an
embedding of $\displaystyle\mathrm{Spin}_{7}$ in
$\displaystyle\mathrm{E}_{6}$. As a matter of fact, we haven’t been able to
identify the subgroup of $\displaystyle\mathrm{SL}_{27}$ generated by
$\displaystyle\mathrm{Spin}_{7}$ and $\displaystyle\mathrm{SL}_{3}$. and we
prove that $\displaystyle\mathrm{J}_{3}(\mathbb{O})$ is prehomogeneous for
this action. We also show that this twisted problem is singular on the union
of the Cartan cubic and a sextic hypersurface. To the best of our knowledge,
this geometry has never appeared before in the literature.

###  3.1 The twisted octonionic eigenvalue problem

Let $\displaystyle A=\begin{pmatrix}\lambda_{1}&c&\overline{b}\\\
\overline{c}&\lambda_{2}&a\\\
b&\overline{a}&\lambda_{3}\end{pmatrix}\in\mathrm{J}_{3}(\mathbb{O})$. We say
that $\displaystyle\begin{pmatrix}x\\\ y\\\ z\end{pmatrix}\in\mathbb{O}^{3}$
is in the <<twisted kernel>> of $\displaystyle A$ if:

$\left\\{\begin{split}\lambda_{1}x+yc+\overline{b}z&\ =0\\\
x\overline{c}+\lambda_{2}y+az&\ =0\\\ bx+\overline{a}y+\lambda_{3}z&\ =0\\\
\end{split}\right.$

We say that $\displaystyle A\in\mathrm{J}_{3}(\mathbb{O})$ is twisted-singular
if the twisted kernel of $\displaystyle A$ is not zero. We notice that
$\displaystyle\begin{pmatrix}\lambda_{1}&c&\overline{b}\\\
\overline{c}&\lambda_{2}&a\\\ b&\overline{a}&\lambda_{3}\end{pmatrix}$ is
twisted-singular if and only if the $\displaystyle 24*24$ matrix
$\displaystyle N_{A}=\begin{pmatrix}\lambda_{1}.I_{8}&R_{c}&{}^{t}L_{b}\\\
{}^{t}R_{c}&\lambda_{2}.I_{8}&L_{a}\\\
L_{b}&{}^{t}L_{a}&\lambda_{3}.I_{8}\end{pmatrix}$ is singular, where
$\displaystyle R_{c}$ is the $\displaystyle 8*8$ matrix representing the right
multiplication by $\displaystyle c$ in $\displaystyle\mathbb{O}$. Our first
result related to the twisted eigenvalue problem is the:

###### Theorem 3.1.1

The degeneracy locus of the matrix $\displaystyle N_{A}$ is given by the
equation:

$\displaystyle\bigg{(}\mathrm{det}_{\mathbb{O}}(A)-2\mathrm{Re}(cab)+2\mathrm{Re}(\overline{c}ba)\bigg{)}^{4}\times\mathfrak{S}^{2}=0,$

where:

$\begin{split}\mathfrak{S}&=\left(\lambda_{1}|a|^{2}+\lambda_{2}|b|^{2}+\lambda_{3}|c|^{2}-\lambda_{1}\lambda_{2}\lambda_{3}\right)^{2}\\\
&\hskip
14.22636pt-4\left(\lambda_{1}|a|^{2}+\lambda_{2}|b|^{2}+\lambda_{3}|c|^{2}-\lambda_{1}\lambda_{2}\lambda_{3}\right)\mathrm{Re}(c)\mathrm{Re}(ab)\\\
&\hskip
14.22636pt+4\left(|b|^{2}|a|^{2}\mathrm{Re}(c)^{2}+|c|^{2}\mathrm{Re}(ab)^{2}-|a|^{2}|b|^{2}|c|^{2}\right),\end{split}$

with $\displaystyle|z|^{2}=z\overline{z}$ and
$\displaystyle\mathrm{Re}(z)=\frac{z+\overline{z}}{2}$.

Note that
$\displaystyle\mathrm{det}_{\mathbb{O}}(A)-2\mathrm{Re}(cab)+2\mathrm{Re}(\overline{c}ba)=\mathrm{det}_{\mathbb{O}}(A^{\prime})$,
where $\displaystyle
A^{\prime}=\begin{pmatrix}\lambda_{1}&\overline{c}&\overline{a}\\\
c&\lambda_{2}&b\\\ a&\overline{b}&\lambda_{3}\end{pmatrix}$. We refrain from
adopting such a notation for the cubic term appearing in the equation of the
degeneracy locus of $\displaystyle N_{A}$ as it may be a source of confusions.
We prove Theorem 3.1.1 using a strategy similar to that used for Theorem
2.1.2.

Proof :
$\displaystyle\blacktriangleright\;$ For $\displaystyle a,b,c\in\mathbb{O}$
and $\displaystyle\lambda_{1},\lambda_{2},\lambda_{3}\in\mathbb{C}$ such that
$\displaystyle|a|,|b|,|c|,\lambda_{1},\lambda_{2},\lambda_{3}\neq 0$ and
$\displaystyle|c|^{2}\lambda_{1}-\dfrac{|c|^{4}}{\lambda_{2}}\neq 0$, we
define the following $\displaystyle 24*24$ matrices in block form:

$\displaystyle S_{1}=\begin{pmatrix}\frac{1}{|c|^{2}}R_{c}&0_{8,8}&0_{8,8}\\\
0_{8,8}&I_{8}&0_{8,8}\\\
0_{8,8}&0_{8,8}&\frac{1}{|a|^{2}}{}^{t}L_{a}\end{pmatrix}\ \ \textrm{and}\ \
S_{2}=\begin{pmatrix}I_{8}&\frac{|c|^{2}}{\lambda_{2}}&0_{8,8}\\\
0_{8,8}&I_{8}&0_{8,8}\\\
\frac{1}{\delta}B&\frac{|a|^{2}}{\lambda_{2}}I_{8}&I_{8}\end{pmatrix},$

where $\displaystyle
B=L_{a}L_{b}R_{c}-\dfrac{|a|^{2}|c|^{2}}{\lambda_{2}}.I_{8}$,
$\displaystyle\delta=|c|^{2}\lambda_{1}-\dfrac{|c|^{4}}{\lambda_{2}}$,
$\displaystyle 0_{8,8}$ is the zero $\displaystyle 8*8$ matrix and
$\displaystyle I_{8}$ the identity matrix of size $\displaystyle 8$. One
easily checks that:

$\displaystyle\begin{pmatrix}\lambda_{1}.I_{8}&R_{c}&{}^{t}L_{b}\\\
{}^{t}R_{c}&\lambda_{2}.I_{8}&L_{a}\\\
L_{b}&{}^{t}L_{a}&\lambda_{3}.I_{8}\end{pmatrix}=S_{1}S_{2}\begin{pmatrix}\delta.I_{8}&0_{8,8}&0_{8,8}\\\
0_{8,8}&\lambda_{2}.I_{8}&0_{8,8}\\\
0_{8,8}&0_{8,8}&\nu.I_{8}-\dfrac{1}{\delta}B{}^{t}B\end{pmatrix}{}^{t}S_{2}{}^{t}S_{1},$

where $\displaystyle\nu=\lambda_{3}|a|^{2}-\frac{|a|^{4}}{\lambda_{2}}$. Since
$\displaystyle\mathrm{det}(S_{1})=\dfrac{1}{|a|^{8}|c|^{8}}$ and
$\displaystyle\mathrm{det}(S_{2})=1$, we get:

$\begin{split}\mathrm{det}\begin{pmatrix}\lambda_{1}.I_{8}&R_{c}&{}^{t}L_{b}\\\
{}^{t}R_{c}&\lambda_{2}.I_{8}&L_{a}\\\
L_{b}&{}^{t}L_{a}&\lambda_{3}.I_{8}\end{pmatrix}&=\dfrac{1}{|a|^{16}|c|^{16}}\mathrm{det}\left(\delta\nu\lambda_{2}.I_{8}-\lambda_{2}.B{}^{t}B\right)\\\
&=\mathrm{det}\left((\lambda_{1}\lambda_{2}\lambda_{3}-\lambda_{1}|a|^{2}-\lambda_{2}|b|^{2}-\lambda_{3}|c|^{2}).I_{8}+(L_{a}L_{b}R_{c}+{}^{t}R_{c}{}^{t}L_{b}{}^{t}L_{a})\right)\\\
&=\mathrm{det}(\mathrm{det}_{\mathbb{O}}(A)-2\mathrm{Re}(cab)).I_{8}+(L_{a}L_{b}R_{c}+{}^{t}R_{c}{}^{t}L_{b}{}^{t}L_{a})).\end{split}$

The expression
$\displaystyle\mathrm{det}(\mathrm{det}_{\mathbb{O}}(A)-2\mathrm{Re}(cab)).I_{8}+(L_{a}L_{b}R_{c}+{}^{t}R_{c}{}^{t}L_{b}{}^{t}L_{a}))$
has no denominators so that the above equality holds for all $\displaystyle
a,b,c\in\mathbb{O}$ and
$\displaystyle\lambda_{1},\lambda_{2},\lambda_{3}\in\mathbb{C}$. We deduce the
following:

The determinant of
$\displaystyle\begin{pmatrix}\lambda_{1}.I_{8}&R_{c}&{}^{t}L_{b}\\\
{}^{t}R_{c}&\lambda_{2}.I_{8}&L_{a}\\\
L_{b}&{}^{t}L_{a}&\lambda_{3}.I_{8}\end{pmatrix}$ vanishes if and only if
$\displaystyle 2\mathrm{Re}(cab)-\mathrm{det}_{\mathbb{O}}(A)$ is a (complex)
eigenvalue of
$\displaystyle(L_{a}L_{b}R_{c}+{}^{t}R_{c}{}^{t}L_{b}{}^{t}L_{a})$.

We shall now determine the (complex) eigenvalue of
$\displaystyle(L_{a}L_{b}R_{c}+{}^{t}R_{c}{}^{t}L_{b}{}^{t}L_{a})$. This means
we are looking for the $\displaystyle\kappa\in\mathbb{C}$ such that:

$\exists x\in\mathbb{O},\ x\neq 0\ \
a.(b.(x.c))+(\overline{b}.(\overline{a}.x)).\overline{c}=\kappa.x$ (5)

Let us consider a four dimensional quaternionic subalgebra of
$\displaystyle\mathbb{O}$ which contains $\displaystyle a$ and $\displaystyle
c$. Let $\displaystyle e\in\mathbb{O}$ be a vector orthogonal to
$\displaystyle M$ such that $\displaystyle\mathbb{O}=M\oplus M.e$. We write
$\displaystyle b=b_{0}+b_{1}.e$ and $\displaystyle x=x_{0}+x_{1}.e$, with
$\displaystyle b_{0},b_{1},x_{0},x_{1}\in M$. Using the relations:

$\displaystyle u.(v.e)=(v.u).e,\ (u.e).(v.e)=-\overline{v}.u,\
u.e=e.\overline{u}\ \textrm{and}\ (u.e).v=(u.\overline{v}).e,$

valid for all $\displaystyle(u,v)\in M^{2}$, we find that equation 5 is
equivalent to the system:

$\left\\{\begin{split}a.b_{0}.x_{0}.c+\overline{b_{0}}.\overline{a}.x_{0}.\overline{c}+a.\overline{x_{1}}.b_{1}.\overline{c}-{a}.c.\overline{x_{1}}.b_{1}&=\kappa.x_{0}\\\
x_{1}.\overline{c}.b_{0}.a+b_{1}.\overline{c}.\overline{x_{0}}.a+x_{1}.\overline{a}.\overline{b_{0}}.c-b_{1}.\overline{x_{0}}.a.c&=\kappa
x_{1}.\\\ \end{split}\right.$

which, by conjugating the second line, is equivalent to:

$\left\\{\begin{split}a.b_{0}.x_{0}.c+\overline{b_{0}}.\overline{a}.x_{0}.\overline{c}+a.\overline{x_{1}}.b_{1}.\overline{c}-{a}.c.\overline{x_{1}}.b_{1}&=\kappa.x_{0}\\\
\overline{a}.\overline{b_{0}}.c.\overline{x_{1}}+\overline{c}.b_{0}.a.\overline{x_{1}}+\overline{a}.x_{0}.c.\overline{b_{1}}-\overline{c}.\overline{a}.x_{0}\overline{b_{1}}&=\kappa.\overline{x_{1}}\end{split}\right.$

which, in turn, is equivalent to:

$\left\\{\begin{split}a.b_{0}.x_{0}.c+\overline{b_{0}}.\overline{a}.x_{0}.\overline{c}+a.\overline{x_{1}}.b_{1}.\overline{c}-{a}.c.\overline{x_{1}}.b_{1}&=\kappa.x_{0}\\\
2\mathrm{Re}(\overline{c}ba)\overline{x_{1}}+\overline{a}.x_{0}.c.\overline{b_{1}}-\overline{c}.\overline{a}.x_{0}\overline{b_{1}}&=\kappa.\overline{x_{1}}\end{split}\right.$
(6)

as $\displaystyle\mathrm{Re}(\overline{c}b_{0}a)=\mathrm{Re}(\overline{c}ba).$
It happens that the matrix form of this system can not be exploited as easily
as in the proof of Theorem 2.1.2. We will nevertheless prove that for all
$\displaystyle a,b,c\in\mathbb{O}$, the complex $\displaystyle
2.\mathrm{Re}(\overline{c}ba)$ is an eigenvalue of multiplicity at least
$\displaystyle 4$ of
$\displaystyle(L_{a}L_{b}R_{c}+{}^{t}R_{c}{}^{t}L_{b}{}^{t}L_{a})$. Let
$\displaystyle V_{1}$ be the space of vectors
$\displaystyle\begin{pmatrix}x_{0}\\\ \overline{x_{1}}\end{pmatrix}$ such
that:

$\displaystyle x_{0}=0\ \textrm{et}\
\overline{x_{1}}.b_{1}.\overline{c}=c.\overline{x_{1}}.b_{1}$

One directly checks with the help of system (6) that for any such vectors:

$\displaystyle(L_{a}L_{b}R_{c}+{}^{t}R_{c}{}^{t}L_{b}{}^{t}L_{a})\begin{pmatrix}x_{0}\\\
\overline{x_{1}}\end{pmatrix}=2\mathrm{Re}(\overline{c}ba)\begin{pmatrix}x_{0}\\\
\overline{x_{1}}\end{pmatrix}.$

The quaternionic structure of $\displaystyle M$ insures that
$\displaystyle\dim V_{1}\geq 2$. Let $\displaystyle V_{2}$ be the space of
vectors $\displaystyle\begin{pmatrix}x_{0}\\\ \overline{x_{1}}\end{pmatrix}$
such that:

$\displaystyle\overline{a}.x_{0}.c=\overline{c}.\overline{a}.x_{0}\
\textrm{and}\ {a}.{b_{0}}.x_{0}.{c}={a}.c.\overline{x_{1}}.b_{1}$

One directly checks with the help of system (6) that for any such vectors:

$\displaystyle(L_{a}L_{b}R_{c}+{}^{t}R_{c}{}^{t}L_{b}{}^{t}L_{a})\begin{pmatrix}x_{0}\\\
\overline{x_{1}}\end{pmatrix}=2\mathrm{Re}(\overline{c}ba)\begin{pmatrix}x_{0}\\\
\overline{x_{1}}\end{pmatrix}.$

The quaternionic structure of $\displaystyle M$ insures that
$\displaystyle\dim V_{2}\geq 2$. We note that $\displaystyle V_{1}\cap
V_{2}=\begin{pmatrix}0\\\ 0\end{pmatrix}$ which implies $\displaystyle\dim
V_{1}\oplus V_{2}\geq 4$. As a consequence, $\displaystyle
2\mathrm{Re}(\overline{c}ba)$ is an eigenvalue of multiplicity at least
$\displaystyle 4$ of
$\displaystyle(L_{a}L_{b}R_{c}+{}^{t}R_{c}{}^{t}L_{b}{}^{t}L_{a})$. In
particular, we may write:

$\displaystyle\mathrm{det}\bigg{(}\kappa.I_{8}-L_{a}L_{b}R_{c}+{}^{t}R_{c}{}^{t}L_{b}{}^{t}L_{a}\bigg{)}=(\kappa-2.\mathrm{Re}(\overline{c}ba))^{4}.T(a,b,c,\kappa),$

where $\displaystyle T(a,b,c,\kappa)$ is weighted-homogeneous of degree
$\displaystyle 12$ with weights $\displaystyle(1,1,1,3)$.

We can now conclude the proof of the Theorem as we remember that the
determinant of
$\displaystyle\begin{pmatrix}\lambda_{1}.I_{8}&R_{c}&{}^{t}L_{b}\\\
{}^{t}R_{c}&\lambda_{2}.I_{8}&L_{a}\\\
L_{b}&{}^{t}L_{a}&\lambda_{3}.I_{8}\end{pmatrix}$ vanishes if and only if
$\displaystyle 2\mathrm{Re}(cab)-\mathrm{det}_{\mathbb{O}}(A)$ is a (complex)
eigenvalue of
$\displaystyle(L_{a}L_{b}R_{c}+{}^{t}R_{c}{}^{t}L_{b}{}^{t}L_{a})$. Hence the
degeneracy locus of $\displaystyle N_{A}$ is scheme-theoretically given by the
equation:

$\displaystyle\bigg{(}\mathrm{det}_{\mathbb{O}}(A)-2\mathrm{Re}(cab)+2\mathrm{Re}(\overline{c}ba)\bigg{)}^{4}\times\mathfrak{T}=0,$

where $\displaystyle\mathfrak{T}$ is the degree $\displaystyle 12$ homogeneous
polynomial in $\displaystyle a,b,c$ defined by
$\displaystyle\mathfrak{T}=T(a,b,c,2\mathrm{Re}(cab)-\mathrm{det}_{\mathbb{O}}(A))$.
One easily proves that $\displaystyle\mathfrak{T}=\mathfrak{S}^{2}$ is a
square and the explicit equation:

$\begin{split}\mathfrak{S}&=\left(\lambda_{1}|a|^{2}+\lambda_{2}|b|^{2}+\lambda_{3}|c|^{2}-\lambda_{1}\lambda_{2}\lambda_{3}\right)^{2}\\\
&\hskip
14.22636pt-4\left(\lambda_{1}|a|^{2}+\lambda_{2}|b|^{2}+\lambda_{3}|c|^{2}-\lambda_{1}\lambda_{2}\lambda_{3}\right)\mathrm{Re}(c)\mathrm{Re}(ab)\\\
&\hskip
14.22636pt+4\left(|b|^{2}|a|^{2}\mathrm{Re}(c)^{2}+|c|^{2}\mathrm{Re}(ab)^{2}-|a|^{2}|b|^{2}|c|^{2}\right),\end{split}$

has been first found by Jonathan Hauenstein using the software Bertini.

$\displaystyle\blacktriangleleft$

###  3.2 A quasi-tautological rank $\displaystyle 4$ sheaf on the Cartan cubic

We shall first explain how $\displaystyle\mathrm{Spin}_{7}$ and
$\displaystyle\mathrm{SL}_{3}$ act on $\displaystyle
N_{A}=\begin{pmatrix}\lambda_{1}.I_{8}&R_{c}&{}^{t}L_{b}\\\
{}^{t}R_{c}&\lambda_{2}.I_{8}&L_{a}\\\
L_{b}&{}^{t}L_{a}&\lambda_{3}.I_{8}\end{pmatrix}.$ The action of
$\displaystyle\mathrm{SL}_{3}$ is again by transpose-conjugation and is the
same as in section 2.2. To describe the action of
$\displaystyle\mathrm{Spin}_{7}$, we shall need a few more facts on triality
(we refer to [DeS01] for more details on the triality principle). We recall
that there is an embedding (the triality embedding) of
$\displaystyle\mathrm{Spin}_{8}$ in
$\displaystyle\mathrm{SO}_{8}\times\mathrm{SO}_{8}\times\mathrm{SO}_{8}$
defined as follows:

$\displaystyle\mathrm{Spin}_{8}=\\{(T_{1},T_{2},T_{3})\in\big{(}\mathrm{SO}_{8}\big{)}^{3},\
T_{1}(x)T_{2}(y)=T_{3}(xy),\ \textrm{for all}\ (x,y)\in\mathbb{O}^{2}\\}.$

There are three copies of $\displaystyle\mathrm{Spin}_{7}$ inside
$\displaystyle\mathrm{Spin}_{8}$ which have remarkable group-theoretic
properties with respect to the triality embedding. We are interested in one of
them:

$\displaystyle\mathrm{Spin}_{7}=\\{(T_{1},T_{2},T_{1})\in\big{(}\mathrm{SO}_{8}\big{)}^{3},\
T_{1}(x)T_{2}(y)=T_{1}(xy),\ \textrm{for all}\ (x,y)\in\mathbb{O}^{2}\\}.$

The projection map:

$\begin{split}&\mathrm{Spin}_{7}\xrightarrow{\hskip
22.76228pt}\mathrm{SO}_{8}\\\ &(T_{1},T_{2},T_{1})\longrightarrow
T_{2}\end{split}$

is a double cover of $\displaystyle\mathrm{SO}_{7}$ embedded in
$\displaystyle\mathrm{SO_{8}}$ as the stabilizer of $\displaystyle
1\in\mathbb{O}$. For any $\displaystyle T\in\mathrm{SO}_{8}$, we denote by
$\displaystyle\mathrm{K}_{T}$ the element of $\displaystyle\mathrm{SO}_{8}$
defined by $\displaystyle\mathrm{K}_{T}(x)=\overline{T(\overline{x})}$. It is
easily checked that if $\displaystyle(T_{1},T_{2},T_{1})$ is a triality, then
$\displaystyle(\mathrm{K}_{T_{1}},T_{1},T_{2})$ and
$\displaystyle(T_{2},\mathrm{K}_{T_{1}},\mathrm{K}_{T_{1}})$ are equally
trialities (but, in general, they lie in different copies
$\displaystyle\mathrm{Spin}_{7}\subset\mathrm{Spin}_{8}$).

We now come back to the twisted eigenvalue problem. Let $\displaystyle
A=\begin{pmatrix}\lambda_{1}&c&\overline{b}\\\ \overline{c}&\lambda_{2}&a\\\
b&\overline{a}&\lambda_{3}\end{pmatrix}\in\mathrm{J}_{3}(\mathbb{O})$ and let
$\displaystyle\begin{pmatrix}x\\\ y\\\ z\end{pmatrix}\in\mathbb{O}^{3}$ is in
the twisted kernel of $\displaystyle A$, namely:

$\left\\{\begin{split}\lambda_{1}x+yc+\overline{b}z&\ =0\\\
x\overline{c}+\lambda_{2}y+az&\ =0\\\ bx+\overline{a}y+\lambda_{3}z&\ =0\\\
\end{split}\right.$

A quick computation shows that for any
$\displaystyle(T_{1},T_{2},T_{1})\in\mathrm{Spin}_{7}$, the vector
$\displaystyle\begin{pmatrix}T_{1}(x)\\\ T_{1}(y)\\\
T_{2}(z)\end{pmatrix}\in\mathbb{O}^{3}$ is in the twisted kernel of
$\displaystyle\begin{pmatrix}\lambda_{1}&T_{2}(c)&\overline{\mathrm{K}_{T_{1}}(b)}\\\
\overline{T_{2}(c)}&\lambda_{2}&T_{1}(a)\\\
\mathrm{K}_{T_{1}}(b)&\overline{T_{1}(a)}&\lambda_{3}\end{pmatrix}$, that is:

$\left\\{\begin{split}\lambda_{1}T_{1}(x)+T_{1}(y)T_{2}(c)+T_{1}(\overline{b})T_{2}(z)&\
=0\\\ T_{1}(x)\overline{T_{2}(c)}+\lambda_{2}T_{1}(y)+T_{1}(a)T_{2}(z)&\ =0\\\
\mathrm{K}_{T_{1}}(b)T_{1}(x)+\overline{T_{1}(a)}y+\lambda_{3}T_{2}(z)&\ =0\\\
\end{split}\right.$

For any $\displaystyle(T_{1},T_{2},T_{1})\in\mathrm{Spin_{7}}$ and any
$\displaystyle A=\begin{pmatrix}\lambda_{1}&c&\overline{b}\\\
\overline{c}&\lambda_{2}&a\\\
b&\overline{a}&\lambda_{3}\end{pmatrix}\in\mathrm{J}_{3}(\mathbb{O})$, we then
define:

$\displaystyle(T_{1},T_{2},T_{1}).A=\begin{pmatrix}\lambda_{1}&T_{2}(c)&\overline{\mathrm{K}_{T_{1}}(b)}\\\
\overline{T_{2}(c)}&\lambda_{2}&T_{1}(a)\\\
\mathrm{K}_{T_{1}}(b)&\overline{T_{1}(a)}&\lambda_{3}\end{pmatrix}.$

This defines an action of $\displaystyle\mathrm{Spin}_{7}$ on
$\displaystyle\mathrm{J}_{3}(\mathbb{O})$ which preserves the twisted
eigenvalue problem. The main result of this section is related to the cokernel
of $\displaystyle\begin{pmatrix}\lambda_{1}.I_{8}&R_{c}&{}^{t}L_{b}\\\
{}^{t}R_{c}&\lambda_{2}.I_{8}&L_{a}\\\
L_{b}&{}^{t}L_{a}&\lambda_{3}.I_{8}\end{pmatrix}$. As in section 2.3, we work
over the polynomial ring $\displaystyle R=\mathbb{C}[x_{1},\ldots,x_{27}]$. We
let $\displaystyle c=x_{1}e_{1}+\ldots+x_{8}e_{8}$, $\displaystyle
b=x_{9}e_{1}+\ldots+x_{16}e_{n}$, $\displaystyle
a=x_{17}e_{1}+\ldots+x_{24}e_{n}$, $\displaystyle\lambda_{1}=x_{25}e_{1}$,
$\displaystyle\lambda_{2}=x_{26}e_{1}$ and
$\displaystyle\lambda_{3}=x_{27}e_{1}$, where $\displaystyle
e_{1},\ldots,e_{8}$ is the canonical basis of $\displaystyle\mathbb{O}$.

###### Theorem 3.2.1

Let $\displaystyle
N_{\mathbb{O}}=\begin{pmatrix}\lambda_{1}.I_{8}&R_{c}&{}^{t}L_{b}\\\
{}^{t}R_{c}&\lambda_{2}.I_{8}&L_{a}\\\
L_{b}&{}^{t}L_{a}&\lambda_{3}.I_{8}\end{pmatrix}$ in
$\displaystyle\mathscr{M}_{24,24}(R)$ and let $\displaystyle E_{\mathbb{O}}$
be the cokernel of:

$\displaystyle\mathbb{C}^{24}\otimes\mathscr{O}_{\mathbb{P}^{26}}(-1)\stackrel{{\scriptstyle
N_{\mathbb{O}}}}{{\xrightarrow{\hskip
22.76228pt}}}\mathbb{C}^{24}\otimes\mathscr{O}_{\mathbb{P}^{26}}.$

  1. 1.

The sheaf $\displaystyle E_{\mathbb{O}}$ is
$\displaystyle\langle\mathrm{Spin}_{7},\mathrm{SL}_{3}\rangle$-equivariant
(where $\displaystyle\langle\mathrm{Spin_{7}},\mathrm{SL}_{3}\rangle$ is the
subgroup of $\displaystyle\mathrm{SL}_{24}$ generated by
$\displaystyle\mathrm{SL}_{3}$ and $\displaystyle\mathrm{Spin_{7}}$ as above).

  2. 2.

It is supported on the union of a cubic hypersurface isomorphic to
$\displaystyle\mathscr{C}_{\mathbb{O}}$ and the degree $\displaystyle 6$
hypersurface given by the equation $\displaystyle\mathfrak{S}=0$. Both
hypersurfaces are
$\displaystyle\langle\mathrm{Spin}_{7},\mathrm{SL}_{3}\rangle$-invariant.

  3. 3.

The restriction of $\displaystyle E_{\mathbb{O}}$ to the cubic hypersurface
has generically rank $\displaystyle 4$ while its restriction to the sextic
$\displaystyle\mathfrak{S}=0$ has generically rank $\displaystyle 2$.

In the following, we shall also denote by $\displaystyle\mathfrak{S}$ the
hypersurface defined by $\displaystyle\mathfrak{S}=0$.

Proof :
$\displaystyle\blacktriangleright\;$ Theorem 3.1.1 insures that the degeneracy
locus of $\displaystyle N_{\mathbb{O}}$ is scheme-theoretically given by
$\displaystyle(\mathrm{Det}_{\mathbb{O}}-\mathrm{Re}(cab)+\mathrm{Re}(\overline{c}ba))^{4}\times\mathfrak{S}^{2}=0$.
We already noted that
$\displaystyle(\mathrm{det}_{\mathbb{O}}(A)-\mathrm{Re}(cab)+\mathrm{Re}(\overline{c}ba))=\mathrm{det}_{\mathbb{O}}A^{\prime}$,
where $\displaystyle
A^{\prime}=\begin{pmatrix}\lambda_{1}&\overline{c}&\overline{a}\\\
c&\lambda_{2}&b\\\ a&\overline{b}&\lambda_{3}\end{pmatrix}$. As a consequence,
the equation defines a cubic isomorphic to the Cartan cubic, which we denote
by $\displaystyle\mathscr{C}^{\prime}_{\mathbb{O}}$. The sheaf $\displaystyle
E_{\mathbb{O}}$ is then supported on the union of
$\displaystyle\mathscr{C}^{\prime}_{\mathbb{O}}$ and the degree $\displaystyle
6$ hypersurface $\displaystyle\mathfrak{S}$. The rank of the restriction of
$\displaystyle E_{\mathbb{O}}$ to
$\displaystyle\mathscr{C}^{\prime}_{\mathbb{O}}$ is exactly $\displaystyle 4$
as
$\displaystyle(\mathrm{Det}_{\mathbb{O}}-\mathrm{Re}(cab)+\mathrm{Re}(\overline{c}ba))$
does not divide $\displaystyle\mathfrak{S}$. The same argument shows that the
rank of $\displaystyle E_{\mathbb{O}}$ restricted to
$\displaystyle\mathfrak{S}$ is $\displaystyle 2$.

The $\displaystyle\mathrm{SL}_{3}$-equivariance of $\displaystyle
N_{\mathbb{O}}$ follows from the definition of the
$\displaystyle\mathrm{SL}_{3}$ action (by transpose-conjugation). Let us prove
that $\displaystyle N_{\mathbb{O}}$ is also
$\displaystyle\mathrm{Spin}_{7}$-equivariant. For any
$\displaystyle(T_{1},T_{2},T_{1})\in\mathrm{Spin}_{7}$, we know that
$\displaystyle(\mathrm{K}_{T_{1}},T_{1},T_{2})$ is also in
$\displaystyle\mathrm{Spin}_{8}$. Hence, for any
$\displaystyle(x,y,z,w)\in\mathbb{O}^{4}$ and any
$\displaystyle(T_{1},T_{2},T_{1})\in\mathrm{Spin}_{7}$, we have:

$\displaystyle R_{\overline{T_{2}(x)}}=T_{1}{}^{t}R_{x}T_{1}^{-1},\
L_{\overline{\mathrm{K}_{T_{1}}(y)}}=T_{1}{}^{t}L_{y}T_{2}^{-1},\
L_{T_{1}(z)}=T_{1}L_{z}T_{2}^{-1}\ \textrm{and}\
L_{\overline{T_{1}}(w)}=T_{2}{}^{t}L_{w}T_{1}^{-1}.$

As a consequence, for any
$\displaystyle(T_{1},T_{2},T_{1})\in\mathrm{Spin}_{7}$, we find:

$\displaystyle(T_{1},T_{2},T_{1}).N_{\mathbb{O}}=\begin{pmatrix}T_{1}&&\\\
&T_{1}&\\\ &&T_{2}\end{pmatrix}N_{\mathbb{O}}\begin{pmatrix}T_{1}^{-1}&&\\\
&T_{1}^{-1}&\\\ &&T_{2}^{-1}\end{pmatrix},$

which proves the $\displaystyle\mathrm{Spin}_{7}$-equivariance of
$\displaystyle N_{\mathbb{O}}$. In fine, we can conclude that $\displaystyle
E_{\mathbb{O}}$, the cokernel of $\displaystyle N_{\mathbb{O}}$, is
$\displaystyle\langle\mathrm{Spin_{7}},\mathrm{SL}_{3}\rangle$-equivariant,
where $\displaystyle\langle\mathrm{Spin_{7}},\mathrm{SL}_{3}\rangle$ is the
subgroup of $\displaystyle\mathrm{SL}_{24}$ generated by
$\displaystyle\mathrm{SL}_{3}$ and $\displaystyle\mathrm{Spin_{7}}$. Since
$\displaystyle E_{\mathbb{O}}$ is
$\displaystyle\langle\mathrm{Spin}_{7},\mathrm{SL}_{3}\rangle$-equivariant,
its support must be
$\displaystyle\langle\mathrm{Spin}_{7},\mathrm{SL}_{3}\rangle$-invariant. This
proves that the sextic $\displaystyle\mathfrak{S}$ is
$\displaystyle\langle\mathrm{Spin}_{7},\mathrm{SL}_{3}\rangle$-invariant.

$\displaystyle\blacktriangleleft$

###### Remark 3.2.2

On easily checks that the jumping locus of the restriction of $\displaystyle
E_{\mathbb{O}}$ to the cubic $\displaystyle\mathscr{C}^{\prime}_{\mathbb{O}}$
is not the singular locus of $\displaystyle\mathscr{C}^{\prime}_{\mathbb{O}}$
(that is a copy of $\displaystyle\mathbb{OP}^{2}$) : it has strictly smaller
codimension. it would be interesting to know if we can give an interpretation
of $\displaystyle E_{\mathbb{O}}|_{A}$ for any $\displaystyle A$ in the smooth
locus of $\displaystyle\mathscr{C}^{\prime}_{\mathbb{O}}$ using only the
octonionic geometry in $\displaystyle\mathbb{O}$ (as we can do for the
tautological sheaves described in the introduction for the associative
exceptional algebras
$\displaystyle\mathbb{R}\otimes\mathbb{C},\mathbb{C}\otimes\mathbb{C}$ and
$\displaystyle\mathbb{H}\otimes\mathbb{C}$).

###  3.3 On the automorphism group of twisted eigenvalue problem

It is well known (see the section 14 of [Har90]) that the subgroup of
$\displaystyle\mathrm{E}_{6}$ generated by $\displaystyle\mathrm{SL}_{3}$ and
$\displaystyle\mathrm{Spin_{8}}$ is $\displaystyle\mathrm{E}_{6}$ itself. In
fact, one can prove that the subgroup of $\displaystyle\mathrm{E}_{6}$
generated by $\displaystyle\mathrm{SL}_{3}$ and
$\displaystyle\mathrm{Spin_{7}}$ is also $\displaystyle\mathrm{E}_{6}$. The
following argument has been communicated to me by Robert Bryant. Let
$\displaystyle h\in\mathrm{SL}_{3}$ defined by:

$\displaystyle h=\begin{pmatrix}0&-1&0\\\ -1&0&0\\\ 0&0&-1\end{pmatrix}.$

Let us denote by $\displaystyle K_{2}$ the copy of
$\displaystyle\mathrm{Spin_{7}}$ embedded in $\displaystyle\mathrm{Spin}_{8}$
we considered in the previous section:

$\displaystyle
K_{2}=\\{(T_{1},T_{2},T_{1})\in\big{(}\mathrm{SO}_{8}\big{)}^{3},\
T_{1}(x)T_{2}(y)=T_{1}(xy),\ \textrm{for all}\ (x,y)\in\mathbb{O}^{2}\\}.$

It is easily checked that we have $\displaystyle hK_{2}h^{-1}=K_{1}$ as
subgroups of $\displaystyle\mathrm{E}_{6}$, where $\displaystyle K_{1}$ is the
copy of $\displaystyle\mathrm{Spin_{7}}$ embedded in
$\displaystyle\mathrm{Spin}_{8}$ as:

$\displaystyle
K_{1}=\\{(T_{2},T_{1},T_{1})\in\big{(}\mathrm{SO}_{8}\big{)}^{3},\
T_{2}(x)T_{1}(y)=T_{1}(xy),\ \textrm{for all}\ (x,y)\in\mathbb{O}^{2}\\}.$

It is known that these two copies of $\displaystyle\mathrm{Spin}_{7}$ are
sufficient to generate $\displaystyle\mathrm{Spin}_{8}$ (see the section
$\displaystyle 14$ of [Har90]) and we are done.

###### Proposition 3.3.1

The action of $\displaystyle\mathrm{Spin}_{7}$ on
$\displaystyle\mathrm{J}_{3}(\mathbb{O})$ we considered in Theorem 3.2.1 does
not factor through the action $\displaystyle\mathrm{E}_{6}$ on
$\displaystyle\mathrm{J}_{3}(\mathbb{O})$.

Proof :
$\displaystyle\blacktriangleright\;$ Assume by absurd that it does. Then the
remark above shows that
$\displaystyle\langle\mathrm{Spin}_{7},\mathrm{SL}_{3}\rangle=\mathrm{E}_{6}$.
One immediately checks that the action of
$\displaystyle\langle\mathrm{Spin}_{7},\mathrm{SL}_{3}\rangle$ we defined on
$\displaystyle\mathrm{J}_{3}(\mathbb{O})$ preserves the rank, when we see
elements of $\displaystyle\mathrm{J}_{3}(\mathbb{O})$ as $\displaystyle 24*24$
matrices. Since any element of $\displaystyle\mathrm{J}_{3}(\mathbb{O})$ can
be diagonalized by the action of $\displaystyle\mathrm{E}_{6}$ (see the
chapter 14 of [Har90]), this would prove that any element of
$\displaystyle\mathrm{J}_{3}(\mathbb{O})$ has rank $\displaystyle 0,8,16$ or
$\displaystyle 24$ as a $\displaystyle 24*24$ matrix. But we know that a
generic point on $\displaystyle\mathfrak{S}=0$ has rank $\displaystyle 22$
(see Theorem 3.2.1), a contradiction.

$\displaystyle\blacktriangleleft$

We now state the main result of this section:

###### Theorem 3.3.2

The action of
$\displaystyle\mathrm{Aut}^{0}(\mathfrak{S})\times\mathbb{C}^{*}$ on
$\displaystyle\mathrm{J}_{3}(\mathbb{O})$ is prehomogeneous.

This result is slightly surprising as the subgroup of
$\displaystyle\mathrm{SL}_{27}$ generated by
$\displaystyle\mathrm{Spin}_{7}(\mathbb{C})$ and
$\displaystyle\mathrm{SL}_{3}(\mathbb{C})$ is not
$\displaystyle\mathrm{E}_{6}$ (see the above proposition). One also notices
that the prehomogeneous space
$\displaystyle(\mathrm{J}_{3}(\mathbb{O}),\mathrm{Aut}^{0}(\mathfrak{S}))\times\mathbb{C}^{*}$
does not appear in the list of Kimura-Sato [KS77]. This is not surprising as
it is unlikely to be an irreducible prehomogeneous space. It thus provides an
interesting example of a non-irreducible prehomogeneous vector space which
does not split as a sum of lower dimensional prehomoegeneous vector spaces.
Theorem 3.3.2 also shows that the geometry of the twisted octonionic
eigenvalue problem differes eminently from that of the untwisted eigenvalue
problem : the former is much more <<symmetric>> than the latter. Indeed, we
noticed in section 2.2 that
$\displaystyle\mathrm{Aut}^{0}(\SS)\times\mathbb{C}^{*}=\mathrm{SO}_{7}\times\mathrm{SL}_{3}\times\mathbb{C}^{*}$
does not act prehomogeneously on $\displaystyle\mathrm{J}_{3}(\mathbb{O})$.

Proof :
$\displaystyle\blacktriangleright\;$ We have shown in the previous section
that the subgroup $\displaystyle G$ of $\displaystyle\mathrm{SL}_{27}$
generated by $\displaystyle\mathrm{SL}_{3}$ and
$\displaystyle\mathrm{Spin}_{7}$ is included in
$\displaystyle\mathrm{Aut}^{0}(\mathfrak{S})$. We are going to prove that the
action of $\displaystyle\mathbb{C}^{*}\times G$ on
$\displaystyle\mathrm{J}_{3}(\mathbb{O})$ is prehomogeneous. We first start
with:

###### Lemma 3.3.3

Let $\displaystyle\\{1,i,j,k,l,m,n,o\\}$ be a basis of
$\displaystyle\mathbb{O}$ over $\displaystyle\mathbb{C}$. We fix a splitting
$\displaystyle\mathbb{O}=\mathbb{H}\oplus\mathbb{H}.l$, with
$\displaystyle\mathbb{H}=\mathrm{Vect}(1,i,j,k)$. The group
$\displaystyle\mathrm{Spin}_{7}$ acts transitively on the following varieties:

$\begin{split}&S^{6}\times\\{(x+y.l)\in\mathbb{H}\oplus\mathbb{H}.l,\ \langle
x,y\rangle=0,x\neq 0,y\neq
0\\}\subset\mathrm{Im}(\mathbb{O})\times\mathbb{O},\\\
&S^{6}\times\left(\mathbb{H}\backslash\\{0\\}\oplus\\{0\\}\right)\subset\mathrm{Im}(\mathbb{O})\times\mathbb{O},\\\
&S^{6}\times\left(\\{0\\}\oplus\mathbb{H}.l\backslash\\{0\\}\right)\subset\mathrm{Im}(\mathbb{O})\times\mathbb{O},\end{split}$

where $\displaystyle S^{6}$ denotes the subvariety of
$\displaystyle\mathrm{Im}(\mathbb{O})$ defined by $\displaystyle|z|^{2}=1$.

This lemma is well-known, we refer to corollary 5.4 in [Ada96]. We recall a
proof for the convenience of the reader since the results of [Ada96] are
stated over $\displaystyle\mathbb{R}$ and there are some minor modifications
to be made over $\displaystyle\mathbb{C}$.

Proof :
$\displaystyle\blacktriangleright\;$ The projection map:

$\begin{split}&\mathrm{Spin}_{7}\xrightarrow{\hskip
22.76228pt}\mathrm{SO}_{8}\\\ &(T_{1},T_{2},T_{1})\longrightarrow
T_{2}\end{split}$

is a double cover of $\displaystyle\mathrm{SO}_{7}$ embedded in
$\displaystyle\mathrm{SO_{8}}$ as the stabilizer of $\displaystyle
1\in\mathbb{O}$. Hence, $\displaystyle\mathrm{Spin}_{7}$ acts transitively on
$\displaystyle S^{6}$ and the stabilizer of any point in $\displaystyle S^{6}$
is $\displaystyle\mathrm{Spin}_{6}$. We know that
$\displaystyle\mathrm{Spin}_{6}\simeq\mathrm{SL}_{4}$ when we make the
identification
$\displaystyle\mathbb{H}\oplus\mathbb{H}.l=\mathbb{C}^{4}\oplus(\mathbb{C}^{4})^{*}$.
Furthermore, for any $\displaystyle(x+y.l)\in\mathbb{H}\oplus\mathbb{H}.l$,
the scalar product $\displaystyle\langle x,y\rangle$ is the perfect pairing
between $\displaystyle\mathbb{C}^{4}$ and $\displaystyle(\mathbb{C}^{4})^{*}$.
This pairing is obviously invariant for the product of the natural and the
contragedient representation of $\displaystyle\mathrm{SL}_{4}$. The action of
$\displaystyle\mathrm{SL}_{4}$ on
$\displaystyle\mathbb{C}^{4}\backslash\\{0\\}$ is transitive and the
stabilizer in $\displaystyle\mathrm{SL}_{4}$ of any non-zero $\displaystyle
x\in\mathbb{C}^{4}$ contains $\displaystyle\mathrm{SL}_{3}$, which acts on
$\displaystyle x^{\perp}\subset(\mathbb{C}^{4})^{*}$. The transitive action of
$\displaystyle\mathrm{SL}_{3}$ on
$\displaystyle\mathbb{C}^{3}\backslash\\{0\\}$ allows us to conclude the proof
of the lemma.

$\displaystyle\blacktriangleleft$

We can now proceed to the proof of Theorem 3.3.2. Recall that the action of
$\displaystyle\mathrm{Spin}_{7}$ on $\displaystyle\mathrm{J}_{3}(\mathbb{O})$
is given as follows. For any
$\displaystyle(T_{1},T_{2},T_{1})\in\mathrm{Spin_{7}}$ and any $\displaystyle
A=\begin{pmatrix}\lambda_{1}&c&\overline{b}\\\ \overline{c}&\lambda_{2}&a\\\
b&\overline{a}&\lambda_{3}\end{pmatrix}\in\mathrm{J}_{3}(\mathbb{O})$, we
have:

$\displaystyle(T_{1},T_{2},T_{1}).A=\begin{pmatrix}\lambda_{1}&T_{2}(c)&\overline{\mathrm{K}_{T_{1}}(b)}\\\
\overline{T_{2}(c)}&\lambda_{2}&T_{1}(a)\\\
\mathrm{K}_{T_{1}}(b)&\overline{T_{1}(a)}&\lambda_{3}\end{pmatrix}.$

We start with $\displaystyle A=\begin{pmatrix}\lambda_{1}&c&\overline{b}\\\
\overline{c}&\lambda_{2}&a\\\ b&\overline{a}&\lambda_{3}\end{pmatrix}$ generic
in $\displaystyle\mathrm{J}_{3}(\mathbb{O})$. By lemma 3.3.3, we can find
$\displaystyle(T_{1},T_{2},T_{1})\in\mathrm{Spin_{7}}$ such that
$\displaystyle T_{2}(c)=r_{1}+r_{2}.i$, with $\displaystyle
r_{1},r_{2}\in\mathbb{C}$ non zeros (by genericity of $\displaystyle A$).
Hence, we have:

$\displaystyle
A_{1}=(T_{1},T_{2},T_{1}).A=\begin{pmatrix}\lambda_{1}&r_{1}+r_{2}.i&\overline{b_{1}}\\\
r_{1}-r_{2}.i&\lambda_{2}&a^{\prime}\\\
b_{1}&\overline{a^{\prime}}&\lambda_{3}\end{pmatrix},$

with $\displaystyle a^{\prime}=T_{1}(a)$ and $\displaystyle
b_{1}=\mathrm{K}_{T_{1}}(b)$. Let us write $\displaystyle
a^{\prime}=a_{0}^{\prime}+a_{1}^{\prime}.l$ with $\displaystyle
a_{0}^{\prime},a_{1}^{\prime}\in\mathbb{H}$. Since $\displaystyle A$ is
assumed generic, we can arrange $\displaystyle T_{1}$ in the previous
transformation so that $\displaystyle\langle
r_{1}-r_{2}.i,a_{1}^{\prime}\rangle\neq 0$. We plan to make the transformation
$\displaystyle a^{\prime}\longleftarrow a^{\prime}-\frac{\langle
a_{0}^{\prime},a_{1}^{\prime}\rangle}{\langle
r_{1}-r_{2}.i,a_{1}^{\prime}\rangle}(r_{1}-r_{2}.i)$. We thus consider
$\displaystyle H=\begin{pmatrix}1&0&\frac{-\langle
a_{0}^{\prime},a_{1}^{\prime}\rangle}{\langle
r_{1}-r_{2}.i,a_{1}^{\prime}\rangle}\\\ 0&1&0\\\
0&0&1\end{pmatrix}\in\mathrm{SL}_{3}$ and we have:

$\displaystyle
A_{2}={}^{t}HA_{1}H=\begin{pmatrix}\lambda_{1}&r_{1}+r_{2}.i&\overline{b_{2}}\\\
r_{1}-r_{2}.i&\lambda_{2}&a^{\prime\prime}\\\
b_{2}&\overline{a^{\prime\prime}}&{\lambda_{3}}^{\prime}\end{pmatrix},$

with $\displaystyle a^{\prime\prime}=a^{\prime}-\frac{\langle
a_{0}^{\prime},a_{1}^{\prime}\rangle}{\langle
r_{1}-r_{2}.i,a_{1}^{\prime}\rangle}(r_{1}-r_{2}.i)$, $\displaystyle
b_{2}\in\mathbb{O}$ and $\displaystyle{\lambda_{3}}^{\prime}\in\mathbb{C}$. If
we put $\displaystyle a^{\prime\prime}=a_{0}^{\prime\prime}+a_{1}^{\prime}.l$
with $\displaystyle a_{0}^{\prime\prime}\in\mathbb{H}$, then we have
$\displaystyle\langle a_{0}^{\prime\prime},a_{1}^{\prime}\rangle=0$ by
construction. By lemma 3.3.3, we can find
$\displaystyle(T_{1},T_{2},T_{1})\in\mathrm{Spin_{7}}$ such that
$\displaystyle T_{2}(r_{1}+r_{2}.i)=r_{1}+r_{2}.n$ and $\displaystyle
T_{1}(a^{\prime\prime})=r_{3}.i+r_{4}.n$, with $\displaystyle
r_{3},r_{4}\in\mathbb{C}$ non zeros (by genericity of $\displaystyle A$).
Hence, we have:

$\displaystyle
A_{3}=(T_{1},T_{2},T_{1}).A_{2}=\begin{pmatrix}\lambda_{1}&r_{1}+r_{2}.n&\overline{b_{3}}\\\
r_{1}-r_{2}.n&\lambda_{2}&r_{3}.i+r_{4}.n\\\
b_{3}&-r_{3}.i-r_{4}.n&{\lambda_{3}}^{\prime}\end{pmatrix},$

with $\displaystyle b_{3}\in\mathbb{O}$. We now plan to make the
transformation $\displaystyle r_{3}i.+r_{4}.n\longleftarrow
r_{3}.i+r_{4}.n+\frac{r_{4}}{r_{2}}(r_{1}-r_{2}.n)$. We thus consider
$\displaystyle H=\begin{pmatrix}1&0&\frac{r_{4}}{r_{2}}\\\ 0&1&0\\\
0&0&1\end{pmatrix}\in\mathrm{SL}_{3}$ and we have:

$\displaystyle
A_{4}={}^{t}HA_{3}H=\begin{pmatrix}\lambda_{1}&r_{1}+r_{2}.n&\overline{b_{4}}\\\
r_{1}-r_{2}.n&\lambda_{2}&r_{5}+r_{3}.i\\\
b_{4}&r_{5}-r_{.}i&{\lambda_{3}}^{\prime\prime}\end{pmatrix},$

with $\displaystyle r_{5}=\frac{r_{4}}{r_{2}}r_{1}\in\mathbb{C}$,
$\displaystyle{\lambda_{3}}^{\prime\prime}\in\mathbb{C}$ and $\displaystyle
b_{4}\in\mathbb{O}$. By lemma 3.3.3, we can find
$\displaystyle(T_{1},T_{2},T_{1})\in\mathrm{Spin_{7}}$ such that
$\displaystyle T_{2}(r_{1}+r_{2}n)=r_{1}+r_{2}.i$ and $\displaystyle
T_{1}(r_{5}+r_{3}.i)=r_{5}+r_{3}.i$. Hence, we have:

$\displaystyle
A_{5}=(T_{1},T_{2},T_{1}).A_{4}=\begin{pmatrix}\lambda_{1}&r_{1}+r_{2}.i&\overline{b_{5}}\\\
r_{1}-r_{2}.i&\lambda_{2}&r_{5}+r_{3}.i\\\
b_{5}&r_{5}-r_{3}.i&{\lambda_{3}}^{\prime\prime}\end{pmatrix},$

with $\displaystyle b_{5}\in\mathbb{O}$. We now plan to make the
transformation $\displaystyle r_{1}+r_{2}.i\longleftarrow
r_{1}+r_{2}.i+\frac{r_{2}}{r_{3}}(r_{5}-r_{3}.i)$. We thus consider
$\displaystyle H=\begin{pmatrix}1&0&0\\\ 0&1&0\\\
\frac{r_{2}}{r_{3}}&0&1\end{pmatrix}\in\mathrm{SL}_{3}$ and we have:

$\displaystyle
A_{6}={}^{t}HA_{5}H=\begin{pmatrix}{\lambda_{1}}^{\prime}&r_{6}&\overline{b_{6}}\\\
r_{6}&\lambda_{2}&r_{5}+r_{3}.i\\\
b_{5}&r_{5}-r_{3}.i&{\lambda_{3}}^{\prime\prime}\end{pmatrix},$

with $\displaystyle r_{6}=r_{1}+\frac{r_{2}}{r_{3}}r_{5}\in\mathbb{C}$,
$\displaystyle{\lambda_{1}}^{\prime}\in\mathbb{C}$ and $\displaystyle
b_{6}\in\mathbb{O}$. By lemma 3.3.3, we can find
$\displaystyle(T_{1},T_{2},T_{1})\in\mathrm{Spin_{7}}$ such that
$\displaystyle T_{1}(r_{5}+r_{3}.i)=r_{7}\in\mathbb{C}$ (and $\displaystyle
T_{2}(r_{6})=r_{6}$, automatically). Hence, we have:

$\displaystyle
A_{7}=(T_{1},T_{2},T_{1}).A_{6}=\begin{pmatrix}{\lambda_{1}}^{\prime}&r_{6}&\overline{b_{7}}\\\
r_{6}&\lambda_{2}&r_{7}\\\
b_{7}&r_{7}&{\lambda_{3}}^{\prime\prime}\end{pmatrix}.$

We now consider:

$\displaystyle H_{1}=\begin{pmatrix}1&0&0\\\ 0&1&0\\\
-\frac{r_{6}}{r_{7}}&0&1\end{pmatrix},H_{2}=\begin{pmatrix}1&0&0\\\
0&1&\frac{-r_{7}}{\lambda_{2}}\\\
0&0&1\end{pmatrix},H_{3}=\begin{pmatrix}-1&0&0\\\ 0&0&1\\\
0&1&0\end{pmatrix}.$

We have:

$\displaystyle
A_{8}={}^{t}H_{3}{}^{t}H_{2}{}^{t}H_{1}A_{7}H_{1}H_{2}H_{3}=\begin{pmatrix}\mu_{1}&\overline{b_{8}}&0\\\
b_{8}&\mu_{2}&0\\\ 0&0&\mu_{3}\end{pmatrix},$

with $\displaystyle b_{8}\in\mathbb{O}$ and
$\displaystyle\mu_{1},\mu_{2},\mu_{3}\in\mathbb{C}$. By lemma 3.3.3, we can
find $\displaystyle(T_{1},T_{2},T_{1})\in\mathrm{Spin_{7}}$ such that
$\displaystyle T_{2}(\overline{b_{8}})=r_{8}+r_{9}.i$, with $\displaystyle
r_{8},r_{9}\in\mathbb{C}$. Hence, we have:

$\displaystyle
A_{9}=(T_{1},T_{2},T_{1}).A_{8}=\begin{pmatrix}\mu_{1}&r_{8}+r_{9}.i&0\\\
r_{8}-r_{9}.i&\mu_{2}&0\\\ 0&0&\mu_{3}\end{pmatrix}.$

We let $\displaystyle H=\begin{pmatrix}0&0&1\\\ 0&-1&0\\\
1&0&0\end{pmatrix}\in\mathrm{SL}_{3}$. We have:

$\displaystyle A_{10}={}^{t}H.A_{9}.H=\begin{pmatrix}\mu_{3}&0&0\\\
0&\mu_{2}&r_{8}+r_{9}.i\\\ 0&r_{8}-r_{9}.i&\mu_{1}\end{pmatrix}.$

By lemma 3.3.3, we can find
$\displaystyle(T_{1},T_{2},T_{1})\in\mathrm{Spin_{7}}$ such that
$\displaystyle T_{1}(r_{8}+r_{9}.i)=r_{10}$, with $\displaystyle
r_{10}\in\mathbb{C}$. Hence, we have:

$\displaystyle A_{11}=(T_{1},T_{2},T_{1}).A_{10}=\begin{pmatrix}\mu_{3}&0&0\\\
0&\mu_{2}&r_{10}\\\ 0&r_{10}&\mu_{1}\end{pmatrix}.$

Notice that $\displaystyle A_{11}$ is symmetric with complex coefficients.
Since $\displaystyle A\in\mathrm{J}_{3}(\mathbb{O})$ was chosen generic, we
have $\displaystyle\mathrm{det}(A_{11})=\mathrm{det}_{\mathbb{O}}(A)\neq 0$
and we deduce that there exists $\displaystyle H\in\mathrm{GL}_{3}$ such that:

$\displaystyle{}^{t}HA_{11}H=\begin{pmatrix}1&0&0\\\ 0&1&0\\\
0&0&1\end{pmatrix}.$

We thus have proved that a generic element $\displaystyle
A\in\mathrm{J}_{3}(\mathbb{O})$ is in the same orbit as
$\displaystyle\begin{pmatrix}1&0&0\\\ 0&1&0\\\ 0&0&1\end{pmatrix}$ for the
action of the subgroup of $\displaystyle\mathrm{GL}_{24}$ generated by
$\displaystyle\mathrm{Spin}_{7}$ and $\displaystyle\mathrm{GL}_{3}$. As a
consequence, the action of
$\displaystyle\mathbb{C}^{*}\times\mathrm{Aut}^{0}(\mathfrak{S})$ on
$\displaystyle\mathrm{J}_{3}(\mathbb{O})$ is prehomogeneous.

$\displaystyle\blacktriangleleft$

##  Appendix A On the dimension the Lie-algebra
$\displaystyle\mathfrak{aut}(\SS)$

In this section, we provide a Macaulay2 code to get an upper bound on the
dimension of $\displaystyle\mathfrak{aut}(\SS)$. We refer to the proof of
Theorem 2.2.1 in section $\displaystyle 2$, where this computation is used to
determine the neutral component of $\displaystyle\mathrm{Aut}(\SS)$.

We first need to write explicitly the equation of $\displaystyle\SS$ in
Macaulay2. This is somehow the main challenge, as the octonion algebra can not
be simulated in Macaulay2.

    kk = ZZ/313

    V = kk[x_1..x_8,y_1..y_27]

    M=matrix{{x_1,-x_2,-x_3,-x_4,-x_5,-x_6,-x_7,-x_8},
    {x_2,x_1,-x_4,x_3,-x_6,x_5,x_8,-x_7},
    {x_3,x_4,x_1,-x_2,-x_7,-x_8,x_5,x_6},
    {x_4,-x_3,x_2,x_1,-x_8,x_7,-x_6,x_5},
    {x_5,x_6,x_7,x_8,x_1,-x_2,-x_3,-x_4},
    {x_6,-x_5,x_8,-x_7,x_2,x_1,x_4,-x_3},
    {x_7,-x_8,-x_5,x_6,x_3,-x_4,x_1,x_2},
    {x_8,x_7,-x_6,-x_5,x_4,x_3,-x_2,x_1}}

    I8 = matrix{{1,0,0,0,0,0,0,0},{0,1,0,0,0,0,0,0},{0,0,1,0,0,0,0,0},{0,0,0,1,0,0,0,0},
    {0,0,0,0,1,0,0,0},{0,0,0,0,0,1,0,0},{0,0,0,0,0,0,1,0},{0,0,0,0,0,0,0,1}}

    N = matrix{{x_1,-x_2,-x_3,-x_4,-x_5,-x_6,-x_7,-x_8},
    {x_2,x_1,x_4,-x_3,x_6,-x_5,-x_8,x_7},
    {x_3,-x_4,x_1,x_2,x_7,x_8,-x_5,-x_6},
    {x_4,x_3,-x_2,x_1,x_8,-x_7,x_6,-x_5},
    {x_5,-x_6,-x_7,-x_8,x_1,x_2,x_3,x_4},
    {x_6,x_5,-x_8,x_7,-x_2,x_1,-x_4,x_3},
    {x_7,x_8,x_5,-x_6,-x_3,x_4,x_1,-x_2},
    {x_8,-x_7,x_6,x_5,-x_4,-x_3,x_2,x_1}}

    L = mutableMatrix(M)

    for k from 0 to 7 do (for i from 0 to 7 do
    (for j from 1 to 8 do L_(k,i)= sub(L_(k,i), x_j=> y_(j+8))))

    S2 = matrix(L)

    L = mutableMatrix(N)

    for k from 0 to 7 do (for i from 0 to 7 do
    (for j from 1 to 8 do L_(k,i)= sub(L_(k,i), x_j=> y_(j+8))))

    R2 = matrix(L)

    L = mutableMatrix(M)
    for k from 0 to 7 do (for i from 0 to 7 do
    (for j from 1 to 8 do L_(k,i)= sub(L_(k,i), x_j=> y_(j+16))))

    S3 = matrix(L)

    L = mutableMatrix(N)

    for k from 0 to 7 do (for i from 0 to 7 do
    (for j from 1 to 8 do L_(k,i)= sub(L_(k,i), x_j=> y_(j+16))))

    R3 = matrix(L)

    L = mutableMatrix(M)

    for k from 0 to 7 do (for i from 0 to 7 do
    (for j from 1 to 8 do L_(k,i)= sub(L_(k,i), x_j=> y_j)))

    S1 = matrix(L)

    L = mutableMatrix(N)

    for k from 0 to 7 do (for i from 0 to 7 do
    (for j from 1 to 8 do L_(k,i)= sub(L_(k,i), x_j=> y_j)))

    R1 = matrix(L)

    L = mutableMatrix(M)

    for k from 0 to 7 do (for i from 0 to 7 do L_(k,i)= sub(L_(k,i),
     x_1 =>y_1*y_17-y_2*y_18-y_3*y_19-y_4*y_20-y_5*y_21-y_6*y_22-y_7*y_23-y_8*y_24))

    for k from 0 to 7 do (for i from 0 to 7 do L_(k,i)= sub(L_(k,i),
    x_2 =>y_1*y_18+ y_2*y_17+y_3*y_20-y_4*y_19+y_5*y_22-y_6*y_21-y_7*y_24+y_8*y_23))

    for k from 0 to 7 do (for i from 0 to 7 do L_(k,i)= sub(L_(k,i),
    x_3 =>y_1*y_19-y_2*y_20+y_3*y_17+y_4*y_18+y_5*y_23+y_6*y_24-y_7*y_21-y_8*y_22))

    for k from 0 to 7 do (for i from 0 to 7 do L_(k,i)= sub(L_(k,i),
    x_4 =>y_1*y_20+y_2*y_19-y_3*y_18+y_4*y_17+y_5*y_24-y_6*y_23+y_7*y_22-y_8*y_21))

    for k from 0 to 7 do (for i from 0 to 7 do L_(k,i)= sub(L_(k,i),
    x_5 =>y_1*y_21-y_2*y_22-y_3*y_23-y_4*y_24+y_5*y_17+y_6*y_18+y_7*y_19+y_8*y_20))

    for k from 0 to 7 do (for i from 0 to 7 do L_(k,i)= sub(L_(k,i),
    x_6 =>y_1*y_22+y_2*y_21-y_3*y_24+y_4*y_23-y_5*y_18+y_6*y_17-y_7*y_20+y_8*y_19))

    for k from 0 to 7 do (for i from 0 to 7 do L_(k,i)= sub(L_(k,i),
    x_7=> y_1*y_23+y_2*y_24+y_3*y_21-y_4*y_22-y_5*y_19+y_6*y_20+y_7*y_17-y_8*y_18))

    for k from 0 to 7 do (for i from 0 to 7 do L_(k,i)= sub(L_(k,i),
    x_8=>y_1*y_24-y_2*y_23+y_3*y_22+y_4*y_21-y_5*y_20-y_6*y_19+y_7*y_18+y_8*y_17))

    S13 = matrix(L)

    for k from 0 to 7 do (for i from 0 to 7 do (for j from 2 to 8 do
    L_(k,i)= sub(L_(k,i), y_j=>-y_j)))

    for k from 0 to 7 do (for i from 0 to 7 do (for j from 17 to 24 do
    L_(k,i)= sub(L_(k,i), y_j=> y_(j-8))))

    Sb12 = matrix(L)

    L = mutableMatrix(Sb12)

    for k from 0 to 7 do (for i from 0 to 7 do (for j from 1 to 8 do
    L_(k,i)= sub(L_(k,i), y_j=> y_(16+j))))

    for k from 0 to 7 do (for i from 0 to 7 do (for j from 9 to 16 do
    L_(k,i)= sub(L_(k,i), y_j=> y_(j-8))))

    for k from 0 to 7 do (for i from 0 to 7 do (for j from 2 to 8 do
    L_(k,i)= sub(L_(k,i), y_j=> -y_j)))

    Sb3b1 = matrix(L)

    for k from 0 to 7 do (for i from 0 to 7 do (for j from 17 to 24 do
    L_(k,i) = sub(L_(k,i), y_j=> y_(j-8))))

    for k from 0 to 7 do (for i from 0 to 7 do (for j from 2 to 8 do
    L_(k,i)= sub(L_(k,i), y_j=> -y_j)))

    Sb21 = matrix(L)

    L = mutableMatrix(Sb12)

    for k from 0 to 7 do (for i from 0 to 7 do (for j from 2 to 8 do
    L_(k,i) = sub(L_(k,i), y_j=> -y_j)))

    S12 = matrix(L)

    for k from 0 to 7 do (for i from 0 to 7 do (for j from 10 to 16 do
    L_(k,i) = sub(L_(k,i), y_j=> -y_j)))

    S1b2 = matrix(L)

    L = mutableMatrix(Sb21)

    for k from 0 to 7 do (for i from 0 to 7 do (for j from 10 to 16 do
    L_(k,i) = sub(L_(k,i), y_j=> -y_j)))

    S21 = matrix(L)

    AA = (S13)*S2 + transpose(S2)*(Sb3b1)

    P0 = AA_(0,0)

    P1 = -y_27*((S1*transpose(S1))_(0,0))

    P2 = -y_26*((S2*transpose(S2))_(0,0))

    P3 = -y_25*((S3*transpose(S3))_(0,0))

    P = y_25*y_26*y_27 + P0 + P1 + P2 + P3

    PHI123 =  (1/2)*((S12-S21)*transpose(S3))_(0,0)

    L = mutableMatrix(S1)

    for k from 0 to 7 do (for i from 0 to 7 do
    L_(k,i) = sub(L_(k,i), y_1=> 0))

    M1 = matrix(L)

    L = mutableMatrix(S2)

    for k from 0 to 7 do (for i from 0 to 7 do
    L_(k,i) = sub(L_(k,i), y_9=> 0))

    M2 = matrix(L)

    L = mutableMatrix(S3)

    for k from 0 to 7 do (for i from 0 to 7 do
    L_(k,i) = sub(L_(k,i), y_17=> 0))

    M3 = matrix(L)

    G = matrix{{(M1*transpose(M1))_(0,0),(M1*M2)_(0,0),(M1*transpose(M3))_(0,0)},
    {(transpose(M2)*transpose(M1))_(0,0),(M2*transpose(M2))_(0,0),
    (transpose(M2)*transpose(M3))_(0,0)},
    {(M3*transpose(M1))_(0,0),(M3*M2)_(0,0),(M3*transpose(M3))_(0,0)}}

    ASS123square = 4*(det(G) - (PHI123)*(PHI123))

    SODM = P^2 - 4*(PHI123)*P - ASS123square

Here SODM is the equation of the sextic $\displaystyle\SS$, P is
$\displaystyle\mathrm{Det}_{\mathbb{O}}$, PHI123 is $\displaystyle\phi(c,b,a)$
and ASS123square is $\displaystyle\bigg{|}[c,b,a]\bigg{|}^{2}$. We have used
the formula (see lemma 6.56 and 6.61 in [Har90]):

$\displaystyle\bigg{|}[c,b,a]\bigg{|}^{2}=4\bigg{(}\mathrm{Gram}(\mathrm{Im}(c),\mathrm{Im}(b),\mathrm{Im}(a))-\phi(c,b,a)^{2}\bigg{)},$

which allows to get the expression of
$\displaystyle\bigg{|}[c,b,a]\bigg{|}^{2}$ without computing triple products
explicitly in $\displaystyle\mathbb{O}$. We recall that differential of the
map $\displaystyle\varphi$ introduced in the proof of Theorem 2.2.1 at a point
$\displaystyle\left[(m)_{i,j}\right]\in M_{6\times 27}$ is given by :

$\begin{split}d\varphi_{\left[m_{i,j}\right]}\,\,\,:\,\,\,&M_{6\times
27}/\mathfrak{aut}(\SS)\xrightarrow{\hskip 22.76228pt}S^{6}\mathbb{C}^{6}\\\
&\left[q_{i,j}\right]\xrightarrow{\hskip
22.76228pt}\sum_{i=1}^{27}\left(\sum_{j=1}^{6}q_{i,j}.z_{j}\right)\times\dfrac{\SS}{\partial
x_{i}}(m_{1,1}.z_{1}+\cdots+m_{1,6}.z_{6},\cdots,m_{27,1}.z_{1}+\cdots+m_{27,6}.z_{6}),\\\
\end{split}$

We produce a matrix representing $\displaystyle d\varphi$ at a random point
$\displaystyle\left[(m)_{i,j}\right]\in M_{6\times 27}$.

    D = mutableMatrix(V,1,27)

    for i from 1 to 27  do D_(0,i-1) = diff(y_i,SODM)

    M = random(ZZ^27,ZZ^6,Height=>50)

    M2 = M**V

    m = M2 * matrix{{x_1},{x_2},{x_3},{x_4},{x_5},{x_6}}

    for i from 0 to 26 do (for j from 1 to 27 do D_(0,i) = sub(D_(0,i), y_j => m_(j-1,0)))

    W = kk[x_1..x_6]

    Dphi = mutableMatrix(W,1,27)

    for j from 0 to 26 do Dphi_(0,j) = sub(D_(0,j),W)

The matrix Dphi represents $\displaystyle d\varphi$ at the random
$\displaystyle\left[(m)_{i,j}\right]\in M_{6\times 27}$ chosen. We are left to
compute the rank of the image of Dphi in $\displaystyle S^{6}\mathbb{C}^{6}$.

    I = (0)

    for j from 0 to 26 do (for i from 1 to 6 do I = I + x_i*ideal(DphiW_(0,j)))

    II = module I

    III = II**kk

    rank III

The answer (=133) is obtained on a portable workstation in less than 15
minutes. The semi-continuity Theorem insures that the rank of $\displaystyle
d\varphi$ at a generic point $\displaystyle\left[(m)_{i,j}\right]\in
M_{6\times 27}$ and in characteristic $\displaystyle 0$ is bigger or equal to
$\displaystyle 133$.

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