Let $K$ be an algebraically closed field of characteristic $0$, and let $\mathbb{O}$ be the Cayley algebra over $K$. Let 
$$
\mathfrak{J}\_{3}=\{A\in\mathcal{M}\_{3}(\mathbb{O}):A\text{ is Hermitian}\},
$$
that may be considered as a $K$-vector space of dimension $27$. We are going to denote $\mathbb{P}^{26}=\mathbb{P}(\mathfrak{J}\_{3})$.


Given $A,B\in\mathfrak{J}\_{3}$ we may define:
$$
A\circ B:=\frac{1}{2}(AB+BA),
$$
$$
A\*B:=A\circ B-\frac{1}{2}(A\cdot tr(B)+B\cdot tr(A))+\frac{1}{2}(tr(A)tr(B)-tr(A\circ B)),
$$
$$
A\times A:=A\circ A-tr(A)\cdot A+\frac {1}{2}(tr(A\circ A)-tr(A)^{2})\cdot Id,
$$
$$
\det A:=\frac{1}{3}tr((A\*A)\circ A).
$$
We say that $rk(A)=1$ if $A\times A= 0$. Then, I want to prove that
$$
X=\{[A]\in\mathbb{P}^{26}:rk (A)=1\},
$$
is an algebraic variety of dimension $16$ such that its secant variety is 
$$
SX=\{[A]\in\mathbb{P}^{26}:\det A=0\},
$$
and that $\dim SX=25$ (i.e. I want to prove that $X$ is the fourth Severi variety).


It is stated in the last parragraph of p.18 of [this](http://arxiv.org/pdf/0812.1904.pdf) text, but I haven't been able to find a sketch of the proof of this fact. 


There is an alternative way via representation theory that is explained in Lazarsfeld's book on Zak's work or in Zak's '*Tangents and secants of algebraic varieties*', but I would like to know if it is not very hard to see it this way.



Looking for Severi varieties
Partial answer:


Let $A = \left( \begin{array}{ccc}
a & b & c \\
\overline{b} & e & d \\
\overline{c} & \overline{d} & f \end{array} \right) $ be the generic hermitian matrix with octonionic coefficients. This means that $b,c,d$ can be written as $X\_1.1 +X\_2.i\_1 + \cdots + X\_8.i\_7$, where the $X\_p$ are abstract variables (which will take value in $K$) and $1,i\_1, \cdots, i\_7$ are basis of $\mathbb{O}$ over $K$.


The symbols $\overline{b}, \overline{c}, \overline{d}$ are the conjugate of $b,c,d$ and $a,e,f$ are self conjugate, that is they are of the form $X\_1.1$, with $X\_1$ a variable which will take value in the field $K$. Overall, you have $27$ abstract variables $X\_1,\cdots X\_{27}$ (which will take value in $K$) coming into the picture.


Now, if you compute de determinant of $A$ (using Sarrus rule for instance), you will find a cubic equation in $X\_1, \cdots, X\_{27}$ which involves only integer numbers (the purely octonionic numbers disappear, which is kind of extraordinary). This computation makes sense because of the pseudo-associativity of $\mathbb{O}$, you don't need it to be commutative or even associatif.


This equation is the equation of a cubic hypersurface in $\mathbb{P}^{26}$ and it is irreducible. Now let $Z$ be the scheme cut out by the $2 \times 2$-minors of $A$. Again one finds that the minors are equations in $X\_1, \cdots, X\_{27}$ with only integers coefficients (again the purely octonionic coefficients vanish).


Each minor is a quadratic equation in $X\_1, \cdots, X\_{27}$ and it is easy to see that they correspond to the partial derivatives of the equation of $det A$ with respect to each variables $X\_1, \cdots, X\_{27}$. Hence, the scheme $Z$ is the singular locus of $det A = 0$. 


The scheme $Z$ is smooth. Indeed, if it had a singular point, this point woud be a triple point of $det A =0$. Being a cubic hypersurface, $det A =0$ would then be a cone, which isn't the case (this is a bit more complicated to check).


Now, $Z$ being the singular locus of $det A=0$ and this hypersurface being cubic, Bezout's theorem insures that every line meeting two points of $Z$ will be included in $det A=0$. This proves that $S(Z) \subset \{ det A = 0 \}$, where $S(Z)$ is the secant variety of $Z$.


Now two things are left to prove :


\_$\dim Z = 16$,


\_ the secant variety $S(Z)$ is actually the whole hypersurface $det A =0$.


I don't remember a simple proof of the first fact. Let's admit it. Since $Z$ is smooth of dimension $16$ and included in $\mathbb{P}^{26}$, ZAK's theorem on tangency gurantees that the dimension of $S(Z)$ must be at least $25$. The hypersurface $det A = 0$ being irreducible, we have $S(Z) = \{det A = 0 \}$.


**EDIT** : There is a mistake in the description of the $2 \times 2$ minors. Three of them will indeed be quadartic equations over $K$ with integer coefficients. The other $3$ will have fully octonionic equations, each of them giving $8$ equations over $K$ with integer coeffcients. At the end, wet $27$ equations over $K$ with integer coefficients, corresponding indeed to the $27$ partial derivatives of the equation of $det A$.


