# Looking for Severi varieties

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 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.

• page 6, here : arxiv.org/pdf/alg-geom/9412012.pdf perhaps can be useful for you. – Holonomia Feb 11 '16 at 19:29
• see also page 28, Proposition 3.4 here arxiv.org/pdf/0801.4076v2.pdf – Holonomia Feb 11 '16 at 21:23
• It is very nice to see that someone is still working on such subjects. I thought Freudenthal was the last one :) – user21230 Mar 31 '16 at 9:17

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 $det A$.

• Thank you, it helps a lot! I'll ask regarding another interpretation of this variety soon. Maybe it is easier to prove the smoothness and find the dimension of $Z$ in this other way. – Srinivasa Granujan Feb 12 '16 at 7:58
• @SrinivasaGranujan Well it depends how much representation theory you want to do. The operation that you defined will be helpful if one wants to see that $\mathbb{P}^{26}$ is pre-homogeous for the action of $SL_3(\mathbb{O}) = E_6$ and that $Z$ and $S(Z)$ are the only non-trivial orbits. But if you really want to avoid representation theory then I have the impression you don't really need the equations you gave. – Libli Feb 12 '16 at 17:09
• @SrinivasaGranujan Zak's theorem on tangency implies a bound on the defect of a smooth projective variety of dimension $n$ embedded in $\mathbb{P}^m$. It's also called Zak's theorem on linear normaility. In the special case here, it implies that a smooth $16$-dimensional variety embedded in $\mathbb{P}^{25}$ cannot have a secant variety of dimension less than $25$. – Libli Feb 13 '16 at 19:39
• @SrinivasaGranujan If $X$ is a hypersurface in projective space, then $X$ is a cone implies that the Hessian determinant of any equation defining $X$ vanishes. I haven't done this computation, but I am pretty sure that the Hessian determinant of $det A$ does not vanish. – Libli Feb 14 '16 at 0:04
• @SrinivasaGranujan Since $Z$ is smooth, if it is not irreducible, then it means that $Z$ is not connected. But then write $Z = Z_1 \cup Z_2$, with $Z_1$ and $Z_2$ be connected components of $Z$. It's easy to check that $S(Z_1)$ and $S(Z_2)$ are contains different irreducible components of $S(Z)$. Since $S(Z)$ is irreducible, this is impossible, $Z$ must be irreducible. – Libli Feb 19 '16 at 23:09