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Let the split group of type $F_4$ act as the automorphism group of the split Albert algebra $A$. Consider the action of $F_4\times \mathbb{G}_m$ on $A$, given by letting $\mathbb{G}_m$ act by scalar multiplication.

Does this action have finitely many orbits? Are the stabilizers known? I am very new to $F_4$ and I am slowly going through the basics, but a reference in this direction would greatly simplify my life.

Edit: I now realize that the answer to my question is negative, because the rational quotient $A_0/F_4$ is $2$-dimensional (where $A_0$ is the trace-zero subspace of $A$), just by looking at diagonal matrices. Possibly the answer below refers to $E_6\times \mathbb G_m$?

So the correct question is: are there finitely many orbits in the locus of matrices not diagonalizable with distinct eigenvalues? What are their stabilizers?

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    $\begingroup$ When reading the title, I was wondering if $F_4$ means the field of order 4 or a free group on 4 generators. I edited to make more clear what $F_4$ denotes. $\endgroup$
    – YCor
    Commented Mar 12, 2019 at 20:00

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I work over the complexe numbers, but the story is similar over any fields (provided you choose the split octonions). Let us identify the Albert algebra $A$ with the algebra of self-adjoint $3*3$ matrices with octonionic coefficients. The hyperplane given by $\mathrm{Tr}(X) = 0$ is stabilized by $F_4 \times \mathbb{G}_m$ and the action of $F_4 \times \mathbb{G}_m$ on this hyperplane can be devided into two types.

First : the orbits located in the discriminant hypersurface (that is the set of matrices of rank less or equal to $2$). They are:

_the zero of $A$,

_the set of matrice of rank $1$ (denote it by $Z_0$, it has dimension $16$),

_an orbit which closure contains the previous one and that has dimension $17$,

_the (open part) of a certain restricted tangent variety to $Z_0$,

_the discriminant hypersurface minus the previous one and the closure of the third one : it has dimension $25$.

Note that the equation of the discriminant hypersurface is $6\mathrm{det}(X)^2-9(\mathrm{Tr}(X^2))^3 = 0$.

Now, let's turn to the orbits located outside of the discriminant hypersurface. There is a one dimensional family of them: they are the hypersurfaces given by the equation $6t\mathrm{det}(X)^2-s(\mathrm{Tr}(X^2))^3 = 0$, for $[s,t] \in \mathbb{P}^1 \backslash [1,9]$.

All of this is discussed at lentgh and proved in details in proposition 3.5 of https://arxiv.org/pdf/math/0306328.pdf

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    $\begingroup$ If the base field is not closed, iv) becomes a union of orbits according to $N(x)\bmod (F^*)^3$, where $N$ is the norm. $\endgroup$
    – Victor Petrov
    Commented Mar 13, 2019 at 16:27
  • $\begingroup$ I think there is a mistake. Are you doing $E_6\times \mathbb G_m$? $\endgroup$
    – A.Garcia
    Commented Mar 18, 2019 at 18:20
  • $\begingroup$ I am sorry but I am very confused. The book Algebraic Geometry of Parshin Shafarevich et al says that the representation of F4 on the trace zero hyperplane has a 2-dimensional rational quotient. How can this reconcile with what you say? $\endgroup$
    – A.Garcia
    Commented Mar 19, 2019 at 6:15
  • $\begingroup$ I agree that $E_6\times \mathbb G_m$ does not act on trace zero matrices, that was my mistake. $\endgroup$
    – A.Garcia
    Commented Mar 19, 2019 at 6:16
  • $\begingroup$ @A.Garcia You are right, I made a mistake in the description of orbits. I will edit. $\endgroup$
    – Libli
    Commented Mar 19, 2019 at 11:18

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