# Orbits of action of the split group of type $F_4$

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?

• 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. – YCor Mar 12 at 20:00

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

• 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. – Victor Petrov Mar 13 at 16:27
• I think there is a mistake. Are you doing $E_6\times \mathbb G_m$? – A.Garcia Mar 18 at 18:20
• 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? – A.Garcia Mar 19 at 6:15
• I agree that $E_6\times \mathbb G_m$ does not act on trace zero matrices, that was my mistake. – A.Garcia Mar 19 at 6:16
• @A.Garcia You are right, I made a mistake in the description of orbits. I will edit. – Libli Mar 19 at 11:18