ESSENTIAL DIMENSION OF ALBERT ALGEBRAS

MARK L. MACDONALD

Abstract. This paper shows that the number of independent parameters required to describe an Albert algebra up to isomorphism is at most 7. In other words, the essential dimension of the split group of type F_{4} over a field of characteristic not 2 or 3 satisfies ed(F_{4}) ≤ 7. This is achieved by reducing the structural group from the full 52-dimensional automorphism group to a subgroup of dimension 10, and exhibiting an 18-dimenional generically free linear representation which remains generically free once projectivized.

An Albert algebra is a certain type of 27-dimensional non-associative algebra over a field (see 1.2 for a definition; also [SV00] [Alb47]). If the field is algebraically closed, then there is only one Albert algebra, but for arbitrary fields there are many open questions about their classification up to isomorphism. The basic question being addressed in this paper is:

Question 0.1. How many algebraically independent parameters do we need to describe an arbitrary Albert algebra up to isomorphism?

To make this precise, let me introduce some terminology. Say we have an inclusion of fields F ⊂ F′ ⊂ F′′. Let A be an algebra over F′′, and let B be an algebra over a subfield F′ such that A ≅ B ⊗_{F′} F′′ (isomorphic as algebras over F′′). Then we say F′ is a field of definition of A. In this situation, A is an Albert algebra if and only if B is. So, with respect to Question 0.1, at most tr. deg_{F} F′ independent parameters are needed to describe A.

Fix a base field F. Consider the smallest number d such that for any Albert algebra over any field extension of F there exists a field of definition whose transcendence degree over F is at most d. This number, which is known to be finite, may be taken as an answer to Question 0.1. In this paper we show that it is at most 7 (see Theorem 0.3).

0.2. Essential dimension. It is known that isomorphism classes of Albert algebras correspond to isomorphism classes of F_{4}-torsors, for the split exceptional group of type F_{4} [KMRT, 37.11]. The essential dimension of an algebraic group G is a non-negative integer which has the property that ed(G) = 0 precisely for the special groups, which were classified by Grothendieck (for a definition, see Section 1). So one could think of ed(G) as measuring how complicated G-torsors can be, up to isomorphism. In fact, the above number d (and hence our answer to Question 0.1) is equal to ed(F_{4}).

It is known that 5 ≤ ed(F_{4}) due to the existence and non-triviality of Serre’s degree 5 cohomological invariant of Albert algebras [GMS, Thm. 22.5] [BF03, Cor. 3.6]. Previously, the best known upper bound was ed(F_{4}) ≤ 19 for fields of characteristic zero [Le04]. The main result of the present paper is the following theorem.

Theorem 0.3. If the base field is of characteristic not 2 or 3, then ed(F_{4}) ≤ 7.

The proof is given at the end of the paper. The essential dimensions of the exceptional algebraic groups remain largely open (see [Re10] or [Me13] for recent surveys). For example, even if the base field is the complex numbers, we only know that 9 ≤ ed(E_{8}) ≤ 231.

Key words and phrases. Essential dimension, Albert algebras, generically free representations.

Corollary 0.4. If the base field is of characteristic not 2 or 3, then ed(E_{6}^{s.c}) ≤ 8.

Proof. We know ed(E_{6}^{s.c.}) ≤ ed(F_{4}) + 1 by [Re00, 11.7] or [Ga09, 9.12]. □

The main difficulty in the proof is showing the generic freeness of a certain representation (Theorem 3.1). Showing that the stabilizer of a point in general position of this representation is finite is not hard, since this can be done on the level of Lie algebras. But to show it is trivial requires careful consideration of the group action.

To finish this introduction, I will mention how the values of the related essential p-dimensions of Albert algebras are already known for all primes p (see [Me13] for a definition and recent results). We know ed(F_{4}; 2)=5, because any Albert algebra becomes reduced after some degree prime-to-2 extension, and an arbitrary reduced Albert algebra only requires 5 parameters to describe it, by Jacobson’s coordinatization [Mac08]. Also, ed(F_{4}; 3) = 3, because any Albert algebra becomes a first Tits construction J(A, λ) after some degree prime-to-3 extension, and they require three parameters to describe them: two parameters for A (the degree 3 central simple algebra, which must be cyclic), and one parameter for λ. Finally, ed(F_{4}; p) = 0 for all primes p ≥ 5, because any Albert algebra splits after a degree 6 extension. As usual, the essential p-dimensions give lower bounds for the essential dimension, but they are otherwise, apparently, not helpful in finding ed(F_{4}).

1. Preliminaries

Unless otherwise specified, F will denote an arbitrary field of characteristic ≠ 2, 3, and F̄, F_{sep} its algebraic and separable closures respectively. An algebraic group G will be a smooth affine group scheme of finite type over a field; in particular all algebraic groups are assumed to be linear [KMRT, 21]. By a subgroup H ⊂ G we will mean a closed subgroup scheme (not necessarily smooth). X(F) will denote the set X(Spec F) of F -rational points of X. For an algebraic group G, the symbols Norm_{G}(H) or Norm_{G}(Y) will denote the scheme-theoretic normalizer of G which preserves a subgroup H or subscheme Y respectively; the normalizer subgroup need not be smooth [Ja03]. In particular, if G acts on X and x ∈ X(F), then the symbol G_{x} := Norm_{G}(x) will denote the subgroup in G of elements fixing x. The notation ⟨{x_{i}}⟩ ⊂ V will mean the vector subspace generated by some collection of vectors {x_{i}} ⊂ V.

We will say that a G-variety X is generically free if there is an open dense subscheme U ⊂ X such that the (scheme-theoretic) stabilizer of each point in U is trivial. By [BF03, 4] it is enough to check G_{x} = {1} for all x ∈ U(F̄). For an algebraic group G, we will define the essential dimension of G to be the minimum value of dim(X) − dim(G), where X is a generically free G-variety and there exists a G-equivariant dominant map V −−→ X (also called a compression), such that V is a generically free linear representation of G. By [BF03, Cor. 6.16], this is equal to Merkurjev’s definition of essential dimension in terms of the étale cohomology functor F′ ↦ H_{et}^{1}(F′, G(F′_{sep})) (since G is assumed to be smooth, this is the same as the functors for Galois or fppf cohomology).

To understand the representation in Section 3, we will find it convenient to view the vectors simultaneously in terms of octonions and in terms of weight spaces.

1.1. Octonions. The split octonion algebra over a field F will be denoted by 𝕆, and is defined as follows. Its elements are pairs of 2 by 2 matrices over F, which also defines its vector space structure. We will define a conjugation operation

\overline{( \begin{matrix}a_{11} & a_{12} \\ a_{21} & a_{22} \end{matrix} )} = ( \begin{matrix}a_{22} & −a_{12} \\ −a_{21} & a_{11} \end{matrix} ),

and then the multiplication on the 8-dimensional vector space is defined by

(a, b)(c, d) = (ac + d̄b, da + bc̄), where a, b, c, d ∈ M_{2}(F).

The norm on the octonions is defined as n((a, b)) := det(a) − det(b). This presentation follows [SV00, 1.8].

We will also denote standard elements

1 := (( \begin{matrix}1 & 0 \\ 0 & 1 \end{matrix} ), 0), i := (( \begin{matrix}1 & 0 \\ 0 & −1 \end{matrix} ), 0), j := (( \begin{matrix}0 & 1 \\ 1 & 0 \end{matrix} ), 0), k := (0, ( \begin{matrix}1 & 0 \\ 0 & 1 \end{matrix} )).

Then the set {1, i, j, ij, k, ik, jk, (ij)k} ⊂ 𝕆 is an F -vector space basis. One checks that i, j, and k pairwise anti-commute and square to the identity 1.

1.2. Albert algebras. Consider the vector space J_{s} of 3 by 3 matrices with entries in the split octonion algebra, which are hermitian (elementwise conjugation equals matrix transpose). As a vector space, we have

J_{s} := \{x ∈ M_{3}(𝕆)|x̄^{t} = x\} = \{(\begin{matrix}λ_{0} & c_{2} & c̄_{1} \\ c̄_{2} & λ_{1} & c_{0} \\ c_{1} & c̄_{0} & λ_{2} \end{matrix}) |λ_{n} ∈ F, c_{n} ∈ 𝕆\}.

We will also sometimes consider elements (c_{0}, c_{1}, c_{2}) ∈ J_{0} ⊕ J_{1} ⊕ J_{2}. Here J_{n} = 𝕆 as vector spaces for each n = 0, 1, 2; we refer to these as the Peirce subspaces.

If we equip J_{s} with the multiplication x ∘ y := \frac{1}{2}(xy + yx), where concatenation is the usual matrix product in M_{3}(𝕆), then we call it the split Albert algebra over F. An algebra J over a field F is an Albert algebra if for some field extension J ⊗_{F} F′ is isomorphic to the split Albert algebra over F′.

Define the 9-dimensional and 3-dimensional subalgebras as follows

E_{s} := \{(\begin{matrix}λ_{0} & c_{2} & c̄_{1} \\ c̄_{2} & λ_{1} & c_{0} \\ c_{1} & c̄_{0} & λ_{2} \end{matrix}) |λ_{n} ∈ F, c_{n} ∈ ⟨1, i⟩\}, L_{s} := \{(\begin{matrix}λ_{0} & 0 & 0 \\ 0 & λ_{1} & 0 \\ 0 & 0 & λ_{2} \end{matrix}) |λ_{n} ∈ F \}.

Then J_{s} ⊃ E_{s} ⊃ L_{s}. We will say an automorphism preserves a subalgebra, if the image of the subalgebra is itself.

The algebra automorphism group Aut(J_{s}) is the split group of type F_{4}. Over an algebraically closed field there is only one Albert algebra up to isomorphism.

Remark 1.3. It is a non-trivial fact from the structure theory of Jordan algebras, that over F̄, the Albert algebra is the only simple Jordan algebra which does not “come from” an associative algebra. While much is known about the classification of Albert algebras over non-algebraically closed fields, several open questions remain; for example, Serre asked whether Albert algebras are classified by their cohomological invariants with coefficients in ℤ/2 and ℤ/3, and this is still open [Se93, 9.4].

1.4. Spin_{8} ⋊S_{3} ⊂ F_{4}. It is well known that the group of automorphisms preserving the diagonal matrices in J_{s} is Norm_{F_{4}}(L_{s}) ≅ Spin_{8} ⋊S_{3} [KMRT, 39.13]. Here transpositions in S_{3} act on J_{s} by matrix transpose composed with swapping diagonal idempotents, and swapping their corresponding Peirce subspaces [Ga05, 3.2]. We will find it convenient to express the subgroup Spin_{8} (as a scheme) as a subgroup of SO_{8}^{3}, defined by its action on J_{0} ⊕ J_{1} ⊕ J_{2} = 𝕆^{3}. In particular, by [KMRT, 35.8]:

Spin_{8}(F) ≅ \{(g_{0}, g_{1}, g_{2}) ∈ SO_{8}^{3} |g_{0}(x · y) = \overline{g_{2}(\overline{x})} · \overline{g_{1}(\overline{y})}, ∀x, y ∈ 𝕆\}.

Here the multiplication is the usual one in 𝕆. Given such a triple (g_{0}, g_{1}, g_{2}), we can extend to a vector space automorphism of J_{s} by taking the identity on L_{s}. One checks this defines an algebra automorphism of J_{s}. Notice that a different presentation of Spin_{8} was used in [Bö11], so in what follows our statements will differ slightly from his.

1.5. SL_{2} ⊂ SL_{3} ⊂ G_{2} ⊂ Spin_{8}. The triples in Spin_{8} such that g_{0}(1) = g_{1}(1) = g_{2}(1) = 1 are exactly the triples such that g_{0} = g_{1} = g_{2} = g on 𝕆, where g ∈ Aut(𝕆) [KMRT, 35.16]. This is the split group of type G_{2}, and we will abuse notation slightly by referring to this subgroup in Spin_{8} as G_{2}. Similarly we will denote the subgroups SL_{3} := Norm_{G_{2}}(i) and SL_{2} := Norm_{SL_{3}}(j) which are smooth subgroups of the given types [SV00, 2.2].

1.6. T_{2} ⊂ Spin_{8}. For λ ∈ F^{∗}, let R_{λ} (resp. L_{λ}) be the linear isomorphism of 𝕆 given by right (resp. left) multiplication by (( \begin{matrix}λ & 0 \\ 0 & λ^{−1} \end{matrix} ), 0). Then one checks (using the Moufang identities) that the following elements of SO_{8}^{3} live inside Spin_{8}.

r_{λ} := (L_{λ}, R_{λ}, L_{λ^{−1}} ∘ R_{λ^{−1}}), s_{μ} := (R_{μ}, L_{μ^{−1}} ∘ R_{μ^{−1}}, L_{μ}).

For example, if we let (g_{0}, g_{1}, g_{2}) = s_{λ}, and x, y ∈ 𝕆, a := ( \begin{matrix}λ & 0 \\ 0 & λ^{−1} \end{matrix} ), then

\overline{g_{2}(\overline{x})} · \overline{g_{1}(\overline{y})} = (x · \overline{a}) · ((a · y) · a) = (((x · \overline{a}) · a) · y) · a = (x · y) · a = g_{0}(x · y).

Furthermore, these elements generate two one-dimensional tori which intersect in the identity and commute. The two-dimensional split torus they generate will be denoted by T_{2} ⊂ Spin_{8}.

1.7. Decomposition of J_{s}. Consider the (split) maximal torus T_{4} ⊂ F_{4} generated by T_{2} above and the 2-dimensional torus in SL_{3} described in [SV00, 2.3]. Then J_{s} has 24 non-zero T_{4}-weights, each of which has a 1-dimensional weight space and corresponds to a short root of F_{4}. So as an F_{4} representation, J_{s} = F ⊕ V_{26}, the sum of a trivial representation, and the 26-dimensional irreducible representation.

As in 1.1, we can write any octonion (and therefore any element of J_{0}, J_{1}, or J_{2}) as a pair of 2 by 2 matrices. It turns out that the non-zero T_{4}-weight spaces are exactly the matrix entries from the pairs of 2 by 2 matrices in J_{0}, J_{1}, and J_{2}. Using the notation for the F_{4}-roots from [Bour], we can label the weight spaces as follows:

J_{0} : ((\begin{matrix}ϵ_{1} & ϵ_{2} \\ −ϵ_{2} & −ϵ_{1} \end{matrix}) , (\begin{matrix}ϵ_{3} & −ϵ_{4} \\ ϵ_{4} & −ϵ_{3} \end{matrix})),

J_{1} : ((\begin{matrix}(− + ++) & (− + −−) \\ (+ − ++) & (+ − −−) \end{matrix}) , (\begin{matrix}(− − +−) & (+ + +−) \\ (− − −+) & (+ + −+) \end{matrix})),

J_{2} : ((\begin{matrix}(− − −−) & (+ + −−) \\ (− − ++) & (+ + ++) \end{matrix}) , (\begin{matrix}(+ − +−) & (− + +−) \\ (+ − −+) & (− + −+) \end{matrix}))

Here the notation (− + ++) means, for example, the weight \frac{1}{2}(−ϵ_{1} + ϵ_{2} + ϵ_{3} + ϵ_{4}).

The roots of Spin_{8} are given by the 24 long roots of F_{4}, which are ±ϵ_{m} ± ϵ_{n}, m < n. With this notation, the above subgroups SL_{2} and SL_{3} have roots ±(ϵ_{3} − ϵ_{4}) and ±(ϵ_{m} − ϵ_{n}), 2 ≤ m < n ≤ 4 respectively. Furthermore, as SL_{3}-representations, each of J_{0}, J_{1}, and J_{2} are a direct sum of a 2-dimensional trivial representation, the standard 3-dimensional representation, and its 3-dimensional dual.

From 1.2, L_{s} is the 3-dimensional zero weight space, and E_{s} is that together with the six weight spaces from the diagonals of the three matrices on the left-hand sides, above.

2. Reduction to a subgroup

We will denote by F_{4} the split group of that type over the base field. To give the new upper bound on ed(F_{4}), we will reduce to a 10-dimensional subgroup, G ⊂ F_{4}, which has more small representations than F_{4} does.

2.1. The subgroup G ⊂ F_{4}. Using the notation from 1.2, we will define G to be the subgroup of automorphisms of J_{s} which preserve both E_{s} and L_{s},

G := Aut(J_{s}, E_{s}, L_{s}) := Norm_{Norm_{F_{4}}(L_{s})}(E_{s}).

The notation (J, E, L) will refer to a triple of algebras defined over F, such that and J ⊃ E ⊃ L. An isomorphism between two such triples is an isomorphism between the largest algebras, which also restricts to respective isomorphisms on the two subalgebras.

Definition 2.2. An algebra A is a form of an algebra B if there is a field extension F′ such that A ⊗ F′ ≅ B ⊗ F′. A triple (J, E, L) will be called a form of (J_{s}, E_{s}, L_{s}), if after some base field extension, it becomes isomorphic to (J_{s}, E_{s}, L_{s}).

Lemma 2.3. Assume (J, E, L) is a triple such that J,E, and L are forms of J_{s}, E_{s}, and L_{s}, respectively. Then (J, E, L) is a form of (J_{s}, E_{s}, L_{s}).

Proof. It is enough to assume the field is separably closed, and then prove that (J, E, L) becomes isomorphic to (J_{s}, E_{s}, L_{s}). Choose an isomorphism ϕ_{0} sending J_{s} to J. Now there is an automorphism ϕ_{1} of J which sends ϕ_{0}(E_{s}) to E, by [KMRT, 37.20]. Finally, there is an automorphism ϕ_{2} of J preserving E which sends ϕ_{1}(ϕ_{0}(L_{s})) to L, by coordinatization together with [KMRT, 37.20]. So ϕ_{2}∘ϕ_{1}∘ϕ_{0} sends the triple (J_{s}, E_{s}, L_{s}) to (J, E, L). □

Lemma 2.4. The Galois cohomology set H^{1}(F, G) is in bijective correspondence with the set of isomorphism classes of forms of (J_{s}, E_{s}, L_{s}).

Proof. Using the arguments of [KMRT, 29.C], this follows from Lemma 2.3. □

The following proposition was proved in [Bö11] over the complex numbers.

Proposition 2.5. G ≅ (((SL_{3} ×𝔾_{m}^{2})/μ_{3}) ⋊ ℤ/2) ⋊ S_{3} as group schemes.

Proof. We have G ⊂ Norm_{F_{4}}(L_{s}) ≅ Spin_{8} ⋊S_{3}, and let G′ be the kernel of the G action on L_{s}, so G′ ⊂ Spin_{8}. Since S_{3} preserves E_{s}, we have G ≅ G′ ⋊ S_{3}.

The subgroup Aut(𝕆, ⟨1, i⟩) := Norm_{G_{2}}(⟨i⟩) is isomorphic to SL_{3} ⋊ℤ/2 ⊂ Spin_{8} [Ja58, Thm. 4]. Notice that both SL_{3} ⋊ℤ/2 and T_{2} are contained in G′, so consider the homomorphism of group schemes f : (SL_{3} ⋊ℤ/2) × T_{2} → G′ induced by the group product. f is surjective, because for any g ∈ G′, we can find λ, μ ∈ F^{∗} such that g · r_{λ}s_{μ} fixes (1, 1, 1) ∈ J_{0} ⊕ J_{1} ⊕ J_{2}. The subgroups SL_{3} ⋊ℤ/2 and T_{2} commute, and the F -points of ker(f) coincide with the F -points of μ_{3}. By [KMRT, 22.13], the result will now follow if df is a surjection of Lie algebras.

By [KMRT, 21.5(7)], we have Lie(G) = {h ∈ Lie(Spin_{8})|h(E_{s}) ⊂ E_{s}}. Assume h ∈ Lie(G), and write h = h_{0} + ∑_{α} h_{α}, where α runs over the Spin_{8}-roots, and h_{0} ∈ Lie(T_{4}).

We claim that if β is a T_{4}-weight of E_{s}, α a Spin_{8}-root, and α + β is a weight of J_{s}, then h_{α} = 0. In this situation, if h_{α} ≠ 0, then h_{α}V_{β} = V_{α+β}, since they are both 1-dimensional weight spaces; they could also be viewed as root spaces of Lie(F_{4}). But this contradicts the fact that h preserves E_{s}, since α + β cannot be a weight of E_{s}, proving the claim.

By considering β = ±ϵ_{1}, this shows that h_{α} ≠ 0 implies α = ±ϵ_{m} ± ϵ_{n}, where 2 ≤ m < n ≤ 4. By arguing similarly using β = ±(− + ++), we see in fact α must be a root of SL_{3}. So Lie(G) ⊂ Lie(SL_{3}) ⊕ Lie(T_{2}), and the reverse containment is obvious. This completes the proof. □

Theorem 2.6. For any field F, the inclusion G ⊂ F_{4} induces a surjection H^{1}(F, G) → H^{1}(F, F_{4}).

Proof. The map to H^{1}(F, F_{4}) is given by the forgetful functor (J, E, L) ↦ J. To see surjectivity, take a pair of elements x, y ∈ J, and let L be the subalgebra generated by x, and let E be the subalgebra generated by x and y. For generic x and y this creates a form of (J_{s}, E_{s}, L_{s}) as required [KMRT, 39.20]. Notice when F is finite, then H^{1}(F, F_{4}) is trivial. □

3. Generic freeness

For each n = 0, 1, 2, let V_{n} ⊂ J_{n} be the 6-dimensional subspace as follows

V_{n} := ((\begin{matrix}0 & ∗ \\ ∗ & 0 \end{matrix}) , (\begin{matrix}∗ & ∗ \\ ∗ & ∗ \end{matrix})) , V := V_{0} ⊕ V_{1} ⊕ V_{2}.

Then the 18-dimensional subspace V ⊂ J_{s} is preserved by G. Each V_{n} is an irreducible SL_{3} ⋊ℤ/2-representation. We could also write V_{n} = ⟨j, ij, k, ik, jk, (ij)k⟩ ⊂ 𝕆.

Theorem 3.1. G acts on V and ℙ(V) generically freely.

Proof. For generic freeness, it is enough to assume the base field is algebraically closed. Let p = (x, y, z) ∈ V_{0} ⊕ V_{1} ⊕ V_{2}. To prove the theorem, we will first reduce to the case where (x, y, z) is of a certain form, and then we will show the (scheme-theoretic) Gstabilizer of such a point is trivial. First we show that for a generic point G_{p}(F̄) = {1}, and then we show that Lie(G_{p}) = {0}, which together imply G_{p} = {1} ([KMRT, 22.5] or [BF03, 4]).

Notice the size of the G-stabilizer is unchanged along G-orbits. Generically we may assume the norm of z (as an element of 𝕆) is non-zero, so we can move it to a scalar multiple of j ∈ V_{2} via an element of SL_{3} ⊂ G. This is because SL_{3} acts transitively on norm one octonions orthogonal to i and 1 [SV00, 2.2]. Next, we can find an r_{λ} ∈ T_{2} ⊂ G (which fixes V_{2}) that sends y ∈ V_{1} to an element with zero ij coefficient. Finally, SL_{2} ⊂ SL_{3} acts transitively on norm one octonions in ⟨k, ik, jk, (ij)k⟩ [SV00, 2.2], so we can assume y is in the span of j and k. In other words, we will assume p = (x, y, z) ∈ V_{0} ⊕ ⟨j, k⟩ ⊕ ⟨j⟩.

Step 1: Prove G_{p}(F̄) = {1}. Assume g ∈ G(F̄) fixes such an element. We need to show that for (x, y, z) generic of this form, this implies g is trivial. Firstly, notice that generically the norms of x, y, and z are all distinct, and any non-trivial contribution from S_{3} ⊂ G would permute the norms. So we may assume g = (g_{0}, g_{1}, g_{2}) ∈ Spin_{8} ∩G ⊂ SO(𝕆)^{3}.

The identity 1.4 together with g_{2}(j) = j and g_{0}(⟨1, i⟩) = ⟨1, i⟩ implies that g_{1} preserves ⟨j, ij⟩ ⊂ V_{1}. Due to our assumption on y ∈ V_{1}, which is fixed by g_{1}, we see that g_{1}(j) = j and g_{1}(k) = k. So by the identity 1.4 we get g_{0}(1) = 1.

Now let us write g = f · r_{λ} · s_{μ} · ι, where (f, ι) ∈ SL_{3} ⋊ℤ/2, and s and r are defined in 1.6. Notice that g_{0}(1) = 1 implies that (r_{λ}s_{μ})_{0}(1) = 1, which implies μ = λ^{−1}.

Switching to the matrix perspective of the octonions (see 1.1), the identity j = g_{1}(j)

becomes (( \begin{matrix}0 & 1 \\ 1 & 0 \end{matrix} ), 0) = (f · r_{λ} · s_{λ^{−1}})_{1}(( \begin{matrix}0 & 1 \\ 1 & 0 \end{matrix} ), 0) = f · (( \begin{matrix}0 & λ^{−1} \\ λ & 0 \end{matrix} ), 0). Since f is an auto-

morphism fixing the diagonal, this implies (( \begin{matrix}0 & 1 \\ −1 & 0 \end{matrix} ), 0) = f((( \begin{matrix}1 & 0 \\ 0 & −1 \end{matrix} ), 0) · (( \begin{matrix}0 & λ^{−1} \\ λ & 0 \end{matrix} ), 0)) = f(( \begin{matrix}0 & λ^{−1} \\ −λ & 0 \end{matrix} ), 0). Now using linearity of f, and a similar argument with g_{1}(k) = k, we get

f((\begin{matrix}0 & a \\ b & 0 \end{matrix}) , 0) = ((\begin{matrix}0 & λa \\ λ^{−1}b & 0 \end{matrix}) , 0), f(0, (\begin{matrix}a & 0 \\ 0 & b \end{matrix})) = (0, (\begin{matrix}λa & 0 \\ 0 & λ^{−1}b \end{matrix})).

In V_{0} we have that (r_{λ}s_{λ^{−1}})_{0}(( \begin{matrix}0 & x_{1} \\ y_{1} & 0 \end{matrix} ), ( \begin{matrix}x_{2} & y_{3} \\ x_{3} & y_{2} \end{matrix} )) = (( \begin{matrix}0 & λ^{2}x_{1} \\ λ^{−2}y_{1} & 0 \end{matrix} ), ( \begin{matrix}λ^{2}x_{2} & λ^{−2}y_{3} \\ λ^{2}x_{3} & λ^{−2}y_{2} \end{matrix} )), for any

x_{i}, y_{i} ∈ F. There are now two cases to consider.

Case: ι = 1. In other words, g_{0}(i) = i. In this case, g = f · r_{λ}s_{λ^{−1}} acts in V_{0} as (( \begin{matrix}0 & a \\ b & 0 \end{matrix} ), 0) ↦ (( \begin{matrix}0 & λ^{3}a \\ λ^{−3}b & 0 \end{matrix} ),). Since a generic x ∈ V_{0} is fixed by g, this implies λ^{3} = 1, which means s_{λ^{−1}}r_{λ} ∈ SL_{3}, and therefore g ∈ SL_{3}. But g_{1}(k) = k and g_{2}(j) = j imply g also fixes j and k, which means g is trivial.

Case: ι ≠ 1. In other words, g_{0}(i) = −i. Let x = (( \begin{matrix}0 & x_{1} \\ y_{1} & 0 \end{matrix} ), ( \begin{matrix}x_{2} & y_{3} \\ x_{3} & y_{2} \end{matrix} )) ∈ V_{0}. By the identity in 1.4 combined with g_{1}(j) = j and g_{1}(k) = k we have that g_{0}(⟨j, ij⟩) = ⟨j, ij⟩ and g_{0}(⟨k, ki⟩) = ⟨k, ki⟩. So we have

g_{0}((\begin{matrix}0 & x_{1} \\ y_{1} & 0 \end{matrix}) , 0) = ((\begin{matrix}0 & λ^{3}y_{1} \\ λ^{3}x_{1} & 0 \end{matrix}) , 0),

g_{0}(0, (\begin{matrix}x_{2} & 0 \\ 0 & y_{2} \end{matrix})) = (0, (\begin{matrix}λ^{3}y_{2} & 0 \\ 0 & λ^{3}x_{2} \end{matrix})).

Since x is generic and we are assuming it is fixed by g_{0}, this implies λ^{3} = x_{1}/y_{1} and λ^{3} = x_{2}/y_{2}. But for generic x these two ratios are not equal, which gives a contradiction, so no such g exists. This proves that generically the stabilizer of G on V has trivial F̄-points.

Step 2: Prove G_{[p]}(F̄) = {1}. Consider a generic element [p] = [(x, y, z)] ∈ ℙ(V)(F̄); we must show no element of G(F̄) sends (x, y, z) to a scalar multiple of itself. Consider the S_{3} action on the triple of norms up to scalar [(n(x), n(y), n(z))], viewed as an element of ℙ^{2}. Since S_{3} acts faithfully on ℙ^{2}, it acts generically freely there. If g ∈ G(F̄) fixes [(x, y, z)], this implies the S_{3} contribution is trivial. So we can assume (x, y, z) is sent to its negative.

As for Step 1, we may assume (x, y, z) ∈ V_{0} ⊕⟨j, k⟩⊕⟨j⟩. So if g = (g_{0}, g_{1}, g_{2}) sends it to its negative then g_{2}(j) = −j, g_{1}(j) = −j, and g_{1}(k) = −k. The identity 1.4 gives us that g_{0}(jk) = jk. Since g preserves orthogonality in 𝕆, this contradicts our assumption that a generic x ∈ V_{0} is sent to its negative, so generically the stabilizer of G on ℙ(V) has trivial F̄-points.

Step 3: Prove Lie(G_{p}) = Lie(G_{[p]}) = {0}. Notice that if [p] ∈ ℙ(V)(F̄), then we have an inclusion of subgroups G_{[p]} ⊂ Norm_{G}(⟨p⟩). In particular, it is enough to show that Lie(Norm_{G}(⟨p⟩)) = {h ∈ Lie(G)|h · p ∈ ⟨p⟩} = {0} for a generic p ∈ V.

So take p ∈ V_{0} ⊕ ⟨j, k⟩ ⊕ ⟨j⟩ as above, and take h ∈ Lie(G) such that h · p ∈ ⟨p⟩. Use the root space decomposition to write h = h_{0} + ∑_{α} h_{α}.

Here (0, 0, j) is supported on two weight spaces, ±(+ + −−), see 1.7. One of these weight spaces lies in the standard SL_{3}-representation inside V_{2}, while the other lies in its dual inside V_{2}. Lie(G) preserves each V_{n}, so h·(0, 0, j) ∈ ⟨(0, 0, j)⟩. If β = ±(++−−), and α is an SL_{3}-root such that β + α is a weight of V, then h_{α} = 0, because otherwise h_{α}V_{β} = V_{α+β}, which contradicts h·(0, 0, j) ∈ ⟨(0, 0, j)⟩. So if h_{α} ≠ 0 then α = ±(ϵ_{3}−ϵ_{4}). By arguing similarly with β = ±(+ − ++), or ±(+ + −+), we see α could not be ±(ϵ_{3} − ϵ_{4}) either. So h ∈ Lie(T_{4}). Therefore h preserves the weight spaces.

Since we have assumed there exists a λ ∈ F such that h · p = λp, considering the weights of V_{0} (from 1.7), we see that ϵ_{2}(h) = λ and −ϵ_{2}(h) = λ. Since char F ≠ 2, we have λ = 0. Looking at other weight spaces we see ϵ_{3}(h) = ϵ_{4}(h) = 0 as well as ϵ_{1}(h) = 0. This implies h = 0, and completes the proof that G_{p} = G_{[p]} = {1}. □

Proof of Theorem 0.3. By Theorem 2.6 we have that ed(F_{4}) ≤ ed(G) for the smooth 10-dimensional group G defined above. By Theorem 3.1 we have a compression V −−→ ℙ(V) from a generically free G-representation to its 17-dimensional generically free projectivization. Therefore ed(F_{4}) ≤ ed(G) ≤ dim ℙ(V) − dim G = 17 − 10 = 7, as required (see also [BF03, Prop. 6.17]). □

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