The Proposition: what \(\Esix/\Ffour\) is

Notation: \(\Jalg\), \(\det\), \(\one\), \(\Jo\) from the Albert algebra page; \(\Esix\), \(\Ffour\) from its group definitions; all symbols in the Notation tab.

Here \(\Lop{T}:\Jalg\to\Jalg\), \(\Lop{T}(X)=T\jo X\). Sphericity is the subject of the next page.

Proof of (a)The \(\SL_3/\SO_3\) transitivity and stabilizer lemmas, for \(\Jalg\).

1. Transitivity [Krutelevich, Prop. 12(i)]: for the split octonions \(C\) over a field of characteristic \(\ne2,3\), the norm-preserving group of \(\mathfrak h_3(C)\) acts transitively on elements of a given nonzero norm. Over \(\C\) every octonion algebra is split, so \(\Oc_\C\) is the split one and the norm is \(\det\). In the classical model this was the elimination algorithm.
2. Stabilizer [Yokota, §2.13, p. 65]: \(\{g\in\Esix:g\one=\one\}=\Ffour\). Classically: \(gg^T=I\).
3. Isomorphism. The orbit map \(\Esix\to\Xe\), \(g\mapsto g\one\), is surjective by 1 and has fibres the cosets of \(\Ffour\) by 2. \(\Xe\) is smooth: by Euler's identity, \(d\det_X(X)=3\det X=3\ne0\). A bijective orbit map onto a smooth (hence normal) variety in characteristic 0 is an isomorphism.
4. Dimension. \(\dim\Xe=27-1=26=78-52\); the tangent space at \(\one\) is \(\Jo\) (computed here).

Proof of (b)Where is the involution?

1. Decomposition [CS; Yokota, Thm 3.2.1]: \(\esix=\Der(\Jalg)\oplus\{\Lop{T}:\tr T=0\}\), \(\dim=52+26=78\); \(\Der(\Jalg)=\ffour\) [Baez, §4.2].
2. Brackets. For \(\delta,\delta'\in\Der\) and \(S,T\in\Jo\): \[[\delta,\delta']\in\Der,\qquad [\delta,\Lop{T}]=\Lop{\delta T},\qquad [\Lop{S},\Lop{T}]\in\Der .\] The first is clear; the second is the Leibniz rule \(\delta(T\jo X)=\delta T\jo X+T\jo\delta X\); the third is the standard fact that \([\Lop{S},\Lop{T}]\) is a derivation of a Jordan algebra (an inner derivation), checked numerically in .
3. Involution. By 2, \(\theta(\delta+\Lop{T})=\delta-\Lop{T}\) respects brackets, \(\theta^2=\Id\), and its \(+1\) eigenspace is \(\kf=\Der(\Jalg)=\ffour\), its \(-1\) eigenspace \(\pf=\{\Lop{T}:T\in\Jo\}\). Classically: \(\slf_3=\so_3\oplus\{\text{symmetric}\}\).
4. Group. \(\Esix\) is simply connected [Krutelevich, Remark 10(iv)], so \(\theta\) integrates to an automorphism of \(\Esix\). \(\Ffour\) is connected with Lie algebra \(\ffour=\kf\), hence \(\Ffour=(\Esix^\theta)^\circ\).

Proof of (c)Why is every multiplicity 8?

1. \(\Jo=\{\text{traceless diagonal}\}\oplus\bigoplus_{i<j}\{\text{octonion at }(i,j)\}\), of dimensions \(2+8+8+8\). Diagonal elements operator-commute, so \(\af\) is abelian and \(\iot H\) acts semisimply (by , \(\mathrm{ad}(\iot H)^2\) is diagonal on \(\pf\)).
2. For \(H\) with \(h_1,h_2,h_3\) distinct, the eigenvalue of \(\mathrm{ad}(\iot H)^2\) on \(\pf\) is 0 only on \(\af\); so \(\af\) is maximal abelian, a Cartan subspace, and \(\rk=2\).
3. As on the roots page, \(\gf_\alpha\oplus\gf_{-\alpha}=\kf_\alpha\oplus\pf_\alpha\) with \(\dim\gf_\alpha=\dim\pf_\alpha\). By the Lemma, \(\pf_{\alpha_{ij}}=\iot\{\text{octonion at }(i,j)\}\), so \(\mult_{\alpha_{ij}}=\dim_\C\Oc_\C=8\), and there are no other roots.
4. Consequently \(\dim\gf_0=78-6\cdot8=30\) and \(\mf=Z_\kf(\af)\) has \(\dim\mf=30-2=28\).

[Klein, §4.1] states the same root data for the compact form: type \(A_2\), all multiplicities 8.

This is type EIV in Cartan's list; the diagram is tabulated in [Helgason, Ch. X, Table VI] (table and page not opened in the book). Folding the diagram of \(E_6\) by its symmetry gives \(F_4\), the fixed group; that is a different operation from reading off the restricted roots, which come from the white nodes.

The script assets/figures/check_albert.py builds \(\Lop{X}\) as \(27\times27\) matrices on the real form \(\mathfrak h_3(\Oc)\) and computes:

A check on one random instance, not a proof; the proofs are above.