The \(A_2\) row: \(\mult=1,2,4,8\)
\(\Esix/\Ffour\) has no infinite family, but it has a finite one: run the construction with each normed division algebra \(\K\in\{\R,\C,\Ha,\Oc\}\). The root picture stays \(A_2\); only the multiplicity \(\mult=\dim_\R\K\) changes.
Four algebras, one recipeWhy exactly four?
By Hurwitz's theorem, \(\R,\C,\Ha,\Oc\) are the only normed division algebras [Baez, §1.1, Thm 1]. For each, let \(\JK{\K}=\mathfrak h_3(\K)\otimes_\R\C\) (Hermitian \(3\times3\) matrices over \(\K\), complexified) with its determinant, and
\[\Xs{\K}=\{X\in\JK{\K}:\det X=1\},\qquad \dim_\C\JK{\K}=3+3d,\quad d=\dim_\R\K .\]\(\Xs{\R}\) is the \(\SL_3/\SO_3\) page and \(\Xs{\Oc}=\Xe\) is the Albert algebra's variety. The family is a list of four, so the widget below is a selector, not a slider.
Predict, then revealWhich numbers change with \(\K\), and which do not?
The generic stabilizer is defined on the spherical page; guess it from the pattern of the first rows. The \(\Ha\) row uses \(\mathfrak h_3(\Ha)\otimes\C\cong\bigwedge^2\C^6\) with the Pfaffian as determinant Context.
The rowWhat is invariant along the row?
| \(\K\) | \(G/K\) | \(\mult\) | \(\dim G/K\) | \(\dim\mf\) |
|---|---|---|---|---|
| \(\R\) | \(\SL_3/\SO_3\) | 1 | 5 | 0 |
| \(\C\) | \((\SL_3\times\SL_3)/\SL_3\) | 2 | 8 | 2 |
| \(\Ha\) | \(\SL_6/\Sp_6\) | 4 | 14 | 9 |
| \(\Oc\) | \(\Esix/\Ffour\) | 8 | 26 | 28 |
All four have rank 2 and restricted roots \(A_2\); the dimension is \(2+3\mult\). Biquard–Delcroix write the \(\C\) row as \(\mathrm{PGL}_3\times\mathrm{PGL}_3/\mathrm{PGL}_3\); the dimensions agree. The \(\dim\mf\) column is the dimension of \(\mf=Z_\kf(\af)\), read off from \(\dim\kf=\dim\mf+3\mult\).
Dictionary: \(\SL_3/\SO_3\to\Esix/\Ffour\)Which step of the classical argument fails to transfer?
| \(\SL_3/\SO_3\) | \(\Esix/\Ffour\) | argument |
|---|---|---|
| \(\mathrm{Sym}_3(\C)\), \(\dim 6\) | \(\Jalg\), \(\dim 27\) | kept: \(3+3d\) |
| \(\det\), multiplicative | cubic \(\det\), not multiplicative on products | repaired: \(\Tr(xy)=\Tr(yx)\) replaces associativity |
| \(\SL_3\) acting by \(gSg^T\) | \(\Esix=\) \(\det\)-preserving maps | changed: defined by the invariant, not by matrices |
| \(\SO_3=\{gg^T=I\}\) | \(\Ffour=\Aut(\Jalg)=\mathrm{Stab}(\one)\) | kept as a statement; proof cited |
| transitivity by elimination | transitivity on \(\det=1\) | cited; elimination with octonionic pivots needs more |
| \(\pf_{\alpha_{ij}}=\C Y_{ij}\) | \(\pf_{\alpha_{ij}}=\) octonion entry at \((i,j)\) | kept, with \(\mult=8\): proof |
| \(\mf=0\) | \(\mf=\so_8\) | new: the black nodes in the Satake diagram |
A coincidence to distrustIs the rank of \(\Esix/\Ffour\) equal to \(6-4\)?
For \(\Esix/\Ffour\), \(\rk\Esix-\rk\Ffour=6-4=2\), the rank of the variety. The selector shows the same for \(\C\) and \(\Ha\), but for \(\K=\R\): \(\rk\SL_3-\rk\SO_3=2-1=1\ne2\). The rank is \(\dim\af\) (definition), and \(\rk G-\rk K\) is a different invariant that happens to agree in three of the four rows.
It is the dimension of the part of a \(\theta\)-stable maximal torus of \(G\) on which \(\theta\) acts by \(-1\), when that torus is chosen to contain a maximal torus of \(K\). The rank of \(G/K\) uses instead a torus containing \(\af\), where the \(-1\) part is as large as possible. They agree exactly when \(\af\oplus\mathfrak t_\mf\) (with \(\mathfrak t_\mf\) a Cartan of \(\mf\)) and a Cartan of \(\kf\) can be fitted in one torus; \(\SL_3/\SO_3\) shows they need not. Heuristic