Why \(\Esix/\Ffour\) is spherical
A Borel subgroup of \(\Esix\) has dimension 42 and \(\Esix/\Ffour\) has dimension 26. Is there room for an open orbit? The restricted root decomposition of the Proposition answers it in one line.
DefinitionWhat does “spherical” ask for?
(spherical variety) [Pezzini, Def. 2.1.1] Definition
Let \(G\) be connected reductive over \(\C\). A Borel subgroup \(\Bor\subset G\) is a maximal connected solvable subgroup. A normal \(G\)-variety \(Y\) is spherical if \(\Bor\) has an open orbit in \(Y\).
Homogeneous spaces \(G/K\) are smooth, hence normal; the condition is the open orbit. Since \(\dim\Bor\cdot y=\dim\Bor-\dim\Bor_y\), an open orbit needs \(\dim\Bor\ge\dim Y\).
The classical model by handDoes the upper-triangular group have an open orbit on \(\Xs{\R}\)?
\(\Bor\subset\SL_3(\C)\) is the upper-triangular group, \(\dim\Bor=5\). What is \(\dim(\Bor\cap\SO_3(\C))\)?
(\(\SL_3/\SO_3\) is spherical) [cf. Pezzini, Ex. 2.2.5, for \(\SL_3/\SO_3\cdot Z(\SL_3)\)] Proved here
\(\Bor\cdot I=\{bb^T:b\in\Bor\}\) has dimension 5, so it is open in the irreducible 5-dimensional \(\Xs{\R}\).
Concretely, \(\Bor\cdot I\) is the set of \(S\in\Xs{\R}\) whose lower-right \(1\times1\) and \(2\times2\) minors are nonzero (a “reverse Cholesky” factorization \(S=bb^T\)) Heuristic.
The key decompositionWhich directions does \(\kf\) miss?
For a symmetric variety with Cartan subspace \(\af\) and restricted roots \(\Sigma\) (roots of \(\SL_3/\SO_3\)), choose a chamber, i.e. positive roots \(\Sigma^+\), and put \(\nf=\bigoplus_{\alpha\in\Sigma^+}\gf_\alpha\).
(\(\gf=\kf+\af+\nf\)) Proved here
\[\gf=\kf+\af+\nf .\]Proof
- \(\gf=\gf_0\oplus\bigoplus_{\alpha\in\Sigma}\gf_\alpha\) (\(\mathrm{ad}\,\af\) is simultaneously diagonalizable), and \(\gf_0=Z_\gf(\af)=\mf\oplus\af\) with \(\mf=Z_\kf(\af)\subset\kf\), because \(Z_\pf(\af)=\af\) by maximality.
- \(\theta=-1\) on \(\af\), so \(\theta\gf_\alpha=\gf_{-\alpha}\). For \(Y\in\gf_{-\alpha}\), \(\alpha>0\): \(Y=(Y+\theta Y)-\theta Y\) with \(Y+\theta Y\in\kf\) and \(\theta Y\in\gf_\alpha\subset\nf\).
- Hence every summand lies in \(\kf+\af+\nf\).
Symmetric varieties are sphericalCan the hand computation be replaced by an argument that works for \(\Esix\)?
[Vust, §1.3, Thm 1 and the remark after it, p. 322] Proved (cited) and proved here
Every symmetric variety \(G/K\) is spherical: some Borel subgroup \(\Bor\) has \(\Bor\cdot eK\) open. In particular \(\Esix/\Ffour\) is spherical.
Proof
- \(\af\oplus\nf\) is a solvable Lie subalgebra (\([\af,\nf]\subset\nf\), \(\nf\) nilpotent), so it lies in the Lie algebra \(\bfr\) of some Borel subgroup \(\Bor\).
- By , \(\bfr+\kf\supseteq\af+\nf+\kf=\gf\).
- The differential at \(e\) of \(\Bor\to G/K\), \(b\mapsto bK\), is \(\bfr\to\gf/\kf\), surjective by 2. So \(\Bor\cdot eK\) has dimension \(\dim G/K\) and is open.
Vust phrases it at the group level: \(K^\circ P\) is open in \(G\) for a \(\theta\)-anisotropic parabolic \(P\), and such \(P\) contain Borel subgroups \(\Bor\) with \(K^\circ\Bor\) open.
How much of \(\Bor\) is left overFor \(\Esix/\Ffour\), \(42-26=16\). What are those 16 dimensions?
Choose \(\Bor\) as in the proof, containing \(\exp\af\) and \(\exp\nf\). Its Lie algebra is \(\bfr=\bfr_\mf\oplus\af\oplus\nf\) with \(\bfr_\mf\) a Borel subalgebra of \(\mf\), so
\[\dim\Bor-\dim G/K=(\dim\bfr_\mf+\dim\af+\textstyle\sum_{\alpha>0}\mult_\alpha)-(\dim\af+\sum_{\alpha>0}\mult_\alpha)=\dim\bfr_\mf .\]This is the dimension of the stabilizer in \(\Bor\) of a point in the open orbit, the generic stabilizer.
| \(G/K\) | \(\dim\Bor\) | \(\dim G/K\) | \(\mf\) | \(\dim\bfr_\mf\) |
|---|---|---|---|---|
| \(\SL_3/\SO_3\) | 5 | 5 | \(0\) | 0 |
| \((\SL_3\times\SL_3)/\SL_3\) | 10 | 8 | torus, \(\dim 2\) | 2 |
| \(\SL_6/\Sp_6\) | 20 | 14 | \(\slf_2^{\oplus3}\) | 6 |
| \(\Esix/\Ffour\) | 42 | 26 | \(\so_8\) | 16 |
\(\dim\Bor=\rk G+\#\{\text{positive roots of }G\}\): \(2+3\), \(4+6\), \(5+15\), \(6+36\). For \(\so_8\): \(4+12=16\).