Let $G=SL_3(\mathbb{C})$ and $X=G/B$ be the associated full flag variety. Fix a non-degenerate symmetric quadratic form $Q$ on $\mathbb{C}^3$. This gives an order $2$ automorphism $F_Q$ of $X$, mapping a flag $(L,P)$ to $(P^{\perp},L^{\perp})$. Is there any explicit description of the quotient $X/F_Q$ ?
$\begingroup$
$\endgroup$
5
-
5$\begingroup$ Take a look at Danny Gillam's thesis, in particular arxiv.org/abs/0708.0842. He verifies the Crepant Resolution Conjecture for this orbifold. $\endgroup$– Dave AndersonCommented Apr 8, 2015 at 2:24
-
$\begingroup$ @Dave: This seems to be most relevant to the question. It may be useful to cite the published version, though it's hard for most people to access and may not differ from the preprint: ams.org/mathscinet-getitem?mr=3011793 $\endgroup$– Jim HumphreysCommented Apr 8, 2015 at 12:59
-
$\begingroup$ I had a look at the paper. It proves many properties of the quotient, but it does not give any explicit description. Moreover, it mainly focus on the case where the quadratic form is degenerate, so that the involution has fixed points and the quotient is singular; I am more interested in the other case. I wonder if there is something in term of classical representation theory, or in term of some Grassmannian $\endgroup$– GiulioCommented Apr 9, 2015 at 12:42
-
1$\begingroup$ Why is this fixed-point free? Can't $L= P^{\perp}$ if $L$ is isotropic? I think the fixed points form a flag variety of $SO(3)_Q$ - a $\mathbb P^1$. $\endgroup$– Will SawinCommented Apr 10, 2015 at 1:58
-
$\begingroup$ you are right! sorry $\endgroup$– GiulioCommented Apr 10, 2015 at 9:38
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
1
Let's coordinatize $G/B = Proj\ \mathbb C[m_{11}\ldots m_{33}]/\langle trace=0, rank \leq 1\rangle$, with $F_Q(M) = M^T$, the image of $(L,P) \mapsto L P^T$ (a column vector times row vector).
Then the $F_Q$-invariants are generated by $n_i := m_{ii}, q_{i < j} := m_{ij}m_{ji}, s_{i < j} := s_{ij}s_{ji}$. They satisfy $${n}_{1}+{n}_{2}+{n}_{3}, \qquad {p}_{12} {p}_{13}+{p}_{12} {p}_{23}+{p}_{13} {p}_{23},\qquad {n}_{2} {p}_{13}+{n}_{2} {p}_{23}+{n}_{3} {p}_{23}$$ $${n}_{3} {p}_{12}+{n}_{2} {p}_{23}+{n}_{3} {p}_{23},\qquad {n}_{3}^{2}+{p}_{13}+{p}_{23},\qquad {n}_{2} {n}_{3}-{p}_{23},\qquad {n}_{2}^{2}+{p}_{12}+{p}_{23}$$ $${n}_{3} {s}_{12}^{2}+{n}_{2} {s}_{13}^{2}-{s}_{12} {s}_{13} {s}_{23}-{n}_{2} {s}_{23}^{2}-{n}_{3} {s}_{23}^{2}+4 {n}_{2} {p}_{23}+4 {n}_{3} {p}_{23}$$ $${n}_{3} {s}_{12} {s}_{13} {s}_{23}+{p}_{13} {s}_{12}^{2}+{p}_{23} {s}_{12}^{2}-{p}_{23} {s}_{13}^{2}-{p}_{13} {s}_{23}^{2}+4 {p}_{13} {p}_{23}$$ $${n}_{2} {s}_{12} {s}_{13} {s}_{23}-{p}_{23} {s}_{12}^{2}+{p}_{12} {s}_{13}^{2}+{p}_{23} {s}_{13}^{2}-{p}_{12} {s}_{23}^{2}+4 {p}_{12} {p}_{23}$$ $${n}_{2} {p}_{23} {s}_{12}^{2}-{n}_{2} {p}_{12} {s}_{13}^{2}-{n}_{2} {p}_{23} {s}_{13}^{2}+{p}_{12} {s}_{12} {s}_{13} {s}_{23}+{p}_{23} {s}_{12} {s}_{13} {s}_{23}+{n}_{2} {p}_{12} {s}_{23}^{2}-4 {n}_{2} {p}_{12} {p}_{23}$$ $${n}_{2} {s}_{13}^{3} {s}_{23}-{s}_{12} {s}_{13}^{2} {s}_{23}^{2}-{n}_{2} {s}_{13} {s}_{23}^{3}-{n}_{3} {s}_{13} {s}_{23}^{3}-{p}_{13} {s}_{12}^{3}-{p}_{23} {s}_{12}^{3}+{p}_{23} {s}_{12} {s}_{13}^{2}+4 {n}_{2} {p}_{23} {s}_{13} {s}_{23}+4 {n}_{3} {p}_{23} {s}_{13} {s}_{23}+{p}_{13} {s}_{12} {s}_{23}^{2}-4 {p}_{13} {p}_{23} {s}_{12}$$ $${s}_{12}^{2} {s}_{13}^{2} {s}_{23}^{2}+{p}_{13} {s}_{12}^{4}+{p}_{23} {s}_{12}^{4}-2 {p}_{23} {s}_{12}^{2} {s}_{13}^{2}+{p}_{12} {s}_{13}^{4}+{p}_{23} {s}_{13}^{4}-2 {p}_{13} {s}_{12}^{2} {s}_{23}^{2}-2 {p}_{12} {s}_{13}^{2} {s}_{23}^{2}+{p}_{12} {s}_{23}^{4}+{p}_{13} {s}_{23}^{4}+8 {p}_{13} {p}_{23} {s}_{12}^{2}+8 {p}_{12} {p}_{23} {s}_{13}^{2}-8 {p}_{12} {p}_{23} {s}_{23}^{2}-8 {p}_{13} {p}_{23} {s}_{23}^{2}+16 {p}_{12} {p}_{23}^{2}+16 {p}_{13} {p}_{23}^{2}$$ $${n}_{3} {p}_{23} {s}_{13}^{3} {s}_{23}+{p}_{13} {s}_{12} {s}_{13}^{2} {s}_{23}^{2}+{p}_{23} {s}_{12} {s}_{13}^{2} {s}_{23}^{2}+{n}_{3} {p}_{13} {s}_{13} {s}_{23}^{3}+{p}_{13}^{2} {s}_{12}^{3}+2 {p}_{13} {p}_{23} {s}_{12}^{3}+{p}_{23}^{2} {s}_{12}^{3}-{p}_{13} {p}_{23} {s}_{12} {s}_{13}^{2}-{p}_{23}^{2} {s}_{12} {s}_{13}^{2}-4 {n}_{3} {p}_{13} {p}_{23} {s}_{13} {s}_{23}-{p}_{13}^{2} {s}_{12} {s}_{23}^{2}-{p}_{13} {p}_{23} {s}_{12} {s}_{23}^{2}+4 {p}_{13}^{2} {p}_{23} {s}_{12}+4 {p}_{13} {p}_{23}^{2} {s}_{12}$$ a quick, if useless, Macaulay 2 computation:
R = QQ[m_(1,1)..m_(3,3)]
M = matrix apply(3,i->apply(3,j->(m_(i+1,j+1))))
I = ideal trace M + ideal exteriorPower(2,M)
S = QQ[n_1..n_3, p_(1,2), p_(1,3), p_(2,3), s_(1,2), s_(1,3), s_(2,3)]
f = map(R/I,S, {m_(1,1),m_(2,2),m_(3,3),
m_(1,2)*m_(2,1), m_(1,3)*m_(3,1), m_(2,3)*m_(3,2),
m_(1,2)+m_(2,1), m_(1,3)+m_(3,1), m_(2,3)+m_(3,2)})
ker f
answered Apr 8, 2015 at 3:53
-
3$\begingroup$ thanks! Do you think something more intrinsic (=without equations) is possible? Or it is just a random smooth Picard rank one Fano 3fold? $\endgroup$– GiulioCommented Apr 8, 2015 at 8:33