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Let Xl = P(Ol ), Xs = P(Os ). 5 Dl is the discriminant of Xl . Ds is the discriminant of Xs . Proof. We identify g and g∗ via the Killing form. Let α ∈ ∆ be any root, eα ∈ g the corresponding root vector. Since [g, eα ]⊥ = geα , we have P(Ad(G) · eα ) ∗ = P(Ad(G) · geα ). Let tα = {x ∈ t | α(x) = 0}, ˆtα = tα + eα . Then, clearly, ˆtα ∈ geα . Moreover, if α is long then Dl |ˆtα = Dl |tα = 0. If α is short then Ds |ˆtα = Ds |tα = 0. Therefore, in order to prove Theorem it suffices to check that dim Ad(G) · ˆtα = n − 1.

Then U acts on l1 . 14 (c). Therefore, U acts trivially on l1 . Consider the action of U on l1 /l1 . Clearly, the function xβ is U -invariant, therefore each U -orbit in l1 /l1 lies on a hyperplane xβ = const. For any root α ∈ ∆1 such that (α, β) > 0 we have α = β + α ˆ , where α ˆ ∈ ∆ˆ0 . Therefore, the tangent space u · (x mod l1 ) is equal to the hyperplane xβ = 0. Since any orbit of a linear unipotent group is closed [OV], it follows that the U -orbit of x mod l1 is the affine hyperplane xβ = 1.

The Pl¨ ucker embedding in fact gives the embedding of Grω (k, V ) in P(Λk0 V ) as the projectivization of the cone of highest weight vectors. This embedding corresponds to the ample generator of the Pickard group. 1 Polarized Flag Varieties 25 Now consider the orthogonal case G = SO(V ). Then we have the Pl¨ ucker embedding GrQ (k, V ) ⊂ P(Λk V ). Suppose first that k < m if n = 2m + 1 or k < m − 1 if n = 2m. Then Λk V is irreducible as G-module and has the highest weight ωk . Therefore, this embedding corresponds to the ample generator of the Pickard group.