Vector fields and deformations of isotropic super-Grassmannians of maximal type

On complex lie supergroups and split homogeneous supermanifolds

“…There is a functor, say gr, from the category of supermanifolds to the category of split supermanifolds. Let us briefly describe this construction (see, e.g., [12,15] Obviously a split Lie supergroup is a Lie supergroup. Furthermore, the following result holds:…”

“…it is a Lie subsupergroup of (G, O G ) and its odd dimension is zero, we show that the corresponding homogeneous supermanifold (G/H, O G/H ) is split. In particular, any complex Lie supergroup is a split supermanifold.It is well known that a complex homogeneous supermanifold may be nonsplit (see, e.g., [15]). We find here necessary and sufficient conditions for a complex homogeneous supermanifold to be split.…”

Vector fields on osp2m|2n(C)- and Vishnyakova1 2018Journal of Algebra3010View full textAdd to dashboardCite

“…From Lemma 2, it follows that Ker P = {0}. [OS2] without proof. The goal of this section is to prove this statement.…”

“…The Lie superalgebra v(IF o m,k 1 |n,l 1 ) of holomorphic vector fields on a connected component IF o m,k 1 |n,l 1 of the isotropic super-Grassmannian IF m,k 1 |n,l 1 of maximal type was calculated in [OS2].…”

“…Moreover, (Z 1 1 ) T = −Z 1 1 , (Y 1 1 ) T = Y 1 1 and H 1 1 = −(Z 1 1 ) T . Details about this construction can be found in [OS2,V6]. Denote by o 1 ∈ B 1 red the point that is given in the chart corresponding to (15) by the following equations X 1 1 = 0, Y 1 1 = 0 and H 1 1 = 0.…”