# Classical Lie group quotient-ed by its maximal parabolic subgroup

Let $$B$$ is a nondegenerate symmetric bilinear form on $$\mathbb{C}^n$$ then the corresponding complex orthogonal group is $$\{g : GL(n, \mathbb{C}): B(gx, gy) =(x,y) \}$$

In particular we use $$B (x,y) = \sum\limits_{1\leq l \leq m}x^ly^l - \sum\limits_{m \leq l \leq n}x^ly^l, \; n=2m \; \text{or}\; 2m+1,$$ and denote $$O(n, \mathbb{C}) = \{g : GL(n, \mathbb{C}): B(gx, gy) =(x,y) \}.$$

A linear subspace $$E \subset \mathbb{C}^n$$ is totally isotropic if $$B(E,E) =0.$$ The parabolic subgroups of $$O(n, \mathbb{C})$$ are the $$P_{E_1, \cdots, E_k} = \{ g \in O(n,\mathbb{C}): gE_l =E_l \; \text{for}\; 1 \leq l \leq k\}$$

where $$0 \neq E_1 \subset \cdots \subset E_k$$ is a sequence of totally isotropic subspaces.

Now the maximal parabolic subgroups of $$O(n, \mathbb{C})$$ are the $$P_E = \{ g \in O(n, \mathbb{C}) : gE=E\},$$ where $$E$$ is nonzero totally isotropic in $$\mathbb{C}^n.$$

We know that $$O(n, \mathbb{C})/P_E$$ is a projective variety. In particular, that should be a flag variety.

My question is:1) What should be the exact presentation of the variety $$O(n, \mathbb{C})/P_E?$$

2) Can we do the similar thing for $$O(n, \mathbb{R})$$ or $$SO(n, \mathbb{R})?$$

Any help we be appreciated.

1) If $$m\leq n/2$$ is the dimension of $$E$$, the $$O(n,\mathbb C)/P_E$$ is the variety of all $$m$$-dimensional isotropic subspaces of $$\mathbb C^n$$. (It is easy to see that $$O(n,\mathbb C)$$ acts transitively on the set of such subspaces and $$P_E$$ is the stabilizer of one of them.)

2) This depends on what you mean by "the similar thing". On the one hand, $$O(n,\mathbb R)$$ and $$SO(n,\mathbb R)$$act transitively on the space of all $$k$$-dimensional subspaces of $$\mathbb R^n$$. Thus, you can view the Grassmannian $$Gr(k,\mathbb R^n)$$ as a homogeneous space of $$O(n,\mathbb R)$$ and $$SO(n,\mathbb R)$$. This leads to the well known presentation $$Gr(k,\mathbb R^n)=O(n)/(O(k)\times O(n-k))$$ and similar for $$SO$$.

However, $$O(n)$$ and $$SO(n)$$ do not have parabolic subgroups since they are compact. If you want a real analog of those (and something that is closer to the complex case), you have to go to the indefinite orthogonal groups $$O(p,q)$$ for $$p+q=n$$. For those you have isotropic subspaces of dimensions $$1\leq k\leq min(p,q)$$ and the stabilizer of such an isotropic subspace is a parabolic subgroup $$P_k\subset O(p,q)$$. The generalized flag variety $$O(p,q)/P_k$$ then is the variety of $$k$$-dimensional isotropic subspaces of $$\mathbb R^{p+q}$$. The closest analogy to the complex case is obtained for the spilt real forms $$O(m,m)$$ and $$O(m,m+1)$$.

Edit (in view of your comment): What you write there about the complex case is not quite correct. In the definition of $$W_m$$ you have to add the condition that $$B(v_1,v_j)=0$$ for all $$i$$ and $$j$$ to make sure that the span of your vectors is totally isotropic. Moreover, the group $$O(m,\mathbb C)$$ should not be used here. The parabolic subgroup $$P_E$$ has a natural quotient (known as the Levi-factor) isomorphic to $$GL(m,\mathbb C)\times O(n-2m,\mathbb C)$$, and the bundle $$W_m$$ is associated to the standard representation of $$GL(m,\mathbb C)$$, so it is just a $$GL(m,\mathbb C)$$ vector bundle.

In the real cases and for a definite bilinear form, the situation is easier. There you have $$Gr(k,\mathbb R^n)\cong O(n)/(O(k)\times O(n-k))$$ and the analog of $$W_m$$ now really is formed by $$k$$-tuples of linearly independent vectos and this is an $$O(k)$$ bundle (so it carries a natural bundle metric). For $$SO$$ and $$SU$$ the situation is very similar.

• Define $W_m(\mathbb{C}^n) = \{(v_1, \cdots, v_m): v_i's \; \text{are linearly independent vectors in} \; \mathbb{C}^n \}.$ Then the map $W_m(\mathbb{C}^n) \to O(n, \mathbb{C})/P_E$ given by span is an $O(m, \mathbb{C})$-bundle. Then what should be the analogous $W_m$ in the cases $O(n, \mathbb{R})$, $SO(n, \mathbb{R}), SU(n)?$ – Surojit Sep 27 '18 at 8:57
• I have edited my answer to address this. – Andreas Cap Sep 27 '18 at 11:34
• Let $X= \{ \text{all m-dimensional isotropic subspaces of} \; \mathbb{C}^n \}.$ I want to construct a $SO$-bundle over $X$. Is it possible to construct such bundle via inputting some extra conditions in my $W_m(\mathbb{C}^n) = \{ (v_1, \cdots, v_m) : v_i's \text{are linearly independent in} \; \mathbb{C}^n , B(v_i, v_j) =0 \; \text{for all}\; i, j \}$. – Surojit Sep 27 '18 at 13:01
• I don't think that the group $SO(m,\mathbb C)$ is nicely related to the quotient $SO(n,\mathbb C)/P_m$. The latter deals with isotropic subspaces that do not carry a natural bilinear form. Of course, you can look at the variety of non-degenerate subspaces of dimension $m$ (which is isomorphic to $O(n,\mathbb C)/(O(m,\mathbb C)\times O(n-m,\mathbb C)$ and carries a natural $O(m,\mathbb C)$-bundle, but this is not a generalized flag variety in the usual sense. – Andreas Cap Sep 27 '18 at 13:35
• I hope that one can have a similar bundle for $Sp(n, \mathbb{C})$ for the antisymmetric bilinear form on $\mathbb{C}^{2n}$ given by $B(x,y) = \sum\limits_{1 \leq l \leq n}(x^ly^{n+l} - x^{n+l}y^l).$ Therefore we get a $Sp(m, \mathbb{C})$-bundle with total space $W_m= \{ (v_1, \cdots, v_m): v_i \text{'s are linearly independent vectors in and} \; B(v_i, v_j) =0 \text{for all} \; i , j \}$. If I put an additional condition $<v_i, v_j> = \delta_{ij}$ in $W_m,$ then I think the fibre is $Sp(n , \mathbb{C}) \cap U(2n) = Sp(n).$ – Surojit Sep 28 '18 at 12:13