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.

Thank you in advance


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.

  • $\begingroup$ 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 $O(n, \mathbb{R})$, $SO(n, \mathbb{R}), SU(n)?$ $\endgroup$ – Surojit Sep 27 '18 at 8:57
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    $\begingroup$ I have edited my answer to address this. $\endgroup$ – Andreas Cap Sep 27 '18 at 11:34
  • $\begingroup$ 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 $W_m(\mathbb{C}^n) = \{ (v_1, \cdots, v_m) : v_i's \text{are linearly independent in} \; \mathbb{C}^n , B(v_i, v_j) =0 \; \text{for all}\; i, j \}$. $\endgroup$ – Surojit Sep 27 '18 at 13:01
  • $\begingroup$ 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. $\endgroup$ – Andreas Cap Sep 27 '18 at 13:35
  • $\begingroup$ 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 $Sp(n , \mathbb{C}) \cap U(2n) = Sp(n).$ $\endgroup$ – Surojit Sep 28 '18 at 12:13

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