3 edited body; edited body

Let $Q$ live in $\mathbb{P}^{2n-1}$, so $Q$ defines a symmetric bilinear form on $\mathbb{C}^{2n}$, and the complex dimension of $Q$ is $2n-2$. Take a flag of isotropic subspaces $0 \subset F_1 \subset F_2 \subset \cdots \subset F_n$. So $\mathbb{P}(F_i) \subset Q$. For $1 \leq i \leq n-1$, the class of $\mathbb{P}(F_i)$ spans $H_{2i-2}(Q)$ or, Poincare dually, spans $H^{4n-2i-2}(Q)$.

Let $F_i^{\perp}$ denote the $Q$-orthogonal to $F_i$. So $F_n=F_n^{\perp} \subset F_{n-1}^{\perp} \subset \cdots \subset F_2^{\perp} \subset F_1^{\perp}$. For $1 \leq i \leq n-2$, the intersection $\mathbb{P}(F_i)^{\perp} \cap Q$ is a smooth hypersurface in $\mathbb{P}(F_i^{\perp})$. It is a basis for $H_{4n-2i-2}(Q)$, or for $H^{2i-2}(Q)$.

The remaining case is middle cohomology; $H_{2n-2}(Q)$. The recipe of the first paragraph would suggest taking $\mathbb{P}(F_n)$; the recipe of the second paragraph would suggest taking $\mathbb{P}(F_{n-1}^{\perp}) \cap Q$. In fact, these two classes together form a basis for the two dimensional space $H_{2n-2}(Q)$, but there is a better way to think about it. The quadratic form $Q$, restricted to $F_{n-1}^{\perp}$, has kernel $F_{n-1}$, and hence descends to a nondegenerate pairing on $F_{n-1}^{\perp}/F_{n-1}$. A symmetric nondegenerate bilinear form on a $2$-dimensional vector space has precisely two isotropic subspaces. One of them is $F_n/F_{n-1}$. Call the other one $F'_n/F_{n-1}$, so $F'_n$ is another isotropic plane sitting between $F_{n-1}$ and $F_{n-1}^{\perp}$. Then $(\mathbb{P}(F_n), \mathbb{P}(F'_n))$ form a basis for $H_{2n-2}(Q)$. From the above description, we see that $\mathbb{P}(F_{n-1}^{\perp}) \cap Q = \mathbb{P}(F_n) \cup \mathbb{P}(F'_n)$.

The Bruhat cells are just formed by taking each Schubert variety and removing the smaller Schubert varieties inside it.

In my opinion, it would be best to define a complete isotropic flag to consist of the data $F_1 \subset F_2 \subset \cdots \subset F_{n-2} \subset F_n, F'_n$. (Note that we can recover $F_{n-1}$ as $F_n \cap F'_n$.) If you notice that the containments between these subspaces look like the $D_n$ Dynkin diagram, that's not a coincidence...

$\def\Span{\mathrm{Span}}$ ADDED It might help to write all of this out for $Q$ given by $x_1 x_8 + x_2 x_7 + x_3 x_6 + x_4 x_5$. Let $F_1 = \Span(e_1)$, $F_2 = \Span(e_1, e_2)$, $F_3 = \Span(e_1, e_2, e_3)$ and $F_4 = \Span(e_1,e_2,e_3, e_4)$. So $F_i^{\perp} = \Span(e_1, e_2, \ldots, e_{8-i})$ and $F'_4 = \Span(e_1, e_2, e_3, e_5)$.

The Bruhat cells (also known as Schubert cells) are $$(1:0:0:0:0:0:0:0)$$ $$(t:1:0:0:0:0:0:0)$$ $$(t:u:1:0:0:0:0:0)$$ $$(t:u:v:1:0:0:0:0) \ \mbox{and} \ (u:v:w:0:1:0:0:0)$$ t:u:v:0:1:0:0:0)(t:u:-wx:w:x:1:0:0)(t:-vy-wx:v:w:x:y:1:0)(-uz-xy-wx:u:v:w:x:y:z:1)$$(-uz-vy-wx:u:v:w:x:y:z:1)$$

2 added 620 characters in body

Let $Q$ live in $\mathbb{P}^{2n-1}$, so $Q$ defines a symmetric bilinear form on $\mathbb{C}^{2n}$, and the complex dimension of $Q$ is $2n-2$. Take a flag of isotropic subspaces $0 \subset F_1 \subset F_2 \subset \cdots \subset F_n$. So $\mathbb{P}(F_i) \subset Q$. For $1 \leq i \leq n-1$, the class of $\mathbb{P}(F_i)$ spans $H_{2i-2}(Q)$ or, Poincare dually, spans $H^{4n-2i-2}(Q)$.

Let $F_i^{\perp}$ denote the $Q$-orthogonal to $F_i$. So $F_n=F_n^{\perp} \subset F_{n-1}^{\perp} \subset \cdots \subset F_2^{\perp} \subset F_1^{\perp}$. For $1 \leq i \leq n-2$, the intersection $\mathbb{P}(F_i)^{\perp} \cap Q$ is a smooth hypersurface in $\mathbb{P}(F_i^{\perp})$. It is a basis for $H_{4n-2i-2}(Q)$, or for $H^{2i-2}(Q)$.

The remaining case is middle cohomology; $H_{2n-2}(Q)$. The recipe of the first paragraph would suggest taking $\mathbb{P}(F_n)$; the recipe of the second paragraph would suggest taking $\mathbb{P}(F_{n-1}^{\perp}) \cap Q$. In fact, these two classes together form a basis for the two dimensional space $H_{2n-2}(Q)$, but there is a better way to think about it. The quadratic form $Q$, restricted to $F_{n-1}^{\perp}$, has kernel $F_{n-1}$, and hence descends to a nondegenerate pairing on $F_{n-1}^{\perp}/F_{n-1}$. A symmetric nondegenerate bilinear form on a $2$-dimensional vector space has precisely two isotropic subspaces. One of them is $F_n/F_{n-1}$. Call the other one $F'_n/F_{n-1}$, so $F'_n$ is another isotropic plane sitting between $F_{n-1}$ and $F_{n-1}^{\perp}$. Then $(\mathbb{P}(F_n), \mathbb{P}(F'_n))$ form a basis for $H_{2n-2}(Q)$. From the above description, we see that $\mathbb{P}(F_{n-1}^{\perp}) \cap Q = \mathbb{P}(F_n) \cup \mathbb{P}(F'_n)$.

The Bruhat cells are just formed by taking each Schubert variety and removing the smaller Schubert varieties inside it.

In my opinion, it would be best to define a complete isotropic flag to consist of the data $F_1 \subset F_2 \subset \cdots \subset F_{n-2} \subset F_n, F'_n$. (Note that we can recover $F_{n-1}$ as $F_n \cap F'_n$.) If you notice that the containments between these subspaces look like the $D_n$ Dynkin diagram, that's not a coincidence...

$\def\Span{\mathrm{Span}}$ ADDED It might help to write all of this out for $Q$ given by $x_1 x_8 + x_2 x_7 + x_3 x_6 + x_4 x_5$. Let $F_1 = \Span(e_1)$, $F_2 = \Span(e_1, e_2)$, $F_3 = \Span(e_1, e_2, e_3)$ and $F_4 = \Span(e_1,e_2,e_3, e_4)$. So $F_i^{\perp} = \Span(e_1, e_2, \ldots, e_{8-i})$ and $F'_4 = \Span(e_1, e_2, e_3, e_5)$.

The Bruhat cells (also known as Schubert cells) are $$(1:0:0:0:0:0:0:0)$$ $$(t:1:0:0:0:0:0:0)$$ $$(t:u:1:0:0:0:0:0)$$ $$(t:u:v:1:0:0:0:0) \ \mbox{and} \ (u:v:w:0:1:0:0:0)$$ $$(t:u:-wx:w:x:1:0:0)$$ $$(t:-vy-wx:v:w:x:y:1:0)$$ $$(-uz-xy-wx:u:v:w:x:y:z:1)$$

1

Let $Q$ live in $\mathbb{P}^{2n-1}$, so $Q$ defines a symmetric bilinear form on $\mathbb{C}^{2n}$, and the complex dimension of $Q$ is $2n-2$. Take a flag of isotropic subspaces $0 \subset F_1 \subset F_2 \subset \cdots \subset F_n$. So $\mathbb{P}(F_i) \subset Q$. For $1 \leq i \leq n-1$, the class of $\mathbb{P}(F_i)$ spans $H_{2i-2}(Q)$ or, Poincare dually, spans $H^{4n-2i-2}(Q)$.

Let $F_i^{\perp}$ denote the $Q$-orthogonal to $F_i$. So $F_n=F_n^{\perp} \subset F_{n-1}^{\perp} \subset \cdots \subset F_2^{\perp} \subset F_1^{\perp}$. For $1 \leq i \leq n-2$, the intersection $\mathbb{P}(F_i)^{\perp} \cap Q$ is a smooth hypersurface in $\mathbb{P}(F_i^{\perp})$. It is a basis for $H_{4n-2i-2}(Q)$, or for $H^{2i-2}(Q)$.

The remaining case is middle cohomology; $H_{2n-2}(Q)$. The recipe of the first paragraph would suggest taking $\mathbb{P}(F_n)$; the recipe of the second paragraph would suggest taking $\mathbb{P}(F_{n-1}^{\perp}) \cap Q$. In fact, these two classes together form a basis for the two dimensional space $H_{2n-2}(Q)$, but there is a better way to think about it. The quadratic form $Q$, restricted to $F_{n-1}^{\perp}$, has kernel $F_{n-1}$, and hence descends to a nondegenerate pairing on $F_{n-1}^{\perp}/F_{n-1}$. A symmetric nondegenerate bilinear form on a $2$-dimensional vector space has precisely two isotropic subspaces. One of them is $F_n/F_{n-1}$. Call the other one $F'_n/F_{n-1}$, so $F'_n$ is another isotropic plane sitting between $F_{n-1}$ and $F_{n-1}^{\perp}$. Then $(\mathbb{P}(F_n), \mathbb{P}(F'_n))$ form a basis for $H_{2n-2}(Q)$. From the above description, we see that $\mathbb{P}(F_{n-1}^{\perp}) \cap Q = \mathbb{P}(F_n) \cup \mathbb{P}(F'_n)$.

The Bruhat cells are just formed by taking each Schubert variety and removing the smaller Schubert varieties inside it.

In my opinion, it would be best to define a complete isotropic flag to consist of the data $F_1 \subset F_2 \subset \cdots \subset F_{n-2} \subset F_n, F'_n$. (Note that we can recover $F_{n-1}$ as $F_n \cap F'_n$.) If you notice that the containments between these subspaces look like the $D_n$ Dynkin diagram, that's not a coincidence...