On page 3 of [this paper](http://www.imsc.res.in/%7Ekapil/papers/topol.pdf), the authors give a Bruhat cell decomposition of a quadric hypersurface $Q$ of complex dimension $n$. This may be a stupid question, but it doesn't seem clear to me what exactly the decomposition is. They only explicitly describe the real dimension $2n-2$, $n$, and $n+2$ cells, but say that there is exactly one cell for each even real dimension. Particularly, I would like to know what the complex codimension 2 cells are.



Bruhat decomposition of a quadric hypersurface
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...




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$\def\Span{\mathrm{Span}}$
**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
$$(1:0:0:0:0:0:0:0)$$
$$(t:1:0:0:0:0:0:0)$$
$$(t:u:1:0:0:0:0:0)$$
$$(t:u:v:1:0:0:0:0) \ \mbox{and} \ (t:u:v:0:1:0:0:0)$$
$$(t:u:-wx:w:x:1:0:0)$$
$$(t:-vy-wx:v:w:x:y:1:0)$$
$$(-uz-vy-wx:u:v:w:x:y:z:1)$$


