Theorem** on page 288 from Geometry of algebraic curves by Arbarello, Cornalba, Griffiths, Harris

Question

1Let $V$ a $2n$ complex vector space and take on $V$ a quadratic form.
Now define $$ \Sigma=\{\Lambda:Q(\Lambda,\Lambda)\equiv0 \} \subset Gr(n,2n)$$ where $\Lambda$ is a maximal subspace i.e it is not contained in other isotropics subspaces
Show that:

1. $\Sigma$ is a smooth complex variety of dimension $n(n-1)/2$,
2. $\Sigma$ has exactly two conected components $\Sigma_1$, $\Sigma_2$,
3. if $\Lambda \in \Sigma_i$ and $\Lambda^{'} \in \Sigma_j$ are two maximal isotropic subspace for $Q$ than $$dim(\Lambda \cap \Lambda^{'}) \equiv n \quad mod(2) \Leftrightarrow i=j$$.
   On the book the authors say that thees facts are standard and put as references Griffiths Harris, Principles of algebraic surfaces, proposition on page 735. I've seen it but i'm so confused about how to use it. Can someone help me how does it work?

Answer 1

If you try the $n=2$ case first, you'll quickly see that all the others are nearly the same.

When $n=2$ we have a four-dimensional vector space, say $$V = \operatorname{span}\{e_0, e_1, e_2, e_3\}.$$ Since every nondegenerate quadratic form on a vector space (in characteristic $\neq 2$) is equivalent to every other, we can just pick a quadratic form and examine it. So take $$Q(x, y) := x_0 y_3 + x_1 y_2 + x_2 y_1 + x_3 y_0.$$ where $$x = \sum_{i=0}^3 x_i e_i, \qquad y = \sum_{j=0}^3 y_j e_j.$$

Suppose we have some $\Lambda \in Gr(2, V)$. Notice that a two-dimensional subspace of $V$ is the same thing as a line in $\mathbb{P}V$. I claim that $Q(\Lambda, \Lambda) = 0$ if, and only if, the line $\mathbb{P}\Lambda \subset \mathbb{P}V$ lies on the surface $$S = \mathbb{P}(\{v \in V\; | \; Q(v,v) = 0\}) = V(2 x_0 x_3 + 2 x_1 x_2) \subset \mathbb{P}V.$$

To see this, suppose first that $Q(\Lambda, \Lambda) = 0$. Then for any $v \in \Lambda$, $Q(v,v) = 0$, so $[v] \in S$. Conversely, suppose $\mathbb{P}\Lambda \subset S$. Let $\Lambda$ be spanned by $v, w \in V$. Then $[v], [w] \in S$, so $Q(v,v) = Q(w,w) = 0$. Further, $[v+w] \in S$, so $$0 = Q(v+w, v+w) = Q(v,v) + 2 Q(v,w) + Q(w,w) = 2 Q(v,w),$$ i.e. $Q(v,w) = 0$. Therefore $Q(\Lambda, \Lambda) = 0.$

Now the remaining statements come down to the fact that $S$ is a surface with two different rulings, which are easy to exhibit directly from the equation $$x_0 x_3 + x_1 x_2 = 0$$ above. The third point, which is probably the most confusingly phrased, says in this case that:

• Two lines on the same ruling do not intersect, and
• Two lines on opposite rulings intersect in a point.

That statement gets a little more subtle in higher dimensions, but the idea is still about the same.