We say $L< (V\oplus V^{\*})\bigotimes \mathbb{C}$ is isotropic when $< X,Y>=0$ for all $X,Y\in L$


Why $O(4n,\mathbb{C})$ (orthogonal group) acts transitively on the space of maximal isotropics of 
$V\bigotimes \mathbb{C}$ ?


(here $V$ is a vector space of finite dimention $2n$)



Why $O(4n,\mathbb{C})$ (orthogonal group) acts transitively on the space of maximal isotropics of  $V\bigotimes \mathbb{C}$ ?
I think a more general statement can be proved along the following line:


Let $V= \mathbb R^n$. Then on $W:= V\oplus V^\*$ the symmetric bilinear form 
$((v,v^\*)|(w,w^\*)) = \langle w^\*,v\rangle + \langle v^\*, w\rangle$ has signature $(n,n)$.
Now we are quite similar to a symplectic vector space.
Given isotropic $L$, choose a basis $w\_1,\dots w\_n$ of $L$. Then $(w\_i|w\_j)=0$ for all $i,j$.
We can extend this to a basis $w\_1,\dots,w\_n, w^1,\dots,w^n$ of $W$ such that 
$(w\_i,w\_j)=0$, $(w^i,w^j)=0$ for all $i,j$ and $(w\_i,w^j)=\delta\_i^j$. 
Then $w^1,\dots, w^n$ spans a complementary isotropic subspace. 
The group $O(n,n,\mathbb R)$ acts transitively on the set of all such bases.
Thus it acts transitively on the set of pairs of complementary isotropic subspaces.
Thus also transitively on the set of isotropic subspaces. 
Now $O(2n,\mathbb C)$ is the complexification of $O(n,n,\mathbb R)$. 


