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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 $2n$)

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  • $\begingroup$ You need to explain what you mean by $V$. Is it $V = \mathbb{R}^{4n}$? $\endgroup$
    – Robert Bryant
    Dec 6, 2012 at 13:17
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    $\begingroup$ I'm afraid that just saying that $V$ is a vector space of finite dimension doesn't give us enough information. You have to tell us how you think $\mathrm{O}(4n,\mathbb{C})$ is acting on $V\otimes\mathbb{C}$. Obviously, this transitivity won't hold for any finite dimensional $V$ and any action of $\mathrm{O}(4n,\mathbb{C})$ on $V\otimes\mathbb{C}$. What are you not telling us? $\endgroup$
    – Robert Bryant
    Dec 6, 2012 at 13:48
  • $\begingroup$ I revised my question $\endgroup$
    – user21574
    Dec 6, 2012 at 15:14
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    $\begingroup$ Would your question be more simply phrased as "How can one prove that $\mathrm{O}(4n,\mathbb{C})$ acts transitively on the maximal isotropic subspaces of $\mathbb{C}^{4n}$?" What I'm getting at is whether it's important for you that your complex vector space be written as $V\otimes\mathbb{C}$ for some (I presume) real vector space $V$ of dimension $2n$ over $\mathbb{R}$. If the 'real structure' is not important to you, then the above version of your question would be adequate. (By the way, in this case, you'd have transitivity for all $n$, so I'm not sure why you are assuming $4n$.) $\endgroup$
    – Robert Bryant
    Dec 6, 2012 at 16:33
  • $\begingroup$ In fact maximal isotropic $L$ corresponds to a generalized complex structure on $V$ and so the dimension of $V$ have to be even, $\endgroup$
    – user21574
    Dec 7, 2012 at 11:14

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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 $O(n,n,\mathbb R)$.

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    $\begingroup$ And this idea can be extended to give a proof of Witt's Theorem, that in a non-degenerate "formed space" (quadratic form, hermitian form, symplectic, etc.) an isometry from one subspace to another extends to an isometry of the whole space. $\endgroup$
    – paul garrett
    Dec 6, 2012 at 17:41
  • $\begingroup$ Dear Paul, if possible for you please more explain your second comment $\endgroup$
    – user21574
    Dec 8, 2012 at 1:42

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