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Consider the canonical symplectic structure $(\omega, J)$ on $\mathbb{R}^{2n}$.

(i) What can be said about the density of the lagrangian grassmannian $L$ (i.e. those rank $n$ totally isotropic linear subspaces of $\mathbb{R}^{2n}$) in the usual Grassmannian $Gr=Gr_{n}(\mathbb{R}^{2n})$ of rank $n$ linear subspaces?

(ii) How stable is $L$ within $Gr$. ie. which homeomorphisms of $Gr$ preserve $L$ (not necessarily pointwise)?

(iii) In what sense can we see $Gr$ as ''swept out'' by $L$ ie. decompose $Gr$ as the orbits of $L$.

(iv) What projections $Gr \to L$ can we produce?

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up vote 7 down vote accepted

jmart has modified the question since I posted my original answer. I'll now modify my answer to correspond:

(i) The Lagrangian Grassmannian ($L$ in your notation) is a closed submanifold of $Gr_n(\mathbb{R}^{2n})$, so it's not dense at all. In fact, it has dimension $\frac12n(n{+}1)$.

(ii) $L$ is homogeneous under the action of $\textrm{Sp}(n)\subset \textrm{GL}(2n,\mathbb{R})$, the subgroup that preserves the symplectic structure $\omega$. (The $J$ is not needed to define $L$.) Conversely, the subgroup of $\textrm{GL}(2n,\mathbb{R})$ that preserves $L$ in $Gr_n(\mathbb{R}^{2n})$ is easily seen to be the subgroup of $\textrm{GL}(2n,\mathbb{R})$ that preserves $\omega$ up to a multiple.

(iii) I don't really know what you mean by 'the orbits of $L$', since $L$ is not a group. Of course, $Gr_n(\mathbb{R}^{2n})$ is homogeneous under $\textrm{GL}(2n,\mathbb{R})$ so it's the $\textrm{GL}(2n,\mathbb{R})$-orbit of any point of $L$. Perhaps you actually want to know the orbits of $\textrm{Sp}(n)$ acting on $Gr_n(\mathbb{R}^{2n})$, with $L$ being the closed $\textrm{Sp}(n)$-orbit. If you consider, for any $n$-plane $E\in Gr_n(\mathbb{R}^{2n})$, the rank $r(E)$ of the pullback of $\omega$ to $E$, then the $\textrm{Sp}(n)$-orbits are the level sets of $r$ (which takes values in nonnegative integers).

(iv) I'm not sure what you mean by 'projections'. If by 'projection', you mean a smooth map $\pi:Gr_n(\mathbb{R}^{2n})\to L$ that is the identity on $L$, such a thing does not exist.

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Dimension counts do not answer questions of density. The dimension of the lagrangian grassmannian $L$ shows only that $L$ is not generic in $Gr_n$. (Silly example: the rationals in $\mathbb{R}$). So there is a question as to how close a linear subspace is to being lagrangian. Now with my original question I should have known better: that $L$ and $Gr$ are homogenous spaces resolves very easily some simple questions. But this is still unsatisfactory. These identifications are only obtained by ''pushing a basis around''. About the stability of $L$ in $Gr$ I had the following in mind: – J. Martel Jan 24 '12 at 21:26
Consider an almost complex structure $J$ on $(\mathbb{R}^{2n}, \omega)$ ie. we require $J$ to be $\omega-$compatible. Then, for $\theta \in \mathbb{R}$, the automorphism $e^{J \theta}$ acts on $L$. But the orbit $\{e^{J\theta} \ell \}_{\theta}$ of any lagrangian $\ell$ will be one-dimensional in $L$. A question: can we describe explicitly a larger family of ''similiar'' automorphisms of $L$? – J. Martel Jan 24 '12 at 21:38
A simpler question (which i'm still uncertain of) is whether $e^{J \theta}$ is periodic in $\theta$. – J. Martel Jan 24 '12 at 21:39
@jmart: You should look up some sources on symplectic linear algebra, which would answer many of your questions. Re your comments: $L$ is a compact submanifold of $Gr_n$, being defined by equations, so it is not dense in $Gr_n$. Since $J^2=-I$, we have $e^{J\theta} = \cos\theta\ I + \sin\theta\ J$, so $e^{J\theta}$ is periodic (of period $2\pi$), defining a free circle action on $L$. Any $\omega$-compatible $J$ will define a free $S^1$-action on $L$, and these are 'similar', but, of course, there are many $1$-parameter subgroups of $\textrm{Sp}(n,\mathbb{R})$ that are not 'similar'. – Robert Bryant Jan 25 '12 at 7:56
Thank you Prof. Bryant. I will get to work. – J. Martel Jan 25 '12 at 17:30

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