forgive that last comment. Under all the usual embeddings we certainly do have $SO_2^n$ contained in $SO_{2n}$. What I obected to was the idea that a linear mapping which restricts to an isometry on certain transverse 2-planes must be an isometry of the ambient space. I should clarify the situation i had in mind:

Actually I don't see the first inclusion as clear at all: this because $SO_2^n$ (as embedded in $SL_2^n \subset Sp_{2n}$) is not contained in $SO_{2n}$. Let us remind ourselves that $Sp_{2n} \cap SO_{2n}=U_n$--the unitary group.

@Noonan: well I don't know how to crank anything out of this. For one, i don't see how this embedding recognizes 2-planes $P,P'$ which are not transverse. And even if I had some idea of how points on the projectivized light cone lay, I'm unaware of any potential invariants. So I'm not sure what to do with this. Thanks all the same though.

Nothing is lost in supposing that $e_1, \ldots f_n$ span a volume 1 parallelopiped. Then their n-volume in $\wedge^2$ will correspond to $|e_1 \wedge \ldots \wedge f_n|$ (ie =1). For fun, in light of the identity dim$Gr_{2,2n}=(2n-2).2=(4n-4).1=$dim$Gr_{1, 4n-3}$ we might try to find a suitable embedding from the Grassmannian of 2-planes in $\mathbb{R}^{2n}$ into $\mathbb{R}P^{4n-3}$ and then take the volume form there. But i don't think this is to be taken seriously.

What I want from this question is a measure which recognizes that "configurations of $2n$ lines in $\mathbb{R}^{2n}$" is different than "configurations of n 2-planes". The volume $|e_1 \wedge \ldots \wedge f_n|$ is insensitive to configurations of 2-planes. The point is to find a form or measure which isn't.

In my own particular situation this measure is not interesting because all the $e_i, f_i$ generate a unimodular lattice----hence their wedge product is 1. This then does nothing to distinguish them.

A simpler question (which i'm still uncertain of) is whether $e^{J \theta}$ is periodic in $\theta$.

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 $L$?

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: