Given $n$ vectors $v_1, \ldots, v_n$ in $\mathbb{R}^n$ of course we all know at least one measure for their relative configuration: $v_1 \wedge\ldots \wedge v_n$. Now suppose one were given $n$ pairs $(e_1,f_1), \ldots, (e_n,f_n)$ in $\mathbb{R}^{2n}$then what could be a meaningful measure for their relative configuration? Suppose moreover we require this measure to be $geometric$ ie. having fixed, say a full rank lattice $\Gamma$ in $\mathbb{R}^{2n}$ what could be a meaningful measure for the relative configuration of n 2planes $\pi_1, \ldots, \pi_n$ in $\mathbb{R}^{2n}$?

$\begingroup$ The $n$volume in $\Lambda^2\mathbb R^{2n}$ of the parallelopiped with sides $e_i\wedge f_i$? $\endgroup$ – Mariano SuárezÁlvarez Jan 30 '12 at 0:25

$\begingroup$ Nothing is lost in supposing that $e_1, \ldots f_n$ span a volume 1 parallelopiped. Then their nvolume in $\wedge^2$ will correspond to $e_1 \wedge \ldots \wedge f_n$ (ie =1). For fun, in light of the identity dim$Gr_{2,2n}=(2n2).2=(4n4).1=$dim$Gr_{1, 4n3}$ we might try to find a suitable embedding from the Grassmannian of 2planes in $\mathbb{R}^{2n}$ into $\mathbb{R}P^{4n3}$ and then take the volume form there. But i don't think this is to be taken seriously. $\endgroup$ – J. Martel Jan 30 '12 at 17:30
I believe the wedgebased measure you mentioned for the $\mathbb{R}^n$ case works here as well. Assuming each of the $e_i$ and $f_i$ are in $\mathbb{R}^{2n}$, then $\pi_i := e_i \wedge f_i$ is a 2vector (a weighted specification of a 2plane in $\mathbb{R}^{2n}$). This wedge quantity will be zero iff $e_i$ and $f_i$ are linearly dependent.
Wedging together all of the $\pi_i$ will give a $2n$vector having analogous geometric meaning as the 2vector. It will zero iff $X := {e_1, \dots, e_n, f_1, \dots, f_n}$ is a linearly dependent set. Then $e_1 \wedge f_1 \wedge \dots \wedge e_n \wedge f_n$ is the norm you are looking for. The geometric quantity this measures should be the signed volume of the simplex spanned by the vectors in $X$ (there may be a normalizing factor, something like $n!$, depending on convention).
This is certainly a natural measure of such objects, but probably not the only one.

$\begingroup$ In my own particular situation this measure is not interesting because all the $e_i, f_i$ generate a unimodular latticehence their wedge product is 1. This then does nothing to distinguish them. $\endgroup$ – J. Martel Jan 29 '12 at 22:36

2$\begingroup$ 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 2planes". The volume $e_1 \wedge \ldots \wedge f_n$ is insensitive to configurations of 2planes. The point is to find a form or measure which isn't. $\endgroup$ – J. Martel Jan 29 '12 at 23:42
This is more of a musing than an answer, but maybe it can be useful.
There is a nice way to think about the Grassmannians of 2planes in $\mathbb{R}^n$; there is an isomorphism between $Gr(2, n)$ and the projectivized light cone in $\mathbb{C}^n$. Here is how it works: start with your 2plane $P$ in $\mathbb{R}^n$ and pick any two perpendicular vectors $v$, $w$ of equal norm which span $P$. From those two real vectors, form the complex vector $q = v + i w$. Letting $\cdot$ denote the standard (not hermitian!) dot product, the conditions on $v$ and $w$ imply that $q \cdot q = 0$, so $q$ is a point on the light cone in $\mathbb{C}^n$. A different choice of spanning vectors will cause $q$ to be multiplied by some complex factor $z$, so $q$ is welldefined up to complex scalars.
From this perspective you are looking for invariants of configurations of $n$ points on the projective light cone in $\mathbb{C}^{2n}$ (perhaps modulo the action of $O(2n)$ or some other group). This can be cranked out more or less mechanically...

$\begingroup$ @Noonan: well I don't know how to crank anything out of this. For one, i don't see how this embedding recognizes 2planes $P,P'$ 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. $\endgroup$ – J. Martel Jan 30 '12 at 17:40
This is not a terribly informed answer  I have just become aware of the subject material (mass and comass) and don't fully understand the geometric significance. Hopefully it will be useful though.
Take a look at http://www.encyclopediaofmath.org/index.php/Mass_and_comass  it gives a different norm than the one already mentioned, and in fact has an inequality involving the aforementioned norm. Reference: Herbert Federer, Geometric Measure Theory, pg 38.

$\begingroup$ Well, here's a special situation: say we have a lattice L and an integral unimodular symplectic form \omega defined on L. Moreover say we have a symplectic basis e_1, .., e_n, f_1, .., f_n \in L. Then each pair e_i, f_i span a rational symplectic subspace P_i of the lattice. As measure on P_i we can take the eccentricity k_i of the convex hull of {e_i, f_i} (an ellipse). This describes an nfold continuous measure {k_1, ..., k_n} on the symplectic bases (e_1, ..., f_n). But again, this a special special little situation. $\endgroup$ – J. Martel Mar 13 '12 at 3:35