TITLE: Geometry of the multilagrangian Grassmannian 
QUESTION [11 upvotes]: Let's introduce the following variety $MG(3,6)$, which is a "multisymplectic" analog of a Lagrangian Grassmannian $LG(3,6)$.
Consider a 3-form $\omega = dx1 \wedge dx2 \wedge dx^3 - dx4 \wedge dx5 \wedge dx^6$ on a 6-dimensional vector space $V$ over an algebraically closed field $K$, $char(K) = 0$ (I feel that it's safe to assume $K = \mathbb C$).
We can consider a subvariety $MG(3,6)$ of Grassmannian $Gr(3,6)$ consisting of 3-planes $E$ satisfying $\omega(E)=0$.
$MG(3,6)$ turns out to be an 8-dimensional smooth hyperplane section in $Gr(3,6)$ w.r.t the Plucker embedding. I believe that $MG(3,6)$ is NOT a homogenous space for a group action.
Now the question is: What can we say about geometry of $MG(3,6)$? More precisely, I'd like to compute the cohomology $H^*(MG(3,6))$ in terms of some "canonical" cycles, like Chern classes of something etc. 
According to the Weak Lefschetz theorem, there's no problem in small codimensions: the map $H^{2k}(Gr(3,6)) \to H^{2k}(MG(3,6))$ is an isomorphism for $k < 4$,
and therefore lower codimensional cohomology groups are generated by Chern classes of canonical bundle coming from Grassmannian.
So the question really is: how do I find a nice description of cycles in the middle dimension?
I know for sure, that there is one cycle which doesn't pull-back from Grassmannian.
Remark. The reason I'm looking for a "canonical" representation of cycles is that in fact I have some twisted form $X/F$ of $MG(3,6)/\bar F$.
What I'm really looking at is cohomology (Chow groups, actually) of $X$, so I hope to find a basis of cycles which would descend to the field of definition.
Thanks.

REPLY [4 votes]: Evgeny, here are some remarks on your variety. First, I think it should be possible indeed to prove that the whole group of symmetries of this variety is $G=SL(3,C)xSL(3,C)$ (plus a finite discrete bit, as Evgeny noted below).
One first needs to show that the stabiliser of your 3-form is G,
this is worked out in details in a very nice paper of Hitchin 
"The geometry of three-forms in six and seven dimensions"
http://arxiv.org/PS_cache/math/pdf/0010/0010054v1.pdf 
You can check pages 3-5. Then it is sufficient to show that all symmetries of your variety extend to symmetries of  $G(3,6)$ an my feeling this will be the case...
As for your idea of constructing vanishing cycles, you can try to deform your variety, which just corresponds to deforming the 3-form. Now, you should take the less degenerate 3-form, maybe this will be $dx_1\wedge (dx_2\wedge dx_3+dx_4\wedge dx_5)$? (If not, Hitchin's article should help I guess). You can study the singularites of the deformed variety and if by a miracle the singularity will be just a double point, that can hint you tovards what is the vanishing cycle...
ADDED. Let me try to give an argument that could prove potentially that all automorphism of $M(3,6)=X$ may well come from $GL(6,C)$.
The Grassmanian $G(3,6)=G$ has a natural (Plucker) embedding to $CP^{19}$, the projectivisation of the third exterior power $\Lamda^3(C^6)$. The subvariety $X$ is given by a hyperplane section (indeed, every exerior 3-form of $C^6$ difines a hyperplane $CP^{18}$ in it.
Let us argue first, that Aut(X) extends to $Aut(CP^{18})$. $X$ is a Fano variety with $Pic(X)=Z$. So the bundle $O(1)$ on $CP^{18}$ should be a positive power of the anti-canonical bundle of X. If $O(1)=-nK(X)$  with n positive integer, then, since the action of $Aut(X)$ naturally lifts to the action on $K(X)$, it also acts on the sections of $-nK(X)=O(1)$, and so on $C^{19}$ as well as on its projectivisation $CP^{19}$. If $n$ is just a positive rational nubmer we don't get an action on $C^{19}$, but we still get an action on its projectivisation, I guess. Now we need to show, that in reality this action of $Aut(X)$ on $CP^{18}$ extends to the action on $CP^{19}$ that moreover preserves $Gr(3,6)$. I don't see how to do it for the moment. Maybe it is worth to ask a specialist on Fano varieties...