Geometry of the multilagrangian Grassmannian  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.
 A: This is a partial answer to your question: I believe there is exactly one class not coming from $G(3,6)$, and that this class will be represented by an algebraic cycle. However, I don't have a description of it.
My reasoning: Let's count points on your variety over a field with $q$ elements. I get
$$1+q+2q^2+3 q^3+ 4 q^4 + 3q^5 + 2 q^6 + q^7 + q^8.$$
(More details below.)
I have checked that your variety is smooth, so the Weil conjectures tell us that $H^4$ is four dimensional. Three of those dimensions come from $H^4(G(3,6))$, leaving one unexplained. Moreover, all the classes from $G(3,6)$ lie in $H^{2,2}$, so $H^{2,2}$ of your variety has dimension at least $3$. But then, by the symmetry of the Hodge diamond, the missing class is also in $H^{2,2}$. Assuming the Hodge conjecture, some multiple of it must be an algebraic class.
The point count: Write a point in your variety as the row span of a $3 \times 6$ matrix $\left( A \ B \right)$. Your multi-Lagrangian condition is that $\det A=\det B$. This means either $A$ and $B$ both have rank $3$, or neither does. But your matrix is required to have full rank, so the ranks of $A$ and $B$ can be either $(3,3)$, $(2,2)$, $(2,1)$ or $(1,2)$. Remember also that we are counting not just $3 \times 6$ matrices obeying these conditions, but orbits of such matrices under the left action of $GL_3$. I get
Rank $(3,3)$: $(q^3-1)(q^3-q)q^2$ points. 
Rank $(2,2)$: $(q^2+q+1)(q^2+q+1)(q^3+q^2-q-1)$ points. 
Rank $(2,1)$: $(q^2+q+1)(q^2+q+1)$ points. 
Rank $(1,2)$: $(q^2+q+1)(q^2+q+1)$ points.

The following is the record of an incorrect solution, left as a warning to others. Let $A$ be the subvariety of all three planes which are of the form $u \wedge v \wedge w$ where $u$ and $v$ are in $\mathrm{Span}(e_1, e_2, e_3)$ and $w$ is in $\mathrm{Span}(e_4, e_5, e_6)$. Let $B$ be the similar subvariety where I switch the roles of $123$ and $456$.
It is easy to show that $[A]-[B]$, in $H^4(MG(3,6))$, is orthogonal to the classes obtained by restriction from $G(3,6)$. This would suggest that $[A]-[B]$ is our missing class. Unfortunately, it turns out that $[A]-[B]=0$. Proof: Let $S$, in $G(3,6)$, be the locus of those $3$-planes which meet $\mathrm{Span}(e_1,e_2,e_3)$ in dimension $2$. Then $S \cap MG(3,6)=A$, and the intersection is transverse. Let $T$ be those $3$-planes which similarly meet  $\mathrm{Span}(e_4,e_5,e_6)$; so $B=T \cap MG(3,6)$. But $S$ and $T$ are homologous in $G(3,6)$, so $A$ and $B$ are homologous in $MG(3,6)$. Grrrr...
A: 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...
A: I don't have a good, explicit description of the missing class. But I want to point out how we could compute anything we want to know about $H^*(MG(3,6))$. The group $SL_3 \times SL_3$ acts on $MG(3,6)$; the first $SL_3$ acting on the first three components of $6$-space and the second $SL_3$ acting on the last $3$ components. Let $T$ be a maximal torus in $SL_3 \times SL_3$; $T$ has dimension $4$.
There are $18$ fixed points for the $T$-action on $MG(3,6)$. Namely, they are of the form $e_i \wedge e_j \wedge e_k$ where $1 \leq i < j < k \leq 6$, and $ijk$ is not $123$ or $456$. There are $72$ one-dimensional orbits; they join $e_i \wedge e_j \wedge e_k$ to $e_i \wedge e_j \wedge e_{k'}$. And $MG(3,6)$ is a smooth projective variety. 
This means that we can compute $H^*(MG(3,6))$ using the "moment graph" method of Goresky, Kottwitz and Macpherson. See Tymoczko's excellent introduction.
In particular, what do you actually need? It seems to me that you need to know how $\mathrm{Aut}(MG(3,6))$ acts on $H^4(MG(3,6))$. Your twisted form comes from a class in the group cohomology $H^1(G_F, \mathrm{Aut}(MG(3,6)))$. (Here $G_F$ is the Galois group of $\overline{F}$ over $F$.) You need to know what that class maps to in $H^1(G_F, \mathrm{Aut}(H^4(MG(3,6))))$. Your question about descending cohomology classes should be some simple computation about this cocycle. If your base is a function field over $\mathbb{C}$, then I know how to do this, but your language makes me suspect you are working over a number field, and I don't know quite the right details there.
My point is that the moment graph method is explicit enough that we can compute the action of $\mathrm{Aut}(MG(3,6))$ on $H^4(MG(3,6))$, and I think that is all that you need.
A: Note: This is not an answer to your question.
Reading this question I was reminded of the paper Isotropic subspaces of polylinear forms by Tevelev, which might point to relevant literature.
