# 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.

-

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...

-
Thanks for this comment. I haven't thought about such an approach. In fact, the variety MG(3,6) can be given a cell decomposition, which implies that all cohomology classes are algebraic. And then, counting cells (which matches counting points) again implies that there's precisely one cycle that doesn't pull-back from Gr(3,6). I didn't mention this fact in the post, since cohomology basis consisting of cells is the one that won't work for me - the cells are not defined rationally for my twisted form X. –  Evgeny Shinder Dec 17 '09 at 22:52
OK, I think I know what the class is now. See my edits above. –  David Speyer Dec 17 '09 at 23:41
Sadly, I was wrong. –  David Speyer Dec 18 '09 at 0:56
David, it's interesting that you noticed the cocycle [A]-[B], since I condsidered it too. And I had the same proof that it doesn't work. :( –  Evgeny Shinder Dec 18 '09 at 17:03

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"

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...

-
Thanks a lot for the reference! I actually thought that the symmetry group of this 3-form was larger than SL_3 * SL_3. In fact, the group IS a little larger, since it has another connected component corresponding to interchanging vectors e_1, e_2, e_3 with e_4, e_5, e_6 respectively (as is noted in Hitchin's paper). The symmetry group of X is larger than that, since we can scale the form, so that there will be a G_m factor. What I'm confused about is why should we try to extend symmetries of X to Gr(3,6)? –  Evgeny Shinder Dec 22 '09 at 17:16
As for your suggestion to deform X, do you mean to try to trace what happens to cycles when we approach the singular fiber? The problem is in a sense very concrete - deformation is given by Lefschetz pencil in a Plucker embedding, so I might try this. Do you a know a nice reference where people are doing such things? –  Evgeny Shinder Dec 22 '09 at 17:30
Yes, you are 100% right that you can exchange e1,2,3 with e4,5,6 (I was not carefull enough). I would like to show that the symmetries of X can be extended to symmetries of G because the symmetries of G is SL(6,C) (this is not hard to prove). Then, if we know that symmetries of X extend to symmetries of Gr, we will know that the symmetries of X are given (modulo finite discrete part) by the subgroup of SL(6,C) that stabilies your form. This stabiliser is calculated in Hitchin's paper. So you will know all the symmetries of X. I can try to add this bit to my answer, if you wish :) –  Dmitri Dec 22 '09 at 18:18
For your second question, I don't know any relevant refference (I was just guessing). Though maybe you will find interesting an article of Richard Thomas NODES AND THE HODGE CONJECTURE arxiv.org/PS_cache/math/pdf/0212/0212216v2.pdf. My idea (maybe a very stupid one) was that if you will get a double point on a deformation, then it might look just like a double point of a singular quadric. And, say, two algebraic cycles that repersent different classes on the smooth guy will collide on the nodal guy. –  Dmitri Dec 22 '09 at 18:43
I'd appreciate if you add more on symmetries. In particular do you mean Aut(X) or the subgroup of GL(6) that keeps X fixed? –  Evgeny Shinder Dec 22 '09 at 19:31

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.

-
I was assuming that $F$ is an aritrary field, since it didn't make any difference so far. The form of $MG(3,6)$ I'm looking at comes from a central algebra $D$ over $F$. What I really would like to understand is whether the algebraic cycle in question is defined over the base field $F$. –  Evgeny Shinder Dec 18 '09 at 20:45
There is an analogy with Severi-Brauer varieties, which are forms of projective spaces. If we consider projective space of dimension p-1 (p-prime), then the cycles that do descend to Severi-Brauer are the fundamental class and p times the other generators. It is a non-trivial theorem due to Merkurjev and Suslin, and I don't think that it could be deduced from the torus action on the projective space using the method you mention. –  Evgeny Shinder Dec 18 '09 at 20:53
I'm unfamiliar with this method though. Is it related to Biyalincki-Birula's cell decomposition coming from Gm-action? –  Evgeny Shinder Dec 18 '09 at 20:54