# When a hyperplane of symmetric forms is determined by a quadric hypersurface?

Let $L$ be a 2D real vector space, $L^*$ its dual, and $\{V,\omega\}$ the symplectic space with $V=L\oplus L^*$ and $\omega$ unambiguously defined by $\omega(l,\lambda):=\lambda(l)$, for all $l\in L$ and $\lambda\in L^*$.

Take now a (nonzero) symmetric tensor $q\in S^2L$, and recall that $S^2L=(S^2L^*)^*$: I'm interested in the hyperplane $\mathcal{E}:=\{q=0\}\subseteq S^2L^*$, which is made of symmetric forms on $L$. More precisely, I'm going to define a quadric hypersurface $\mathcal{Q}:=\{Q=0\}\subseteq V$, such that $\mathcal{E}$ will be the family of Lagrangian subspaces of $V$ which project non-degenerately to $L$ (horizontal, for short), and have nontrivial intersection with $\mathcal{Q}$.

To realize what I mean by the fact that $\mathcal{E}$ is made of horizontal Lagrangian subspaces of $V$, recall that the collection of such subspaces, denoted by $\mathcal{L}_0(V,\omega)$, is an open and dense subset in the whole Lagrangian Grassmannian $\mathcal{L}(V,\omega)$, and it is canonically identified with $S^2L^*$.

Now I have to cook out this $Q\in S^2V^*$, just by using $q$ and $\omega$. To this end, regard $q$ as a homomorphism: $q: L^*\to L$, and take also its adjoint $q^*:L\to L^*$. Since $L$ and $L^*$ are pieces of $V$ (and of $V^*$ as well), the sum $q+q^*$ makes sense, and it can be regarded as an endomorphism of $V$ (or as an enfdomorphism of $V^*$). If $\omega$ is regarded as a map from $V$ to $V^*$, then the commutator $$Q:=[\omega,q+q^*]$$ makes senses, and it produces the form $Q$ I was looking for (direct coordinate computations shows that the above defined homomorphism from $V$ to $V^*$ is symmetric).

So, I managed to prove this result:

any hyperplane $\mathcal{E}$ of symmetric tensors on $L$ can be written as $$\mathcal{E}=\{L_0\in \mathcal{L}_0(V,\omega)\mid L_0\cap \mathcal{Q}\neq 0\},$$ where $\mathcal{Q}$ is a quadric hypersurface in $V=L\oplus L^*$ canonically associated with $q$.

Actually, I just paraphrased a result from the geometric theory of Monge-AmpĂ©re equations in 2 independent variables, without mentioning jet spaces, PDEs, and contact forms. As many (though not all) people interested in the geometry of PDEs, I do not no much about Algebraic Geometry, and I'm afraid to re-discover the wheel. Since I plan to generalize the above result, I'd like to know if something like that already exists in Algebraic Geometry. More precisely,

QUESTION: given a symplectic real vector space $(V,\omega)$ and a symmetric tensor $q$ on an its fixed Lagrangian subspace $L$, it there any canonical way to associate to $q$ a symmetric form $Q$ on $V$, in such a way that the hyperplane of equation $q=0$ in $S^2L^*$ coincides with the subset of $\mathcal{L}_0(V,\omega)$ made of subspaces nontrivially intersecting the quadric hypersurface $Q=0$?

As I explain above, the answer is positive for $V=L\oplus L^*$, and $\dim L=2$, and I also have a clue on how to prove it in general... yet I have a feeling that these things are already known in a field I'm no expert in, and I'd appreciate if anyone suggested a reference for me!

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Is your construction canonical in $dim L=2$? It seems that it requires a choice of splitting, rather than just a Lagrangian subspace. And if you have a splitting in higher dimension, then presumably the same construction works? – Lev Borisov Sep 3 '13 at 11:57
@LevBorisov: at a first sight it looks like one needs a splitting to make it canonical (for any dimension of $L$), but I'm convinced that, in the end, the contributions of the splittings cancel out (one needs two of them). My idea is to use the short exact sequence $0\to L\to V\to L^*\to 0$ associated to any Lagrangian subspace (here $L^*=V/L$ is NOT a subspace of $V$), whose dual is $0\to L\to V^*\to L^*\to 0$, and try to combine $q$ and $q^*$ (acting crosswise between the endpoints) into a linear map between the central points of the sequences... but this is precisely the point I'm stuck to! – Giovanni Moreno Sep 3 '13 at 13:44