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Let $(M,[\theta])$ be a contact manifold, $\dim M=2n+1$, and denote by $\mathcal{I}^\theta$ the differential ideal generated by the contact form $\theta$.

An exterior differential system on $M$ of the form $$ (M,\mathcal{I}^\theta+\mathcal{I}) $$ can be called a contact EDS on $M$.

A contact EDS on $M$ is naturally interpreted as a PDE imposed on Legendrian submanifolds of $M$: in local contact coordinates, a Legendrian (i.e., $\mathcal{I}^\theta$-integral and $n$-dimensional) submanifold is the graph of the first jet of a function $u$ in $n$ independent variables, and the condition of being also $\mathcal{I}$-integral reads as a second-order PDE on $u$. For instance, if $\mathcal{I}$ is generated by an $n$-form, then one gets a Monge-Ampere equation. (If $\mathcal{I}=0$ one gets the trivial PDE $0=0$, whose solutions are all the Legrendrian submanifolds.)

QUESTION: what is the class of $n$-dimensional scalar second-order PDEs which corresponds to contact EDS on $(2n+1)$-dimensional contact manifolds? Is it larger than the class of Monge-Ampere equations, or not?

I'm really curious, provided that "my" definition of contact EDS above is correct, where it can be found in the literature. (The sum $\mathcal{I}^\theta+\mathcal{I}$ is just a fancy way to say that I'm taking an ideal containing the contact ideal.)

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If one wants the ideal to represent a single second-order PDE on a single unknown function of $n$ variables well-defined up to contact transformations, then the ideal should just be $\mathcal{I}^\theta$ with a single $\mathcal{I}^\theta$-primitive $n$-form $\Upsilon$ adjoined (i.e., a Monge-Ampère system). However, one can have larger ideals than this that represent systems of second-order PDE on a single unknown function of $n$ variables. (These can be either involutive or not.)

One way to see this is to consider any hypersurface $\Sigma$ in the space $\mathcal{V}_n(\mathcal{I}^\theta)$ (which is a smooth manifold of dimension $\tfrac12(n^2{+}5n{+}2)$). You can ask when $\Sigma$ is the space of integral elements of some ideal $\mathcal{J}$ on $M^{2n+1}$, and one finds that this happens if and only if $\mathcal{J}$ is a Monge-Ampère system.

As for involutive second-order systems that contain the contact ideal, there are many (too many to classify).

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  • $\begingroup$ I see: you regard $\mathcal{V}_n(\mathcal{I}^\theta)$ as a bundle over $M$ and, since its fibre at $x\in M$ is the Lagrangian Grassmannian of the $2n$-dimensional space $\ker\theta_x$, which has dimension $\frac{1}{2}(n^2+n)$, you get $\frac{1}{2}(n^2+5n+2)=\frac{1}{2}(n^2+n)+(2n+1)$ overall. Now we have a one-codimensional sub-bundle $\Sigma$, and we wish to show that, if $\Sigma$ is the space of the integral elements of some ideal, then, fibre by fibre, $\Sigma_x$ is an hyperplane section of the Lagrangian Grassmannian (i.e. a Monge-Ampere equation): no idea how to do that - is it classical? $\endgroup$
    – Giovanni Moreno
    Commented Mar 1, 2016 at 14:28
  • $\begingroup$ @GiovanniMoreno: Yes, it's essentially classical. The point is that the Lagrangian Grassmannian in $\ker(\theta_x)$ is the intersection of the full Grassmannian with the space of $\mathrm{d}\theta_x$-primitive $n$-vectors in $\ker(\theta_x)$ at each point $x\in M$. Its hyperplane sections are then given by the $\mathrm{d}\theta_x$-primitive $n$-forms in $\ker(\theta_x)^\ast$, which follows from a dimension count, once you know the dimension of this space, which is a classical fact from symplectic linear algebra. $\endgroup$
    – Robert Bryant
    Commented Mar 1, 2016 at 14:57
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    $\begingroup$ @GiovanniMoreno: (continued). Further, it follows from symplectic linear algebra that, if there is a $k$-form $\psi$ that is not contained in $\mathcal{I}^\theta$ then $k\le n$ and, if $k < n$, then there exists an $(n{-}k)$-form $\phi$ such that $\psi\wedge\phi$ is an $n$-form that is not contained in $\mathcal{I}^\theta$. This shows that any ideal $\mathcal{J}$ that properly contains $\mathcal{I}^\theta$ must contain a nonzero $\mathrm{d}\theta$-primitive $n$-form, implying that the integral elements of $\mathcal{J}$ must lie in a hyperplane section of the Lagrangian Grassmannian. $\endgroup$
    – Robert Bryant
    Commented Mar 1, 2016 at 17:37

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