Why is there no symplectic version of spectral geometry?

Question

First, recall that on a Riemannian manifold $(M,g)$ the Laplace-Beltrami operator $\Delta_g:C^\infty(M)\to C^\infty(M)$ is defined as $$ \Delta_g=\mathrm{div}_g\circ\mathrm{grad}_g, $$ where the gradient of a function $f\in C^\infty(M)$ is defined by $$\iota_{\mathrm{grad}_g(f)}g=df,$$ $\iota_\bullet$ being the contraction with a vector field, and the divergence of a vector field $X\in \Gamma^\infty(TM)$ is defined by $$ \mathcal{L}_X\,\mathrm{vol}_g=\mathrm{div}_g(X)\,\mathrm{vol}_g, $$ where $\mathcal{L}_\bullet$ is the Lie derivative and $\mathrm{vol}_g$ the Riemannian volume density on $M$.

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On a symplectic manifold $(M,\omega)$ we can proceed by complete analogy and define the symplectic Laplacian $\Delta_\omega:C^\infty(M)\to C^\infty(M)$ as $$ \Delta_\omega=\mathrm{div}_\omega\circ\mathrm{grad}_\omega, $$ where the symplectic gradient of a function $f\in C^\infty(M)$ is defined by $$\iota_{\mathrm{grad}_\omega(f)}\omega=df,$$ and the symplectic divergence of a vector field $X\in \Gamma^\infty(TM)$ is defined by $$ \mathcal{L}_X\,\mathrm{vol}_\omega=\mathrm{div}_\omega(X)\,\mathrm{vol}_\omega, $$ where $\mathrm{vol}_\omega=\frac{1}{n!}\omega^n$ is the symplectic volume form on $M$ (here $2n=\mathrm{dim}\,M$).

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Now, a fundamental difference between Riemannian manifolds and symplectic manifolds is that on the latter we have the Darboux theorem, and a simple computation shows that in Darboux coordinates $q_1,\ldots,q_n,p_1,\ldots,p_n$, the symplectic Laplacian defined above is given by the formula $$ \Delta_\omega = \sum_{i=1}^n\Big(\frac{\partial^2}{\partial p_i\partial q_i}-\frac{\partial^2}{\partial q_i\partial p_i}\Big). $$ By the theorem of Schwarz, this means that $$ \Delta_\omega=0. $$ This seems to be a first hint why we don't have a symplectic version of spectral geometry.

I would like to know:

Can we prove or refer to a theorem of the form:

The Darboux theorem is equivalent to the statement that there is no second order differential operator on $M$ that is invariant under all symplectomorphisms ?

A broader question is:

Is there a deeper reason behind the non-existence of a symplectic version of spectral geometry?

Indeed, recall that on a Riemannian manifold the Laplace-Beltrami is not the only natural differential operator that is invariant under isometries, see

Laplace-Beltrami and the isometry group

https://math.stackexchange.com/questions/833517/is-every-scalar-differential-operator-on-m-g-that-commutes-with-isometries-a/833672#833672

and the related question

https://math.stackexchange.com/questions/1348788/why-is-a-diffeomorphism-an-isometry-if-and-only-if-it-commutes-with-the-laplacia

It would be nice to have an argument why in fact there cannot be symplectic analogues of any of the natural differential operators commuting with isometries of a Riemannian manifold.

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Some comments on the answers received so far:

I have accepted Ben McKay's answer because it comes closest to what I hoped for. It indeed seems to settle the question for scalar differential operators.

As Igor Rivin pointed out, there is a non-trivial study of operators acting on symplectic spinors. It would be interesting to know whether there are more extra structures apart from symplectic spinor bundles that do not correspond to introducing a Riemannian structure but lead to natural differential operators whose study one could call "spectral symplectic geometry".

I am not sure if I really agree with Dan Fox' point of view that a smooth setting in which there are no local invariants should be regarded as part of differential topology rather than differential geometry. Maybe it boils down to the question if global analysis is to be considered part of differential geometry or rather of differential topology. I would say it is part of both.

Answer 1

The characteristic variety (i.e. vanishing locus of the symbol) of a symplectomorphism invariant scalar differential equation is a real projective hypersurface invariant under the group of projectivized linear symplectic transformations. This group acts transitively on the real points of projective space, so preserves no hypersurface.

Answer 2

From a certain point of view the premise of the question is wrong. The study of sympletic manifolds with no additional structure is akin to differential topology rather than differential geometry. From this point of view, that there is no spectral theory of symplectic manifolds is no more surprising than that there is no spectral theory of smooth manifolds. Generally speaking, the symmetries of a second order differential operator, such as a Laplacian, are finite-dimensional, and its existence requires a geometric context.

The distinction here between topology and geometry is the following ill-defined and overly crude dichotomoy. Topology refers to settings (e.g. categories) in which the infinitesimal symmetries (however understood) of the defining structures are infinite-dimensional, whereas geometry refers to setting in which the infinitesimal symmetries are finite-dimensional (maybe infinitesimal is not quite the right word). Proper formulations of such statements require notions like pseudo-groups, sheaves, fiber bundles, prolongations, etc. Put another way, when the defining conditions yield no local invariants, the context is topological, when there are local invariants, the context is geometric. So there is continuous topology, differential topology, symplectic topology, contact topology, etc., in which the symmetries are homeomorphisms, diffeomorphisms, symplectomorphisms, contactomorphisms, etc. and there is always a basic theorem that any two objects in the category are locally equivalent, hence there are no local invariants.

Geometry requires some further structure, generally speaking a connection of some sort. Note that a Laplacian like differential operator is not far from being a connection. In [1] Connes has shown that a smooth compact Riemannian manifold is determined by a spectral triple, which consists of a Hilbert space, an algebra acting on the Hilbert space, and a self-adjoint differential operator (that will be the Dirac operator, a square-root of the Laplacian) that satisfy a series of conditions. In some very rough sense, having the context to make sense of spectral theory determines geometry rather than topology. In one of the other answers there is mentioned the notion of a sympletic spinor. While the bundle of symplectic spinors is defined using what is here being called topological data (spin and metaplectic structures are just refinements of smooth and symplectic structures), the definition of the operator on its sections analogous to the usual Dirac operator requires a connection in some way compatible with the underlying symplectic structure. Thus this notion requires a geometric context too.

From a slightly different point of view, $Sp(n, \mathbb{R})$ is the analogue of $GL(n, \mathbb{R})$, and geometry starts upon passing to a compact subgroup of $Sp(n, \mathbb{R})$, analogous to $O(n)$. The smooth and symplectic frame bundles of a smooth or symplectic manifold are principal bundles for infinite-dimensional groups of diffeomorphisms and symplectomorphisms that contain $GL(n, \mathbb{R})$ and $Sp(n, \mathbb{R})$, but their reductions to $O(n)$ or $U(n)$ bundles are not preserved by the actions of these infinite-dimensional groups.

Unfortunately, because people used to view a symplectic form as an antisymmetric analogue of a Riemannian metric, what here is called "symplectic topology" is often called "symplectic geometry". To my mind this is misleading because it is like calling "differential topology" by the name "differential geometry". When one speaks of differential geometry, one generally has in mind a smooth manifold equipped with a metric. Similarly, symplectic geometry ought to mean a symplectic manifold equipped with some further geometric structure (the most direct analogues with the smooth setting being Hermitian or Kähler metrics; the issue of integrability means that there is no longer a single analogue, rather several).

Not everyone will agree with the terminological distinction being made here nor with how it is made, and I have intentionally avoided making it precise, but it seems to me the key is the distinction between infinite-dimensional and finite-dimensional "symmetries", however one chooses to make it precise.

Answer 3

I don't know that the premise is correct. It seems that there is symplectic spectral geometry. See, for example:

Vassilevich, Dmitri, Spectral geometry of symplectic spinors, J. Math. Phys. 56, No. 10, 103511, 10 p. (2015). ZBL1325.81105.

Answer 4

We talk about spectral geometry because the length spectrum (i.e. the set of lengths of all closed geodesics) on the Riemannian manifold is intimately related to the spectrum of the corresponding Laplace operator (which is an elliptic operator and on the compact manifolds its spectrum is discrete).

However, there is no length on the symplectic manifolds, and 'symplectic Laplacian' not only is not elliptic, but it is most likely its spectrum coincides with $\mathbb{R}$. It seems to me that 'spectral geometry without spectra' is a bit of a stretch

Answer 5

You've probably heard the condition $d\omega=0$ referred to as an "integrability" condition, because it's most analogous to the condition that your Riemannian metric is flat. So I'd respond to your main question by asking what are the isometries of a flat manifold? These are the analogous restrictions you are hoping to place on a second order differential operator which is symplectomorphism invariant. As Ben points out, this is too much to ask for.

A different view on the question is as follows: the geometry of your manifold coming from a choice of Riemannian metric is entirely determined by understanding the analysis of the associated Laplace-Beltrami operator. Yes, $\Delta_g$ is defined in terms of $g$, but it also determines $g$ in turn: in any coordinate patch $U\subset M^m$, given by functions $\{x^1,\ldots,x^n\}$ you can recover the matrix coefficients of the metric by differentiating the appropriate functions, i.e. $\Delta_g(x^ix^j)=g_{ij}(x)$.

Until you've selected a metric, you haven't specified what "geometry" your manifold is endowed with. And understanding that metric is equivalent to understanding it's Laplacian. For this, and the other reasons he mentions, I'd agree with Dan that Symplectic topology is a better name for the subject.

Answer 6

There is not only symplectic Dirac operator of Habermann, but also a symplectic Dirac operator on complex symplectic spinors which exist on any symplectic manifolds and not only on those whose first Chern class is even. They are parallel to the Dirac operators on Spin^c-structures. When the manifold is a homogeneous space, the spectral problem reduces to representation theory. From its point of view, in the case of complex symplectic spinors, the spectra of complex symplectic Diracs are connected to the spectra of the classical Diracs - conformal weight action.