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 $M$.

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 $2n=\mathrm{dim}\,M$).

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


and the related question


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.

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.


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.

  • 3
    $\begingroup$ I am sorry but I don't know what it means that "the symbol is a real projective hypersurface". To me, the principal symbol of a scalar differential operator of order m on a manifold is a function on the cosphere bundle (with the zero section removed) that is fiber-wise positively homogeneous of order m. Do you mean that its level surfaces are fiber-wise projective hypersurfaces? Or do you refer to a local situation where we have a total symbol and not only a principal symbol? $\endgroup$ – B K Oct 19 '17 at 17:01
  • $\begingroup$ @BK: Maybe he meant the projective algebraic variety corresponding to the symbol. $\endgroup$ – Bombyx mori Oct 19 '17 at 18:08
  • 1
    $\begingroup$ How does it correspond to the symbol? $\endgroup$ – B K Oct 19 '17 at 18:14
  • $\begingroup$ @BK: The principal symbol is homogeneous of degree $n$. So it is cutting out a hypersurface in the projective space. $\endgroup$ – Bombyx mori Oct 19 '17 at 18:20
  • 2
    $\begingroup$ Yes: an operator has a symbol, but if we are only looking for an equation, then we can rescale and get an equivalent equation, so we want the hypersurface of zeroes of the symbol function. This hypersurface is more precisely called the characteristic variety. $\endgroup$ – Ben McKay Oct 19 '17 at 19:33

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.


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.

  • 3
    $\begingroup$ I was about to suggest this. Even if there is no analogue for the Laplace operator, there is a symplectic version of the Dirac operator. But beware! The symplectic spinor representation is infinite-dimensional! $\endgroup$ – Vít Tuček Oct 19 '17 at 16:42
  • 1
    $\begingroup$ Thank you for the reference. While it looks very interesting, one needs to assume that the symplectic manifold admits a metaplectic structure in order to study symplectic spinors. This is not exactly in the spirit of my question because assuming an extra structure besides the symplectic form can of course lead the question ad absurdum: For example, assume that M is Kaehler. Then it is both symplectic and Riemannian and one could argue that there is indeed a spectral geometry of symplectic manifolds, as long as they are Kaehler :-) $\endgroup$ – B K Oct 19 '17 at 17:12
  • 1
    $\begingroup$ @B K: reading the answer by @Dan Fox, it would seem that a metaplectic structure is of "topological" type rather than properly "geometrical". $\endgroup$ – Qfwfq Sep 9 '18 at 20:45

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.

  • $\begingroup$ There are many examples of isospectral manifolds which are not isometric. en.wikipedia.org/wiki/Isospectral#Isospectral_manifolds $\endgroup$ – Vít Tuček Oct 25 '17 at 11:43
  • $\begingroup$ This is true, but this only demonstrates that the metric is not determined by the spectrum of the $\Delta_g$ alone. What I said above still holds. It would be very convenient if linear operators on Hilbert spaces could be understood entirely from the data of their spectrum, but this is too much too hope for in general. $\endgroup$ – Hadrian Quan Oct 26 '17 at 12:49
  • $\begingroup$ @Hadrian Quan I've never seen that formula $\Delta_g(x^ix^j)=g_{ij}(x)$ for the metric in terms of the Laplacian before. I can see why this would be true for normal coordinates about $x$ (which is a little circular), but not more generally. Do you have a reference? $\endgroup$ – Christian Bueno Jun 26 '18 at 6:31

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.

  • $\begingroup$ Thanks for the intriguing answer. Could you add some precise references? $\endgroup$ – j.c. Sep 9 '18 at 20:39
  • 1
    $\begingroup$ For existence of Mp^c structures - Forger, M., Hess, H., Universal metapl. structures and geometric quantization, Comm. Math. Phys. 64 (1979), No. 3, 269–278. For some spectra computation: real Mp - the cited diploma of Wyss, for complex Mp structures - Cahen, M., Gutt, S., La Fuente Gravy, L., Rawnsley, J., On Mpc-str. and sympl. Dirac ops. J. Geom. Phys. 86 (2014) (explicit realization of symplectic spinors by Bargmann spaces, from RT-point of view, (too?) closely related to results of Habermann, Habermann in their monograp on Sympl. Spinors and sympl. Diracs ops by Springer.¨ $\endgroup$ – Svata Sep 10 '18 at 22:08

Your Answer

By clicking "Post Your Answer", you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.