One way to view a symplectic manifold $(M,\omega)$ is as a real line bundle $\pi_1: M\times \mathbb{R}\to M$ equipped with a flat connection $d: \Omega^{k}(M, M\times\mathbb{R})\to \Omega^{k+1}(M, M\times \mathbb{R})$ and a form $\omega\in \Omega^2(M,M\times \mathbb{R})$ which is closed and non degenerate, i.e. $d\omega=0$ and the map $v\mapsto \omega(v,\cdot)$ is an isomorphism $T_pM\to (M\times\mathbb{R})_p\otimes T^*_pM$ for each $p\in M$. With this alternative way of thinking, one might define a generalization of symplectic manifolds in the language of line bundles.

I will define a "line-symplectic manifold" as a quadruple $(L, M, d_{\nabla}, \omega)$ where $\pi: L\to M$ is a real line bundle over $M$, $d_{\nabla}: \Omega^{k}(M,L)\to \Omega^{k+1}(M,L)$ is a (possibly flat) connection on $L$ and $\omega\in \Omega^{2}(M,L)$ is a non-degenerate $L$ valued $2$-form with $d_{\nabla}\omega=0$.

We might define a morphism of line-symplectic manifolds $(M,L,d_{\nabla},\omega)\to (M', L', d_{\nabla'}, \omega')$ as a bundle-map $F:L\to L'$, $f: M\to M'$ such that $d_{\nabla}(F^*\alpha)=F^*(d_{\nabla'}\alpha)$ for all $\alpha\in \Omega^k(M',L')$ and $F^*\omega'=\omega$.

There are a few natural questions to ask:

(1) Do there exist non-trivial line-symplectic manifolds? i.e. are there any line-symplectic manifolds that are not isomorphic to $(M, M\times \mathbb{R}, d, \omega)$ for some symplectic manifold $(M,\omega)$?

(2) If there are non-trivial line-symplectic manifolds, does the existence of a line-symplectic structure guarantee that the underlying manifold has a symplectic structure?

(3) If there are non-trivial line-symplectic manifolds, do they share any properties with traditional symplectic manifolds? e.g. can we define a reasonable generalization of the Lie derivative $\hat{\mathcal{L}}$ such that for the line-Hamiltonian vector field of a section $\sigma\in \Gamma(M,L)$, $X_{\sigma}\in \mathfrak{X}(M)$, do we have $\hat{\mathcal{L}}_{X_{\sigma}}\omega=0$? How do the dynamics of line-Hamiltonian vector fields compare to hamiltonian vector fields?


1 Answer 1


Let's assume that your connection is flat. This is a reasonable assumption, since more generally, you would at least need that your isomorphisms preserve the connection 2-form, which is a pretty stringent (non-topological) requirement in dimensions $\geq 4$, so unless you have a good reason to prescribe a specific curvature form, it's best to stick to the flat setting. Then, you're defining what is known as locally conformal symplectic (LCS) geometry. Much of the literature also takes $L$ to be oriented implicitly, though it's not necessary. There are a few definitions of LCS geometry in the literature, so when you look, you'll have to be careful about translating, but let me at least explain the terminology and provide a hint of part of the translation.

LCS geometry essentially falls out of work of Cartan on pseudo-groups (though historically, it really wasn't taken up until work of Vaisman in the 70's and 80's), the idea being that you can define a geometry in terms of atlases but in which transition maps are allowed to be symplectic (i.e. we're talking about legitimate symplectic forms $\Omega$) up to homothety ($\phi^*\Omega = c\Omega$ for $c \neq 0$ locally constant). You may ask how your definition is legitimately symplectic ($d\Omega = 0$) as opposed to twisted-symplectic ($d_{\nabla}\omega = 0$). The basic translation is as follows. The dual bundle $L^*$ is foliated by the graphs of parallel sections, and each leaf comes with a canonical 2-form, with value at $\phi \in L^*_p$ given by $\Omega_\phi := \langle \phi, \omega \rangle$. Then each leaf, aside from the zero section, is legitimately symplectic, but since we don't have a preferred choice of leaf, we get that we are locally working at all scales.

From this perspective, your connection is really defining a holonomy class in $H^1(M;\mathbb{R}^*)$ which is enough to determine an isomorphism of flat bundles. You can then ask for LCS structures for that given holonomy. You can think that if you take the universal cover $\widetilde{M}$, then it comes with a legitimate symplectic form (up to global scale) such that deck transformations act by homotheties. If you require all homotheties to be positive, so that your holonomy class is in $H^1(M;\mathbb{R}_+) \cong H^1(M;\mathbb{R})$, you're in the setting in which $L$ is oriented.

To answer your questions:

(1) Yes. For the simplest example, if $(Y,\xi = \ker \alpha)$ is a contact manifold (meaning $\alpha \wedge d\alpha^n \neq 0$), then $(\mathbb{R} \times Y, d(e^t\alpha) = e^t(dt \wedge \alpha + d\alpha))$ is symplectic, and the $\mathbb{Z}$-action $t \mapsto t + T$ is a symplectic transformation up to homothety. Hence, the quotient $\mathbb{R}/T\mathbb{Z} \times Y$ comes with a natural LCS structure with a nontrivial holonomy class.

(2) No. The answer to (1) implies already that $S^1 \times S^3$ admits such a structure, even though it cannot be symplectic since $H^2(S^1 \times S^3) = 0$. In fact, recent work of Bertelson and Meigniez proves an existence h-principle for LCS structures on any almost symplectic manifold in any nontrivial holonomy class (and in fact you can also specify the LCS structure to further live in any specified twisted second cohomology class).

(3) There's a whole story for the ways in which LCS geometry and symplectic geometry are similar and different, and it would take a long time to state all of the interactions. For classical examples and discussion about the group of Hamiltonian diffeomorphisms, see e.g. work of Banyaga and Haller and Rybicki, the latter of which proves that the group of LCS-Hamiltonian diffeomorphisms is simple.

In response to a comment below (this was too long to comment back), my understanding of the historical motivations is as follows (see also this survey of Bazzoni for more):

(1) E. Cartan was interested in classifying pseudo-groups (1909), which might be considered models for geometry. From his classification naturally popped out (locally) conformal symplectic geometry. (Remark: This may be considered pre-history, since I'm not sure that Cartan himself really studied anything about the field itself, and his work isn't really mentioned in the LCS literature at all.)

(2) The term "symplectic" had already been coined in the late 1930's (e.g. Weyl), but the corresponding "flat" geometry that we now call symplectic geometry had not yet been invented until, as far as I'm aware, Hwa-Chung Lee's 1943 paper. Incredibly, Lee defines not just the “flat” geometry, but the “conformally flat” geometry. Hence, it is actually a historical accident that LCS geometry has remained relatively hidden under the shadows of standard symplectic geometry – the notion of an LCS manifold was developed at the same time!

(3) The resurgence of LCS geometry in the mid-70’s was largely due to work of Vaisman in a series of papers starting with one from 1976, and tended to focus quite a bit on extra metric structure (e.g. locally conformal Kaehler (LCK) manifolds). It is worth noting that such metric aspects, which are a little askew from the more differential topological discussion above, still maintains a vibrant research community.

(4) It is worth saying that some of the most basic tools in symplectic geometry have been imported into the LCS world. I mentioned the Haller-Rybicki work on the simplicity of the LCS Hamiltonian group as an example. The Moser trick was imported by Bande and Kotschick, and the Moser trick can be made relative and parametric so as to give corresponding neighborhood theorems, as in work of Lê and Oh (for coisotropics) and Otiman and Stanciu (more generally).

(5) More recent renewed interest (including my own) was largely motivated by work of Eliashberg and Murphy, where they prove a precursor to the aforementioned Bertelson-Meigniez result, which uses many of the same tools. One hope for the future is to be able to import both rigid and flexible techniques from symplectic geometry into LCS geometry.

  • $\begingroup$ Thanks for the info. Do you know the historical motivations behind studying LCS geometry? $\endgroup$
    – J.V.Gaiter
    May 2 at 20:03
  • 1
    $\begingroup$ I've added an incomplete and biased discussion of the history of LCS geometry to my answer above. $\endgroup$
    – KSackel
    May 3 at 1:43

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