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Stream: learning: questions

Topic: Darboux theorem and atlases

John Onstead (Feb 03 2026 at 12:35):

If we let $C$ be an arbitrary site, $\mathrm{Atlas}(C)$ is the subcategory of $\mathrm{Sh}(C)$ on sheaves that have an atlas (a covering from the representables satisfying the atlas conditions). This makes $C$ a dense subcategory of $\mathrm{Atlas}(C)$. Common examples are if $C$ is open subsets of Euclidean space with continuous, differentiable, or smooth maps, whose categories of atlases are topological, differentiable, and smooth manifolds respectively.

John Onstead (Feb 03 2026 at 12:36):

In symplectic geometry, Darboux's theorem has the implication that every symplectic manifold looks locally like $\mathbb{C}^n$ with its usual symplectic form, directly like how a smooth manifold looks locally like $\mathbb{R}^n$. This made me wonder if an analogous condition to the smooth manifold atlas holds. That is, true or false: if we let $C$ be the site of open subsets of $\mathbb{C}^n$ with the usual symplectic form and symplectic structure preserving smooth maps between them, will we get an equivalence of categories $\mathrm{Atlas}(C) \cong \mathrm{SympMan}$?

(I tried looking into it myself, but I ran into a roadblock when even trying to define what the morphisms of $\mathrm{SympMan}$ are in the first place. Some sources say symplectomorphisms, but that would make the category a groupoid. Others say lagrangian correspondences, but those aren't always composable. Shouldn't it just be "smooth maps between the underlying smooth manifolds that also preserve the symplectic structure" or is that not a sensible definition?)

John Baez (Feb 03 2026 at 21:42):

If you want noninvertible maps between smooth maifolds, you may want Poisson maps - physicists often use those. Every symplectic manifold has an underlying Poisson manifold, and an invertible Poisson map between symplectic manifolds is the same as a symplectomorphism.

Alonso Perez-Lona (Feb 03 2026 at 21:59):

John Onstead said:

Others say lagrangian correspondences, but those aren't always composable.

For this approach, there has been significant work in constructing the "category" of Lagrangian correspondences as envisioned by Weinstein. As one might guess, the point ends up going to infinity categories to resolve those problematic intersections.

David Michael Roberts (Feb 04 2026 at 08:39):

Sadly John B's link points to a page without the string "Poisson map", but the relevant definition is

A morphism h:M→N of Poisson manifolds is a smooth function between the underlying smooth manifolds which preserves the Poisson brackets

I guess I or someone else here should add the terminology....

John Onstead (Feb 04 2026 at 09:56):

Alonso Perez-Lona said:

For this approach, there has been significant work in constructing the "category" of Lagrangian correspondences as envisioned by Weinstein. As one might guess, the point ends up going to infinity categories to resolve those problematic intersections.

That's interesting! This makes me wonder then- if Weinstein's category is indeed an $(\infty, 1)$ category, then what would be an example of a model category that presents this particular infinity category? That's certainly something I'll have to look more into!

John Onstead (Feb 04 2026 at 10:00):

John Baez said:

If you want noninvertible maps between smooth maifolds, you may want Poisson maps - physicists often use those. Every symplectic manifold has an underlying Poisson manifold, and an invertible Poisson map between symplectic manifolds is the same as a symplectomorphism.

Thanks, that's very useful! So maybe a more appropriate category to consider would be $\mathrm{PoisMan}$ of Poisson manifolds and the Poisson maps between them. Then the question would be to evaluate the properties of the embedding into this category from the category of open subsets of $\mathbb{C}^n$ along with their usual symplectic form and the Poisson maps between them. It seems like it's no longer a dense embedding, but if every Poisson manifold has foliations that are symplectic, then perhaps one can hope the embedding is a strong generator or even a regular generator.

David Michael Roberts (Feb 04 2026 at 10:09):

That's a nice question to ponder: given two non-disjoint Darboux charts on a symplectic manifold, what kind of map is the coordinate change diffeomorphism on the overlap? Note that you can think about things like [[pseudogroup]] of such maps between open subsets of a standard model symplectic space, which gives a certain category where the maps are partially-defined isomorphisms (a more technical version again would be something like the [[Haefliger groupoid]], where you have germs of symplectomorphisms or similar)

John Onstead (Feb 04 2026 at 10:26):

David Michael Roberts said:

given two non-disjoint Darboux charts on a symplectic manifold, what kind of map is the coordinate change diffeomorphism on the overlap

Would it just be a symplectomorphism? Since Darboux' theorem says two symplectic manifolds of the same dimension are locally symplectomorphic with one another.

John Onstead (Feb 04 2026 at 10:26):

John Onstead said:

So maybe a more appropriate category to consider would be $\mathrm{PoisMan}$

Also, I could restrict my attention from the category of all Poisson manifolds to the subcategory on Poisson manifolds that can be equipped with a symplectic form. But how different would those objects be from symplectic manifolds? Some sources list symplectic manifolds as a special case of Poisson manifolds (as if they were Poisson manifolds with extra property), but the fact the morphisms are different means that symplectic manifolds have to be Poisson manifolds with extra structure. Is it possible to have a Poisson manifold and two non-symplectomorphic symplectic structures on it? I guess I'm asking if this is a "property-like structure" or not.

John Baez (Feb 04 2026 at 17:04):

@John Onstead wrote:

if every Poisson manifold has foliations that are symplectic,

Yes, every Poisson manifold has a god-given symplectic foliation; this might be considered the fundamental theorem of Poisson geometry (though I'm not sure people use that phrase).

Is it possible to have a Poisson manifold and two non-symplectomorphic symplectic structures on it?

No.

I guess I'm asking if this is a "property-like structure" or not.

Yes. Ordinary geometers say that to be symplectic is a property of a Poisson manifold. Here's how it goes:

A Poisson manifold can be defined to be a manifold $M$ with a bivector field $\Pi$ obeying

$[\Pi,\Pi] = 0$

where the bracket here is the [[Schouten bracket]]. This is equivalent to whatever other definitions you've seen of Poisson manifold. We call $\Pi$ the Poisson tensor.

A bivector field on $M$ is a section of $\Lambda^2 TM$, and such a thing is equivalent to a smooth antisymmetric bilinear form $\pi$ on $T^\ast M$. Given a Poisson manifold $(M,\Pi)$ whose bilinear form $\pi$ is nondegenerate, $\pi$ determines an isomorphism $TM \cong T^\ast M$ in the usual way, so it gives a smooth nondegenerate antisymmetric form $\omega$ on $TM$. Less obviously,

$[\Pi,\Pi] = 0 \implies d \omega = 0$

Thus, any Poisson manifold whose Poisson tensor is nondegenerate becomes a symplectic manifold.

We can also run this argument backward, taking a symplectic manifold $(M,\omega)$, using the isomorphism $TM \cong T^\ast M$ defined by $\omega$ to get a Poisson tensor $\Pi$, and using

$d \omega = 0 \implies [\Pi, \Pi] = 0$

to get a Poisson manifold $(M,\Pi)$.

In the end we say a symplectic manifold is a Poisson manifold whose Poisson tensor is nondegenerate.

John Onstead (Feb 04 2026 at 19:43):

John Baez said:

Yes. Ordinary geometers say that to be symplectic is a property of a Poisson manifold.
...
Thus, any Poisson manifold whose Poisson tensor is nondegenerate becomes a symplectic manifold.

Thanks for the explanation! So in that case the full subcategory of Poisson manifolds on those whose tensor is nondegenerate really is a category of symplectic manifolds. That makes things a lot easier to think about!

Alonso Perez-Lona (Feb 04 2026 at 22:25):

John Baez said:

Thus, any Poisson manifold whose Poisson tensor is nondegenerate becomes a symplectic manifold.

This is surely obvious, but can one drop the non-degeneracy condition symmetrically? That is, is a Poisson manifold the same as a manifold with a presymplectic structure? Or is there some additional subtlety?

John Baez (Feb 05 2026 at 18:04):

John Onstead said:

So in that case the full subcategory of Poisson manifolds on those whose tensor is nondegenerate really is a category of symplectic manifolds. That makes things a lot easier to think about!

Right! An interesting thing is that bivectors naturally push forward, so the definition on a Poisson map $f: X \to Y$ is that pushing forward the Poisson tensor on $X$ gives you that on $Y$. Only when the Poisson tensor is nondegenerate can we turn it into a 2-form $\omega$, namely a symplectic structure. 2-forms naturally pull back, so the definition of a symplectomorphism is that pulling back the symplectic structure on $Y$ gives the symplectic structure on $Y$. This switcheroo seems weird. But it's not a paradox: symplectomorphisms are automatically diffeomorphisms, locally, so we can reinterpret pulling back along $f$ as pushing forward along $f^{-1}$, locally.

But all this trickery only works when the Poisson tensor is nondegenerate. It's probably less confusing to focus on the category of Poisson manifolds and Poisson maps - or maybe even better, the category of [[Poisson algebras]].

John Baez (Feb 05 2026 at 18:08):

Alonso Perez-Lona said:

John Baez said:

Thus, any Poisson manifold whose Poisson tensor is nondegenerate becomes a symplectic manifold.

This is surely obvious, but can one drop the non-degeneracy condition symmetrically? That is, is a Poisson manifold the same as a manifold with a presymplectic structure?

No, they're completely different beasts!

Or is there some additional subtlety?

There's just no natural way, in general, to turn a 2-form (a section of $\Lambda^2 T^\ast M$) into a bivector field (a section of $\Lambda^2 T M$) or vice versa, unless the 2-form or bivector field is nondegenerate - since the nondegeneracy condition says precisely that it gives an isomorphism $TM \cong T^\ast M$.

Alonso Perez-Lona (Feb 05 2026 at 19:11):

John Baez said:

There's just no natural way, in general, to turn a 2-form (a section of $\Lambda^2 T^\ast M$) into a bivector field (a section of $\Lambda^2 T M$) or vice versa, unless the 2-form or bivector field is nondegenerate - since the nondegeneracy condition says precisely that it gives an isomorphism $TM \cong T^\ast M$.

Hmm, I see, that's unfortunate. I've always found formulations with bivector fields confusing. I like closed differential 2-forms because this provides the starting point for quantization as reinterpreting it (when possible) as the curvature of some prequantum line bundle, which has more of a global flavor. But bivector fields, I don't know how to make sense of in this "lift to a global structure" kind of thing -- and yet they are needed for Poisson-Lie symmetries, which I feel like I don't really understand exactly because of this.

John Baez (Feb 05 2026 at 22:58):

I think it's useful to realize that the algebra of multivector fields is just as rich and interesting as the algebra of differential forms! Either one forms a supercommutative graded algebra under wedge product. The algebra of differential forms also has the differential $d$ making it into a differential graded algebra. Some people say that makes differential forms to multivector fields.

But the algebra of multivector fields has its own special structure, the [[Schouten bracket]] making it into a [[Gerstenhaber algebra]]! Many people learn the Lie bracket of vector fields and stop there, but that's just a special case of the Schouten bracket of multivector fields. The Schouten bracket is the only way to extend the Lie bracket of vector fields that obeys a bunch of nice identities like

$[a, b \wedge c] =$ some obvious thing

Key point: the Schouten bracket of multivector fields contains essentially the same information as the differential on differential forms, but in dual form.

The multivector fields and the differential forms fit together in a larger structure with some more operations. For example a bunch of folks know that you can feed a vector field in a $p$-form and get an $(p-1)$-form. This is a special case of feeding an $n$-vector field into a $p$-form and getting a $(p-n)$-form.

This larger structure with more operations has been called a 'calculus'. I'm having trouble remembering a reference - the closest I can get right now is

• Johannes Huebschmann, Lie-Rinehart algebras, Gerstenhaber algebras and Batalin-Vilkovisky algebras, Annales de l'Institut Fourier, 48 (1998), 425-440.

but I'm thinking that some Russian guy also wrote about this.

Another earlier source is

• Jean-Louis Koszul, Crochet de Schouten-Nijenhuis et cohomologie (Schouten-Nijenhuis bracket and cohomology), in Élie Cartan et les mathématiques d'aujourd'hui - Lyon, 25-29 juin 1984, Astérisque, no. S131 (1985), pp. 257-271.

John Baez (Feb 05 2026 at 23:01):

One last thing: on a Poisson manifold, the Poisson tensor $\Pi$ makes the multivector fields into not merely a Gerstenhaber algebra but something richer: a [[Batalin-Vilkovisky algebra]].

John Baez (Feb 05 2026 at 23:09):

Again, all this is jargon for a bunch of identities that are individually completely easy to believe. There are, however, some more sophisticated ways to think about BV-algebras.

Alonso Perez-Lona (Feb 06 2026 at 15:04):

John Baez said:

One last thing: on a Poisson manifold, the Poisson tensor $\Pi$ makes the multivector fields into not merely a Gerstenhaber algebra but something richer: a [[Batalin-Vilkovisky algebra]].

This is nice! Though I think it is the fact that the nontrivial results go in this direction what makes me not understand their global relevance, given that BV is primarily concerned with perturbative physics (I do however recall a paper by Luigi Alfonsi taking a step in the direction of global BV, to be fair).

John Baez (Feb 06 2026 at 15:30):

I don't know what you mean by "global" here - maybe "nonperturbative"?

Anyway, I don't understand BV algebras in any depth and don't intend to; I really just brought them up because it's hard to get certain information about how multivector fields and differential forms on any manifold form a 'calculus' without looking at papers on BV algebras, which are unfortunately concerned with the case of a Poisson manifold and have more ambitious goals.

To reduce the mystery for anyone reading this, the multivector fields on a manifold form a Gerstenhaber algebra, which is a graded vector space that has a graded-commutative product $\wedge$ and graded-anticommutative bracket $[-,-]$ of degree $-1$ such that

• $\wedge$ is associative
• $[-,-]$ obeys the graded Jacobi identity
• for any $a$, $[a,-]$ is a derivation of the $\wedge$ product.

Alonso Perez-Lona (Feb 06 2026 at 20:11):

John Baez said:

I don't know what you mean by "global" here - maybe "nonperturbative"?

Yes, both really, I was referring to the fact that by interpreting the symplectic form as the curvature of a globally-defined object, namely a line bundle, one may then start the process of geometric quantization, which is nonperturbative. This jump from some element of an algebra (of differential forms) to a groupoid (of line bundles) is what I don't see for vector fields, which is why I mentioned I find them obscure.

Alonso Perez-Lona (Feb 06 2026 at 20:12):

But yeah, perhaps I am effectively asking what are the relevant objects for nonperturbative BV!