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A Poisson structure on a smooth manifold $M$ is a map $C^\infty(M)\times C^\infty(M)\times C^\infty(M)$ satisfying certain conditions.

For a vector space $V$, we can talk about a Poisson structure on $V$ (seeing $V$ as $\mathbb{R}^n$ for some $n$, if necessary).

A Poisson structure on $V$ would then be a map $\{-,-\}_V:C^\infty(V)\times C^\infty(V)\times C^\infty(V)$ satisfying certain conditions. The vector space structure would let us consider a special subset of $C^\infty(V)$, namely the dual space $L(V,\mathbb{R})=V^*$.

One can ask if the map $\{-,-\}_V$ restricts to give a map $[-,-]:V^*\times V^*\rightarrow V^*$. If $\{-,-\}_V$ is such a map, then, we call the Poisson structure $\{-,-\}_V$ to be a linear Poisson structure. The Jacobi condition satisfied by $\{-,-\}_V$ will make sure that $[-,-]$ satisfies Jacobi condition. Thus, gives a Lie algebra structure on $V^*$. So, a linear Poisson structure on a vector space gives a Lie algebra structure on $V^*$.

Calling a vector space with a linear Poissons structure as a linear Poisson vector space, we see that there is a mapping from the collection of linear Poisson vector spaces to the collection of Lie algebras.

Let $\mathfrak{g}$ be a Lie algebra, and $\mathfrak{g}^*$ be the dual of $\mathfrak{g}$ (ignoring the Lie bracket structure). For any vector space $V$, we have the natural (evaluation) map $V\rightarrow V^{**}=L(V^*,\mathbb{R})$. The same would be the case fo $\mathfrak{g}$, we have the map $\varphi:\mathfrak{g}\rightarrow L(\mathfrak{g}^*; \mathbb{R})$. Seeing $L(\mathfrak{g}^*; \mathbb{R})$ as subspace of $C^\infty(\mathfrak{g}^*)$, we can think of $\varphi$ as a map $\varphi:\mathfrak{g}\rightarrow C^\infty(\mathfrak{g}^*)$.

One can ask if there exists a Lie algebra structure on $C^\infty(\mathfrak{g}^*)$ so that, the map $\varphi:\mathfrak{g}\rightarrow C^\infty(\mathfrak{g}^*)$ becomes a morphism of Lie algebras. One can further ask if that structure satisfies the Leibniz condition to become a Poisson structure on $\mathfrak{g}^*$. It turns out such structure exists. By the very definition (asking $\varphi$ to be a Lie algebra morphism), the Poisson structure on $\mathfrak{g}^*$ is seen to be a linear Poisson structure. Thus, given a Lie algebra $\mathfrak{g}$, we get a linear Poisson vector space. Thus, there is a one-one correspondence

$$\left\{\text{linear Poisson vector spaces}\right\}\leftrightarrow\left\{\text{Lie algebras}\right\}.$$

A vector bundle over a manifold $M$ can be seen as a collection of vector spaces that fit together nicely. So, if we want to talk about a linear Poisson structure on a vector bundle $E\rightarrow M$, one possible thing to do is to ask that on each fiber there is a linear Poisson structure that fit together nicely.

Let $E\rightarrow M$ be a vector bundle. Start with a Poisson structure $\pi:C^\infty(E)\times C^\infty(E)\rightarrow C^\infty(E)$. With no extra conditions, this gives a map $\pi_m:C^\infty(E_m)\times C^\infty(E_m)\rightarrow C^\infty(E_m)$ for each $m\in M$.

  • we ask that, these $\pi_m$ are linear Poisson structures on $E_m$ for each $m\in M$.

By above discussion, this gives a Lie algebra structure on $E_m^*$ for $m\in M$.

Thus, starting with a Poisson structure on $E$, assuming some conditions, we get a collection of Lie algebras $\{E_m^*\times E_m^*\rightarrow E_m^*\}$ indexed by elements of $M$. As the vector space structures on fibers $\{E_m\}_{m\in M}$ are nicely fit together, we would want the same from the Lie algebra structures $\{[-,-]_m:E_m^*\times E_m^*\rightarrow E_m^*\}_{m\in M}$. One way to ask this is that, these Lie brackets combine to give a Lie algebroid $E^*\rightarrow M$.

I am thinking about the following question:

How can one formalize the above mentioned idea of asking for some extra conditions to be satisfied by a Poisson structure $\{-,-\}_E$ on $E$ to give a Lie algebroid $E^*\rightarrow M$.

In literature there are already some definitions (I could not fully appreciate them) of linear Poisson structures on vector bundles. Is the definition proposed above differ very much from the existing notion or, is there any obvious relation between existing notion and the one mentioned above?

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  • $\begingroup$ As a Poisson structure on a manifold can be described in terms of a bivector field, some authors describe linear Poisson structure in terms of some extra condition on the bivector field. I am also trying to come up with the a reasonable notion of linearity in terms of bivector field. I see that there is already a definition, but, I could not fully appreciate. So, I am trying to come up with my own version and see if my version matches with the existing definition of a linear bivector field. $\endgroup$ Commented Aug 26, 2021 at 11:04
  • $\begingroup$ Do you want the canonical symplectic structure on the total space of the cotangent bundle to be an example? $\endgroup$ Commented Aug 26, 2021 at 11:18
  • $\begingroup$ @მამუკაჯიბლაძე I am very new to this subject. So, I am not sure what I should be expecting.. If there is any existing notion of canonical Poisson structure on total space of the cotangent bundle, then I would like that to be an example for the definition I proposed... Not very sure why Poisson structure has to be symplectic.. $\endgroup$ Commented Aug 26, 2021 at 11:25

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I don't think that a Poisson structure on $E$ gives you a Poisson structure on the fibers.

If you want to generalize linear Poisson structures I suggest thinking about degree $-1$ Poisson structures on $E$. Recall that functions in $C^{\infty}(E)$ are graded according to their weight under scalar multiplication. So $C^{\infty}(M)$ has degree $0$, and $\Gamma(Sym^{n}(E^{*}))$ has degree $n$. So a weight $-1$ Poisson structure gives you maps

$\Gamma(E^*) \times C^{\infty}(M) \to C^{\infty}(M), \ \ \ \ \ \Gamma(E^*) \times \Gamma(E^*) \to \Gamma(E^*)$ ,

and by Leibniz property these maps determine the rest. I believe if you unpack the Leibniz rule and Jacobi identity for the Poisson structure, you will see that you get a map $\rho: E^{*} \to TM$ and Lie bracket on $\Gamma(E^*)$. In other words, you get the structure of a Lie algebroid on $E^*$.

You get in this way an extension of the bijection you noted above. Now it says that given a vector bundle $E \to M$, there is a bijection between degree $-1$ Poisson structures on $E$ and Lie algebroid structures on $E^*$. The classic example of this is the canonical symplectic structure on the cotangent bundle $T^*M$, which is dual to the Lie algebroid structure on $TM$.

Going back to the original point, you can see that you get a Poisson structure on a fiber $E_{m}$ exactly when $E^{*}_{m}$ has a Lie bracket. And this happens when $\rho$ vanishes at $m$.

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  • $\begingroup$ This and more is explained in this short paper: iopscience.iop.org/article/10.1070/RM1997v052n02ABEH001802 $\endgroup$ Commented Aug 26, 2021 at 11:54
  • $\begingroup$ That paper is behind the paywall.. Can you kindly suggest some free version . $\endgroup$ Commented Aug 26, 2021 at 12:00
  • $\begingroup$ "I don't think that a Poisson structure on $E$ gives you a Poisson structure on the fibers." I am not very happy about it :D I thought it should.. I do not have enough reputation to change the definition so I have to go with the existing definition :) $\endgroup$ Commented Aug 26, 2021 at 12:01
  • $\begingroup$ A general submanifold of a Poisson manifold does not inherit a Poisson structure (related: a subspace of a Lie algebra is not generally a Lie algebra). So I don't think you should expect a Poisson structure on the fibers. But if you want to force it, then you can assume that $\rho = 0$. The result is a bundle of linear Poisson structures, dual to a bundle of Lie algebras. $\endgroup$ Commented Aug 26, 2021 at 12:07
  • $\begingroup$ Ok. It is too much to ask a subspace of Lie algebra to be Loe algebra, but, I thought I was asking something reasonable :P. If I give up that point of view, I have absolutely no idea how can one start thinking about poisson structure on vector bundle.. This idea of weight -1 I am seeing for first time.. Can you suggest some freely available reference.. $\endgroup$ Commented Aug 26, 2021 at 12:50
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It turns out that linear Poisson structure need not (or should not) give a Poisson structure on the fibers. But, the conclusion that a linear Poisson structure on a vector bundle $E\rightarrow M$ should give a Lie algebroid structure on $E^*\rightarrow M$ is still true.

So, let us trace back and see what we get.

A Lie algebroid structure on $E^*\rightarrow M$ comes with a pair of maps;

  • a morphism of vector bundles $E^*\rightarrow TM$,
  • a map $\Gamma(M,E^*)\times \Gamma(M,E^*)\rightarrow \Gamma(M,E^*)$.

Considering the section of vector bundle map gives a map $\Gamma(E^*)\rightarrow \Gamma(M,TM)$. As vector fields are just nice derivations, we can actually see the map $\Gamma(E^*)\rightarrow \Gamma(M,TM)$ as $$\Gamma(E^*)\rightarrow \text{Derivations}(C^\infty(M)\rightarrow C^\infty(M)),$$ or, just as a map $\Gamma(E^*)\times C^\infty(M)\rightarrow C^\infty(M)$.

So, a Lie algebroid structure gives maps $\Gamma(E^*)\times C^\infty(M)\rightarrow C^\infty(M)$, and $\Gamma(M,E^*)\times \Gamma(M,E^*)\rightarrow \Gamma(M,E^*)$.

The above answer by unknownymous says this is what a weight $-1$ Poisson structures come with. I am yet to figure out more about the notion of weighted Poisson structures and motivation for defining it that way.

Please add any references (open access) which you think might be relevant to understand about weighted Poisson structures.

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