Let $X$ be a smooth variety over $\mathbb{C}$ and $\mathscr{A}$ a sheaf of twisted differential operators on $X$. The latter comes equipped with a natural filtration and the associated graded algebra $\text{gr } \mathscr{A}$ is identified with $\text{Sym } \mathscr{T}\_X$, the sheaf of functions on the cotangent bundle $T^\*X$.


Both of these algebras have a Poisson structure: $\text{gr } \mathscr{A}$ the one induced by the commutator, and $\text{Sym } \mathscr{T}\_X$ the one arising from the canonical symplectic structure on $T^\*X$. Can anyone explain to me why these Poisson structures coincide under the aforementioned isomorphism?


**Edit:** When I say "sheaf of twisted differential operators" I just mean a quasicoherent sheaf of filtered associative algebras $\mathscr{A}$ on $X$ together with an isomorphism $\text{gr } \mathscr{A} \cong \text{Sym } \mathscr{T}\_X$. The answer probably doesn't involve any difficult computations but I'm pretty murky about what's happening on the cotangent bundle. Any general remarks in that direction, especially from an algebraic perspective, would be helpful.



Poisson structure on the cotangent bundle
Both of these constructions are étale local on $X$. I'm using étale since you seem to want to work in algebraic geometry; it's probably better to say it's local in the classical/analytic topology.


In either case, I only need to show this for $\mathbb{C}^n$. In that case, it's simply a matter of writing things down; my TDO is trivial now, so it's just algebraic differential operators on C^n. It's associated graded is the polynomial ring $\mathbb C[x\_1,\dots, x\_n,\partial\_1,\dots, \partial\_n]$ with the rule that $\{\partial\_i,x\_j\}=\delta\_{ij}$.


On the other hand, if you trace through the definition of the Poisson bivector on $T^\*\mathbb{C}^n$, then what happens? If you want to compute $\{w\_i,x\_j\}$, where $w\_i$ is the dual coordinate to $x\_i$ on $T^\*\mathbb{C}^n$: you turn these into vector fields using the symplectic form (you get $d/dx\_i$ and $-d/dw\_j$), and you pair these using the symplectic form. If you chose the symplectic form $\sum\_i dw\_i\wedge dx\_i$ then you get a cookie, because you exactly get $\{w\_i,x\_j\}=\delta\_{ij}$, and all is right. You'll note, I ignored the Poisson brackets of "like" coordinates, but these follow easily from the same principle.


If you want to do this in a coordinate-free way, what you should do is realize the computations you need to do are:


* $\{f,g\}=0$ for functions on $X$ in both cases (easy)
* $\{Y,f\}=Y(f)$ for $f$ a function on $X$ and $Y$ a vector field on $X$, thought of as a function on $T^\*X$ linear on the fibers.
* $\{Y,Z\}=[Y,Z]$ for $Y$ and $Z$ both vector fields (here $[Y,Z]$ is Lie bracket); at that point you'll be set, since the symmetric powers of the tangent bundle are generated by these.


These are essentially defining properties for a TDO, so we just need to check them for the canonical structure.


The latter 2 equations follow from a basic property: that if you think of a vector field as a function on the cotangent bundle, and turn it into a Hamiltonian vector field using the symplectic form, you'll get the vector field itself on the base and a natural extension of it to the cotangent bundle which you can think of as something like geodesic flow. (In any set of coordinates, it's clear what to do, but the coordinate free version isn't coming to mind). Since we can think of $\{Y,f\}$ or $\{Y,Z\}$ as Lie derivative with respect to this vector field, the desired formulas follow.


