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Sep 16, 2022 at 7:43 comment added Jan Grabowski Taking $A$ and the zero Poisson bracket, $A_h$ is just the ("undeformed") formal power series algebra over $A$, $A[[h]]$. That is true no matter what properties $A$ has. If $A$ is commutative and the Poisson bracket is non-zero, you will typically get a noncommutative algebra (i.e. a deformation). Does that help?
Sep 14, 2022 at 7:03 comment added Anil Bagchi. But if $A$ is commutative so is $A_h.$ Right? Then what about the commutator $[a_1, a_2],$ for $a_1, a_2 \in A_h\ $? Will it not be zero and so is it's coefficient of degree $1$ term? If yes, then that will imply $A$ has a trivial Poisson structure. Right? But I still think I am misinterpreting something which is actually being meant to say by the two resourceful authors. I am wondering on where did I mess things up. It will be great if you please clarify it to me. Thanks a lot for your help.
Sep 13, 2022 at 7:07 history answered Jan Grabowski CC BY-SA 4.0