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I am going through the chapter Quantization of Lie bialgebras from the book A Guide to Quantum Groups by Chari and Pressley. There I found a notion called Quantization which deals with deformations of Hopf algebras of certain kind. To begin with, the authors first described what is called a quantization of a commutative Poisson-Hopf algebra $A.$ As I said, first of all it is a deformation $A_h$ of $A$ as a Hopf algebra. Secondly, it has to satisfy the following condition $:$

$$\{x_1, x_2 \} \equiv \frac {a_1 a_2 - a_2 a_1} {h}\ (\text {mod}\ h),$$

if $a_1, a_2 \in A_h$ reduce to $x_1, x_2 \in A\ (\text {mod}\ h).$

The second condition is something which I am unable follow properly. First of all if we are begin with a commutative Poisson-Hopf algebra then doesn't that imply that the RHS of that modularity condition is zero or am I misinterpreting of what is actually being meant to say? Secondly, what do the authors mean by the phrase "... if $a_1, a_2 \in A_h$ reduce to $x_1, x_2 \in A\ (\text {mod}\ h)$"? Could anyone please give me some insight on what is happening here? Any help in this regard would be warmly appreciated.

Thanks for your time.

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1 Answer 1

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You didn't give the definition of $A_h$ but if you look there, you should see that elements of it are formal power series in the parameter $h$ with coefficients from $A$. Then "mod $h$" means "take the constant term of the power series, i.e. if $a=\sum a_i h^i$, $a \mod h=a_0$.

Then the condition you give means that if you look at two formal power series in the deformation, the commutation of their degree 1 coefficients with respect to $h$ is controlled by the Poisson bracket (of their degree parts). (Degree 1, from dividing through by $h$ and then taking mod $h$.)

In particular, if the Poisson bracket chosen is globally zero, then commuting of the degree 1 coefficients is undeformed.

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  • $\begingroup$ 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. $\endgroup$ Commented Sep 14, 2022 at 7:03
  • $\begingroup$ 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? $\endgroup$ Commented Sep 16, 2022 at 7:43

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