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.


1 Answer 1


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.

  • $\begingroup$ 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. $\endgroup$ Sep 14 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? $\endgroup$ Sep 16 at 7:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.