Let $A$, $B$ be two commutative and associative $\mathbb k$-algebras and let $A_\hbar:=A[[\hbar]]$, $B_\hbar:=B[[\hbar]]$ be the corresponding ring of formal series. Of sense [Deformation theory and quantization I; F. Bayen, M. Flato, C. Frønsdal, A. Lichnerowicz, D.Sternheimer; 1978] or [Deformation quantization of Poisson manifolds, I; M. Kontsevich; 1997; http://arxiv.org/abs/q-alg/9709040 ] we consider two star products $*_1$ and $*_2$ on $A$ and $B$. Then $(A_\hbar, *_1)$ and $(B_\hbar, *_2)$ are two associative $\mathbb k [[\hbar]]$-algebras. One can say that $*_1$ and $*_2$ are "weak equivalents" if there is an algebra $\mathbb k [[\hbar]]$-isomorphism between $(A_\hbar, *_1)$ and $(B_\hbar, *_2)$. We also can say that $*_1$ and $*_2$ are "equivalents" if there is a sequence $\{T_r:A\to B\}_{r\in \mathbb N}$ of $\mathbb k$-maps such that

i) $T_0:A\to B$ is an algebra isomorphism.

ii) For all $a,b\in A_\hbar$, $T(a *_1 b)=T(a)*_2T(b)$, where $T:=T_0+T_1\hbar+T_2\hbar^2+\cdots$

It is clear (for me I guess) that "equivalent" implies "weak equivalent" and if we have that $A=B$, then the notion of "equivalent" coincide with the usual one given by [BFFLS].

I have two questions:

1) "Weak equivalence" implies "equivalence" ?

2) If a Lie group $G$ acts on $A$ and $B$, so this action extend $\mathbb k[[\hbar]]$-linearly on $A_\hbar$ and $B_\hbar$. Suppose that $*_1$ and $*_2$ are equivalents, $T:(A_\hbar,*_1)\to (B_\hbar,*_2)$, and suppose also that $T$ is a $G$-morphism. What we can say about the $T_r$ of $T$ ? Are they $G$-morphisms ?

Any comment(s) would be highly appreciated. I'll be so glad if you know and point me some references about this subject. Thank you in advance.


1 Answer 1


to 1) A $\mathbb{k}[[\hbar]]$-linear map between $A[[\hbar]]$ and $B[[\hbar]]$ is necessarily of the form $T = T_0 + \hbar T_1 + \cdots$ with $T_r\colon A \longrightarrow B$ being $\mathbb{k}$-linear and extended $\mathbb{k}[[\hbar]]$-linearily as usual. The proof is rather straightforward, you can find it eg. in my book :)

Then, a $\mathbb{k}[[\hbar]]$-linear map $T$ is invertible iff $T_0$ is invertible. So 1) and 2) are equivalent.

The notion of equivalence of deformations, however, is usually taken to be more strict in the case where $A = B$. Then one requires that $T_0$ is even the identity and not just invertible. The usual classification of star products on Poisson manifolds uses this version, originally developt by Gerstenhaben in his "Deformation of rings and algebras"-papers.

to 2) Since the group $G$ acts in each order of $\hbar$ with the classical action, all the $T_r$ have to be $G$-invariant themselves, once $T$ is invariant. There are a lot of papers on the classification of $G$-invariant star products where you can find these kind of statements, eg. by Bieliavsky-Cahen-Gutt, Neumaier, Fedosov, ...


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