I would like to ask a simple question. Let $A=\mathbb{C}\langle x\_{1},\dots,x\_{n} \rangle/I$, where $I$ is the two-sided ideal generated by $x\_{i}x\_{j}=a\_{ij}x\_{j}x\_{i}$ for $1\le i,j\le n$. We say a $\mathbb{C}$-algebra $B$ is n Calabi-Yau if the category of left $B$-modules is a Calabi-Yau n category, i.e. $Ext\_{B}^{i}(M,N)\cong Ext\_{B}^{i-n}(N,M)^{\*}$ for any left $B$-modules and $i \in \mathbb{Z}$. I am not sure if this is equivalent to Ginzburg's definition of CY algebra, but the above definition seems simpler and practical. 


My questions is, when is the above algebra $A$ a $n$ Calabi-Yau? More precisely, what condition on $a\_{ij}$s do we need, other than $a\_{ij}=a\_{ji}^{-1}$? 


Also, if we consider bi-module category (with similar definition), are things much harder? 


Thank you in advance. 



When is a quantum affine space $\mathbb{A}^{n}$ Calabi-Yau? 
Note as in my notes mentioned in B. Bischof's response, there is a nontrivial condition required to be Calabi-Yau instead of merely twisted Calabi-Yau.


Finite global dimension would follow from your condition since in particular it would imply that any $\operatorname{Ext}$ vanishes above degree $n$.


Your definition is, however, missing a finiteness condition. For instance, in the simplest case where $B$ is the polynomial algebra and $B=M=N$, then $\operatorname{Ext}^0(M,N)=B$ but $\operatorname{Ext}^n(M,N)=0$. 


You can fix this problem by requiring that $M$ and $N$ be finite-dimensional. However, the conditions are then not equivalent, but rather Ginzburg's definition implies (but is not implied by) your condition. For example, if $B$ is a Weyl algebra (which is Calabi-Yau) there are no finite-dimensional modules at all.


Here is a proof that Ginzburg's condition implies yours: Suppose B is $n$-CY and $M$ and $N$ are finite-dimensional modules. Let $P^\bullet$ be a projective bimodule resolution of $A$. Thus $P^{\bullet} \otimes\_A M$ is an $A$-module resolution of $M$ and similarly for $N$. Then 


$\operatorname{Ext}^i(M,N)^\* \cong \operatorname{H}^i(\operatorname{Hom}\_{A}(P^{\bullet} \otimes\_A M, N))^\* \cong \operatorname{H}^i(\operatorname{Hom}\_{A^e}(P^{\bullet}, A) \otimes\_{A^e} (M^\* \otimes N))^\*$,


and now using the CY condition, we get 


$\cong \operatorname{H}^i(P^{n-\bullet} \otimes\_{A^e} (M^\* \otimes N))^\*
\cong \operatorname{H}^i(\operatorname{Hom}\_{A}(P^{n-\bullet} \otimes\_A N, M)) \cong \operatorname{Ext}^{n-i}(N,M).$


In the proper setting where $B$ is replaced by a dg algebra which has finite-dimensional homology, then the converse does hold provided you allow $M$ and $N$ to be modules with finite-dimensional homology. Note that it is important here that we work in the dg setting, since otherwise we again have $\operatorname{Ext}^n(B,B)=0$ for all $n \geq 1$.


