# When is a quantum affine space $\mathbb{A}^{n}$ Calabi-Yau?

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?

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Are $r_{ij}$ the same as $a_{ij}$ in your question? –  Vladimir Dotsenko Jul 19 '12 at 9:45
Thanks, Vladimir. I corrected the typo. –  Muon Jul 19 '12 at 10:18

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 $n \geq 1$.

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Thanks for your answer, Travas. You are right, the finite global dimension follows from the definition. I still don't quite understand your exercise: Is it obvious that the condition implies n CYness? –  Muon Jul 20 '12 at 20:39
Not immediately obvious to me, but one can prove that it implies n-CY by noticing that the condition is exactly what is needed to have a cyclically supersymmetric superpotential as in the sense of, e.g., Bocklandt-Schedler-Wemyss, and then one can see that the corresponding complex is indeed a projective bimodule resolution of your algebra A, just as in the case where all the $a_{ij}$ are one. –  travis schedler Jul 21 '12 at 8:15