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

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

2012-07-19T08:45:49Z

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.

Answer by B. Bischof for When is a quantum affine space $\mathbb{A}^{n}$ Calabi-Yau?

2012-07-19T09:58:45Z

The answer to the question "is the functions on quantum affine space a Calabi-Yau algebra?" is yes.

It is presented here (in particular 5.2) in Travis Schedler's notes from MSRI.

However, if this coincides with your alternate definition of CY I do not know. The definition of CY and twisted CY as used in this are 3.27. If I remember correctly the discussion, this is supposed to coincide in some sense with the categorical definition. In particular, by making the Ext condition with respect to Hochschild Homology, but also adding a resolution length condition.

Answer by Travis Schedler for When is a quantum affine space $\mathbb{A}^{n}$ Calabi-Yau?

2012-07-20T18:16:22Z

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 $n \geq 1$.