Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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.

share|improve this question
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

2 Answers 2

up vote 7 down vote accepted

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 $n \geq 1$.

share|improve this answer
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

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.

share|improve this answer
Thanks! This is exactly what I am looking for, as long as the definitions coincide. I wonder whether this is easy to prove or not. You are right: I may want to impose "smoothness", or finiteness of global dimension. I think that my example is OK. –  Muon Jul 19 '12 at 10:29

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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