User travis schedler - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T05:54:14Z http://mathoverflow.net/feeds/user/25244 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/102628/when-is-a-quantum-affine-space-mathbban-calabi-yau/102760#102760 Answer by Travis Schedler for When is a quantum affine space $\mathbb{A}^{n}$ Calabi-Yau? Travis Schedler 2012-07-20T18:16:22Z 2012-07-21T13:24:41Z <p>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.</p> <p>Finite global dimension would follow from your condition since in particular it would imply that any $\operatorname{Ext}$ vanishes above degree $n$.</p> <p>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 $\operatorname{Ext}^n(M,N)=0$. </p> <p>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.</p> <p>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 </p> <p><code>$\operatorname{Ext}^i(M,N)^* \cong \operatorname{H}^i(\operatorname{Hom}_{A}(P^{\bullet} \otimes_A M, N))^* \cong \operatorname{H}^i(\operatorname{Hom}_{A^e}(P^{\bullet}, A) \otimes_{A^e} (M^* \otimes N))^*$</code>,</p> <p>and now using the CY condition, we get </p> <p><code>$\cong \operatorname{H}^i(P^{n-\bullet} \otimes_{A^e} (M^* \otimes N))^* \cong \operatorname{H}^i(\operatorname{Hom}_{A}(P^{n-\bullet} \otimes_A N, M)) \cong \operatorname{Ext}^{n-i}(N,M).$</code></p> <p>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 $n \geq 1$.</p>