The first Weyl algebra is Calabi-Yau

Question

Why is the Weyl algebra $A_1(k)$ over a field $k$ Calabi-Yau?

(My definition of Calabi-Yau is Ginzburg's)

Answer 1

Consider the Koszul resolution $K_\bullet$ of $A=A_1(k)$ as an $A$-bimodule —this was probably first constructed in [Sridharan, R. Filtered algebras and representations of Lie algebras. Trans. Amer. Math. Soc. 100 1961 530--550. MR0130900 (24 #A754)] (which you can find here); you can find a description of the complex here.

Since $K_\bullet$ is of finite length and its homogeneous components are finitely generated free $A$-bimodules, we see that $A$ satisfies the required finiteness conditions.

To compute $Ext_{A^e}^\bullet(A,A\otimes A)$ we have to find the homology of the complex $\hom_{A^e}(K,A\otimes A)$. If you look at it for a while, you will see that there is in fact an isomorphism of complexes of $A$-bimodules $$\hom_{A^e}(K,A\otimes A) \cong K_\bullet$$ so we immediately see that $A$ is Calabi-Yau.