Are all smooth affine algebras over a field Calabi-Yau?
I'm thinking yes since they satisfy Van den Bergh duality with dualizing module themselves (have I made a mistake in this reasoning)/
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Sign up to join this communityAre all smooth affine algebras over a field Calabi-Yau?
I'm thinking yes since they satisfy Van den Bergh duality with dualizing module themselves (have I made a mistake in this reasoning)/
If $X$ is smooth projective and $D = \cup D_i \subset X$ is an ample divisor so that $Y = X \setminus D$ is affine, then there is an exact sequence $$ \oplus {\mathbb Z}D_i \to Pic X \to Pic Y \to 0. $$ The canonical class of $Y$ is the image of the canonical class of $X$, so it is trivial if and only if $K_X$ lieas in the subgroup of $Pic X$ generated by the irreducible components $D_i$ of $X$. Clearly, this is not always the case. For example, if $X = P^3$ and $D$ is an irreducible cubic hypersurface then $Y$ is not Calabi-Yau.
I don't think so. If A is commutative and of finite global dimension, say n, finitely generated as k-algebra, then by HKR you have $HH_n(A)\cong\Omega^n(A)$, and you want it to ve isomorphic to $A$ as an $A$-module.This is more or less the same as saying "every manifold is orientable". However, I would like to see an example
$k[x,y]/(x^2,y^2,xy)$'s dualizing module is the injective hull of $k[x,y]/(x^2,y^2,xy)/(x+y)\cong k$ which is just $k$.
So this fails even in dimension $0$.
(But van den Bergh duality does hold so the story is not that sad..)