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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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    $\begingroup$ This is very far from true it seems to me. For example, why not take a smooth hypersurface $D$ in $P^n$ of very high degree and remove it? I've never read Van den Bergh's paper but presumably it is a precursor to more modern categorical notions of Calabi-Yau which reduce to the ordinary commutative notions in the case of affine varieties. $\endgroup$ Jul 16, 2014 at 2:48
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    $\begingroup$ Why should the dualizing module be trivial?? $\endgroup$
    – abx
    Jul 16, 2014 at 4:24

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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.

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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

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  • $\begingroup$ Sasha's example is perfectly clear. $\endgroup$
    – abx
    Jun 2, 2019 at 4:17
  • $\begingroup$ Is not so clear to me. Isn't it of infinite global dimension? It's Koszul dual is the free algebra on 2 generators. Or I'm missing something? $\endgroup$ Jun 2, 2019 at 4:27
  • $\begingroup$ Sorry. I got confussed with the other example. For Sasha's example, I'm not good in geometría, I can check / compute HH for some álgebra given by generators and relaciona, but $\endgroup$ Jun 2, 2019 at 4:33
  • $\begingroup$ and relations, but I 'm not able to remember the definition of ample divisor, etc $\endgroup$ Jun 2, 2019 at 4:34
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$k[x,y]/(x^2,y^2,xy)$'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..)

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