# Smooth Affine algebras are Calabi-Yau

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)/

• 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. Jul 16, 2014 at 2:48
• Why should the dualizing module be trivial??
– abx
Jul 16, 2014 at 4:24

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

• Sasha's example is perfectly clear.
– abx
Jun 2, 2019 at 4:17
• 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? Jun 2, 2019 at 4:27
• 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 Jun 2, 2019 at 4:33
• and relations, but I 'm not able to remember the definition of ample divisor, etc Jun 2, 2019 at 4:34

$k[x,y]/(x^2,y^2,xy)$'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..)