# Why should noncommutative CYs be dgas?

For a (commutative) CY manifold of dimension $n$, Serre duality implies that there are ifunctorial isomorphisms $$\operatorname{Hom}_{D^b(X)}(E,F)\cong\operatorname{Hom}_{D^b(X)}(F,E[n])^*$$ in the bounded derived category of coherent sheaves and $[n]$ denotes the $n$-fold shift.

Kontsevich and Soibelman define a differential graded algebra $A$ to be Calabi-Yau of dimension $n$ if its (perfect) derived category satisfies the formula analogous to those of $D^b(X)$ above. Ginzburg gives an alternative (but inequivalent) definition of the Calabi-Yau property by requiring an $A$-bimodule isomorphism $A\to A^![n]$ where $A^!=\mathbf{R}\operatorname{Hom}_{A^e}(A,A^e)$. Either way, in both of these definitions, a Calabi-Yau algebra is always required to be a differential graded algebra.

Question: Why should a noncommutative CY space be differential graded?

Any thoughts or explanations or references would be much appreciated.

• What else would you expect it to be? – Fernando Muro Mar 29 '14 at 16:08
• I think of Calabi-Yau as being a property of a dga rather than differential graded being a property of a CY algebra... – Sam Gunningham Mar 29 '14 at 16:57
• CY is a structure of a dga not a property. – user36931 Mar 31 '14 at 1:12
• I get that CY is an additional structure on a dga. What I am seeking is some historical/geometric reason that noncommutative CY's are modeled on dgas as opposed to ordinary algebras. In the commutative world, the coordinate ring of a CY variety need not be a dga (I think?), and much noncommutative geometry is done with ordinary algebras. – Steve Mar 31 '14 at 15:37
• @Steve you shouldn't look at ordinary algebras as opposed to dgas, you should rather look at the former contained in the latter. The coordinate ring of a variety $X$ must be understood as a dga $A$ such that $D^b(X)\simeq D(A)$. Of course $A$ is in generaly only defined up to derived Morita equivalence. If $X$ is an affine variety, $A$ is the classical coordinate ring. But for many varieties you need an honest dga (think of projective varieties). – Fernando Muro Mar 31 '14 at 17:54

For any reasonable smooth scheme the category $D^b(X)$ is equivalent to perfect DG-modules over a DG-algebra. So, the world of schemes is embedded into the world of DG-algebras. This is why we think about DG-algebras as about noncommutative spaces.
• Do you know where one could find a precise statement of this fact? I'm familiar with Block's result embedding $D^b(X)$ into the derived category of the Dolbeault dga of $X$, but the notion of a CY algebra predates his result. – Steve Mar 31 '14 at 15:39