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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.

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    $\begingroup$ What else would you expect it to be? $\endgroup$
    – Fernando Muro
    Commented Mar 29, 2014 at 16:08
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    $\begingroup$ I think of Calabi-Yau as being a property of a dga rather than differential graded being a property of a CY algebra... $\endgroup$
    – Sam Gunningham
    Commented Mar 29, 2014 at 16:57
  • $\begingroup$ CY is a structure of a dga not a property. $\endgroup$
    – user36931
    Commented Mar 31, 2014 at 1:12
  • $\begingroup$ 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. $\endgroup$
    – Steve
    Commented Mar 31, 2014 at 15:37
  • $\begingroup$ @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). $\endgroup$
    – Fernando Muro
    Commented Mar 31, 2014 at 17:54

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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.

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  • $\begingroup$ 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. $\endgroup$
    – Steve
    Commented Mar 31, 2014 at 15:39
  • $\begingroup$ This scholar.google.ru/… is a good reference. $\endgroup$
    – Sasha
    Commented Mar 31, 2014 at 15:48

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