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.

  • 1
    $\begingroup$ What else would you expect it to be? $\endgroup$ – Fernando Muro Mar 29 '14 at 16:08
  • 2
    $\begingroup$ 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 Mar 29 '14 at 16:57
  • $\begingroup$ CY is a structure of a dga not a property. $\endgroup$ – user36931 Mar 31 '14 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 Mar 31 '14 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 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.

  • $\begingroup$ 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 Mar 31 '14 at 15:39
  • $\begingroup$ This scholar.google.ru/… is a good reference. $\endgroup$ – Sasha Mar 31 '14 at 15:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.