User m4tth3w - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T03:55:42Z http://mathoverflow.net/feeds/user/23316 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/95510/mathcald-quasi-isomorphisms-and-coherent-omega-modules $\mathcal{D}$-quasi-isomorphisms and coherent $\Omega$-modules M4tth3w 2012-04-29T15:16:44Z 2012-04-30T21:45:42Z <p>Let $X$ be a smooth $\mathbf{C}$-scheme of finite type. In section 7.2 of their preprint "Quantization of Hitchin's Integrable System and Hecke Eigensheaves", Beilinson and Drinfeld define an adjunction <code>$\mathcal{D}:Z^0(\mathbf{dgMod}_{\mathrm{qcoh}}(\Omega^{\cdot}_{X/\mathbf{C}}))\leftrightarrows \mathbf{Cplx}(\mathbf{Mod}^{\mathrm{r}}_{\mathrm{qcoh}}(\mathcal{D}_X))):\Omega$</code> between the category of (closed morphisms of) quasi-coherent dg-modules over the de Rham algebra of $X/\mathbf{C}$ and that of complexes of quasi-coherent right $\mathcal{D}_X$-modules. Their aim is to obtain a derived equivalence between a certain localization of this category of dg-modules over <code>$\Omega^{\cdot}_{X/\mathbf{C}}$</code> and the usual derived category of the abelian category <code>$\mathbf{Mod}^{\mathrm{r}}_{\mathrm{qcoh}}(\mathcal{D}_X)$</code>. The functor $\mathcal{D}$ does not send all quasi-isomorphisms to quasi-isomorphisms and Beilinson and Drinfeld's solution is to invert the morphisms of dg-modules that are sent to quasi-isomorphisms by $\mathcal{D}$, which they call $\mathcal{D}$-<i>quasi-isomorphisms</i>. In 7.2.6(iii) (and also in 2.1.10 of their "Chiral algebras"), Beilinson and Drinfeld assert that any quasi-isomorphism of bounded $\mathcal{O}_X$-coherent dg-<code>$\Omega^{\cdot}_{X/\mathbf{C}}$</code>-modules is a $\mathcal{D}$-quasi-isomorphism (the converse is true without the finiteness hypothesis). In light of Kapranov's paper ("On dg-modules over the de Rham complex and the vanishing cycles functor"), this sounds reasonable.</p> <p>On the other hand, as suggested by 6.23 in <a href="http://wiki.math.toronto.edu/TorontoMathWiki/images/c/ce/MAT1191_Lecture_Notes.pdf" rel="nofollow">these lecture notes</a>, if we take $X=\operatorname{Spec}(\mathbf{C}[t])$ and consider the left $\mathcal{D}_X$-module $\mathcal{O}_X\operatorname{e}^t$ given by the free $\mathcal{O}_X$-module of rank $1$ generated by $\operatorname{e}^t$, i.e. the integrable connection $f\mapsto (f+f')\mathrm{d}t:\mathbf{C}[t]\to\mathbf{C}[t]\mathrm{d}t$, then its de Rham complex appears to be acyclic, $\mathcal{O}_X$-coherent and bounded. I think this means that the image of the corresponding right $\mathcal{D}_X$-module under the functor $\Omega$ is acyclic, bounded and $\mathcal{O}_X$-coherent, hence $\mathcal{D}$-acyclic. As the adjuction morphism $\mathcal{D}\Omega\to\operatorname{id}$ is a quasi-isomorphism by [BD, 7.2.4], the right $\mathcal{D}_X$-module corresponding to $\mathcal{O}_X\operatorname{e}^t$ is zero, which sounds absurd.</p> <p>Question: Have I made a stupid error somewhere and, if not, how does one reconcile this example with Beilinson-Drinfeld's description of $\mathcal{D}$-quasi-isomorphisms for $\mathcal{O}_X$-coherent <code>$\Omega^{\cdot}_{X/\mathbf{C}}$</code>-modules? </p> http://mathoverflow.net/questions/95510/mathcald-quasi-isomorphisms-and-coherent-omega-modules/95610#95610 Comment by M4tth3w M4tth3w 2012-04-30T22:02:48Z 2012-04-30T22:02:48Z Thank you! Great answer!