most recent 30 from http://mathoverflow.net 2013-05-25T19:15:59Z http://mathoverflow.net/feeds/question/82142 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/82142/recollements-and-global-dimension C Zhu 2011-11-29T04:29:03Z 2012-04-28T19:57:54Z

<p>Let $A, B, C$ be algebras. Suppose that $D^b(A)$ (the bounded derived category of $A$) admits a recollement relative to $D^b(B)$ and $D^b(C)$.</p> <p>Then, by a result of Alfred Wiedemann's paper "On stratifications of derived module categories," the algebra $A$ has a finite global dimension if and only if so are $B$ and $C$.</p> <p>Now, suppose that the bounded derived categories of $A$ and $B$ are equivalent. By Rickard's result, $D^b(A)$ admits a recollement relative to $D^b(B)$ and 0. Hence, the global dimension is a derived invariant.</p> <p>But, BGS's Koszul duality give an equivalent between the bounded derived categories of the symmetric algebra and the exterior algebra. Obviously, their global dimension are quite different. So, what's the reason?</p>

http://mathoverflow.net/questions/82142/recollements-and-global-dimension/82149#82149 Chris Brav 2011-11-29T07:35:32Z 2011-11-29T07:35:32Z

<p>As Mariano points out, global dimension of non-commutative algebras is not derived invariant, which surprised me at first since the dimension of regular projective varieties is preserved by derived equivalence. </p> <p>On the other hand, being homologically smooth is an invariant of the differential-graded derived categories. Homologically smooth means that the diagonal bimodule is perfect (has a finite projective resolution by projective bimodules). Alternatively, this means that the identity functor is perfect in the dg category of nice endofunctors of your category. Since perfection of the identity functor is a categorical property, it is preserved by equivalence. (See Toen's The homotopy theory of dg categories and derived Morita theory.)</p> <p>About the symmetric algebra/exterior algebra example. As dg algebras with trivial grading and trivial differential, the symmetric algebra is homologically smooth and the exterior algebra is not, so they can't be derived equivalent. The BGG/BGS equivalence includes a grading on the modules, together with some boundedness conditions, and the Goresky-Macpherson type equivalence involves considering the symmetric algebra and exterior algebra as dg algebras with trivial differential but non-trivial grading.</p>