most recent 30 from http://mathoverflow.net 2013-05-21T19:37:31Z http://mathoverflow.net/feeds/user/19598 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 C Zhu 2011-11-30T12:23:07Z 2011-11-30T12:23:07Z

@ Mariano Su&#225;rez-Alvarez I have make a mistake! I intended to say the following. Is the finiteness of global dimension a derived invariant?

http://mathoverflow.net/questions/82142/recollements-and-global-dimension C Zhu 2011-11-29T04:57:41Z 2011-11-29T04:57:41Z

It seems that this contradicts to statement of the global dimension on the recollements of the bounded derived categories of algebras. So, where is the mistake?