User c zhu - MathOverflow 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 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 Comment by C Zhu 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 Comment by C Zhu 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?