# Recollements and global dimension

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 $D^b(C)$.

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 $C$.

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.

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?

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The global dimension is not derived invariant: there are non-hereditary algebras derived equivalent to path algebras. –  Mariano Suárez-Alvarez Nov 29 '11 at 4:33
(There are concrete examples, including the one you are interested, at mathoverflow.net/questions/57437/…) –  Mariano Suárez-Alvarez Nov 29 '11 at 4:38
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? –  C Zhu Nov 29 '11 at 4:57
Your derivation of the invariance of global dimension is not very convincing: I'd look there :) –  Mariano Suárez-Alvarez Nov 29 '11 at 5:01
@ Mariano Suárez-Alvarez I have make a mistake! I intended to say the following. Is the finiteness of global dimension a derived invariant? –  C Zhu Nov 30 '11 at 12:23