Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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?

share|improve this question
5  
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
3  
(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
show 1 more comment

1 Answer

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.

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.)

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.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.