5
$\begingroup$

Let A be a positively graded algebra such that A_0 has finite global dimension (see "On a common generalization of Koszul duality and tilting equivalence" by Dag Madsen, arXiv.org > math > arXiv:1007.3282)

I am not an expert in this kind of algebras but is it true that, like in the classical case of Koszul algebras, the generalized Koszul algebras are of finite global dimension?

In the classical case where A_0 is semisimple (or of global dimension zero) we have that the global dimension of A is finite.

What happens in the generalized case? Does the same property hold, namely that the global dimension of A is again finite?

Improve this question
$\endgroup$
10
  • 3
    $\begingroup$ Please spell my name correctly :) and the global dimension of $A$ does not have to be finite, neither in the classical nor the generalized case. $\endgroup$
    – Dag Oskar Madsen
    Oct 7, 2013 at 15:40
  • $\begingroup$ I apologize for the incorrect spelling. $\endgroup$
    – Aleksa
    Oct 7, 2013 at 15:59
  • $\begingroup$ But in the thesis "Tilting objects in derived categories of equivariant sheaves" (Christopher Ira Brav) Theorem 4.2.3 claims that in the classical case when A is locally finite and A is noetherian then A has finite global dimension. $\endgroup$
    – Aleksa
    Oct 7, 2013 at 16:03
  • $\begingroup$ With the same assumptions maybe there is finite global dimension in the generalized case? $\endgroup$
    – Aleksa
    Oct 7, 2013 at 16:05
  • 3
    $\begingroup$ Here is where the global dimension comes into play in the classical case: For $B$ Koszul, $B$ (resp. $B^!$) finite dimensional iff $B^!$ (resp. $B$) has finite global dimension. May be this is the thing you are looking for? $\endgroup$
    – Aaron Chan
    Oct 7, 2013 at 19:17

0

Reset to default

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Browse other questions tagged or ask your own question.