4
$\begingroup$

Let $A$ and $B$ be finite-dimensional algebras with finite global dimension over some field (in fact I am thinking of rational incidence algebras of finite posets).

Suppose we know that $A$ and $B$ are derived equivalent (for instance using Ladkani's results on universally derived equivalent posets). Then by Zhao: A note on the equivalence of $m$-periodic derived categories the $2$-periodic derived categories of finite-dimensional modules over $A$ respectively $B$ are triangle-equivalent.

Question: Is there a way to conclude that the $2$-periodic derived categories of countable-dimensional modules over $A$ respectively $B$ are triangle-equivalent as well?

In the non-periodic setting, a derived equivalence gives a perfect tilting module that should yield equivalences of derived categories of modules of any desired size.

The $2$-periodic derived category of $A$-modules is also the derived category of the dg algebra $A[\beta^{\pm 1}]$ with $A$ in degree $0$ and an invertible "Bott" element $\beta$ of degree $2$, and with trivial odd entries and trivial differentials. But at least for dg rings, a derived equivalence need not be induced by a tilting module (see Theorem 3.11 and Proposition 3.14 in Keller's ICM address). Since the counterexample involves torsion and non-trivial differentials, one could still hope that things work out in the nice case at hand where we work with rational graded algebras vanishing in odd degrees.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.