Let $A$ be a finite dimensional quiver algebra and $M$ a finite dimensional $A$-module. Assume we want to test whether $M$ is a periodic module, meaning that $\Omega^n(M) \cong M$ for some $n \geq 1$. The smallest such $n$ is then called the period of the module. Assume that $A$ has the property that a module is periodic iff it is bounded (meaning that the dimensions of $\Omega^k(M)$ have a finite bound). This condition was first investigated (I think) by Rainer Schulz, see https://www.sciencedirect.com/science/article/pii/0021869386902048 . I think this condition is for example satisfied for all group algebras and I think there are no known examples over finite fields where this is not true. It can be tested with the computer whether a given module has period at most $k$ when one enters $M$ and a number $k$ (for example using the GAP-package QPA). But there is the problem: In case $M$ is periodic, the program eventually stops but if $M$ is not periodic the computer cant say for sure and might calculate forever.

This motivated the following questions:

Is there a useful "bound" s(A,M) that just depends on the vector space dimension of $A$ and $M$ such that $M$ is not periodic in case there exists a $k$ such that $\Omega^k(M)$ has dimension at least $s(A,M)$?

In case such a theoretical result would be available, testing whether a module is periodic would be a finite problem.

I have not much experience about that, but I would guess that $s(A,M)=2 dim(A) dim(M) +30$ might do?

  • 2
    $\begingroup$ Why would testing whether a module is periodic be a finite problem then? Are you working over finite fields? $\endgroup$ – Jabby Sep 29 '18 at 21:34
  • $\begingroup$ @Jabby You are right that there are evil algebras where this criterion does not work. I think over finite fields no such algebra is known. Thanks I added this to the question. $\endgroup$ – Mare Sep 29 '18 at 21:48
  • $\begingroup$ If $s$ is allowed to depend on the ground field $K$, there is a theoretical bound $s$ for finite fields $K$. That's just because for fixed dimensions $a, m$ there are up to isomorphism only finitely many $K$-algebras of dimension $a$ and for fixed $A$ there are up to isomorphism only finitely many $A$-modules of dimension $m$. --- So a criterion for usefulness of the bound $s$ in the case of finite fields might be, that $s$ doesn't depend on the ground field. $\endgroup$ – tj_ Oct 1 '18 at 8:36

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