# Finite test for periodicity of a module

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?

• Why would testing whether a module is periodic be a finite problem then? Are you working over finite fields? – Jabby Sep 29 '18 at 21:34
• @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. – Mare Sep 29 '18 at 21:48
• 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. – tj_ Oct 1 '18 at 8:36