3
$\begingroup$

Let $\Lambda$ be a finite-dimensional self-injective algebra (over an algebraically closed field, if necessary). Let $Pic(\Lambda)$ be the group of natural isomorphism classes of self-equivalences $mod(\Lambda)\rightarrow mod(\Lambda)$ of the category $mod(\Lambda)$ of finite-dimensional right $\Lambda$-modules. Similarly, let $StPic(\Lambda)$ be the group of natural isomorphism classes of self-equivalences $\underline{mod}(\Lambda)\rightarrow \underline{mod}(\Lambda)$ of the stable category $\underline{mod}(\Lambda)$ of finite-dimensional right $\Lambda$-modules. What can we say about the kernel of the obvious morphism $Pic(\Lambda)\rightarrow StPic(\Lambda)$? Is there any known example of non-trivial element in the kernel? I'm particularly interested in the case of $\Lambda$ being of finite representation type.

$\endgroup$
2
  • 3
    $\begingroup$ Probably not the kind of example you're looking for, but can't you just take $\Lambda=k[x]/(x^2)$, and the self-equivalence induced by the algebra automorphism $x\mapsto\lambda x$ for $\lambda\in k\setminus\{0,1\}$? $\endgroup$ Commented May 19, 2018 at 9:20
  • $\begingroup$ @JeremyRickard nice! Yes, I this looks exactly like what I was looking for. Could you post as an answer so that I can accept it? $\endgroup$ Commented May 20, 2018 at 21:29

1 Answer 1

3
$\begingroup$

Let $\Lambda=k[x]/(x^2)$, which has finite representation type, and take the self-equivalence of the module category induced by the algebra automorphism $x\mapsto\lambda x$ for some $\lambda\in k\setminus\{0,1\}$.

This is non-trivial, since the automorphism is not inner. But the self-equivalence of the stable module category that it induces is trivial.

I don't know any more interesting examples.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .