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Let $A$ and $B$ be Frobenius algebras that are stable equivalent.

In case $A$ is symmetric, is $B$ also symmetric? (no, see the comment of Jeremy Rickard) Does it hold in case $A$ and $B$ are quiver algebras over an algebraically closed field?

Can we characterise when the stable module category of a Frobenius algebra is symmetric (meaning coming from a symmetric algebra) using some local information? One idea might be looking when $\tau(M) \cong \Omega^2(M)$ for all indecomposables $M$, but this is not enough.

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    $\begingroup$ Maybe you want the ground field to be algebraically closed, but there's a counterexample over a non-algebraically closed field at the end of Ohnuki, Yosuke, Takeda, Kaoru, and Yamagata, Kunio. "Symmetric Hochschild extension algebras." Colloquium Mathematicae 80.2 (1999): 155-174. $\endgroup$
    – Jeremy Rickard
    Commented Apr 4, 2020 at 14:17
  • $\begingroup$ @JeremyRickard Thanks, that counts. I added whether it also holds over algebraically closed fields. $\endgroup$
    – Mare
    Commented Apr 4, 2020 at 15:15

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