# $D(A) \otimes D(A)= A$?

Can finite dimensional algebras (over a field $K$) A with $D(A) \otimes_A D(A) \cong A$ as $A$-bimodules be somehow classified? I think taking for A a selfinjective algebra with nakayama permutation of order at most two should work. Are there nonselfinjective examples? Here $D(A)=Hom_K(A,K)$ is the dual of the regular module. Here a little motivation:

Symmetric algebras are characterised by $D(A) \cong A$ as bimodules.

More generally define the cyclic monoid generated by $D(A)$ with multiplication $\otimes_A$, when is this a group? That is, which algebras satisfy $A \cong D(A)^{\otimes i}$ for some $i \geq 1$? Are there non-selfinjective examples? It might also be interesting to look just at one-sided isomorphisms instead of bimodule isomorphisms, where this notion might generalise frobenius algebras.

• This may be obvious for specialists, but for the sake of others, what is $D(A)$? The derived category? And what kind oh isomorphism is allowed?Thanks. – Joël Jun 24 '17 at 0:09
• @Joël sorry, I edited the question. – Mare Jun 24 '17 at 8:22

If $A\cong D(A)^{\otimes i}$as $A$-bimodules, for some $i\geq1$, then $-\otimes_AD(A)$ is a self-equivalence of the module category, and so takes projectives to projectives. But it takes $A$ to $A\otimes_AD(A)\cong D(A)$, so $A$ must be self-injective.
• You mean for some i instead for any $i \geq 1$ or? Well this looks embarrassingly easy. So the answer is all Frobenius algebras with nakayama autormorphism $\psi$ such that $D(A) \cong A_{\psi}$ and $\psi$ with $\psi^i=id$ are exactly those with \$D(A)^{\otimes i}=A? – Mare Jun 24 '17 at 12:05