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A finite dimensional (connected if needed) $K$-algebra $A$ over a field $K$ is called symmetric when $A \cong Hom_K(A,K)$ as $A$-bimodules. Symmetric algebras are Frobenius algebras and include for example any group algebra of a finite group.

Being symmetric implies that all terms of Hochschild cohomology and homology are isomorphic (as $k$-vector spaces). Is the other direction also true, that is:

Question: When all terms of Hochschild cohomology and homology are isomorphic for an algebra $A$, is $A$ symmetric? Is this at least true when $A$ is a Frobenius algebra?

If this is true, maybe one can use this to generalise the definition of symmetric algebras to more general rings.

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    $\begingroup$ I don't have an answer to your good question (which only concerns finite-dimensional algebras). However, I believe that this criterion solely wouldn't be a good generalization for algebras in general, since $k[x]$ is a noetherian ring that satisfies $HH_i(k[x])=HH^i(k[x])$, but it is not self-injective (so, it would be strange to call it symmetric). $\endgroup$
    – Guillerme C. Cruz
    Commented Apr 18, 2022 at 22:09
  • $\begingroup$ @GuillermeC.Cruz True, maybe one should restrict to semiperfect noetherian rings or some sort of that. $\endgroup$
    – Mare
    Commented Apr 19, 2022 at 7:58

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