Given a nonsemisimple symmetric algebra B and a non-selfinjective algebra A (all algebras are finite dimensional over a field and connected). Can A and B have isomorphic Hochschild-cohomology rings?
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$\begingroup$ You want $A$ and $B$ connected, presumably? $\endgroup$– Jeremy RickardCommented Jun 13, 2017 at 10:00
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$\begingroup$ yes, they should be connected. I added it to the question. $\endgroup$– MareCommented Jun 13, 2017 at 10:01
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$\begingroup$ As a motivation: I think if the answer to the question is "no", this would give a proof of the Tachikawa conjecture. But I have close to zero experience with Hochschild cohomology. $\endgroup$– MareCommented Jun 13, 2017 at 10:07
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1 Answer
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There are examples of non-semisimple algebras with trivial Hochschild cohomology: for example, the path algebra $C$ of a quiver whose underlying graph is a tree.
Also, for finite dimensional algebras, the Hochschild cohomology algebra of a tensor product of algebras is the tensor product of their Hochschild cohomology algebras.
So take any symmetric algebra $B$, and let $A=B\otimes_kC$ for $C$ as above.
answered Jun 13, 2017 at 11:10