# Are algebras with invertible linear duals always Frobenius?

Let $$A$$ be a finite dimensional algebra over a ground field $$k$$. The linear dual $$A^* = Hom_k(A,k)$$ is naturally an $$A$$-$$A$$ bimodule. I am interested in those algebras such that $$A^*$$ is an invertible $$A$$-$$A$$ bimodule. That is, there is another $$A$$-$$A$$ bimodule $$L$$ and $$A$$-$$A$$ bimodule isomorphisms $$L \otimes_A A^* \cong A \cong A^* \otimes_A L$$.

One class of algebras which has this property are the Frobenious algebras. One of the classical definitions of a Frobenius algebra is that it is an algebra with an isomorphism of right $$A$$-modules $${A^*}_A \cong A_A$$. If this is an isomorphism of bimodules then this is a symmetric Frobenius algebra. More generally we have $${}_A{A^*}_A \cong {}_A{}^\sigma A_A$$, where the right-hand side is simply $$A$$ as a bimodule but where the left action is twisted by the Nakayama isomorphism $$\sigma$$. In particular since the Nakayama isomorphism is an isomorphism, $$A^*$$ is an invertible bimodule.

Question: If $$A$$ is an algebra such that $$A^*$$ is an invertible bimodule, does $$A$$ admit the structure of a Frobenius algebra?

Upon reviewing some old notes to myself, apparently at one time I believed that the answer to the above question is yes. However I don't remember the reasoning and didn't record a reference. Further, I am suspicious of my old self because in general there are certainly invertible bimodules which do not come from twisting the left action of the trivial bimodule. I would be happy to understand a counterexample or to find out that my old self was right.

One motivation for studying these algebras is that they arise naturally in extended topological field theory. There is a certain variant of 2D framed tqfts (the "non-compact" variant) and these algebras are in bijection with those tqfts with values in the Morita 2-category. So I would also be interested in anything further that could be said about these algebras, even with further assumptions like $$k$$ being characteristic zero.

• Note that the bimodule $A^*$ induces a self-equivalence of the category $A$-Mod which sends ${}_AA$ to ${}_AA^*$. This means that ${}_AA$ and ${}_AA^*$ share all categorically defined properties. In particular since ${}_AA^*$ is injective, this means ${}_AA$ is both injective and projective, and it follows that $A$ is a quasi Frobenius algebra. Also projective and injective modules coincide. This is not my area and I am having trouble finding a concrete example of a finite dimensional quasi Frobenius algebra which is not Frobenius, but that might be a place to start. Oct 13, 2020 at 20:29

For a finite dimensional algebra $$A$$, $$A^{\ast}$$ being an invertible bimodule is equivalent to $$A$$ being self-injective (which is the same as quasi-Frobenius for finite dimensional algebras).

One implication has already been covered in comments. If $$A^{\ast}$$ is invertible, then $$-\otimes_{A}A^{\ast}$$ is a self-equivalence of the right module category, and so sends projectives to projectives. So $$A^{\ast}$$ is projective.

For the other implication, assume $$A$$ is self-injective. Then $$-\otimes_{A}A^{\ast}$$ is left adjoint to $$\operatorname{Hom}_{A}(A^{\ast},-)$$, and it is easy to check that the unit $$A\to \operatorname{Hom}_{A}(A^{\ast},A\otimes_{A}A^{\ast}),$$ which is given by $$a\mapsto[\varphi\mapsto a\otimes\varphi]$$ for $$a\in A$$, $$\varphi\in A^{\ast}$$, is an isomorphism.

But $$\operatorname{Hom}_{A}(A^{\ast},-)$$ is exact and therefore isomorphic to $$-\otimes_{A}L$$, where $$L=\operatorname{Hom}_{A}(A^{\ast},A)$$, by the Eilenberg-Watts theorem. So $$A^{\ast}\otimes_{A}L\cong A$$ as $$A$$-bimodules.

The same argument with left modules shows that $$A^{\ast}$$ has a left inverse, and so $$A^{\ast}$$ is invertible.

For a typical example of a self-injective algebra that is not Frobenius, start with a Frobenius algebra $$A$$ with an indecomposable projective right module $$P$$ such that $$P\otimes_{A}A^{\ast}\not\cong P$$, and take a Morita equivalent algebra $$B$$ that is the endomorphism algebra of a progenerator that contains $$P$$ and $$P\otimes_{A}A^{\ast}$$ as direct summands with different multiplicities.

The simplest example is where $$A$$ is the path algebra of a quiver with two vertices $$v_{1}$$ and $$v_{2}$$, with an arrow $$a$$ from $$v_{1}$$ to $$v_{2}$$ and an arrow $$b$$ from $$v_{2}$$ to $$v_{1}$$, modulo the relations $$ab=0=ba$$. Let $$e_{i}$$ be the idempotent corresponding to vertex $$v_{i}$$, and $$P_{i}=e_{i}A$$ the corresponding indecomposable projective right module.

Then $$B=\operatorname{End}_{A}(P_{1}^{2}\oplus P_{2})$$ is self-injective (since it's Morita equivalent to $$A$$) but not Frobenius. The indecomposable projective corresponding to $$P_{1}$$ under the Morita equivalence occurs with multiplicity two as a summand of $$B$$, but with multiplicity one as a summand of $$B^{\ast}$$.

• This example of a quasi-Frobenious non-Frobenius algebra is isomorphic to an example known already in 1939, see page 624 of Nakayama, Tadasi (1939), "On Frobeniusean algebras. I", Annals of Mathematics, Second Series, Annals of Mathematics, 40 (3): 611–633, doi:10.2307/1968946. This was pointed out by Skowronski and Yamagata. Oct 14, 2020 at 14:23