# On nearly Frobenius algebras

Let $$A$$ be a quiver algebra over a field $$k$$ with multiplication $$m$$. By https://arxiv.org/pdf/1705.10222.pdf definition 6, $$A$$ is called nearly Frobenius in case there exists a (non-zero? Was this forgotten in the definition?) linear map $$\Delta : A \rightarrow A \otimes_k A$$ such that the following holds:

(1) $$\Delta m = (1 \otimes m) (\Delta \otimes 1)$$

(2) $$\Delta m = (m \otimes 1) (1 \otimes \Delta)$$.

The Frobenius dimension is defined as the vector space dimension of the space of such $$\Delta$$ (so $$A$$ is nearly Frobenius iff the Frobenius dimension is non-zero?).

This a generalisation of Frobenius algebras.

Question 1: For Frobenius algebras there are many equivalent characterisations, for example $$A \cong D(A)$$, is there also a characterisation of nearly Frobenius algebras by some isomorphism of modules (or any other non-obvious characterisation)?

Question 2: Is there a way to calculate the Frobenius dimension using the GAP-package QPA?

I realized that I didn't look at the definition careful enough. So I have erased my first answer, since it didn't say anything useful.

The requirements actually says that

$$(1)\quad \Delta(ab) = \Delta(a)b$$

and

$$(2)\quad \Delta(ab) = a\Delta(b).$$

In other words, the linear homomorphism $$\Delta\colon A \to A\otimes_k A$$ should be a homomorphism of $$A$$-$$A$$-bimodules. Therefore the Frobenius dimension of an algebra is nothing else than the dimension of the space of homomorphisms

$$\operatorname{Hom}_{A\otimes_k A^{\operatorname{op}}}(A,A\otimes_k A).$$

This can be easily computed in QPA by the following line of commands:

Aenv := EnvelopingAlgebra( A );
P := DirectSumOfQPAModules( IndecProjectiveModules( Aenv ) );
M := AlgebraAsModuleOverEnvelopingAlgebra( A );
Length( HomOverAlgebra( M, P ) );


The last number is the Frobenius dimension of the algebra $$A$$.

So, Frobenius dimension = $$\dim_k \operatorname{Hom}_{A\otimes_k A^{\operatorname{op}}}( A, A\otimes_k A )$$.

The QPA-team.

• Wow, thanks. This really simplifies it alot. Do we have $Hom_{A^e})(A,A^e) \cong Hom_A(D(A),A)$? If I remember correctly we have $Ext_{A^e}^i(A,A^e) \cong Ext^i_A(D(A),A)$ for $i \geq 1$ and it should probably also hold for $i=1$. – Mare Jan 27 at 22:31
• In Cartan-Eilenberg, Chapter 9, Corollary 4.4 says that $\operatorname{Ext}^n_{A^e}(A,\operatorname{Hom}_k(B,C)) \simeq \operatorname{Ext}^n_A(B,C)$ for a $k$-algebra $A$. Since $\operatorname{Hom}_k(B,C) \simeq \operatorname{Hom}_k(B,k)\otimes_k C$, your claim follows. – Oeyvind Solberg Jan 28 at 7:39
• Thanks, it seems that Cartan-Eilenberg is still the best place to search for such isomorphisms. Your observation actually makes many results on nearly Frobenius quite easy to show and completely answers my question. I wonder whether the authors of the article are aware of this? – Mare Jan 28 at 9:01