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Frobenius algebras are finite-dimensional algebras together with a compatible inner product. Commutative Frobenius algebras have attracted recent interest because they're equivalent to 2D oriented TQFTs.

6 votes
1 answer
167 views

Commutative Frobenius algebra with non-invertible window element, but not square zero

For any commutative Frobenius algebra $A$ there is an associated window element $\omega \in A$. If $\mu: A \otimes A \to A$ denotes the multiplication, $1 \in A$ the unit, $b: A \otimes A \to k$ the n …
Chris Schommer-Pries's user avatar
9 votes
Accepted

Are there examples of finite-dimensional complex non-semisimple non-commutative symmetric Fr...

Given any finite dimensional algebra $A$, consider the linear dual $\hat{A}= \hom(A, k)$ as an $A$-$A$-bimodule. Then $R = A \oplus \hat{A}$ may be equipped with an algebra structure as follows: $$(a, …
Chris Schommer-Pries's user avatar
11 votes
1 answer
230 views

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 …
Chris Schommer-Pries's user avatar
15 votes
Accepted

Cohomology rings and 2D TQFTs

These 2D TQFTs do not come from extended theories (unless X is discrete). I interpret this as saying that these theories are non-local (in the 2D bordism) and so you will have trouble interpreting the …
Chris Schommer-Pries's user avatar
8 votes
1 answer
446 views

Separable and finitely generated projective but not Frobenius?

Let R be a commutative ring, and $A$ an $R$-algebra (possibly non-commutative). Then $A$ is separable if it is finitely generated (f.g.) projective as an $(A \otimes_R A^{\mathrm{op}})$-algebra. Suppo …
Chris Schommer-Pries's user avatar