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Results tagged with frobenius-algebras
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user 184
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.
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 …
- 27.5k
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 …
- 27.5k
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 …
- 27.5k
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, …
- 27.5k
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 …
- 27.5k