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4 votes
1 answer
348 views

Do Frobenius subalgebras form a lattice?

A finite-dimensional, unital, associative algebra $A$ over a field $k$ is termed a Frobenius algebra if it is endowed with a nondegenerate bilinear form $\sigma : A \times A \to k$ satisfying the ...
Sebastien Palcoux's user avatar
2 votes
0 answers
114 views

How many minimal relations are needed to obtain a Frobenius algebra?

Let $A_n:=K \langle x_1,x_2,...,x_n \rangle$ be the non-commutative polynomial ring in $n$-variables over the field $K$ and let $J=\langle x_1,...,x_n \rangle$ be the ideal spanned by the $x_i$. An ...
Mare's user avatar
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5 votes
1 answer
224 views

Convolution algebra associated to a finite dimensional algebra

Given a finite dimensional $k$-algebra $A$ (we can assume it is given by a connected quiver with relations). One can form its trivial extension $T(A)$ (see for example https://math.stackexchange.com/...
Mare's user avatar
  • 26.5k
2 votes
1 answer
404 views

Example of a Frobenius algebra that is not projective over a Frobenius subalgebra

I'd like to know an example of a Frobenius algebra $A$, with a subalgebra $B$ that is itself a Frobenius algebra, such that $A$ is not projective as a left $B$-module. I don't require any ...
Alistair Savage's user avatar
4 votes
2 answers
657 views

Central idempotents from characters in Frobenius algebras (generalizing Lusztig arXiv:math/0208154v2 §19)

$\newcommand{\refone}{\textbf{(1)}} \newcommand{\Hom}{\operatorname{Hom}} \newcommand{\tr}{\operatorname{Tr}} \newcommand{\kk}{\mathbf{k}}$ Let $\kk$ be a field. Let $A$ be a $\kk$-algebra which is ...
darij grinberg's user avatar
5 votes
1 answer
524 views

Self-injective basic algebras

Do you know of any self-injective basic algebra $A$ over a field $k$ such that its enveloping algebra $A^{\mathrm{op}}\otimes_k A$ is not self-injective? The algebra $A$ cannot be finite-dimensional, ...
Fernando Muro's user avatar