All Questions
6 questions
4
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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 ...
2
votes
0
answers
114
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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 ...
5
votes
1
answer
224
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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/...
2
votes
1
answer
404
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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 ...
4
votes
2
answers
657
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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 ...
5
votes
1
answer
524
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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, ...