All Questions
7 questions
3
votes
0
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73
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Turning a Frobenius algebra into a symmetric algebra via tensor products
Let $A$ be a finite dimensional Frobenius algebra over a field $K$, which means that $A \cong D(A)$ as right $A$-modules. Being symmetric means that $A \cong D(A)$ as $A$-bimodules. Here $D(-)=Hom_K(-,...
- 26.5k
3
votes
0
answers
102
views
Frobenius algebras associated to posets and coalgebra structures
Let $P$ be a finite poset that we assume for simplicity to be bounded (that is it has a global maximum M and minimum m).
Let k be a field, then the classical incidence algebra $kP$ has $k$-vector ...
- 26.5k
5
votes
0
answers
83
views
It there an algebra of the form $B_T$ with global dimension 3?
Let $A$ be the (symmetric Frobenius) algebra $A=K[x]/(x^3) \otimes_K K[x]/(x^3)$ over a field $K$, which is isomorphic to the group algebra of $C_3 \times C_3$, with $C_3$ cyclic of order 3, when $K$ ...
- 7,893
5
votes
1
answer
226
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Frobenius algebras from symmetric polynomials
Let $K$ be a field of characteristic 0 (maybe it works for more general fields) and $K[x_1,...,x_n]$ the polynomial ring in $n$ variables. Let $e_1,e_2,...,e_n$ denote the elementary symmetric ...
- 50.8k
2
votes
0
answers
85
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Algebras from a basis of a Frobenius algebra
Let $A$ be a commutative Frobenius algebra over a field $K$ (we can assume that $A$ is local).
We can assume $A=K[x_1,...,x_r]/I$ for an ideal $I$ with $J^n \subseteq I \subseteq J^2$ where $J=<x_i&...
- 26.5k
5
votes
1
answer
365
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2TQFT and commutative Frobenius algebras
There is an equivalence between the category of commutative finite dimensional Frobenius algebras and 2 dimensional topological quantum field theories, see for example the book by Joachim Kock, which ...
- 3,830
2
votes
0
answers
81
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Characterisation of Frobenius algebras via sequences
Given a commutative Frobenius algebra, finite dimensional over a field $k$.
We assume that the algebra is connected and in fact given by quiver and relations. Let $S$ be the unique simple modules of ...
- 26.5k