10
$\begingroup$

Recall that a finite dimensional algebra $A$ over a field $K$is called a Frobenius algebra in case $A \cong D(A)$ as right modules, where $D(A) \cong Hom_K(A,K)$. In case $A \cong D(A)$ as bimodules, then $A$ is called a symmetric algebra. It is well known that a finite dimensional Hopf algebra is a Frobenius algebra.

Question: Is a finite dimensional local Hopf algebra a symmetric algebra?

Examples of local finite dimensional Hopf algebras are group algebras of $p$-groups and it was shown in When is the exterior algebra a Hopf algebra? that for the exterior algebra, it is indeed true that being a Hopf algebra implies that it is symmetric.

Improve this question
$\endgroup$
3
  • $\begingroup$ Could you restate your question in English, please? $\endgroup$
    – Bugs Bunny
    Commented Dec 5, 2018 at 11:07
  • $\begingroup$ @BugsBunny I wrote a new formulation of the question. $\endgroup$
    – Mare
    Commented Dec 5, 2018 at 11:16
  • $\begingroup$ I think the answer is NO. By Oberst-Schneider symmetricity is equivalent unimodularity and $S^2$ being inner. Something like $u_q (n)$ at a root of unity should give a counterexample. $\endgroup$
    – Bugs Bunny
    Commented Dec 10, 2018 at 14:13

0

Reset to default

You must log in to answer this question.