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Is it known when a monoid algebra over a field is symmetric and representation-finite? For groups the answer is nice, so maybe there is a nice generalisation giving conditions on the field and the monoid?

Answer for groups is that the group algebra is always symmetric and representation-finite if the characteric doesnt divide the order or equals p and divides the order and G has only cyclic p-Sylow groups.

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  • $\begingroup$ This is too hard to do. It is equivalent to asking the same for all finite categories. Can you answer this for all monimial algebras $\endgroup$
    – Benjamin Steinberg
    Commented Jun 29, 2017 at 13:21
  • $\begingroup$ Yes thats easy. A monomial algebra is symmetric iff it is a symmetric nakayama algebra. $\endgroup$
    – Mare
    Commented Jun 29, 2017 at 13:50
  • $\begingroup$ Is that easy to see from the relations? I know how to characterize these if the monoid is regular. $\endgroup$
    – Benjamin Steinberg
    Commented Jun 30, 2017 at 11:56
  • $\begingroup$ I should say symmetric not representation finite. $\endgroup$
    – Benjamin Steinberg
    Commented Jun 30, 2017 at 12:00
  • $\begingroup$ @benjaminSteinberg The result is the same. Selfinjective forces indecomposable projectives to have simple socles. For monomial algebras, thats only possible if the monomial algebra is a nakayama algebra, meaning its quiver is a circle. $\endgroup$
    – Mare
    Commented Jun 30, 2017 at 13:57

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If $M$ is a von Neumann regular finite monoid (for each $a\in M$ there is a $b$ with $aba=a$) and $K$ is a field then $KM$ is symmetric and representation finite iff its maximal subgroups have representation finite algebras over $K$ and its sandwich matrices are invertible over the algebras of its maximal subgroups. In this case $KM$ is isomorphic to a direct product of matrix algebras over the group algebras of its maximal subgroups.

The general case seems to me impossible to say much about.

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