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Recall that a finite dimensional quiver (always assume connected quiver) algebra is selfinjective iff $P(S) \rightarrow P(socP)$ is a permutation (called the nakayama permutation), when S is simple and P(-) denotes the projective cover. Is there a nice class of $n!$ connected quiver algebras with n simples such that their nakayama permutations realise every of the n! permutations? If possible they should all have the same quiver and just differ by their relations.

Example of "nice class" would be for example selfinjective Nakayama algebras, but they realise only n out of n! many permutations. I can not formally define nice but somehow the algebras should be very similar, like in the nakayama case: All have the same quiver and just different Loewy length mod n. Bonus question: Can you find such a nice class realising all permutations among the representation-finite selfinjective algebras? If not, is there a particular nice class of selfinjective algebras realising the permutations, which also appeared somewhere else (maybe even in applied geometric representation-theory).

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Take a quiver with $n$ vertices, and one arrow $i\to j$ for each pair $(i,j)$ of vertices, including those where $i=j$.

Impose the following relations:

  • all paths of length greater than two are zero
  • all paths of length two are zero unless they are paths from $i$ to $\pi(i)$ for some vertex $i$
  • for each vertex $i$, all paths of length two from $i$ to $\pi(i)$ are equal.
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  • $\begingroup$ This looks like a nice but very forced solution. I leave the topic open for possibly more suggestions and my bonus question. $\endgroup$ – Mare Jan 12 '17 at 10:07
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You may not like this answer -- I am just using (some!) Nakayama algebras.

Let $Q$ be the quiver with vertices $1$ to $n$, and arrows from $i$ to $\pi(i)$. Then kill all relations of length 2.

Now soc($P(S_i)$)=$S_{\pi(i)}$.

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  • $\begingroup$ This is not connected or not selfinjective, or ? $\endgroup$ – Mare Jan 11 '17 at 23:59
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    $\begingroup$ It is self-injective, but it's not connected. (I hadn't noticed that you included connectedness as a condition.) $\endgroup$ – Hugh Thomas supports Monica Jan 12 '17 at 13:42

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