# $n$-fold tensor products of $D(A)$ for finite dimensional algebras

Let $$A$$ be a finite dimensional quiver algebra over a field $$K$$ and let $$D(-):=Hom_K(-,K)$$ denote the natural duality (assume algebras are connected).

Define $$\psi_A:=sup \{ n \geq 1 | D(A)^{\otimes n} \neq 0 \}$$ and $$W_A:=D(A)^{\otimes \psi_A}$$.

Here for an $$A$$-bimodule $$M$$, $$M^{\otimes n}$$ denotes the $$n$$-fold tensor product of $$M$$ with itself over $$A$$.

It seems horribly complicated, but in special cases those definitions might have nice properties. Were they studied before?

Note that Nakayama algebras with a linear quiver and $$n$$ simple modules are counted by the Catalan nubmers $$C_{n-1}$$ and thus are in bijection with 321-avoiding permutations on $$n-1$$-symbols. (Nakayama algebras with a linear quiver are exactly the quotient algebras of the ring of upper triangular matrices over the field $$K$$ by an admissible ideal)

Here some observations/guesses with the computer for Nakayama algebras with a linear quiver:

1. $$W_A$$ is injective.

2. There are exactly $$2^{n-2}$$ algebras with $$\psi_A = 2$$ and all of those algebras seem to have global dimension at most 3.

3. The generating function of the statistic $$A \rightarrow \psi_A$$ (http://www.findstat.org/StatisticsDatabase/St001290/) seems to coincide with the generating function on 321-avoiding permutations given by $$\pi \rightarrow f(g(\pi))$$, where $$g$$ is the first fundamental transformation on permutations (http://www.findstat.org/MapsDatabase/Mp00086) and $$f$$ number of right-to-left minima of a permutation (http://www.findstat.org/StatisticsDatabase/St000991).

Now it looks rather horribly to try to prove those things by direct computations but maybe there is a trick or a nice interpretation of $$\psi_A$$ and $$W_A$$ for Nakayama algebras and maybe even more general algebras (maybe QF-3 algebras)?

edit: I found a good reason for 1. and 2. by direct computations with the Nakayama functor applied to the minimal faithful projective-injective module. But maybe there is an argument for 1. that avoids any calculations? 3. remains mysterious.