Let $A=kQ$ be a path algebra over a field $k$ for a finite acyclic quiver with enveloping algebra $A^e$.

Question: When is it true that $\tau_{A^e}(A) \cong A$ as a left and as a right $A$-modules? (which should mean that $\tau_{A^e}(A)$ is an invertible bimodule)

This holds when $Q$ is the quiver with 3 vertices 1,2,3 and arrows 1->2 and 3->2. Curiously, I found no other examples. I think this is related to the fact that $\tau_{A^e}^2(A) \cong A$ as $A$-bimodules for this example, and it feels like this is the unique example where this holds.

Let $A=kQ$ be a path algebra over a field $k$ for a finite acyclic quiver with enveloping algebra $A^e$.

Question: When is it true that $\tau_{A^e}(A) \cong A$ as a left and as a right $A$-modules? (which should mean that $\tau_{A^e}(A)$ is an invertible bimodule)

This holds when $Q$ is the quiver with 3 vertices 1,2,3 and arrows 1->2 and 3->2. Curiously, I found no other examples. I think this is related to the fact that $\tau_{A^e}^2(A) \cong A$ as $A$-bimodules for this example, and it feels like this is the unique example where this (or even $\tau_{A^e}^k(A) \cong A$ for some $k \geq 1$) holds.

Let $A=kQ$ be a path algebra over a field $k$ for a finite acyclic quiver with enveloping algebra $A^e$.

Question: When is it true that $\tau_{A^e}(A) \cong A$ as a left and as a right $A$-modules? (which should mean that $\tau_{A^e}(A)$ is an invertible bimodule)

This holds when $Q$ is the quiver with 3 vertices 1,2,3 and arrows 1->2 and 3->2. Curiously, I found no other examples.

Let $A=kQ$ be a path algebra over a field $k$ for a finite acyclic quiver with enveloping algebra $A^e$.

Question: When is it true that $\tau_{A^e}(A) \cong A$ as a left and as a right $A$-modules? (which should mean that $\tau_{A^e}(A)$ is an invertible bimodule)

This holds when $Q$ is the quiver with 3 vertices 1,2,3 and arrows 1->2 and 3->2. Curiously, I found no other examples. I think this is related to the fact that $\tau_{A^e}^2(A) \cong A$ as $A$-bimodules for this example, and it feels like this is the unique example where this holds.