For a given algebra $B$ over a field $K$ the trivial extension $T(B)$ of $B$ is defined as follows:
The underlying vectorspace is $T(B)=B \oplus D(B)$ where $D(B)=Hom_K(B,K)$ and the multiplication is given by
$(a,q)(a^\prime,q^\prime)=(aa^\prime,aq^\prime+qa^\prime)$ for $a,a^\prime\in B$ and $q,q^\prime\in D(B)$.
Assume you have a symmetric quiver algebra $A=KQ/I$ over a field $K$. Here symmetric means that the algebra is a symmetric Frobenius algebra, or equivalently $ A \cong D(A)$ as $A$-bimodules.
Question: What are easy ways to know whether $A$ is derived equivalent to a trivial extension algebra?
For example $K[x]/(x^2)$ is the trivial extension of the field $K$, but $K[x]/(x^3)$ is not derived equivalent to a trivial extension.
(one might also ask this question with "isomorphic" instead of "derived equivalent", which makes it a bit easier)
For representation-finite symmetric algebras there are two nice ways:
-The number of indecomposable modules is equal to the number of roots for a simply laced Dynkin type for a given number of simples.
-We have $Ext_A^1(M,M)=0$ for any indecomposable $A$-modules $M$.
Probably the first way cant be generalised but maybe the second way has a genealisation by checking an Ext-condition for certain modules. It would be especially interesting to have a good finite test, that can be used with QPA for example (and maybe gives a $B$ with $T(B)$ derived equivalent to $A$).
