For any commutative Frobenius algebra $A$ there is an associated window element $\omega \in A$. If $\mu: A \otimes A \to A$ denotes the multiplication, $1 \in A$ the unit, $b: A \otimes A \to k$ the non-degenerate pairing, and $c: k \to A \otimes A$ the copairing, then the window element is given by

$$\omega = \mu \circ c(1)$$

The window element is important from a TQFT perspective.

Let's suppose we are working over an algebraically closed characteristic zero field. If the window element is a unit, then the Frobenius algebra is semisimple. I am interested in non-semisimple examples. For example $A = k[x]/x^{n+1}$ is a commutative Frobenius algebra where the trace picks off the coefficient of $x^n$. In this case the window element is $\omega = (n+1)x^n$. This element is nilpotent, but it squares to zero.

I have tried a number of other examples, but in all cases I have tried the window element squares to zero. For example this will be the case if $A$ is a graded algebra which satisfies Poincare duality.

My question is whether this must always be the case? Are there commutative Frobenius algebras where the window element is not a unit, but also does not square to zero?