For any commutative [Frobenius algebra](https://en.wikipedia.org/wiki/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?



Commutative Frobenius algebra with non-invertible window element, but not square zero
Assume $A$ is a connected (not necessarily commutative) non-semisimple Frobenius algebra that is finite dimensional over a field of characteristic 0 and given by quiver and relations. (for the commutative case all this reduced to be a local commutative Frobenius algebra that is not a field).


We should have $c(1)= \sum\_{i} y\_i \otimes x\_i$, where $x\_i, y\_i$ for $i=1,...,dim A$
are defined by the condition $b (x\_i \otimes y\_j )= \delta\_{i,j}$ (the existence of such $x\_i, y\_i$ is equivaleng to $A$ being a Frobenius algebra, see lemma 2.11 in the book of Lorenz on representation theory)
Now $\mu c(1)= \sum y\_i x\_i$ (see section 9.1.4 in the book by Lorenz) and
$b (c(1))= \sum b(y\_i \otimes x\_i)=dim A$.
This implies (see Proposition 1.10.18 in the book on representation theory by Zimmermann) that $\mu c(1) /dim A$ is in the socle and thus squares to zero.


Thus the statement is true in the case of commutative local Frobenius algebras when the algebra is not a field extension.


In the non-local case it should be wrong by taking for example $K \times K[x]/(x^2)$ as suggested in the comments by Will Sawin.


By the way for local Frobenius algebras this window element has a purely combinatorial interpretation up to a scalar, namely it is the unique longest (in terms of product of arrows) non-zero element in the quiver algebra.


