# Commutative Frobenius algebra with non-invertible window element, but not square zero

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?

• Can you define the widow element using just the Frobenius form of trace rather than the copairing. Nov 10, 2022 at 21:03
• Did you try avoiding local rings? A Frobenius local commutative ring has a unique minimal ideal and it must square to zero and I guess your element lives there. Probably something similar happens if your ring has no field direct factors Nov 10, 2022 at 21:07
• Can you take the product of an algebra where it is a unit and an algebra where it squares to zero? Nov 10, 2022 at 21:37
• Every finite dimensional algebra is a direct product of local ones. The local rings are the indecomposable ones. So you probably get it is square zero iff no local direct factor is a field. Nov 10, 2022 at 22:02
• If A is a commutative Frobenius local ring over an algebraically closed field then the socle is one dimensional. In the case that A is not a field it is a square zero ideal by nilpotence if the radical. A form on A is a Frobenius form iff it doesn't vanish on the socle. This is why I guess your window element is in the socle but I am not very familiar with the category definition of Frobenius algebra Nov 10, 2022 at 22:05

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.
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.