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?

  • $\begingroup$ Can you define the widow element using just the Frobenius form of trace rather than the copairing. $\endgroup$ Nov 10, 2022 at 21:03
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    $\begingroup$ 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 $\endgroup$ Nov 10, 2022 at 21:07
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    $\begingroup$ Can you take the product of an algebra where it is a unit and an algebra where it squares to zero? $\endgroup$
    – Will Sawin
    Nov 10, 2022 at 21:37
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    $\begingroup$ 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. $\endgroup$ Nov 10, 2022 at 22:02
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    $\begingroup$ 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 $\endgroup$ Nov 10, 2022 at 22:05

1 Answer 1


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.


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