Is the unit in the definition of a symmetric Frobenius algebra necessary?

Question

Consider a symmetric Frobenius algebra without unit, that is, a finite-dimensional complex associative algebra $\delta$ with a linear functional $\epsilon$, such that $\epsilon\circ \delta$ is a non-degenerate symmetric bilinear form, and $$ \epsilon\circ\delta\circ(\delta\otimes \operatorname{id})=\epsilon\circ\delta\circ(\operatorname{id}\otimes\delta)\;.$$ Can every such algebra be equipped with a unique unit to make it a full symmetric Frobenius algebra?

Answer 1

Denote by $V$ the underlying vector space of your algebra and by $\bullet$ the product. Then the nondegenerate bilinear form $\eta$ identifies $V$ with $V^*$. I claim that the preimage of the linear form $\varepsilon$ under this identification is the unit. Denote it by 1. Then, for any $a,b \in V$, we have $$ \eta(1 \bullet a , b) = \varepsilon(1 \bullet a \bullet b) = \eta(1, a \bullet b) = \varepsilon(a \bullet b) = \eta(a, b), $$ where we use the definition of 1 in the third equality.

It follows that $1 \bullet a = a$.