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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?

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    $\begingroup$ Do you require the symmetric bilinear form to be non-degenerate? The 1-dimensional non-unital algebra with zero multiplication (i.e. A is generated by x and x^2=0), satisfies your conditions with any linear functional. $\endgroup$ Mar 11, 2021 at 17:44
  • $\begingroup$ Oh yes, sorry, I forgot to say non-degenerate! $\endgroup$
    – Andi Bauer
    Mar 11, 2021 at 19:27

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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 $1 \bullet a = a$.

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  • $\begingroup$ Thanks! In the last expression of the equation the dot should be a comma, right? $\endgroup$
    – Andi Bauer
    Mar 11, 2021 at 20:16
  • $\begingroup$ Had to draw it in pictures to understand it, but yes that works, thanks! Am I correct that the symmetry requirement is necessary for the left- and right unit to be the same? $\endgroup$
    – Andi Bauer
    Mar 11, 2021 at 20:32
  • $\begingroup$ Yes, to both. I replaced the dot with a comma. For the left and right units to be the same you need the bilinear form $\eta$ to be symmetric. $\endgroup$ Mar 11, 2021 at 20:54

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