that $\operatorname{tr}$ is a [*trace*]{} on the algebra $A$: that is, it vanishes on all commutators $[a,a'] = aa' - a'a$. Conversely, given any linear map $\operatorname{tr}: A \rightarrow {\mathbb C}$ which vanishes on all commutators, the formula $b(a,a') = \operatorname{tr}(aa')$ defines a bilinear form $b$ giving a map of $(A,A)$-bimodules $A \rightarrow A^{\vee}$. We say that $\operatorname{tr}$ is [*nondegenerate*]{} if this map is an isomorphism. A pair $(A, \operatorname{tr})$ where $A$ is a finite dimensional algebra over ${\mathbb C}$ and $\operatorname{tr}$ is a nondegenerate trace on $A$ is called a [*Frobenius algebra*]{}.

We can summarize the above discussion as follows: giving a fully dualizable object of ${\mathcal{A}lg}$ which is fixed under the action of the group $\operatorname{SO}(2)$ is equivalent to giving a semisimple Frobenius algebra $(A,\operatorname{tr})$. Theorem \[bdo\] implies that every such pair $(A,\operatorname{tr})$ determines an extended TQFT $F: {\mathbf{Bord}}_{2}^{\operatorname{SO}} \rightarrow {\mathcal{A}lg}$ such that $F(\ast) \simeq A$. The value of $F$ on the circle $S^1$ can be identified with the $( {\mathbb C},
{\mathbb C})$-vector space given by the tensor product $A \otimes_{ A \otimes
A^{op} } A$: in other words, the quotient of $A$ by the subspace $[A,A]$ generated by all commutators. Using the self-duality of $F(S^1)$ provided 