${\mathbb C}$-module and as an $A \otimes
A^{op}$-module. The first condition amounts to the requirement that $A$ be finite dimensional over ${\mathbb C}$, while the second condition requires that the algebra $A$ be semisimple.

The circle group $\operatorname{SO}(2)$ acts on the classifying space of the $2$-groupoid of fully dualizable objects of ${\mathcal{A}lg}$. In more concrete terms, this means that every fully dualizable object $A \in {\mathcal{A}lg}$ determines a functor from the fundamental $2$-groupoid of $\operatorname{SO}(2)$ into ${\mathcal{A}lg}$ which carries the identity element of $\operatorname{SO}(2)$ to $A$. Applying this to a generator of the fundamental group $\pi_1 \operatorname{SO}(2)$, we get an automorphism of $A$ in ${\mathcal{A}lg}$: this automorphism is given by the vector space dual $A^{\vee}$, regarded as an $(A,A)$-bimodule. To realize $A$ as a fixed point for the action of $\operatorname{SO}(2)$, we need to choose an identification of $A$ with $A^{\vee}$ as $(A,A)$-bimodules. In other words, we need to choose a nondegenerate bilinear form $b: A \otimes A \rightarrow {\mathbb C}$ which satisfies the relations $$b(xa,a') = b(a,a'x) \quad \quad b(ax, a') = b(a,xa').$$ If we set $\operatorname{tr}(a) = b(1,a)$, then the first condition shows that $b(a,a') =
\operatorname{tr}(a'a)$, while the second condition shows that $b(a,a') = \operatorname{tr}(aa')$. It follows 