uniquely determined by  $C$.

It is possible to state a more precise version of Theorem \[bd\] by describing the class of fully dualizable objects $C \in {\mathcal{C}}$ without mentioning the $n$-category ${\mathbf{Bord}}_{n}^{\operatorname{fr}}$. If $n=1$, this condition is easy to state: the object $C$ should admit a dual $C^{\vee}$, so that there exist evaluation and coevaluation maps $$C \otimes C^{\vee} \stackrel{e}{\rightarrow} {\bf 1} \quad \quad { \bf 1} \stackrel{c}{\rightarrow} C^{\vee} \otimes C$$ (here ${\bf 1}$ denotes the unit object of ${\mathcal{C}}$) which are compatible in the sense that the compositions $$C \rightarrow C \otimes C^{\vee} \otimes C \rightarrow C$$ $$C^{\vee} \rightarrow C^{\vee} \otimes C \otimes C^{\vee} \rightarrow C^{\vee}$$ both coincide with the identity. For $n > 1$, we need to assume that analogous finiteness assumptions are satisfied not only by the object $C$, but by the $1$-morphisms $e$ and $c$. We refer the reader to [@L] for a more complete discussion.

Theorem \[bd\] has a curious consequence: since the orthogonal group $\operatorname{O}(n)$ acts on the $n$-category ${\mathbf{Bord}}_{n}^{\operatorname{fr}}$ (by change of framing), we get an induced action of $\operatorname{O}(n)$ on the collection of fully dualizable objects of any symmetric monoidal $n$-category ${\mathcal{C}}$ (more precisely, the orthogonal group $\operatorname{O}(n)$ can be made to 