TITLE: Categorification of determinant
QUESTION [29 upvotes]: The notion of trace of a matrix can be generalized to trace of an endomorphism of a dualizable objects in a symmetric monoidal category. (See Ponto & Shulman for a nice description.)
Is there a categorification of the notion of determinant as well? If it exists, where can I read about it? If it doesn't exist, what is the conceptual obstruction to it and what is special about the trace that makes it amenable to categorification in such generality?

REPLY [8 votes]: Let me expand a bit on David C's answer. A Picard groupoid is a symmetric monoidal category in which every object is invertible (together with certain commutativity and associativity constraints), so one can apply the trace formalism there.
Consider the Picard groupoid $\mathrm{Pic}^\mathbb{Z}(X)$ whose objects are graded lines, that is, $(\mathcal{L},\alpha)$ for $\mathcal{L}$ a line bundle on $X$ and $\alpha:X\to\mathbb{Z}$ a continuous function. For the morphisms, we set $\mathrm{Hom}((\mathcal{L},\alpha),(\mathcal{L}',\alpha'))$ to be isomorphisms $\mathcal{L}\to\mathcal{L}'$ if $\alpha=\alpha'$ and the empty set otherwise. Given any vector bundle $V$ on $X$, we can define an object $\mathrm{det}(V)\in\mathrm{Pic}^\mathbb{Z}(X)$.
In particular, when $X=\mathrm{Spec}(k)$ and $V$ is a finite dimensional vector space, an automorphism $f:V\to V$ yields a map $\mathrm{det}(f):(\mathrm{det}(V),\dim(V))\to (\mathrm{det}(V),\dim(V))$ in $\mathrm{Pic}^\mathbb{Z}(\mathrm{Spec}(k))$, whose categorical trace is given by the usual determinant of $f$.
Deligne constructed, for any exact category $\mathcal{E}$, a Picard groupoid $\mathcal{P}(\mathcal{E})$ such that $\pi_i(\mathcal{P}(\mathcal{E}))=K_i(\mathcal{E})$ for $i=0,1$, together with a universal determinant functor, and one could possibly play the same game as in the previous example. That is, to get a notion of the determinant of an automorphism $f:V\to V$ in $\mathcal{E}$ we can apply the universal determinant $\mathrm{det}(f):\mathrm{det}(V)\to\mathrm{det}(V)$ to obtain an endomorphism of $\mathrm{det}(V)\in\mathcal{P}(\mathcal{E})$, and we can look at $\mathrm{tr}_{\mathcal{P}(\mathcal{E})}(\mathrm{det}(V))$ which lives in $K_1(\mathcal{E})$.