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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?

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    $\begingroup$ For the category of $\mathbb{Z}/2$-graded vector spaces, you can consider the Berezinian (en.wikipedia.org/wiki/Berezinian). The Berezinian is rather complicated and its formal properties are mediocre, despite the fact that the category is very simple. This is not promising for further generalisation. $\endgroup$
    – Neil Strickland
    Commented Mar 16, 2021 at 17:56
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    $\begingroup$ @NeilStrickland Emphasizing up on Andre's comment, the superdeterminant aka Berezinian seems to this category theorist to have perfectly good formal properties: it is the universal map of super Lie groups $GL(m|n) \to GL(1)$, etc. What properties do you find mediocre? $\endgroup$
    – Theo Johnson-Freyd
    Commented Mar 16, 2021 at 22:26
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    $\begingroup$ With regard to the original question, note that nice properties of the determinant require commutativity of the underlying ring, so it is reasonable to expect a categorification to take a symmetric monoidal ($\infty$-)category $C$ as its input. In this case the $K$-theory spectrum acquires a $E_\infty$-ring spectrum structure, and I'd guess that $gl_1(K(C))$ should be the universal recipient of a determinant map; this is at least consistent with the example of super vector spaces and the fact that the higher $K$-groups are recipients of determinant-like invariants like Reisemeister torsion. $\endgroup$
    – Bertram Arnold
    Commented Mar 16, 2021 at 22:53
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    $\begingroup$ I suspect a more fruitful way of considering this problem is to deal with the characteristic polynomial rather than the determinant per se. Recall (en.wikipedia.org/wiki/Characteristic_polynomial#Properties) that the characteristic polynomial of the matrix $A$ can be written as $p_{A}(t)=\sum _{k=0}^{n}t^{n-k}(-1)^{k}\operatorname {tr} (\wedge ^{k}A)$, where exterior powers are indicated. The traces in this sum can in turn be expressed as determinants, as the Wikipedia link shows. $\endgroup$
    – Steve Huntsman
    Commented Mar 17, 2021 at 12:35
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    $\begingroup$ Ganter and Kapranov have defined symmetric and exterior powers of k-linear categories. Perhaps using their definitions both the determinant and the characteristic polynomial can be defined in these categories. $\endgroup$
    – Nalan
    Commented Mar 17, 2021 at 13:28

3 Answers 3

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There is a notion of determinant functor, they were introduced for abelian and exact categories by P. Deligne in his paper "Le déterminant de la cohomologie" (https://publications.ias.edu/sites/default/files/Number58.pdf). There is an extension to categories of bounded complexes by F. Knudsen and D. Mumford.

More recently you also have versions for triangulated categories, cf this paper by M. Breuning "Determinant functors on triangulated categories" and also by F. Muro, A. Tonks and M. Witte "On determinant functors and K-theory".

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You can try to define the determinant of an $n \times n$ matrix with entries in a bipermutative (or symmetric bimonoidal) category $R$ by an analogue of the usual signed sum of $n$-fold products. However, it will usually not be a monoidal functor, and the inclusion $BGL_1(R) \to K(R)$ does generally not admit a retraction, which you might expect to get from a determinant. There is a counterexample in

and a discussion of what more might be needed (to circumvent this obstruction) in

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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 $K_1(\mathcal{E})$.

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