This question about the determinant of a perfect complex reminded me of an old question that I had.

The construction of the determinant (as in MR1914072 or MR0437541) is a difficult piece of mathematics. However, the definition of the determinant of a single complex is easy. Indeed, Jonathan Wise wrote down the definition in his (2-sentence) question. I have read that the difficult part of the construction lies in defining the determinant of a quasi-isomorphism in a functorial way. Reading the papers, the construction certainly looks difficult, but it would be nice to have a concrete example to convince skeptics.

Recall how functoriality works. To every quasi-isomorphism $f \colon E_\bullet \to F_\bullet$, one wants to assign an isomorphism $\operatorname{det}(f) \colon \operatorname{det}(E_\bullet) \to \operatorname{det}(F_{\bullet})$ in such a way that $\operatorname{det}$ preserves composition. *I would like a concrete computation of $\operatorname{det}(f)$ for a non-trivial example.*

To make life easy, let's just work with complexes of vector spaces over $\mathbb{C}$.

Here are cases that I can work out:

**1)** Both $E_{\bullet}$ and $F_{\bullet}$ are both concentrated in a fixed degree.

Say $E_{\bullet} = F_{\bullet}$ are both complexes concentrated in degree $0$. Then $f$ is a genuine automorphism, and the map $\operatorname{def}(f)$ is just the usual determinant that I learned in linear algebra. In other words, if we fix bases and represent $f$ as a matrix, then, with respect to the natural bases, $\operatorname{det}(f)$ is given by multiplication by the determinant of that matrix.

If we instead consider the case where both complexes are concentrated in degree $d$, then I think $\operatorname{det}(f)$ is $(-1)^{d}$ times the map just described.

**2)** The complex $E_{\bullet}$ is the zero complex and $F_{\bullet}$ is a 2-terms complex.

Then $\operatorname{det}(E_{\bullet}) = \mathbb{C}^1$, $\operatorname{det}(F_{\bullet})= \operatorname{Hom}(\bigwedge F_{1}, \bigwedge F_{0})$, and $\operatorname{def}(f)$ is the map that sends $1$ to the top exterior power of the differential $\partial$ of $F_{\bullet}$. In other words, $\operatorname{det}(f)$ is basically the determinant of the differential for $F_{\bullet}$

**Question:** Say that $f \colon E_{\bullet} \to F_{\bullet}$ is a quasi-isomorphism of 2-term complexes. Fix bases for all of the relevant vector spaces and represent $f$ and the two differentials by 4 different matrices. How do you explicitly write down the determinant map $\operatorname{det}(f) \colon \operatorname{det}(E_{\bullet}) \to \operatorname{det}(F_{\bullet})$ in terms of these matrices?

*Note:* Tracing through the construction in MR1914072 or MR0437541, I think that one gets a description of $\operatorname{def}(f)$ in terms of standard triangulated category constructions. I am not looking for an explanation of that, I really want a formula that I could show to, say, an undergraduate linear algebra student.

**Edit** @darij grinberg, thanks for the response. You are correct that Wise only defined the determinant of a complex. The answers to that question discussed the fact that there is a natural functor from the category of vector bundles (with morphisms taken to be quasi-isomorphisms) to the category of line bundles with the property that the map on objects is given by Wise's formula. I am asking for a description of the map on morphisms.

The relevant functor is defined in the two articles that I referred to as well as the slides that "YLB" linked to. I edited the post to include a link to the mathsci review of the articles (and fixed a typo).