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).

  • $\begingroup$ Sorry, but I don't see a definition in Wise's posting. You are talking about the determinant of a quasi-isomorphism between complexes, while Wise talks about the determinant of a complex (which goes back to Cayley, cf. Gelfand/Kapranov/Zelevinsky Appendix A, where an explicit formula is given). $\endgroup$ Apr 25 '10 at 14:31

Denote the basis of $E_0$ (resp. $E_1$, $S_0$, $S_1$) (sorry that I changed the notation because I do not want to have too many f's) by $e_{0,1},\cdots,e_{0,n_0}$ (resp. $e_{1,1},\cdots,e_{1,n_1}$, $s_{0,1},\cdots,s_{0,m_0}$, $s_{1,1},\cdots,s_{1,m_1}$, and the matrix corresponds to $f_0$ (resp. $f_1$) by $F_0$ (resp. $F_1$). It is always possible to choose bases of $E_1$ and $S_1$ such that the matrices corresponding to differentials are of the form $$\begin{pmatrix} I & 0 \end{pmatrix}$$ or its transpose.

Case 1: if $f$ is split injective, then there is an exact sequence $$0\rightarrow E_.\rightarrow S_. \rightarrow H_. \rightarrow 0$$ with $H_.$ acyclic. Then the determinant of $f$ can be obtained by the combine the isomorphisms $\det E_. \otimes\det H_. \xrightarrow{\sim} \det S_.$ and $\det S_. \xrightarrow{\sim} \mathbb{1}$.

In this case, you just have invertible matrices $F_0$ and $F_1$, and the basis of $S_.$ can be chosen suitable such that the first half is mapped from $E_.$ and the lower half of basis of $S_.$ contributes to the basis of $H_.$. Also, the map from $S_.$ to $H_.$ is projection. By above, $\det f$ is just a scaler multiple whose scaler is given by $\det F_0 (\det F_1)^{-1}$.

Case 2: otherwise, we can construct an intermediate complex, namely $C_.$, which is defined as $$C_i:=E_i\oplus S_i \oplus E_{i-1}$$ with differentials $$d_{C,i}:= \left( \begin{array}{ccc} d_{E,i} & 0 & -1\\ 0 & d_{S,i} & f_{i+1} \\ 0& 0& -d_{E,i-1} \end{array} \right)$$ and the maps $$ \begin{aligned} &\alpha: E_. \rightarrow C_., e \mapsto (e,0,0)\\ &\beta: S_.\rightarrow C_., s \mapsto (0,s,0)\\ &\gamma: C_.\rightarrow S_., (e,s,e') \mapsto f(e)+s. \end{aligned} $$ Then it is easy to check that $\gamma\beta=\operatorname{id}.$ and $\gamma\alpha=f$. So we can apply the method in case 1 to construct $\det f$, which is $\det^{-1}\beta \det\alpha$.

In your case, by choosing the bases wisely, the only non-trivial part during computing $\det f$ is the determinant of $\alpha$, which is also not too complicated when written as matrix. And the final conclusion is that $$\det f=\det F_0(\det F_1)^{-1}.$$

  • 1
    $\begingroup$ i don't know why this looks like a mess, sorry for the inconveniece :( $\endgroup$
    – Yujia Qiu
    Apr 29 '11 at 16:33
  • $\begingroup$ @Yujia Qiu: Thanks for the response! I have not thought about this question in a bit. I will try to think about your answer over the weekend. $\endgroup$
    – jlk
    May 6 '11 at 4:24
  • $\begingroup$ @Yuija Qiu: Sorry for the delayed response. I am slightly confused by your your final conclusion. The vector spaces $F_0$ and $S_0$ could have different dimensions (and similarly for $F_1$ and $S_1$). The fact that the complexes are quasi-isomorphic only tells us that $\operatorname{dim}(F_1) - \operatorname{dim}(F_0) = \operatorname{dim}(S_1) - \operatorname{dim}(S_0)$. When the dimensions are different, what is meant by $\operatorname{det}(F_0)$? $\endgroup$
    – jlk
    May 21 '11 at 6:08

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