## 15.122 Determinants of complexes

In Section 15.119 we have seen how to a perfect complex $K$ over a ring $R$ there is associated an isomorphism class of invertible $R$-modules, i.e., an element of $\mathop{\mathrm{Pic}}\nolimits (R)$. In fact, analogously to Section 15.118 it turns out there is a functor

$\det : \left\{ \begin{matrix} \text{category of perfect complexes} \\ \text{morphisms are isomorphisms} \end{matrix} \right\} \longrightarrow \left\{ \begin{matrix} \text{category of invertible modules} \\ \text{morphisms are isomorphisms} \end{matrix} \right\}$

Moreover, given an object $(L, F)$ of the filtered derived category $DF(R)$ of $R$ whose filtration is finite and whose graded parts are perfect complexes, there is a canonical isomorphism $\det (\text{gr}L) \to \det (L)$. See for the original exposition. We will add this material later (insert future reference).

For the moment we will present an ad hoc construction in the case of perfect objects $L$ in $D(R)$ of tor-amplitude in $[-1, 0]$. Such an object may be represented by a complex

$L^\bullet = \ldots \to 0 \to L^{-1} \to L^0 \to 0 \to \ldots$

with $L^{-1}$ and $L^0$ finite projective $R$-modules, see Lemma 15.74.2. In this case we set

$\det (L^\bullet ) = \det (L^0) \otimes _ R \det (L^{-1})^{\otimes -1} = \mathop{\mathrm{Hom}}\nolimits _ R(\det (L^{-1}), \det (L^0))$

Let us say a complex of this form has rank $0$ if $L^{-1}_\mathfrak p$ and $L^0_\mathfrak p$ have the same rank for all primes of $R$. If $L^\bullet$ has rank $0$, then we have seen in Section 15.118 that there is a canonical element

$\delta (L^\bullet ) \in \det (L^\bullet )$

which is simply the determininant of $d : L^{-1} \to L^0$. Note that $\delta (L^\bullet )$ is a trivialization of $\det (L^\bullet )$ if and only if $L^\bullet$ is acyclic.

Consider a map of complexes $a^\bullet : K^\bullet \to L^\bullet$ such that

1. $a^\bullet$ is a quasi-isomorphism,

2. $a^ n : K^ n \to L^ n$ is surjective for all $n$,

3. $K^ n$, $L^ n$ are finite projective $R$-modules, nonzero only for $n \in \{ -1, 0\}$.

In this situation we will construct an isomorphism

$\det (a^\bullet ) : \det (K^\bullet ) \longrightarrow \det (L^\bullet )$

Using the exact sequences $0 \to \mathop{\mathrm{Ker}}(a^ i) \to K^ i \to L^ i \to 0$ we obtain isomorphisms

$\gamma ^ i : \det (\mathop{\mathrm{Ker}}(a^ i)) \otimes \det (L^ i) \to \det (K^ i)$

for $i = -1, 0$ by Lemma 15.118.2. Since $a^\bullet$ is a quasi-isomorphism the complex $\mathop{\mathrm{Ker}}(a^\bullet )$ is acyclic and has rank $0$. Hence the canonical element $\delta (\mathop{\mathrm{Ker}}(a^\bullet ))$ is a trivialization of the invertible $R$-module $\det (\mathop{\mathrm{Ker}}(a^\bullet ))$, see above. We define $\det (a^\bullet ) : \det (K^\bullet ) \to \det (L^\bullet )$ as the unique isomorphism such that the diagram

$\xymatrix{ \det (K^\bullet ) \ar[rr]_{\det (a^\bullet )} \ar[dr]_{\delta (\mathop{\mathrm{Ker}}(a^\bullet ))} & & \det (L^\bullet ) \\ & \det (K^\bullet ) \otimes \det (\mathop{\mathrm{Ker}}(a^\bullet )) \ar[ru]_{\gamma ^0 \otimes (\gamma ^{-1})^{\otimes -1}} }$

commutes.

Lemma 15.122.1. Let $R$ be a ring. Let $a^\bullet : K^\bullet \to L^\bullet$ be a map of complexes of $R$-modules satisfying (1), (2), (3) above. If $L^\bullet$ has rank $0$, then $\det (a^\bullet )$ maps the canonical element $\delta (K^\bullet )$ to $\delta (L^\bullet )$.

Proof. Write $M^ i = \mathop{\mathrm{Ker}}(a^ i)$. Thus we have a map of short exact sequences

$\xymatrix{ 0 \ar[r] & M^{-1} \ar[r] \ar[d]_{d_ M} & K^{-1} \ar[r] \ar[d]_{d_ K} & L^{-1} \ar[r] \ar[d]_{d_ L} & 0 \\ 0 \ar[r] & M^0 \ar[r] & K^0 \ar[r] & L^0 \ar[r] & 0 }$

By Lemma 15.118.3 we know that $\det (d_ K)$ corresponds to $\det (d_ M) \otimes \det (d_ L)$ as maps. Unwinding the definitions this gives the required equality. $\square$

Lemma 15.122.2. Let $R$ be a ring. Let $a^\bullet : K^\bullet \to L^\bullet$ be a map of complexes of $R$-modules satisfying (1), (2), (3) above. Let $h : K^0 \to L^{-1}$ be a map such that $b^0 = a^0 + d \circ h$ and $b^{-1} = a^{-1} + h \circ d$ are surjective. Then $\det (a^\bullet ) = \det (b^\bullet )$ as maps $\det (K^\bullet ) \to \det (L^\bullet )$.

Proof. Suppose there exists a map $\tilde h : K^0 \to K^{-1}$ such that $h = a^{-1} \circ \tilde h$ and such that $k^0 = \text{id} + d \circ \tilde h : K^0 \to K^0$ and $k^1 = \text{id} + \tilde h \circ d : K^{-1} \to K^{-1}$ are isomorphisms. Then we obtain a commutative diagram

$\xymatrix{ 0 \ar[r] & \mathop{\mathrm{Ker}}(b^\bullet ) \ar[r] \ar[d]_{c^\bullet } & K^\bullet \ar[r]_{b^\bullet } \ar[d]_{k^\bullet } & L^\bullet \ar[r] \ar[d]^{\text{id}} & 0 \\ 0 \ar[r] & \mathop{\mathrm{Ker}}(a^\bullet ) \ar[r] & K^\bullet \ar[r]^{a^\bullet } & L^\bullet \ar[r] & 0 }$

of complexes, where $c^\bullet$ is the induced isomorphism of kernels. Using Lemma 15.118.3 we see that

$\xymatrix{ \det (\mathop{\mathrm{Ker}}(b^ i)) \otimes \det (L^ i) \ar[r] \ar[d]_{\det (c^ i) \otimes 1} & \det (K^ i) \ar[d]^{\det (k^ i)} \\ \det (\mathop{\mathrm{Ker}}(a^ i)) \otimes \det (L^ i) \ar[r] & \det (K^ i) }$

commutes. Since $\det (c^\bullet )$ maps the canonical trivialization of $\det (\mathop{\mathrm{Ker}}(a^\bullet ))$ to the canonical trivializatio of $\mathop{\mathrm{Ker}}(b^\bullet )$ (Lemma 15.122.1) we see that we conclude if (and only if)

$\det (k^0) = \det (k^{-1})$

as elements of $R$ which follows from Lemma 15.118.6.

Suppose there exists a direct summand $U \subset K^{-1}$ such that both $a^{-1}|_ U : U \to L^{-1}$ and $b^{-1}|_ U : U \to L^{-1}$ are isomorphisms. Define $\tilde h$ as the composition of $h$ with the inverse of $a^{-1}|_ U$. We claim that $\tilde h$ is a map as in the first paragraph of the proof. Namely, we have $h = a^{-1} \circ \tilde h$ by construction. To show that $k^{-1} : K^{-1} \to K^{-1}$ is an isomorphism it suffices to show that it is surjective (Algebra, Lemma 10.16.4). Let $u \in U$. We may choose $u' \in U$ such that $b^{-1}(u') = a^{-1}(u)$. Then $u = k^{-1}(u')$. Namely, both $u$ and $k^{-1}(u')$ are in $U$ and $a^{-1}(u) = a^{-1}(k^{-1}(u'))$ by a calculation1 Since $a^{-1}|_ U$ is an isomorphism we get the equality. Thus $U \subset \mathop{\mathrm{Im}}(k^{-1})$. On the other hand, if $x \in \mathop{\mathrm{Ker}}(a^{-1})$ then $x = k^{-1}(x) \bmod U$. Since $K^{-1} = \mathop{\mathrm{Ker}}(a^{-1}) + U$ we conclude $k^{-1}$ is surjective. Finally, we show that $k^0 : K^0 \to K^0$ is surjective. First, since $a^0 \circ k^0 = b^0$ we see that $a^0 \circ k^0$ is surjective. If $x \in \mathop{\mathrm{Ker}}(a^0)$, then $x = d(y)$ for some $y \in \mathop{\mathrm{Ker}}(a^{-1})$. We may write $y = k^{-1}(z)$ for some $z \in K^{-1}$ by the above. Then $x = k^0(d(z))$ and we conclude.

Final step of the proof. It suffices to find $U$ as in the preceding paragraph, but this may not always be possible. However, in order to show equality of two maps of $R$-modules, it suffices to do so after localization at primes of $R$. Hence we may assume $R$ is local. Then we get the following problem: suppose

$\alpha , \beta : R^{\oplus n} \longrightarrow R^{\oplus m}$

are two surjective $R$-linear maps. Find a direct summand $U \subset R^{\oplus n}$ such that both $\alpha |_ U$ and $\beta |_ U$ are isomorphisms. If $R$ is a field, this is possible by linear algebra. In general, one takes a solution over the residue field and lifts this to a solution over the local ring $R$. Some details omitted. $\square$

Lemma 15.122.3. Let $R$ be a ring. Let $a^\bullet : K^\bullet \to L^\bullet$ and $b^\bullet : L^\bullet \to M^\bullet$ be maps of complexes of $R$-modules satisfying (1), (2), (3) above. Then we have $\det (b^\bullet ) \circ \det (a^\bullet ) = \det (b^\bullet \circ a^\bullet )$ as maps $\det (M^\bullet ) \to \det (K^\bullet )$.

Lemma 15.122.4. Let $R$ be a ring. The constructions above determine a functor

$\det : \left\{ \begin{matrix} \text{category of perfect complexes} \\ \text{with tor amplitude in }[-1, 0] \\ \text{morphisms are isomorphisms} \end{matrix} \right\} \longrightarrow \left\{ \begin{matrix} \text{category of invertible modules} \\ \text{morphisms are isomorphisms} \end{matrix} \right\}$

Moreover, given a rank $0$ perfect object $L$ of $D(R)$ with tor-amplitude in $[-1, 0]$ there is a canonical element $\delta (L) \in \det (L)$ such that for any isomorphism $a : L \to K$ in $D(R)$ we have $\det (a)(\delta (L)) = \delta (K)$.

Proof. By Lemma 15.74.2 every object of the source category may be represented by a complex

$L^\bullet = \ldots \to 0 \to L^{-1} \to L^0 \to 0 \to \ldots$

with $L^{-1}$ and $L^0$ finite projective $R$-modules. Let us temporarily call a complex of this type good. By Derived Categories, Lemma 13.19.8 morphisms between good complexes in the derived category are homotopy classes of maps of complexes. Thus we may work with good complexes and we can use the determinant $\det (L^\bullet ) = \det (L^0) \otimes \det (L^{-1})^{\otimes -1}$ we investigated above.

Let $a^\bullet : L^\bullet \to K^\bullet$ be a morphism of good complexes which is an isomorphism in $D(R)$, i.e., a quasi-isomorphism. We say that

$\xymatrix{ L^\bullet \ar[rr]_{a^\bullet } & & K^\bullet \\ & M^\bullet \ar[lu]^{b^\bullet } \ar[ru]_{c^\bullet } }$

is a good diagram if it commutes up to homotopy and $b^\bullet$ and $c^\bullet$ satisfy conditions (1), (2), (3) above. Whenever we have such a diagram it makes sense to define

$\det (a^\bullet ) = \det (c^\bullet ) \circ \det (b^\bullet )^{-1}$

where $\det (c^\bullet )$ and $\det (b^\bullet )$ are the isomorphisms constructed in the text above. We will show that good diagrams always exist and that the resulting map $\det (a^\bullet )$ is independent of the choice of good diagram.

Existence of good diagrams for a quasi-isomorphism $a^\bullet : L^\bullet \to K^\bullet$ of good complexes. Choose a surjection $p : R^{\oplus n} \to K^{-1}$. Then we can consider the new good complex

$M^\bullet = \ldots \to 0 \to L^{-1} \oplus R^{\oplus n} \xrightarrow {d \oplus 1} L^0 \oplus R^{\oplus n} \to 0 \to \ldots$

with the projection map $b^\bullet : M^\bullet \to L^\bullet$ and the map $c^\bullet : M^\bullet \to K^\bullet$ using $a^{-1} \oplus p$ in degree $-1$ and using $a^0 \oplus d \circ p$ in degree $0$. The maps $b^\bullet : M^\bullet \to L^\bullet$ and $c^\bullet : M^\bullet \to K^\bullet$ satisfy conditions (1), (2), (3) above and we get a good diagram.

Suppose that we have a good diagram

$\xymatrix{ L^\bullet \ar[rr]_{\text{id}^\bullet } & & L^\bullet \\ & M^\bullet \ar[lu]^{b^\bullet } \ar[ru]_{c^\bullet } }$

Then by Lemma 15.122.2 we see that $\det (c^\bullet ) = \det (b^\bullet )$. Thus we see that $\det (\text{id}^\bullet ) = \text{id}$ is independent of the choice of good diagram.

Before we prove independence in general, we think about composition. Suppose we have quasi-isomorphisms $L_1^\bullet \to L_2^\bullet$ and $L_2^\bullet \to L_3^\bullet$ of good complexes and good diagrams

$\vcenter { \xymatrix{ L_1^\bullet \ar[rr] & & L_2^\bullet \\ & M_{12}^\bullet \ar[lu] \ar[ru] } } \quad \text{and}\quad \vcenter { \xymatrix{ L_2^\bullet \ar[rr] & & L_3^\bullet \\ & M_{23}^\bullet \ar[lu] \ar[ru] } }$

We can extend this to a diagram

$\xymatrix{ L_1^\bullet \ar[rr] & & L_2^\bullet \ar[rr] & & L_3^\bullet \\ & M_{12}^\bullet \ar[lu] \ar[ru] & & M_{23}^\bullet \ar[lu] \ar[ru] \\ & & M_{123}^\bullet \ar[lu] \ar[ru] }$

where $M_{123}^\bullet \to M_{12}^\bullet$ and $M_{123}^\bullet \to M_{23}^\bullet$ have properties (1), (2), (3) and the square in the diagram commutes: we can just take $M_{123}^ n = M_{12}^ n \times _{L_2^ n} M_{23}^ n$. Then Lemma 15.122.3 shows that

$\xymatrix{ \det (L_2^\bullet ) & \det (M_{23}^\bullet ) \ar[l] \\ \det (M_{12}^\bullet ) \ar[u] & \det (M_{123}^\bullet ) \ar[l] \ar[u] }$

commutes. A diagram chase shows that the composition $\det (L_1^\bullet ) \to \det (L_2^\bullet ) \to \det (L_3^\bullet )$ of the maps associated to the two good diagrams using $M_{12}^\bullet$ and $M_{23}^\bullet$ is equal to the map associated to the good diagram

$\xymatrix{ L_1^\bullet \ar[rr] & & L_3^\bullet \\ & M_{123}^\bullet \ar[lu] \ar[ru] }$

Thus if we can show that these maps are independent of choices, then the composition law is satisfied too and we obtain our functor.

Independence. Let a quasi-isomorphism $a^\bullet : L^\bullet \to K^\bullet$ of good complexes be given. Choose an inverse quasi-isomorphism $b^\bullet : K^\bullet \to L^\bullet$. Setting $L_1^\bullet = L$, $L_2^\bullet = K^\bullet$ and $L_3^\bullet = L^\bullet$ may fix our choice of good diagram for $b^\bullet$ and consider varying good diagrams for $a^\bullet$. Then the result of the previous paragraphs is that no matter what choices, the composition always equals the identity map on $\det (L^\bullet )$. This clearly proves independence of those choices.

The statement on canonical elements follows immediately from Lemma 15.122.1 and our construction. $\square$

[1] $a^{-1}(k^{-1}(u')) = a^{-1}(u') + a^{-1}(\tilde h(d(u'))) = a^{-1}(u') + h(d(u')) = b^{-1}(u') = a^{-1}(u)$

Comment #5958 by Tuomas Tajakka on

The two versions of det(a) induced by a map of complexes a: K -> L is quite confusing. When it is first defined for maps satisfying the conditions (1), (2), and (3), the map goes in the "wrong" direction -- det(a): det(L) -> det(K). However, in the proof of Lemma 0FJM, det(a) is defined again for more general maps, and this time it goes in the "right" direction -- det(a): det(K) -> det(L).

Since this discrepancy seems intentional, it might be a good idea two use different symbols for the two different maps, and remark that the first definition, using conditions (1), (2), and (3), is auxiliary and goes in the wrong direction. (Or I might be missing something and there is no discrepancy after all.)

Comment #6140 by on

Yes, this is awful. I fixed it by changing the direction of the initial determinant. I hope I didn't miss any arrows which should be reversed! The changes are here. Thanks!

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