There is an interesting question on mathoverflow by Damian Rössler.

His question reminds me of an approach to constructing determinants of perfect complexes that I’ve been meaning to think about. The goal is to do a minimum of computation. For the impatient: the outcome of the discussion below is that you can probably construct determinants in this manner, but that it is of very limited use — in particular it wouldn’t answer Damian’s question and related questions.

Aside: Currently, the Stacks project discusses determinants of perfect complexes only when they have tor amplitude in [-1, 0], see here and here.

Let S be a scheme. Denote Pic(S) the category whose objects are invertible O_S-modules and whose morphisms are isomorphisms of invertible modules. An object E of the derived category D(O_S) is called perfect if it is locally on S quasi-isomorphic to a bounded complex of finite free O_S-modules. Denote Perf-Iso(S) the category whose objects are perfect objects of D(O_S) and whose morphisms are isomorphisms E → E’ of perfect objects of D(O_S). If s ∈ S is a point we denote E ⊗ κ(s) the derived pullback of E to s viewed as the spectrum of its residue field. We often think of E ⊗ κ(s) as an object of D(κ(s)), i.e., as a complex of κ(s) vector spaces.

Let E be an object of D(O_S). We will say E is *mighty fine* if (a) E is perfect, (b) the cohomology sheaves of E are finite locally free, and (c) for any point s ∈ S the restriction E ⊗ κ(s) of E to s has no two consecutive nonzero cohomology spaces. In other words, if H^n(E ⊗ κ(s)) is nonzero, then both H^{n – 1}(E ⊗ κ(s)) and H^{n + 1}(E ⊗ κ(s)) are zero. Conditions (a) and (c) actually imply condition (b) as follows from the following lemma.

**Lemma:** Let E be a perfect object of D(O_S). Let s ∈ S be a point such that E ⊗ κ(s) of E to s has no two consecutive nonzero cohomology spaces. Then there is an open neighbourhood U of s in S such that E|_U is mighty fine.

To prove the lemma you can use Lemma 068U to split the complex into a direct sum of vector bundles in a neighbourhood of s in S. (The lemma applies because a perfect complex is pseudo-coherent, see Lemma 0658.)

If E is mighty fine, then we define det(E) as the alternating tensor product of the determinants of the cohomology sheaves of E. This makes sense because these sheaves are finite locally free. The construction is clearly a functor on the category of mighty fine complexes and isomorphisms in D(O_S) between these. Also, it is compatible with pullbacks.

Having made these definitions we impose the following requirements on our determinant: A *determinant* is a rule that to every scheme S assigns a functor det : Perf-Iso(S) → Pic(S) satisfying the following requirements

- det is compatible with pullbacks,
- if E is mighty fine, then det(E) is as above,
- if a : E → E’ is an isomorphism of mighty fine objects, then det(a) is as above.

We claim that this already almost completely pins down the determinant. To explain this we need another definition.

Let E be an object of D(O_S). We will say E is *workable* if there exists a quasi-compact open immersion j : U → S which is dense and schematically dense such that E|_U is mighty fine. Observe that by the first axiom of determinants above there always is a canonical map

det(E) ⟶ j_*det(E|_U)

and by our assumptions on j we find that this map is injective! Since we have already constructed the right hand side, we conclude, for example, that if a : E → E’ is an isomorphism in D(O_S), then there is absolutely no choice for the map det(a) : det(E) → det(E’). It must be the unique map det(a) : det(E) → det(E’) compatible with the aready defined map det(a|_U) provided it exists. Moreover, still in the workable case, we’re going to define det(E) as a submodule of j_*det(E|_U) constructed affine locally (see below).

Thus it is useful to have many workable perfect complexes.

**Lemma:** Let E be a perfect object of D(O_S). For every s ∈ S there exists an open neighbourhood U of s in S and a morphism f : U → V and a workable perfect object E’ in D(O_V) and an isomorphism a : E|_U → f^*E’.

**Proof.** Locally on S the object E is given by a bounded complex of finite free O_S-modules. Thus locally on S the object E is the pullback of the universal complex on the variety of complexes V = V(r). Here r = (r_i) is a sequence of nonnegative integers almost all zero and V(r) is the scheme parametrizing sequences of size r_{i + 1} × r_i matrices M_i with M_i M_{i – 1} = 0 for all i. The variety of complexes V is a finite type scheme over Z which is reduced by Theorem 10.2 in a paper by Musili and Sheshadri of 1983. The irreducible components of the variety of complexes V are known to be Cohen-Macaulay and normal (although this doesn’t seem to be particularly helpful for us) and moreover these components are distinguished by the ranks of the maps in the complex at their generic points. Finally, it is easy to see that at each generic point one obtains a mighty fine complex — because it is easy to show that over a field any bounded complex of finite dimensional vector spaces which has two consecutive nonzero cohomology spaces can be deformed to a complex with different (smaller) betti numbers for its cohomology. **EndProof**

**Lemma.** Let U be a scheme. Consider morphisms f_i : U → V_i, i = 1, 2 where V_i is a variety of complexes with universal complex E_i and an isomorphism a : f_1^*E_1 → f_2^*E_2 in D(O_S). Then, Zariski locally on U, there exists a morphism of schemes g : U → W and smooth morphisms g_i : W → V_i, i = 1, 2 and a map of complexes b : g_1^*E_1 → g_2^*E_2 which is a quasi-isomorphism such that f_i = g_i o g and a = g^*b. (In particular the complexes g_i^*E_i are workable as the pullback of a workable complex by a flat morphism of schemes.)

**Proof sketch.** Say V_1 = V(r1) and V_2 = V(r2) for some rank vectors r1, r2. Zariski locally on U the isomorphism a is given by a map of complexes between the pullbacks of the universal complexes. Thus we can let W be the scheme of finite type over Z parametrizing maps of complexes with rank vectors r1 and r2 which are moreover quasi-isomorphisms. Then deformation theory of complexes shows that both forgetful maps W → V_1 and W → V_2 are smooth (this uses that we are versally deforming E_1 on V_1 and E_2 on V_2). **EndProof**

These lemmas show that it suffices to produce a sufficiently canonical determinant for the pullback of the universal complexes over the varieties of complexes by smooth morphisms. I think this reduces us to the algebra problem formulated below. The main point I am trying to make with this post is that now we’re looking for the existence of certain lattices in modules and we are not checking any diagrams commute!

**Algebra problem:** Given a ring A, a nonzerodivisor f in A, a bounded complex of finite free A-modules M such that M_f is mighty fine, construct a rank 1 locally free A-submodule det(M) ⊂ det(M_f) where det(M_f) is the alternating tensor product of the determinants of the cohomology modules of M_f over A_f which is compatible with pullbacks and with quasi-isomorphisms between such complexes.

For example, if A is a dvr and f is a power of the uniformizer, then we have the lattices H^i(M)/torsion ⊂ H^i(M_f) and we can take the corresponding lattice in the determinant. A similar construction works if A is a valuation ring. If the varieties of complexes are semi-normal (I don’t know if this is true, do you?), then we’d only have to look at Noetherian semi-normal rings A and our choice for dvrs would be enough.

**Drawback of this method:** One of the properties you want is that given an object (E, F) of the filtered derived category of S such that E and gr E are perfect, then we have a canonical isomorphism det(E) = ⊗ det(gr_i E). However, I do not think it is true that a “versally deformed” pair (E, F) has E and gr E workable. (I think I have a counter example.) Thus the method fails to produce the desired isomorphism in a sort of straightforward manner. Similarly, this method also wouldn’t answer Damian Rössler question… Sigh!

I thought that Max Lieblich already have a technical meaning to “mighty fine.” Is your definition the same as his?

Typo correction: have -> gave.