TITLE: Determinantal identities for perfect complexes
QUESTION [17 upvotes]: Let $S$ be a noetherian scheme. Let $V,W$ be vector bundles on $S$. There is a canonical isomorphism of line bundles
$$
{\rm det}(V\otimes W)\cong{\rm det}(V)^{\otimes{\rm rk}(W)}\otimes{\rm det}(W)^{\otimes{{\rm rk} V}}\,\,\,\,(\ast)
$$
which is invariant under any base change. See eg 
https://math.stackexchange.com/questions/571839/determinant-of-a-tensor-product-of-two-vector-bundles/571906
for this. 
There are similar identities for other tensor operations, eg
$$
{\rm det}(\Lambda^2(W))\cong{\rm det}(W)^{\otimes({\rm rk}(W)-1)}
$$
My question is: are there similar identities for perfect complexes in place of $V,W$ ?
Recall that an object in the derived category of ${\cal O}_S$-modules is called perfect if is Zariski locally 
isomorphic to a bounded complex of vector bundles. 
One can show that if $V^\bullet$ and $W^\bullet$ are bounded complexes of vector bundles (on all of $S$) 
then there is an isomorphism
$$
{\rm det}(V^\bullet\otimes W^\bullet)\cong{\rm det}(V^\bullet)^{\otimes{\rm rk}(W^\bullet)}\otimes{\rm det}(W^\bullet)^{\otimes{{\rm rk}(V^{\bullet})}}\,\,\,\,(\ast\ast)
$$
where now ${\rm det}(\cdot)$ is the Knudsen-Mumford determinant of perfect complexes. This follows from identity $(\ast)$. However the isomorphism is not canonical. In other words, I don't know how to construct an isomorphism $(\ast\ast)$, which is functorial for isomorphisms in the derived category (or more concretely, for quasi-isomorphisms of bounded complexes of vector bundles on $S$). 
In particular, I don't know whether there is an isomorphism $(\ast\ast)$ (even a non canonical one) when $V^\bullet$ and $W^\bullet$ are only assumed to be perfect.
I would be grateful if anyone could share ideas, or direct me to references on this kind of problem. 
I am aware of Deligne's work on Picard categories and axiomatic descriptions of determinants but this seems to be of little help. One could try to prove an identity like $(\ast\ast)$  by showing that both sides satisfy the axiomatic properties of determinants (fixing $V^\bullet$ or $W^\bullet$) but such a verification seems difficult and tedious. Another way to proceed might be to write down an isomorphism $(\ast\ast)$ applying $(\ast)$ term by term and to verify functoriality for quasi-isomorphisms directly on the definition of the functoriality of the Knudsen-Mumford determinant but this again is difficult because this functoriality is defined in a very indirect way (see proof of Th. 1 in the paper of Knudsen-Mumford https://www.mscand.dk/article/view/11642 or 
How to write down the determinant of a quasi-isomorphism?
). One would expect all the 
determinantal identities that are valid for vector bundles to be valid automatically for perfect complexes. There should be a way to show this.

REPLY [12 votes]: The formula also holds for perfect complexes. This can be deduced from the case of vector bundles, although it requires a lot of structure in that case. Namely, we need to use the fact that the determinant of vector bundles can be promoted to a morphism of $E_\infty$-semirings in algebraic stacks
$$
\det \colon \mathrm{Vect}\to \mathrm{Pic}^\mathbb{Z}.
$$
Here, $\mathrm{Vect}$ is the stack of vector bundles, with the ring structure given by $\oplus$ and $\otimes$, and $\mathrm{Pic}^\mathbb{Z}$ is the stack of pairs $(L,n)$ where $L$ is a line bundle and $n$ a locally constant integer. The additive structure on $\mathrm{Pic}^\mathbb{Z}$ comes from the fact that it is the $\infty$-groupoid of invertible objects in the symmetric monoidal stable $\infty$-category $\mathrm{QCoh}(S)$ (or more explicitly, $(L,n)+(L',n') = (L\otimes L', n+n')$ with a sign $(-1)^{nn'}$ in the braiding). The multiplicative structure is more subtle: abstractly it is a square-zero extension of the constant sheaf of rings $\underline{\mathbb Z}$ classified by a certain derivation $\underline{\mathbb Z} \to B\mathrm{Pic}$ induced by the sign map $\pi_1\underline{\mathbb S}=\underline{\mathbb Z}/2 \to \mathbb G_m$.
Once we have this morphism $\det$, it factors through the Zariski sheafification of the group completion of $\mathrm{Vect}$, which coincides with the Zariski sheafification of algebraic K-theory, so we get a morphism of $E_\infty$-rings
$$
\det \colon K\to \mathrm{Pic}^\mathbb{Z},
$$
giving in particular the desired formula for $\det(P\otimes Q)$ for $P,Q\in \mathrm{Perf}(S)$. (Note: this works also for perfect complexes on algebraic stacks, since $\mathrm{Pic}^\mathbb{Z}$ is an fpqc sheaf.)
In principle it is possible to construct the above-mentioned structure by hand, since everything takes place in the 2-category of presheaves of groupoids, but the details are probably quite tedious. Fortunately there is also a very easy way to construct all this structure: the stack $\mathrm{Pic}^\mathbb{Z}$ is simply the $1$-truncation of $K$ as a Zariski sheaf, and $\det$ is the canonical map to the truncation. Indeed, if $R$ is a local ring, the determinant map $\det\colon K(R) \to \mathrm{Pic}^\mathbb{Z}(R)$ exhibits $\mathrm{Pic}^\mathbb{Z}(R)$ as the $1$-truncation of $K(R)$, since it is an isomorphism on $\pi_0$ and $\pi_1$. Since $1$-truncation preserves finite products, it preserves $E_\infty$-ring structures.