10
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ One simple-minded comment is that $det$ is a homomorphism from $K_0(Perf(X))$ to $Pic(X)$, hence $\endgroup$ – Evgeny Shinder Mar 5 at 21:47
  • $\begingroup$ ...and under not very restrictive conditions on $X$, e.g. quasiprojective we have $K_0(Perf(X)) = K_0(VB(X))$ ($VB$ stands for vector bundles). Hence if an additive identity is checked on vector bundles, it seems to follow for perfect complexes. $\endgroup$ – Evgeny Shinder Mar 5 at 21:55
  • $\begingroup$ In the article "Tamagawa numbers for motives with (non-commutative) coefficients" by Burns and Flach emis.de/journals/DMJDMV/vol-06/21.html the authors extend some determinant functor from the category of projective $R$-modules to perfect complexes. I don't know whether this gives something in the direction you want, but maybe the approach could be useful. $\endgroup$ – François Brunault Mar 5 at 22:35
  • $\begingroup$ @Evgeny Shinder. Thank you for your comment. Yes I agree. These mild conditions (eg quasi-projective) imply that perfect complexes are isomorphic in the derived category to finite complexes of vector bundles and then one can reduce to vector bundles (as in the example). But even then, I need a canonical iso., invariant under base change. $\endgroup$ – Damian Rössler Mar 5 at 22:50
  • $\begingroup$ @François Brunault. Thank you for the reference. The Knudsen-Mumford does just that though. My question concern the KM determinant. $\endgroup$ – Damian Rössler Mar 5 at 22:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.