Given a projective flat morphism $p: X \rightarrow Y$ of integral noetherian schemes of relative dimension one.

For a coherent sheaf $F$ on $Y$ we can define a line bundle $det(F)$ on $Y$ and for a coherent locally free sheaf $E$ on $X$ we define a line bundle $Det(E)$ on $Y$ by $Det(E):=det(p_{*}E)\otimes det(R^1p_{*}E)^{-1}$.

Now we want to define a pairing $<.,.> : Pic(X)\times Pic(X)\rightarrow Pic(Y)$ by $<L,M>:=Det(L\otimes M)\otimes Det(L)^{-1}\otimes Det(M)^{-1}\otimes Det(\mathcal{O}_X)$

How can I see that this pairing satisfies $<L_1\otimes L_2,M>\cong <L_1,M>\otimes <L_2,M>$?

I think this pairing is also called Deligne bracket and this definition here is equivalent to the usual definition using local sections $s,t$ of $L$ and $M$ to get a local section $<s,t>$ of $<L,M>$. With this one can easily see that the pairing is compatible with tensor products. But i cannot seem to prove this using the determinant definition.

Is there some trick that helps? Or is it just not possible?

Edit: We may assume that every coherent sheaf on $X$ and $S$ has a finite locally free resolution. Furthermore i would like this isomorphism to be invariant under base change, but if this is not possible, what is the correct definition of $Det$ for this to be possible?

  • $\begingroup$ What is the definition of $\det(F)$? $F$ might not be a perfect complex... $\endgroup$ Commented Nov 25, 2014 at 21:05
  • $\begingroup$ I was using the definition given in Kobayashi, CH. V, §6. For example if $F$ is torsion free one can take $det(F)=(\Lambda^{r}(F))^{**}$. Otherwise one can choose local resolutions of $F$ by vector bundles and then take the alternating product of their determinants. $\endgroup$
    – DonD
    Commented Nov 26, 2014 at 14:40
  • $\begingroup$ Do you want a canonical isomorphism (inv. under base-change) ? Or would you be happy with any isomorphism ? $\endgroup$ Commented Nov 26, 2014 at 17:26
  • $\begingroup$ But $F$ might not have a finite resolution by vector bundles unless $X$ is smooth projective... $\endgroup$ Commented Nov 26, 2014 at 18:54
  • $\begingroup$ I added some comments regarding your questions. $\endgroup$
    – DonD
    Commented Nov 27, 2014 at 7:40

1 Answer 1


The fact is that Deligne pairing is bilinear, see 6.2 here

P. Deligne. Le d ́eterminant de la cohomologie. Contemporary Mathematics , 67:93–177, 1987.

So we have $$<L_1\otimes L_2,M>\cong <L_1,M>\otimes <L_2,M>$$

This isomorphism of line bundles becomes an isometry if we equip each line bundle with the Deligne metric


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.