2
$\begingroup$

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?

Improve this question
$\endgroup$
6
  • $\begingroup$ What is the definition of $\det(F)$? $F$ might not be a perfect complex... $\endgroup$
    – Piotr Achinger
    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
    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 ? $\endgroup$
    – Damian Rössler
    Nov 26, 2014 at 17:26
  • $\begingroup$ But $F$ might not have a finite resolution by vector bundles unless $X$ is smooth projective... $\endgroup$
    – Piotr Achinger
    Nov 26, 2014 at 18:54
  • $\begingroup$ I added some comments regarding your questions. $\endgroup$
    – DonD
    Nov 27, 2014 at 7:40

1 Answer 1

Reset to default
2
$\begingroup$

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

Improve this answer
$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .