# How to see that this pairing of line bundles is multiplicative?

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?

• What is the definition of $\det(F)$? $F$ might not be a perfect complex... – Piotr Achinger Nov 25 '14 at 21:05
• 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. – DonD Nov 26 '14 at 14:40
• Do you want a canonical isomorphism (inv. under base-change) ? Or would you be happy with any isomorphism ? – Damian Rössler Nov 26 '14 at 17:26
• But $F$ might not have a finite resolution by vector bundles unless $X$ is smooth projective... – Piotr Achinger Nov 26 '14 at 18:54
• I added some comments regarding your questions. – DonD Nov 27 '14 at 7:40

## 1 Answer

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