0
$\begingroup$

I'm reading a part of Mumford's Abelian Varieties, and in the Chapter The theorem of the cube: II he claims that some "Künneth formula" tells us that if $L_1$ is a line bundle on a product $X \times Y_1$ such that $L_1 \cong p_2^*(M_1)$ for some line bundle $M_1$ on $Y_1$, then $p_{2,*}(L_1) \cong M_1$.

Here $X$ is a complete variety over an algebraically closed field $k$, and $Y_1$ is a scheme of finite type over $k$.

The Künneth formula I know relates (co-)homology on product spaces with the (co-)homology on the base spaces, for example as in the stacks-project, tag 0BEC. I don't see how the claim should follow from this.

Does the claim follow from this Künneth formula, or is something else going on here?

$\endgroup$
  • $\begingroup$ Certainly the projection formula is going on here. Perhaps to apply it properly, one needs to use Kunneth. $\endgroup$ – aginensky Mar 21 at 14:38
  • $\begingroup$ @aginensky Yeah by the projection formula we get: $p_{2,*} p_2^* M_1 = p_{2,*}(\mathcal{O} \otimes p_2^*M_1) = p_{2,*}(\mathcal{O}_{X\times Y_1}) \otimes M_1$, so we reduce this to show that $p_{2,*}\mathcal{O}_{X \times Y_1} \cong \mathcal{O}_{Y_1}$. I think I can show this using "Cohomology and Base Change" (Hartshorne, p.290+291), but I still don't see any connection to the Künneth formula. $\endgroup$ – red_trumpet Mar 21 at 14:58
  • $\begingroup$ Perhaps in saying that the structure sheaf of the product is the tensor product of the pullbacks of the two structure sheaves. $\endgroup$ – aginensky Mar 21 at 14:59
1
$\begingroup$

It's always hard to reconstruct someone else's train of thought, but here's one attempted explanation.

As aginensky says, the first step seems to be to use the projection formula. This shows that $(p_2)_* L_1 = M_1 \otimes (p_2)_* O_Z$ where I have written $Z$ to denote the product (to save typing).

So now it is enough to show that $(p_2)_* O_Z = O_{Y_1}$. To do this, recall that $(p_2)_* O_Z$ is defined by

\begin{align*} U &\mapsto H^0((p_2)^{-1}(U), O_{(p_2)^{-1}(U)})\\ &= H^0(U \times X, O_{U \times X}) \end{align*} Now Künneth tells us that this is isomorphic to \begin{align*} &H^0(U,O_U) \otimes_k H^0(X,O_X) =H^0(U,O_U)\end{align*} using the fact that $X$ is complete.

$\endgroup$
  • $\begingroup$ Seems correct. Two notes: In the last equation, it should be $\otimes$ instead of $\times$, and afaik $U \mapsto H^0(U\times X, \mathcal{O})$ already is a sheaf, so no need to sheafifiy here. $\endgroup$ – red_trumpet Mar 21 at 20:54
  • $\begingroup$ You are right about both things. $\endgroup$ – Asal Beag Dubh Mar 22 at 9:37

Your Answer

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

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