7
$\begingroup$

Let $(X, \mathcal O_X)$ be a ringed topological space. Consider two $\mathcal O_X$ modules, $\mathcal F$ and $\mathcal G$. First we define the tensor product presheaf $\mathcal F \otimes_{p,\mathcal O_X} \mathcal G$, which assigns every open set $U$ in $X$ the $\mathcal O_X(U)$-module $\mathcal F(U) \otimes_{\mathcal O_X(U)} \mathcal G(U)$. Now, $\mathcal F \otimes_{p,\mathcal O_X} \mathcal G$ is a presheaf of abelian groups. We take the the sheafication of the $\mathcal F \otimes_{p,\mathcal O_X} \mathcal G$ to obtain a sheaf of abelian groups, $\mathcal F \otimes_{\mathcal O_X} \mathcal G$. My question is for any open set $U$ in $X$ what is the $\mathcal O_X(U)$-module on $(\mathcal F \otimes_{\mathcal O_X} \mathcal G)(U)$?

$\endgroup$
15
$\begingroup$

If $(X,\mathcal O_X)$ is a scheme and if $U\subset X$ is is an open affine subscheme, then we have for all quasi-coherent sheaves $\mathcal F,\mathcal G$ of $\mathcal O_X$-modules the extremely pleasant equality of $\mathcal O_X(U)$-modules:

$$(\mathcal F \otimes_{ \mathcal O_X} \mathcal G)(U)= \mathcal F(U) \otimes_{\mathcal O_X(U)} \mathcal G(U) $$

However if $U$ is not affine all hell can break loose:
For example if $X=\mathbb P^1_\mathbb C$ is the complex projective line and if $\mathcal F=\mathcal O_X(1), \mathcal G=\mathcal O_X(-1)$, then for these quasi-coherent sheaves $\mathcal F \otimes_{ \mathcal O_X} \mathcal G=\mathcal O_X$, so that for $U=X$: $$(\mathcal F \otimes_{ \mathcal O_X} \mathcal G)(X)= \mathcal O_X(X) =\mathbb C \neq \mathcal F(X)\otimes _{\mathcal O_X(X)}\mathcal G(X)=\mathbb C^2\otimes_\mathbb C 0=0$$

Reference
The first displayed equality is proved in Qing-Liu, chapter 5, Proposition 1.12 (b), page 162.

$\endgroup$
  • $\begingroup$ Great answer, thanks. Just to double check, $\mathcal F(X)$ is the space of quotients of homogenous polynomials in two variables, with the top one degree more than the bottom, and the bottom shouldn't vanish on $\mathbb P^1$. So the bottom ends up being a nonzero constant (by projective Nullstellensatz), and the top is just a degree 1 homogenous polynomial, which can be identified with $\mathbb C^2$ as you said. $\endgroup$ – hwong557 Oct 18 '16 at 21:08
  • 2
    $\begingroup$ Dear hwong, your comment is perfectly correct. More genrally the set of global sections $\mathcal O_{\mathbb P^n}(d)(\mathbb P^n)$ of $\mathcal O_{\mathbb P^n}(d)$ consists of the polynomials in $n+1$ indeterminates which are homogeneous of degree $d$. (In particular if $d\lt 0$ the only global section is zero!) The dimension of that vector space of sections $\mathcal O_{\mathbb P^n}(d)(\mathbb P^n)$ is $\binom {n+d}{d}$, which is indeed equal to $2$ for $n=d=1$ $\endgroup$ – Georges Elencwajg Oct 18 '16 at 21:38
3
$\begingroup$

Are you asking for the $\mathcal O_X(U)$ action on $(\mathcal F \otimes_{\mathcal O_X} \mathcal G)(U)$? If so, it works on the level of stalks. Let $f \in \mathcal O_X(U)$ and let $s \in (\mathcal F \otimes_{\mathcal O_X} \mathcal G)(U)$. $s$ can be thought of as a collection of compatible germs $s_p \in \mathcal F_p \otimes_{\mathcal O_{x, p}} \mathcal G_p$. Then $f \cdot s$ is the collection of germs $f_p \cdot s_p$, where the dot is the $\mathcal O_{x, p}$-module action on $\mathcal F_p \otimes_{\mathcal O_{x, p}} \mathcal G_p$. One just needs to verify that the collection of $f_ps_p$ is still a collection of compatible germs, which they are.

$\endgroup$
  • $\begingroup$ Yes. I actually just want to prove $(\mathcal F \otimes_{\mathcal O_X} \mathcal G)(U)$ is a $\mathcal O_X(U)$. Is there any way to see that without going to the level of stalks? $\endgroup$ – grok Oct 18 '16 at 21:32
  • $\begingroup$ Dear grok: given any presheaf of $\mathcal O_X$-modules, its sheafification is automatically an $\mathcal O_X$-module too. Apply this to the presheaf of $\mathcal O_X$-modules $\mathcal F \otimes_{p,\mathcal O_X} \mathcal G$ whose sheafification is $\mathcal F \otimes_{\mathcal O_X} \mathcal G$ $\endgroup$ – Georges Elencwajg Oct 18 '16 at 21:41
  • $\begingroup$ @grok. I agree with Georges' comment above, but as far as I know, the only way to see that $(\mathcal F \otimes_{\mathcal O_X} \mathcal G)$ is an $\mathcal O_X(U)$-module is to go through stalks. The sheafification is in terms of stalks so I don't know how to get around it. $\endgroup$ – hwong557 Oct 18 '16 at 21:51

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.