Tensor product of $\mathscr{O}_X$-modules which results in a presheaf.

Background: Over a locally ringed space $X$, if we define the tensor product of two $\mathscr{O}_X$-modules $\mathscr{F}$ and $\scr{G}$ naively as $U \mapsto \mathscr{F}(U) \otimes \mathscr{G}(U)$, we won't necessarily get a sheaf and we sheafify this presheaf to get the actual definition of tensor product of two $\mathscr{O}_X$-modules.

What I would like to know is what is the intuitive reason why the naive definition is not a sheaf. For example, on $\mathbb{P}^n = \mathrm{Proj}\,k[x_0,\dots, x_n]$, if we take the hyperplane $H=V(x_0)$, then $\Gamma(\mathbb{P}^n, \mathscr{O}_{\mathbb{P}^n}(1)\otimes \mathscr{O}_H) \ne \Gamma(\mathbb{P}^n, \mathscr{O}_{\mathbb{P}^n}(1))\otimes_k \Gamma(H,\mathscr{O}_H)$. This is because on the left hand side, the global section $x_0$ got killed by $\mathscr{O}_H$, but not on the right hand side. However, I have a bad intuitive understanding still of why the two sides aren't equal. Also, I would like to know about more exotic cases as well.

Question: What causes the definition $U \mapsto \mathscr{F}(U) \otimes \mathscr{G}(U)$ to fail to be a sheaf?

Your help is greatly appreciated!

Solution 1:

When $\mathscr{F}$ and $\mathscr{G}$ are both invertible sheaves of $\mathcal{O}_X$-modules, the failure of the presheaf $U \mapsto \mathscr{F}(U) \otimes_{\mathcal{O}(U)} \mathscr{G}(U)$ to satisfy the sheaf axiom can be viewed as a failure of the $\mathscr{F}$ and $\mathscr{G}$ to be trivialized on a common open cover of $X$ "in a compatible manner".

What do I mean by this? Suppose there is an open set $U \subset X$ on which both $\mathscr{F}$ and $\mathscr{G}$ are trivialized, and moreover we have that for any open subset $V \subset U$, we have isomorphisms $\mathcal{O}(V) \to \mathscr{F}(V)$ and $\mathcal{O}(V) \to \mathscr{G}(V)$, and these respect restrictions. Then, one can check that the naive definition of $(\mathscr{F} \otimes_{\mathcal{O}_X} \mathscr{G})(U) = \mathscr{F}(U) \otimes_{\mathcal{O}(U)} \mathscr{G}(U)$ will satisfy the sheaf axiom.

This will unfortunately not happen in general, so sheafification tells us that we should instead define the tensor product sheaf $\mathscr{F} \otimes_{\mathcal{O}_X} \mathscr{G}$ by the rule $$ (\mathscr{F} \otimes_{\mathcal{O}_X} \mathscr{G})(U) := \left\{ (s_i) \in \prod_i \mathscr{F}(U \cap U_i) \otimes_{\mathcal{O}(U \cap U_i)} \mathscr{G}(U \cap U_i) \colon s_i |_{U \cap U_i \cap U_j} = s_j |_{U \cap U_i \cap U_j } \right\}, $$ where $\{ U_i \}$ is an open cover of $X$ on which $\mathscr{F}$ and $\mathscr{G}$ are both trivialized (but not necessarily with the compatibility conditions).

Another (quite spectacular) example to keep in mind, which demonstrates that the tensor product of sheaves is not just taking the tensor at the level of sections, can be found in this post, where Matt E. remarks in his answer that, for $n > 0$, $$ k = \Gamma(\mathbb{P}^d_{k}, \mathcal{O}_{\mathbb{P}^d}) = \Gamma(\mathbb{P}^d_k, \mathcal{O}(n) \otimes \mathcal{O}(-n)) \not= \Gamma(\mathbb{P}^d_k, \mathcal{O}(n)) \otimes_k \underbrace{\Gamma(\mathbb{P}^d_k, \mathcal{O}(-n))}_{=0} = 0. $$