# Tensor product of invertible sheaves

Given two invertible sheaves $\mathcal{F}$ and $\mathcal{G}$, one can define their tensor product, but in this definition $\mathcal{F} \otimes \mathcal{G} (U)$ is (apparently) not simply equal to $\mathcal{F} (U) \otimes \mathcal{G}(U)$ for an open set $U$. This latter structure is a presheaf; can someone give me an example for when it is not a sheaf? Is there a characterization for when this actually does give a sheaf?

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If for example $\mathcal F = \mathcal O_X,$ then $\mathcal F(U)\otimes_{\mathcal O_X(U)}\mathcal G(U) = \mathcal O_X(U)\otimes_{\mathcal O_X(U)} \mathcal G(U) = \mathcal G(U)$, so in this case we see that the presheaf $\mathcal F\otimes \mathcal G$ is equal to $\mathcal G$, and so is a sheaf.

But this is not typical. A more illustrative example is $\mathcal F = \mathcal O(n)$ for $n > 0$ (on some positive dimensional projective space $\mathbb P^d$) and $\mathcal G = \mathcal O(-n) = \mathcal F^{-1}$. Then if $U = \mathbb P^d$, we have $\mathcal G(\mathbb P^d) = 0$, and so the values of the presheaf $\mathcal F\otimes\mathcal G$ on $\mathbb P^d$ is equal to $0$. On the other hand, the actual (sheaf) tensor product $\mathcal F \otimes \mathcal G$ is equal to the structure sheaf $\mathcal O_{\mathbb P^d}$ (because $\mathcal F$ and $\mathcal G$ are mutually inverse), which has a one-dimensional space of global sections.

So rather than trying to find situations when the presheaf tensor product is actually a sheaf, you will be better off getting an intuition for the sheafification process that goes into forming the sheaf tensor product from the presheaf version.

If $\mathcal F$ is a presheaf and $\overline{\mathcal F}$ the corresponding sheaf, then two key points are:

• there is a canonical map $\mathcal F(U) \to \overline{\mathcal F}(U)$ for any open set $U$;

• the natural map on stalks $\mathcal F_x \to \overline{\mathcal F}_x$ is an isomorphism at every point $x$.

Thus, in the context of tensor products, there is always a canonical map $\mathcal F(U)\otimes \mathcal G(U) \to (\mathcal F\otimes\mathcal G)(U)$ (where the target denotes the sheaf tensor product), and the stalk of $\mathcal F\otimes\mathcal G$ at any point is the tensor product of the corresponding stalks of $\mathcal F$ and $\mathcal G$.

Finally, if $X =$ Spec $A$ is affine and $\mathcal F$ and $\mathcal G$ correspond to $A$-modules $M$ and $N$ respectively, then $\mathcal F\otimes\mathcal G$ is the invertible sheaf associated to the product $M\otimes_A N$. In particular, if $U =$ Spec $A_f$ is a distinguished open associated to $f \in A$, then the sections of $\mathcal F\otimes \mathcal G$ are equal to $(M\otimes_A N)_f = M_f\otimes_{A_f}N_f,$ which is the tensor product of the sections of $\mathcal F$ and $\mathcal G$ on Spec $A_f$. So this is one case when one can work with the naive presheaf picture and get the correct answer, which is often useful in calculations (and also psychologically helpful in keeping a down-to-earth perspective on the general formalism).

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I'm not an algebraist and maybe Matt will bail me out, but here is an intuitive take on this question.

An invertible sheaf is usually associated to a divisor, and if $D$ is a divisor, then sections of the invertible sheaf $\mathcal{O}(D)$ are equivalent to meromorphic functions with pole divisor dominated by $D$. So sections of $\mathcal{O}(D) \otimes \mathcal{O}(E)$ are meromorphic functions with pole divisor dominated by $D+E$.

Hence any product of a section of $\mathcal{O}(D)$ with a section of $\mathcal{O}(E)$ will give a section of $O(D+E)$, but there is no reason to expect that every meromorphic function with pole divisor dominated by $D+E$ is such a product.

E.g. on an elliptic curve, there are non constant meromorphic functions with 2 poles at any two points $P$, $Q$, but there are only constant functions with pole divisor $P$. Thus we cannot factor a global non constant meromorphic function with pole divisor $P+Q$, as a product of functions each having only one pole. We can however factor it locally on proper open sets.

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