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?


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 quasi-coherent 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).


I'm not an algebraist, and maybe Matt will bail me out, but here is an intuitive take on this question, (from the complex geometry viewpoint if you like, or just substitute "rational" for "meromorphic").

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.

edit: I apologize if my example is too special to be helpful. I think Matt has hit it on the head by emphasizing the process of forming a sheaf from a presheaf. From my naive viewpoint, a sheaf is formed from a presheaf of objects having some property P, by taking all objects locally having property P. In this case then, if I understand it, the key point is that often the property of an element (e.g. a stalk - valued function) belonging to the tensor product of global sections is more restrictive than that of locally belonging to the tensor product of sections. My elliptic curve example was meant as an explicit example of this phenomenon.

  • 1
    $\begingroup$ I am assuming you are a complex geometer? You should mention so in the answer if this is the case, that it is the point of view you chose. $\endgroup$ – Patrick Da Silva Sep 6 '14 at 15:15

Sorry for resurrecting such an old thread yet again...

This following paragraph is basically subsumed in Professor Emerton's and Smith's terrific answers, albeit in my own words. Let $p$ be a point of $\mathbb{P}^1$, and let $\mathcal{O}_{\mathbb{P}^1}(-p)$ denote the sheaf of functions which have a zero at $p$ (in other words, this is the ideal sheaf of $p$ in $\mathbb{P}^1$). Then $\mathcal{O}_{\mathbb{P}^1}(-p)$ has no global sections, since the only global sections are constants, so that the naïve definition of the tensor product has no global sections. However, the tensor product of these two sheaves is nothing more than $\mathcal{O}_{\mathbb{P}^1}$, the sheaf of holomorphic functions (the idea being that the two conditions, pole at $p$ and zero at $p$ cancel each other), which has plenty of global sections (the constants).

This following paragraph is not subsumed, though. For a very explicit example of where the naïve definition of the tensor product of two sheaves is not necessarily a sheaf, let $X = \{a, b\}$ be a two point space, with open sets $\{\emptyset, \{a\}, \{b\}, \{a, b\}\}$. Define $\mathcal{F}$ by$$\mathcal{F}(\{a\}) = \mathbb{Z}/2\mathbb{Z},\text{ }\mathcal{F}(\{b\}) = \mathbb{Z}/2\mathbb{Z},\text{ }\mathcal{F}(\{a, b\}) = \mathbb{Z}/2\mathbb{Z} \oplus \mathbb{Z}/2\mathbb{Z},$$and $\mathcal{G}$ by$$\mathcal{G}(\{a\}) = \mathbb{Z}/2\mathbb{Z},\text{ }\mathcal{G}(\{b\}) = \mathbb{Z}/3\mathbb{Z},\text{ }\mathcal{G}(\{a, b\}) = \mathbb{Z}/2\mathbb{Z} \oplus \mathbb{Z}/3\mathbb{Z}.$$Then the naïve definition of the tensor product would give$$(\mathcal{F} \otimes \mathcal{G})(\{a\}) = \mathbb{Z}/2\mathbb{Z},\text{ }(\mathcal{F} \otimes \mathcal{G})(\{b\}) = 0,\text{ }(\mathcal{F} \otimes \mathcal{G})(\{a, b\}) = \mathbb{Z}/2\mathbb{Z} \oplus \mathbb{Z}/2\mathbb{Z},$$which is clearly not a sheaf.

  • 2
    $\begingroup$ This does not make sense unless you explain what the restriction maps are. $\endgroup$ – Georges Elencwajg Apr 23 '16 at 16:52

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.