Let \((X, \mathcal{O}_X )\) be a locally ringed topological space. It is well known that the tensor product of sheaves of \(\mathcal{O}_{ X }\)-modules is given by the sheafification of the section-wise tensor product. I was wondering if there was any situation in which this section-wise tensor-product is already a sheaf. Due to my background, I will mostly concentrate on the case when \(X\) is a scheme.

Notation

Before starting with the maths, let’s fix some notation. Throughout the post \(U\) will denote an open subset of \(X\). With \(\mathcal{F}\) and \(\mathcal{G}\) we will refer to elements of \(\mathsf{QCoh}(X)\). Denote

\[ \mathcal{F} \widetilde{\otimes} \mathcal{G}(U) : = \mathcal{F}(U) \otimes_{ \mathcal{O}_{ X }(U) } \mathcal{G}(U) \]

the section-wise tensor product. Let’s also employ the following slight abuse of notation

\[ \mathcal{F} \otimes \mathcal{G} : = \mathcal{F} \otimes_{\mathcal{O}_{ X }} \mathcal{G}. \]

Affine schemes

We will later need to understand the local situation, therefore let’s start with an affine scheme \(X : = \operatorname{Spec}(A)\). One can show the following.

Lemma 1

Take \(V \subset X\) affine open, then for all \(\mathcal{F}, \mathcal{G} \in \mathsf{QCoh}(X)\) we have

\[ \mathcal{F} \otimes \mathcal{G}(V) \simeq \mathcal{F}(V) \otimes_{\mathcal{O}_{ X }(V)} \mathcal{G}(V). \]

Instead of proving this lemma here, I will just cite tag 01IB of the stacks project as a reference for this fact.

Moreover, through the wonders of stackexchange one can also learn that the previous isomorphism does not hold for general open subsets1!

Free sheaves

Claim 2

The section-wise tensor product with a free \(\mathcal{O}_{ X }\)-module of finite rank is a sheaf.

\[ \mathcal{F}(U) \otimes_{ \mathcal{O}_{ X }(U) } \mathcal{O}_{ X }^n(U) \simeq \bigoplus_{ i = 1 }^n \mathcal{F}(U) \simeq \prod_{ i = 1 }^n \mathcal{F}(U). \]

Products of sheaves are computed section-wise2, therefore the right-hand side defines a sheaf.

Locally-free sheaves

For locally-free sheaves, instead, the above claim already fails. Indeed, let’s show the following.

Claim 3

The section-wise tensor product of locally-free sheaves of \(\mathcal{O}_{ X }\)-modules need not be a sheaf.

It is enough to provide an example to show this claim. Choose \((X, \mathcal{O}_{ X }) : = \mathbb{P}^1_{ \mathbb{C}}\). It is well known that \(\mathrm{Pic}(X) \simeq \mathbb{Z}\), i.e. all line bundles on \(X\) are of the form \(\mathcal{O}(n)\) for some \(n \in \mathbb{Z}\). Moreover, we know how the global sections of these sheaves look. For our purposes, we only need that \(\mathcal{O}(n)(X)\) is non-trivial if \(n \geq 0\) and \(\mathcal{O}(n)(X) = 0\) if \(n < 0\). Finally, from the isomorphism of groups \(\operatorname{Pic}(X) \simeq \mathbb{Z}\), we get that \(\mathcal{O}(n) \otimes \mathcal{O}(m) \simeq \mathcal{O}(n+m)\).

We now pick \(n=2\) and \(m=-1\) in the above equality. Then

\[ \mathcal{O}(n) \otimes \mathcal{O}(m) (X) \simeq \mathcal{O}(1)(X) \neq 0 \]

Moreover, \(\mathcal{O}(1)\) is a sheaf. Taking the standard open cover of \(\mathbb{P}^1\) by \(\mathbb{A}^1\), we get

\[ \mathcal{O}(1)(X) \simeq \ker \left( \prod_{ i=1 }^2 \mathcal{O}(2) \otimes \mathcal{O}(-1) (\mathbb{A}^1) \to \mathcal{O}(2) \otimes \mathcal{O}(-1) (\mathbb{G}_m)\right) \]

It is possible to show3 that, for \(i\colon \mathbb{A}^1 \hookrightarrow \mathbb{P}^1\),

\[ \mathcal{O}(2) \otimes \mathcal{O}(-1) (\mathbb{A}^1) \simeq i^{-1}\mathcal{O}(2) \otimes_{ \mathcal{O}_{ \mathbb{A}^1 } } i^{-1} \mathcal{O}(-1) (\mathbb{A}^1), \]

where the first tensor product is taken in \(\mathsf{QCoh}(X)\) and the second in \(\mathsf{QCoh}(\mathbb{A}^1)\). Here we can invoke the lemma on affines and obtain

\[ i^{-1}\mathcal{O}(2) \otimes_{ \mathcal{O}_{ \mathbb{A}^1 } } i^{-1} \mathcal{O}(-1) (\mathbb{A}^1) \simeq \mathcal{O}(2)(\mathbb{A}^1) \otimes_{ \mathcal{O}_{ \mathbb{A}^1 }(\mathbb{A}^1) } \mathcal{O}(-1) (\mathbb{A}^1). \]

The same argument, applied to \(\mathbb{G}_m\), shows that

\[ \mathcal{O}(2) \otimes \mathcal{O}(-1) (\mathbb{G}_m) \simeq \mathcal{O}(2)(\mathbb{G}_m) \otimes_{ \mathcal{O}_{ \mathbb{G}_m }(\mathbb{G}_m) } \mathcal{O}(-1) (\mathbb{G}_m). \]

Putting this all together, this means that the sheaf condition reads

\[ \mathcal{O}(1)(X) \simeq \ker \left( \prod_{ i=1 }^2 \mathcal{O}(2)(\mathbb{A}^1) \otimes_{ \mathcal{O}_{ \mathbb{A}^1 }(\mathbb{A}^1) } \mathcal{O}(-1) (\mathbb{A}^1) \to \mathcal{O}(2)(\mathbb{G}_m) \otimes_{ \mathcal{O}_{ \mathbb{G}_m }(\mathbb{G}_m) } \mathcal{O}(-1) (\mathbb{G}_m)\right). \]

If \(\mathcal{O}(2) \widetilde{\otimes} \mathcal{O}(-1)\) was to be a sheaf, the left hand side should read

\[ \mathcal{O}(2) \widetilde{\otimes} \mathcal{O}(-1) (X) \simeq \mathcal{O}(2)(X) \otimes_{ \mathbb{C} } 0 \simeq 0. \]

We finally have a contradiction.

  1. The given example is not flat though, so one could wonder if adding some hypothesis on the sheaves could save the day. I still have not worked out if the tensor product of locally-free sheaves on affines can be computed section-wise. ↩︎

  2. Heuristically: products are limits, and the sheaf condition is a limit. It is therefore reasonable that they commute, hence that product of sheaves are computed section-wise. Rigorously: \(\operatorname{for}\colon \mathsf{Sh}(X) \to \mathsf{PSh}(X)\) admits a left adjoint (sheafification), therefore it commutes with limits. ↩︎

  3. This boils down to the following fact. Take a presheaf \(\mathcal{P}\) on \(X\), and denote by \(\widetilde{\mathcal{P}}\) its sheafification. Write \(i\colon U \hookrightarrow X\). Then \(\widetilde{\mathcal{P}}(U) \simeq \widetilde{i^{-1} \mathcal{P}}(U)\), where the second sheafification is done on \(U\). One can show this for example by proving that the espace étalé associated to \(i^{-1}\mathcal{P}\) is endowed with the induced topology with respect to the obvious inclusion into the espace étalé associated to \(\mathcal{P}\). ↩︎