This question is similar to

Conditions such that taking global sections of line bundles commutes with tensor product?


Tensor product of invertible sheaves

except that I am concerned with smooth manifolds. If $M$ is a smooth manifold and $V$ and $W$ are (smooth, real) vector bundles on $M$, are there reasonable conditions that ensure that the map $\Gamma(M,V)\otimes_{\mathscr{O}_M(M)}\Gamma(M,W)\rightarrow\Gamma(M,V\otimes W)$ is an isomorphism? Equivalently, are there any reasonable conditions under which taking global sections of finite locally free $\mathscr{O}_M$-modules commutes with tensor product?

I don't really know much smooth manifold theory, and my intuition is mostly derived from algebraic geometry, where the analogous question, as is indicated in the questions I linked to, is not a simple one, even for invertible sheaves. I'd always assumed the situation for smooth manifolds would be the same, but I was just looking at the Wikipedia page on vector-valued differential forms, and it seems (unless I'm misunderstanding which is certainly possible) to claim there that for $V$ an arbitrary vector bundle and $W$ the $p$-th exterior power of the cotangent bundle, $p\geq 1$, the answer to my question is "yes," with the justification being that "$\Gamma(M,-)$ is a monoidal functor." Looking at the definition of a monoidal functor, it seems to me that all this implies is that there is a map from the tensor product of the global sections to the global sections of the tensor product (which is clear from the definitions) and not that this map is an isomorphism.


1 Answer 1


This is true without any further assumptions on $M,V,W$ and discussed quite thoroughly in chapter 7 of Conlon's wonderful book "Differentiable Manifolds". The proof given in the text is quite elegant so I can't resist recalling the argument.

There is an obvious $C^{\infty}(M)$-bi-linear map $\alpha \colon \Gamma(V) \times \Gamma(W) \rightarrow \Gamma(V \otimes W)$ given by $\alpha(\xi,\eta)_p = \xi_p \otimes \eta_p$ which gives rise to a map $\alpha \colon \Gamma(V) \otimes_{C^{\infty}(M)} \Gamma(W) \rightarrow \Gamma(V \otimes W)$. We want to show that $\alpha$ is an isomorphism.

When $V$ and $W$ are trivial bundles of rank $n$ and $k$ respectively, one can choose $n$ pointwise linearly independent sections $\xi_i \in \Gamma(V)$ and $k$ pointwise linearly independent sections $\eta_j \in \Gamma(W)$ and then $\Gamma(V \otimes W)$ is easily seen to be a free $C^{\infty}(M)$ module with basis $\{ \xi_i \otimes \eta_j \}_{i,j}$ and so $\alpha$ is an isomorphism of $C^{\infty}(M)$ modules.

If $V$ and $W$ are not trivial, we interpret them as a direct summands of some trivial bundles. That is, we find vector bundles $V^{\perp}$ and $W^{\perp}$ such that $V \oplus V^{\perp}$ and $W \oplus W^{\perp}$ are trivial bundles. Then, by looking at the diagram

$$ \require{AMScd} \begin{CD} \Gamma((V \oplus V^{\perp}) \otimes (W \oplus W^{\perp})) @<{\tilde{\alpha}}<< \Gamma(V \oplus V^{\perp}) \otimes_{C^{\infty}(M)} \Gamma(W \oplus W^{\perp}) \\ @AAA @AAA \\ \Gamma(V \otimes W) @<{\alpha}<< \Gamma(V) \otimes_{C^{\infty}(M)} \Gamma(W) \end{CD} $$

we see that $\tilde{\alpha}$ is injective by the previous paragraph, and we can check directly that both vertical arrows (that are defined with the help of the injective bundle maps $v \mapsto (v,0)$ of $V \rightarrow V \oplus V^{\perp}$ and $w \mapsto (w,0)$ of $W \rightarrow W \oplus W^{\perp}$) are injective, showing that $\alpha$ is injective. Similarly, by reversing the direction of the vertical arrows and replacing them with projections, we see that $\alpha$ is also surjective.

The crux of the proof is the result that every bundle $V$ over $M$ can be realized as a subbundle of a trivial bundle $F$ (stated in terms of the module of global sections, this means that $\Gamma(V)$ is a $C^{\infty}(M)$-projective module). This can be shown by constructing an epimorphism $\psi \colon F \rightarrow V$ of vector bundles from a trivial bundle $F$ onto $V$ and using a fiber metric to split $F$ as an inner direct sum $F = \ker(\psi) \oplus \ker(\psi)^{\perp}$ with $\ker(\psi)^{\perp} \cong V$.

The construction of $\psi$ uses partition of unity. When $M$ is compact, one takes a partition of unity $\{\lambda_i\}_{i=1}^n$ subordinate to a cover $\{U_i\}_{i=1}^n$ of $M$ over which $V$ trivializes with generating global sections $\xi_i^j \in \Gamma(U_i,E|_{U_i})$. Then, one can define global sections $\sigma_i^j = \lambda_i \xi_i^j$ by zero extension outside $U_i$. We obtain finitely many sections, and by taking the trivial bundle with the vector space $X = \mathrm{span} \{ \sigma_i^j \} \subseteq \Gamma(V)$ as fiber, and defining $\psi \colon M \times X \rightarrow V$ as $\psi(p,\sum \sigma_i^j) = \sigma_i^j(p)$ we obtain the required map.


  1. One can use the ideas described in the proof above to show Swan's theorem about the equivalence of categories between the category of (smooth, finite rank) vector bundles over $M$ and the category of projective finitely generated $C^{\infty}(M)$ modules.
  2. In the algebraic / holomorphic setting, this fails badly, at least for projective varieties. While in the smooth category, you can identify a vector bundle with the module of global sections, in other settings the module of global sections doesn't hold enough information about the vector bundle and one should consider the sheaf of sections instead. This failure also provides a geometric example of why one needs to sheafify when taking tensor products (one expects that the sheaf of sections of the tensor product will be the tensor product of the sheaves of sections and so $\Gamma(\mathcal{O}(-1) \otimes \mathcal{O}(1)) \cong \Gamma(E(1) \otimes E(-1)) \cong \Gamma(E(0)) = \mathbb{C}$ which obviously cannot hold if you don't sheafify).
  3. The fact that every (continuous) finite rank vector bundle is a direct summand of a trivial bundle is true even if we merely assume that $M$ is compact and Hausdorff. One can get rid of the assumption that $M$ is compact when $M$ is a manifold, but cannot get rid of it in general. In Hatcher's "Vector Bundles and K-Theory", it is shown that the tautological line bundle over $M = \mathbb{RP}^{\infty}$ is not a direct summand of a trivial bundle using characteristic classes.
  • 1
    $\begingroup$ I think this does not follow from Serre Swan theorem because the equivalence of categories is just a plain equivalence (as it is stated in the theorem). The tensor product of modules is not a colimit so the equivalence itself is not enough to conclude. The equivalence is monoidal only a posteriori right? $\endgroup$
    – elidiot
    Oct 4, 2019 at 10:36
  • $\begingroup$ @elidiot: I think you're right. I've edited my answer, thanks! $\endgroup$
    – levap
    Oct 28, 2019 at 16:27
  • $\begingroup$ In the last part where you show that $\Gamma(V)$ is a projective $C^\infty(M)$-module, you're assuming that $M$ is compact; is the statement also true for non-compact $M$? ("no restrictions on $M$" seems to imply it is) $\endgroup$
    – philip98
    Aug 21, 2021 at 20:54
  • $\begingroup$ @philip98: Yes, but you need to use dimension theory. See page 236 in Conlon's book. $\endgroup$
    – levap
    Aug 21, 2021 at 21:07
  • $\begingroup$ Is my understanding correct : For tensor product of two holomorphic bundles which are in their own right smooth bundles also, any (holomorphic)section of the tensor product is a tensor product of two smooth sections but they needn't be holomorphic ? $\endgroup$ Mar 2, 2022 at 13:42

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