Definition Let $E_{1} \xrightarrow{\pi_{1}} M_{1}$, $E_{2} \xrightarrow{\pi_{2}} M_{2}$ be two vector bundles over $M_{1}$ and $M_{2}$ with fibers $V_{1}$, $V_{2}$ respectively. The exterior tensor product $E_{1} \boxtimes E_{2}$ is defined as the vector bundle over $M_{1} \times M_{2}$, whose fiber over $(x, y) \in M_{1} \times M_{2}$ is $E_{1 x} \otimes E_{2 y}$, where $E_{1 x}$ is the fibre of $E_{1}$ over $x$ and $E_{2 y}$ is the fibre of $E_{2}$ over $y$.

In the article Brouder, Dang, Laurent-Gengoux, and Rejzner - Properties of field functionals and characterization of local functionals they have the so called fundamental result on the projective tensor product of sections of a vector bundle

Proposition III.8. Let $\Gamma(M, B)$ be the space of smooth sections of some smooth finite rank vector bundle $B \rightarrow M$ on a manifold $M$. Then $\Gamma(M, B)^{\hat{\otimes}_{\pi} k}=\Gamma\left(M^{k}, B^{\boxtimes k}\right)$.

Where $\boxtimes$ means exterior tensor product and $\otimes_\pi$ is the projective tensor product.

I am trying to prove this theorem. My first step is to prove the algebraic isomorphism. This is how I am trying to prove the algebraic isomorphism by following the same steps as in A nonlinear theory of generalized tensor fields on Riemannian manifolds by Eduard Nigsch at page 89:

Let $V$ be the fiber of the trivial vector bundle $B \xrightarrow{\pi} M$. We know there is a bilinear map from $\alpha :V\times V \rightarrow V\otimes V$ with $$\alpha(v_1,v_2)=v_1 \otimes v_2$$

where $v_1,v_2 \in V$.

Now let $\psi :B \rightarrow U\times V$ be a local trivialization map with $U \subset M$.

Now we define the local sections $\sigma_i$ by $\sigma_i(x) =\psi^{-1}(x,v_i)$.

We also define the local sections $\gamma_{ij}$ by $\gamma_{ij} =\phi^{-1}(x,y,v_i\otimes v_j)$ where $\phi :B{\boxtimes}B \rightarrow U^2 \times V\otimes V$ is a local trivialization map of the bundle $B{\boxtimes}B \rightarrow M \times M$.

Define the map $g :\Gamma\left(M, B\right) \times \Gamma\left(M, B\right) \rightarrow \Gamma\left(M\times M, B{\boxtimes}B\right)$ by $$g=\sum_{i, j} m \circ\left(i d \times m\left(\cdot, \gamma_{i j}\right)\right) \circ\left(\sigma_{i}^{*} \times \sigma_{j}^{*}\right)$$

where $m: C^{\infty}(M) \times \Gamma(B\otimes B)\rightarrow \Gamma(B\otimes B) $ is the module multiplication on $\Gamma(B\otimes B)$ and $\sigma_{i}^{*}$ is the dual of $\sigma_{i}$.

By the properties of the tensor product there is a bilinear map $f: \Gamma(M, B)\otimes \Gamma(M, B)\rightarrow \Gamma\left(M\times M, B{\boxtimes}B\right)$ such that if $g$ is the map $h:\Gamma\left(M, B\right) \times \Gamma\left(M, B\right) \rightarrow \Gamma(M, B)\otimes \Gamma(M, B)$ we have that $$g=f \circ h.$$

For the inverse define $h^{-1}: \Gamma\left(M\times M, B{\boxtimes}B\right)\rightarrow \Gamma(M, B)\otimes \Gamma(M, B)$ $h^{-1}(s)=\sum_{i, j} \gamma_{i j}^{*}(s) \alpha_{i} \otimes \beta_{j}$ for $s \in \Gamma\left(M\times M, B{\boxtimes}B\right)$

Now we have that $$ \begin{aligned} &h^{-1}\left(f\left(t \otimes u\right)\right)=h^{-1}\left(h \left(t^{i} \alpha_{i}, u^{j} \beta_{j}\right)\right)=h^{-1} \left(t^{i} u^{j} \gamma_{i j}\right)=t^{i} u^{j} \alpha_{i} \otimes \beta_{j}\\ &=t^{i} \alpha_{i} \otimes u^{j} \beta_{j}=t \otimes u \text { and }\\ &h(h^{-1}(s))=h\left(s^{i j} \alpha_{i} \otimes \beta_{j}\right)=s^{i j} g\left(\alpha_{i}, \beta_{j}\right)=s^{i j} \gamma_{i j}=s . \end{aligned} $$

Is my proof correct?

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    $\begingroup$ It suffices to establish the result for trivial line bundles, namely, $C^\infty(M\times N)\cong C^\infty(M)\otimes C^\infty(N)$ for two manifolds $M,N$. $\endgroup$
    – Z. M
    Commented Apr 23, 2022 at 10:27
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    $\begingroup$ That's right, there is no reason to get your hands dirty with local trivializations. When $B$ is trivial with fiber $V$, $\Gamma(M,B) = C^\infty(M)\otimes V$, so then it is enough to use the associativity and commutativity (up to isomorphism) of $\otimes$. In addition, every vector bundle $B$ is the image of an idempotent projection acting fiber-wise on some trivial vector bundle. So complete the argument by distributing the tensor product over direct sums. $\endgroup$ Commented Apr 23, 2022 at 10:40
  • $\begingroup$ Could you extend your comments as an answer? $\endgroup$ Commented Apr 23, 2022 at 10:48
  • $\begingroup$ @Igor Khavkine You did not say if my proof is correct $\endgroup$ Commented Apr 24, 2022 at 8:20
  • $\begingroup$ Sorry, I myself also don't want to deal with local trivializations, so I didn't read your proof carefully. If any proof for you is fine, I can turn my comment into an answer a bit later. $\endgroup$ Commented Apr 24, 2022 at 19:31

1 Answer 1


The tensor product (either one) is symmetric $A \otimes B \cong B \otimes A$, associative $(A\otimes B) \otimes C \cong A \otimes (B \otimes C)$ and distributive over direct sums $A \otimes (B \oplus C) \cong (A\otimes B) \oplus (A\otimes C)$. Spaces of sections preserve direct sum decompositions $\Gamma(M, E_1 \oplus E_2) \cong \Gamma(M,E_1) \oplus \Gamma(M,E_2)$. Any vector bundle $E\to M$ can be realized as a summand of a trivial vector bundle $(V\times M \to M) \cong (E\oplus E' \to M)$, where we can realize the sub-bundles $E$ and $E'$ and the image and kernel of a fiber-wise projection $P$ on $V\times M$ (it is sufficient that $M$ is contractible to a compact space). The completed tensor product satisfies the identity $C^\infty(M_1) \otimes C^\infty(M_2) \cong C^\infty(M_1 \times M_2)$ (for details see Trèves opological Vector Spaces, Distributions and Kernels 1970).

With the above background, the proof is straightforward. Start with the observation that for a trivial bundle $\Gamma(M, V\times M) \cong C^\infty(M) \otimes V$. Hence $$\Gamma(M_1,V_1\times M_1) \otimes \Gamma(M_2,V_2\times M_2)\cong C^\infty(M_1\times M_2) \otimes (V_1\otimes V_2) \cong \Gamma(M_1\times M_2, (V_1 \times M_1) \boxtimes (V_2 \times M_2)).$$ Applying direct sum decompositions to both sides gives $$ (\Gamma(M_1,E_1) \oplus \Gamma(M_1,E'_1)) \otimes (\Gamma(M_2,E_2) \oplus \Gamma(M_2,E'_2)) \cong \Gamma(M_1\times M_2, (E_1\oplus E'_1) \boxtimes (E_2\oplus E'_2)) . $$ Finally, the desired isomorphism $$ \Gamma(M_1,E_1) \otimes \Gamma(M_2,E_2) \cong \Gamma(M_1\times M_2, E_1 \boxtimes E_2) $$ follows because both sides coincide with the image of the projection $P_1\otimes P_2$ (interpreted as acting fiber-wise or on spaces of sections as needed).

  • $\begingroup$ Could you give me a source that proof that $\Gamma(M, V\times M) \cong C^\infty(M) \otimes V$ for trivial bundle ? $\endgroup$ Commented Apr 26, 2022 at 1:41
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    $\begingroup$ @amiltonmoreira It's hard to give a precise reference, since this observation is at the level of a textbook exercise (fix a basis for $V$ and work in components). $\endgroup$ Commented Apr 26, 2022 at 10:57

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