# Fundamental result on the projective tensor product of sections of a vector bundle

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?

• 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 $M,N$.
– Z. M
Commented Apr 23, 2022 at 10:27
• 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. Commented Apr 23, 2022 at 10:40
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).
• Could you give me a source that proof that $\Gamma(M, V\times M) \cong C^\infty(M) \otimes V$ for trivial bundle ? Commented Apr 26, 2022 at 1:41
• @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). Commented Apr 26, 2022 at 10:57