3
$\begingroup$

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?

$\endgroup$
6
  • 2
    $\begingroup$ 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
  • 1
    $\begingroup$ 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

1
+150
$\begingroup$

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).

$\endgroup$
2
  • $\begingroup$ 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
  • 1
    $\begingroup$ @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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.