Let $G_i\to P_i\to M$ be principal fiber bundles with representations $\rho_i\colon G_i\to\mathrm{GL}(V_i)$ and associated vector bundles $E_i\to M$. Given local sections $s_i\colon U\to P_i$, I expect that a bijection \begin{equation} C^\infty(U,V_1\otimes\cdots\otimes V_n)\to\Gamma(U,E_1\otimes\cdots\otimes E_n) \end{equation} can be constructed.

Here's my guess:

For all $m\in U\subset M$, \begin{align} V_i&\to (E_i)_m\\ v&\mapsto[s_i(m),v] \end{align} is a bijection$^1$ and we define \begin{equation} \Phi_m\colon V_1\otimes\cdots\otimes V_n\to(E_1)_m\otimes\cdots\otimes(E_n)_m \end{equation} to be the unique isomorphism s.t. $\Phi_m(v_1\otimes\cdots\otimes v_n)=[s_1(m),v_1]\otimes\cdots\otimes[s_n(m),v_n]$.

The isomorphism \begin{equation} \Phi\colon C^\infty(U,V_1\otimes\cdots\otimes V_n)\to\Gamma(U,E_1\otimes\cdots\otimes E_n) \end{equation} is then defined by \begin{equation} (\Phi(f))(m)=(\Phi_m\circ f)(m). \end{equation}

$^1$This follows from the definition of the equivalence classes and the fact that $G\ni g\to pg\in P$ is a bijection for all $p\in M$ (according to the definition of principal fiber bundles).

  • $\begingroup$ What exactly is the question? Your $\Phi$ seems to me like a good guess... $\endgroup$ – nicrot000 Jun 7 at 15:21
  • $\begingroup$ @nicrot000 Since it was only a guess, I wished to have a confirmation. $\endgroup$ – Filippo Jun 7 at 19:26

So, since you asked in the comments above, here is your confirmation.

Maybe I can add, that since you have sections $s_i$ on $U$ in each of your principal bundles, they are all trivializable over that same $U$, consequently each of the associated vector bundles is trivializable over this $U$.

On the other hand, $C^\infty(U,V_1\otimes\cdots\otimes V_n)$ is a section space over $U$ in the trivial bundle $U\times(V_1\otimes\cdots\otimes V_n)$ and applying the inverse of the respective trivialization to each factor should simply give you the isomorphism $$ C^\infty(U,V_1\otimes\cdots\otimes V_n)\to\Gamma(U,E_1\otimes\cdots\otimes E_n). $$ The (inverse) trivilization in each factor is then, as you remarked in the footnote, given by $v\mapsto[s_i(m),v]$.

  • $\begingroup$ Please let me know if I understand your answer correctly: Consider a vector bundle $\pi\colon E\to M$ with general fiber $V$ and a bundle atlas $A$. Each bundle chart $\phi\colon\pi^{-1}(U)\to U\times V$ in $A$ is bijective and therefore defines a bijection $\Gamma(U,E)\to\Gamma(U,M\times V)$. Since $\Gamma(U,M\times V)$ is isomorphic to the set of functions from $U$ to $V$, a bundle chart allows us to identify local sections with vector valued functions. My question is about the special case where $E$ is the tensor product bundle of associated vector bundles. $\endgroup$ – Filippo Jun 8 at 14:31
  • $\begingroup$ It doesn't matter if $V$ is just a generic space, or if $V$ is assumed to be a space of the shape $V=V_1\otimes...\otimes V_n$ with some other spaces $V_1,\dots,V_n$. I think I can't point out where exactly your confusion lies in. $\endgroup$ – nicrot000 Jun 8 at 17:42
  • $\begingroup$ For the first part, yes this is correct, up to the detail that the bijectivity of $\phi$ alone is not enough, you really need that $\phi$ is an isomorphism of the bundles $\pi^{-1}(U)$ and $U\times V$. $\endgroup$ – nicrot000 Jun 8 at 17:46

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