Let $E,F \to M$ be two smooth vector bundles over a compact manifold $M$. It is well-known that the homomorphism fields $Hom(E, F) \to M$ are a smooth vector bundle, too. In fact, this bundle can be thought of as $E^* \otimes F$. Denote by $\Gamma( \_)$ the smooth sections of a vector bundle. There is an isomorphism
$$ \Psi: \Gamma(Hom(E,F)) \to Hom(\Gamma(E), \Gamma(F)) $$
Here, $Hom(\Gamma(E), \Gamma(F))$ denotes the linear maps between the real vector spaces $\Gamma(E)$ and $\Gamma(F)$. This isomorphism is explicitely given by:
$$ \forall \alpha \in \Gamma(Hom(E, F)): \forall s \in \Gamma(E): \forall p \in M: (\Psi(\alpha))(s)(p) = \alpha_p(s_p) $$
Is an analogous statement true, if one replaces the smooth sections $\Gamma(\_)$ by the $L^2$-sections? More precisely, assume in addition that $E$ and $F$ carry Riemannian fibre metrics and that $M$ is a compact Riemannian manifold. These metrics induce a canonical fibre metric on $Hom(E, F)$ as well (pointwise it is given by the operator norm). Does $\Psi$ induce an isometry (or at least an isomorphism)
$$L^2(Hom(E, F)) \to Hom(L^2(E), L^2(F)) ?$$
Somehow, I don't get the exponents right in order to show that $\Psi(\alpha)$ is even a map from $L^2$ to $L^2$. :(
