I have the following exercise:
Let $E \rightarrow M$ be a vector bundle, $\nabla, \tilde{\nabla}$ two connections on $E$. Show that there exists $A \in \Omega^1(M,End(E)) := \Gamma(T^{*}M \otimes End(E))$, s.t. $\tilde{\nabla}=\nabla+A$.
So far, I have that $\Gamma(T^{*}M \otimes End(E))\cong Mult_{C^{\infty}(M)}(X(M) \times \Gamma(E) \times \Gamma(E^{*}))$. Now it's obvius to define $A:= \tilde{\nabla}-\nabla$ and to show that $\tilde{\nabla}-\nabla \in Mult_{C^{\infty}(M)}(X(M) \times \Gamma(E) \times \Gamma(E^{*}))$. Is that correct?
Now I know that $A(X,s) \in Hom_{C^{\infty}(M)}(\Gamma(E^{*}, C^{\infty}(M)) \cong \Gamma(E)$ but I'm not sure how to shows multilinearity here.
Can anybody give me a hint?
In what follows, let $(\eta,e) \in C^\infty(M)$ denote the pairing of $\eta \in \Gamma(E^\ast)$ with $e \in \Gamma(E)$. Suppose that $X \in \mathfrak{X}(M)$, $e \in \Gamma(E)$, $\eta \in \Gamma(E^\ast)$, and consider $$ A(X,e,\eta) := (\eta,\tilde{\nabla}_Xe-\nabla_X e). $$ That $A$ is $C^\infty(X)$-linear in the third argument is trivial; that $A$ is $C^\infty(X)$-linear in the first argument follows from $C^\infty(X)$-linearity of a covariant derivative $\nabla_X$ or $\tilde{\nabla}_X$ in the vector field $X$. Can you use the Leibniz rule for covariant derivatives to prove $C^\infty(X)$-linearity of $A$ in the second argument?