Why is the space of all connection on a vector bundle an affine space?

Question

I think this result is very well known, but I don't understand its proof.

Let E a vector bundle over a manifold M, and $\Omega^i(E):=\Gamma(\Lambda^iT^*M\otimes E)$ the space of E-valued differential forms of degree i. A connection on E is a map $\nabla: \Omega^0(E)=\Gamma(E)\rightarrow\Omega^1(E)$. The claim is that for two arbitrary connetions $\nabla_1$ and $\nabla_2$ on E, their difference $A=\nabla_1-\nabla_2$ is in $\Omega^1(End(E))$, so the space of connections is an affine space for the vector space $\Omega^1(End(E))$.

My problem is the appearance of $\Omega^1(End(E))$. The key point of the proof is to show that for any section $s\in\Gamma(E)$, the expression $A(s)(p) = ((\nabla_1-\nabla_2)s)(p)$ depends only on s(p). The conclusion is:

Since $(A(s))(p)$ depends on $s$ only through $s(p)$, it follows that $A\in\Omega^1(End(E))$.

Why can one make this conclusion?

I have been given more details on that:

The following map is $\mathbb{R}$-linear: $$\begin{align} E_p&\stackrel{A_p}{\longrightarrow}T^*_pM\otimes E\\ E_p\ni s&\mapsto (A(s))(p) \end{align}$$ therefore $$\begin{align} A_p &\in Hom(E_p, T^*_pM\otimes E_p)\\ &=E^*_p\otimes T^*_pM\otimes E_p\\ &=(E^*_p\otimes E_p)\otimes T^*_pM\\ &=End(E_p)\otimes T^*_pM \end{align}$$ Here, I don't understand the second line: why is $(A(s))(p)$ in $T^*_pM\otimes E$? In other words, why is $A$ of the form $$ A(s) = \omega\otimes s'(s),\qquad\omega\in\Omega^1(M),\enspace s'(s)\in\Gamma(E) $$ and $\omega$ does not depend on $s$?

EDIT: what I have learned from the answer of Jeremy Daniel

Let $E\rightarrow M$ a vector bundle, $\nabla$ a connection on E and $A=\nabla_1-\nabla_2$. Then locally, for any $s\in\Gamma(E)$, we have the assignment: $$\begin{align} \nabla:s&\mapsto\sum_i\left(\omega_i(s):p\mapsto(\omega_is)(p)\right)\otimes dx^i&(\nabla s):M\rightarrow T^*M\otimes E\\ A:s&\mapsto\sum_i\left(\alpha_i:p\mapsto\alpha_i(s(p))\right)\otimes dx^i&(As):M\rightarrow T^*M\otimes E \end{align}$$ Although both $(\nabla s)$ and $(As)$ are maps $M\rightarrow T^*M\otimes E$, the difference is in the maps $\omega_i$ and $\alpha_i$, which are of different nature. To assign a value to $p$, $\omega_i$ has to consider information in a neighbourhood of $p$, e.g. derivatives of a given section $s$ at p, only with this data $(w_is)(p)$ can be computed. In contrast, $\alpha_i$ does not need this information and depends only on the value $s(p)$ at that given point, thus it is a tensor. It maps points in a fiber $E_p$ to points in the same fiber, thus $\left.\alpha_i(s)\right|_{E_p}\in End(E_p)$.

Comments to this are appreciated!

Answer 1

There is a useful lemma, that is often called "fundamental lemma of differential geometry" or "tensoriality lemma" that says the following:

If $E$ and $F$ are two vector bundles over $X$, and $A$ is a linear map from $\Gamma(E)$ to $\Gamma(F)$ (hence defined on global sections) then, $A$ is $C^\infty(M)$- linear iff there exists a vector bundle map $\alpha: E \rightarrow F$, such that $A(f)(x) = \alpha(f(x))$, where $f$ is a section of $E$ (hence, $A$ is defined pointwise).

This is proved by using smooth functions with little compact support in a neighborhood of $x$, which is equal to $1$ in a neighborhood of $x$.

Given this lemma, you have $A:= \nabla_1 - \nabla_2: \Gamma(E) \rightarrow \Gamma(\Omega^1(E))$. Compute $\nabla_1 - \nabla_2(fs)$ where $f$ is in $C^\infty(M)$ and $s$ is in $\Gamma(E)$. You find:

$$ (\nabla_1 - \nabla_2) (fs) = df \otimes s + f \nabla_1 s - df \otimes s - f \nabla_2 s = f(\nabla_1 - \nabla_2)s. $$

This proves that $A$ is canonically associated to a vector bundle map $E \rightarrow \Omega^1(E)$, that is a tensor in $\Omega^1(End(E))$.