Connection on a bundle

Question

Let $\nabla_A$ and $\nabla_B$ be two connections on a vector field $E$ over a compact manifold $M$, we know that $\nabla_B=\nabla_A+a$ for some $a\in\Omega^1(X;\text{End}(E))$. And for any smooth section $s$ of $E$ we can write $\nabla_B(s)=\nabla_A(s)+a\cdot s.$ I want to understand the action of $a$ on $s$. Let's say in a local coordinate neighborhood around a point $p\in M,e_i$s form ortho-normal covectors such that $a=\sum\limits_i e_i\otimes a_i$ where $a_i\in\Omega^0(X;\text{End}(E)).$ Then is this how the action works? \begin{align*} a\cdot s=\sum\limits_i e_i\otimes a_i(s) \end{align*}

Answer 1

Question: "Then is this how the action works?"

Answer: If $R:=\mathcal{O}_M$ is the sheaf of realvalued smooth funtions on $M$ and $E$ is the sheaf of sections of your finite rank real smooth vector bundle on $M$, it follows $End_R(E) \cong E^*\otimes_R E$. Let $T$ denote the sheaf of vector fields on $M$. A connection is a map of $R$-modules

$$\nabla: T \rightarrow End(E)$$

and given two connections $\nabla_1,\nabla_2$ there is an $R$-linear map

$$\phi: T \rightarrow E^*\otimes_R E$$

with $\nabla_2=\nabla_1 +\phi.$ On an open set $U$ it follows the map $\phi_U :=\sum\phi_i \otimes e_i$ with $\phi_i \in E^*(U),e_i \in E(U)$ acts as follows: Given $e\in E(U)$ (a local section of $E$ over $U$) we get

$$\phi_U(e):=\sum_i \phi_i(e)e_i \in E(U).$$

Your formula $F1$: "Then is this how the action works? a⋅s=∑iei⊗ai(s)"

Note: The map

$$\rho: E^* \otimes_R E \rightarrow End_R(E)$$

is defined by

$$\rho(\phi \otimes e)(v):= \phi(v)e.$$

There is no "tensor product" involved when you evaluate as in your formula $F1$. Hence it seems your formula $F1$ is wrong.

Answer 2

If you consider two sections $\nabla_1, \nabla_2$, such that
$$\nabla_i (fs) = df \otimes s + f \nabla_i s,$$ then $\nabla_1 - \nabla_2: \Gamma(E) \to \Omega^1(E) \cong \Omega^1(M) \otimes \Gamma(E)$ is a $C^\infty$-linear map. If $e = (e_i)$ if a local frame, any section can be written as $s = s^ie_i$ for $s^i \in C^\infty$. Therefore, for $A \equiv \nabla_1 - \nabla_2$, $$A(s) = s^iA(e_i) = s^i(a^j_{~i} \otimes e_j)$$ for $a^j_{~i} \in \Omega^1(M)$. Note that I used Einstein summation convention. By remembering $\text{End}(E) \cong E^* \otimes E$, you may define $$a = a^j_{~i} \otimes (e_i)^*\otimes e_j \in \Omega^1(M) \otimes \Gamma(E^*) \otimes \Gamma(E) \cong\Omega^1(\text{End}(E)),$$ where $e^*$ is the dual frame of $e$. You clearly have $$A(s) = a\cdot s.$$