Any linear connection $\nabla^E$ has a unique extension to a tensor derivation on $E$, i.e. a connection (which I'll just write $\nabla$) on each tensor product $E\otimes\cdots\otimes E \otimes E^* \cdots \otimes E^*$ satisfying
- $\nabla \xi =\nabla^E \xi$ for $\xi \in \Gamma(E)$,
- the Leibniz rule $\nabla(\xi \otimes \zeta) = \nabla \xi \otimes \zeta + \xi \otimes \nabla \zeta$ and
- $\nabla(C(\xi)) = C(\nabla(\xi))$ for any contraction $C$.
I think this is a fact that is well worth your time to prove and internalize, and it makes your desired claim easy: Since an endomorphism is just a section of $E \otimes E^*,$ we thus have $$\nabla_X(\Phi(\xi)) = \nabla_X(C(\Phi \otimes \xi)) =(\nabla_X \Phi)(\xi) + \Phi(\nabla_X \xi),$$ so the the commutator is just $[\nabla_X^E, \Phi] = \nabla_X\Phi\in\Gamma(\operatorname{End}(E)).$
If you want to prove your claim more directly, here are a couple of approaches to try:
- Show directly that $[\nabla^E_X, \Phi] : \Gamma(E) \to \Gamma(E)$ is $C^\infty(M)-$linear, and thus an endomorphism by the bundle homomorphism characterization lemma. This should be a very neat calculation using the Leibniz rule.
- Fix a local frame for $E$ and work in components/coordinates. You should be able to manipulate $[\nabla^E_X, \Phi]\xi$ and end up with some linear function of the components $\xi^b$.
Let $E \rightarrow M$ a smooth vector bundle on a manifold $M$. Let $\Gamma(E)$ be the space of sections. A connection on $E$ is an $\mathbb{R}$-linear map
$$
\nabla : \Gamma(E) \rightarrow \Gamma\left(E \otimes T^{*} M\right)
$$
satisfyng Leibniz rule
$$
\nabla(f \sigma )=\sigma \otimes d f+f \cdot \nabla \sigma
$$
This way, if $X$ is a vector field one can define a notion of derivative of a section
$$
\nabla_{X} : \Gamma(E) \rightarrow \Gamma(E)
$$
by means of $\nabla_{X} \sigma=(\nabla \sigma)(X)$.
If we think of sections like a generalization of functions (I personally call them twisted functions) the connection is a device that let us derive this functions.
This concept is nothing but a particular case of a connection on a fiber bundle when the bundle is a vector bundle.
I think that what you are looking for is that every derivative operator gives rise to horizontal subspaces in $TE$.
For every $e\in E$, with $\pi(e)=x\in M$, we have a "natural" linear space $T_e E_x$ that is called the vertical space. The union in $e\in E$ gives us a subbundle $V\subseteq TE$ called the vertical bundle.
At a first glance, there is no natural choice for an horizontal bundle, but $\nabla$ determines one. We will define
$$
H_e:=ds_x (T_x M)
$$
where $s$ is a section such that $s(x)=e$ and $\nabla_X s=0$ for every $X\in T_x M$.
This way, the vector bundle connection determines an Ehresmann connection.
I have written more ideas on vector bundle connections here.
Edited
Maybe, what you are looking for is what follows.In a general setup (not necessarily a vector bundle) a way to specify a connection consists of a projection, i.e., a map $v:TE\mapsto TE$ such that $v^2=v$, and with $v(T_e E)=V_e$. In this case, $H_e=ker(v|_{T_e E})$. Remember that in any vector space $W$, a map such tat $p^2=p$ lets you decompose $W=ker(p)\oplus im(p)$ (see projection). The idea behind is that you can comprise $W$ to $p(W)$ in so many ways as horizontal subspaces choices.
So equivalently, the horizontal subbundle can be described by a $TE$-valued 1-form on $E$, called the connection form.
One more interpretation of your doubt
On the other hand, I guess you could define $(\ker \nabla)_e$ as a vector subspace of $T_e E$ which is tangent at $e\in E$ to the image of any section $s:M\to E$ which is horizontal in $e$, i.e., such that $\nabla_X s=0$ for every $X\in T_{\pi(e)}X$.
Best Answer
$\Gamma(E \otimes T^*M)$ is the set of sections of the vector bundle $E \otimes T^*M = \bigcup_{p \in M} E_p \otimes T_p^*M$. The usual second definition (that I will be using) is
Now I will sketch one way to get between the two definitions.
($D \to \nabla$): For fixed $X \in \Gamma(TM)$, define $\nabla_X(s) := i_X(Ds)$ where, for $\omega \otimes s \in \Gamma(T^*M \otimes E)$, $i_X(\omega \otimes s) = \omega(X) s \in \Gamma(E)$. Then it's a straightforward exercise to check that this has the desired properties.
($\nabla \to D$): Let $x^1, \dots, x^n$ be local coordinates for some chart $U \subseteq M$. Define in local coordinates $$D(s) = \sum_{i=1}^n dx^i \otimes \nabla_{\frac{\partial}{\partial x^i}} s.$$ Then you can check that this definition has the right behaviour under a change of coordinates to define a global object. Finally, it is an easy coordinate based calculation to check the desired properties.
Finally, by a calculation in local coordinates, you can check that these constructions are inverse to each other so that the definitions are equivalent.