I was trying to understand the kernel of the connection on a smooth vevtor bundle $E$.
Let $\pi:E\to M$ be a smooth vector bundle, we can associate a connection on $E$, which defines $$\nabla:\Gamma(E)\to \Gamma(T^*_{M}\otimes E)$$ that satisfies the Leibniz rule.
Is $\ker \nabla \subset E$ sub vector bundle of $E$? why it's a distribution on $E$?
Best Answer
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 $X\in T_{\pi(e)}X$.