I am trying to get more familiar with Ehresmann-connections by looking at how they look like in local trivializations. I am roughly aware of the abstract theory, but not everything is immediately clear, hence this "investigation".

Now, let $M$ be a smooth $n$ dimensional (real) manifold and let $\pi:E\rightarrow M$ be a rank $k$ vector bundle. Local trivializations are identified with local frames of $E$. A generic local section is written as $\psi=\psi^Ae_A$ and the index notation conventions of Einstein and Schouten are employed.

A bundle chart for $E$ then looks like $(x^\mu,\psi^A)_{\mu=1,...,n;\ A=1,...,k}$, with a common domain $U$ for both the local chart on $M$ and the local trivialization of $E$ (this domain will be left implicit in this post).

If a Koszul connection on $E$ is given locally by $$ \nabla_\mu\psi^A=\partial_\mu\psi^A+\omega_{\mu\ \ B}^{\ A}\psi^B, $$ then this connection determines an Ehresmann-connection on $E$ by letting $$ H_{(x,\psi)}=\text{span}(D_\mu), $$ where $$ D_\mu=\frac{\partial}{\partial x^\mu}-\omega_{\mu\ \ B}^{\ A}\psi^B\frac{\partial}{\partial\psi^A}. $$

This much is clear. However, something's unclear if I try to work the other way around.

We can define the canonically existing vertical distribution as $$ V_{(x,\psi)}=\text{span}\left\{\frac{\partial}{\partial\psi^A}\right\}, $$ and is easily verified that this is independent of the local trivialization. Now, let's try to define a horizontal distribution locally as $$ H_{(x,\psi)}=\text{span}\{D_\mu\}, $$ where I look for $D_\mu$ in the form $$ D_\mu=\frac{\partial}{\partial x^\mu}+Q^A_\mu\frac{\partial}{\partial\psi^A}.$$

The reason I am looking for the horizontal frame in this form is that, the presence of $\partial/\partial x^\mu$ will ensure that $D_\mu$ can never be vertical.

Question 1: If I am interested in the most general Ehresmann connection on $E$ (linearity is not assumed), is this true that any horizontal distribution has a frame in this form?

Now, going further, I want to ensure that my connection is linear, in the sense that parallel transport is linear. It is natural, then that I should demand $Q^A_\mu(x,\psi)$ to have only linear dependence on the fiber coordinates, so $Q^A_\mu=-\omega^{\ A}_{\mu\ \ B}(x)\psi^B$. This will obviously give back what I started from (the covariant derivative).

My problem is that Lee (Jeffrey, author of Manifolds and Differential Geometry, not to be confused with John Lee) states in the aforementioned book that the only condition for an Ehresmann connection on $E$ to be a linear connection is that it respects scalar multiplication, eg. that $(S_\lambda)_*H_{(x,\psi)}=H_{(x,\lambda\psi)}$ where $S_\lambda$ is multiplication by the scalar $\lambda$.

I guess that in my own approach/notation, this means that $Q^A_\mu(x,\psi)$ is a homogenous function (of degree 1) of $\psi$. However, as far as I am aware, homogenity does not necessarily imply linearity.

Question 2: What is the necessary condition for $H$ to be a linear Ehresmann-connection? Is Lee right or wrong? If he is right, then how does the form $D_\mu=\partial_\mu-\omega_{\mu\ \ B}^{\ A}\psi^B\partial_A$ come from a homogenous $Q^A_\mu$? Am I doing everything correctly?


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