Explicit derivation of the horizontal subspace, connection and covariant derivative. I created two posts on more or less the same subject, and after a bounty with no responses I am trying to post a fresh question while being the clearest I could possibly be.
I am trying to derive a concrete example of fiber bundle where I can :


*

*Directly compute the vertical subspace of the tangent space of the bundle.

*Being confronted to an explicit choice for the horizontal subspace.

*Compute the connection one-form. This will require the explicit computation of the Vertical projection and the isomorphism $i:T_eG \rightarrow V_pP$ as well as the inverse $i^{-1}:V_pP \rightarrow T_eG$.

*Derive from the connection the covariant derivative of a section and then of a vector field.

*Validate from the derivation the infamous formula of directional derivative in $\mathbb{R}^2$ where we use a flat connection.


Basically I want to find the undergraduate formula of directional derivative by using the general framework of fiber bundle. Do any of you know a ressource, exercise, or textbook where I can get hints on how to proceed for all those points ?
What I've tried ?
I tried to consider the Frame bundle $F\mathbb{R} \rightarrow \mathbb{R}$ endowed with the group $GL(1,\mathbb{R})$. I managed to compute the vertical subspace, but I'm stuck at the horizontal one where I find an explicit formula that describe vectors in the horizontal subspace but still can't find out where the " choice " comes in the picture. I also had difficulties computing $i^{-1}$. Since my derivation are quite long on paper I don't want to clutter the post with my work.
Thanks for any help.
 A: For the tangent/frame bundle, this is pretty easy.
First, off, we should start with the easier-to-grasp notion of a Koszul-connection.
Assume that $M$ is a smooth $n$-manifold, and $\nabla$ is a linear connection (given via a covariant derivative) on $M$. Further assume that a local frame $(\mathcal{U},e_a)$ is given on $M$, and the local connection forms are given by $\omega^a_{\ b}$. Now the covariant derivative of a vector field is given by $d^\nabla V^a=dV^a+\omega ^a_{\ b}V^b$. If we also introduce a coordinate neighborhood, we have $$ \nabla_\mu V^a=\partial_\mu V^a+\omega^{\ a}_{\mu\ b}V^b. $$
Here greek indices are for coordinate frames, latin indices are for general frames.
Tangent bundle:
I know you asked for principal bundles, but this is really easier to do for vector bundles, so I'm gonna first do it on the tangent bundle.
The induced coordinates on the tangent bundle are given by $(x^\mu,v^a)$, the local coordinate frame on the tangent bundle of the tangent bundle is given by $(\partial/\partial x^\mu,\partial/\partial v^a)$.
You can easily check that the vertical spaces are generated by the $\partial/\partial v^a$ vectors, and during a change of both the coordinate system on $M$ and a change of local trivialization for $TM$, these vectors get mixed among one another, so their span is invariant, so we can write $$V_{(x,v)}TM=\text{span}(\partial/\partial v^a|_{(x,v)}).$$
Horizontal spaces:
Let $\gamma:[t_0,t_1]\rightarrow M$ be a smooth curve and let $V(t)$ be a vector field along this curve. Let this vector field be parallel along the curve. Then we have a curve $\bar{\gamma}:[t_0,t_1]\rightarrow TM$, $\bar{\gamma}(t)=(\gamma(t),V(t))$ on $TM$. If the vector field $V$ is parallel, then we want this resultant curve to be horizontal. The curve's tangent vector (in $TTM$) is $$ \frac{d\bar{\gamma}}{dt}=\frac{d\gamma^\mu}{dt}\frac{\partial}{\partial x^\mu}+\frac{dV^a}{dt}\frac{\partial}{\partial v^a}, $$ but because $V$ is parallel along the curve we have $$ \frac{dV^a}{dt}+\omega^{\ a}_{\mu\ \ b}V^b\frac{d\gamma^\mu}{dt}=0, $$ substituing this into the curve's tangent vector gives $$ \frac{d\bar{\gamma}}{dt}=\frac{d\gamma^\mu}{dt}\frac{\partial}{\partial x^\mu}-\frac{d\gamma^\mu}{dt}\omega^{\ a}_{\mu\ \ b}V^b\frac{\partial}{\partial v^a}=\frac{d\gamma^\mu}{dt}\left(\frac{\partial}{\partial x^\mu}-\omega^{\ a}_{\mu\ \ b}V^b\frac{\partial}{\partial v^a}\right)=\frac{d\gamma^\mu}{dt}D_\mu, $$ where $D_\mu$ is given by the expression in the round bracket (with the added caveat that if you do not evaluate $D_\mu$ along a curve just at some generic point then you need to replace $V^b\equiv V^b(t)$ with the generic fiber coordinate $v^b$).
We see then, that the horizontal spaces are given by $$ H_{(x,v)}TM=\text{span}(D_\mu). $$
Connection form:
The horizontal spaces can be altenatively given by giving a set of $\dim(V_{x,v}TM)$ 1-forms, which annihilate the horizontal spaces. It is easy to check that the annihilator of the horizontal spaces is then given by $$ \phi^a=dv^a+\omega^{\ a}_{\mu\ \ b}v^b dx^\mu. $$
it must be noted that if we want the span of $D_\mu$ or the kernel of $\phi^a$ to be invariant, the $\omega^{\ a}_{\mu\ \ b}$ coefficients must transform precisely as you know it already from the Koszul-way of doing connections. If this is true, then the combination $$\Phi=\frac{\partial}{\partial v^a}\otimes\phi^a$$ is invariant and provides a $VTM$-valued 1-form on $TM$ that projects onto the vertical spaces. We can call this the connection form of the connection.
Horizontal lifting and vertical isomorphism:
There are more elegant, invariant ways to characterize these, which you can find in books such as Kobayashi/Nomizu, Kolár/Michor/Slovák, Jeff Lee etc. Because you asked for more concrete stuff, I will provide more concrete realizations.
Let $X$ be a smooth vector field on the base space $M$, which is locally decomposed with respect to the coordinate frame as $$X=X^\mu\frac{\partial}{\partial x^\mu}.$$
The horizontal frame on $TM$, given by $D_\mu$ transforms the same way as $\partial/\partial x^\mu$, so the horizontal lift of $X$ can be calculated as $$ X^h=X^\mu D_\mu. $$ It should be noted that the components $X^\mu$ are the same as the components of the vector field on the base space, and they do not depend on the fiber coordinates $v^a$ at all.
In a vector bundle we also have a convenient isomorphism on the vertical spaces. Because $V_{(x,v)}TM=T_v(T_xM)$ (it is the tangent space to the fiber at $x$), and because $T_xM$ is a vector space, we have an isomorphism $\kappa:V_{(x,v)}TM\mapsto T_xM$. Because the vertical frame $\partial/\partial v^a$ transforms the same way as $e_a$, this isomorphism is given by $V^a(\partial/\partial v^a)|_{(x,v)}\mapsto V^a e_a|_{x}$.
Covariant derivative:
Let $\gamma:[t_0,t_1]\rightarrow M$ be a smooth curve. Let $V(t)$ be a smooth vector field along this curve. If $V$ is parallel along $\gamma$, then the lifted curve $\bar{\gamma}(t)=(\gamma(t),V(t))$ is horizontal in the sense that its tangent vector is horizontal. We can characterize the failure of this section-along-a-curve to be parallel by projecting it vertically, so we take $\Phi\circ d\bar{\gamma}/dt$. This is now a section (along a curve) of the bundle $VTM$ (seen as a fibration over $M$), but we can use the vertical isomorphism to make it $TM$-valued (note that $\kappa$ can produce multivalued functions but it is avoided here because values for different $t$ are located in different fibers), so we take $$ \frac{d^\nabla V}{dt}=\kappa\circ\Phi\circ\frac{d\bar{\gamma}}{dt}. $$
Let us calculate this in a local trivialization!
We have $d\bar{\gamma}/dt=(d\gamma^\mu/dt)\partial_\mu+(dV^a/dt)\partial_a$ ($\partial_\mu=\partial/\partial x^\mu$ and $\partial_a=\partial/\partial v^a$), now we let the connection form act on it:
$$ \Phi(d\bar{\gamma}/dt)=\partial_c\otimes\left(dv^c+\omega^{\ c}_{\mu\ \ b}V^bdx^\mu\right)(\dot{\gamma}^\nu\partial_\nu+\dot{V}^a\partial_a)=\dot{V}^c\partial_c+\dot{\gamma}^\mu\omega^{\ c}_{\mu\ \ b}V^b\partial_c. $$
Now the map $\kappa$ will simply exchange $\partial_c$ to $e_c$, so we get $$\kappa(\Phi(d\bar{\gamma}/dt))=(\dot{V}^a+\dot{\gamma}^\mu\omega^{\ a}_{\mu\ \ b}V^b)e_a, $$ which is precisely the formula for the covariant derivative on the base.
What about total covariant derivative for a section $V$? The natural generalization then, of the above formula is to replace $d\bar{\gamma}/dt$ with $V_*(X)$, where $*$ denotes pushforward, and $X$ is a vector field on $M$. This gives us $$ \nabla_XV=\kappa\circ\Phi\circ V_*(X) $$ and $$ \nabla V=d^\nabla V=\kappa\circ\Phi\circ V_*. $$
The frame bundle:
Note: I ran out of time but I don't want this answer to be possibly wasted, so I'll stop here and post it incompletely. This should already give you some pointers, but I intend to edit this post in a few hours and also do the frame bundle.
