I am trying to write down all the connections on the simple frame bundle $F\mathbb{R} \rightarrow \mathbb{R}$ as a choice of horizontal subspaces, but I get stuck. Please feel free to correct any of my statements, also I provide only results as I feel confident with the following derivations.
Let $F\mathbb{R} \rightarrow \mathbb{R}$ be the frame bundle where $F\mathbb{R}\xrightarrow{\sim} \mathbb{R}\times\mathbb{R}_*$ and the projection map $\pi : \mathbb{R}\times\mathbb{R}_* \rightarrow \mathbb{R}$ the obvious projection onto the first factor. We consider the group $GL(1,\mathbb{R})$ acting on the left.
We say that a tangent vector is in the vertical subspace if it is in $ker(\pi_*)$ where $\pi_*$ is the push forward map. Or similarly if the tangent vector is in the form $X^A_pf = \frac{\partial}{\partial t}\big|_0 f(p.exp(tA))$ where A is in the Lie algebra of the group. So :
1) I found that for $X_p = f(a,b)^i\frac{\partial}{\partial x^i}\big|_{(a,b)} \implies \pi_*(X_p) = f(p)^1\frac{\partial}{\partial t}\big|_a$ So we can conclude that $X_p$ is Vertical if and only if $f(p)^1 = 0$.
To verify this result I tried to calculate $X^A_p$. For this I need to be able to compute $exp(tA)$ which is defined in my textbook as : $exp: T_eG \rightarrow T_pP$ which in my case is $exp: T_o\mathbb{R} \rightarrow T_{(a,b)}(\mathbb{R}\times \mathbb{R})$ and $exp(A) = \gamma^A(1)$ with $\gamma^A(1)$ the integral curve of the Left invariant vector field generated by A. So I started off by calculating the left invariant vector field :
Let $A = a\frac{\partial}{\partial t}\big|_0$ be a vector in the Lie algebra. Define $L_g$ to be the left action : $L_g(a) = a + g$. I found out that $L_{g*}(A) = a\frac{\partial}{\partial t}\big|_g$. Now to calculate the integral curve we should have : $\gamma'(t) = a$ which leads me to having $\gamma(t) = at$ so that $\gamma(1) = exp(A) = a$
So we proceed with $X^A_{(b,c)}f = \frac{\partial}{\partial t}\big|_0 f((b,c).exp(tA)) = \frac{\partial}{\partial t}\big|_0 f((b,c).ta) = \frac{\partial}{\partial t}\big|_0 f((b,cta) = ca\partial_2\big|_{(b,c)}f $ which is indeed equivalent to $f(p)^1 = 0$ since only the second component is non-zero.
So since the horizontal subspace is : \begin{align} T_{(a,b)}\mathbb{R}\times \mathbb{R}_* = Hor \oplus Ver\\ R_{g*}Hor_{(a,b)} = Hor_{(a,bg)} \end{align}
I set $X_{(a,b)} = f(a,b)^i\frac{\partial}{\partial x^i}\big|_{(a,b)} \in Hor \implies R_{g*}(X_{(a,b)}) = f(a,b)^1\frac{\partial}{\partial x^1}\big|_{(a,bg)} + gf(a,b)^2\frac{\partial}{\partial x^2}\big|_{(a,bg)}$ So we have the following constraints :
$f(a,bg)^1 = f(a,b)^1 \implies f(a,b)^1 = f(a,1)^1$
$f(a,bg)^2 = gf(a,b)^2 \implies f(a,g)^2 = gf(a,1)^2$
From here I do not know what does it mean to : 1) Make a specific choice for Hor , for me Hor is completely determined bu the previous constraints. 2) How to project a tangent vector to Ver and Hor ? and How does that depend on the specific choice of Hor ? ( I need this to compute an explicit equation for the connection form $\omega_p = i^{-1}\circ Ver$.
Thanks.
EDIT : What is p in the third paragraph ?
p is actually a point in the frame bundle $F\mathbb{R}$, so concretely $p\in \mathbb{R}\times \mathbb{R}_*$. p was a bad choice of notation.
What is $exp(tA)$
This is a lie group element generated by the lie algebra element A. This is the Lie algebra of the Lie group $GL(1,\mathbb{R})$. We consider the left invariant vector field generated by A, then we find the integral curve $\gamma$ where the tangent vector of the curve at the identity is A, and then $exp(tA) = \gamma(t)$
What is $f(a,b)^i$
Yes this f is different from the f in the previous paragraph. It is the coordinates of the vector in the basis $\frac{\partial}{\partial x^i}\big|_{(a,b)}$. $(a,b)$ is point in the Frame bundle. And yes $p$ is a point in the frame bundle as well, I realize the notation is messy... But they are all point in the Frame bundle. $f(p)^1$ is the first coordinate.