I'm desperately trying to understand connections on a principal fibre bundles and the definition I've been given has to do with vertical and horizontal vector spaces. I am just starting to studying these objects and I find very frustrating that there are so few examples around (at least I can't find any real computation or examples) so I tried to make one myself. Let's consider the Hopf bundle $S^1 \hookrightarrow S^3 \rightarrow S^2$ with projection
$$ \pi(x,y)=(2y \bar{x},|x|^2-|y|^2)\ \ x,y\in S^3 $$
The vertical vector are defined as the kernel of the tangent map $T\pi$. Writing $x$ and $y$ in exponential notation we get
$$ x=re^{i\theta_1} y=(1-r)e^{i\theta_1} \text{ being that $x$ and $y$ are on $S^3$} \\ \pi(x,y)=(2(1-r)e^{i\theta_2}re^{-i\theta_1},r^2-(1-r)^2)=(2r(1-r)e^{i(\theta_2-\theta_1)},2r-1) $$ Let's take a generic curve on $S^3$ $\gamma(s)=(r(s)e^{i\theta_1(s)},(1-r(s))e^{i\theta_2(s)})$ and compute $D(\pi \circ \gamma)$ (which denotes the derivative) $$ \frac{d}{ds}\pi \circ \gamma = \frac{d}{ds} (2r(s)(1-r(s))e^{i(\theta_2(s)-\theta_1(s))},2r(s)-1)\text{ (omitting the dependance on $s$)}\\ (2r'(1-r)e^{i(\theta_2-\theta_1)}-2r\cdot r'e^{i(\theta_2-\theta_1)}+2r(1-r)e^{i(\theta'_2-\theta'_1)},2r') $$ if we want that the image is zero then we must have $2r'=0$ in second component and therefore $r=\text{const}$ which substituting in the derivative formula leads to $$ (2r(1-r)e^{i(\theta'_2-\theta'_1)},0) $$ which is zero iff $r=1 \vee r=0$. This means that the only curves that generate vertical vectors are curves like $$ \gamma(s)=(e^{i\theta_1(s)},0) \text{ and } \gamma(s)=(0,e^{i\theta_2(s)}) $$ So I have all the vertical vectors that I need.
Is any of this correct? Are there other ways to do this or examples I can look at? I stress that all of this is new to me so please be gentle :)
$\newcommand{\Cpx}{\mathbf{C}}$Here's a way to say/see the same thing that seems to me simpler and more geometrically natural. As you know, the three-sphere is the set of unit vectors in $\Cpx^{2}$, $$ S^{3} = \{(x, y) \in \Cpx^{2} : |x|^{2} + |y|^{2} = 1\}. $$ (In your first displayed equation, $x$ and $y$ appear to be complex numbers, not points of the three-sphere.) The unit circle $S^{1} = \{\lambda \in \Cpx : |\lambda| = 1\}$ acts on $S^{3}$ by scalar multiplication. If we write $\lambda = e^{it}$ with $t$ real, this action may be written $$ \phi_{t}(x, y) = e^{it}(x, y). $$ For each $(x, y)$ in $S^{3}$, the velocity $\frac{d}{dt}\big|_{t=0}\phi_{t}(x, y) = (ix, iy)$ spans the vertical space at $(x, y)$, and the orthogonal complement in $T_{(x,y)} S^{3}$ is the horizontal space.
This picture can be realified by writing $x$ and $y$ in terms of real and imaginary parts, stereographically projected to $\mathbf{R}^{3}$ if desired, and so forth.