Show that the set of all phase curves of the spherical pendulum for a specific energy form themselves a 2-sphere

Question

In V.I.Arnold's book about ordinary differential equations the question in the title is posed. I already know that the phase curves on a specific energy level lie on a 3-sphere in the four dimensional phase space and that the phase curves are great circles of the 3-sphere. The spherical pendulum's differential equations are $x_1'=x_2,x_2'=-x_1,y_1'=y_2,y_2'=-y_1$. The phase curves are of the following kind: $x(t)=\cos(t)\pmatrix{a\\b\\c\\d}+\sin(t)\pmatrix{b\\-a\\d\\-c}$ with $a,b,c,d$ setting the initial conditions. By searching the internet I found that the Hopf fibration might help, but I'm not really familiar with it. Can anybody help out or recommend some literature?

Answer 1

Well, combine the solutions: $z(t) = x_1(t) + i \, x_2(t)$ and $w(t) = y_1(t) + i \, y_2(t)$ both curves in on the complex line $$z : \mathbb{R} \to \mathbb{C}$$ $$w : \mathbb{R} \to \mathbb{C}$$ Then $$z' = x_1' + i \, x_2' = x_2 - i \, x_1 = -i (x_1 + i \, x_2) = -i \, z$$ $$w' = y_1' + i \, y_2' = y_2 - i \, y_1 = -i (y_1 + i \, y_2) = -i \, w$$ so solving the linear pendulum's equations is equivalent to solving the linear complex equations above, so $$\big(z(t), w(t)\big) = \big(e^{-it} z, \, e^{-it}w\big)$$ Notice that if we define the energy hypersurface $|z|^2 + |w|^2 = c_0^2$ for a real positive constant $c_0>0$, then $$|z(t)|^2 + |w(t)|^2 = |e^{-it} z|^2 + | e^{-it}w|^2 = |z|^2 + |w|^2 = c_0^2.$$ In other words, a solution of the system lies on a three dimensional sphere of radius $c_0$, as long as it start from a point $(z,w)$ lying on that same sphere.

Now define the map (this is one version of the Hopf bundle/Hopf fibration map) $$\pi \,\, : \,\, \mathbb{S}^3 = \{(z,w) \, : \, |z|^2 + |w|^2 = c_0^2\} \,\, \to \,\, \mathbb{CP}^1 = \mathbb{C} \cup \{\infty\}$$ $$ \pi \, : \, (z,w) \mapsto [z : w] = \frac{z}{w} \, \in \, \mathbb{C} \,\, \text{ or } \,\, \frac{z}{w} = \infty \,\, \text{ if } w=0$$ Then each solutions is mapped to $$\pi(e^{-it}z,\,e^{-it}w) = \frac{e^{-it}z}{e^{-it}w} = \frac{z}{w}$$ the same point on the complex line $\mathbb{C}$ or to infinity. However, $\mathbb{C} \cup \{\infty\} \cong \mathbb{S}^2$ is also known as the Riemann sphere.