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

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?

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.
• Thank you! Do you have any hint on how to prove that, if the 3-sphere is represented as $\mathbb{R}^3\cup \{\infty\}$, all those circles intersect each other exactly once? Sep 28 '16 at 6:39
• @Matthew Oh, actually these circles never intersect each other. Any pair of circles (trajectories of the system, fibers of the Hopf fibration) are linked (interlocked). Usually the passage from the three sphere to $\mathbb{R} \cup \{\infty\}$ is made by means of stereographic projection (form three sphere to three space). Just wiki Hopf fibration. I think there should be some pictures there. Sep 29 '16 at 2:25