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Rewriting the ODE in the title, we get $$\dot{x_1}=x_2,\dot{x_2}=-x_1, \dot{x_3}=x_4, \dot{x_4}=-x_3.$$ It is easy to show that $$x_1=A\cos(t)+B\sin(t),x_2=B\cos(t)-A\sin(t),\\x_3=C\cos(t)+D\sin(t),x_4=D\cos(t)-C\sin(t).$$ In Ordinary Differential Equations by V. I. Arnold, there are several problems on this equation. For example, it can be shown that each phase curve is on a 3-sphere, and is a great circle of that.

The next problem is more difficult: show that the phase curves on a given 3-sphere form a 2-sphere.

I attempt to find out which 3-dimensional subspace the 2-sphere lies in, but in vain.

And the last question is about the linking number. Since 3-sphere can be regarded as $\mathbb R^3\cup \{\infty\}$, a partition of 3-sphere into circles determines a partition of $\mathbb R^3$ into circles and nonclosed circles. Show that any two of the circles of this partition are linked with linking number 1.

I am not sure how to visualize this partition, and what the circles stand for.

Can anyone give some advice? Thanks in advance.

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  • $\begingroup$ @Amzoti Yes, and I have edited it. $\endgroup$ Mar 11 '15 at 14:39
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    $\begingroup$ I'm not an expert on this, but it seems to be connected with the so called Hopf fibration; see here: en.wikipedia.org/wiki/Hopf_fibration $\endgroup$ Mar 11 '15 at 15:00
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    $\begingroup$ It seems that every property you are interested in derives from the fact that each solution of this differential system stays on a surface $$[x_1^2+x_2^2=a^2,\,x_3^2+x_4^2=b^2],$$ which, in turn, is included in the 3-sphere $$x_1^2+x_2^2+x_3^2+x_4^2=r^2$$ for some suitable $r$. $\endgroup$
    – Did
    Mar 13 '15 at 15:02
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    $\begingroup$ The two-sphere (the set of orbits) turns out not to be embedded in the three-sphere. It may help to introduce complex coordinates $z_{1} = x_{1} + ix_{2}$, $z_{2} = x_{3} + ix_{4}$; the flow of your system is $$(t, z_{1}, z_{2}) \mapsto (e^{it}z_{1}, e^{it}z_{2}).$$The three-sphere is invariant (see also Did's comment), and the quotient of $S^{3}$ by this circle action is the Hopf map (see Christian's comment); the quotient space is the set of complex lines through the origin of $\mathbf{C}^{2}$. $\endgroup$ Mar 13 '15 at 16:04
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    $\begingroup$ By not rewriting the ODE we find that: $$ x=A\cos(t)+B\sin(t)\\ y=C\cos(t)+D\sin(t) $$ Solving for $\cos(t)$ and $\sin(t)$ and summing the squares of these then yields a Lisajous Ellipse, degenerated eventually (if $AD-BC=0$): $$ \left(\frac{Cx-Ay}{AD-BC}\right)^2+\left(\frac{Dx-By}{AD-BC}\right)^2 = 1 $$ That's all "interesting" I can see. What more is there to be said? $\endgroup$ Mar 14 '15 at 9:44
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$\newcommand{\Cpx}{\mathbf{C}}\newcommand{\Proj}{\mathbf{P}}$This answer introduces well-known constructions (the equivalence of the complex projective line with the Riemann sphere, and the Hopf map from the $3$-sphere to the $2$-sphere) that are accessible given the presumed background of readers of Arnol'd, but may or may not occur naturally in the course of approaching this problem about pairs of harmonic oscillators.

The Riemann Sphere: For present purposes, the Riemann sphere is the holomorphic curve obtained by taking two copies $U_{0}$ and $U_{1}$ of the complex line, with respective coordinates $z$ and $w$, and identifying $w$ in $U_{1}$ with $z = 1/w$ in $U_{0}$. The origin ($w = 0$) in $U_{1}$ is regarded as the point at infinity in $U_{0}$, and vice versa.

The Complex Projective Line: If $z_{0}$ and $w_{0}$ are complex numbers, not both zero, the complex line they determine is the set of (complex) scalar multiples of $(z_{0}, w_{0})$. Throughout, the term "line" refers to a complex line through the origin in $\Cpx^{2}$. (A complex line is an oriented, real $2$-plane. However, most real $2$-planes through the origin of $\Cpx^{2}$ are not complex lines.)

Every non-zero vector in $\Cpx^{2}$ lies on a unique line. Two non-zero vectors $(z_{1}, w_{1})$ and $(z_{2}, w_{2})$ lie on the same line if and only if there exists a complex number $\lambda$ (necessarily non-zero) such that $$ (z_{2}, w_{2}) = \lambda (z_{1}, w_{1}). $$

The set lines is, by definition, the complex projective line, $\Cpx\Proj^{1}$. This space acquires the structure of a holomorphic curve, equivalent to the Riemann sphere $\Cpx \cup \{\infty\}$, as follows: A complex number $z$ in $U_{0}$ corresponds to the line through $(z, 1)$. A complex number $w$ in $U_{1}$ corresponds to the line through $(1, w)$. If $w \neq 0$ and $z = 1/w$, then $$ (1, w) = w(1/w, 1) = w(z, 1) \sim (z, 1); $$ that is, each non-zero number $w$ in $U_{1}$ is identified with $z = 1/w$ in $U_{0}$. This is precisely the gluing that defines the Riemann sphere.

The Hopf Fibration: Define a (holomorphic) mapping $\Pi:\Cpx^{2} \setminus\{(0, 0\} \to \Cpx\Proj^{1}$ by sending each (non-zero) pair $(z, w)$ to the line through $(z, w)$. The restriction of $\Pi$ to the $3$-sphere $$ S^{3} = \{(z, w) : |z|^{2} + |w|^{2} = 1\} $$ is the Hopf map $\pi:S^{3} \to \Cpx\Proj^{1}$. The preimage of each point is the intersection of $S^{3}$ with a line, a great circle called a Hopf fibre. Since $\pi$ induces a bijection between Hopf fibres and points of $\Cpx\Proj^{1}$, the set of Hopf fibres is a $2$-sphere.

To see that distinct fibres of the Hopf map link once inside $S^{3}$, note that if $L_{1}$ and $L_{2}$ are distinct lines, then $\Cpx^{2} = L_{1} \oplus L_{2}$. Projection to the second summand is surjective, and the Hopf fibre in $L_{2}$ "winds around" $L_{1}$ (since this circle winds around the origin in $L_{2}$).


Consider the first-order linear system $$ \dot{x}_{1} = x_{2},\quad \dot{x}_{2} = -x_{1},\quad \dot{x}_{3} = x_{4},\quad \dot{x}_{4} = -x_{3}. $$ Introducing complex coordinates $z = x_{1} + ix_{2}$ and $w = x_{3} + ix_{4}$ (and therefore identifying $\mathbf{R}^{4}$ with $\Cpx^{2}$, the preceding system becomes $$ \dot{z} = iz,\quad \dot{w} = iw. \tag{1} $$ The solution of (1) with initial conditions $z(0) = z_{0}$, $w(0) = w_{0}$, is $$ \left[\begin{array}{@{}cc@{}} z(t) \\ w(t) \\ \end{array}\right] = e^{it} \left[\begin{array}{@{}cc@{}} z_{0} \\ w_{0} \\ \end{array}\right] = \left[\begin{array}{@{}cc@{}} e^{it} & 0 \\ 0 & e^{it} \\ \end{array}\right] \left[\begin{array}{@{}cc@{}} z_{0} \\ w_{0} \\ \end{array}\right]. \tag{2} $$

Since scalar multiplication maps each line to itself, each solution curve (2) is the Hopf fibre in the line determined by the unit vector $(z_{0}, w_{0})$. Since solutions of (1) are precisely Hopf fibres, the set of solutions is a $2$-sphere, and distinct solution curves link once in $S^{3}$.

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