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Let $A=\text{antidiag}(\omega_2,-\omega_2,\omega_1,-\omega_1)$ with $\omega_i\in\mathbb R^+$ and $\theta=\frac{\omega_2}{\omega_1}$. Show that the Orbit $\mathcal O(x)=\{\exp(tA)x:t\in\mathbb R\}$ of $x$ is closed if $\theta\in\mathbb Q$.

I would be happy if someone could point out the error in my thoughts: First of all, $$\exp(tA)=\text{diag}(e^{-\alpha t},e^{-\alpha t},e^{\alpha t},e^{\alpha t})$$ where I set $\alpha:=i\sqrt{\omega_1\omega_2}$ to shorten things a bit. And therfor we can write the orbit as $$\mathcal O(x)=\left\{\begin{pmatrix}e^{-\alpha t}x_1\\e^{-\alpha t}x_2\\e^{\alpha t}x_3\\e^{\alpha t}x_4\end{pmatrix}:t\in\mathbb R\right\}$$ Now here is where I think I made a mistake: Let $T:=\frac{2\pi }{\omega_1\cdot\sqrt\theta}\in\mathbb R$, then $$\omega_1\sqrt\theta T=2\pi\\ \Leftrightarrow \\iT\sqrt{\omega_1^2\theta}=2\pi i\\\Leftrightarrow\\\alpha T=2\pi i$$ and thus $\exp(TA)x=x$, hence $x$ is a periodic point of $\mathcal O(x)$, making the orbit closed. As you can see I didn't use the condition $\theta\in\mathbb Q$ at all so this can't be correct. Where is the mistake and how can I correct it?

Buh
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  • The matrix $A$ is real, thus you should also get a real exponential matrix. And indeed you get from $$A^2=-ω_1ω_2I$$ that $$\exp(tA)=\cos(\sqrt{ω_1ω_2}t)I+\frac{\sin(\sqrt{ω_1ω_2}t)}{\sqrt{ω_1ω_2}}A.$$ But still this gives $\exp(TA)=I$ for $T=2\pi/\sqrt{ω_1ω_2}$. – Lutz Lehmann Jun 09 '18 at 09:54
  • If however the two oscillations were decoupled, $$A=\pmatrix{0&ω_1&0&0\-ω_1&0&0&0\0&0&0&ω_2\0&0&-ω_2&0},$$ then indeed the claim is true, you need commensurable frequencies to get closed orbits. – Lutz Lehmann Jun 09 '18 at 10:03
  • Okay, I see that your expression for $\exp(tA)$ is the correct one. But I thought that if $A$ is diagonalizable with eigenvalues $\lambda_1,\dots,\lambda_n$, then $\exp(tA)=\text{diag}(e^{t\lambda_1},\dots,e^{t\lambda_n})$ and I just plugged $A$ into wolfram alpha to get the jordan decomposition. So where is the error in that? – Buh Jun 09 '18 at 10:56
  • No, if $A=UDU^{-1}$ then $\exp(tA)=U\exp(tD)U^{-1}$. You are missing the reverse basis transformation in the exponential formula. – Lutz Lehmann Jun 09 '18 at 12:54
  • But when $D$ is diagonal (because $A$ is diagonalizable) then $\exp(tD)$ is diagonal as well and should commutate with $U$? Maybe that is my error? – Buh Jun 09 '18 at 16:24
  • Yes, that is an error. General diagonal matrices do not commute with general other matrices. – Lutz Lehmann Jun 09 '18 at 16:50
  • I have now talked to the professor and indeed the given matrix was wrong. And it is supposed to be the matrix given by you (just multiplied with $-1$). Now, let's start again by calculating the matrix exponential. For your matrix after some calculation I get $$\exp(tA)=\begin{pmatrix}\cos(\omega_1 t)&-\sin(\omega_1 t)&0&0\\sin(\omega_1 t)&\cos(\omega_1 t)&0&0\0&0&\cos(\omega_2 t)&-\sin(\omega_2 t)\0&0&\sin(\omega_2 t)&\cos(\omega_2 t)\end{pmatrix}$$. Before I go any further, is this correct? – Buh Jun 10 '18 at 08:35
  • That is correct for the sign changed matrix, the corrected version of your professor. Just compare the Taylor expansions on the right with $\exp(tA)=I+tA+O(t^2)$. – Lutz Lehmann Jun 10 '18 at 10:20
  • Okay, then the problem is easily solved here. But since I've got your attention now: Apparently one can show that if we have $\theta\notin\mathbb Q$ instead, the orbit $\mathcal O(x)$ is dense in the torus ${y\in\mathbb R^4:y_1^2+y_2^2=a,\ y_3^2+y_4^2=b}$ for suitable $a,b$. I can show that it is contained and we should have $a=x_1^2+x_2^2$ and $b=x_3^2+x_4^2$. Any idea though how to show it's dense? – Buh Jun 10 '18 at 12:59
  • This goes back to a theorem named after Kronecker or Minkowski, that $θ$ is irrational if and only iff ${mθ+n:m,n\in\Bbb Z}$ is dense in $\Bbb R$. – Lutz Lehmann Jun 10 '18 at 13:05
  • I see. This sounds familiar, but unfortunately topology is a big weakness of me. Would you be so kind to shortly sketch the proof? – Buh Jun 10 '18 at 13:26
  • Essentially, this set is a grid/lattice and thus is either dense or has some smallest positive element. In the second case one can find very quickly that that element has to be rational, the whole set is rational and thus also $θ$. See also https://math.stackexchange.com/q/450493/115115, https://math.stackexchange.com/q/272545/115115 and similar. – Lutz Lehmann Jun 10 '18 at 14:02
  • You got me wrong there. I can see that $\mathbb Z+\theta\mathbb Z\subset\mathbb R$ is dense. But how does this help me regarding the orbit? – Buh Jun 10 '18 at 16:14
  • Any cross-section of the torus transversal to the orbit has this structure. Consider the points where the first rotation returns to the identity, $t=kT$, $T=2\pi/ω_1$. The second rotation is at angle $kθ\cdot 2\pi+2\pi\Bbb Z$ and the density of this set in the second circle relates directly to the density of the orbit in the torus. – Lutz Lehmann Jun 10 '18 at 17:15
  • Alright, thanks! You really helped me a lot here. If you just write some random answer I can mark it as answered and you get the points if you want to. – Buh Jun 10 '18 at 18:50

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