Let \(X=\mathbb C/\Lambda\) be a complex torus and define \[ F:X\to X, \qquad [z]\mapsto[2z]. \]
Show that \(F\) has a dense forward orbit: there exists \(p\in X\) such that \[ \overline{\{F^n(p):n>0\}}=X. \]
Let \(R\) be the rational function in the Weierstrass double-angle formula \[ \wp(2z)=R(\wp(z)). \] Prove that \(R\) has a dense forward orbit on \(\widehat{\mathbb C}\).
The source’s parenthetical description of a dense orbit writes \(\{F^n(p):n>0\}=X\) without closure. Literal equality is impossible because the forward orbit is countable while the torus is uncountable; density is the stated and intended property.
Solution
Choose a real basis \(\lambda_1,\lambda_2\) of the lattice. Then \[ \Phi:\mathbb R^2/\mathbb Z^2\longrightarrow X, \qquad (s,t)\longmapsto[s\lambda_1+t\lambda_2] \] is a homeomorphism, and under this identification \(F\) is coordinatewise doubling modulo \(1\).
1. There is a point of \(\mathbb R^2/\mathbb Z^2\) with dense orbit under\[ D(s,t)=(2s,2t)\pmod{\mathbb Z^2}. \]
Let \[ \mathcal A=\{(0,0),(0,1),(1,0),(1,1)\}. \] Form an infinite word \[ \mathbf u=(u_1,u_2,u_3,\ldots), \qquad u_j=(a_j,b_j)\in\mathcal A, \] by concatenating, for \(m=1,2,3,\ldots\), all words of length \(m\) over the alphabet \(\mathcal A\). Then every finite word over \(\mathcal A\) occurs in \(\mathbf u\), in fact arbitrarily far out.
Define \[ s=\sum_{j\ge1}\frac{a_j}{2^j}, \qquad t=\sum_{j\ge1}\frac{b_j}{2^j}. \] The sequence \(\mathbf u\) is not eventually constant in either coordinate, so these binary expansions cause no terminal-expansion ambiguity.
Multiplication by \(2\) modulo \(1\) shifts binary digits to the left. Thus \[ D^n(s,t) \] has binary digit pairs \[ u_{n+1},u_{n+2},u_{n+3},\ldots. \]
Now let \[ Q=I_{\varepsilon_1\cdots\varepsilon_m} \times I_{\delta_1\cdots\delta_m} \] be a dyadic rectangle, where the first factor consists of numbers whose first \(m\) binary digits are \(\varepsilon_1,\ldots,\varepsilon_m\), and similarly for the second factor. The length-\(m\) word \[ ((\varepsilon_1,\delta_1),\ldots,(\varepsilon_m,\delta_m)) \] occurs somewhere in \(\mathbf u\), arbitrarily far from the beginning. Hence for some \(n>0\), \[ D^n(s,t)\in Q. \] Every nonempty open set in the torus contains a dyadic rectangle, so the forward orbit of \((s,t)\) is dense.
2. Part (a) follows for every complex torus \(X=\mathbb C/\Lambda\).
1. Since\[ F\circ\Phi=\Phi\circ D, \]we have\[ F^n(p)=\Phi(D^n(s,t)). \]The homeomorphism \(\Phi\) carries the dense orbit from 1 to a dense forward orbit of \(F\) on \(X\).
3. The Weierstrass map semiconjugates torus doubling to the rational duplication map:\[ \wp\circ F=R\circ\wp. \]
This is exactly the double-angle identity \[ \wp(2z)=R(\wp(z)). \] Iterating gives \[ \wp(F^n(p))=R^n(\wp(p)) \qquad(n\ge0). \]
4. The rational map \(R\) has a dense forward orbit on \(\widehat{\mathbb C}\).
2. Let\[ V\subset\widehat{\mathbb C} \]be any nonempty open set. The map\[ \wp:X\to\widehat{\mathbb C} \]is continuous and surjective, so\[ \wp^{-1}(V) \]is a nonempty open subset of \(X\). Density of the \(F\)-orbit gives some \(n>0\) with\[ F^n(p)\in\wp^{-1}(V). \]Using 3,\[ R^n(q) =\wp(F^n(p)) \in V. \]Thus the forward orbit of \(q\) under \(R\) meets every nonempty open subset of the sphere and is therefore dense.