Exercise 5.7.20* (Density of $\langle P \rangle$ on $\mathscr{C}_f(\mathbb{R}$)

Question

Let $\mathscr{C}_f(\mathbb{R})$ be defined by (5.50) where $a,b,c$ are real numbers, and suppose that the polynomial on the right side of (5.50) has only one real root (so that the curve $\mathscr{C}_f(\mathbb{R})$ lies in one connected component). Show that if $P \in \mathscr{C}_f(\mathbb{R})$ has infinite order, then the points $nP$ are dense on $\mathscr{C}_f(\mathbb{R})$.

Richard Ganaye · Accepted Answer

ewcommand{\Q}{\mathbb{Q}}

ewcommand{\Z}{\mathbb{Z}}

ewcommand{\N}{\mathbb{N}}

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ewcommand{\C}{\mathbb{C}}

I am forced to use knowledges outside this book to provide the framework for a proof. I don't know if there exists some elementary solution, but such a solution may use the explicit isomorphism of Problem 23.
\begin{proof} 
When the curve $\mathscr{C}_f(\mathbb{R})$ lies in one connected component, the group $E_f(\R)$ is a compact connected real Abelian Lie group of dimension $1$, so is isomorphic to $S_1 = \R/\Z$. The subgroups of $S_1$ are finite, or dense on $S_1$. If $P$ has infinite order in $E_f(\R)$, then $\langle P angle = \{n P, n \in \Z\}$ is an infinite subgroup of $E_f(\R)$, thus its image in $S_1$ is dense on $S_1$. This shows that $\langle P angle$ is dense on $E_f(\R)$, because an isomorphism of Lie groups is also an homeomorphism.
\end{proof}

Exercise 5.7.20* (Density of $\langle P \rangle$ on $\mathscr{C}_f(\mathbb{R}$)

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Exercise 5.7.20* (Density of $\langle P \rangle$ on $\mathscr{C}_f(\mathbb{R}$)

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