A dense curve on the Torus

Question

Let $\Gamma_a:\mathbb{R}\to S^1\times S^1 = \mathbb{T}^2$ be the winding curve defined by $\Gamma_a(t)=(e^{it},e^{iat})$ where $a\in\mathbb{R}$.

Prove that $\Gamma_a(\mathbb{R})$ is a regular submanifold of $\mathbb{T}^2$ if and only if $a\in\mathbb{Q}$.

I know that:

$\Gamma_a$ is an embedding $\Rightarrow$ $\Gamma_a(\mathbb{R})$ is a regular submanifold.

For $\Gamma_a$ to be an embedding:

• $\Gamma_a$ is immersion (I already proved. $\Gamma_a'(t)\neq0\space\forall t$)
• $\Gamma_a$ is injective (I could only prove if $a\in\mathbb{R}\setminus\mathbb{Q}$. If $\Gamma_a(t)=\Gamma_a(s)$ then $\begin{cases}t-s=n \\ a(t-s)=m\end{cases}$ with $n,m\in\mathbb{Z}$. This can only be true if $n=t-s=0$)
• $\Gamma_a^*:\mathbb{R}\to\Gamma_a(\mathbb{R})$ is an homeomorphism. (I checked continuity. I couldn't see the injectivity. And so I couldn't see the continuity of the inverse either)

To get the converse I know that, if $a\in\mathbb{R}\setminus\mathbb{Q}$:

$\Gamma_a(\mathbb{R})$ is dense in $\mathbb{T}^2$.

But I don't know how can it be useful to prove that $\Gamma_a(\mathbb{R})$ is not a regular submanifold.

Answer 1

Here's a hint: if you already know that $\Gamma_a(\mathbb R)$ is dense in $\mathbb T^2$, then perhaps you can leap to a generalization, which as it turns out is pretty easy to prove:

For any manifold $M$ of dimension $m$ and any submanifold $N \subset M$ of dimension $n < m$, $N$ is not dense in $M$.