Immersion of a map and regular submanifold

Question

I know $f:\mathbb{R}\rightarrow S^1\times S^1$ with $f(t)=(e^{iat}, e^{ibt})$ with fixed $a$ and $b$ is an immersion. What kind of constraint can we impose on $a$ and $b$ so that the image of $f$ is a regular sub-manifold of $S^1\times S^1$?

Answer 1

The case where both $a$ and $b$ are equal to zero then $f$ is a constant function. In such a case you have that the image is a point. A point is a $0$ manifold and it's inclusion is an injective immersion which is homeomorphic to it's image so we can consider the case where $b\neq 0$. $f$ will be an immersion in this case. But You have that if $a/b \notin \mathbb{Q}$ then the image will be dense. In such case it can't be a submanifold since given any chart $\varphi:U\rightarrow \mathbb{R}^2$ will send $f(\mathbb{R})\cap U$ to a dense subset of $\mathbb{R}^2$ Which can't be a subspace since it's all of $\mathbb{R}^2$. So one must have $a/b\in\mathbb{Q}$ If $a/b\in\mathbb{Q}$ then the image is homeomorphic to $S^1$. Identifying the torus as being $\mathbb{R}^2/\mathbb{Z}^2$ you have that $f$ can be lifted to a function $\tilde{f}$ to $\mathbb{R}^2$ which will have as image a line with slope $a/b$ passing through the origin. Since it's rational we have $\tilde{f}$ will pass through a point with integer coordinates at $n\cdot b$ where $n\in \mathbb{Z}$. This means that $f$ will have period $b$. By identifying $\mathbb{R}/b\mathbb{Z}\equiv S^1$ one can define a homeomorphism from $S^1$ to the image of $f$ through the quotient map. This map will be an embedding that has the same image of $f$. So the only condition is that either $a/b\in \mathbb{Q}$ or $b/a\in\mathbb{Q}$ or $a=0$ and $b=0$. (There's probably a neater way to write this)