Is the image of $f: R \to S^1 \times S^1, f(t) = (e^{it}, e^{\sqrt 2it})$ a submanifold?

Question

Let $f: R \to S^1 \times S^1$ be $f(t) = (e^{it}, e^{\sqrt 2it})$,is $f(\mathbb R)$ a submanifold? By easy verification, $f$ is an injective immersion. Then if I want to show that the image is a submanifold, I will want to show that $f$ is a smooth embedding. It remains to show that the inverse is continuous. Can the inverse be written down explicitly?

Answer 1

Let us prove that $f(\mathbb{R})$ is dense in $T = S^1 \times S^1$. This will show that $f(\mathbb{R})$ is not a submanifold of $T$ (since a submanifold $S$ of a manifold $M$ is never dense).

It is a well-known result that any infinite subgroup of $S^1$ (endowed with complex multplication) is dense. See for examaple Subgroup of the unit circle under complex multiplication.

Consider $(x,y) \in T$ and $\epsilon > 0$. Choose $t \in \mathbb{R}$ such that $e^{it} = x$. Then for each $t_k = t + 2k\pi$, $k \in \mathbb{Z}$, we also have $e^{it_k} = x$.

But now the map $g : \mathbb{Z} \to S^1, g(k) = e^{2\sqrt 2 \pi i k}$, is an injective group homomorphism. Hence its image is an infinite subgroup of $S^1$ and therefore dense in $S^1$. Choose $k$ such that $\lvert x g(k) - y \rvert < \epsilon$ (note that $\mu_x : S^1 \to S^1, \mu_x(z) = xz$, is a homeomorphism, hence $\mu_x g(\mathbb{Z})$ is dense in $S^1$). Then $$e^{\sqrt 2i t_k} = e^{\sqrt 2i t} e^{2\sqrt 2\pi i k} = xg(k)$$ and $$\lvert e^{\sqrt 2i t_k} - y \rvert < \epsilon .$$ This show that each neighborhood of $(x,y)$ in $T$ contains an element of $f(\mathbb{R})$.