Showing a subset of the torus is dense

Question

Let $T^2\subset\mathbb{C}^2$ denote the (usual) torus. Let $a\in\mathbb{R}$ be an irrational number and define a map $f(t)=(e^{2\pi it}, e^{2\pi i at})$. Prove that:

(a) $f$ is injective

(b) $f$ is an immersion, but not an embedding

(c) $f(\mathbb{R})$ is dense in $T^2$

(d) $f(\mathbb{R})$ is not an embedded submanifold of $T^2$

I've already shown (a) and (b), but (c) has me stumped. (I haven't tried my hand at (d) yet.)

For ease I've just been working with a square (with edges identified). I took an open set (wlog, an open ball) in the square, but it's not clear what will happen if the image of $f$ doesn't hit this ball. It seems like we want to somehow get that the image "wraps" back onto itself, which would imply that $a$ is rational, but... I don't see how to get there.

Any help is appreciated!

Answer 1

I think you want to show directly that the image does hit any ball. For any ball with center $(x,y)$ in the unit-square-with-edges-identified, translate the ball down along the direction $(1,a)$ such that its center sits at, say, the vertical edge of the square, with center $(x,y)-x(1,a)=(0,y-ax)$. If the translated ball is hit, the original ball will be hit $x$ time later.

Thus, it is enough to show that every ball centered on the vertical edge is hit. What can you say about the intersection of the image and the vertical edge?

Answer 2

It suffices to prove that for each $w \in S^1$ the space $$T_w = f(\mathbb R) \cap (\{w\} \times S^1)$$ is dense in $\{w\} \times S^1$. Let us write $w = e^{2\pi i \theta}$.

Clearly $f(t) \in T_w$ if and only if $e^{2\pi i t} = w = e^{2\pi i \theta}$ which is equivalent to $e^{2\pi (t - \theta) i} = 1$. The latter means $t - \theta \in \mathbb Z$. Hence $T_w = \{w\} \times S_w$ with $$S_w = \{ e^{2\pi \alpha (\theta + k) i} \mid k \in \mathbb Z \} = \{ e^{2\pi \alpha k i}w \mid k \in \mathbb Z \}.$$

Therefore it suffices to prove that $S_w$ is dense in $S^1$.

Let us begin with $w = 1$. We have $$S_1 = \{ e^{2\pi \alpha k i} \mid k \in \mathbb Z \} .$$ This is a subgroup of $S^1$ (with respect to complex multiplication) for each $\alpha$. It is infinite if and only if $\alpha$ is irrational. A well-known theorem says that infinite subgroups of $S^1$ are dense. For a proof see for example Subgroup of the unit circle under complex multiplication. Therefore $S_1$ is dense in $S^1$.

Let us next consider an arbitrary $w$. The map $$\mu_w : S^1 \to S^1, \mu_w(z) = zw$$ is a homeomorphism (with inverse $\mu_{1/w}$). Clearly $\mu_w(S_1) = S_w$. Therefore $S_w$ is dense in $S^1$.