Showing a dense curve in the torus is weakly embedded

Question

This is Lee's problem 5-11 in his smooth manifolds book.

I want to show that a particular curve $C$ on the torus $T$ (a line with irrational slope in the square model of the torus) is weakly embedded. Weakly embedded here means that if there is any smooth map $f:M\mapsto T$, with $f(M)\subset C$, then the associated map $F:M\mapsto C$ is also smooth. In other words, it's always valid to restrict the range.

I know it's enough to just show $F$ is continuous. But I don't see how to do that. I was hoping I could use connectivity (since an open subset of $T$ has a bunch of arcs from $C$), but I can't seem to make it work.

Does anyone have a good hint (no full solution please)?

EDIT: I think I can see the basic idea after looking at the proofs of Proposition 19.16 and Theorem 19.17 in Lee's book.

We can take a small open set $U$ of $T$ around a point in $C$, and then $C$ inside $U$ looks like a countable collection of line segments. If we then take a small connected neighborhood in $M$, it has to be mapped into this countable collection, and so only ends up in one. So it seems my idea of using connectedness was correct. It's probably more work to make this whole idea formal.

Answer 1

Let $\gamma:\mathbb{R}\to \mathbb{T}^2$ be the smooth immersion $\gamma(t)=(e^{it},e^{i\alpha t})$, where $\alpha$ is any irrational number, and let $S$ denote the immersed submanifold of $\mathbb{T}^2$ determined by this map. Since the sets of the form $\gamma((t_0-\varepsilon,t_0+\varepsilon))$ for $t_0\in \mathbb{R}$ and $0<\varepsilon <2\pi$ form a basis for the topology of $S$, to prove the continuity of $f$ it suffices to show that the inverse images of these sets under $f$ are open in $M$.

Given $t_0\in \mathbb{R}$ and $0<\varepsilon <2\pi$, let $U$ denote the open set of $\mathbb{T}^2$ given by the formula $$U=\{(e^{it},e^{is})\mid -\varepsilon<t<\varepsilon,\ s\in\mathbb R\}.$$ And then let $\varphi:U\to(-\varepsilon,\varepsilon)\times\mathbb{S}^1$ denote the homeomorphism $\varphi(e^{it},e^{is})=(t,e^{is-\alpha t})$. (Geometrically, $\varphi$ "twists" and "straighten" $U$ so that the image of $\gamma$ will be just the horizontal lines on the cylinder. ) Now we have $$\varphi(U\cap S)=\bigcup_{n\in\mathbb{Z}}(-\varepsilon,\varepsilon)\times \{e^{2n\pi\alpha i}\},$$ and it is easy to check that this equation represents the partition of $\varphi(U\cap S)$ into its path components. It follows that the path components of $U\cap S$ are $\varphi^{-1}((-\varepsilon,\varepsilon)\times \{e^{2n\pi\alpha i}\})=\gamma(2n\pi-\varepsilon,2n\pi+\varepsilon)$. In particular, $\gamma(-\varepsilon,\varepsilon)$ is a path component of $U\cap S$. Therefore, $f^{-1}(\gamma(-\varepsilon,\varepsilon))$ is a union of some subcollection of path components of $f^{-1}(U)$, and is therefore open in $M$.

Answer 2

If $f(M)$ is a submanifold, then it has an intrinsic metric $d$ from torus.

And if $c: (-\infty,\infty)\rightarrow T$ is an immersion whose image is $C$, then $c([-n,n])$ has an intrinsic metric $d_n$. From $d_n$, we have an intrinsic metric $d_\infty$ on the whole $C$.

Note that $d=d_\infty$ on $f(M)$ so that continuity of $f: M\rightarrow C$ is followed.