This is Example 15.9 of Tu's Introduction to Manifolds.
Ler $L$ be a line with irrational slope in $\mathbb{R}^2$ and $H$ the image of $L$ under the projection $\pi: \mathbb{R}^2 \rightarrow \mathbb{R}^2/\mathbb{Z}^2$. I see how if $L$ has rational slope it will trace out a closed circle. Likewise, if $L$ has irrational slope I see how it will not ever close up and will instead "fill up" the torus by winding around infinitely often. I have a few questions based on this example.
- Tu says that
The image $H$ of $L$ under the projection $\pi$ is a closed curve if and only if the line $L$ goes through another lattice point, say $(m, n) \in \mathbb{Z}^2$. This is the case if and only if the slope of $L$ is $n/m$, a rational number or $\infty$; then $H$ is the image of finitely many line segments on the unit square.
I do not understand the last part of the above. Shouldn't $H$ just be a single closed loop? Why would $H$ be the image of multiple line segments and why must there be finitely many?
- In a exercise, he says
Suppose $H \subset \mathbb{R}^2/\mathbb{Z}^2$ is the image of a line $L$ with irrational slope in $\mathbb{R}^2$. We call the topology on $H$ induced from the bijection $f: L \rightarrow H$ the induced topology and the topology on $H$ as a subset of $\mathbb{R}^2/\mathbb{Z}^2$ the subspace topology. Compare these two topologies: is one a subset of the other?
He give the solution to this exercise as:
A basic open set in the induced topology on $H$ is the image under $f$ of an open interval in $L$. Such a set is not open in the subspace topology. A basic open set in the subspace topology on $H$ is the intersection of $H$ with the image of an open ball in $\mathbb{R}^2$ under the projection $\pi$; it is a union of infinitely many open intervals. Thus, the subspace topology is a subset of the induced topology, but not vice versa.
I did not fully understand the answer he gave. An alternate answer is given in this post. Going off the answer in that post, if we choose a ball $B \subset \mathbb{R}^2$ that intersects $L$, this will be homeomorphic to an open interval and so it is open. Thus $\pi(B \cap L)$ will be open in the induced topology. In the answer of the linked post, he says any neighborhood in the subspace topology of any point in $\pi(B \cap L)$ contains the image of points in $L$ that are not in $\pi(B \cap L)$. I assume this follows from the denseness of the image of $L$? How does this show that $\pi(B \cap L)$ is not open in the subspace topology? Furthermore, how do we know that reverse inclusion holds, that the subspace topology is necessarily a subset of the induced topology?
Best Answer
Yes. $H$ is a single closed loop which is the image of finitely many line segments.
Assume there is the square grid on the plane. For simplicity, let $m,n\ne0$. Consider the line segment from $(0,0)$ to $(m,n)$. It crosses a finite number of squares (definitely no more than $mn$). The grid lines divide it into a finite number of segments (sharing some endpoints), which are all mapped to segments on the square torus $\mathbb{R}^2/\mathbb{Z}^2$ (glued at all endpoints).
2.
Correct.
The linked answer in turn links another answers, which
explains thisexplains why the two topologies must be different. To show that $\pi(B\cap L)$ is not open in the subspace topology, we can consider finite intersections (and unions) of the basic sets in the subspace topology and show that they always consist of the infinite number of segments, so you cannot get $\pi(B\cap L)$ that way.Well, if a basic open set in the subspace topology is a union of (infinitely many) open intervals which are subsets of $H$, i. e. basic open sets in the induced topology, then it is also open in the induced topology, and so are other open sets in the subspace topology.