How is a group visually represented?

Question

I don't understand an example on the wkipedia article for Lie groups:

The group given by $H = \left\{ \left( \begin{array} { c c } { e ^ { 2 \pi i \theta } } & { 0 } \\ { 0 } & { e ^ { 2 \pi i a \theta } } \end{array} \right) | \theta \in \mathbb { R } \right\} \subset \mathbb { T } ^ { 2 } = \left\{ \left( \begin{array} { c c } { e ^ { 2 \pi i \theta } } & { 0 } \\ { 0 } & { e ^ { 2 \pi i \phi } } \end{array} \right) | \theta , \phi \in \mathbb { R } \right\}$ with $a \in \mathbb { P } = \mathbb { R } \backslash \mathbb { Q }$ a fixed irrational number,is a subgroup of the torus $\mathbb { T } ^ { 2 }$ that is not a Lie group when given the subspace topology.

I don't understand why the group is represented by all these lines in a $2\pi$ length square. A portion of the group $H$ inside $\mathbb {T^2}$ Small neighborhoods of the element $h \in H$ are disconnected in the subset topology on $H$

Can someone clarify why we have this image ?

Thank you.

Answer 1

I think the picture is meant to illustrate the image of $\mathbb R$ under the map $$\theta\mapsto \tau(\theta)=(\theta\bmod1,a\theta\bmod1),$$ or rather, the image of $[0,A]$ under $\tau$, where $A\approx 16$. The image of all of $\mathbb R$ under $\tau$ would visually fill up the unit square, even though it does not do so set-theoretically. The closure of the image of $\mathbb R$ would equal the unit square, however. The point $h$ in the picture shows how the set of $\theta$ such that $\tau(\theta)$ is close to $h$ is disconnected: every now and then a big $\theta$ makes $\tau(\theta)$ come close to $h$, and so on.

Why $\theta \bmod 1$ and not $\exp(2\pi i\theta)$? Because it is hard to put pixels on the page whose row and column indices are complex numbers. You are supposed to imagine the unit square as representing the torus $\mathbb T^2$ by thinking the point $(x,y)$ on the page represents the point $(\exp(2\pi i x),\exp(2\pi i y))$ on the torus.