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.

  • $\begingroup$ See math.stackexchange.com/questions/1134829/… $\endgroup$ Jan 22, 2019 at 23:02
  • $\begingroup$ It's kind of like rational multiples of an irrational on the number line - they form a kind of "fine dust" which is dense. One could similarly form, for instance, $\Bbb Q\sqrt{2}\times\Bbb R$ within $\Bbb R^2$, and it would be a dense foliage of lines, "parametrized" by the fine dust on the number line. The striations on a torus coming from an irrational slope are the same - indeed if you were to intersect it with any closed circle through the torus, the intersection would be a fine dust on the circle, just like on the number line. $\endgroup$
    – anon
    Jan 23, 2019 at 2:06

1 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.

  • $\begingroup$ Why do we have $A\approx16$ in this image ?If I understand the mod 1 is what separates each parallel line,and it is done because we could not see in the filled out unit square the disconnections,correct ? $\endgroup$
    – user159729
    Jan 23, 2019 at 0:44
  • $\begingroup$ I counted (maybe with some error) 16 line segments crossing the right hand margin, which is where $\approx 16$ comes from. Without the $A$ cutoff the square would be a solid block of ink. $\endgroup$ Jan 23, 2019 at 0:51

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