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I'm currently working through Jeff Lee's 'Manifolds and Differential Geometry'. He defines a Lie Subgroup, $H$, to be an abstract subgroup of a Lie Group $G$, such that the inclusion map $H\hookrightarrow G$ is an immersion. He then lists an example of a Lie Subgroup which I am trying to understand.

Defining $S^1$ to be the unit circle in $\mathbb{C}$, he writes:

'The image in the torus $T^2=S^1\times S^1$ of the map $\mathbb{R}\rightarrow S^1\times S^1$ given by $t\mapsto (e^{2\pi i t},e^{2\pi i a t})$ is a Lie Subgroup. This map is a homomorphism. If $a$ is rational, then the image is an embedded copy of $S^1$ wrapped around the torus several times depending on $a$. If $a$ is irrational, then the image is still a Lie subgroup but it is now dense in $T^2$.

I am struggling to understand this example at all. I can see why the map is a homomorphism, but I do not understand his argument about the rationality/irrationality of $a$. I also wouldn't know how to show the image of the map is a Lie subgroup. That is, I wouldn't know how to show the inclusion map is a submersion.

Any help with this would be very much appreciated!

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Draw the torus as a square and start drawing the lines (the embedding of $S^1$. The parameter $a$ gives you the slope of a line in the plane where the square is embedded. Start at zero with that slope. Once you reach the border, use the (usually) unique other boundary point to start over again drawing a new line with the same slope.

Now: If you have a rational slope, you will end up again at you starting point after some (how many?) repitions of drawing. If you have a irrational slope you will keep on drawing new lines and you will figure, that you come arbritrary close to every point in that square.

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  • $\begingroup$ This was really meant to be an "explanation" not a rigorous attempt to prove the result. Once you figure what I mean (and get the necessary intuition for the problem) you will find it easy to come up with a proof. $\endgroup$ – Daniel Valenzuela Feb 10 '15 at 0:25
  • $\begingroup$ Thanks for your comment. I don't follow what you are saying at all. I don't know what you mean by 'the lines' or what you mean by 'start at zero'. $\endgroup$ – beedge89 Feb 10 '15 at 6:05
  • $\begingroup$ Note that $T^2 = \mathbb R^2 / \mathbb Z^2$ . But $\mathbb R^2$ is a vectorspace and you can consider lines (lines are per definition for me images of linear mappings from $\mathbb R$) with slope $a$. $\endgroup$ – Daniel Valenzuela Feb 10 '15 at 6:22

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