Discrete subgroups of $\mathbb R^1\times S^1$ such that the quotient is compact

Asked Sep 10 '20 at 08:17

Active Sep 10 '20 at 10:16

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Let $\Gamma\leq\mathbb R\times S^1$ be a discrete subgroup such that $\frac{\mathbb R^1\times S^1}\Gamma$ is compact. Then what can $\Gamma$ be? Also, what are the possible results for the quotient space.

Intuitively, I think that for any $\Gamma$ with above properties, $\frac{\mathbb R\times S^1}\Gamma$ will always be diffeomorphic to the torus $S^1\times S^1$. Is that true?

asked Sep 10 '20 at 08:17

Chetan Vuppulury

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Hint: This quotient is orientable, connected and has zero Euler chadacteristic. – Moishe Kohan Sep 10 '20 at 13:59
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Pull back to the universal cover. Do you know the classification of discrete subgroups of $\mathbb{R}^2$? – Qiaochu Yuan Sep 10 '20 at 16:10
@MoisheKohan: How do you compute the Euler characteristic without knowing its action? – C.F.G Sep 10 '20 at 16:27
Perhaps by $\dfrac{\mathbb R^1\times S^1}\Gamma \to \mathbb R^1\times S^1$? – C.F.G Sep 10 '20 at 16:30
If you are looking for a related post that you have answered, it is here – C.F.G Sep 10 '20 at 16:35
@C.F.G By constructing an invariant nonzero vector field. – Moishe Kohan Sep 10 '20 at 16:59
Interesting! Now how do you know that is invariant? – C.F.G Sep 10 '20 at 17:02
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Alternative (but related) hint to what Moishe gave: $\Gamma$ is normal, so $(\mathbb{R}\times S^1)/\Gamma$ is a compact abelian 2-dimensional Lie group. – Jason DeVito - on hiatus Sep 10 '20 at 17:27
@JasonDeVito: As usual, Common trick. and the rest is similar to this – C.F.G Sep 10 '20 at 17:34
@JasonDeVito: I know that there is a theorem that asserts that "the quotient of a Lie group by a normal Lie subgroup is also a Lie group" but why it is compact? – C.F.G Sep 10 '20 at 17:39
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@C.F.G: That's the assumption on $\Gamma$ given by Chetan. – Jason DeVito - on hiatus Sep 10 '20 at 17:39
Aha. I am thinking around 1 hour that how to solve this!! So the compactness is in assumption. So in general it is not compact? any proper example? – C.F.G Sep 10 '20 at 17:42
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@C.F.G: Take $\Gamma = {e}$. Less trivially, the $\Gamma$ where the quotient is not compact are precisely those which have trivial projection to $\mathbb{R}$. – Jason DeVito - on hiatus Sep 10 '20 at 17:47

Discrete subgroups of $\mathbb R^1\times S^1$ such that the quotient is compact

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