Let's say that $G$ is the group of isometries of $\mathbb{R}^2$ with the usual metric, and $H < G$ is a subgroup generated by one element. Prove that the quotient $\mathbb{R}^2/H$ is either noncompact or not Hausdorff (anyway, I think it is always noncompact...) The thing is, I can visualize this when $H$ leaves the origin fixed. But when $H$ is made of compositions of translations and rotations it's really bad, I don't see how the quotient turns out to be... Any help would be appreciated!
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@user79365: yes, the point is that when the generator is the composition of a translation and a rotation, and the rotation is an irrational multiple of $\pi$, I don't see it. Since if the origin is fixed (when the generator is $z \mapsto e^{i \theta}z$, with $\theta$ irrational multiple of $\pi$), you can always see that it is non-Hausdorff by considering a neighborhood of $[1]$ and of $[e^{i/3}]$. But you cannot do the same when there is also a translation... – Second Jun 01 '13 at 09:44
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2@user79365: that's not true; there are plenty of things in $\operatorname{Aut} \mathbb{R}^2$ other than pure translations and pure reflections. – Daniel McLaury Jun 01 '13 at 09:45
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All right, this is easily solved by characterization of plane isometries. I am confused though: can the composition of a rotation and a translation be viewed as a "glide reflection"? How? – Second Jun 01 '13 at 11:11
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1Translation and rotation are orientation-preserving, whereas reflection is orientation-reversing... – Daniel McLaury Jun 01 '13 at 18:48
1 Answers
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A single isometry, with the exception of the identity, can be classified as one of the following:
- translation
- rotation
- reflection
- glide reflection
A composition of a translation and a rotation is again a rotation, just around a different center. For discrete groups, your quotient space will be compact if and only if $H$ contains two linearily independent translations. This would be the case of wallpaper groups. But since your $H$ is generated by a single transformation, that shouldn't worry you. So let's go through the alternatives.
Translation
A group generated by a single translation will be the frieze group p1. Its quotient space is an infinite strip, with edges identified, but since it extends towards infinity it is not compact.
Rotation
A rotation by a rational fraction of $2\pi$ will give you a cyclic group as symmetry group, and a glued up pizza slice (also known as a cone) as quotient space. Again non-compact.
For an irrational fraction of $2\pi$ your pizza slice becomes arbitrarily thin (see comments below for more thoughts on that), but there are still infinite sequences along a ray in radial direction, so this is again non-compact.
Reflection
This is an involution, the quotient space is a half plane. Not compact.
Glide reflection
The symmetry group is frieze group p11g, and you can again choose a strip of the plane as fundamental domain. Same argument as for translations tells you the quotient space is again non-compact.
Identity
The quotient space of the identity transformation is the original space, very clearly non-compact.
MvG
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Yes, thanks, I was just confused about compositions of rotations and translations. Anyway, I believe that the quotient under a rotation of an irrational multiple of $\pi$ is not a single line. – Second Jun 03 '13 at 16:06
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@Second: at least if the angle of rotation is a transcendental multiple of $\pi$, then every point on a circle around the center of rotation will eventually be mapped onto any other point on the same circle, so equivalence classes will in fact be such concentric circles, and you can choose representants on a single ray. But you are probably right in the case where the ratio between rotation angle and $2\pi$ is neither rational nor transcendental. I'm not exactly sure what you'd get in those cases, but infinity will always be a part of it. – MvG Jun 03 '13 at 16:18
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if an equivalence class was a whole circle, that would give us an enumeration of the real numbers between $0$ and $2\pi$, but they are uncountable. This is why I think that the quotient is larger than a single line. – Second Jun 03 '13 at 22:08
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@Second: Valid point. I guess I'll have to update my answer in some fashion, even though the main result still stands. – MvG Jun 04 '13 at 04:43
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I just found the question “What is known about the quotient group ℝ/ℚ?” in this context. Interesting. – MvG Jun 04 '13 at 08:53