This is Question 1 from Example 3.1.10 in Burago, Burago, and Ivanov’s A Course in Metric Geometry, though fair warning I haven't actually read any of that book. A friend of mine asked about it in a grad student discord and I got invested in it.

What topological space do you get when you quotient $\mathbb{R}^2$ by $(x,y) \sim (-y,2x)$?

I think I have an answer for this, but it's sufficiently weird that I wanted some outside confirmation that I've done it right.

So, if we let $\mathbb{Z}$ act on $\mathbb{R}^2$ where the generator acts by $T = \begin{pmatrix} 0 & -1 \\ 2 & 0 \end{pmatrix}$, it's clear that the orbits are exactly the desired equivalence classes. There's a fixed point at $0$, which is a bit annoying, so to start let's work with the punctured plane instead.

A fair amount of thought (and some sage code) lets us find a fundamental domain for this action:

Edit (Apr 27): I've still been working on this problem, and I found a much nicer fundamental domain than my original one. I've left my original details commented in the post body so they won't distract potential answerers, but are still available to people who want to see them.

fundamental domain

You can see the orbits of the purple and blue regions tile the plane, so taking one of each gives a fundamental domain. We can read the identifications off this picture as well, and we see that our region is a torus.

Of course, we still have the origin to deal with. The way that I think this works is as a kind of generic point in the quotient space. Notice every neighborhood of the origin contains a tail of copies of the fundamental domain. I think this should mean that, in the quotient, the closure of the origin should be the entire space. This makes sense as well since the equivalence relation isn't closed, so the quotient space shouldn't be hausdorff, but I'm not used to reasoning about quotient spaces that aren't manifolds (maybe with a few singular points).

So, the tl;dr is that I think the quotient space is a torus with one "generic" point whose closure is everything. Does this seem correct?

Thanks in advance ^_^

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    $\begingroup$ I think the proper thing to say here is that any neighborhood of the origin contains the whole quotient space. The origin by itself still makes a closed set. So the quotient space is $T_0$ but not Haussdorff. $\endgroup$ Apr 27, 2021 at 16:37
  • $\begingroup$ @EthanDlugie -- That does sound better. $\endgroup$ Apr 27, 2021 at 16:44
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    $\begingroup$ Also the interesting thing about this problem (or rather why it's in the metric geometry book) is that in the quotient pseudometric, all points are distance 0 from each other. So the topology induced by the quotient pseudometric is quite different from the quotient topology. $\endgroup$ Apr 27, 2021 at 17:55
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    $\begingroup$ @EthanDlugie: I think, your 1st comment should be an answer. $\endgroup$ Apr 28, 2021 at 23:18

1 Answer 1


Formatting my comment as an answer:

Your identification of the fundamental domain, along with some bookkeeping of the boundary identifications, shows that $\mathbb{R}^2-\{0\}$ descends to a topological torus in the quotient space. The origin is a fixed point of the $\mathbb{Z}$ action, so it descends to a single point. Thus the quotient space is partitioned into two subsets, one homeomorphic to the torus and one a single point.

But the topology is not just a disjoin union $\mathbb{T^2} \sqcup \{0\}$. Every point in the torus does have a neighborhood separating it from $0$. You can see this by taking a small open ball around any point in $\mathbb{R}^2-\{0\}$ and noting that the images of these balls under (positive and negative) powers of $T$ have diameters roughly proportional to the distance between their centers and the origin.

On the other hand, the image of the origin in the quotient space has the strange property that any open neighborhood of it contains the whole torus, and so it contains the whole quotient space. This follows because any open set around the origin in $\mathbb{R}^2$ contains representatives of every single orbit of your $\mathbb{Z}$ action.

In the language of the separation axioms of point set topology, your quotient space is $T_0$ but not $T_1$ (and not $T_2$, i.e. not Hausdorff).


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