This is perhaps a soft question.

Let $X=\mathbb{S}^1 \times \mathbb{S}^1$. Let $\mathbb{Z}_2$ act on $X$ by setting $(-1) \cdot (\theta,\psi)=(\theta+\pi,\psi+\pi)$. Consider the quotient space $X/ \mathbb{Z}_2$ which is obtained after identifying $ (\theta,\psi) \sim(\theta+\pi,\psi+\pi)$.

Is there a succinct description of $X/ \mathbb{Z}_2$ as some product or twisted/fibered product or something like that?

Are there other "simple" descriptions of this space? Is it related to some projective space?

I feel like there should be a "right" terminology to describe it, or a way to recognize it as some familiar space, but I fail to see it.

I understand that identifying antipodal points on the $2$-torus embedded in $\mathbb{R}^3$ results in a Klein bottle- but this is not the same identification we are doing here:

Here we identify $(\theta,\psi)=(\theta+\pi,\psi+\pi)$, and in the embedded description we identify $(\theta,\psi)=(\theta+\pi,-\psi)$.

  • 5
    $\begingroup$ It is a closed, connected, oriented surface of zero Euler characteristic, hence.... $\endgroup$ Commented Feb 27, 2020 at 17:04

2 Answers 2


There are many ways to prove this.

  1. One, is to realize the torus as the quotient of ${\mathbb C}$ by the group of translations $\Gamma$ generated by the translations
    $$ a: z\mapsto z+ 2\pi, b: z\mapsto z+2\pi i. $$ Lifting you involution $\tau: (\theta, \psi)\mapsto (\theta+\pi, \psi+\pi)$ yields the translation $$ c: z\mapsto z+ (1+i)\pi. $$ The group $\hat{\Gamma}\subset {\mathbb C}$ generated by $a$ and $c$ is an index 2 extension of $\Gamma$. By drawing the fundamental parallelogram of $\hat\Gamma$, you should be able to convince yourself that ${\mathbb C}/\hat\Gamma$ is diffeomorphic to the torus $T^2$. (Actually, this is a general fact that if $\Gamma < {\mathbb R}^2$ generated by translations along two linearly independent vectors then ${\mathbb R}^2/\Gamma$ is diffeomorphic to $T^2$.)

  2. An alternative argument relies upon the classification of surfaces. The involution $\tau$ has no fixed points in $T^2$. Hence, $T^2\to S=T^2/\langle \tau\rangle$ is a covering map. The involution $\tau$ preserves orientation (for instance, since it is isotopic to the identity, but there are many other ways to see this). Thus, $S$ is a compact connected oriented surface and $$ \chi(S)=\frac{1}{2}\chi(T^2)=0. $$ Hence, by the classification of surfaces, $S$ is diffeomorphic to $T^2$. In fact, every topological spaces covered by $T^2$ is homeomorphic to $T^2$ or the Klein bottle.

  • $\begingroup$ Is my argument correct?: The quotient space of each horizontal circle (a torus on table) of torus by the above action, is $\Bbb RP^1\sim \Bbb S^1$. So patching together these quotient spaces yield the desired space i.e. $\mathbb{S}^1 \times \mathbb{S}^1$!! $\endgroup$
    – C.F.G
    Commented Mar 1, 2020 at 7:26
  • 1
    $\begingroup$ @C.F.G: What your argument proves is that the quotient space (a surface) fibers over $S^1$ with circle fibers. Now, you have to argue that there are exactly two such surfaces: $T^2$ and the Klein bottle. Then argue that the Klein bottle cannot occur. $\endgroup$ Commented Mar 1, 2020 at 12:57
  • $\begingroup$ OMG, Klein bottle is also circle fibers over circle!! Thanks for remarking this. So it is enough to prove that the action described in the question is orientation preserving. Now how to prove this one? $\endgroup$
    – C.F.G
    Commented Mar 1, 2020 at 19:50
  • 1
    $\begingroup$ @C.F.G: As I said in the 2nd proof, you can use the fact that $\tau$ is isotopic to the identity: A diffeomorphism of an oriented manifold isotopic to the identity always preserves orientation. To see the isotopy, note that $\tau$ is given by a pair of $\pi$-rotations on each factor $S^1$ of $T^2$. Thus, use the fact that every rotation of $S^1$ is isotopic to the identity (through rotations). $\endgroup$ Commented Mar 4, 2020 at 1:16

The way I think about it is from the perspective of Lie theory. In each dimension, there is a unique compact abelian Lie group, namely, the torus $T^k$.

Now, $\mathbb{Z}/2\mathbb{Z}\subseteq T^2$ generated by $\langle (\pi, \pi)\rangle$ is normal (since $T^2$ is abelian, so we can form the quotient $Y:=T^2/(\mathbb{Z}/2\mathbb{Z})$). Being the continuous homomorphic image of $T^2$, $Y$ must be a compact abelian Lie group, so it must be isomorphic to $T^2$ as a Lie group. In particular, $Y$ is diffeomorphic to $T^2$.

$ \ $

Alternatively, you can explicitly write a diffeomorphism from $T^2$ to $Y$. Define $f:T^2\rightarrow T^2$ by $f(\theta, \psi) = (\theta + \psi, \theta - \psi)$.

Note then that $$f(\theta + \pi, \psi + \pi) = (\theta + \psi + 2\pi, \theta - \psi + 0 \pi) = (\theta + \psi, \theta - \psi) = f(\theta, \psi),$$ so $f$ descends to a map $\overline{f}:Y\rightarrow T^2$.

One can easily prove that $f$ is surjective, that $f$ is two-to-one (in such a way as $\overline{f}$ is injective), $f$ is smooth, and that $d_p f$ has full rank everywhere. It follows from this that $\overline{f}$ is a diffeomorphism.

  • $\begingroup$ Is it always true that two isomorphic Lie groups are diffeomorphic? $\endgroup$
    – C.F.G
    Commented Apr 11, 2020 at 13:36

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .