The surface obtained by identifying antipodal points of $\mathbb{S}^1 \times \mathbb{S}^1$

algebraic-topologydifferential-topologygeneral-topologyquotient-spacessurfaces

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 $(\theta,\psi)=(\theta+\pi,-\psi)$.

Best Answer

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.

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