Quotient space of torus under coordinate reversion and conjugation

Question

We consider the torus as $X=\mathbb{T}^2$ where $\mathbb{T}=\{z\in \mathbb{C}: |z|=1\}$. Let $G=\mathbb{Z}_2$ act on $X$ by $1 \cdot (z,w)\mapsto (\bar{w}, \bar{z})$ and the identity map $0 \cdot (z, w)\mapsto (z,w)$. What is, up to homeomorphism, the quotient space $Y=X/G$?

Is it possible to deduce what this space is without using homology? I studied other cases where the action only reverses the coordinates (in that case $X/G$ is the Mobius strip) and where the action only conjugates the coordinates (in that case $X/G \cong [-1,1]^2$) but I am stumped on how to handle both at once. When dealing with just reversion, we could identify the torus as a square with appropriate sides glued and carry across the action to this space and figure out the resulting space after a little work. In the conjugate only case, we could find a direct mapping immediately and use the universal property of quotient maps. Now, a direct mapping does not seem possible here and I am unsure of how the conjugation part of the action carries over to the glued-square representation of $\mathbb{T}^2$. Any help or hint is appreciated (even one with homology as I am gradually studying introductions into the topic).

(I like to think about these problems as a refreshing break from probability theory but I am no topologist so I apologize if this is trivial).

Answer 1

We identify $\mathbb T^2$ as the quotient obtained from $[0,1]^2$ by identifying $(x,0)\sim (x,1)$ and $(0,y)\sim (1,y)$ for all $0\leq x,y\leq 1$. Under this homeomorphism, the equivalence class $[ (x,y)]$ maps to $(e^{2\pi i x}, e^{2\pi i y})\in \mathbb T^2$. So if we identify $(z,w)\sim (\bar w, \bar z)$ in $\mathbb T^2$, we are identifying $(x,y)\sim (1-y,1-x)$ in $[0,1]^2$ in addition to the previous identifications as well.

Now observe that $(x,y) \mapsto (1-y,1-x)$ is reflection about the line $X+Y=1$

So basically you take a square sheet of paper, fold it across the diagonal and paste it to get a right angled triangle. Now you have to glue the 2 right angled sides and the three vertices but the right angled sides you have to glue in the correct orientations. So you land up with a simplicial complex.

Here's a diagram.

Using Van Kampen, we get $\pi_1(X/G)=\langle a, b | a^2b \rangle \cong \mathbb Z$ and hence $H_1 (X/G)= \mathbb Z$

Answer 2

Option 1: Continue from where the other answer left off. Cut along the $x=y$ line, and reassemble gluing the two "a" arrows -- you get a picture of the Mobius band i.e. a square with (only) two opposite sides identified in opposite directions. This is basically identical with the answer to the "reverse coordinates" case you have linked.

Option 2: The fact that the two cases are identical can be argued directly: the automorphism (smooth map with smooth inverse) $\mu:T^2\to T^2$ $\mu(z,w)= (z,\bar{w})$ conjugates the two actions: the first one $q_1(z,w)=(w,z)$ and $q_2(z,w)=(\bar{w}, \bar{z})$ i.e. $q_1(\mu(z,w))=(\bar{w}, z)=\mu(q_2(z,w))$. This implies that the quotients are identified for general reasons: just send the class of any $p$ in $T^2/q_2$ to the class of $\mu(p)$ in $T^2/q_1$ (this is well-defined by "conjugation" equation above, and invertible by invertibility of $\mu$).

Option 3: You can show that the quotient is the Mobius band as follows:

Consider the map $\pi:T^2\to S^1$, $(z,w)\to z\bar{w}$. The points $(z, w)$ and $(\bar{w}, \bar{z})$ are mapped to the same image, so it descends to the quotient $X/G$. Parametrize the base $S^1$ as $\alpha=e^{i\theta}, \theta\in[0, 2\pi]$

Preimage of $\alpha$ under $\pi$ in the torus is a circle $(\alpha w, w)$, and it is cut by $p_1=(\sqrt{\alpha}, 1/\sqrt{\alpha})$ and $p_2=(-\sqrt{\alpha}, -1/\sqrt{\alpha})$ into two arcs each of which is a lift of the preimage of $\alpha$ in $X/G$. Thus $X/G$ is fibered over the circle with fibers being intervals. It is also reasonably clear that if we take $w=e^{i\phi}$ with $\phi \in [-\theta/2, \theta/2]$ as the lifting arc, then as one goes around the circle the endpoints of the preimage segment get switched, i.e. the fibration $\pi:X/G\to S^1$ is the one of the Mobius band.

The case where $G$ acts by $(z,w)\to(w,z)$ instead is handled the same way, by taking the map $\pi(z,w)=zw$ (note that these maps are also "conjugate" by $\mu$).