Suppose we have a $\mathbb{Z}/2\mathbb{Z}$ action on torus $\mathbb{T} \times \mathbb{T}$ by $(\xi,\theta)$ goes to $(-\xi,\bar{\theta})$. Then what is the quotient space?
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1$\begingroup$ I think it might be prudent to mention what realisation of torus you're talking about. I assume that is $(S^1)^2\subseteq {\bf C}^2$ in this case? $\endgroup$– tomaszApr 3, 2014 at 2:19
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$\begingroup$ Also, you should probably note when you cross-post this to mathoverflow as well... $\endgroup$– Simon RoseApr 3, 2014 at 2:19
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$\begingroup$ @SimonRose I posted first to MO, then I realized it might not be appropriate for MO. $\endgroup$– sionnApr 3, 2014 at 2:22
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$\begingroup$ @tomasz yes exactly. $\endgroup$– sionnApr 3, 2014 at 2:23
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2 Answers
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Since the action is (presumably) the one inherited from $\mathbb{C}$, you should be able to see that it has no fixed points. In particular, the map from $\mathbb{T} \times \mathbb{T}$ to its quotient is a covering map. Edit: We can now compute the Euler characteristic of the quotient by the usual formula for the Euler characteristic of a cover: $$ 0 = \chi(\mathbb{T} \times \mathbb{T}) = 2\chi(\mathbb{T} \times \mathbb{T}/\sim) $$ since the degree of the quotient map is 2. Hence the Euler characteristic of the quotient is also zero, and so it is either a torus or a Klein bottle.
Since the map in question does not preserve orientation, it must be a Klein bottle.
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1$\begingroup$ nice. Can you please say a line what Riemann-Hurwitz says? $\endgroup$– sionnApr 3, 2014 at 2:47
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$\begingroup$ It's a statement that lets you compute the genus of a curve given the data of a map from another curve and the ramification data. However, it might not be the simplest way to look at it, so I will edit the answer. $\endgroup$ Apr 3, 2014 at 13:16
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The quotient space is the Klein bottle.
I'm going to assume that the definition of the Klein bottle we're using is the usual square with two sides identified in the same direction and the other sides identified in the opposite direction.
Replace the torus you have above with its quotient model coming from the square, in which the opposite sides are identified in the same direction. The first question to address is how your map acts in this model. I don't know how to insert figures so I'll try to give an adequate description in words.
The map that takes a point on the circle to its negative is just a rotation by $\pi$ radians, which corresponds to a translation by $\frac{1}{2}$ via the map $x\rightarrow e^{2\pi ix}$. Thus the map is first shifting the square to the right by $\frac{1}{2}$. If we split the square down the middle along the line $x=\frac{1}{2}$ and take into account the identifications, the left and right halves are "swapped" without any flips.
Now, think of the circle as being a line segment with identified endpoints corresponding to the point $1$ in the complex plane. Conjugation fixes the points $1$ and $-1$ on the circle in the complex plane, and the latter is the midpoint of our interval. Conjugation works symmetrically to interchange points on either side of this in the plane, and so corresponds to flipping our interval upside down around the midpoint. So the second part of our map is just a flip of the square around the line $y=\frac{1}{2}$.
In full our map interchanges the left and right halves of our square and then flips it upside-down. Thus the left side is mapped onto the right side and vice versa. Since in our new space, we are identifying points with their images, we can effectively forget about the right half of our square, except for the right edge of the left half, at $x=\frac{1}{2}$. But the left edge of the square maps to this boundary by a translation followed by a flip. That is, on this rectangle, the left edge is identified with the right edge by a flip, and the top and bottom edges are already identified, since we started with the torus. This matches the description of the Klein bottle originally given above, after stretching in the $x$ direction to get a perfect square.
