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From the Wikipedia article on projective planes:

[...] consider the unit sphere centered at the origin in $\mathbb{R}^3$. Each of the $\mathbb{R}^3$ lines in this construction intersects the sphere at two antipodal points. Since the $\mathbb{R}^3$ line represents a point of $\mathbb{RP}^2$, we will obtain the same model of $\mathbb{RP}^2$ by identifying the antipodal points of the sphere. The lines of $\mathbb{RP}^2$ will be the great circles of the sphere after this identification of antipodal points.

My question is:

When the construction of the real projective plane is essentially identifying antipodal points of the sphere, what is its analogue when identifying antipodal points of the torus?

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What are you thinking of as the "antipodal map" on the torus? There are a few reasonable candidates, and the answer will depend on which one you choose... –  Micah Dec 5 '12 at 23:54
    
@Micah: I am thinking of a dummy's antipodal map: the torus = circle $\times$ circle, antipodal points being antipodal on one of these circles. –  Hans Stricker Dec 5 '12 at 23:59
    
@Micah: What other reasonable candidates do you have in mind? –  Hans Stricker Dec 6 '12 at 0:20
    
The obvious one that's topologically different is the one in the answer, where you take the antipodal map on both circles. Depending on how explicit you're being, you might also consider (the antipodal map on the first circle) $\times$ (the identity on the second) to be different from (the identity map on the first circle) $\times$ (the antipodal map on the second)... –  Micah Dec 6 '12 at 0:36
    
Note, however, that whatever choice you make you'll have to end up with either a Klein bottle or a torus -- you'll get a surface that has half the Euler characteristic of the torus, the torus has Euler characteristic zero, and the only other surface with Euler characteristic zero is the Klein bottle. –  Micah Dec 6 '12 at 0:37
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1 Answer

It is not absolutely clear what antipodal points on a torus are. However, if you take a standard torus in $\mathbb{R}^3$, obtained by rotating a circle $(x-a)^2+z^2 = b^2$ with $a>b>0$ about the $z$-axis, and if you call $(-x,-y,-z)$ the antipodal point of $(x,y,z)$, I believe you get the Klein bottle.

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If you are right, I find this a really great - so to say - finding: "The Klein bottle is to the torus, what the real projective plane is to the sphere." –  Hans Stricker Dec 6 '12 at 0:01
    
This is right -- a fundamental domain is just half of the torus. –  Aaron Mazel-Gee Dec 6 '12 at 0:04
    
Note that this is not "the dummy's antipodal map", though; then you could take the same fundamental domain, but you'd just get a torus back. –  Aaron Mazel-Gee Dec 6 '12 at 0:06
    
@Aaron: Thanks - explicitly - for this hint. –  Hans Stricker Dec 6 '12 at 0:08
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