What is an atlas for a torus manifold?

[aheight]

I don't quite understand some work I'm doing creating the normal Riemann surface for the function $f(z)=\frac{A}{\sqrt{(1-z^2)(k^2-z^2)}}$. I can use Schwarz-Christoffel transforms to map the function to a rectangular polygon in the zeta-plane then map this rectangle onto a torus. But I don't understand what becomes the atlas to make the surface of the torus a manifold. I was wondering if someone could help me understand this.

Thanks

 

[fresh_42]

Maybe this helps: http://www.math.ubc.ca/~feldman/m428/manifolds.pdf

 

[mathwonk]

That link gives a nice little explicit way to map the plane with cartesian coordinates theta and phi, onto an explicit torus, but it seems to omit one tiny point. I.e. when we map the plane onto a torus by wrapping it around in two directions, we have to be slightly careful to obtain an atlas. Namely, this map is not injective of course, and in the case considered in the link, to obtain an injective restriction, one must restrict, as suggested there, to open rectangles of side length 2pi (or less). However when this is done, the image is of course not all of the torus, but is the torus minus a pair of perpendicular and intersecting circles.

This is why he says to use two such rectangles which omit different pairs of circles. If you look at it however you will note that the two pairs of omitted circles intersect in 2 points. Hence even using two such charts will still not give you an open cover of the whole torus, and one must use another 3rd such open rectangular chart to get an atlas, i.e. to get an open covering of the torus by coordinate charts. I.e. if one chart has theta and phi running from 0 to 2π, then it omits all points of the torus where theta = 0. e.g. it omits the point where (theta,phi) = (0,pi). But if the second chart has theta and phi running from -pi to pi, then that second chart omits all points where phi = pi, in particular it also omits the point (theta,phi) = (0,pi).

Can you find the second omitted point? In any case it only requires one more rectangular chart to complete the open cover, i.e. to give an atlas, for the torus. Just take any rectangle of side length 2pi, and starting at a point other than 0 or pi (and not equivalent to them mod 2pi). E.g. take the rectangle whose sides run over (-pi/2, 3pi/2). I hope I got this right since I am doing it in my head. But if you just draw any pair of perpendicular circles on a torus, and then slide the circles around a bit, you will see that there must always be at least 2 points that lie on both pairs of circles.

The moral is that the best way to obtain an atlas on a torus is indeed to use the map from the plane onto the torus, and restrict it to rectangles of the plane to get individual charts, as you originally said. But you must be careful you use enough rectangles to cover the torus.

But if we give the author the benefit of the doubt and assume he understood this point, there is a way out. Although he was not clear on this, let's assume he meant for us to use all 4 rectangles we can form from the intervals (0,2pi) and (-pi,pi), i.e. include also (0,2pi)x(-pi,pi) and (-pi,pi)x(0,2pi). Then we do get an atlas, but one that requires 4 charts instead of 3. Again check me on this, as I am just visualizing it in my head.

 

[Orodruin]

Apart from what has been said already, it is possible to define an atlas using only two charts. You can construct it by glueing two cylinders together (there exist global coordinate charts for cylinders).

Edit: I should specify "to define an atlas for the torus using only two charts". Even if it may be clear from context in this thread, the sentence taken out of context would not be generally true.

 

[aheight]

Thanks guys,

Is it correct to say once I implement this mapping onto the torus and create basically a (conformal) coordinate system of the double cover over the torus, I thus create the normal Riemann surface for the function $\frac{A}{\sqrt{(1-z^2)(k^2-z^2)}}$?