3 ms·

After some thought (and before I read your comment), I am indeed mistaken. Take the long/lat mapping we put onto Earth - it contains trapezoids. However, the s

After some thought (and before I read your comment), I am indeed mistaken. Take the long/lat mapping we put onto Earth - it contains trapezoids. However, the sections on a torus are not always parallelograms. Take your donut and lay it on the table. It has an equator of sorts where the oil didn't cook it as dark. Imagine little rectangles around the donut, long sides resting on this outer equator, short sides touching adjacent rectangles. We continue stacking rectangles over the donut and toward the inner equator. If we maintain the same number of rectangles circling around the central axis (the one that passes through the center, perpendicular to the equatorial plane, but never intersects the donut), then the long sides of the rectangles must get shorter as we approach the inner equator. If we look at one of these rectangles, we can see that the long side toward the outer equator is longer than the side toward the inner equator. And that the short sides are no longer parallel. Thus, these sections are not parallelograms.

tgb 9y ago
You're right that if you take a trapezoid and identify the sides, you get something that is topologically a torus. However, we were trying to measure distances on the torus and hence need it to be geometrically a torus - i.e. we need distances to preserved at least locally when we map from the domain (the square or trapezoid) to the torus. You can see that this isn't done with the trapezoid: you have to identify two edges that have different lengths, hence lengths must get messed up! Or in your example, each trapezoid on the donut has different size and shape so it's not useful for measuring distances. Though getting this to work for the donut is impossible, so you had a difficult task. (If your domain is in the flat plane, then the resulting torus must be flat as well. Which means it's not the kind of torus you can eat since it can't fit in 3 dimensional (flat) space along side us. But Pac Man and friends live on flat torii.) Nonetheless, you're right that the domains don't always have to be parallelograms. Each covering has a parallelogram domain but in general coverings have an infinite number of possible domains. They satisfy some conditions though: essentially they're a parallelogram where the edges don't have to be straight lines so long as the left and the right edges (and the top and bottom edges) are the same curve just translated. So generally the parallelogram domain is easiest to work with (and in fact there are multiple parallelogram domains you can choose so one usually chooses the one with the smallest edge lengths).

tgb 9y ago

You're right that if you take a trapezoid and identify the sides, you get something that is topologically a torus. However, we were trying to measure distances on the torus and hence need it to be geometrically a torus - i.e. we need distances to preserved at least locally when we map from the domain (the square or trapezoid) to the torus. You can see that this isn't done with the trapezoid: you have to identify two edges that have different lengths, hence lengths must get messed up! Or in your example, each trapezoid on the donut has different size and shape so it's not useful for measuring distances. Though getting this to work for the donut is impossible, so you had a difficult task. (If your domain is in the flat plane, then the resulting torus must be flat as well. Which means it's not the kind of torus you can eat since it can't fit in 3 dimensional (flat) space along side us. But Pac Man and friends live on flat torii.) Nonetheless, you're right that the domains don't always have to be parallelograms. Each covering has a parallelogram domain but in general coverings have an infinite number of possible domains. They satisfy some conditions though: essentially they're a parallelogram where the edges don't have to be straight lines so long as the left and the right edges (and the top and bottom edges) are the same curve just translated. So generally the parallelogram domain is easiest to work with (and in fact there are multiple parallelogram domains you can choose so one usually chooses the one with the smallest edge lengths).