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stoutfellowI've completely solved the problem of simple closed geodesics on rectangular prisms. The solution, phrased as non-technically as possible, is under the cut. (No proofs or computations, I promise.)
First, let's recall what the question is. In essence, we're looking at all possible ways of wrapping a ribbon around a rectangular box, where we insist that a) when the ribbon crosses an edge, it does so smoothly - no kinking; b) the ribbon never crosses itself; and c) the ribbon closes up - it returns to where it began and lines up smoothly with its beginning. If the box is cubical, there are three distinct ways of doing this: wrapping the ribbon parallel to one edge, so it runs around four faces and returns; wrapping the ribbon at a 45-degree angle to one edge, so it "cuts a corner" six times before returning; and wrapping it so that its slope, relative to one edge, is 1/2 - it cuts two corners, runs across the next face, cuts two more corners, and runs across back to its beginning. The first one is possible on any box, but the other two impose limitations on the dimensions of the box.
There are other possibilities, none of which work on the cube. There's a family which I've been calling "barberpoles": take a box which is much longer in one direction than in the others, and stand it vertically. Run the ribbon at an angle across the top; let it spiral down to the bottom, cross the bottom, and spiral back up. The longer the box is (proportional to the other dimensions) the more times the ribbon can spiral around. Barberpoles aren't the only possibilities, however; when the ribbon spirals back up, it may not return to its initial point. That's OK; let it cross again, and spiral down and up again. It may cross the top many times before finally closing up. (Actually, if you pick the angle at random, it's not going to close up; you have to be lucky to get a perfect fit.)
Now, suppose the top and bottom are mxn, and that the crossings on top all stretch between the edges of length m. The bottom crossings may run between the edges of length m - that's Case IIIA - or between the other two edges - Case IIIB. A IIIA example can be characterized by two numbers: a, which is the number of times it crosses the top, and b, which counts the number of times it spirals around. (Actually, b is roughly 1/4th of the total number of times it crosses the vertical sides.) a has to be positive; b has to be non-negative; and the two numbers have to have no common factor greater than 1. A IIIB example can likewise be given by a and b, with similar meanings; the only difference is that b has to be positive in this case.
I started working on this problem December 19 of last year, on my holiday flight to California, and I've been wrestling with it pretty steadily for almost three months. It's done; now I have to write the article.
WOO-HOO!
First, let's recall what the question is. In essence, we're looking at all possible ways of wrapping a ribbon around a rectangular box, where we insist that a) when the ribbon crosses an edge, it does so smoothly - no kinking; b) the ribbon never crosses itself; and c) the ribbon closes up - it returns to where it began and lines up smoothly with its beginning. If the box is cubical, there are three distinct ways of doing this: wrapping the ribbon parallel to one edge, so it runs around four faces and returns; wrapping the ribbon at a 45-degree angle to one edge, so it "cuts a corner" six times before returning; and wrapping it so that its slope, relative to one edge, is 1/2 - it cuts two corners, runs across the next face, cuts two more corners, and runs across back to its beginning. The first one is possible on any box, but the other two impose limitations on the dimensions of the box.
There are other possibilities, none of which work on the cube. There's a family which I've been calling "barberpoles": take a box which is much longer in one direction than in the others, and stand it vertically. Run the ribbon at an angle across the top; let it spiral down to the bottom, cross the bottom, and spiral back up. The longer the box is (proportional to the other dimensions) the more times the ribbon can spiral around. Barberpoles aren't the only possibilities, however; when the ribbon spirals back up, it may not return to its initial point. That's OK; let it cross again, and spiral down and up again. It may cross the top many times before finally closing up. (Actually, if you pick the angle at random, it's not going to close up; you have to be lucky to get a perfect fit.)
Now, suppose the top and bottom are mxn, and that the crossings on top all stretch between the edges of length m. The bottom crossings may run between the edges of length m - that's Case IIIA - or between the other two edges - Case IIIB. A IIIA example can be characterized by two numbers: a, which is the number of times it crosses the top, and b, which counts the number of times it spirals around. (Actually, b is roughly 1/4th of the total number of times it crosses the vertical sides.) a has to be positive; b has to be non-negative; and the two numbers have to have no common factor greater than 1. A IIIB example can likewise be given by a and b, with similar meanings; the only difference is that b has to be positive in this case.
I started working on this problem December 19 of last year, on my holiday flight to California, and I've been wrestling with it pretty steadily for almost three months. It's done; now I have to write the article.
WOO-HOO!
